# Mixed Quantitative Aptitude Questions Set 44

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. How much more would Rs. 20,000 fetch, after two years, if it is put at 20% p.a.Compound Interest payable half yearly than if it is put at 20% p. a. compound interest payable yearly?
A)  Rs.400
B)  Rs.410
C) Rs.455
D) Rs.482
E) Rs.440
Option  D
Solution:
Required amount = 20,000 [(1+10/100)^4 – (1+20/100)^2] = Rs.482
2. A and B entered into a partnership with capitals in the ratio 5 : 7 . After 4 months, A withdraw (1/3)rd of the capital and Q withdraw (1/4)th of the capital. The profit at the end of the year was 60,000. Find the share A in the profit.
A) Rs.22150
B) Rs.24000
C) Rs.22450
D) Rs.24580
E) Rs.25500
Option  B
Solution:
Let the capitals of A and B be 5x and 7x resp.
Ratio of A’s and B’s share of profit = (5x * 4)+(2/3 * 5x * 8) : (7x * 4) + (3/4 *7x * 8)
= 2 : 3
A’s share = (2 * 60,000)/ 5 =  Rs.24000
3. The circumference of  a circular field is 20 m less than the perimeter of a square field . If the radius of the circular field is 9 m less than the side of the square field , what is the cost of gravelling the circular field at the rate of Rs. 50 per square metre?
A) Rs.7100
B) Rs.6500
C) Rs.7700
D) Rs.8000
E) Rs.5400
Option C
Solution:
Let radius be r , side of the square field = (r +9) metre
According to the question,
4x – 2 *pi*r = 20
=> 4(r+9) – 2 *(22/7)*r = 20
=> 4r + 36 – (44/7)r = 20
=> r = 7 metre
Therefore, area of the circular field = pi * r^2
= (22/7)*7*7 = 154 m^2
cost of gravelling = 154*50 = Rs.7700
4. There are six 40 W lamps which are on for 5 hours a day and three 80 W fans which are on for 10 hours a days. If electricity costs Rs. 2 per kilowatt hours what is the monthly electricity bill?
A) Rs.198
B) Rs.250
C) Rs.200
D) Rs. 216
E) Rs.220
Option D
Solution:
Electricity consumption per day =6 * 40 * 5 + 3 * 80 * 10
= 1200 + 2400 = 3600W
= 3.6 kW
Monthly electricity bill = 108*2 = Rs. 216
5. A bottle contains (3/4) of solution A  and the rest solution B. How much of the mixture must be taken away and replaced by an equal quantity of solution B so that the mixture has half solution A and half solution B?
A) 25(1/2)%
B) 24(1/3)%
C)22(1/8)%
D) 32(1/2)%
E) 33(1/3)%
Option E
Solution:
Let the mixture be 100 units
solution A = (3/4)*100 = 75
Ratio of solution A and solution B = 3 : 1
solution B = (1/4)*100 = 25
Let x L of mixture is taken away , then quantity of solution A left.
=(3 – 3x/4) = and solution B = (1 – x/4)  + x
given , 3 – (3x/4) = 1 – (x/4) + x
=> x = 4/3
Required % = (1/3)*100 = 33(1/3)%
6. The average age of 70 girls in a classroom is 20 years . The average age of a group of 15 girls in the classroom is 18 years and the average age of another 30 girls in the classroom is 15 years . What is the average age of the remaining girls in the classroom?
A) 27.2 years
B) 30 years
C) 25 .5 years
D) 20.12 years
E) 19 years
Option A
Solution:
70 * 20 = 15*18 + 30*15 + 25*x
=> 1400 = 270 + 450 + 25*x
=> 1400 – 720 = 25*x
=> 680 /25 = x = 27.2 years
7. 12 boys can complete a piece of work in 24 days and 15 men can do the same work in 16 days . How much work will be remaining after 6 boys and 5 men work for 16 days ?
A) 1/6
B) 1/3
C) 1/8
D) 1/5
E) 1/4
Option  B
Solution:
12 * 24 boys == 15 * 16 men
=> 6 boys == 5 men
6 boys + 5 men = 10 men
=> M1 D1 = M2 D2
=> 15 * 16 = 10 * D2
=> D2 = (15 * 16)/10 = 24 days
Therefore , work done in 16 days = 16/24 = 2/3
Remaining work = 1 –(2/3) = 1/3
8. The distance between the starting and ending point is 36km.  A boat rows in still water at 6 km/h . It takes 8 hours less to cover this distance in downstream than in upstream. What is the rate of the stream ?
A) 2 kmph
B) 5 kmph
C) 4 kmph
D) 8 kmph
E) 3 kmph
Option E
Solution:
Speed of the current = x kmph
Rate of downstream = (6+x) kmph
Rate of upstream = (6 – x) kmph
Therefore,
36/(x-6) – 36/(x+6) = 8
=> x^2 + 9x – 36 = 0
=> (x – 3)(x + 12) =0
=> x = 3 kmph
9. Find the different ways in which the letters of the word CORPORATION be arranged in such a way that the vowels always come together?
A) 51000
B) 35025
C) 45250
D) 50400
E) 52000
Option D
Solution:
Vowels always comes together = O, O, A, I, O
Letters = C, R, P, R, T, N
Required ways = (7!/2!) * (5!/3!) = 50400
10. A bag contains 4 red , 6 yellow and 7  blue balls. Two balls are drawn at random . What is the probability that none of the balls drawn are red in colour ?
A) 24/33
B) 11/45
C) 25/32
D) 16/34
E) 39/68
Option E
Solution:
Total number of balls in the bag = 4+ 6 + 7 = 17
Total possible outcomes = 17C2
Total favourable outcomes = 13C2
Required probability = 39/68

## 11 Thoughts to “Mixed Quantitative Aptitude Questions Set 44”

1. chinnu

ma’am 10th questn the given answer is reciprocal of probability…8th questn solution is not wrt questn

1. chinnu

also in 2nd questn instead of 12 months 16 mnths is considered…in 7th questn 10 days instead of 16 days

2. deepali

10 th ka 39/68 na?

2. deepali

8th ans ?

1. jaga
1. deepali

thnkyouu:))

2. deepali

wese option k sath jldi hojaega 😛

1. jaga

hmmm

3. jaga

q2,,,,,ans mai 12 months nahi hoga ..8months hoga,,,
q1 mai compounded annualy 20% diya hua hai question mai ….ans mai 10% ??

1. Banya

yes .. corrected.

4. jaga

thank u mam