Math

Question 1

A certain tin contains exactly 10 identically shaped gumdrops, of which 2 are cherry, 2 are lemon, 2 are grape, 2 are cinnamon and 2 are strawberry. If 2 gumdrops are randomly selected, one at a time and without replacement, what is the probability that the second gumdrop is the same flavor as the first?

a) 1⁄45
b) 1⁄25
c) 1⁄9
d) 1⁄5
e) 2⁄5

Answer 1

C

The correct answer is . We’ve got a tin with 10 gumdrops, 2 in each of 5 flavors. We want to know the probability of randomly choosing 2 gumdrops of the same flavor. If we select the first gumdrop at random, we leave 9 gumdrops in the tin. Now, on the second draw, we want to select the gumdrop that’s the same flavor as the first. How many gumdrops are the same flavor as the first? Only 1.

The question is basically asking, “If there’s a tin with 9 gumdrops, where 8 of the gumdrops are paired with another gumdrop of the same flavor, what’s the probability that you’ll select the 1 gumdrop that has no pair in the tin?” So 1 out of 9 gumdrops have the characteristic we’re looking for, which gives us a probability of .

Incorrectly, many people will find the probability of selecting any gumdrop of any one of the 5 flavors (2/10, or 1/5) and then multiply that by the probability of selecting the remaining gumdrop of that flavor (1/9), which would give us a probability of (which is, of course, one of the trick answers). This is the probability of selecting two gumdrops of a specific flavor. But the question didn’t ask about a specific flavor.

However, if we think about it as asking the question, “What is the probability of getting the two lemons or the two grapes or the two strawberries or the two cinnamons or the two cherries?”, we’d need to find the individual probabilities of each specific flavor and add them. Well, each scenario would have the same probability of , so if we add up for each of the 5 scenarios, we’d have , which is . Either way, we end up with .

Question 2

39 − (25 − 17)

Answer 2

31

Question 3

15 x 3 / 9

Answer 3

5

Question 4

How to know if integer is divisible by 3?

Answer 4

if the sum of the integer’s digits is divisible by 3.

Take the number 147. Its digits are 1, 4, and7. Add those digits to get the sum: 1 + 4 + 7 = 12. The sum, 12,is divisible by 3, so 147 is divisible by 3.

Question 5

How to know if integer is divisible by 9?

Answer 5

9 if the sum of the integer’s digits is divisible by 9. This rule is very similar to the divisibility rule for 3. Take the number 288. Add the digits: 2+ 8 + 8 = 18. The sum, 18, is divisible by 9, so 288 is divisible by 9.

Question 6

Prime numbers to 20

Answer 6

Memorize the smaller primes: 2, 3, 5, 7, 11, 13, 17, and 19.

Question 7

Approximately what percent less than the sales of brand B clothing by store T in the year 2000 ($750,000) were the sales of brand E clothing by store T in the year 2000 ($350,000) ?

Answer 7

The formula for percent decrease = (original - new)/ original * 100%. From the first graph, which is a bar graph, the sales of brand B clothing were approximately $750,000, and the sales of brand E clothing were approximately $300,000. So the percent decrease is approximately 
= (750-350)/750 * 100% 
= (450/750) * 100% 
= (45/75) * 100% 
= (3/5) * 100% 
= 3 * 20% 
= 60%. 

The correct answer is D.

Question 8

If 7y is divisible by 210, must y be divisible by 12 ?

Answer 8

No: For y to be divisible by 12, it would need to contain all of the prime factors of 12: 2, 2, and 3. Does it? The problem states that 7y is divisible by 210, so 7y contains the prime factors 2, 3, 5, and 7 (because 2 × 3 × 5 × 7 = 210). However, the question asks about y, not 7y, so divide out the 7. Therefore, y must contain the remaining primes: 2, 3, and 5. Compare this to the prime factorization of 12: y does have a 2 and a 3, but it does not necessarily havetwo 2’s. Therefore,y could be divisible by 12, but it doesn’t have to be.

Alternatively, you could start by dividing out the 7. If 7y is divisible by 210, y is divisible by 30. Therefore,y contains the prime factors 2, 3, and 5, and you can follow the remaining reasoning from above. Alternatively, since y is divisible by 30, y could be 30, which is not divisible by 12, or y could be 60, which is divisible by 12.

Question 9

If integer a is not divisible by 30, but ab is, what is the least possible value of integer b? ​
(A) ​20
​(B) ​15 ​
(C) ​2

Answer 9

(C) 2:For integer a to be a multiple of 30, it would need to contain all of the prime factors of 30: 2, 3, and 5. Since a is not divisible by 30, it must be missing at least one of these prime factors. So if ab is divisible by 30, b must supply any missing prime factors. The least possible missing prime is 2. If b = 2 and a = 15 (or any odd multiple of 15), then the initial constraints will be met:ab will be divisible by 30, but a by itself will not be.

You can also solve this problem by testing the answer choices. Determine whether there is a value of athat would make abdivisible by 30 for each answer choice. Since you’re asked for the least possible value, start with the least answer choice. If bis 2, a could be 15. In that case, ais not divisible by 30, but abis. Answer (C) is correct. Answers (A) and (B) are possible values of b, but not the least possible values.

Question 10

14 and 3 divide evenly inton. Is 12 a factor of n?

Answer 10

Maybe. The two sets of factors don’t overlap, so you can keep them all. For 12 to be a factor of n, n must contain all of the prime factors of 12 (2 × 2 × 3). In this case,n contains a 3, but only contains one 2 for sure, so 12 could be a factor of n but does not have to be.

