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This mock test of Test: Numeric Entry- 2 for GRE helps you for every GRE entrance exam.
This contains 15 Multiple Choice Questions for GRE Test: Numeric Entry- 2 (mcq) to study with solutions a complete question bank.
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*Answer can only contain numeric values

QUESTION: 1

A sum of money becomes 8 times itself in 15 years when placed at compound interest. In how many years will it double itself?Correct

Solution:

Let P, R and T be the principle, rate and time and A be the amount.

A = P(1+R/100)^{T}

8P=P(1+R/100)^{15}

8= (1+R/100)^{15}

2^{3}= [(1+R/100)^{5}]^{3}

2 = (1+R/100)^{5}

Hence, the amount doubles itself in 5 years.

[R^T=R*R*...*T times]

*Answer can only contain numeric values

QUESTION: 2

If (x^{3}+1/x^{3}) = 52, then find the value of (x+1/x). [x^{3}=x*x*x]

Solution:

(x+1/x)^{3} = (x^{3}+1/x^{3}) + 3(x+1/x) (x+1/x)^{3}

= 52 + 3(x+1/x) Let (x+1/x)

= y We have y^{3}

=52+3y y^{3} - 3y - 52

= 0 Clearly

y = 4

Hence,

(x+1/x) = 4

*Answer can only contain numeric values

QUESTION: 3

If x+1/x = 2, then find the value of x^{3}+1/x^{3}. [x^{3}=x*x*x]

Solution:

x+1/x = 2 (x+1/x)^{3}

= x^{3}+1/x^{3} + 3*x*1/x(x+1/x)

Substituting the value of (x+1/x),

we get 2^{3} = x^{3}+1/x^{3} + 3(2) x^{3}+1/x^{3}

= 8 - 6

= 2

*Answer can only contain numeric values

QUESTION: 4

Find the value of k for which the equations 4x+5y = 3 and kx + 15y = 9 have infinitely many solutions

Solution:

For the equations to have infinitely many solutions,

4/k = 5/15 = 3/9

k = 4*15/5 = 4*3 = 12

The equations have infinitely many solutions for

k = 12

*Answer can only contain numeric values

QUESTION: 5

The difference between two numbers is 2 and the difference between their squares is 24. Find the larger number.

Solution:

Let the smaller number be x and the greater number be y.

y - x = 2 ...(1)

y^{2} - x^{2} = 24 ...(2)

Substituting y = x+2 from equation (1) in equation (2) we get

(x+2)^{2} -x^{2} = 24 x^{2}+4x+4 -x^{2}

= 24 4x+4=24 4x = 24-4 = 20

x = 20/4 = 5

y = x+2 = 5+2 = 7

The larger number is 7.

[x^{2}=x*x]

*Answer can only contain numeric values

QUESTION: 6

Find the value of (bx-ay), if x/a+y/b = 2 and ax -by = a^{2}-b^{2}. [a^{2}=a*a]

Solution:

x/a+y/b = 2 bx + ay = 2ab ...(1)

and

ax-by = a^{2}-b^{2} ...(2)

Multiplying (1) by b and (2) by a, we get

(b^{2})x+aby + (a^{2})x -aby

= 2ab^{2} + a^{3} -ab^{2} (a^{2}+b^{2})x

= a(a^{2}+b^{2}) x = a Putting x = a in (1),

we get

ba+ay = 2ab

y = 2b-b=b

Hence,

x = a and y = b bx-ay = ba-ab

= 0

*Answer can only contain numeric values

QUESTION: 7

Find the first term of a GP whose second term is 2 and the sum to infinity is 8.

Solution:

Let the first term of the GP be a and the common ratio be r

The second term is given by ar and the sum to infinity is given by

S=a/(1-r) ar = 2 and a/(1-r) = 8

r = 2/a and a = 8 - 8r

Hence, a = 8 - 8*2/a a^2 = 8a -16 a^2-8a+16=0 (a-4)^2=0 a = 4

Hence,

the first term of the GP = a = 4 [a^2=a*a]

*Answer can only contain numeric values

QUESTION: 8

How many elements does the power set of A contain if A = {x,y}?

