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 ?
4/9
35/69
82/265
36/129
1/5
Option 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 ?
32000
48000
30000
52000
45000
Option 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 days
12 9/11 days
8 2/5 days
6 3/5 days
26 3/8 days
Option 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.
5
8
2
6.5
2.5
Option 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’ ?
4000
2000
1500
3000
5600
Option 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 > Y
X < Y
X ≤ Y
X ≥ Y
X = 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 > Y
X < Y
X ≤ Y
X ≥ Y
X = 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 = 4096
X > Y
X < Y
X ≤ Y
X ≥ Y
X = 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 = 0
X > Y
X < Y
X ≤ Y
X ≥ Y
X = 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 > Y
X < Y
X ≤ Y
X ≥ Y
X = 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
