## GRE Problem Solving Select Many Sample Questions

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1. Question:

Which of the following are the roots of the equation sqrt(7)x^2-6x-13*sqrt(7)=0? Indicate all correct choices.

A. – sqrt(7)

B. 13/sqrt(7)

C. 13*sqrt(7)/7

D. -13*sqrt(7)

E. 13-sqrt(7)

[x^2=x*x]

A, B and C

Explanation:

sqrt(7)x^2-6x-13*sqrt(7)=0

sqrt(7)x^2 -13x+7x-13*sqrt(7)=0

x[sqrt(7)*x-13]+sqrt(7)[sqrt(7)*x-13]=0

[x+sqrt(7)][sqrt(7)*x-13]=0

x=-sqrt(7), 13/sqrt(7)

Also, 13/sqrt(7)=13*sqrt(7)/7

Options A, B and C are true.

2. Question:

Which of the following is true for the first 24 natural numbers? Indicate all correct options.

A. Their sum is given by 25*24/2

B. Half of them are even

C. Half of them are odd

D. There are 8 prime numbers

E. There are 5 co-prime numbers

A, B and C

Explanation:

There are 24 numbers

Sum of the first n natural numbers is given by n(n+1)/2

Sum of the first 24 natural numbers=24(24+1)/2=24*25/2

Option A is true.

Since even and odd numbers alternate, half of the numbers are even and half are odd.

Options B and C are true.

Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23

There are 9 prime numbers and hence option D is false.

There can be more than 5 pairs of co-prime numbers and hence option E is false.

3. Question:

The arithmetic mean of two numbers is 5 and the geometric mean is 4.
Which of the following are the numbers? Indicate all correct options.

A. 2

B. 4

C. 6

D. 8

E. 10

A and D

Explanation:

Let the two numbers be x and y

(x+y)/2=5 and sqrt(xy)=4

x+y=5*2 and xy=4^2=16

Put x=10-y in xy=16

(10-y)y=16

10y-y^2=16

y^2-10y+16=0

y^2-8y-2y+16=0

y(y-8)-2(y-8)=0

(y-2)(y-8)=0

The numbers are 2 and 8

Options A and D are true. All other options are wrong.

[4^2=4*4]

## GRE Problem Solving Select Many Sample Questions

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