Solution:

A = {x,y} The set A contains 2 elements.

The number of element in the power set of a set containing n elements is given by

2^n. Power set of A contains 2^2 = 4 elements.

[2^2=2*2]

*Answer can only contain numeric values

QUESTION: 9

The radius of the base of a solid cylinder is x cm and its height is 3 cm. It is re-cast into a cone of the same radius. Find the height of the cone in cm.

Solution:

Volume of a cylinder = pi*r^2*h,

where pi = 22/7 and r and h are the radius and height of the cylinder respectively.

Volume of cone = 1/3*pi*r^2*h

where pi = 22/7, and r and h are the radius and height of the cone respectively.

Volume of the given cylinder = pi*x^2*3

= 3*pi*x^2

Volume of cone = 1/3*pi*x^2*h = pi*x^2*h/3

Since the two volumes are equal

3*pi*x^2 = pi*x^2*h/3 3=h/3 h = 3*3 = 9 cm

The height of the cone is 9 cm.

[pi=22/7, r^2=r*r]

*Answer can only contain numeric values

QUESTION: 10

Cards numbered 1 through 10 are placed in an urn. One ticket is drawn at random. In how many chances out of 20 shall the card have a prime number written on it?

Solution:

There are 10 cards in the urn. Favourable numbers = 2, 3, 5, 7

Required probability

= 4/10 = 8/20

There are 8 chances out of 20 of drawing a prime number.

*Answer can only contain numeric values

QUESTION: 11

Find the value of a if (x-a) is the g.c.d. of x^2-x-6 and x^2+3x-18. [x^2=x*x]

Solution:

Since (x-a) is the g.c.d of the given polynomials, (x-a) is a factor of both the polynomials.

Putting the value a in place of x in the two polynomials,

we get a^2-a-6=0

and

a^2+3a-18=0 a^2-a-6

=a^2+3a-18 -a-6

=3a-18

3a+a =18-6

4a=12

a=3

*Answer can only contain numeric values

QUESTION: 12

The amount of money returned by Sam to Pam after two years was Rs.5500. How much money did Pam lend at 5% rate of simple interest?

Solution:

Let P, R, T and SI be the principle, rate, time and simple interest.

Amount = SI+ P = P*R*T/100 + P 5500

= P*5*2/100 + P 5500

= (P+10P)/10 P = 5500*10/11

= 5000 Rs.

5000 were lent

*Answer can only contain numeric values

QUESTION: 13

In a two-digit number, the sum of the digits is 12. The smaller digit subtracted from the larger digit gives us 4. How many such numbers are possible?

Solution:

Let the two digits be x and y such that x>y.

According to the conditions,

x+y = 12...(1)

x - y = 4...(2)

Adding (1) and (2), we get

x+y +x -y = 12+4 2x = 16

x = 8 y

= 12-8 = 4

The possible numbers are 84 and 48.

Hence, two such numbers are possible.

*Answer can only contain numeric values

QUESTION: 14

One man can complete a work in 25 days and one woman can complete it in 10 days. In how many days can 5 men and 3 women complete the work?

Solution:

One man can complete the work in 25 days.

Work completed by one man in one day = 1/25

One woman can complete the work in 10 days.

Work completed by one woman in one day

= 1/10

Work completed by 5 men and 3 women in one day

= 5*1/25+3*1/10

= 1/5+3/10 = (2+3)/10

= 5/10 =1/2

5 men and 3 women complete the work in 2 days.

*Answer can only contain numeric values

QUESTION: 15

Find n when n>0 and P(n,4) = 20*P(n,2).

Solution:

P(n,4) = 20*P(n,2) n!/(n-4)!

= 20*n!/(n-2)! (n-2)!

= 20*(n-4)! (n-2)(n-3)(n-4)!

= 20*(n-4)! (n-2)(n-3)

=20 n^2-5n+6-20=0

n^2-5n-14=0

n^2-7n+2n-14=0

n(n-7)+2(n-7)=0

n=-2,7

Since n>0,

we have n = 7

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