CHAPTER-1

Analytical Reasoning Analytical Reasoning forms an important part of generally all Campus Placement Test. The questions in this section can either test analytical or logical reasoning. This booklet will concentrate on the former. The thinking process associated with such problems resembles solving a case, wherein the clues here are in the form of certain conditions, which may or may not be mutually exclusive. The best way to get a hang of these problems is to work on them continuously. Initially, they may deter you but once the student follows the rules given below and solves many problems, it will appear simple. Games1 in this section can both be verbal and numerical. The first part of this booklet will consist of the basics behind solving such problems, such as the elements forming the problem and the rules to follow to crack such problems. The second part will deal with verbal analytical reasoning wherein a few varieties of games / puzzles / problems will be provided, which will be followed by a discussion on solving the same. Questions may not be provided for a few of these games, as solving the conditions directly will give us the solution. The third part will deal with numerical reasoning, which basically requires a strong sense of arithmetic. The methodology for solving each game is only suggestive and the student can use his own discretion to solve it in an easier fashion.

Conditions As mentioned above, conditions are a set of clues, which when analysed, provide the solution for the problem. These conditions cannot be violated and all these conditions have to be analysed before attempting any question relating to the problem. These conditions are not necessarily in order and therefore, it becomes imperative for the student to go through all of them before identifying the important ones.

Rules of the Game There are a few rules of this game, which help in solving AR problems systematically. These rules help you frame the problem in the right perspective, organise your thoughts and finally, crack the problem and its associated set of questions. A few such rules have been given below. 1.

2.

Identify the most important condition, if any.

3.

Picturise the information.

4.

Organise the information.

Reading the problem and its conditions is a pre-requisite. The student has to spend sometime reading and analysing the statements. Read the conditions carefully. Please donâ&#x20AC;&#x2122;t over-read into the conditions and statements. Its also important to study the implications of the conditions, i.e. try to analyse it from a what-if angle (though not the case in all the problems) Since the conditions are not given in any specific order, the student has to peruse through all of them and identify the key condition, which will provide the platform from where the problem can be solved. If there are any restrictions in the problem, like for example, one of the elements has always to be at one place and then start by putting that element there. In case of ordering problems (wherein an order of the elements has to be ascertained), this becomes especially important.

If, for example, the problem deals with 5 people standing at a bus stop, the student must be able to juxtapose it in his analysis. Since there wont be sufficient place to work out these kinds of problems, picturisation helps in arranging the information Picturisation is the first step in organising information. Once the student has an idea of the problem, he has to organise the information in an easily comprehensible manner, i.e. he should try to associate it with whatever he feels comfortable. The student should put every bit of detail on paper, say, make notes. Some ways of organising information are

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Grids/Tables: Once the student has identified a particular structure of relationships, all the information can be arranged in the form of a multi-columnar grid. The different elements of the problem, as given by the conditions, must be entered onto the grid to facilitate a logical thinking process. It also helps in eliminating some of the answers provided in the solution.

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Symbols/Notations: If the student is comfortable substituting symbols for the different conditions, he should do so, as, this, like tables, provides a simplistic view of the problem. These may differ according to the type of problem and the comfort level of the student.

5.

Elimination of Answer choices. Some of the above rules do help one in eliminating answer choices. It is generally not advisable in case of AR to start off by eliminating choices. As you will see, many problems require a bottom-up approach, but, that is only after a careful perusal and analysis of the conditions. Eliminate choices when the conditions are insufficient to answer the questions.

6. 7.

Be careful of the language used. Certain words or phrases are oft-repeated and these can well form important clues in framing the solution. Finally,

answer all questions at the same time, if the game demands

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SOLVED EXAMPLES The World This Week This game deals with scheduling a program over a time period, say a week or a year. It consists of a few conditions, from which the order of telecast has to be identified. These are problems, which require the identification of a particular order or a schedule. Other variants of this game include 

Identifying people staying in an apartment complex,



Ordering of books in a shelf etc.

Example A TV channel is going to telecast 6 soppy soaps over the week – Sob, Cry, Weep, Wail, Lament and Moan. One of these will be telecast each day from Monday to Saturday, with Sunday being a FunDay. The conditions specified by the Boss are as follows: 

Sob must be telecast earlier in the week than Lament.



Weep must be telecast on Tuesday



Wail must be telecast on the day immediately before or immediately after the day on which Cry is telecast

Questions Q1.

If Cry is telecast on Thursday, the earliest day on which Lament can be featured is a) Monday b) Tuesday c) Wednesday d) Friday e) Saturday

Q2.

If Moan is to be telecast on Friday, Sob must be telecast on a) Monday b) Tuesday c) Wednesday d) Thursday e) Saturday

Q3.

If Wail is to be featured on Thursday, the latest day on which Sob can be telecast is a) Monday b) Tuesday c) Wednesday d) Friday e) Saturday

Q4.

Which of the following soaps can be telecast on Monday? a) Cry b) Moan c) Lament d) Weep e) Wail

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Q5.

If Moan is to be telecast on Thursday, which of the following is true? a) Sob must be telecast on Wednesday b) Cry must be telecast on Saturday c) Wail must be telecast exactly two days after Lament is telecast d) Lament must be telecast on Wednesday e) Lament must be telecast later in the week than Moan

Q6.

If Moan is to be telecast on Friday, what is the total number of acceptable schedules available to the TV channel? a) 1 b) 2 c) 3 d) 4 e) 5

Solution/Discussion We shall begin discussing the solution by following the rules that have been specified above. A thorough reading of the problem provides us with the following leads – It is imperative in these kinds of ordering problems that we begin with conditions that fix the exact position of one or more elements, and then work towards narrowing the possibilities for other elements. When the order cannot be determined by the condition, then, attack the questions and proceed with eliminating choices. There are 3 conditions, of which, the second condition that Weep should be telecast on Tuesday can be considered to be the important condition since it provides an ideal base to work the problem. The first condition suggests that Sob can be telecast on any day except Saturday, and Lament can be telecast on any day except Monday. Since Tuesday is already ruled out, the days of telecast for these two soaps are down to four. From the third condition, we can ascertain that Cry cannot be telecast on Monday because Weep is to be telecast on Tuesday. Therefore, Cry can be telecast on any day from Wednesday to Saturday. Let us now make a table to help organise this information

Days Soaps

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Weep

We can proceed in the following manner Calculations Monday: No Lament and Cry….so it’s either Sob, Wail or Moan Wednesday to Friday: All except Weep Saturday: All except Weep and Sob Since there can be many combinations of the telecast of these soaps, we can proceed with the problem by reading the questions. This approach will also help in eliminating answer choices.

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Solution 1: For every question, we can fill the table. As for this question, if Cry is telecast on Thursday, the table would look like this –

Days

Monday

Soaps

Tuesday

Wednesday

Weep

Thursday

Friday

Saturday

Cry

From our calculation earlier, we have seen that the earliest day that Lament can be telecast is on Wednesday. WE now check whether it is feasible with Cry being telecast on Thursday.

Days

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Soaps

Sob

Weep

Lament

Cry

Wail

Moan

We find that it is indeed feasible and therefore, the answer to this question is Wednesday

Solution 2: Moan has not figured in our calculations so far. So if we assume that Moan is telecast on Friday, then we have a table that looks like this

Days

Monday

Soaps

Tuesday

Wednesday

Thursday

Weep

Friday

Saturday

Moan

Moan’s telecast on Friday provides us with only one combination of days for the telecast of Cry and Wail, Wednesday and Thursday, since they have to be telecast on consecutive days. And from the first condition, which specifies that Sob must be telecast earlier in the week than Lament, Sob must be telecast on Monday.

Days

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Soaps

Sob

Weep

Cry

Wail

Moan

Lament

The answer to the second question is Monday

Solution 3: We make a table again with Wail being telecast on Thursday.

Days

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Soaps

Sob

Weep

Cry

Wail

Moan

Lament

This table is similar to the one used to arrive at Solution 2, but the question has been framed differently. Here we are supposed to find the latest day on which sob can be telecast and not the earliest day. From our earlier calculation, we find that Sob can be telecast on Monday and on any day from Wednesday to Friday. Since Wail is telecast on Thursday, Cry should be telecast on Wednesday or Friday. Since there is no condition relating to Moan’s telecast, Moan can be telecast on Monday and Sob on Friday. Do not forget that Lament has to be telecast after Sob.

Days

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Soaps

Moan

Weep

Cry

Wail

Sob

Lament

The answer to the third question is Friday

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Solution 4: From our earlier calculation, we can easily say that Moan is the only soap, among the answer choices, that can be telecast on Monday.

Solution 5: This question requires us to visualise different alternatives and ascertain their veracity. Days

Monday

Soaps

Tuesday

Wednesday

Weep

Thursday

Friday

Saturday

Moan

We can do this by analysing each of the answer choices. Never forget the calculations Sob must be telecast on Wednesday: If Sob is telecast on Wednesday, Lament will have to be telecast on wither Friday or Saturday, which is not possible because Cry and Wail have to be telecast on consecutive days. Therefore, this is untrue. Cry must be telecast on Saturday: Cry can be telecast on Friday or Saturday. Though this is not entirely untrue, it is an option we should keep in mind till we arrive at the exact answer. Wail must be telecast exactly two days after Lament is telecast. Since, in this case, Lament has to be telecast on Wednesday, Wail can be telecast either on Friday or Saturday, i.e. either 2 or 3 days after lament is telecast. So, even this option is not entirely untrue and should be kept in abeyance till we arrive at the final answer. Lament must be telecast on Wednesday: From the argument for the earlier answer choices, we find that Lament has to be telecast on Wednesday alone. Lament must be telecast later in the week than Moan: this is not possible since Cry and Wail have to be telecast on consecutive days and only Friday and Saturday are available for them. Therefore, after examining the answer choices, we find that the correct answer is (d).

Solution 6: We again make use of a table here, with Moan appearing on Friday

Days

Monday

Soaps

Tuesday

Wednesday

Thursday

Weep

Friday

Saturday

Moan

The different schedules can be tabulated as under Days

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Soaps

Sob

Weep

Cry

Wail

Moan

Lament

Sob

Weep

Wail

Cry

Moan

Lament

Since there are only 2 acceptable schedules, the answer is 2.

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Round Table In this game, there are a few people sitting around a table. The basic theme here is to identify, from the conditions, the exact placement of a person around the table.

Example: There are six people, A, B, C, D, E and F sitting around a table. 1. The person sitting to the left of A faces D. 2. B and E sit on either side of C 3. A and B do not face each other.

Questions generally relate to the seating arrangement around the table and therefore are very simple. For example, Who sits in front of B? Who faces B if A and D swap places? Solution/Discussion: This is a very simplistic version of the problem. The best and the simplest method to solve round-table problems is to start by drawing a table and seating the students according to the conditions. For example, from the first condition, by seating A at the top of the table, we can ascertain the exact seating position of D. A

D

Since the identity of the person seated next to A cannot be ascertained with one condition alone, we move on to the second condition. After analysing the second condition along with the diagram above, we can ascertain the exact seating position of E, as illustrated by the following diagram. A

C

D

The third condition that A and B do not face each other means that A is sitting to the left of A and facing D and E is facing A (from the second condition above). And therefore the final seating arrangement would like this A B F

C

D E

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Different Strokes This game is a variant of the above, with more number of elements. One thing to remember, the more the conditions, the easier it is to solve the problem. There are many variants of this game, like people standing in a bus-stop/railway station etc. Example There are 5 houses coloured Red, Green, Black, White and Blue in a row. Each of the houses is occupied by bachelors of different professions (Singer, Accountant, consultant, Software engineer and lawyer) and each of them has a different vehicle in which to commute to office. The colour of the vehicle is different from not only that of his house but also of his neighbours house. 1) A lives in a Blue House, which has the same number of houses on either side. 2) B is an accountant and the colour of his car is similar to the colour of the lawyer’s house. 3) E has a black car and his house is next to the Green house 4) The person living in the White House is a consultant 5) D is a software engineer living in the Black House 6) The lawyer drives a white car and he stays next to the Red House 1.

Who is the lawyer? a) A b) E c) D d) C

2.

Which of the following gives an orderly description of the houses? a) GRBWB b) BGWBR c) GRBBW d) WBBGR

3.

What is the colour of the consultant’s car? a) Black b) Blue c) Green d) Red

4.

Which of the following lives in the house next to B? a) A b) D c) C d) E

Solution/Discussion The best method to solve these types of problems is to organise the information in the form of a multi-dimensional grid. The dimensions as provided in the case are: People, Profession, Houses and Cars. We begin by filling in the information known to us in the correct boxes. First, we make individual grids connecting each of the dimensions and then consolidate them in one single grid based on our analysis.

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From condition 1, we know that A lives in the blue house and that it has two houses on either side. Grid 1 shows the relationship Though there are two parts to the second condition, we take them one at a time. We match B with that of an accountant. Grid 3 shows the relationship E has a black car and his house is next to the Green house, which means that the colour of E’s house is neither black (the case specifically mentions that the house and a vehicle of a person cannot be of the same colour) nor green nor blue. Grid 2 shows this relationship D is a software engineer living in the black house. From this, we ascertain that D is not the neighbour of E since E has a black car (refer to the case which specifically mentions that the colour of the vehicle is not similar to the colour of the neighbour’s house). We can now say that E and D live on either side of A. Now for the remaining two conditions: The lawyer drives a white car and stays next to the red house. Therefore the colour of the lawyer’s house is not white, black or red (which, in turn, means that the colour of B’s car is not white, black, or red) The person living in the White House is a consultant. Therefore, the consultant is not the neighbour of the lawyer.

If we combine all the dimensions into one single grid, Grid 6, we can make the following guesses: Either A or C should be a lawyer Either C or E is a consultant. E should be staying either in the white house or the red house. The colour of B’s car is either Blue or Green. The colour of the Lawyer’s house is Blue or Green. Since the lawyer stays next to the red house and drives a white car, his house can neither be red nor white in colour. This combined with the above mean that B is not the neighbour of the lawyer and A.

Since E’s house is not green, black or blue in colour, he should be the neighbour of B. Since he stays next to the Green house, the colour of B’s house is Green, which means that B drives a blue car. This, in turn, implies that the colour of the lawyer’s house is blue, i.e. A is the lawyer. Therefore, the singer lives in the red house. The lawyer stays next to the red house and therefore, the colour of E’s house is red, which means that he is the singer and C is the consultant. Grid 1 House

Green

Red

Blue

Black

White

Name

B

E

A

D

C

Car

White

Blue

Green

Red

Black

Name

A

B

C

D

E

Profession

Lawyer

Accountant

consultant

Engineer

Singer

Name

A

B

C

D

E

Grid 2

Grid 3

Grid 4

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Profession

Singer

Accountant

Lawyer

Consultant

Engineer

House

Red

Green

Blue

White

Black

Profession

Singer

Accountant

Lawyer

Consultant

Engineer

Car

Black

Blue

White

Green

Red

Grid 5

Grid 6 Name

House

Car

Profession

A

Blue

Green

Lawyer

B

Green

Blue

Accountant

C

White

Green

Consultant

D

Black

Red

Engineer

E

Red

Black

Singer

To show the same information in a multi-dimensional grid, House

GREEN

RED

BLUE

BLACK

WHITE

Person

B

E

A

D

C

Profession

ACCOUNTANT

SINGER

LAWYER

ENGINEER

CONSULTANT

Vehicle

BLUE

BLACK

WHITE

RED

GREEN

With the above grid, all the questions can be answered. This question can also be solved using symbols or notations. For example, the fact that the colour of B’s car is similar to the lawyer’s house can be recorded as B’s car = Lawyer’s house, and so on. But, it is required that students be comfortable using notations in place of grids etc.

