GRE Problem Solving Select Many Sample Questions

---

---

1. Question:

Which of the following is equal to [a^(1/2)+a^(-1/2)]/(1-a) + [1-a^(-1/2)]/[1+a^(1/2)]? Indicate all correct options.

A. -2/(a-1)

B. 1/(a-1)

C. 1/a-1

D. 2/a-1

E. 2/(1-a)

[a^(1/2)=sqrt(a)]

Correct Answer:

A and E

Explanation:

[a^(1/2)+a^(-1/2)]/(1-a) + [1-a^(-1/2)]/[1+a^(1/2)]

=[a^(1/2)+1/a^(1/2)]/(1-a) + [1-1/a^(1/2)]/[1+a^(1/2)]

=[a^(1/2)*a^(1/2)+1]/[a^(1/2)(1-a)] + [a^(1/2)-1]/[a^(1/2)(1+a^(1/2))]

={(a+1)[1+a^(1/2)] + [a^(1/2)-1](1-a)]} / [a^(1/2)(1-a)(1+a^(1/2))]

={a+1+a^(3/2)+a^(1/2) + a^(1/2)-a^(3/2) -1+a} / [a^(1/2)(1-a)(1+a^(1/2))]

=2a^(1/2)[1+a^(1/2)]/[a^(1/2)(1-a)(1+a^(1/2))]

=2/(1-a)

=-2/(a-1)

Options A and E are true.

2. Question:

The population of a town is 32000. It increases 15% annually. Which
of the following statements is true? Indicate all correct options.

A. It will be 42320 in two years

B. It will be 42320 in three years

C. It will be 36000 after one year

D. It will 68000 after one year

E. It will be 36800 after a year

Correct Answer:

A and E

Explanation:

This problem is similar to a compound interest problem

Population after a year=32000(1+15/100)

=32000(100+15)/100

=32000(115/100)

=36800

The option E is true. Options C and D are false.

Population after two years=36800(1+15/100)

=36800(100+15)/100

=36800*115/100

=42320

The option A is true and B is false.

3. Question:

The average of 5 consecutive natural numbers is n. Which of the following is true? Indicate all such options.

A. The sum of the numbers is 5n.

B. The product of the numbers is 25n^3.

C. If the next two numbers be included, the average would increase by 1.

D. If the next two numbers be included, the average would increase by 1.4.

E. If the next five numbers be included the average would become (n+5)

[n^3=n*n*n]

Correct Answer:

A and C

Explanation:

The average of 5 numbers is n

Average=sum of numbers/number of numbers

n=sum of numbers/5

Sum of numbers=5n

Option A is true.

Let the first number be a

[a+(a+1)+(a+2)+(a+3)+(a+4)]/5=n

5a+10=5n

a+2=n

a=n-2

Product of the numbers=a(a+1)(a+2)(a+3)(a+4)

=(n-2)(n-1)(n)(n+1)(n+2)

=n(n^2-1)(n^2-4) which is not equal to 25n^3

Option B is false.

If the next 3 numbers are included, the average becomes

Average=(a+a+1+a+2+a+3+a+4+a+5+a+6)/7

=(7a+21)/7

=(7n+-14+21)/7

=(7n+7)/7=n+1

Option C is true and D is false.

If the next 5 numbers are included, the average becomes

Average=[a+a+1+a+2+a+3+a+4+a+5+a+6+a+7+a+8+a+9)/10

=(5n+5a+35)/10

=(5n+5n-10+35)/10

=(10n+25)/10=n+2.5

Option E is false.

GRE Problem Solving Select Many Sample Questions

---

---