Sample SAT Math – Answers and Explanations See how you did on our SAT Sample Math Questions. Read below for the Answers and Explanations.
Number properties 1.
E
We’re asked for the number of odd integers between 10 and 62, so let’s be more 3 3 clear about his range. 10 is the same as 3 ⅓ and 62 is the same as 20 ⅔. We count 3 3 the odd integers between 3 ⅓ and 20 ⅔. We can’t include 3 since 3 is less than 3 ⅓. Similarly we can’t include 21 since it’s larger than 20 ⅔. This leaves the odd integers of 5, 7, 9, 11, 13, 15, 17 and 19. That’s a total of 8. 2.
E
We need to find the one choice that isn’t always true. To find it, let’s test each choice. Choice A is always true: Since P ÷ 9 has a remainder of 4, P is 4 greater than some multiple of 9, so the number 4 less than P is a multiple of 9. And if P – 4 is a multiple of 9, then the next multiple of 9 would be (P – 4) + 9, or P + 5; thus choice B is also true. With Choice C, we know that since P – 4 is a multiple of 9, it is also a multiple of 3. By adding 3s, we know that (P – 4) + 3, or P – 1, and (P – 4) + 3 + 3, or P + 2, are also multiples of 3. Choice C must be true. And since P – 1 is a multiple of 3, when P is divided by 3, it will have a remainder of 1 and choice D is always true. This leaves only choice E. In simpler terms, choice E states that P is always odd. Since multiples of 9 are alternately odd and even (9, 18, 27, 36…), P – 4 could be either even or odd, so P also could be either even or odd. Choice E is not always true, so it is the correct answer choice. If you have trouble thinking this way, try picking a number for P that fits the description (“a remainder of 4 when divided by 9…”). You could pick 13, but if you try 13 in each answer choice you’ll find that all five seem to be true. You’ll have to try a different number for P. How about 22? It fits the description, and it disproves the statement in choice E. Picking numbers is not always the fastest method – sometimes you’ll have to try many numbers before you find what you need – but it’s a good tactic if you don’t know what else to do. 3.
1
152 = (15)(15) = 3 (3x5)2. So 3 and 5 are the two distinct prime factors of 152. 643 = (2x2x2x2x2x2)3. So 2 is the single prime factor of 643. Don’t forget to do the last part of the problem – find the positive difference between the number of distinct prime factors, 2 – 1 = 1.
Averages 1.
93
Jerry’s average is 85. His total number of points for the 3 tests is the same as if he had scored 85 on each of the tests: 85 + 85 + 85, or 225. He wants to average 87 over 4 tests, so his total must be 87 + 87 + 87 + 87 = 348. The difference between his total score after 3 tests and the total that he needs after 4 tests is 348 – 255, or 93. Jerry needs a 93 to raise his average over the 4 tests to 87. Another way of thinking about the problem is to “balance” the average around 87. Imagine Jerry has 3 scores of 85. Each of the first 3 is 2 points below the average of 87. So together, the first 3 tests are a total of 6 points below the average. To balance the average at 87, the score on the fourth test will have to be 6 points more than 87, or 93. 2.
C
If the average of r and s is 40, r + s = 80. If the average of t and s is 25, t + s = 50. Subtract the equation t + s = 50 from the equation r + s = 80: r + s = 80 - (t + s = 50) r – t = 30
Ratios and Rates 1.
600
The ratio of parts is 4: 3: 2, making a total of 9 parts. Since 9 parts are equal to 1,800 votes, each part represents 1,800, or 200 votes. Since Ms Frau represents 3 parts, 900 She received a total of 3 x 200, or 600 votes. (Another way to think about it: Out of every 9 votes, Ms Frau gets 3, which is 3 or ⅓ of the total number of votes. 9 ⅓ of 1,800 is 600). We could also have solved this algebraically by setting up a proportion with F as Ms. Frau’s votes: 3= F 9 1,800 3 x 1,800 = F 9 600 = F
2.
D
We can solve this algebraically. Let the number of yellow balls received be x. Then the number of white balls received is 30 more than this, or x + 30. So # of white balls = 6 = x + 30 # of yellow balls 5 x Cross multiply:
6x = 5(x + 30)
Solve for x:
6x = 5x +150 x = 150
Percents 1.
D
The number that is 400 percent greater than 15 is 500 percent of 15, or 5 x 15 = 75. 150 percent of 30 means (1.5)(30) = 45. So 75 is what percent greater than 45? Percent change = actual change = 30 = 2 = 66 â…” % original amount 45 3
Powers and Roots 1.