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Conditions Galore These games are slightly different from the others in that the problem does not specify the exact number of places to assign elements, rather, some elements have to be assigned for some questions and some others for other questions. These problems are of the selection variety wherein you are asked to distribute elements into groups. The strategy for this is similar to that of the earlier problems, namely ordering/scheduling problems, but because of the “if-then” relationship (or the probability relationship) prevalent, it requires a more organised solution process, especially with the use of symbols. Some of these relations are given below. Please note the implications of the a single statement. Eg 1: If A is in the team, then B also has to be in the team. If the team consists of A, then it will also have B, i.e. A, therefore B. But, if B is in the team, it doesn’t mean that A is in the team, i.e. B, therefore A is not a correct way of reasoning. Similarly if B is not in the team, it also means that A is not in the team, i.e. not B, therefore not A. The converse, however, isn’t true, i.e., if A is not in the team, it doesn’t mean that A is not in the team, i.e. not A, not B is a wrong way of reasoning. Eg 2: In a vote for a resolution, A and B never both aye. This means that both never vote aye, but they can both vote nay. Please don’t jump into conclusions that since they cannot both vote aye, they can also never both vote nay.

Example: A track coach is deciding which and how many of her athletes – L, M, N, O, P, R and S – will compete in an upcoming track meet. She will decide according to the following guidelines: 1. If L competes, M must compete 2. If M and N both compete, O cannot compete 3. If N and O both compete, R cannot compete 4. If O competes, either P or S must compete 5. Either P or R must compete, but they cannot both compete 6. P and S cannot both compete.

Questions 1.

If only 3 athletes can compete in the track meet, which of the following could be that group of athletes? a) LMN b) MPS c) MPR d) NOP e) NOR

2.

If O and S both compete in the track meet, which of the following must be true? a) N competes b) P competes c) R competes d) L does not compete e) M does not compete

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3.

If O and R both compete in the track meet, which of the following cannot be true? a) M competes b) N competes c) S competes d) L does not compete e) P does not compete

4.

If L and N both compete in the track meet, what is the maximum number of athletes who can compete a) 3 b) 4 c) 5 d) 6 e) 7

5.

If S competes in the track meet, which of the following combinations of 3 athletes can be among those who also compete? a) LMP b) LNO c) LOP d) MOR e) NOR

Solution/Discussion This problem requires the simplest of tables. The idea here is to identify who can compete together and who cannot and the best way to do that would be to draw a table which separates one from the other, i.e

Compete

Not compete

Another simple method would be to symbolise the clues. But, as mentioned before, only people who are comfortable with symbolising should do so to avoid any confusion. Some of the symbols used are: Arrows: to indicate that if L competes, M must compete. L → M Boxes: To indicate that if M and N both compete, O cannot compete.MN

→ ~O

Not equal: To indicate that P and S cannot both compete. P/S A combination of symbols can also help in arriving at the solution. As per the tabular method illustrated above, we can answer each question by deciding on whether the athletes compete together or not. This is another problem which can be solved using the questions, and thereby, the process of elimination. For example, if we consider the second question in the problem,

2.

If O and S both compete in the track meet, which of the following must be true? a) N competes b) P competes

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c) R competes d) L does not compete e) M does not compete

Using the table and the clues given above, we can arrive at the answer. The last clue clearly mentions that P and S cannot both compete. Therefore, put P in the Not compete column. Another clue specifies that either P or R must compete and not both. Since P is not competing, R can compete. So put R in the Compete column. Using these two clues alone, we can arrive at the answer choice, which is, R competes. We can also see that N can also compete in this case, but, if we read the question again – it says – which of the following must be true, and not can or may be true. Also, this is the logical answer, which appears on the face of the conditions. In such cases, it is advisable to take things at face value and leave the implications behind.

Compete

Not compete

OSR

P

Similarly, for the other questions.

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Liar Liar These are games which contain statements by a few people, which are either true or false. From these statements, the student has to answer a few questions depending on the kind of condition given in the problem. This problem requires considerable amount of time and the student has to be ready to invest it during the exam. It may seem confounding in the beginning but once the thinking process is in place and the approach methodical, it will appear simple. Its time for the What-What island, where the inhabitants answer any question with two sentences; one of which is true and the other is false. You are looking for Venkat’s house and you meet 3 people – Anand, Ravi and Som. You ask them, “Which is Venkat’s house?” Anand says: Venkat’s house is No.9. I am his neighbour. Ravi says: Anand is not my neighbour. Anand and som live in the same house. Som says: Venkat’s house is not No.9. Anand is Ravi’s neighbour. There are only two houses and four people in What-what. Two people live in each house.

Q1.

From the above, you can decide that a) Venkat stays in house no.9 b) Venkat does not stay in house no. 9 c) Venkat does not stay in what-what d) Ravi and Som stay together.

Q2.

Who stays with Anand? a) Ravi b) Venkat c) Som d) Cant say

Solution/Discussion This question is very different from the previous ones and requires some time on the part of the student to understand and solve it. A methodical process would be useful in this case. To begin with, start with the first speaker and label one sentence each as true or false, similarly for the other speakers. Ascertain whether the labeling makes sense at the end of the third speaker. For example, We label Anand’s first sentence as true (T) and the second as false (F) and similarly for Ravi’s and Som’s sentences. The final output will look like: Venkat’s house is no.9 and Anand is not his neighbour Anand is not Ravi’s neighbour and Anand and som do not live in the same house. Venkat’s house is not No.9 and Anand is not Ravi’s neighbour. We can see the contradiction evident in the first and third sentences relating to Venkat’s house. Since this doesn’t make much sense, we re-label the sentences accordingly, taking care of the initial contradiction. If we label Anand’s first sentence as true, Som’s first sentence has to be false, and thereby Som’s second sentence is true. This means that Anand is Ravi’s neighbour and therefore, Ravi’s first sentence is false. But here lies another contradiction – if Anand and Som live in the same house, then Ravi and Venkat live

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in the other house, but Anand;s second sentence specifically says that He is not Venkat’s neighbour. Therefore, even this is a wrong choice. Now, we try the other way, by labeling Anand’s first sentence as false and Som’s first sentence as true. By the chain of thoughts, it is evident that Anand is not Ravi’s neighbour and therefore Anand and Ravi live in the same house, which is also true in case of the second sentence of Anand. Therefore, Venkat does not stay in House no. 9 and Ravi stays with Anand.

Alpha-Numeric Here, numbers are coded as words or vice-versa and from the rules of this game, exact codes have to be identified. Variants of this game are – 

the usage of words only as codes



substituting symbols for relations, i.e. if a*b means a is the father of b, and a@b means a is the sister of b, then how does one denote a paternal aunt

Two English words are codified as follows. Each number represents only one letter and each letter is represented by only one number. Word 1: 8 3 7 6 3 2 9 Word 2: 3 6 7 5 8 4 1 6 The following rules are known to the person decoding them. I. Letters T and R occur exactly three times II. Letters S and A occur exactly two times. III. Letters E, P, O and N occur exactly once. IV. One of the words starts with T and the other with S. V. E occurs only in word 1. Solution/Discussion We proceed methodically analysing one condition at a time. From the first condition, it is clear that T or R should stand for 3 or 6. From the second condition, it is clear that S or A should stand for 8 or 7 We will ignore the third condition for sometime because we cannot conclude anything from it. From the fourth condition, we can conclude that T stands for 3 and S stands for 8. Therefore, R stands for 6 and A stands for 7. E stands for either 2 or 9. O,P and n can be substituted for 5,4 and 1 in Word 2. Decoding the numbers individually, we find the following Word 1: 8 3 7 6 3 2 9 S T ART Word 2: 3 6 7 5 8 4 1 6 T RA S

R

In this particular question, the words are not those that make some meaning. Its more open-ended than a word that has a meaning attached to it. I say so because it is not possible to decipher all the numbers correctly, and therefore, its left to the student as to what he makes of the numbers and words which cannot be correctly identified.

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Absolute Relatives In this game, again a variant of the first 2 games, from the data given, relationships have to be identified. A, B, C, D, E, F, G and H are people who are related as below I. A is the father of 2 children C (male) and D (female) II. H is the mother of two children E (male) and F (female) III.

B is E’s mother-in-law

IV. D is the daughter-in-law V. E’s wife is F’s sister-in-law VI. E’s son will also be A’s grandson and C’s daughter will also be H’s grand-daughter Q1.

Who is A’s wife? a) H b) F c) B d) E e) NONE OF THESE

Q2.

Who is H’s husband? a) B b) A c) D d) F e) NONE OF THESE

Q3.

Who is D’s mother-in-law? a) B b) H c) F d) E e) NONE OF THESE

Q4.

A’s daughter will also be A. B’s daughter in law B. E’s wife C. H’s daughter a) A only b) B only c) C only d) A and B only e) B and C

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Q5. D is a) Wife of A b) Wife to E c) Daughter of G d) F’s sister e) None of these

Q6.

E’s mother is also a) D’s mother b) C’s grandmother c) B’s sister d) G’s wife e) A’s wife

Q7.

If X is C’s son and Y is E’s daughter then X and Y must be a) Brother & Sister b) Husband and Wife c) D’s children d) H’s grand children e) None of these

Q8.

Out of 8 people, how many must be male? a) 4 b) 3 c) 5 d) 2 e) 6

Q9. Out of 8 people, how many must be female? a)

2

b)

3

c)

4

d)

5

e)

6

Q10. Which of the following represent a correct pair of husband and wife? a)

A&H

b)

B&G

c)

E&D

d)

A&F

e)

C&D

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Solution/Discussion How better to solve a family problem than by using a family tree. We can use symbols to denote the different relationships. Males and Females can be denoted using m and f respectively. The following symbols can be used: (Indicative symbols) Father: up-arrow Mother: up-arrow Mother-in-law: diagonal up arrow Father-in-law: diagonal up arrow Daughter-in-law: diagonal down arrow Son: down-arrow Daughter: down-arrow Using the conditions given above, the relationships can be symbolised as follows:

Am

Cm Df

Bf

Hf

Em Ff

E’s wife

F’s sister-in-law

Am

H

Em

Cm

Son

daughter

From the symbols above, we can infer the following: Since A has 2 children, C & D, E must be the son-in-law. Since H has 2 children, E & F, C must be the daughter-in-law (which is also mentioned in the conditions) E’s wife therefore is D, who is the brother of C, who is F’s husband. Since G is the only missing link, G must be the husband of H, just as B is the wife of A The families are : AB, GH, ED and CF. There are an equal number of males and females in the family. Therefore, X and Y are the grand-children of AB and GH. The other questions can be solved similarly.

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Map Games These types of games describe connection between the elements specified in the game. Symbols are very much useful in this type of games. In a message relay system there are exactly seven terminals â&#x20AC;&#x201C; F,G,H,J,K,L and M. A terminal can transmit any messages initiated by that terminal as well as any messages received from others, but only according to specific rules: Messages can be transmitted in either direction between G and H, in either direction between J and M, and in either direction between K and L. Messages can be transmitted from F to K, from H to J, from K to G, From M to F, and from M to H.

Q1. Which of the seven terminals can transmit messages directly to the greatest number of terminals? a) F

b) H

c) J

d) K

e) M Q2.

If a message initiated by G is to reach K, and is to be transmitted to no more terminals than necessary, it must be transmitted to a total of how many terminals, other than G and K? a) 1

b) 2

c) 3

d) 4

e) 5 Q3.

A message from H that eventually reaches L must have been transmitted to all of the following terminals except a) F

b) G

c) J

d) K

e) M Q4.

If J is removed from the message relay system for a day, it is still possible for a message to be transmitted on that day all the way along a route from a) F to H

b) G to K

c) G to M

d) H to K

e) L to M Q5.

If K is removed from the message relay system for a day, which of the following terminals cannot receive any messages from any other terminal on that day a) F

b) G

c) H

d) J

e) L Q6.

A message can travel along two alternative routes that have no terminal in common except the initiating terminal and the final recipient terminal if the initiating terminal and the final recipient terminal, respectively, are a) G and J

b) G and L

c) H and L

d) K and M

e) M and G Q7.

A message being transmitted along which of the following routes must reach each of the seven terminals at least once? a) F to G and then to M

b) J to H and then to L

c) L to H and then to M

d) M to G and then to K

e) M to L and then to F

Solution/discussion Arrows can be used to show the connection between the different elements. One-way or Two-way arrows can be used to differentiate between the connections.

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G

H

I

M

K

L

F

K

H

J

K

G

M

F

M

H

Using these connections, we make a link connecting these elements.

M

F

J

H

G K

L

With this combined map, we can proceed to answering the questions. The process of elimination can also be used in this case. If there arises a need to redraw the “connection” chart for a particular question, go ahead and do it.

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Card Games Card games require a basic knowledge of playing cards, the suits in cards, etc. For the starter, there are 4 different suits, in 2 different colours – Spades and Clubs, in black, and Hearts and Diamonds, in red. There are 13 cards in all, in each suit. The cards are Ace, 2,3,4,5,6,7,8,9, 10, Jack, Queen and King. The last three are picture cards. Though card and die games appear more in quantitative ability in the form of probability questions, they have known to appear in reasoning questions also.

From a normal pack of playing cards, 20 cards were used in a game. These were all the four suits of Ace, King, Queen, Jack and number ten. A, B, C and D are the players. Each player has all the five cards, in one or different suits. C’s five cards were in 3 different suits and consisted of 3 red and 2 black cards. D’s five cards were also in 3 different suits, his ace being in the same suit as his queen, and his king in the same suit as his jack. B held more than one black car. A’s five cards were all in the same suit. C held the king of Spades, and D, the ten of diamonds.

Q1.

Between them, B,C and D held at least …..black cards a) 6 b) 7 c) 4 d) 3

Q2.

Who held the queen of hearts? a) C b) D c) A d) B

Q3. Who held the Ace of diamonds? a) A b) B c) C d) D Q4. A’s five cards are all a) Spades b) Clubs c) Hearts d) Diamonds

Solution/Discussion Firstly, we organise the given information. For our convenience, lets denote the different suits as S, D, H and C. Secondly, we start with identifying the important condition, if any. Lets begin by putting all the information together. C’s cards were in 3 different suits and consisted of 3 red and 2 black cards. C held the King of spades. D’s cards were also in 3 different suits, with ace and queen of the same suit and the king and jack of another suit. He has the ten of diamonds. Because 10 is of a different suit, the other cards may belong to spades, hearts or clubs.

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B held more than one black card.

All of A’s cards were in the same suit. Since C and D have one each of spades and diamonds, he must have either all hearts or all clubs.

Since we haven’t been able to glean more out of the conditions, we follow a bottom-up approach, and analyse the questions.

To answer the first question, we know that C has 2 black cards, B has more than one black card. 4 black cards are already accounted for. From what we have discussed above, we know that 2 of D’s cards must be black (either spades or clubs). Therefore, B,C and D account for at least 6 black cards (Please note the word “at least”, 6 may not be the ultimate number of black cards accounted for by them).

Since they account for 6 black cards, they must include at least 1 spade and 1 club. Therefore, all of A’s cards are hearts. So, the other 2 suits held by D are Spades and Clubs. Since C takes the King of spades, King and Jack must belong to clubs and Ace and Queen must belong to spades. Now we can make a small table with what we know.

