Problem Solving Select Many GRE Sample Questions 10
GRE Problem Solving Select Many Sample Questions
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1. Question:C(n,8) = C(n,6). Which of the following is true? Indicate all correct choices.
- A. n = 14
- B. n = 1
- C. n< 8
- D. C(n,2) = 91
- E. C(n, 2) = 12
Explanation:
C(n,8) = C(n,6)
n!/[8!(n-8)!] = n!/[6!(n-6)!]
6!(n-6)! = 8!(n-8)!
6!(n-6)(n-7)(n-8)!=8*7*6!(n-8)!
(n-8)(n-7)=7*8
n^2-13n+42=56
n^2-13n-14=0
n^2-14n+n-14=0
n(n-14)+1(n-14)=0
n=-1, 14
Since n cannot be nagative, n = 14
Option A is true and B and C are false.
C(n,2) = C(14,2)
= 14!/[2!(14-2)!]
= 14!/(2!12!)
= 14*13/2 = 7*13 = 91
Option D is true and E is false.
2. Question:
There are 5 letters to be put into 5 envelops. Each letter and each envelop bears the name of the receipent. Which of the following is true? Indicate all correct options.
- A. There is only one way of putting the letters in the correct envelops
- B. There are total 120 ways of putting the letters in the envelops
- C. There are 119 wrong ways of putting the letters in the envelops
- D. There are 25 ways of putting the letters in the envelops
- E. There are 24 wrong ways of putting the letters in the envelops
Explanation:
The first letter can be put in the envelops in 5 ways
The second letter canbe put in the envelops in 4 ways
The third letter can be put in the envelops in 3 ways
and so on
Total number of ways = 5*4*3*2*1
= 120
There is only 1 correct way of putting the letters in the envelops
There are 120-1 = 119 incorrect ways.
Options A, B and C are true.
3. Question:
Two dice are tossed together. Which of the following is true? Indicate all correct choices.
- A. The probability that the sum of the two numbers is divisible by 3 is 1/12
- B. The probability that the sum of the two numbers is divisible by 4 is 1/12
- C. The probability that the sum of the two numbers is divisible by 3 is 1/3
- D. The probability that the sum of the two numbers is divisible by 4 is 1/4
- E. The probability that the sum of the two numbers is divisible by 3 or 4 is 5/9
Explanation:
Let the event of divisibility by 3 be A and that of divisibility by 4 be B.
Total possible outcomes are 6*6=36
Elements in the sample space for which sum of numbers is 3: (1,2),(2,1),(1,5),(5,1),(2,4), (4,2), (4,5)
(5,4),(3,6)(6,3)(6,6)
P(A) = 12/36 = 1/3
Elements in the sample space for which sum of numbers is 4: (2,2),(1,3),(3,1),(3,5), (5,3), (4,4)
(2,6)(6,2)(6,6)
P(B) = 9/36 = 1/4
Options A and B are false and C and D are true.
Number of events nwhich the sum of numebrs is divisible by 3 and 4 both= 1 (6,6)
P(sum of numbers is divisible by 3 or 4) = P(A)+P(B) - P(sum is divisible by 3 and 4)
= 1/3+1/4-1/36 = 5/9
Option E is true.
GRE Problem Solving Select Many Sample Questions
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