Test: Numeric Entry- 2 - GRE MCQ

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
2³= [(1+R/100)⁵]³
2 = (1+R/100)⁵
Hence, the amount doubles itself in 5 years.
[R^T=R*R*...*T times]

(x+1/x)³ = (x³+1/x³) + 3(x+1/x) (x+1/x)³
= 52 + 3(x+1/x) Let (x+1/x)
= y We have y³
=52+3y y³ - 3y - 52
= 0 Clearly
y = 4
Hence,
(x+1/x) = 4

x+1/x = 2 (x+1/x)³
= x³+1/x³ + 3*x*1/x(x+1/x)
Substituting the value of (x+1/x),
we get 2³ = x³+1/x³ + 3(2) x³+1/x³
= 8 - 6
= 2

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

Let the smaller number be x and the greater number be y.
y - x = 2 ...(1)
y² - x² = 24 ...(2)
Substituting y = x+2 from equation (1) in equation (2) we get
(x+2)² -x² = 24 x²+4x+4 -x²
= 24 4x+4=24 4x = 24-4 = 20
x = 20/4 = 5
y = x+2 = 5+2 = 7
The larger number is 7.
[x²=x*x]

x/a+y/b = 2 bx + ay = 2ab ...(1)
and
ax-by = a²-b² ...(2)
Multiplying (1) by b and (2) by a, we get
(b²)x+aby + (a²)x -aby
= 2ab² + a³ -ab² (a²+b²)x
= a(a²+b²) 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

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]

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]

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]

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.

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

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

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.

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.

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