Person

Cards

A

A K Q J 10 – all hearts

B C

Ks

D

10 d A Q s K J c

The cards that remain in each suit are: Spades: J and 10 Clubs: A, Q and 10 Diamonds: A, K, Q and J C has 3 red and 2 black colour cards. The only remaining cards in red are the ones in diamonds above. Since C already has a K, he must have the A,Q and J of diamonds, which means that he also has the 10 of clubs. Now that C is over, all the remaining cards are assigned to B. The final table will look as follows:

Person

Cards

A

A K Q J !0 h

B

K d J 10 s A Q c

C

K s A Q J d 10 c

D

10 d A Q s K J c

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Number Game This game requires a good knowledge of simple arithmetic, mainly LCM and factorisation. The problems will generally be in the form of a grid, wherein based on the conditions given, the grid has to be filled with the appropriate numbers.

One such game is illustrated below:

There are nine letters A to I, each represented by a different number from 1 to 9. The grid is positioned as below:

A

B

C D E

F

G H I

Each of the combinations of letters, A+B+C, C+D+E, E+F+G and G+H+I is equal to 13. Match the letters to the numbers.

Solution/discussion

We can begin by making a grid similar to that made by the letters above.

A

B

C

D E

F

G H I

Here, distinction has to be made between A,B, D, F, H and I on one side and C,E, and G on the other, because the latter occur in two sums of numbers.

Since the numbers have to total 13, we can try different ways in which the number can be arrived at by adding three numbers.

1+3+9

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1+4+8 1+5+7 2+3+8 2+4+7 2+5+6 3+4+6

We can safely say that one of A,B,D,F,H and I can assume the number 9, and 9 can occur only with 1 and 3. So, if we assume A to be 9, then we can assume B as 3 and C as 1, which means that D and E should stand for either 8 and 4 or 5 and 7.

Using the first combination of 8 and 4, we find that E cannot be 8 because the only other combination possible with 8 is 2 and 3, and 3 is already present. So E has to be 4, which means that F and G should stand for 7 and 2. G, again, cannot stand for 7, because the only other possible combination with 7 is 1 and 5, and 1 is already present. If G stands for 2, then H and I stand for 5 and 6.

On the other hand, if D and E stand for 5 and 7, then, assuming E stands for 7, we find that the only other combination with 7 is F and G being 4 and 2. If G is 4, then H and I should be 3 and 6, but 3 has already been used, so, this combination is not possible. If G is 2, then H and I should be 3 and 8 or 4 and 7, and again, we find that in both cases, a number has already been used. From these two alternatives, we find that the first one is a feasible alternative and therefore, one possible outcome of matching numbers can be,

A 9; B 3; H 5;

C 1; I 6

D 8;

E 4; F 7;

G2;

Please note that there can be other solutions based on the arrangement of numbers, say, for example, if a student assigns 9 to I, then the grid would look different. But it is preferable to begin with a number that can appear only once on the grid, in this case, 9. Variations of this game include Â&#x153;

Finding numbers for a similar grid, where products, instead of sums, are given.

Â&#x153;

Identifying missing numbers in a sum/multiplication/division problem.

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Bulls Eye This game basically involves scoring a certain number of points on the dartboard. This may involve more of trial and error than method. But, even in the former, a proper understanding of multiplication tables and factorisation is necessary. As the following problem will indicate, its never easy to solve such problems in a fraction of a minute unless one can calculate quickly. The following numbers appear on a dartboard â&#x20AC;&#x201C; 46, 44, 42, 33, 31, 13 and 11, with 46 on the Bulls eye and the rest of the numbers appearing on either side of it. Ascertain the least number of attempts required for a person to score exactly 100 points.

11

13 31

33

44 46

42

Solution/Discussion This problem requires the student to have quick calculation skills, especially multiplication. A good way to attack the problem is to use the bottom-up approach, i.e. start from 100 onwards and identify the numbers on the darts that will satisfy the given condition. From the board, we can easily identify that the score before the dart hits the final 11 points to score 100 should be 89. Similarly, it should be 87 for 13, and so on. Solving along these lines for all kinds of possible combinations, such as, 11+13=24, therefore 76 to be scored beforehand 11+31=44, therefore 56 to be scored beforehand 13*2 = 26, therefore 74 to be scored beforehand, and so on, till we can arrive at a proper match. In this case, after a few combinations of numbers and their multiples, we identify that a person needs 8 attempts to score exactly 100 points. The 8 attempts can be broken down into 6 attempts at the target 13 and 2 at 11, which is equal to 13*6=78 11*2=22, totaling 100.

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An important point to note would be that the answer would definitely not be a very small number, because the analysis required for it would be very minimal.

Who’s the winner??? This game is a variant of the number game illustrated above. These games generally consist of a table, wherein some numbers are missing or are substituted by letters. There are no explicit conditions in these games, and so require some trial and error to solve them. Its Copa America time and the South American football teams are playing each other for top honours. The table below gives the results at a certain point during the tournament. A team gets 2 points for a win, 1 point for a draw and no points for losing. Each letter below represents a distinct integer value. No two letters represent the same integer. Each team plays the other only once. Team

Played

A

Y

Won

Drawn

Lost

Points

a

Y

B

Q

C

X

Y

p

a

Q

D

Q

A

E

Y

A

The questions may be w.r.t. number of games played by a team, points taken, whom they still have to play against and values of the letters. A couple of them are given below: Q1. What is the value of q? a) 0 b) 1 c) 2 d) 4 Q2. How many games has E played till now? a) 0 b) 1 c) 2 d) 3

Solution/Discussion A team could have played its 4 matches, won all of them and scored 8 points, or a team would have lost all its matches. Therefore, the numbers can range from 0 to 8. All columns of the table are filled w.r.t team C, so, we use C as the starting point. At this point, it is advisable to look through the questions, and look at a question, which can help us solve the row pertaining to C. In this case, only the first question is relevant. When we analyse the first question, we find that q cannot take the value of 3. Also, C’s row consists of 5 different numbers, and he should have won at least one point. We can make a simple assumption that q is equal to 4 and start working the other numbers. If q is 4, then C must have played at least 2 games and won them both or played 3, won one, drawn two, or won two and lost one. C couldn’t have played 4 games because the value of q is 4. Also, C couldn’t have played 2 and won them both because 2 different letters, x and y, represent the columns of played and won respectively. So, he must have played 3 games. Now, we assume that C has won 2 matches and lost the other. Then, y represents 2 , p represents 0 and a represents 1.

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Once we have arrived at the numbers, we’ll proceed to solving the whole table. A has played 2 matches, has 2 points from them, and has lost a match. Therefore, he must have won the other match. So, we fill in A’s row with the relevant numbers. Since only one column in team B and E has been filled, we shall ignore them for the time being and proceed with team D. We find that D has played 4 matches and has only one point. Therefore, he must have drawn a game and lost the rest. We fill in the columns with the respective numbers. At this stage, the table will look as follows:

Team

Played

Won

Drawn

Lost

Points

A

Y

A

p

a

Y

B

Q

C

X

Y

p

a

Q

D

Q

P

a

x

A

E

Y

Since team D has drawn a game and none of the other teams identified, A or C, has drawn a game, B or E should have drawn a game with D. B must have played at least 2 games since it’s score is 4. Now, if we analyse the “Lost” column, we find that 5 games have been lost but on comparison with the “Won” column, we find that only 3 games have been won. Therefore, we can be sure that B must have won at least one game. It can also happen that B would have drawn two of its games, one each with D and E. (For the time being, we keep the number of losses constant and work the problem). So, the other game must have been won by E. If we look at the table now,

Team

Played

Won

Drawn

Lost

Points

A

Y

A

p

a

Y

B

X

A

y

p

Q

C

X

Y

p

a

Q

D

Q

P

a

x

A

E

Y

A

a

p

X

Converting the table into its numerical form – Team

Played

Won

Drawn

Lost

Points

A

2

1

0

1

2

B

3

1

2

0

4

C

3

2

0

1

4

D

4

0

1

3

1

E

2

1

1

0

3

From the table above, we can find that D has drawn with B and lost to the other 3 teams. B and C have to play one game each but they can’t play themselves since the other 2 teams A and C have 2 games left to play. Therefore, since B has already played C, it must have beaten it, which, in turn, means that C must have beaten A for its only other win and is yet to play E. Similarly, B is yet to play A, and A and C have to play each other.

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Variants of this game include: 

Identifying the correct answer choices given by students



Ascertaining the number of runs scored, wickets taken etc. by players in a cricket match

We have tried to cover different problems that students can encounter during entrance examinations. Though it is not an exhaustive list of problems, students should find it easier to follow a methodical approach on solving analytical reasoning problems. The common mistakes made by students are

Hurrying through the conditions and later, finding it difficult to answer the questions

 Following a set method of solving every game, irrespective of its type. Say, using symbols to solve every game and every question in the game. 

Wasting time reading too much into the conditions, especially in the “conditional” problems.

Before signing off, some things to remember: 

Read the conditions and the accompanying questions carefully



Organise the information according to your convenience. If you are comfortable using symbols, go ahead, but please take some time to identify the easiest approach. Its just a matter of a few seconds to read the conditions and identify the correct and easiest approach.



Use the bottom-up approach if the conditions are not sufficient to solve the questions.



“Conditional” problems take some time. Understand the implications of the conditions while solving them. In case of questions that may have two answers, go for the most logical one or the one that appears correct on the face of it, rather than working out its implications.



Last, but not the least, if nothing strikes you during the examination or if you feel that no approach will help, try Trial and Error. It should work.

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CHAPTER-2

Numerical Puzzles Recent papers of the Campus Placement Test have had a number of group problems in the quantitative sections. In this chapter, the basic approach to these group problems has been discussed with examples. Each group of problems contains the directions that are nothing but the description of a numerical problem situation. The problem situations are from any of the already discussed areas in the quantitative ability fundamentals such as Interest Rates, Profit/Loss, Time, Speed, Functions, Geometry etc. We have called these situations as puzzles because all the unknown information has to be unveiled from one or two links provided generally at the end of the description of the situations. As can be appreciated from the solved examples and exercise problems given under, two different approaches are required to solve these puzzles depending on the type of questions. In the first type, all the unknown but related details have to be determined before finding the best answer choice for the questions (Please refer to solved example 1). The questions only pertain to the unknown variables in the problem situation. In the second type each question contains additional data and the problem needs to be solved with this information separately for each question. In this type it is necessary that each question be treated independently. Also the information provided in one question pertains only to that question and should not be used in subsequent problems, unless otherwise mentioned.

Solved Examples Example 1 : DIRECTIONS for questions 1 to 5. Jeshwanth, Krishna and Lokesh invested in a certain business. As Lokesh was to manage the business alone, it was decided that Lokesh would take 20% of the profit earned by the business and the remaining amount would be shared by all the three in proportion to their investment. Thus at the end of the first year each of them earned Rs.10,000/- from the business. While Krishna withdrew 50% of his investment in the second year, Jeshwanth invested his first year’s profit also into the business. The business earned the same amount of profit in the second year as in the first year. Jeshwanth got a share which was 2.5 times the share of Krishna. 1.

What is Krishna’s total profit for two years? 1. Rs.17,102

2.

3. Rs.21,903.60

4. Rs.14,788

What is Krishna’s investment in the second year? 1. Rs.20,000

3.

2. Rs.15,581.40

2. Rs.18,000

3. Rs.28,000

4. Rs.15,000

3. Rs.16,000

4. Rs.14,000

What is Lokesh’s investment in the business? 1. Rs.17,500

2. Rs.22,350

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4.

What is the % profit on investment for Lokesh in the second year? 1. 38%

5.

2. 65%

4. 47%

What is the profit earned by the business as a percentage of investment in the first year? 1. 42.75%

Ans:

3. 29%

2. 31.25%

3. 52.22%

4. 35.35%

To answer all these questions, it is first necessary to organize the given information. From the data given it is clear that both Jeshwanth and Krishna get the same share of profit in the first year. Hence they must have invested the same amount in the first year. Let the investment made by Jeshwanth and Krishna in the first year be X. The total profit for the first year is Rs.30,000/- and 20% of this amount, that is Rs.6,000/- goes to Lokesh for managing the business alone. As Lokesh totally received a share of Rs.10,000/- in the first year, his share in proportion to his investment must be Rs.10,000 - Rs.6,000 = Rs.4,000. Clearly this amount is 40% of the share received by the other two partners. Therefore Lokesh’s investment in the first year should be 40% of Jeshwanth’s or Krishna’s investment i.e., 0.4X.

Now the data can be represented for two years as below. Jeshwanth

Krishna

Lokesh

Investment

X

X

0.4X

Profit

Rs.10,000

Rs.10,000

Rs.10,000

Investment

X + 10,000

0.50X

0.40X

Profit

2.5Z

Z

30,000 - 3.5Z

I Year

II Year

(assuming Z as Krishna’s Profit) As the ratio of the profit shared by Jeshwanth to that of Krishna is 2.5, the ratio of their investments must also be equal to 2.5.

(X + 10, 000) Therefore

0.5X

= 2.5

On solving the above equation X = 40,000/The investments of Jeshwanth, Krishna and Lokesh for the second year are respectively Rs.50,000, Rs.20,000 and Rs.16,000. The profit earned by the business in the second year is Rs.30,000. Out of this Rs.6,000 is paid to Lokesh for managing the business. The balance amount of Rs.24,000/- is to be shared among the three in proportion to their investment. The final table looks as depicted below.

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Jeshwanth

Krishna

Lokesh

Investment

Rs.40,000

Rs.40,000

Rs.16,000

Profit

Rs.10,000

Rs.10,000

Rs.10,000

Investment

Rs.50,000

Rs.20,000

Rs.16,000

Profit

Rs.13,953.50

Rs.5,581.40

Rs.10,465.10

I Year

II Year

1.

Krishna’s total profit for two years is Rs.15,581.40.

2.

Krishna’s investment in the second year is Rs.20,000.

3.

Lokesh’s investment in the business is Rs.16,000.

4.

10,465.10 Lokesh’s profit to investment ratio is Rs. 16,000 . This is equal to approx. 65%.

5.

The profit earned by the business is Rs.30,000 and the total investment is Rs.96,000. The profit as a percentage of the investment is 30/96 = 31.25%. Answer is 2.

Example 2: DIRECTIONS for questions 6 to 10.

The PNV group is interested in investing in a High Technology Venture. The project details have been worked out and are as below. Investment required = 250% of the annual sales desired. Profit = 25% Working capital requirement = 35% of cost of goods. 6.

What should be the investment if PNV wants to earn a profit of Rs.2 crore in the first year? 1. Rs.25 cr.

Ans:

2. Rs.35 cr.

3. Rs.28 cr.

4. Rs.18 cr.

% Profit = (Sales - Cost) / Cost x 100 25 = (S - C)/ C x 100 Or S = 1.25 C Therefore profit = S - C = S - S/1.25 = 0.25S/1.25 Required profit is Rs.2 crores.

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â&#x2C6;´ Sales = Rs.10 crores.

It is given that the investment is 250% of annual sales desired. Hence the investment would be Rs.25 crores. Answer is 1. 7.

If PNV would like to restrict its investments to Rs.10 crores, what will be the annual profit? 1. Rs.1 cr.

2. Rs.80 lakhs

3. Rs.1.2 cr.

4. Rs.2.2 cr.

Ans:

The annual sales possible with an investment of Rs.10 crores is Rs.4 crores. The profit with this sales would be Rs.0.8 crores or Rs.80 lakhs. Answer is 2.

8.

