# Mixed Quantitative Aptitude Questions Set 35

Quantitative Aptitude Questions for IBPS RRB/PO/Clerk, SBI PO, NIACL, NICL, RBI Grade B/Assistant, BOI, Bank of Baroda and other competitive exams

1. The population of a village increases by 10% during the first year but during the next year, it decreases by 15%. If the population at the end of the second year was 56100. Find the population at the beginning of the first year.
A) 55,000
B) 70,000
C) 60,000
D) 75000
E) 90,000
Option C
Solution:
Required population = 56100 *[100/(100 + 10)]*[100/(100 – 15)]
= 60,000
2. A tank can be filled by two pipes in 20 minutes and 30 minutes resp. The two pipes are opened when the tank was empty. The first pipe was closed after some time. The tank was filled in 18 minutes. Find after how much time from the start, was the first pipe closed ?
A) 10mins.
B) 8mins.
C) 15mins.
D) 20mins.
E) 13mins.
Option B
Solution:
Total capacity = 60 units
first pipe = 3 units/mins.
second pipe = 2 units/mins.
Total time required by both the pipe to fill the empty tank = 60/5 = 12mins.
By second pipe it is filled in 18 mins. = 36 litre
Remaining = 60 – 36 = 24 litre
By first pipe = 24/3 = 8mins.
3. 7 years ago ,the average age of a husband and wife was 25 years at the time of  their marriage . Now the average age of the family including husband ,wife and a child  born during the interval is 22 years. What is the present age of the child?
A) 6 years
B) 4 years
C) 8 years
D) 2 years
E) 3 years
Option D
Solution:
Total age 7 years ago = 25 * 2 = 50 years
Total age ,now = 50 +7*2  = 64 years
Total age after child is born = (64 + x)years
Average = (64+x)/3 = 22
= > x = 2 years
4. Neha had “n” apples . She distributed them among 4 children in the ratio of (1/2): (1/3) : (1/5) : (1/8) .If she gave them each one a complete apple , find the minimum number of chocolates ?
A) 110
B) 120
C) 155
D) 139
E) 172
Option D
Solution:
LCM of 2,3,5,8 = 120
(1/2)*120 = 60
(1/3)*120 = 40
(1/5)*120= 24
(1/8)*120 = 15
Therefore ,minimum number of apples = 60 + 40 + 24 + 15 = 139
5. The radius of a circle wheel is 70cm .A cyclist takes 22hours to reach a destination at the speed of 20kmph.How many revolutions will the wheel make during this journey ?
A) 1lakh
B) 5lakh
C) 50,000
D) 30,000
E) 2lakh
Option A
Solution:
Circumference of the wheel = 2*pi*r = 2*(22/7)*70 = 440 m
Therefore ,number of revolutions = (2000000 * 22)/440 = 1,00,000
6. The sum of side of a square and length of a rectangle is 30m and the sum of the side of the square and breadth of the rectangle is 24m. If the length of the rectangle is twice its breadth ,what is the respective ratio between the area of the square and the area of the rectangle ?
A) 11:7
B) 9:8
C) 9:2
D) 8:5
E) 10:9
Option C
Solution:
Let the width of rectangle be x
then length = 2x
Let the side of square = y
Now ,
2x + y = 30 ———–(1)
x + y = 24————-(2)
subtracting (2) from (1) equations ,we get
=> x = 6 m
=> y = 18 m
Therefore ,
area of the square : area of the rectangle = (18*18) : (6*12) = 9 : 2
7. 7 men , 5 women and 8 children were given an assignment of distributing 2000 books to students in a school over a period of 3 days. All of them distributed books on the first day. On the second day 2 women and 3 children remained absent and on the third day 3 men and 5 children remained absent . If the ratio of the number of books distributed in a day by a man , a woman  and a child was 5 : 4 : 2 respectively, a total of approximately how many books were distributed on the second day ?
A) 750
B) 600
C)784
D) 655
E) 525
Option D
Solution:
Let no. of books distributed by a man ,a woman and a child per day be 5x , 4x and 2x resp.
Books distributed on the first day = 7*5x + 5*4x + 8*2x = 71x
Books distributed on the second day = 7*5x + 3*4x + 5*2x = 57x
Books distributed on the third day = 4*5x + 5*4x + 3*2x = 46x
Therefore ,
71x + 57x + 46x = 2000
=> x = 2000/174
Total books distributed on the second day = 57x = 655
8. A certain sum is invested for 2 years in scheme X at 20% p.a. compound interest compounded annually . Same sum is also invested for the same period in scheme Y at x% p.a. at as simple interest . The interest earned from scheme X is twice of that earned from scheme Y. What is the value of x ?
A) 10
B) 13
C) 9
D) 8
E) 11
Option E
Solution:
Let principal be P.
C.I. = P[(1+ (R/100)^2) -1] = P[(1 + (20/100)^2) – 1]
=P[(6/5)^2 – 1] = P[(36/25) – 1]
= 11P/25
S.I. = (P*T*R)/100 = (P * 2 * x )/100 = 2Px/100
Now,
11P/25 = 2*(2Px/100)
=> x = 11
9. An urn contain 5 red balls , 6 green balls and 7 blue balls.If three balls are drawn random from the urn.What is the probability that two balls are red and one ball is ball?
A) 41/408
B) 22/407
C) 35/408
D) 30/407
E) 31/400
Option C
Solution:
Total possible outcomes = 18C3 = 816
Favourable outcomes = 5C2 * 7C1 = 10 * 7 = 70
Required probability = 70/816 = 35/408
10. A committee of 5 is to be formed from a group of 12 students consisting of 8 boys and 4 girls . In how many ways can the committee be formed if it consists of exactly 3 boys and 2 girls ?
A)260
B) 336
C)300
D) 257
E) 470
Option B
Solution:
Required ways = 8C3 * 4C2 = 56 * 6 = 336

## 6 Thoughts to “Mixed Quantitative Aptitude Questions Set 35”

1. Ritesh

2. \$~\$PrAdeEP\$K\$~\$~\$

10k u so mch mam..:))

3. ADK

Thank you
Q.2. t == 20(1-18/30)==8 mins

4. Mera bhi din ayega :)))

Ty mam

5. Lee64

Thnks AZ

6. _/_

ty