GRE Practice Test  Numeric Entry GRE Practice Test 2
Q. 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
 Answer: 5 years

EXPLAINATION: 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]
Q. 2
If (x^3+1/x^3) = 52, then find the value of (x+1/x). [x^3=x*x*x]
 Answer: 4

EXPLAINATION: (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
Q. 3
If x+1/x = 2, then find the value of x^3+1/x^3. [x^3=x*x*x]
 Answer: 2

EXPLAINATION: 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
Q. 4
Find the value of k for which the equations 4x+5y = 3 and kx + 15y = 9 have infinitely many solutions
 Answer: 12

EXPLAINATION: 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
Q. 5
The difference between two numbers is 2 and the difference between their squares is 24. Find the larger number.
 Answer: 7

EXPLAINATION: 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 = 244 = 20 x = 20/4 = 5 y = x+2 = 5+2 = 7 The larger number is 7. [x^2=x*x]
Q. 6
Find the value of (bxay), if x/a+y/b = 2 and ax by = a^2b^2. [a^2=a*a]
 Answer: 0

EXPLAINATION: x/a+y/b = 2 bx + ay = 2ab ...(1) and axby = a^2b^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 = 2bb=b Hence, x = a and y = b bxay = baab = 0
Q. 7
Find the first term of a GP whose second term is 2 and the sum to infinity is 8.
 Answer: 4

EXPLAINATION: 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/(1r) ar = 2 and a/(1r) = 8 r = 2/a and a = 8  8r Hence, a = 8  8*2/a a^2 = 8a 16 a^28a+16=0 (a4)^2=0 a = 4 Hence, the first term of the GP = a = 4 [a^2=a*a]
Q. 8
How many elements does the power set of A contain if A = {x,y}?
 Answer: 4

EXPLAINATION: 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]
Q. 9
The radius of the base of a solid cylinder is x cm and its height is 3 cm. It is recast into a cone of the same radius. Find the height of the cone in cm.
 Answer: 9cm

EXPLAINATION: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]
Q. 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?
 Answer: 8

EXPLAINATION: 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.
Q. 11
Find the value of a if (xa) is the g.c.d. of x^2x6 and x^2+3x18. [x^2=x*x]
 Answer: 3

EXPLAINATION: Since (xa) is the g.c.d of the given polynomials, (xa) is a factor of both the polynomials. Putting the value a in place of x in the two polynomials, we get a^2a6=0 and a^2+3a18=0 a^2a6=a^2+3a18 a6=3a18 3a+a=186 4a=12 a=3
Q. 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?
 Answer: Rs.5000

EXPLAINATION: 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
Q. 13
In a twodigit 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?
 Answer: 2

EXPLAINATION: 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 = 128 = 4 The possible numbers are 84 and 48. Hence, two such numbers are possible.
Q. 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?
 Answer: 2 days

EXPLAINATION: 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.
Q. 15
Find n when n>0 and P(n,4) = 20*P(n,2).
 Answer: 7

EXPLAINATION: P(n,4) = 20*P(n,2) n!/(n4)! = 20*n!/(n2)! (n2)! = 20*(n4)! (n2)(n3)(n4)! = 20*(n4)! (n2)(n3)=20 n^25n+620=0 n^25n14=0 n^27n+2n14=0 n(n7)+2(n7)=0 n=2,7 Since n>0, we have n = 7