What is the working capital required for a sale of 15 crores? 1. Rs.3.7 cr.

2. Rs.2.9 cr.

3. Rs.5.3 cr.

4. Rs.4.2 cr.

Ans:

The profit with a sales turnover of Rs.15 crores is Rs.3 crores. Hence the cost of goods will be Rs.12 crores. The working capital requirement is given to be 35% of the cost. Therefore the working capital required is 0.35 x 12 = Rs.4.2 crores. Answer is 4.

9.

In problem 8 above, if 40% of the investment is borrowed from Financial Institutions at an interest rate of 18%, what will the profit after interest earned by the company be? 1. 0.3 cr.

2. 2.2 cr.

3. 0.85 cr.

4. 1.15 cr.

Ans:

As the sale is Rs.15 crores, the investment would be 2.5 x 15 = Rs.37.5 crs. The amount borrowed from financial institutions is 0.40 x 37.5 = Rs.15 crores. The annual interest on this borrowing is 18% of 15 crs. = Rs.2.7 crs. The profit earned is Rs.3 crs. and the profit after interest will be Rs.0.3 crs. Answer is 1.

10.

If an annual net profit (profit after interest) of Rs.2 crores is to be achieved and PNV wants to borrow 40% of the investment at an interest rate of 18%, what is the total investment required? 1. Rs.200 cr.

Ans:

2. Rs.20 cr.

3. Rs.250 cr.

4. Indeterminate.

If the investment is I, the sales possible annually is I/2.5. The gross profit will be I/2.5 x 0.25/1.25. The amount to be borrowed from the financial institutions is Rs.0.41. The annual interest on this amount is 0.18 x 0.41. The profit after interest is I/2.5 x 0.25/1.25 - 0.18 x 0.41 = 2 crores. On solving I = Rs.250 crores. Answer is 3.

Example 3: DIRECTIONS for questions 11 to 16.

XYZ Co. Ltd earned a profit of 20% in 1989. In 1990 the sales of the company improved by Rs.6,000/- and the profit increased to 25%. Capitalising on the high demand for its product, the company earned the highest profit of 50% in 1991 but maintained the sales at the same level as

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1990. In spite of the severe recession during 1992, the companyâ&#x20AC;&#x2122;s sales improved impressively by 20% but the profit slumped to the lowest ever figure of 12.5%. The amount of profit earned in 1992 was the same as the amount of profit earned in 1989. 11.

What was the sales of XYZ in 1990? 1. Rs.42,000

12.

2. Rs.36,000

2. Rs.8,000

2. 20%

2. Rs.24,000

3. 28%

4. 30%

3. Rs.41,000

4. Rs.18,000

If the cost of each item produced in 1991 was the same as that in 1990, the percentage decrease in quantity sold in 1991 as compared to 1990 was 1. 18%

16.

4. Rs.11,000

The total profit earned for four years is 1. Rs.20,000

15.

3. Rs.10,000

The average percentage profit for four years is 1. 25%

14.

4. Rs.30,000

What was the profit earned in 1990? 1. Rs.6,000

13.

3. Rs.27,500

2. 16.66%

3. 22.5%

4. 25%

In problem 15 above the percentage increase in the price in 1991 as compared to 1990 was 1. 20%

Solution:

2. 25%

3. 30%

4. 35%

The given information can be compiled as follows (assuming sales in 1989 as S) Sales

1989

1990

1991

1992

S

S + 6,000

S + 6,000

1.20 (S + 6,000)

% profit

Profit

20%

0.20S 1.2

25%

0.25(S + 6,000) 1.25

50%

0.50(S + 6,000) 1.5

12.5%

0.125 x 1.2(S + 6,000) 1.125

It is given that the profit earned in 1992 is equal to that in 1989.

0.125 x 1.2 (S + 6000) 0.20 S = 1.2 , 1.125 Therefore, On solving S = 24,000

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The details can now be written as follows: Sales

% profit

Profit

1989

24,000

20%

4,000

1990

30,000

25%

6,000

1991

30,000

50%

10,000

1992

36,000

12.5%

4,000

11.

Sales in 1990 was Rs. 30,000. Answer is 4.

12.

Profit earned in 1990 was Rs. 6,000. Answer is 1.

13.

The total sales for four years is Rs. 1,20,000 and the total profit is Rs. 24,000. Therefore the total cost is Rs. 96,000. The % profit is 25. Answer is 1.

14.

Total profit earned for four years is Rs. 24,000. Answer is 2.

15.

The cost of goods in 1990 was 30,000 - 6,000 = 24,000. If cost of each item is Rs. 200 (assume), the quantity sold in 1990 was 24,000/200 = 120. Similarly the cost of goods in 1991 was 30,000 - 10,000 = 20,000. As the cost of each item remains same (at Rs. 200 in this case), the quantity sold in 1991 was 20,000/200 = 100. Hence the percentage decrease in quantity sold in 1991 as compared to 1990 was (120 100)/120 = 16.66%. Answer is 2.

16.

From the above answer, the price of each item in 1990 was 30,000/120 = 250. The price of each item in 1991 was 30,000/100 = 300. The increase in price from 1990 to 1991 is (300 - 250)/250 = 20%. Answer is 1.

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Practice Exercise 1 DIRECTIONS for questions 1 to 5.

Both Amar and Prem individually invested certain amounts in a bank at a rate of interest of 5%. Interest at the end of each year will be accumulated into the principal amount. At the end of the first year, Amar withdrew 50% of the accumulated amount, while Prem withdrew Rs.11,500. At the end of the third year, Prem had an accumulated amount equal to that which Amar had at the end of the second year. Amar withdrew Rs.4,050 at the end of the second year. The total interest earned by Amar is Rs.3,950 for three years. 1.

What is the amount initially invested by Amar? 1. 35,000

2.

2. 40,000

3. 30,000

4. 48,000

2. 3,950

3. 1,975

4. 4,350

3. 1050

4. 1800

Interest earned by Amar in the third year is 1. 1235

5.

4. 30,000

The total interest earned by Prem for three years is 1. 3,550

4.

3. 31,500

The amount initially deposited by Prem is 1. 35,000

3.

2. 40,000

2. 900

If Prem had not withdrawn any amount in three years he would have had an accumulated amount of 1. 34,725

2. 43,475

3. 45,447

4. 39,912

DIRECTIONS for questions 6 to 10.

Ramkumar invests 60% of his retirement money in fixed deposits that earn an interest of 15% at the end of every year. He decides to take out this interest amount at the end of every year to meet his personal expenses. He invests the balance 40% of the retirement money in shares. At the end of the first year he sells off all the shares and realises a profit of 25%. Inspired by the return on investment on shares, Ramkumar decides to take out 25% of the money in fixed deposits and invest it in shares along with the balance money. At the end of the second year he sells all the shares and incurs a loss of 10%. Taken aback by the loss, Ramkumar puts the amount withdrawn by him earlier from fixed deposits back into it. He reinvests the balance amount again in shares. After selling all the shares at the end of the third year he realises a profit of 10%. The profit earned by Ramkumar in the three years of his transactions in shares is Rs.18,900. 6.

What is the retirement benefit that Ramkumar obtained? 1. 2.5 lakhs

7.

2. 2 lakhs

3. 3.2 lakhs

4. 1.6 lakhs

What is the total amount of interest earned by Ramkumar on fixed deposits? 1. 49,500

2. 27,000

3. 52,000

4. 38,900

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8.

What is the loss incurred by Ramkumar in transacting in shares in the second year? 1. 16,500

9.

3. 13,000

4. 18,700

What was Ramkumarâ&#x20AC;&#x2122;s investment in shares in the third year? 1. 87,000

10.

2. 15,000

2. 1,05,000

3. 98,500

4. 72,250

The average annual return on investment earned by Ramkumar in three years is 1. 10.8%

2. 15.4%

3. 19.2%

4. 9.2%

DIRECTIONS for questions 11 to 15:

A company manufactures two products X and Y. Both these products are to be processed on machines A, B, C in that order. Product X requires 10 hours on machine A, 20 hours on machine B and 15 hours on machine C. Product Y requires 12 hours on machine A, 15 hours on machine B and 24 hours on machine C. There are two machines of type A, 3 machines of type B and 4 machines of type C. The company works for 8 hours a day. 11.

If the company makes only product X, how many numbers of X can be made in a month consisting of 25 working days? 1. 55

12.

2. 41

3. 55

4. 72

2. 33

3. 13

4. 16

If products X and Y are to be dispatched in pairs, how many pairs of X and Y can be produced in a month of 25 days? 1. 17

15.

4. 35

The company has to produce 20 numbers of product X in a month. If there are 25 working days in the month, how many numbers of product Y can be made with the remaining capacity? 1. 24

14.

3. 72

In problem 11 above, how many numbers of product Y alone can be produced in the month? 1. 33

13.

2. 30

2. 18

3. 12

4. 27

In problem 14 above, what is the percentage unutilized capacity? 1. 9.33%

2. 21%

3. 10.5%

4. 8%

DIRECTIONS for questions 16 - 20:

Ghosh Babu deposited a certain amount of money in a bank in 1986. The bank calculated interest on the balance in the account at 10 percent simple interest, and credited it to the account once a year. At the end of the first year, Ghosh Babu withdrew the entire interest and 20 percent of the initial amount. Again, at the end of the second year, he withdrew the interest and 50 percent of the remaining amount. At the end of the third year, he withdrew the interest and 50 percent of the remaining amount. Finally, at the end of the fourth year, Ghosh Babu closed the account and collected the entire balance of Rs.11,000.

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16.

The initial amount in rupees, deposited by Ghosh Babu was 1. 25,000

17.

2. 75,000

2. 20,000

2. Second

4. 11,000

3. Third

4. Fourth

The year, at the end of which, Ghosh Babu collected the maximum interest was 1. First

20.

3. 4,000

The year, at the end of which, Ghosh Babu withdrew the maximum amount was 1. First

19.

4. None of these.

The total interest, in rupees, collected by Ghosh Babu was 1. 12,000

18.

3. 50,000

2. Second

3. Third

4. Fourth

The year, at the end of which, Ghosh Babu withdrew the smallest amount was 1. First

2. Second

3. Third

4. Fourth

DIRECTIONS for questions 21 - 25:

Prakash has to decide whether or not to test a batch of 1,000 widgets before sending them to the buyer. In case he decides to test, he has two options : (a) Use test I (b) Use test II. Test I costs Rs.2 per widget. However, the test is not perfect. It sends 20% of the bad ones to the buyer as good. Test II costs Rs.3 per widget. It brings out all the bad ones. A defective widget identified before sending can be corrected at a cost of Rs.25 per widget. All defective widgets are identified at the buyerâ&#x20AC;&#x2122;s end and a penalty of Rs.50 per defective widget has to be paid by Prakash. 21.

22.

23.

24.

Prakash should not test if the number of bad widgets in the lot is 1. less than 100

2. more than 200

3. between 120 and 190

If the number of defective widgets in the lot is between 200 and 400, Prakash 1. Should use either Test I or Test II

2. Should use Test I only

3. Should use Test II only

4. Cannot decide.

If there are 200 defective widgets in the lot, Prakash 1. Should use either Test I or Test II

2. Should use Test I or not use any test

3. Should use Test II or not use any test

4. Cannot decide.

If Prakash is told that the lot has 160 defective widgets he should 1. Use Test I

25.

2. Use Test II only

3. No test

4. Use Test I or No test.

If there are 120 defective widgets in the lot, Prakash

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1. Should use either Test I or not test

2. Should use Test II or not test

3. Should use Test I or Test II

4. Should use Test I only.

DIRECTIONS for questions 26 - 30:

On 27-11-92, Sanjay bought 500 shares of company A at a price of Rs. 80/-, 100 shares of company B at a price of Rs. 45 and 200 shares of company C at a rate of Rs. 50/-. The sensitive index on 2711-92 was 2,000. Sanjay knows that the price of share A is directly proportional and share price of B is inversely proportional to the sensitive index. The share price of C is proportional to the square root of the share price of A. 26.

If Sanjay again buys 500 shares of A at an index of 2400, what will be his average price of share A? 1. Rs.92/-

27.

2. 3,825

3. 1,110

4. 2,000

2. Rs.82,000

3. Rs.48,250

4. Rs.72,000

Sanjay predicts that the index would reach a peak of 2400 and would fall down to 1800 in the coming month. What will be the profit earned by Sanjay if he sells all his shares at the peak index and buys back the same shares at the lowest index? 1. Rs.14,800

30.

4. Rs.12/-

What is the total value of shares possessed by Sanjay at an index of 3,000? 1. Rs.75,250

29.

3. Rs.96/-

If Sanjay sells 200 shares of C when the index is 2,500, what is his gain in rupees? 1. 2,500

28.

2. Rs.98/-

2. Rs.12,700

3. Rs.13,850

4. Rs.18,750

3. 1,000

4. 6,400

3. 57/9

4. 65/7

At what index will the price of C double? 1. 4,000

2. 8,000

DIRECTIONS for questions 31 - 35:

x ⊕ y = x2 + x2 y2

x x∅y= 31.

What is the value of (2 ⊕ 3) ∅ 4? 1. 47/13

32.

2. 13/2

Given x = 5, for what value of y does (y ⊕ 3) ∅ x become equal to 5? 1. 27/7

33.

y2 + y

2. 0

3. 17/3

4. 19/4

If x ⊕ y = m , x ∅ y = n, what is m ⊕ n for x = 2, y = 1?

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1. 412 34.

3. 197

4. -4

In problem 33 above if y = 1, then for what value of x will m be equal to n? 1. 2

35.

2. 278

2. -2

3. 1

4. -4

When will x â&#x160;&#x2022; y be equal to x + y? 2. Only when y is equal to x2.

1. Only when y is zero. 3. When y is equal to either zero or

x.

4. None of these.

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CHAPTER-3

Blood Relationships In the study of Blood relations we come across three major types of problems. ¾

The relation will be described with you as a centre and is given in a round about manner i.e. one needs to under go a series of relationships before arriving at the conclusion.

¾

The relation will be given between the two people in a round about manner.

¾

Here certain codes are used to indicate certain relations. One needs to decode it and come to the conclusion.

In order to solve the problems related to blood relation we need to know the following relations Mother’s mother

:

Grand mother

Father’s mother

:

Grand mother

Mother’s father

:

Grand father

Father’s father

:

Grand father

Grandmother’s brother

:

Grand uncle

Grandmother’s sister

:

Grand aunt

Grandfather’s brother

:

Grand uncle

Grandfather’s sister

:

Grand aunt

Father’s son

:

Brother

Mother’s son

:

Brother

Mother’s daughter

:

Sister

Father’s daughter

:

Sister

Mother’s brother

:

Uncle

Father’s brother

:

Uncle

Mother’s sister

:

Aunt

Son’s wife

:

Daughter-in-law

Daughter’s husband

:

Son-in-law

Husband’s sister

:

Sister-in-law

Wife’s sister

:

Sister-in-law

Husband’s brother

:

Brother-in-law

Wife’s brother

:

Brother-in-law

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Sister’s son

:

Nephew

Brother’s son

:

Nephew

Sister’s daughter

:

Niece

Brother’s daughter

:

Niece

Uncle or Aunt’s son

:

Cousin

Uncle or Aunt’s daughter

:

Cousin

Sister’s husband

:

Brother-in-law

Brother’s wife

:

Sister-in-law

Children of same parents and sister)

:

Siblings (brother and brother, sister and sister, brother

Spouse means either husband’s wife or wife’ husban4. The graph given below will help us to get the concepts properly.