Question 11

The sum of the positive integers x and y is 17. If x has only two factors and y is divisible by 5, which of the following is a possible value of x? ​
(A) ​3 ​
(B) ​7
​(C) ​12

Answer 11

(B) 7:Both x and y are positive integers and they sum to 17, so both must be less than 17. Since x has only two factors, it must be a prime number; its factors are itself and 1. Since y is divisible by 5, it must be 5, 10, or 15. List out the possible scenarios; start with y since you know there are only three possible values. y + x= 17
5 + 12 = 17 No good: 12 isn’t a prime number. 10 + 7 = 17 Bingo! 10 is a multiple of 5, and 7 is a prime number.

You could write out the third scenario, but wait! Check the answers whenever you have a possible solution. There is a 7 in the answers, so you’re done. (It turns out that 15 + 2 = 17 also fits the criteria given in the problem, but 2 is not among the answer choices.)

Question 12

2^-3
2^-2
2^-1
2^0
2^2
2^3
2^4
2^5
2^6

Answer 12

(1/2^3) = (1/8) = 0.125
(1/2^2) = (1/4) = 0.25
(1/2^1) = (1/2) = 0.5
4
8
16
32
64

Each power of 2 is 2 times the previous power of 2.

Question 13

3^2
3^3
3^4

Answer 13

9
27
81

Question 14

4^2

4^3

Answer 14

4^2 = 2^4 = 16
4^3 = 2^6 = 64

Question 15

5^2

5^3

Answer 15

25

125

Question 16

10^2

10^3

Answer 16

100

1,000

Question 17

Squares
10^2
11^2
12^2
15^2
20^2
30^2

Answer 17

100
121
144
225
400
900

Question 18

Cubes 
2^3
3^3 
4^3 
5^3 
10^3

Answer 18

8
27
64
125
1,000

Question 19

Alice and David are two of the ten people in a group. In how many different ways can this group of 10 people be divided into a group of 7 people and a group of 3 people if Alice and David are to be in the same group?

48
56
64
80
128

Answer 19

64.

In mathematics, nCk is sometimes used to denote the number of different subgroups (or “combinations”) containing k different objects that can be selected from a group of n different objects, where n is a positive integer, k is a nonnegative integer, and 0 < k < n. Given nCk, we calculate the number of combinations as
nCk = n!/ (k! (n-k)!)

The two people Alice and David can be in the group of 7 people or the group of 3 people.

If Alice and David are in the group of 7 people, then we can select the other 5 people in that group from the remaining 10 - 2 = 8 people. So if Alice and David are in the group of 7 people, then we can select the other 5 people in 8C5 ways. Using the formula with n = 8 and k = 5. Therefore if Alice and David are in the group of 7 people, then the number of ways to divide the 10 people into a group of 7 and a group of 3 is 56.

If Alice and David are in the group of 3 people, then the third person in that group can be any of the other 8 people. The number of ways to choose 1 person from 8 people is 8C1. Without needing to use the formula, it is clear that the number of ways to choose 1 person from 8 people is 8, so 8C1 = 8. Once the third person for the group of 3 people is selected, the other 7 people will form the group of 7 people. So if Alice and David are in the group of 3 people, then the number of ways to divide the 10 people into a group of 7 and a group of 3 is 8.

We now know that if Alice and David are in the group of 7 people, then the number of ways to divide the 10 people into a group of 7 and a group of 3 is 56, while if Alice and David are in the group of 3 people, then the number of ways to divide the 10 people into a group of 7 and a group of 3 is 8. The total number of ways to divide the 10 people into a group of 7 and a group of 3, with Alice and David in the same group, is 56 + 8 = 64. Choice (C) is correct.

Question 20

Square root of 2

Square root of 3

Answer 20

approx. 1.4 (Feb 14th Valentines)

approx. 1.7 (Mar 17th St Patricks)

Question 21

The average (arithmetic mean) of 5 positive numbers is 8. If the average of 4 of these numbers is 5.5, what is the value of the other number?

2.5
18
20
22
30

Answer 21

To solve this question, we must use the average formula to set up an algebraic equation that incorporates the information provided to us in the stem. Once we do this, we can solve for the remaining number.

We are told that the average of four positive numbers is 5.5. Using the average formula, we find that the sum of these four terms must be 22. Because we are told that the average of all 5 terms is 8, the sum of all five terms must be 40. Consequently, we know that our unknown term equals 40 - 22, or 18.

Choice (B) is correct.

Question 22

Data Sufficiency

AD
BCE

1
2
T - together
E - either
N - neither

Answer 22

If statement 1 is sufficient cross out BCE.
» If statement 2 is sufficient then D
» If statement 2 is not sufficient than A

If statement 1 is not sufficient cross out AD.
» If both sufficient C
» If statement 2 also not sufficient E
» If statement 2 is sufficient B

Question 23

The 1999 hot dog eating champion averaged 25 bites every 20 seconds.  At this rate approximately how many bites would she take in 90 seconds?    
6
72
90
113
220

Answer 23

To solve this it is easiest to figure out how many bites she takes per second.

Bites/second is determined by 25/20, which can be simplified to by removing a 5 factor from both the numerator and denominator to 5/4 , which equals 1.25 bites per second.

Now that we know she takes 1.25 bites every second, to answer the question we need to multiply that by 90 seconds.

At this rate she would take 1.25 * 90 = 112.5 bites or roughly 113.

The correct answer is D.

Question 24

1/10
2/10 = 1/5
3/10
4/10 = 2/5
5/10 = 1/2
6/10 = 3/5
7/10 
8/10 = 4/5
9/10 
10/10 = 1
11/10
12/10

Answer 24

0. 1 = 10%
1. 2 = 20%
2. 3 = 30%
3. 4 = 40%
4. 5 = 50%
5. 6 = 60%
6. 7 = 70%
7. 8 = 80%
8. 9 = 90%
9. 0 = 100%
10. 1 = 110%
11. 2 = 120%