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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeGiven below is the table which shows the ratio of Efficiency of X and Y in completing five different works and time taken by X alone to complete these five works. Line Graph shows the number of days Y actually worked on these five works.
After Y had worked for the given number of days on work C and work D, X completes the remaining of work C and work D. Time taken by X in completing the remaining of work C is what percent more or less than time taken by him in completing the remaining of work D.
Correct
Answer: 2) 500/9 %
Explanation:
Y alone will complete work C in= (8 *1)/2= 4 days
Y alone will complete work D in= (6 * 7)/6=7 days
Part of work C and work D completed by Y in given time= 2/4, 4/7
Remaining of work C and work D is completed by X
So,
X will complete remaining work of C= (12/4)*8= 4 days
X will complete remaining work of D= (14/7)*6= 18/7 days
Required %= ((4(18/7)/ (18/7))* 100= 500/9%Incorrect
Answer: 2) 500/9 %
Explanation:
Y alone will complete work C in= (8 *1)/2= 4 days
Y alone will complete work D in= (6 * 7)/6=7 days
Part of work C and work D completed by Y in given time= 2/4, 4/7
Remaining of work C and work D is completed by X
So,
X will complete remaining work of C= (12/4)*8= 4 days
X will complete remaining work of D= (14/7)*6= 18/7 days
Required %= ((4(18/7)/ (18/7))* 100= 500/9% 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeGiven below is the table which shows the ratio of Efficiency of X and Y in completing five different works and time taken by X alone to complete these five works. Line Graph shows the number of days Y actually worked on these five works.
X and Y together started working on work D but both left after working for 2 days. Remaining work is completed by M and N together in 4 days. If ratio of efficiency of X and M in completing work D is 7 : 3, then in how much time N alone will complete the work D.
Correct
Answer: 5) 42 days
Explanation:
Y can complete work D= 6*7/6= 7 days
Part of work D completed by X and Y in 2 days= 2/6 + 2/7
= 1/3 + 2/7
= 13/21
Time taken by M in completing work D= (6/3) * 7= 14 days
So,
In 4 days M will complete= 4/14
= 2/7 part
M and N together complete= (113/21)= 8/21days
But M completes 2/7 of work D.
Remaining (8/212/7) = 2/21 is completed by N in 4 days
So, N alone will complete in work D= 4÷ 2/21= 42 daysIncorrect
Answer: 5) 42 days
Explanation:
Y can complete work D= 6*7/6= 7 days
Part of work D completed by X and Y in 2 days= 2/6 + 2/7
= 1/3 + 2/7
= 13/21
Time taken by M in completing work D= (6/3) * 7= 14 days
So,
In 4 days M will complete= 4/14
= 2/7 part
M and N together complete= (113/21)= 8/21days
But M completes 2/7 of work D.
Remaining (8/212/7) = 2/21 is completed by N in 4 days
So, N alone will complete in work D= 4÷ 2/21= 42 days 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeGiven below is the table which shows the ratio of Efficiency of X and Y in completing five different works and time taken by X alone to complete these five works. Line Graph shows the number of days Y actually worked on these five works.
If percentage of work C completed by X in 4 days is equal to the percentage of work C completed by 4 women in 5 hours and ratio of efficiency of a woman and a child in completing work C is 5 : 3, then in how much time work C will be completed by 6 children
Correct
Answer: 1) 100/9 hours
Explanation:
% of work C Completed by X in 4 days
= 4/8 * 100
= 50%
This is equal to work C completed by 4 women in 4 days
So, one woman will complete= 40 days
One child will complete= (40/3) * 5
6 children will complete= ( (40*5) / (3*6))= 100/9 daysIncorrect
Answer: 1) 100/9 hours
Explanation:
% of work C Completed by X in 4 days
= 4/8 * 100
= 50%
This is equal to work C completed by 4 women in 4 days
So, one woman will complete= 40 days
One child will complete= (40/3) * 5
6 children will complete= ( (40*5) / (3*6))= 100/9 days 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeGiven below is the table which shows the ratio of Efficiency of X and Y in completing five different works and time taken by X alone to complete these five works. Line Graph shows the number of days Y actually worked on these five works.
If another person Z can complete work B in (Q – P) days where P and Q are the times taken by X and Y together to complete work B and C respectively, then what is the ratio of efficiency of Y and Z in completing work B?
Correct
Answer: 3) 1:9
Explanation:
Y will complete in work B= (4*5) / 5= 4 days
Y will complete in work C= (8*1) / 2= 4 days
P= (5*4)/9 = 20/9days
Q= (8*4)/12= 8/3 days
Z will complete in work B= 8/3 – 20/9= (2420)/9 = 4/9 days
Ratio of time taken by Y and Z in completing work B
= 4: 4/9
=9: 1
Ratio of efficiency= 1:9Incorrect
Answer: 3) 1:9
Explanation:
Y will complete in work B= (4*5) / 5= 4 days
Y will complete in work C= (8*1) / 2= 4 days
P= (5*4)/9 = 20/9days
Q= (8*4)/12= 8/3 days
Z will complete in work B= 8/3 – 20/9= (2420)/9 = 4/9 days
Ratio of time taken by Y and Z in completing work B
= 4: 4/9
=9: 1
Ratio of efficiency= 1:9 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeGiven below is the table which shows the ratio of Efficiency of X and Y in completing five different works and time taken by X alone to complete these five works. Line Graph shows the number of days Y actually worked on these five works.