Draw family tree in which stem represents mother and father, roots, grant parents, branches, children, and leaves, the children of the children. Aunt can be mother’s or father’s sisters. Uncles are father’s or mother’s brothers. These are common relations and one can easily spot out the relationship.

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Practice Exercise 2 1.

2.

If a+b means a is the husband of b, a ÷ b means a is the sister of b and a × b means a is the son of b which of the following shows that A is the daughter of B? 1. A ÷ D × B

2. D × B ÷ C ÷ A

3. B x C × A

4. C + B ÷ A

The man who is receiving the hat is the brother of uncle’s daughter. What is the relation of the speaker and the man getting the hat. 1. Son in law

3.

3. Cousin

4. Niece

2. Son in law

3. Brother

4. Husband

2. Aunt

3. Sister in law

4. Niece

2. Brother

3. Uncle

4. Nephew

2. Uncle

3. Sister

4. Daughter in law

Pointing to a lady Mohan said “She is the daughter of a woman who is the mother of the husband of my mother. How is the lady related to Mohan? 1. Aunt

10.

2. Brother

Introducing a man to her husband a wife said that his brother’s father is the only son of my grand father. How is the wife related to the man? 1. Mother in law

9.

4. Grandmother

Pointing to his son’s photograph Ralph said to a woman “His mother is the only daughter of your mother.” How is the woman related to Ralph? 1. Wife

8.

3. Aunt

Introducing a man a woman said, “he is the only son of my mother’s mother”. How is the woman related to the man? 1. Mother

7.

2. Sister

Pointing to a lady in a photograph John said, “She is the daughter of my grandfather’s only son”. How is John related to the lady? 1. Father in law

6.

4. Nephew

Dennis said to Raman that the boy who has won the game is the younger of the two brothers of the daughter of my father’s wife.” How is the boy related to Dennis? 1. Son in law

5.

3. Brother

Pointing to a girl in the photograph Umesh said “Her mother’s brother is the only son of my mother’s father.” How is girl’s mother related to Umesh? 1. Mother

4.

2. Cousin

2. Mother

3. Daughter in law

4. Sister in law

Pointing to a man in group Sheela said “He is the brother of the daughter of the wife of my husban4.” How is the man in group related to Sheela? 1. Son

2. Uncle

3. Sister

4. Niece

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CHAPTER-4

Calendars General Information about Calendars 1.

A period of one year comprises of 12 months.

2.

Total number of weeks in a year is 52.

3.

A normal year (Non Leap Year) consist of 365 days, where as a Leap Year consist of 366 days.

4.

Number of days in a month is either 31 days or 30 days depending upon what month we are taking into consideration but only in the month of February we find 28 days and if it is a leap year then the month of February contains 29 days.

Concept of Odd Days Consider that 2006 February 22nd was a Thursday. If it is required to be find the day on which the next 5th would fall, first calculate the total number of days from February 22. It happens to be 11 days. Then divide this number by 7 to obtain the reminder. Here, we have 11/7 = 4 as the reminder. Then count the 4th day starting from Thursday ∴ The day on March 5th is Monday

Odd days are basically the remainder obtained by dividing the number of days under consideration by 7.

Concept of Leap Year Every year consists of 365 day and quarter of a day, but for all practical consideration we take the year to be comprising of 365 days only. That quarter of a day is taken into account every year and at the end of 4 year we say that one extra day is added to the total number of days. Hence a leap year is going to have 366 days A leap year is a year, which is divisible by 4 and if it’s a century it must be divisible by 400. 366 The number of odd days in a leap year =

7

= 2 odd days

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365 The number of odd days in a non leap year =

7

= 1 odd day

As such, if June 19th of one particular year is Monday then June 19th of the very next year (if it is a non leap year) happens to be Tuesday because a non leap year has one odd day. Whereas, if the next year is a leap year then June 19th is going to be Wednesday as a leap year is going to be having 2 odd days. In a Century

100 Odd days in a period of 100 years =

4

= 25

As such, in a century, there must be 25 leap years and 75 non leap years. But the 25th year happens to be 100th year, which is not a leap year. ∴ Totally 24 leap years and 76 non leap years, in a century.

We know that each leap year has 2 odd days and non leap year 1 odd day. So, a period of 100 years is going to have (48+76) = 124 odd days 124×2 i.e. effectively

7

=3 odd days.

Similar Cases

124×3 A period of 300 years is going to have

7 124× 4

A period of 400 years is going to have

7

=1 odd day.

= 6 odd days

But the 400th year happens to be a leap year hence it would contain an extra day or odd day. ∴ Total number of odd days is, 6+1 = 7 odd days or effectively ‘0’ odd days.

This concept will help us to calculate the days on any year, any month. We need to remember a table so as to find out the day depending upon the number of odd days. If we get 0 as the odd day then it is Sunday. Similarly, 1 ⇒ Monday

2 ⇒ Tuesday

3 ⇒ Wednesday

4 ⇒ Thursday

5 ⇒ Friday

6 ⇒ Saturday

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Example: What was the day on 28 November, 1982. Step 1: Consider up to 1900

It can be split as 1600 + 300 We know that 1600 is a multiple of 400, so the total number of odd days is zero. A period of 300 years will have 1 odd days. So effectively up to 1900 we have just 1 odd day. Now consider 81 years. By dividing 81 by 4, we get 20 leap years and 61 non leap years. i.e. 40 + 61 = 101 odd days (20×2+61×1 = 101) Up to 1981 we have (1 + 101) odd days The month of January has 31 days ⇒ 3 odd days.

(4 weeks + 3 extra days)

February has 28 days ⇒ 0 odd days. March has 31 days ⇒ 3 odd days. April has 30 days ⇒ 2 odd days. May has 31 days ⇒ 3 odd days. June has 30 days ⇒ 2 odd days. July has 31 days ⇒ 3 odd days. August has 31 days ⇒ 3 odd days. September has 30 days ⇒ 2 odd days. October has 31 days ⇒ 3 odd days. So up to 1981 Oct 31st we have (1+101+3+0+3+2+3+2+3+3+2+3) odd days Finally consider month of November up to 28 days When you add 28 days to the consisting odd days You will get total number of odd days up to 28th November 1982 = 126 + 28 = 154 odd days or 0 odd days ∴ From the table we can make out that it is Sunday.

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Sometimes, it may be required to find out which year has or will have the same calendar as this year. If it is a leap year it is going to repeat after 28 years and if it is a non leap year, the method is explained in the example below. Example: Which year will have the same calendar as that if 2001. 2001 is a non leap year hence 1 odd day. 2002 is a non leap year hence 1 odd day. 2003 is a non leap year hence 1 odd day. 2004 is a leap year hence 2 odd day. 2005 is a non leap year hence 1 odd day. 2006 is a non leap year hence 1 odd day. Once we get the total number of odd days to be 7 or 0, the next year is going to have the same calendar as the year under consideration. i.e. year 2007 will have the same calendar as 2001 year. Note: If the year after 7 odd day count happens to be a leap year then we have continue the same procedure will we get a non leap year by taking multiple of 7 into consideration.

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Practice Exercise 3 1.

If December 25th 1965 was a Friday then what day was Christmas in 1966? 1. Saturday

2.

7.

3. 7

2. 9 birthdays

3. 12 birthdays

2. 1971

3. 1972

2. 1992 November 15th

3. 1992 November 16th

4. 1992 November 11th

4. 8

4. 10 birthdays

4. 1973

If August 15th, 1947 was a Friday and is celebrated as India’s Independence Day. How many ‘Fridays’ will India celebrate as Independence Day in this century? 2. 10 Fridays

3. 11 Fridays

4. 12 Fridays

3. Wednesday

4. Thursday

3. Wednesday

4. Thursday

3. Wednesday

4. Saturday

What is the day on 19th July 1983? 2. Tuesday

What is the day on 14th April 1982? 1. Monday

10.

2. 6

1. 1992 November 14th

1. Monday 9.

4. Saturday

November 16th, 1998 was Monday. When was November 16th a Monday the last time ?

1. 9 Fridays 8.

3. Sunday

If June 7th, 1965 was a Thursday then find the year when June 7th was Thursday the next time. 1. 1970

6.

2. Monday

Mrs. Lily was born on February 29th, 1940. She joined a school as a teacher in 1958, June 1st. She retired on March 31st, 1995. How many birthdays did she celebrate in her school? 1. 8 birthdays

5.

4. Tuesday

Former Prime Minister of India Mr. Morarji Desai was born on February 29th in 1896. What was his age when he celebrated his first birthday? 1. 5

4.

3. Monday

If November 16th 1987 was a Thursday what will be November 16th 1988. 1. Tuesday

3.

2. Sunday

2. Tuesday

What is the day on 10th April 1982? 1. Monday

2. Tuesday

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Practice Exercise 4 1.

What is the day on 14th May 1981? 1. Monday

2.

4. Thursday

2. Tuesday

3. Wednesday

4. Thursday

2. Tuesday

3. Wednesday

4. Thursday

2. Tuesday

3. Wednesday

4. Thursday

2. Tuesday

3. Wednesday

4. Thursday

2. Tuesday

3. Wednesday

4. Thursday

3. Sunday

4. Thursday

3. Wednesday

4. Thursday

What is the day on 10th April 1991? 1. Monday

10.

3. Friday

What is the day on 4th June 1981 ? 1. Monday

9.

2. Saturday

What is the day on 25th March 1980? 1. Monday

8.

4. Thursday

What is the day on 14th April 1982? 1. Monday

7.

3. Wednesday

What is the day on 19th July 1983? 1. Monday

6.

2. Tuesday

What is the day on 18th July 1979? 1. Monday

5.

4. Thursday

What is the day on 8th October 1977? 1. Monday

4.

3. Wednesday

What is the day on 7th June 1982? 1. Monday

3.

2. Tuesday

2. Tuesday

What is the day on 10th February 1992? 1. Monday

2. Tuesday

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CHAPTER-5

Clocks The Clocks, we take under consideration, are all analog ones and 12 hour clock. We consider only minute-hand and hour-hand for all calculations and seconds-hand is not considered.

Concept of Relative Speed Consider the time to be 2 pm. For it to be 3 pm, it takes 60 minutes. In a matter of 60 minutes, the minute-hand makes one entire revolution i.e 360° ∴Distance traveled by minute-hand in one minute is given by 360°/60 = 6°

If we divide the entire clock into four parts, each section constitutes for 90°. Between any two number which are at the end of quadrants there are two more numbers. ∴ Angle between any two numbers is 90/3 = 30° ∴ The hour-hand moves by an angle of 300 while traveling between 2 pm to 3 pm ∴ Distance traveled by hour-hand in one minute is given by 30°/60° = ½°

Since both the hands are moving in the same direction the relative speed is going to be the difference of the speed i.e (6 – ½)° = 5½° If the same thing needs to be defined in terms of minute in one hour, the minute-hand covers 60 minute division while the hour-hand covers just 5 minute division. Hence relative speed is = 60 – 5 = 55 minute.

To find out the angle when the time is given or when the angle is the following relation can be used: θ = 30h θ=

11 2

11 2

m (when 30h >

m - 30h ( when

11 2

11 2

m)

m > 30h)

where, θ is the angle formed between the hands of the Clock. m is the minutes h is the hour

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Worked Examples 1.

What is the angle between the minute-hand and hour-hand when the clock shows 2 hours and 30 minutes? Solution:

θ=

11 2

11 2

- 30h

×30 = 165

30h = 30×2 = 60 ∵

11 2

m > 30h

∴θ = 2.

11 2

m - 30h = 165 - 60 = 105

Find the angle between 3 ‘o’ Clock and 4 ‘o’ Clock when the angle formed between the hands of the clock is 45°. Solution:

θ=

11 2

m - 30h

∴ 45 = ∴m =

11 2

270 11

m - 30 × 3 ⇒ 135 = = 24

11 2

m

6 11 24

∴ An angle of 45° is formed at 3 past 3.

6 11 min.

The Clock was slow by 10 minutes at 6 am. But when I saw the watch at 6 pm in the evening it was 20 minutes fast. Find the time when it had shown the correct time. Solution:

When the clock was 10 minute slow initially, instead of showing 6 am it was actually showing 5:50 am and started gaining time from that moment. It was showing 20 minutes more than the correct time at 6 pm. That implies at 6 pm, it was showing 6:20 pm.

In a matter of 12 hours (6 am in the morning to 6 pm in the evening), the clock has gained totally 10 + 20 = 30 minutes. It will show the correct time when it gains 10 minutes, from 6 am onwards.

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To gain 30 minutes it has taken 12 hrs. 10 ×12 ∴ In order to gain 10 minutes it will take,

30

= 4 hrs

⇒ 4 hours from the 6 am. Hence, the clock had shown the correct time at 10 am in the morning.

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Practice Exercise 5 1.

Find the angle between the hour-hand and the minute-hand when the time is 3 hours 40 minutes. 1. 36°

2.

3. 90°

4. 25°

2. 15°

3. 130°

4. 2.5°

2. 157.5°

3. 130°

4. 2.5°

Find the angle between hour-hand and minute-hand at 9’o clock. 1. 36°

6.

2. 15°

Find the angle between hour-hand and minute-hand at 8 hours 15 minutes. 1. 36°

5.

4. 65°

Find the angle between hour-hand and minute-hand at 7 hours 5 minutes. 1. 36°

4.

3. 130°

What is the angle between the hour-hand and minute-hand at 6.30 p.m.? 1. 36°

3.

2. 72°

2. 157.5°

3. 90°

4. 2.5 °

2 clocks are set at exactly 12 o’clock. One clock gains 5 minutes every hour while the other loses 5 minutes every hour. After how many hours will both the minute-hands point at the same number? What is time showing by both the clocks? 1. Clock - I shows 6.30 and Clock -II shows 6.30 2. Clock - I shows 6.30 and Clock -II shows 5.30 3. Clock - I shows 6.30 and Clock -II shows 7.30 4. Clock - I shows 6.30 and Clock -II shows 8.30

7.

2 clocks are set exactly at 6.00 a.m. Clock - I gains 6 mins every 1/2 hour and Clock - II loses 6 minute every 1/2 hour. When Clock I shows 8.24, what is the exact time? What is the time given by Clock II? 1. 8.36 a.m.

8.

2. 9.36 a.m.

3. 7.36 a.m.

3. 6.36 a.m.

A watch loses 5 minutes every half an hour. It is set to the correct time at 4 p.m. What is the exact time when the time indicated by the watch is 8.00 p.m.?

9.

1. 8 hours 48 minutes.

2. 9 hours 48 minutes.

3. 10hours 48 minutes.

4. 11 hours 48 minutes.

A watch gain 3 minutes in every 20 minutes. It is set exactly to correct time at 10.00 a.m. What is the time when it is actually 3.00 p.m. ? 1. 3.45 p.m.

2. 6.45 p.m.

3. 8.45 p.m.

4. 9.45 p.m.

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10.

Two clocks chime together at 1.00 p.m. After that clock-I gains 5 minutes every half an hour. Next when the clocks chime together, what is the time indicated by clock-I and clock-II. What is the exact time? 1. 4.00 p.m.

2. 6.00 p.m.

3. 7.00 p.m.

4. 8.00 p.m.

Practice Exercise 6 1.

A clock gains 5 mins per hour.

It is set right at 2 noon.

After what time will it show the

correct time again? 1. 36 hrs 2.

2. 72 hrs

3. 144 hrs

4. 96 hrs

A clock gains 2 minutes for every minute it covers. It is set right at 12 noon. What is the correct time when it shows 9 p.m., on the same day? 1. 2 p.m

3.

