Test: Numeric Entry- 1 - GRE MCQ

Let the fraction be x/y
According to the conditions y = x - 4 x - y = 4 ...(1)
8*(y-2) = (x+1) 8y-16=x+1 x-8y = -17...(2)
Subtracting (1) from (2),
we get x - 8y - x + y
= -17 - 4 -7y = 21 y
= 3 x = 4 + y
= 4+ 3 = 7
The numerator is 7

Let the age of the son and father be x and y respectively. 5 years back, their ages were (x-5) and (y-5)
According to the given conditions
4x = y 4x-y = 0 ...(1)
7(x-5) = (y-5) 7x - 35
= y -5 7x -y = 30 ...(2)
Subtracting (1) from (2),
we get 7x - y - 4x + y = 30 - 0 3x = 30 x = 10 years
y = 4*10 = 40 years
The present age of the father is 40 years.

Let the number be x.
20% of x = 35% of 144
= 35*144/100
= 50.4 20% of x = 50.4 20x/100 = 50.4 x
= 50.4*100/20 = 252
The required number is 252

AB = 5
AB =
5=
Squaring both sides,
we get
25 = 16 + x^2/4+1+x x^2/4 + x -8
=0 x²+4x-32=0 x²+8x-4x-32
=0 x(x+8)-4(x+8)=0 (x-4)(x+8)=0
x = 4 [x² = x*x]

When two lines are parallel, their slopes are equal. The differene of their slopes is equal to 0.

For the line to be parallel to the x axis, its slope is 0 Slope = - coefficient of x/coefficient of y = -(k-3)/(4-k²) = 0
Hence, k-3=0 k=3 For k=3,
the given line is parallel to the x axis.

x²+y²-12x+11=0
(x²-12x)+y²+11=0
(x-6)²+(y+0)² -36 +11 = 0
(x-6)²+(y+0)² = 36-11=25
(x-6)²+(y+0)² = 5²
The radius of the circle is 5 units.

Let the second number be 3x.
The first number will be 2*3x = 6x
The third number will be
1/3*6x = 2x 6x+3x+2x = 132
11x=132 x
= 132/11
= 12
The first number will be 6x = 6*12 = 72

[27^(2/3)]/[64^(-4/3)] =
[(3³)^(2/3)]/[(4³)^(-4/3)]
= [3^(3*2/3)]*[4^(3*4/3)]
= (3²)*(4⁴) = 9*256 = 2304

Let the number be x.
According to the conditions,
3x/4 -150
= 3x/14 x/4 - 50
= x/14 x*14 - 50*4*14
= x*4 14x - 2800
= 4x 10x = 2800 x
= 280

a/(a+b) = 17/23
Hence,
when a+b = 23,
a = 17
Thus,
b = 23
- a = 23 - 17
= 6 (a+b)/(a-b)
= (17+6)/(17-6) = 23/11
Hence, the number to be filled in the blank is 23.

Work done by one man in 2 days = 1/5
Work done by one man in one day = 1/(5*2) = 1/10
Work done by one woman in one day = 1/(4*3) = 1/12
Work done by one child in one day = 1/(5*3) = 1/15
Work done by one man, one child and one woman in one day
= 1/10+1/12+1/15 = (6+5+4)/(5*2*2*3 )
=15/(15*4) = 1/4
They will take 4 days to complete the work.

When A completes 1000 meters, B covers 950 meters.
When B completes 1000 meters, C covers 960 meters.
Distance covered by A/ distance covered by B = 1000/950
Distance covered by B/distance covered by C = 1000/960
Distance covered by A = 1000/950 * distance covered by B = 1000/960 * 1000/950*distance covered by C Distance covered by C,
when A covers 2 km =
(950*960)/(1000*1000)*2000
= 1824 metres
A gives C a start of
2000-1824 = 176 meters

Let's denote the principal amount as P. We will first calculate the compound interest and simple interest separately after 3 years at 5% per annum.

Simple Interest (SI):
The formula for calculating simple interest is:
SI = (P * R * T) / 100
Where P is the principal amount, R is the rate of interest, and T is the time period in years.

In this case, R = 5% and T = 3 years. So,
SI = (P * 5 * 3) / 100
SI = (15 * P) / 100

Compound Interest (CI):
The formula for calculating compound interest is:
CI = P * (1 + R / 100)^T - P
Where P is the principal amount, R is the rate of interest, and T is the time period in years.

In this case, R = 5% and T = 3 years. So,
CI = P * (1 + 5 / 100)^3 - P
CI = P * (1.05)^3 - P

Now, we are given that the difference between CI and SI is Rs. 61. So,
CI - SI = 61
(P * (1.05)^3 - P) - (15 * P) / 100 = 61

Let's simplify the equation:
P * (1.05)^3 - P - (15 * P) / 100 = 61
P * ((1.05)^3 - 1 - 0.15) = 61

Now we can solve for P:
P = 61 / ((1.05)^3 - 1 - 0.15)
P = 61 / (1.157625 - 1 - 0.15)
P = 61 / 0.007625
P = 8000

So, the principal amount (P) is Rs. 8000.