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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeThe average earning of a group of persons is Rs. 100 per day. The difference between the highest earning and lowest earning persons of the group is Rs. 90. If those two people are excluded the average earning of the group decreases by Re. 2. If the minimum earning of the person in the group lies between 71 and 77 and the number of persons initially in the group was equal to a prime number, with both its digits prime. The number of persons in the group initially was?
Correct
Answer :2. 23
Solution:
Total earnings = 100N
HL = 90
(100N(H+L))/(N2) = 98
2N = (H+L)196
2N = 2L106
L = 76 satisfies above conditions so N = 23Incorrect
Answer :2. 23
Solution:
Total earnings = 100N
HL = 90
(100N(H+L))/(N2) = 98
2N = (H+L)196
2N = 2L106
L = 76 satisfies above conditions so N = 23 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA sum of Rs.195100 is to be paid back in 3 equal annual installments how much is each installment if rate is 4% per annum compounded annually?
Correct
Answer :1. Rs. 70304
Solution:
x/ 1.04 + x/1.04^2 + x/1.04^3 = 195100
x = 70,304Incorrect
Answer :1. Rs. 70304
Solution:
x/ 1.04 + x/1.04^2 + x/1.04^3 = 195100
x = 70,304 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeRamu decided to marry 2 years after he gets a job. He was 16 yr old when he passed class 12th. After passing class 12th, he had completed his graduation course in 4 yr and PG course in 2 yr. He got the job exactly 1 yr after completing his PG course. At what age will he get married?
Correct
Answer :5. None
Solution:
16+4+2+1+2= 25Incorrect
Answer :5. None
Solution:
16+4+2+1+2= 25 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe area of a rectangle is 4 times of the area of a square. The length of the rectangle is 234 meters and the breadth of the rectangle is 2/3 rd of the side of the square. What is the area of the rectangle in m^2?
Correct
Answer :2. 6084
Solution:
234*(2/3) = 4s
s = 39
b = 2s/3 = 26
area = 234*26 = 6084Incorrect
Answer :2. 6084
Solution:
234*(2/3) = 4s
s = 39
b = 2s/3 = 26
area = 234*26 = 6084 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeFrom 2010 onwards, till 2012 the price of computers increased every year by 10%. After that due to government subsidy the price of computers decreases every year by 10%. The price of a computer in 2014 will be approximately how much per cent less than the price in 2010?
Correct
Answer :3. 2%
Solution:
Let price in 2010 be 100
2011 – 110 and 2012 – 121
2013 108.9 and 2014 – 98.01
10098.01 = 2%Incorrect
Answer :3. 2%
Solution:
Let price in 2010 be 100
2011 – 110 and 2012 – 121
2013 108.9 and 2014 – 98.01
10098.01 = 2% 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA, B, C have 80, x and y balls respectively. If B gives 40 balls to A, he is left with half as many balls as C. If together they had 120 more balls, then average would be 200. What is the ratio of x to y?
Correct
Answer :1. 2:3
Solution:
A = 80 B = X C =Y
B = X40 C =Y
X40 = Y/2
(80+X+Y+120)/3 = 200
X+Y = 400
2X80 =Y
X =160 Y = 240 ratio = 2:3Incorrect
Answer :1. 2:3
Solution:
A = 80 B = X C =Y
B = X40 C =Y
X40 = Y/2
(80+X+Y+120)/3 = 200
X+Y = 400
2X80 =Y
X =160 Y = 240 ratio = 2:3 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe price of an article reduces to Rs. 2304 after two equal successive discounts. The markup is 80% above the cost price of Rs. 2000.What is the new profit percentage if instead of two equal successive discounts the markup price was further increased successively two times by the same percentage?
Correct
Answer :4. 125%
Solution:
SP = 2304
MP = 2000*180/100 = 3600
2304 = 3600 (1x/100)^2
x =20%
now instead giving discount
3600*120/100*120/100= 5184
5184 = 2304* ((100+P)/100)
P =125%Incorrect
Answer :4. 125%
Solution:
SP = 2304
MP = 2000*180/100 = 3600
2304 = 3600 (1x/100)^2
x =20%
now instead giving discount
3600*120/100*120/100= 5184
5184 = 2304* ((100+P)/100)
P =125% 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeJeevan drives his car at 15 m/s. Moving ahead for some hours he finds some problem in headlights of the car. So he takes 3 minutes and 20 seconds in Changing the bulb of the headlight by stopping the car. Mean while he notices that another car which was 1600 m back is now 1400 m ahead of his car. What is the speed of this car?
Correct
Answer :1. 54 km/hr
Solution:
In 2 minutes 20 sec another car covered 1600+1400 = 3000 m
Speed = 3000/200 = 15 m/s = 54 km/hrIncorrect
Answer :1. 54 km/hr
Solution:
In 2 minutes 20 sec another car covered 1600+1400 = 3000 m
Speed = 3000/200 = 15 m/s = 54 km/hr 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeMary was travelling in her boat when the wind blew her hat off and the hat started floating back downstream. The boat continued to travel upstream for twelve more minutes before Mary realized that her hat had fallen off and turned back downsteam. She caught the moving hat exactly at the starting point. Find the speed of river if Mary’s hat flew off exactly 3 km from where she started?
Correct
Answer :3. 7.5 Km/hr
Solution:
Distance covered by boat Up stream = 3 + (SR)*0.2 ……1
Distance covered by boat down stream = (S+R)*t ……..2
Hat distance covered = 3 = (t+0.2)*R
t = (30.2R)/R……3
Equation 1 and 2 same distance
then
(S+R)*t = 3 + (SR)*0.2
Substitute eqn 3 in above
(30.2R)/R * (S+R) = 3+0.2*(SR)
R =7.5 Km/hrIncorrect
Answer :3. 7.5 Km/hr
Solution:
Distance covered by boat Up stream = 3 + (SR)*0.2 ……1
Distance covered by boat down stream = (S+R)*t ……..2
Hat distance covered = 3 = (t+0.2)*R
t = (30.2R)/R……3
Equation 1 and 2 same distance
then
(S+R)*t = 3 + (SR)*0.2
Substitute eqn 3 in above
(30.2R)/R * (S+R) = 3+0.2*(SR)
R =7.5 Km/hr 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative Aptitude18996, 9486, 3152, 780, 150, 21, ?
Correct
Answer :2. 1
Solution:
18996/2 12 = 9486
9486/3 10 = 3152
3152/4 – 8 = 780
780/5 – 6 = 150
150/6 4= 21
21/7 2 = 1Incorrect
Answer :2. 1
Solution:
18996/2 12 = 9486
9486/3 10 = 3152
3152/4 – 8 = 780
780/5 – 6 = 150
150/6 4= 21
21/7 2 = 1
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