2. 2:30 p.m

3. 3 p.m

4. 6 p.m

Let x be the angle between the minute-hand and the hour-hand of the clock when the time is 1:05 p.m. At which of the following times is the angle between the hands equal to 12x? 1. 1:00 p.m

4.

2. 2:30 p.m

Albert Einstein is a hyperactive kid.

3. 4:52 p.m

4. 11:38 p.m

One day, when he is feeling bored with life, he starts

playing with a clock. It is 7:00 in the evening. He moves the minute-hand backward, and stops it at the first position when the angle between the minute-hand and the hour-hand becomes what it was at 7:00. What time is the clock showing now? 6 : 05 1. 5.

5 11 p.m.

6 : 07 2.

2 8 p.m.

6 : 03 3.

4 17 p.m.

6 : 23 4.

1 9 p.m.

Rutherford wants to measure the height of a skyscraper. Having no other alternative, he hits upon an ingenious solution. He takes a stopwatch at the top of the skyscraper, starts it and drops it.

Then he rushes down to the ground, only find that his beloved stopwatch has

shattered on impact with the ground.

What would the stopwatch have been showing just

before it broke, if the top f the skyscraper subtends an angle of 300 at a point 173 m. from its base? (Assume g = 10 m/sec2) 1. 3:8 sec 6.

2. 4:9 sec

3. 6:1 sec

Gaurav decides to develop a calendar program on a computer.

4. 4.4 sec When given a date as an

input, the program tells the day corresponding to that date. However, Gaurav makes an error while programming, so all the leap year are taken as non-leap years, while all the non-leap years are taken as leap years by the program. What day does the program give for 1 Jan 2001? 1. Sunday 7.

2. Monday

3. Tuesday

4. Friday

Amit is a firm believer in astrology, so one Tuesday he goes to his family astrologer to find the next time that he is going to make a windfall. The astrologer predicts, â&#x20AC;&#x153;It is going to be a

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Thursday, and the number of days between that day and today is going to be a multiple of 40.â&#x20AC;? Can you tell Amit after how many days does his auspicious day arrives? 1. 120 8.

2. 160

3. 200

4. 240

The Kingdom of Eccentricity has a 6 day week, i.e., there is no concept of a Saturday in Eccentricity. Today, the 15th of August 1947 is a Sunday in Eccentricity. What day will lie on 15th of August 2001? 1. Tuesday

9.

2. Thursday

3. Sunday

4. Monday

A clock is set right at 12 noon. It loses a different amount of time every subsequent hour, the different times being related by the fact that they form an arithmetic progression. Let t1 be the time lost in first hour, t2 in second hour and so on. If t3 = 7 min and t13 = 27 min, then what does the clock show when it is 5 a.m. the next day? 1. 1:04 p.m.

10.

2. 11:37 a.m.

3. 11:37 p.m.

4. 10:23 a.m.

Yogesh designs a clock which moves anti-clockwise instead of the normal clockwise routine. If today 31st March 2000 and the clock is started, what will be the date and year as shown by the clock (they also go back in time) after 100 days? 1. 23 Dec 1999

2. 22 Dec 1999

3. 21 Dec 1999

4. 23 Dec 2000

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CHAPTER-6

Direction of Movement •

The concept involves a person (in general) moving certain distances in specified directions.

Then we are asked to find out the net distance traveled between the initial position and final position.

Best approach to solve these problems is to draw a diagram as per the information given in the question.

Let the diagram reflect all the possible data given.

In general terms north direction is referred as the vertical direction and south direction is referred as the downward direction.

For solving the problems on direction one must be thoroughly one needs to be quiet aware of the all possible directions.

The problems related to direction sense are aimed to find out two major aspects, 1.

Total distance traveled between the initial and the final position

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2.

Final direction in which he is moving in the end or after some time

Worked Examples 1.

Rajeev travels a distance of 3 metres towards east from his house; he then travels a distance of 8 metres southwards and then travels a distance of 3 metres towards east and finally travels a distance of 11 metres southwards. What is his vertical distance from his house? Solution:

From the data that is given The total vertical distance traveled 11+8=19m. 2.

Sridhar starts from his house and travels 10 meters towards east, then 8 meters towards right, and then travels 8 meters towards east and 3 meters towards south after that. Finally he turns right and travels 1 meter. What is the total distance he has traveled from his house in the north-south direction? Solution:

The distance he traveled from his house in north-south direction is equal to 8 + 3 = 11 meters. 3.

Anil travels 7 meters towards east, then he turns right and travels 3 metres; then travels 5 meters towards left and then proceeds 3 metres northwards and finally travels 2 meters westwards. How far is he from his house in the vertical direction?

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Solution:

The distance covered by Anil in the north-south direction from his house is equal to 3 - 3 = 0 meters 4.

From A, Sampath travels a distance of 6 metres southwards, then travels a distance of 7 metres leftwards, then travels 6 metres northwards and finally Sol travels 6 meters eastwards to reach a new location. What is the distance he traveled from his previous location? Solution:

The distance traveled vertically is 6 - 6 = 0 metres and the distance traveled horizontally is equal to 7 + 6 = 13 m. Therefore, the distance traveled from his original location is also equal to 13 m. 5.

Lucas starts from his house and goes 2 metres towards east, then turns towards right and goes 25 metres and again goes towards east travelling 15 metres and then turns left and travels for 18 metres. He then goes towards east and travels 7 metres. Howâ&#x20AC;&#x2122;far is he from his house? Solution:

Distance traveled in horizontal direction is : 2 + 15 + 7 = 24 m Distance traveled in vertical direction is : 25 - 18 = 7 m Hence the total distance traveled will be equal to : â&#x2C6;&#x161;(242 + 72) = 25 m

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Practice Exercise 7 1.

Namita travels a distance of 4 m northwards, then travels 3 m westwards, then travels 12 metres leftwards and finally travels 15 metres rightwards. What is the approximate distance of the place she reached from his original place? 1. 17 m

2.

3.

1. 7 m towards east

2. 1 m towards west

3. 7 m towards south

4. 1 m towards east

3. 22 m

4. 12 m

and goes towards his house. To reach her house, she has to Travel 5 km towards north-east from the office followed by 5 travel 5 km towards south-west followed by 5 km towards far is the office from her house?

2. 10 m

3. 0 m

4. 20 m

2. South

3. South-West

4. North-East

One evening two friends Ramya and Sowmya were talking to each other face to face. If Sowmya's shadow was exactly to her right side, which direction was Ramya facing? 2. South

3. West

4. East

Nandy going 50 m to the south of her house, turns left and goes another 20 m then turning to the north, he goes 30 m and then starts walking to her house. In which direction is she walking now? 1. North-West

9.

4. 8 m

After walking 6 km, Dipankar turned right and traveled a distance of 2 km, then turned left and covered a distance of 10 km. In the end Dipankar was moving towards the north. From which direction did Dipankar start my journey?

1. North 8.

3. 12 m

2. 32 m

Sharmila starts from his office follow the following directions: km towards ,south-east; then north-west. Approximately how

1. East 7.

2. 2 m

Krishna starts from his house and travels a distance of 10 m southwards and then travels a distance of 12 m rightwards, then travels a distance of 10 m rightwards and finally travels a distance of 10 m in the eastern direction. At what horizontal distance is he from his house?

1. 5 m 6.

4. 34 m

Laxmi starts from his house, travels a distance of 12 m westwards, then travels a distance of 10 m northwards, then a distance of 10 m eastwards, then a distance of 10 m southwards. What is her distance from his house presently?

1. 2 m 5.

3. 20 m

Sangam travels a distance of 12 m northwards and then travels a distance of 5 m westwards, then a distance of 3 m leftwards and again 6 m leftwards and finally travels 15 m towards the south. What is the present horizontal distance from the place he had started? Is he to the west or east of the starting point?

1. 22 m 4.

2. 42 m

2. North

3. South-East

4. South- West

One evening, two friends Ganga and Sunil were talking to each other with their backs towards each other. If Gangaâ&#x20AC;&#x2122;s shadow was exactly to the right of her, which direction was Sunil facing?

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1. South 10.

4. North-West

2. North-East

3. North

4. South-West

2. 8 km

3. 22 km

4. 17 km

2. 8 km

3. 12 km

4. 151.6 km

A person starts from his house and travels a distance of 10 m southwards and then travels a distance of 12 m rightwards, then travels a distance of 10 m rightwards and finally travels a distance of 10 m in the eastern direction. At what horizontal distance is he from his house? 1. 2 m

15.

3. West

A man goes northwards and travels 5 km and then goes 5 km towards the east, then travels 15 km towards the right and finally travels 17 km towards the right1. Approximately how far is he from the original place? 1. 221.8 km

14.

2. South-East

A person travels 7 km towards the east, then turns right and travels 7 km, then travels 15 km towards the left, then goes 12 km towards the left again and finally goes 5 km towards the north. How far is he from his original place in the horizontal direction? 1. 24 km

13.

4. West

Gagan started walking towards north to his office which is 3 km from his house. From there he turns 135° in the anti-clock wise direction and then 180° in the clock-wise direction. Which direction is he facing now from his office? 1. North-West

12.

3. East

A watch shows 8.30. If the minute hand points towards east, in what direction will the hour hand point? 1. South-West

11.

2. North-west

2. 32 m

3. 22 m

4. 12 m

Sadanada walks 8 km towards North1. Then he turns right and walks a further 8 km. How far and in what direction is he from the starting point? 1. 10 km towards North

2. 11 km towards North-East

3. 16 km towards North-East

4. None of these

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CHAPTER-7

Letter Series Techniques to Tackle

1. Such questions depends on accurate observation. 2.

Sometime we have to arrange the alphabet in two rows

A Z

B Y

C X

D W

E V

F U

G T

H S

I R

J Q

K P

L O

M N

And observe the pattern shown in the series. 3. Sometimes alphabets are to be arranged in four rows thus

A N O

B M P Z

C L Q Y

D K R X

E J S W

F I T V

G H U

4.

Some series are formed by skipping letters backward or forward.

5.

The positional value of the alphabets can be useful to a great extent.

6.

The series of arrangement may be arithmetic series where each alphabet differs by a common difference or may be a geometric series where the ratio of position values of alphabets is always equal.

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Practice Exercise 8 1.

Make a meaningful word with the 6th, 8th, 9th, 10th and 14th letters of the word CORRESPONDENCE. Point out the third letter of the new word. 1. E

2.

2. K

3. M

4. G

2. 6th

3. 5th

4. 8th

2. E

3. S

4. T

2. I

3. M

4. L

If the letters of the alphabet are arranged in the reverse order which will be the 14th letter to the left of the third letter from the right? 1. Q

8.

4. T

Which of the following will be fourth to the right of sixteenth position from the right end? 1. X

7.

3. R

Which letter in the alphabet is the 8th letter to the right of the letter which is 20th from the left? 1. D

6.

2. E

If every fourth letter starting from F is replaced by days of a week starting from Sunday what will be the position of Thursday counting from your right? 1. 9th

5.

4. X

Which letter will be midway between the 14th letter from the left end and 19th letter from the right end in the alphabet? 1. I

4.

3. S

If letters on alternating positions, while starting from D, are dropped, which of the following letters will be second to the right of the seventh position from your left? 1. L

3.

2. M

2. J

3. R

4. K

Which letter is midway between the eighth letter from the right and seventh letter from the left in the following alphabet? ABCDEFGHIJKLMNOPQRSTUVWXYZ 1. L

9.

3. M

4. N

If the alphabet are written in the reverse order which will be the fourth letter to the right of the 13th letter from the left? 1. J

10.

2. G

2. L

3. K

4. M

Which letter in the word APPLES (other than A) occupies the same position as it does in alphabet? 1. E

2. P

3. L

4. T

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CHAPTER-8

Number Series Techniques to Tackle

The series are formed by: 1.

Skipping numbers backward or forward.

2.

Multiplying first number by some number to get the next number.

3.

Squaring one number to get the next.

4.

Sometimes alternate numbers are related.

5.

Sometimes square roots of numbers give the next number.

Each series is constructed on the basis of one and/or two functions. Series based on the Difference between the Numbers.

This can be further classified in to two types. Type 1:- Series in which the difference between the numbers or alphabets position value is constant.

There is always a constant difference between two successive numbers. Example: the numbers of the series 1, 3, 5, 7, 9, ..... Here next number is obtained by adding a constant figure of 2 to the preceding term of the series. Type 2:- Series in which the difference between the numbers or alphabets position value will keep on increasing or decreasing.

The difference between successive terms keeps increasing or decreasing, as per the given series. Example: the numbers of the series 1, 3, 6, 10, ..... Here, the difference between the first two terms of the series is 1; the difference between the second and third terms is 2; the difference between the third and the fourth terms is 3 and so on. That is, the difference between any two successive terms is exactly 1 more than the difference between the first number of this pair and the number immediately preceding this number.

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Practice Exercise 9 1.

25, 13, 8, 6, ? 1. 5.5

2.

2. 7, 4

3. 7, 6

4. 8, 7

2. 29

3. 26

4. 35

2. 36, 41

3. 37, 42

4. 35, 40

2. 677

3. 635

4. 648

2. 7, 4

3. 5, 4

4. 3, 6

2. 254

3. 246

4. 234

2. 648, 646

3. 607, 708

4. 702, 507

4, 12, 28, 60, 122, ? 1. 264

10.

4. 37

729, 243, 81, 27, ? 1. 9, 3

9.

3. 35

2. 34

0, 1, 2, 5, 26, ? 1. 627

8.

4. 18, 26

7, 12, 17, 22, 27, 32, ? 1. 33, 38

7.

3. 18, 32

1, 2, 4, 7, 11, 16, 22, ? 1. 31

6.

2. 28, 36

35, 28, 22, 17, 13, 10,? 1. 9, 8

5.

4. 4.5

76, 72, 67, 61, 54, 46, ? 1. 33

4.

3. 2.5

1, 3, 6, 10, 15, 2,? 1. 16, 12

3.

2. 3.5

203, 304, 405, 506, ? 1. 504, 507

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CHAPTER-9

Data Sufficiency (DS) In this section, you are required to classify each problem according to four or five fixed answer choices, rather than find a solution. Each problem consists of a question (or an argument) and is followed by two or three statements depending on the type of data sufficiency problem discussed below. Three types of data sufficiency problems that are commonly tested are outlined in this section. The section also includes problem-solving strategies and solved examples. The data sufficiency problem usually consists of four parts viz: 1. The directions 2. The given (original) information 3. The question that is asked based on the given information in part 2. 4. 2 or 3 statements (depending on type) from which the answer has to be drawn. Part 1: There would be a set of directions, based on which each of the following questions need to be answered. These directions would specify which option should be selected as response while answering the questions. Part 2: The information given at the beginning of the question including any geometrical diagram forms Part 2 of the DS question format. For instance, “Find the area of a rectangle if its length is 20 cm“ or ”If 10 workers lay 100 bricks on a floor measuring ...sq. ft., how many workers are needed to complete the job in ....days?” In other words, the information given is in the form of a definite statement. Part 3: A question is either given in the information sentence itself as in the above example or followed afterwards by the information part of the format. Questions normally pertain to mathematical problems. Generally there are two forms in which the question part appears.

1.

Asking a specific number in an answer, e.g., “What is the area of the rectangular figure shown in the diagram below?” In such cases the answers must be a single numerical value.

