Quantitative Aptitude Mixed Questions for IBPS Clerk Prelims: Set-2



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Quantitative Aptitude Mixed Questions for IBPS Clerk Prelims: Set-2

1.Three friends A, B and C started a company in January 2016. All of them invested Rs. 200000 each at the start. In the start of April 2016, B invested an additional investment of Rs. 100000, which he withdrew at the end of September 2016. In the start of October 2016, C made an additional investment of Rs. 100000. At the end of December 2016, the company distributed a total profit of Rs. 202500. What was A’s share in the profit?
a. Rs. 61500
b. Rs. 60000
c. Rs. 63000
d. Rs. 65000
e. Rs. 6750

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B. Rs. 60000

A’s investment = 200000 for 12 months
B’s investment = 200000 for 12 months and 100000 for 6 months
C’s investment = 200000 for 12 months and 100000 for 3 months
Ratio of A’s investment to B’s investment to C’s investment
[200000*12] : [200000*12+100000*6] : [ 200000*12+100000*3] = 2400000 : 3000000 : 2700000
A’s profit share = [8/(8+10+9)]*202500 = Rs. 60000

2.Ten years ago Ramesh’s age was four times his son’s age. Six years hence, Ramesh’s age will be twice his son’s age. What is the average of their current age?
a. 45 years
b. 30 years
c. 34 years
d. 32 years
e. 56 years

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B. 30 years

Let Ramesh’s age is ‘r’ and his son’s age is ‘s’
r-10 = 4x(s-10)
r+6 = 2x(s+6)
On solving the two equation we get,
r = 42, s = 18.
Average = (42+18)/2 = 30 years

3.A man travelled half the distance of his journey at the speed of 30kmph. He then travelled one-third of the remaining distance at the speed of 40kmph. He further covers half of the remaining distance at the speed of 60kmph. He covered the remaining distance at 40kmph. What was his average speed?
a. 34 kmph
b. 32 kmph
c. 36 kmph
d. 38 kmph
e. 42 kmph

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C.  36 kmph

Let the total distance be ‘d’
He travelled d/2 at the speed of 30kmph.
(1/3)*(d-d/2) = d/6 at the speed of 40kmph
(1/2)*(d-d/2-d/6) = d/6 at the speed of 60kmph
d/6 at the speed of 40kmph
Let Average speed be A
Balancing the time taken to travel
(d/2)*(1/30)+ (d/6)*(1/40)+(d/6)*(1/60)+(d/6)*(1/40) = d/A
=> d/36 = d/A
Therefore, A = 36kmph

4.A garden of half the radius of a circular park is built at the center of the park. There is a one-meter running track around the garden. The remaining portion of the park is allotted for planting big trees. If area of the garden is 154 sq.m. What is the ratio of area of running track to the area of the portion allotted for planting big trees?
a. 7:147
b. 15:147
c. 11:132
d. 15:132
e. 17:112

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D. 15:132

5.Two distinct numbers are selected from first 100 natural numbers and are multiplied together. What is the probability that the resultant number is divisible by 3?
a. 33/50
b. 83/150
c. 33/150
d. Cannot be determined
e. None of these

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B. 83/150

6.A shopkeeper makes a profit of 15% by selling a video camera for Rs. 6440. If the cost price increases by 30% and selling price is increased by 20%. The new profit of the shopkeeper is what percentage of the old profit made?
a. 64.89%
b. 36.49%
c. 53.33%
d. 45.94%
e. None of these

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C. 53.33%

CP of Video Camera = (100/(100+Profit%))*SP
= (100/(100+15))*6440= 5600
Old profit = 6440 – 5600 = 840
After cost increases by 30%
New CP =1.3*5600 = 7280
New SP = 1.2*6440 = 7728
New Profit = 7728-7280 = 448
Required percentage = (448/840) *100 = 53.33%

7.At simple interest, an amount of money invested becomes Rs. 17472 in three years and Rs. 21356 in seven years. Find the money invested.
a. Rs. 14595
b. Rs. 13652
c. Rs. 14559
d. Rs. 15562
e. Rs. 16751