E
You can easily raise 8 to the 5th power using your calculator and see that it equals 32,768. Plugging each of the 3 statements into your calculator, being wary of the order of operations, you will see that each statement also equals 32,768. So answer choice E is correct. Alternatively, look at this question as a good review of the rules for the product of exponential expressions. In order to make the comparisons easier, transfer 85 and each of the 3 options so that they have a common base. Since 2 is the smallest base among the expressions to be compared, let it be your common base. Since 85 = (23)5 = (23 x 5) = 215. I: 25 x 45 = 25 x (22)5 = 25 x 22x5 = 25 x 210 = 25 + 10 = 215. Okay II: 215. Okay III: 25 x 210 = 25 + 10 = 215. Okay Again, all three are equivalent to 215 or 85.
Advanced Algebra 1.
C
There are a few common quadratic expressions that occasionally appear on the SAT. If you learn to recognize them, a problem like this is easy. Any expression in the form of a2 – b2, which is referred to as “the difference of two squares” can be factored into (a+b)(a-b). The numerator of the given fraction is in this form, so you can rewrite the fraction as (x+y)(x-y) x+y Cancel a factor of (x+y) from numerator and denominators, leaving x-y. So x-y = 13. 2.
C
First, plug in the first set of values (f=10 and r=300) to find k. Then use k and the new value of r, (which is 750), to find the new value of f. 10 = k(300) 10 = 1 = k 300 30 f – 1 (750) = 25 30
General Word Problems 1.
A
Liza starts with 5n photographs. If she gives n photos to each of her 3 friends, she gives away 3 x n, or 3n, photos. She is left with 5n – 3n, or 2n photos. 2.
80
The student has done 30 of the 60 problems, and has used up 20 of the 60 minutes. Therefore he has 60 – 30, or 30 problems left, to be done in 60 – 20, or 40 minutes. We find his average time per problem by dividing the time by the number of problems. Time per problem = 40 minutes 30 problems = 4 minutes per problem 3 Each minute has 60 seconds. So we multiply by 60 to find the number of seconds. 4 minutes x 60 seconds = 80 seconds
3 3.
minute E
Let the amount of John’s money = J and let the amount of Allen’s money = A. John gives $5 to Allen so now he has J – 5 and Allen has A + 5. Allen gives $2 to Frank so now he has $2 less, or A + 5 – 2 = A + 3. They all have the same amount of money now, so A+3=J–5 Or A+8=J Since Allen needs $8 to have the same as John, John has $8 more.
Logic Word Problems 1.
D
We will get the maximum number of groups by making each group as small as possible. Each group must have at least 3 people in it, so divide 40 by 3 to find the number of 3-person groups: 40 = 13 with remainder of 1. 3 So we have 13 groups with 1 person left over. Since each group must have at least 3 people, we must throw the extra lonely student in with one of the other groups. So we have 12 groups with 3 students each, and 1 group with 4 students, for a maximum total of 13 groups.
Triangles 1.
D
The area of a right triangle is ½ (leg1 x leg2). Since an isosceles right triangle has legs of the same length, this equals½ (leg)2. So ½ (leg)2 = 32, (leg)2 = 64, and leg = √64 = 8. Therefore, the triangle has legs of length 8. In an isosceles right triangle, the ratio of either leg to the hypotenuse is 1:√2, so here the hypotenuse must be 8√2 in length.
Quadrilaterals and Other polygons 1.
C
The sum of the degree measures of the angles of a quadrilateral is 360. Since the angles of the quadrilateral DEFG are in a ratio of 2: 3: 5: 6, you can set up an equation in which x represents part of the ratio: 2x + 3x + 5x + 6x = 360 16x = 360 x = 22.5 Now find the difference in the degree measures of the largest and smallest angles: 6x – 2x = 4x = 4(22.5) = 90
Solids: 1.
18
We can determine the volume of the rectangular solid since we’re given all its dimensions: 4, 8, and 16. The volume of a rectangular solid is equal to l x w x h. So the volume of this solid is16 x 8 x 4, and this must equal the volume of the cube as well. The volume of a cube is the length of an edge cubed, so we can set up an equation to solve for e: e3 = 16 x 8 x 4 To avoid the multiplication, let’s break the 16 down into 2 x 8: e3 = 2 x 8 x 8 x 4 We can now combine 2 x 4 to get another 8 e3 = 8 x 8 x 8 e=8 The length of an edge of the cube is 8. 2.
50
When the rectangular solid was cut into 2 identical cubes, 2 new faces were formed: 1 on each cube along the line of the cut. So the difference between the original surface area and the combined surface area of the resulting cubes is equal to the surface area of the 2 new faces. To find the area of each of these faces, you need to find the length of an edge of the cube. Since the rectangular clock was divided into 2 equal cubes, the volume of each of these cubes is equal to ½ the volume of the original solid, or 250 cubic inches ÷ 2 – 125 cubic inches.
So an edge of one of these cubes has a length equal to the cube root of 125, which is 5. Therefore, the area of 1 face of the cube equals 5 x 5, or 25. So 2 of these faces have a total area of 2 x 25, or 50 square inches.