2. Requiring “Yes” or “No” as an answer. In both instances the answers have to be drawn from the given statements only. Part 4: The question is followed by two or three statements (depending on the type), in each of which will be given a particular relationship or fact and sometimes even two facts/relationships.

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Data Sufficiency Problem Type 1 The general instructions for this type of problem are as under: Directions: Each of the following problems has a question followed by two statements, which are marked A and B. Use the data given in A and B together or separately and mark

1.

If statement A alone is sufficient to answer the question.

2.

If statement B alone is sufficient to answer the question.

3.

If both statements together are needed to answer the question but neither statement alone is sufficient.

4.

If neither statement is sufficient to answer the question.

This type of data sufficiency problem therefore has 4 alternatives. The problem can be well illustrated by the following flow chart. Is A alone sufficient to answer the tiNo Is B alone sufficient to answer the

Yes

Yes

No Are A & B together sufficient to answer the No Answer 4

Yes

Solved Examples on Type 1 Data Sufficiency 1.

What is the first term in the series? a.

The 3rd term in the series is 2.

b.

The 2nd term in the series is twice the first and the 3rd term is three times the 2nd.

It is clear that statement A alone would not be of any help in answering the question. Statement B gives the relationships among the first three terms in the series, but does not provide the absolute value of the first term. The question can only be answered from statement A and B together. Hence the answer is 3.

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2.

Is x2 > y? a.

x3 - 8x - 27 < y

b. y < 0

Statement A gives only a relationship between x and y but fails to answer the question. Statement B indicates that y is negative. As x2 is always positive, it is clear that x2 is greater than y. Hence statement B alone can answer the question. Answer is 2. 3.

What is the area of triangle ABC? a. AB = 4cms, BC = 7cms, AC = 11 cms

b. Angle C < 800

Statement A gives the dimensions of the three sides of the triangle. In a triangle the sum of any two sides will always be greater than the third side. In statement A, the sum of AB and BC is equal to the third side AC. Hence the given figure is a straight line and the area is zero. Answer is 1. 4.

How many books can XYZ publishing co. print in a day? a. XYZ publishing co. has 12 presses. b. Each press can print 1052 pages everyday. Here neither statement A alone nor statement B alone can answer the question. Using both statements together one can only get the total number of pages that can be printed in XYZ co., in a day but not the number of books printed as required in the question. Hence the answer is 4.

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Practice Problems Set 1 For each of the following questions mark:

1.

If statement A alone can answer the question.

2.

If statement B alone can answer the question.

3.

If statements A and B together are needed to answer the question, but neither statement alone is sufficient.

4. If neither A nor B can answer the question. 1.

At what time did Mike arrive at the party? a. Sarah arrived at 8 p.m., Jack arrived at 8.15 p.m. and Tony arrived at 8.20 p.m. b. Mike arrived at least 15 minutes before any of the three - Sarah, Jack and Tony.

2.

What is the area of the shaded portion between the two circles A and B? B

a. Length of q = 7 cms.

A

O

b. Radius of the outer circle A = 21 cms. q

3.

In triangle PQR, QR = 7cms, â&#x2C6; Q = 500. What is the value of angle P? P

a. QR = PR. b. The bisector of angle R is perpendicular to QP. 4.

Q

R

Is the number A an odd integer? a. 5x = A. x is an integer. b. 8y = A. y is an integer.

5.

Is the value of p greater than 6 given q is a positive integer? a.

6.

1 1 < q 2

p >4 q

In the equation z = x2 + x + yx, z is even. Is x even? a. y is even.

7.

b.

b. xy is even.

What is the value of a - b?

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a. 4a + 3b = 14 8.

Is

b. a + b = 4

1 1 > ? p q

a. p > q 9.

10.

b. p > 1

What is Jackâ&#x20AC;&#x2122;s percentage share of profit in the partnership firm PQR? a.

Michael got \$12000 more than Allen did.

b.

As per the partnership arrangement Michael got 40% of the profit while Allen and Jack shared the remaining amount equally.

Is p > q? a. 0 < p < 0.50

11.

b. 0.25 < q < 1.5

What is the area of the circle with centre 0 and chord AC? C

a. OA = 6.5 cm

0

b. AC = 5 cm 12.

A

What is the value of angle a?

d

a. Angle b = 300

b c

a

b. Angle d = 200 A 13.

700

What is the value of angle BDC? a. BD = DC

D

b. BD and DC are bisectors of angles B and C respectively. 14.

b. 1 > y > 0

One of the values of x is? a. x2 - 1 = 0

16.

C

Is xy > yx? a. x > 0

15.

B

b. x > 0

What was the profit percentage on the sale of bread loaves? a. The baker made a profit equal to the selling price of 4 loaves. b. The baker made a profit of \$4.80.

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17.

Is x > 0? a.

x is not an imaginary number.

b. The numerical value of x is greater than 1/2. 18.

19.

What is the value of angle x? a. x = y

x

b. x = p

y

q z

p r

The average temperature of a certain week was 260C. What was the temperature on the fourth day? a. The average temperature on the first four days was 26.50C. b. The average temperature on the last four days was 27.50C.

20.

If F = 12, what is the value of T? a. Z = 40

21.

Is 1/m2 = 2? a. m2 + 2m > 4

22.

b. T = F[9 x Z (6 Ăˇ 2 x 3)] + F2 - 9ZF

b. m > 0.75

What is the area of the shaded portion (O is the centre of the circle)? P

Q

a. Chord PQ = 12cm. b. Chord OA = 7 cm. 23.

0 A

When will car A overtake car B? a. Car A is moving 22 kmph faster than car B. b. If car B reduces its speed by 50% it would take 44 minutes for car A to overtake car B.

24.

25.

26.

How many days does it take for 12 men to complete the work? a.

If there were 3 men less, the same work would have taken 4 days more.

b.

2 more men have to work in order to complete the work as per schedule.

What is the length of the train? a.

The train takes 27 seconds to cross a bullock cart moving in the same direction.

b.

Moving at 40 kmph the train crosses a platform of 40m length in 12 seconds.

What is the volume of the rectangular box?

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27.

a.

The total surface area is 122 sq. cm.

b.

One of the dimensions is 7 cm.

What is the area of triangle xyz? a.

xy = 12cm

b.

Angle z = 90Â°

X

Y

Z 28.

29.

Is x < y? a.

x2 - 36 = 0

b.

y>7

Is p > q? a. p3 q2 > 0 b. p2 q5 < 0

30.

What is the value of x? a. xy = 6 b. x2 + y2 = 22

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The approach to this type of problem is explained in the following flow chart:

Is (1) alone sufficient to answer the question?

yes

Is (2) alone sufficient to answer the question?

no

yes

no

Is (2) alone sufficient to answer the question?

n

Are(1) & (2) together sufficient to answer the question?

yes

n

yes

Flow chart explaining approach to type 2 data sufficiency problem

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Solved Examples on Type 2 Data Sufficiency 1.

Is xy < 0? 1. x2 y3 < 0

2. xy2 > 0

It is not possible to answer the question with statement (1) alone. Statement (2) only indicates that x is positive (y2 is always positive and x has to be positive for xy2 to be greater than 0) and is not sufficient to answer the question. It is only possible to answer the question with statements (1) and (2) together. (From statement (1) it is clear that y is negative and hence xy should always be negative). The answer is C. 2.

Is x > y? 1. x2 > y2

2. x - y > 0

Statement 1 alone will not answer the question as either x or y can be negative. Statement 2 alone will answer the question. The answer is B. 3.

What is the value of angle x? a. p = 120°

b. q = 60° a

d

p x 60

0

b

q c

Each statement alone is sufficient as angle x = angle q (alternate angles) and angle x = 180 (180 - p) - 60 = p - 60° (angles in a triangle). The answer is D. 4.

What is the perimeter of the rectangle with an area of 60 sq. cms? 1. The length of the diagonal is 13cms. 2. The ratio of the length to the breadth of the rectangle is 2:4. Each statement alone is sufficient to answer the question. The answer is D.

5.

What is the selling price of item p? 1.

If the selling price was increased by Rs. 10, the profit percentage would have increased by 12%.

2.

The cost price of the item is Rs. 117.

Neither statement 1 alone nor statement 2 alone can answer the question. Both statements together can answer the question. The answer is C.

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Practice Problems Set 2 31.

32.

Is PQ parallel to RS? 1.

a=b

2.

a + b = 2 right angles

Is 1/x > 1? 1. x > 1

33.

x<2

2.

a=2

Is a + 1/a > 2? 1.

34.

2.

a>1

What is the speed of the river? 1. A boatman can row a distance of 24 kms. from the starting point and come back in 10 hours. 2. A boatman can row at a rate of 4 kmph. with the river and at a rate of 3 kmph against it.

35.

36.

37.

38.

39.

Is x a positive integer? 1.

x3 is a positive integer.

2.

x2 + 1 is positive.

How many runs did Venkat score in a five match cricket series? 1.

If he had scored 1 run less, his average would have dropped by 0.15.

2.

If he had scored 1 run more, his average would have increased by 0.045.

Did Socrates speak the truth? 1.

Socrates says “All Greeks are liars”.

2.

Socrates was a Greek.

When would `Jim’ catch up with the tortoise? 1.

The tortoise has a 100 m head start.

2.

Jim runs 10 times as fast as the tortoise.

Is p a positive integer? 1.

40.

px = 1

2.

x=0

What are the possible values of a and b?

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1. 41.

44.

45.

46.

48.

49.

x2 - 1 = 0

2.

x-1=0

1 1 > x y

2.

y is positive.

What is the radius of the circle? 1.

AB = BC = AC = 3 cms

2.

A

3 3 cms 2

B

1.

AB = AC

2.

BD = DC

D

C

A

Is A equidistant from both B and C?

B

x

D

x

C

Is m > n? 1.

m2 > n2

2.

m < -2

Is x > y? 1.

47.

(a - b) 2 = 25

Is x > y? 1.

43.

2.

Is x positive? 1.

42.

a + b = 15

(x + y) 2 is positive.

2.

x is positive.

Is x positive? 1.

x3 + 1 = 0

2.

x2 - 1 = 0

What speed did John run at in the marathon? 1.

Jim ran 2 kmph faster than Mike whose speed was 4 kmph greater than Terry’s.

2.

John’s speed was 8 kmph greater than Terry’s speed.

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2. 50.

Mirzad’s uncle Barty is Frooda’s brother.

What is the perimeter of triangle XYZ? 1.

Triangle XYZ is a right angled triangle.

2.

The length of the hypotenuse is 13 cms.

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Data Sufficiency Problem Type 3 Directions: Each of the following problems has a question followed by three statements, which are marked a, b and c. Use the data and mark 1. If one statement itself is sufficient to answer the question. 2. If any two of all three possible pairs of statements are sufficient, but none of the statements by itself is sufficient to answer the question. 3. If any two statements but not all three statements are sufficient to answer the question. 4. If all three statements are required to answer the question. 5. If none of the statements individually or jointly answer the question.

This type of data sufficiency problem therefore has 5 alternatives, as illustrated by the following chart. Is A alone sufficient Yes to answer the question? No Is B alone sufficient to answer the question?

Yes

No Is C alone sufficient to answer the question?

Yes

No Are A & B together sufficient to answer the question?

Yes

Are B & C together No sufficient ANSWER 3 to answer the question?

No

Yes

Yes

No Are A & C together sufficient to answer the question?

Are A & C together sufficient to answer the question?

Yes

Are B & C together sufficient to answer the question?

Yes

No Are B & C together sufficient to answer the question?

Yes

No

No

Are A, B & C sufficient to answer the question? No

Yes

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Solved Examples on Type 3 Data Sufficiency 1.

How many people were there in the stadium to watch the cricket match? a.

The total number of tickets and passes issued was 46,000.

b.

2000 people with valid tickets were not allowed inside.

c.

Many people without tickets and passes were in the stadium.

It is very clear that the statements A, B and C alone will not answer the question. It is not possible to answer the question by using any pairs of these statements. All the statements together also will not give any answer. The best answer is 5. 2.

How long did it take for the bus to reach Baroda? a.

If the average speed was more by 50% it would have taken 2 hours less.

b.

The bus moved at a constant speed of 55 kmph.

c.

The bus reached Baroda half an hour late.

Here it is possible to answer the question only with statement A. Statements B and C cannot answer the question. Hence the answer is 1. 3.

What is the value of angle x? C

A

a.

The diameter of circle O is 5.66 cms.

b.

AC = 4 cms.

0 B

c.

BC = 4 cms.

Angle subtended by the diameter at the circumference is 900. Hence it is possible to answer the question if any two dimensions of the triangle ABC are known. The question can therefore be answered by any two of the statements. The answer is 2.

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Practice Problems Set 3 51.

52.

Who got the highest share of profit among A, B and C? a.

A got half as much profit as B and C together got.

b.

B received Rs. 850 as profit.

c.

C got half as much profit as A and B together got.

AB is the tangent to the circle with center O. What is the value of angle ACO? a.

The diameter of the circle is 11 cms.

b.

OB = 26 cms.

A

C 0

c. 53.

54.

55.

56.

B

Angle CAB = 105°

What is the average age of a family of 5 ? a.

The average age of the first three persons is 37 years.

b.

The average age of the last three persons is 26 years.

c.

The age of the third person in the family is 28 years.

What is the distance between A and B? a.

Distance between A and C is 157 kms.

b.

Distance between B and C is 87 kms.

c.

Nearest distance from C to the road joining A and B is 45 kms.

What is the area of triangle ABC? a.

Angle B = 90°

b.

Perimeter of the triangle is 16 cms.

c.

ABC is not an isosceles triangle.

What is the area of the rhombus ABCD? a.

The length of diagonal AC = 12 cms.

b.

The perimeter of the rhombus is 28 cms.

c.

The distance between parallel sides is 4 cms.

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Data Sufficiency Problems Type 4 Directions : Each of the following problems has a question followed by two statements which are marked A and B. Use the data and mark 1. If only one statement (A) or (B) is sufficient to answer the question. 2. If either (A) or (B) is independently sufficient to answer the question. 3. If neither of them can answer the question. 4. If both (A) and (B) are together sufficient to answer the question.

This type of data sufficiency problem, therefore has 2 alternatives, as illustrated by the following chart. Is A alone sufficient to answer the question? Is B alone sufficient to answer the question? Is either A or B independentl y sufficient to Are A and B together sufficient to answer the question?

Yes

Yes

Yes

Yes

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Solved Examples on Type 4 Data Sufficiency 1.

Does x + y have a minimum value? a.

x,y>0

b.

xy = 1

It is clear that both statements together answer the question since

1 x= y

x+y=

2

⎡ 1 ⎤ 1 - y⎥ = + y - 2 ⎢ y ⎢⎣ y ⎥⎦ ⎤ ⎡ 1 - y⎥ ⎢ ⎥⎦ ⎣⎢ y

2

⎡ 1 ⎤ - y⎥ ⎢ ⎢⎣ y ⎥⎦

2

=x+y-2

+2

implies minimum value is 2. Hence the answer is (4) 2.

What is the area of triangle ABC? a.

ABC is a right angled triangle.

b.

The hypotenuse is 13 cms.

The area cannot be determined since only one side is known. Note that (A) and (B) do not imply that the triangle has sides 13, 12, 5 cms unless it is specifically mentioned that the sides have integral values. The answer here is (3). 3.

What is the principal amount if a certain sum of money trebles itself in ten years? a.

The rate of interest is 20% p.a.

b.

The sum would double itself in five years.

Both statements (A) and (B) give the same value of r (r = rate) as given by the question itself. Since both statements do not give any additional information, the principal cannot be found. Hence the answer is (3).