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C. Rs. 14559

Let the simple interest on the principal for 3 years be ‘x’ and for 7 years it will be ‘y’
x = P*R*3/100 and y = P*R*7/100
=> y = (7/3)x
Now, P+x = 17472 ………(i)
P+y = 21356
or, P+(7/3)x = 21356 ………(ii)
Operate [(7/3)*Eq.(i) – Eq.(ii)] (7/3)*P + (7/3)*x – P – (7/3)x = (7/3)*17472 – 21356
or, 4P/3 = 19412
or, P = 14559
Hence, option c
Alternate solution
According to the question, it is given that,
P + (P * R * 3)/100 = 17472
or, P (1 + 3R/100) = 17472 …….(i)
and P + (P * R * 7)/100 = 21356
or, P (1 + 7R/100) = 21356 ……(ii)
Equation (ii) – (i) we get,
4PR/100 = 3884
or, P [3R/100] = 3/4 (3884)
or, P [3R/100] = 2913……….(iii)
Putting the value of equation (iii) in equation (i) we get,
P + 2913 = 17472
or, P = Rs. 14559

8.Pipe A can fill a tank in 2 hours while Pipe B can empty the tank in 6 hours. If both the pipes are opened simultaneously, the empty tank can be filled in time ‘T’. Ram opened both the pipes and came back to check the tank after time T and found that only 75% tank was filled because pipe A got blocked some time ago. For how long the pipe A had been blocked?
a. 24 minutes
b. 25 minutes
c. 30 minutes
d. 36 minutes
e. 45 minutes

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C. 30 minutes

Let us assume the volume of tank = 12 ltr
Pipe A can fill the tank in 2 hours i.e., 6 ltr/hr
Pipe B can empty the tank in 6 hours i.e., 2 ltr/hr
If both the pipe are opened simultaneously the tank will fill Time ‘T’
T= 12/(6-2) = 3 hr
Let us suppose that pipe A got blocked after time ‘t’
Pipe A filled the tank for t hr and pipe B filled the tank for T hr and hence 75% of tank got filled
=> 6*t – 2*T = 75% of 12
=> 6t – 6 = 9 => t = 2.5 hr
The pipe A had been blocked for ‘T-t’ time i.e., 3-2.5 = 0.5 hr = 30 min

9.Mixture A contains liquid B&C in the ratio of 2:3 while mixture D contains liquid C&E in the ratio of 3:7. The two mixtures are mixed in such a way that for every portion of B there is 2 portions of E in the resultant mixture. Find the ratio of C to B&E together in the resultant mixture.
a. 11:12
b. 12:11
c. 14:11
d. 11:14
e. 17:2

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D. 11 : 14

In the final mixture, for 7 unit of liquid E there has to be 3.5 units of B
To mix 3.5 units of liquid B, 8.75 unit of mixture A needs to be taken
=> Units of liquid C from mixture A= 8.75 – 3.50 = 5.25
Units of liquid C from mixture D=3
Total Units of liquid C in the final mixture=5.25 + 3=8.25
Total Units of liquid B & E together=3.5+7=10.5
Required Ratio=8.25/10.5=11:14

10.Farukh worked for 2 hours and then Rushdi joined him. Farukh then worked for four more hours and then left the work. After Farukh left, Rushdi took another 20 hours to complete the remaining portion of the work. If both of them work together they can complete the work in 18 hours. How many hours will Farukh and Rushdi take respectively to complete the work if they work separately?
a. 20 hours, 35 hours
b. 24 hours, 36 hours
c. 36 hours, 24 hours
d. 54 hours, 27 hours
e. 57 hours, 24 hours

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D. 54 hours, 27 hours

Number of hours Farukh worked = 2 + 4 = 6 hours
Rushdi worked for (4 + 20) = 24hours.
Let Farukh can complete the work alone in F hours and Rushdi can complete the work alone in R hours Per hour work of Farukh = 1/F
And that of Rushdi = 1/R.
1/F + 1/R = 1/18
or, 18/F + 18/R = 1 ……..(1)
As per question, Farukh worked for 6 hours and Rushdi for 24 hours to complete the work
=> 6*(1/F) + 24*(R)= 1 ………(2)
Operate (1) – 3x(2)
18/F + 18/R – 18/F – 72/R = 1 – 3
=> -54/R = -2 => R = 27
On substituting the value of R in eq.(1) we get F = 54

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