 # Quant Test for IBPS Clerk 2018 Main Exam Set – 10

Directions(1-5): Study the data carefully and answer the questions that follow :
Number of passed student in graduation in different universities and their ratio in B.tech, MCA and PGDM : Percentage of Female students in B.tech, MCA and PGDM in different universities : 1. What is the approximate number of female students who passed in MCA discipline from JNU ?
389
350
392
400
415
Option E
Required number of female students = 2590*4/90*40% == 415

2. What is the respective ratio between passed female students in MCA from BHU to passed male students in PGDM from IP ?
494:935
413:900
455:933
483:936
400:931
Option D
Required ratio = 2300*5/10*42% : 2400*6/10*65% = 483:936

3. Total number of students in BHU is what percent of the total no. students from IP ?
75%
100%
99%
80%
96%
Option E
Required% = 2300/2400*100 == 96%

4. What is the respective ratio of male students in B.Tech, MCA and PGDM in JNU ?
87 : 77 : 33
80 : 80 : 33
85 : 83 : 35
71 : 80 : 30
80 : 82 : 31
Option B
Required ratio = 80×3 : 60×4 : 3×33 = 240 : 240 : 99 = 80 : 80 : 33

5. Number of female in PGDM in JNU is what percent of the female students in MCA in BHU ?
108%
91%
100%
85%
98%
Option A
Required% = 2590* 3/10*67/100== 520
And 2300* 5/10*42% == 483 = 520/483*100 = 108%

6. The sum of areas of two rectangles (R1 and R2) of same length is 693 cm^2 and the ratio of breadth of rectangle R1 and breadth of rectangle R2 is 6:5 resp. If the area of square having length of side equal to the length of the rectangle is 441 cm ^2, then find the value of a cone having radius equal to the breadth of rectangle R2and height equal to the length of rectangle.
3400 cm^3
4950 cm^3
4050 cm^3
3330 cm^3
4000 cm^3
Option B
Length of rectangle = side of square = (441)^2 = 21 cm
sum of areas of two rectangles = R1+R2 = 693
Now,21*6x + 21*5x = 693
=>x = 3
Breadth of rectangle R2 = 5*3 = 15 cm
Required volume of the cone = 1/3*pi*15^2 *21 = 4950 cm^3

7. Ganesh and Gita invested some amounts in schemes offering simple interest at rate of 8% and 12% per annum resp. Ganesh and Gita deposited their respective amounts for 4 years and 2 years resp. and the interest earned by both of them found to be same. If the amount invested by Ganesh is Rs. 7000 less than the amount invested by Gita, find the average of the amounts invested by both of them.
Rs. 27000
Rs. 26500
Rs. 24500
Rs. 25000
Rs. 20000
Option C
Let the amount invested by Gita be x and Ganesh be (x-7000).
x*12% *2 = (x-7000)*8% *4
=> x= 28000
So, the amount invested by Ganesha and Gita is Rs. 28000 and Rs. 21000 resp.
Required average = (28000+21000)/2 = Rs. 24500

8. In a vessel, there is a mixture of apple, orange and mango juices in the ratio of 3 : 5 : 4 respectively. A quantity of 12 litres from the mixture is replaced by 8 litres of apple juice. Thereafter the quantities of apple and orange juices in the resultant mixture become same. Find out the initial quantity of mixture in the vessel.
50 Litres
42 Litres
38 Litres
55 Litres
60 Litres
Option E
After change in mixture, quantity of orange left in the mixture
= (5x – 5/12*12) ——-(1)
Quantity of apple left = 3x – 3/12*12+8 = 3x+5 ——(2)
Equating (1) and (2), we get 3x+5 = 5x – 5
=> x = 5 Initial quantity of the mixture = 60 Litres

9. BCCI has to select a team of 6 batsmen and 5 bowlers to be sent for England tour, while Dhoni(Batsman) and Ashwin (Bowler) is always selected. If there are total 10 batsmen and 8 bowlers in the team. Find the total numbers of ways of selection.
3000
3500
4500
4000
4410
Option E
As one batsman and one bowler is selected.
Then we have to select only (6-1) = 5 batsmen out of 9 batsmen and (5-1) = 4 bowlers out of 7 bowlers.
Required number of ways = 9C5*7C4 = 126*35 = 4410

10. A bag contains (x+1) red, (x+4) blue and 2x green balls. Two balls are drawn from the box and the probability that a blue ball and a red ball are drawn is 1/6. Find the total number of balls in the bag.
25
20
33
45
40
Option C
Total number of balls in the bag = x+1+x+4+2x = 4x+5
Probability of red and blue ball are drawn
=> (x+1)C1*(x+4)C1/(4x+5)C2 = 1/6 (x+4)/(4x+5) = 1/3
=> x= 7
Total number of balls in the bag = 4*7+5 = 33

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