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Practice Problems Set 4 57.

58.

59.

60.

61.

62.

63.

64.

What is 5/6th of a number? a.

One/fifth of the number is 63.

b.

The number is a multiple of 3.

What is the cost of 7 pens and 6 pencils? a.

One pen costs Rs. 35.

b.

21 pens and 18 pencils cost Rs. 506.

What would be the cost of fencing a circular field at Rs. 65/- per metre? a.

The radius of the field is 10m.

b.

The circumference to area ratio is 6:5.

What is the length of the train which crosses another train in 25 seconds? a.

Both the trains travel with the same speed and are of the same length.

b.

The speed of the train is 45 km/hr.

Is x an integer? a.

x is a number between 1 and 5.

b.

x=

y where y is a number between 2 and 5. 2

In the XY plane if the point (x, 0) is on the line I, what is the value of x? a.

I is parallel to the Y - axis.

b.

The point (5, -6) lies on the line I.

Is x + y an odd integer? a.

x and y are distinct prime numbers.

b.

xy is an even integer.

What is the value of the three digit number ∆ ⊗ ⊗ if ∆ and ⊗ denote the digits of the number? a.

⊗=4

b.

∆⊗⊗+

∆∆ 843

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65.

66.

67.

68.

69.

70.

71.

72.

Anu had an average score (A.M) of 86 on three tests. What was her lowest score in the three tests? a.

Anu’s highest score is 97.

b.

The average (A.M) of Anu’s 2 highest scores was 92.

Typists A and B type at constant rates. If the time it takes typist A to type 100 pages is 2.5 hours how long does it take typist B to type 100 pages? a.

It takes typist A and typist B working independently and at the same time 1.5 hours to type a total of 100 pages.

b.

The time it takes typist A to type 100 pages is 1.25 hours less than the time it takes typist B.

How many of the 10,000 acres of Jyothi’s farm were cultivated this season? a.

Exactly 1/15th of the acreage could not be cultivated.

b.

Exactly 1/3rd of the acreage that could be cultivated was not cultivated.

If both the numerator and denominator of a fraction are positive and each is increased by 5 what is the value of the resulting fraction? a.

In the resulting fraction the denominator is thrice the numerator.

b.

In the original fraction the denominator was 6 less than the numerator.

What would be the cost of surrounding a rectangular field by a road 30 metres wide? a.

The diagonal of the rectangular field is 750 metres.

b.

The unit cost of construction is Rs.900 per sq. metre.

A trader wants to make 10% profit after allowing a discount of 15% on the marked price. What should be the marked price of the article? a.

The cost price is Rs.500.

b.

The selling price is Rs.650.

In a school of 750 students what is the ratio of boys to girls? a.

The school runs in two shifts.

b.

The number of boys exceeds the number of girls by 25.

If A and B can complete a job in x days in how many days would B alone finish the job? a.

x = 20.

b.

B is faster than twice the rate of A.

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82

73.

74.

75.

76.

77.

78.

79.

80.

81.

82.

Is triangle PQR congruent to triangle XYZ? a.

PQ = XY and QR = YZ.

b.

∠R = ∠Z.

Is triangle ABC similar to triangle DEB? a.

∠A = ∠D.

b.

AB = DE and AC = DF.

What is the remainder when (n + 5) 2 is divided by 4? a.

n is odd.

b.

n is a multiple of 5.

Is x a positive number? a.

x = x2

b.

x 3 = x5

Is xy > 30? a.

2≤x≤4

b.

3>y

Can a, b, c, d be the adjacent sides of a square ABCD? a.

a = b and c = d.

b.

∠A = ∠C, ∠B = ∠D

Is sinθ - cosθ > 0? a.

θ is an acute angle.

b.

Sinθ and Cosθ < 1.

What is the profit percentage of a dealer if he sells two articles A and B? a.

The selling price of both the articles is Rs. x.

b.

One is sold at a loss of 20% and the other at a profit of 20%.

Is x > y? a.

2x + 3y = 5

b.

4x + 7y = 10

Is n > 1? a.

n + |n| = 0

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83

b. 83.

84.

85.

86.

87.

n3 – 4|n| = 0

What is the age of the teacher? a.

The average age of students and teacher together is 15 years.

b.

The average age of students alone is 12 years.

Is p < q? a.

1 −1 <p < 7 7

b.

1 −1 < q< 6 6

What is the average of three consecutive odd integers? a.

One of the integers is -3.

b.

The sum of the integers is -15.

Is x - y > 0? a.

x5 > y5.

b.

x 2 > y2 .

Is x - y < 0? a.

x4 > y4.

b.

x 2 > y2 .

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10. SOLUTIONS Answers to DS Practice Problems

1–d

2–c

3-a

4–b

5-c

6-d

7-c

8–d

9–b

10 - d

11 – a

12 - a

13 - b

14 - d

15 – a

16 – d

17 - a

18 – c

19 - c

20 - b

21 - d

22 – c

23 – d

24 - a

25 – b

26 - d

27 - d

28 - c

29 – c

30 – d

31 - a

32 – a

33 - d

34 - b

35 - a

36 – e

37 – e

38 - e

39 – e

40 - c

41 - b

42 - c

43 – a

44 – a

45 - c

46 – e

47 - a

48 - e

49 - a

50 – e

51 – c

52 - a

53 – d

54 - d

55 - e

56 - b

57 – a

58 – a

59 - b

60 – d

61 - c

62 - d

63 - d

64 – a

65 – d

66 - b

67 – d

68 - a

69 - c

70 - b

71 – a

72 – c

73 - c

74 – d

75 - a

76 - c

77 - d

78 – c

79 – c

80 - d

81 – d

82 - b

83 - c

84 - c

85 – a

86 – a

87 - c

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1-2

2–3

3-1

4–2

5-1

6-2

7-1

8-3

9–1

10 - 1

11 – 2

12 - 1

13 - 3

14 - 1

15 - 1

16 – 3

17 - 1

18 – 2

19 - 1

20 - 4

21 - 1

22 - 3

23 – 1

24 - 1

25 – 4

26 - 3

27 - 3

28 - 1

29 - 3

30 – 2

31 - 2

32 – 2

33 - 4

34 - 3

35 - 4

Solutions to Practice Exercise: Numerical Puzzles Questions 1 - 5:

AMAR Start-amounts invested

PREM

X

Y

End of 1 year - interest

0.05X

0.05Y

Amount at the beginning of II year

1.05X 2

1.05Y - 11,500

(withdrew 50%)

(withdrew Rs. 11,500)

1.05X 2 0.05 x

0.05 x (1.05Y - 11,500)

Interest during the II year

(1.05)2 X 2

Amount at the end of II year

1.05(1.05Y - 11,500)

(1.05)2 X Beginning of III year

(

2

- 4,050)

1.05(1.05Y - 11,500)

(1.05)2 X Interest during the III year

0.05(

2

- 4,050)

0.05 x 1.05(1.05Y-11,500)

(1.05)2 X Total amount at III year end

1.05 (

2

- 4,050)

1.052 x (1.05Y - 11,500)

Given : Total amount with Prem at the end of third year = Total amount with Amar at the end of second year.

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(1.05)2 X 2

⇒ (1.05)2 x (1.05Y - 11,500) =

- (1)

Also given : Total interest earned by Amar for 3 years = Rs.3,950.

(1.05)2 X 0.05 x 1.05X 2 2 + 0.05 - 4050) = 3950 ⇒ 0.05X + Solving X = Rs.40,000. Using this value of X in (1), we get the value of Y = Rs.30,000 1.

(2)

Rs. 40,000

2.

(3)

Rs. 30,000

3.

(1)

Total interest earned by Prem = 0.05Y + 0.05(1.05Y - 11,500) + 0.05 x 1.05(1.05Y - 11,500) = Rs. 3550

(1.05)2 X 2

4.

(2)

Interest earned by Amar in the III year is 0.05 (

5.

(1)

Prem would have had Rs. 30,000 x 1.053 = Rs. 34,728

- 4,050) = Rs. 900

Questions 6 - 10: M = Retirement money

Fixed Deposits

Shares

Initial

0.60M

0.40M

Return during 1st year

0.15 x 0.60M = 0.09M

0.25 x 0.40M = 0.10M

Total at start of 2nd year

0.60M - 0.25 x 0.60M

0.40M + 0.10M + 0.15M

= 0.60M – 0.15M

= 0.65M

= 0.45M Return in the 2nd year

0.15 x 0.45M

-(0.1 x 0.65M) = -(0.065M)

Net amount at start of 3rd year

0.45M + 0.15M (returned)

0.65M – 0.065M

= 0.60M

= 0.595M

0.15 x 0.60M

0.1 x 0.595M = 0.0595M

Return during the 3rd year

Net profit earned through shares in three years = 0.10M - 0.065M + 0.0595M = 18,900 ∴ M = Rs.2,00,000

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87

6.

(2)

7.

(1)

Amount of interest earned = 0.15 x 0.6M + 0.15 x 0.45M + 0.15 x 0.6M = 0.15 x 1.65M = Rs.49,500

8.

(3)

Loss incurred = 0.1 x 0.65 M = Rs.13,000

9.

(1)

Investment in shares in the third year = 0.435M = Rs.87,000

(1)

Total interest + Profit on shares 200000 x 3 Average annual return =

10.

49500 + 15700 = 200000 x 3 = 10.8% Questions 11 - 15:

A

B

C

X

10 hrs

20 hrs

15 hrs

Y

12 hrs

15 hrs

24 hrs

No. of machines

2

3

4

Working hours per day = 8 11.

(2)

Total number of working hours per machine per month = 25 x 8 = 200 Total number of hours available on A = 2 x 200 = 400 hrs, B = 3 x 200 = 600 hrs, C = 4 x 200 = 800 hrs No. of units of X that can be produced on

400 600 800 10 20 A= = 40, B = = 30, C = 15 = 53 Maximum number of pieces that can be produced is the least of these three i.e., 30.

12.

(1)

No. of Y’s total that can be produced on

600 400 800 15 12 A= = 33, B = = 40, C = 24 = 33 Max. No. of Y’s that can be produced is 33. 13.

(3)

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No. of hrs required to produce 20 units of X

Balance hours left in the month

A

10 x 20 = 200 hrs

400 - 200 = 200 hrs

B

20 x 20 = 400 hrs

600 - 400 = 200 hrs

C

15 x 20 = 300 hrs

800 - 300 = 500 hrs

No. of Y’s that can be produced in the balance hours. A: 200/12 = 16

B: 200/15 = 13

C: 500/24 = 20

Max. No. of Y’s that can be produced = 13 14.

(1)

Time required to make 1 unit each of X and Y on A

10 + 12 = 22 hrs

B

20 + 15 = 35 hrs

C

15 + 24 = 39 hrs

No. of pairs of X and Y that can be produced on A: 400/22 = 18

B: 600/35 = 17

C: 800/39 = 22

Max. no. of pairs that can be produced = 17 15.

(1)

Time required to make 17 pairs of X and Y on A

17 x 22 = 374

B

17 x 35 = 595

C

17 x 39 = 663

Unused capacity = available - used = (400 + 600 + 800) - (374 + 595 + 663) = 168

168 % unused capacity = 1800 = 9.33% Questions 16 - 20:

1986 Interest 1987

M 0.1M M - 0.2M

(withdrew 20% of initial money plus interest)

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89

Interest

0.1 x 0.8M

1988

0.8M - 0.4M = 0.4M

Interest

0.1 x 0.4M

1989

0.4M - 0.2M = 0.2M

Interest

0.1 x 0.2M

Total

0.2 + 0.02M = 0.22M

Total interest

0.1M + 0.1 x 0.8M + 0.1 x 0.4M + 0.1 x 0.2M = 0.24M

Given: Total = 11,000 ⇒ 0.22M = 11,000 ∴ M = Rs.50,000 16.

(3)

17.

(1) Total interest = 0.24M = Rs.12,000

18.

(2)

19.

(1)

20.

(4)

Questions 21 - 25: 21.

(1)

Cost if the no. of defectives is 100 Test 1 1000 x 2 + 20 x 50 + 80 x 25 Test 2 1000 x 3 + 100 x 25 = 5,500 No test 100 x 50 = 5000

22.

(3)

If more than 200 are defective, Test II would prove to be economical.

23.

(1)

Cost when the no. of defectives is 200 Test 1 1000 x 2 + 160 x 25 + 40 x 50 = Rs. 8000 Test 2 1000 x 3 + 200 x 25 = Rs. 8000

24.

(1)

25.

(4)

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90

26.

(3)

Price of A α Index,

A1 I = 1 A2 I2 C1

27.

(3)

A,

Price of C α

A1

=

80 2000 = A2 2400

, A2 = Rs. 96

C2 A2

2500 A2 at I = 2500 = 2000 x 80 50 C2 =

80 x

100 = 55.55 = 100

Gain in rupees on selling 100 shares of C = 55.55 x 200 - 50 x 200 = Rs.1111.11

3000 Price of A at I = 3000 = 2000 x 80 = 120, Price of B at I = 3000 2000 x 45 3000 = 30

28. (1)

=

50 Price of C at I = 3000 =

80 x

120 = 61.25

Total value of shares = 500 x 120 + 100 x 30 + 200 x 61.25 = Rs.75,250 29.

(3)

30.

(2)

31.

(2)

2 ⊕ x = 22 + 22 x 32 = 4 + 36 = 40

40 40 ∅ 4 = 4

32.

(2)

2

40

40 + 64 104 52 13 16 16 8 + 4 = 16 + 4 = = = = 2

y 2 + 9y 25 y ⊕ x = y2 + 9y, (y2 + 9y) - 5 = -5=5 y(y + 9) = 0, y = 0 or y = -9

33.

(4)

m = 2 ⊕ 1 = 22 + 22 x 12 = 8

2 n=2∅1= 1 +1=3 m ⊕ n = 82 + 82 x 32 = 64 + 576 = 640

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34.

m = x ⊕ 1 = x2 + x2 = 2x2

(3)

n=x∅1=x+1 m = n or 2x2 = x + 1 2x2 - x - 1 = 0, 2x2 - 2x + x - 1 = 0 2x (x - 1) + (x - 1) = 0

1 (x - 1)(2x + 1) = 0, x = 1 or x = - 2 35.

(4)

Answer Keys for Practice Exercise 2 1

1

6

4

2

2

7

1

3

1

8

3

4

2

9

1

5

3

10

1

Answer Keys for Practice Exercise 3 1

1

6

3

2

4

7

1

3

4

8

2

4

1

9

2

5

2

10

4

Answer Keys for Practice Exercise 4 1

4

6

2

2

1

7

2

3

2

8

4

4

3

9

3

2

10

1

5

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Answer Keys for Practice Exercise 5 1

3

6

2

2

2

7

3

3

4

8

1

4

2

9

1

5

3

10

1

Answer Keys for Practice Exercise 6 1

3

6

3

2

3

7

4

3

1

8

1

4

1

9

3

5

4

10

2

Answer Keys for Practice Exercise 7 1 2 3

3 2 4

6 7 8

2 2 1

11 12 13

2 3 4

4

4

1

14

1

5

2

9 1 0

2

15

2

Answer Keys for Practice Exercise 8 1 2 3 4

2 1 2 3

5

4

6 7 8 9 1 0

2 1 1 1 1

Answer Keys for Practice Exercise 9 1 2 3 4 5

1 2 4 4 2

6 7 8 9 10

3 2 1 3 3

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Quantitative aptitude vol 2