- The number of boys in a class are x and the number of girls are 4 less than the number of boys. The sum of weight of boys is 630 and the average weight of boys is 45 kg. If 2 students are selected for a exam then what will be the probability that the number of boys the numbers of girls are equal ?
4/935/6982/26536/1291/5Option B

Total number of boys = x

number of girls = x – 4

the number of boys = 630/45 = 14

the number of girls = 14 – 4 = 10

probability = 14C1 * 10C1 / 24C2

= 14 * 10 * 2/24 * 23 = 35/69 - Difference between CI received in first 1.5 years at 20% per annum compounded annually and CI received in last 1.5 years at same rate of interest in compounded half yearly on the same sum is rs.495, then find the sum ?
3200048000300005200045000Option E

Let sum = 100x

total rate of in 1st 1.5 years = 20 + 10 + 20 * 10 /100 = 32%

CI received = 100x * 32/100 = 32x

CI received in last 1.5 years = 100x * 1.1 * 1.1 * 1.1 – 100x = 33.1x

difference = 33.1x – 32x = 495

1.1x = 495

x = 450

sum = 100x = 100* 450 = 45000 - X and Z alone can do a piece of work in 25 days and 30 days respectively, while Y takes as half time as X and Z take together. If they start working alternatively starting by Y, followed X and Z respectively, then find in how many days work will be completed ?

5 8/17 days12 9/11 days8 2/5 days6 3/5 days26 3/8 daysOption B

LCM of 25 and 30 = 150

efficiency of X = 150/25 = 6

efficiency of Z = 150/30 = 5

efficiency of Y = 22

when all three works alternatively

3 days work = 22 + 6 + 5 == 33 work

in total 12 days = 12/3 * 33 = 132 work

remaining work = 150 – 132 = 18

remaining work completed by Y = 18/22 = 9/11 day

total days = 12 9/11 days - A boat takes total 10 hours to cover the distance of 84 km in upstream and 84 km in downstream. If the speed of boat is increased by 12km/hr, then the new upstream speed is doubled of its usual speed. Find the time taken by boat to cover 140 km in downstream.
5826.52.5Option A

Let speed of boat = x km/hr

speed of stream = y km/hr

downstream speed = x + y

upstream speed = x – y

new speed of boat = x + 12

upstream speed = x + 12 – y

2 (x – y ) = x + 12 – y

x – y = 12

upstream speed = 12

time taken by boat to cover 84 km in upstream 84/12 = 7

time taken by boat in downstream = 10 – 7 = 3

downstream speed = 84/3 = 28 km

time taken by boat to cover 140 km in downstream = 140/28 = 5 hours - A and B started a partnership business with investment (x + 500) and (x – 1000) respectively. After 6 months a withdrew 40% of his amount. If profit received by A at the end of the year is 2800 out of the total profit rs.4800, then what is the value of ‘x’ ?
40002000150030005600Option D

Investment of A = (x + 500)* 6 + ( x + 500)* .6 * 6 = 6(1.6x + 800)

investment of B = (x – 1000) * 12

profit of A = 2800

profit B = 4800 – 2800 = 2000

6( 1.6x + 800)/(x – 1000)12 = 2800/2000

8x + 4000 = 14x – 14000

x = 3000 - I. 2x^2 – 13x + 15 = 0

II. y^2 – 6y + 8 = 0

X > YX < YX ≤ YX ≥ YX = Y or no relation.Option E

I. 2x^2 – 10x – 3x + 15 = 0

2x = 10, 3

x = 5 , 1.5

II. y^2 – 4y – 2y + 8 = 0

y = 4, 2 - I. x^2 – 27x + 92 = 0

II. y^2 – 17y + 60 = 0

X > YX < YX ≤ YX ≥ YX = Y or no relation.Option E

I. x^2 – 23x – 4x + 92 = 0

x = 23, 4

II. y^2 – 12y – 5y + 60 = 0

y = 12 , 5 - I. x^2 = 196

II. y^3 = 4096X > YX < YX ≤ YX ≥ YX = Y or no relation.Option B

I. x^2 = 196

x = 14, -14

II. y^3 = 4096

y = 16 - I. 5x^2 – 7x + 2 = 0

II. y^2 + 12y + 32 = 0X > YX < YX ≤ YX ≥ YX = Y or no relation.Option A

I. 5x^2 – 5x – 2x + 2 = 0

5x = 5, 2

x = 1, .4

II. y^2 + 8y + 4y + 32 = 0

y = -8, -4 - I. 3x^2 – 19x + 20 = 0

II. 4y^2 – 6y + 2 = 0

X > YX < YX ≤ YX ≥ YX = Y or no relation.Option A

I. 3x^2 – 15x – 4x + 20 = 0

3x = 15, 4

x = 5, 1.33

II. 4y^2 – 4y – 2y + 2 = 0

4y = 4, 2

y = 1. .5