After Y had worked for the given numbers of days on work C, D and E, what is the sum of times taken by X in completing the remaining of work C, D and E?
Correct
Answer: 5) 60/7 days
Explanation:
Y alone will complete work C, D and E will be 4, 7 and 9/2 days respectively.
Part of work of C, D and E done by Y = 2/4, 4/7 and 2/3 days respectively.
Remaining of work C, D and E is completed by X= (1/2) * 8, (3/7) *6, (1/3) * 6 days respectively.
Required sum= 4 + 18/7 + 2
= 6 + 18/7
= 60/7 daysIncorrect
Answer: 5) 60/7 days
Explanation:
Y alone will complete work C, D and E will be 4, 7 and 9/2 days respectively.
Part of work of C, D and E done by Y = 2/4, 4/7 and 2/3 days respectively.
Remaining of work C, D and E is completed by X= (1/2) * 8, (3/7) *6, (1/3) * 6 days respectively.
Required sum= 4 + 18/7 + 2
= 6 + 18/7
= 60/7 days 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe following piechart shows the percentage of passed candidates in SBI exam from cities X,Y,Z,K,L and M out of the total passed candidates from all six cities together in year 2010.
Table shows the percentage of fresher candidates who passed from each city out of the total passed from that city in year 2010.
If in year2010, total number of freshers passed from city K was320, then how many fresher candidates passed the SBI exam from city L?
Correct
Answer – 1) 384
Explanation
Let the total number of candidates passed in exam for city K be x.
25 % of x = 320
X = 320x 100/25 = 1280
Total passed candidates from all city K is1280
Total passed candidates from all city be y
10% of y =1280
Y= 12800
Total passed from city L
=25/100 x12800 = 3200
Total passed candidates from city L
=12/100 x 3200 = 384Incorrect
Answer – 1) 384
Explanation
Let the total number of candidates passed in exam for city K be x.
25 % of x = 320
X = 320x 100/25 = 1280
Total passed candidates from all city K is1280
Total passed candidates from all city be y
10% of y =1280
Y= 12800
Total passed from city L
=25/100 x12800 = 3200
Total passed candidates from city L
=12/100 x 3200 = 384 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe following piechart shows the percentage of passed candidates in SBI exam from cities X,Y,Z,K,L and M out of the total passed candidates from all six cities together in year 2010.
Table shows the percentage of fresher candidates who passed from each city out of the total passed from that city in year 2010.
If in year 2010, total passed candidates from all cities was 1250, then what is the number of the nonfresher candidates from cityX who passed the SBI exam in same year?
Correct
Answer – 4)280
Explanation
Non fresger candidates who passed from city X
=1250 x 28/100 x 80/100
=280Incorrect
Answer – 4)280
Explanation
Non fresger candidates who passed from city X
=1250 x 28/100 x 80/100
=280 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe following piechart shows the percentage of passed candidates in SBI exam from cities X,Y,Z,K,L and M out of the total passed candidates from all six cities together in year 2010.
Table shows the percentage of fresher candidates who passed from each city out of the total passed from that city in year 2010.
If the nonfresher candidates passed from city Y in year 2010 were 180, then how many total candidates passed the SBI exam from all cities together?
Correct
Answer 3)1500
Explanation
Toatl passed candidates from city Y
=180 x 100/75 = 240
A total candidates passed from all cities = 240×100/16 = 1500Incorrect
Answer 3)1500
Explanation
Toatl passed candidates from city Y
=180 x 100/75 = 240
A total candidates passed from all cities = 240×100/16 = 1500 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe following piechart shows the percentage of passed candidates in SBI exam from cities X,Y,Z,K,L and M out of the total passed candidates from all six cities together in year 2010.
Table shows the percentage of fresher candidates who passed from each city out of the total passed from that city in year 2010.
If there is an increase of 10% and 20% in the number of passed candidates in city X and Y in year 2011 respectively from year 2010 and total passed candidates from city Z in 2010 was 770. Then what would be the difference in no. of passed candidates from city X and Y in year 2011?
Correct
Answer – 2)812
Explanation
Total no. of candidates passed in 2010 = 770 x 100/11 = 7000
No. of candidates passed from city X in 2010
=28/100 x 7000 = 1960
Candidates passed in 2011 from city X
=110/100 x 1960 = 2156
No. of candidates passed from city Y in 2010
=16/100 x 7000 = 1120
Candidates passed in 2011 from city Y
=120/100 x 1120 = 1344
Required difference = 21561344
=812Incorrect
Answer – 2)812
Explanation
Total no. of candidates passed in 2010 = 770 x 100/11 = 7000
No. of candidates passed from city X in 2010
=28/100 x 7000 = 1960
Candidates passed in 2011 from city X
=110/100 x 1960 = 2156
No. of candidates passed from city Y in 2010
=16/100 x 7000 = 1120
Candidates passed in 2011 from city Y
=120/100 x 1120 = 1344
Required difference = 21561344
=812 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe following piechart shows the percentage of passed candidates in SBI exam from cities X,Y,Z,K,L and M out of the total passed candidates from all six cities together in year 2010.
Table shows the percentage of fresher candidates who passed from each city out of the total passed from that city in year 2010.
If total passed candidates from city Y in year 2010 was 320, then what is the ratio between the no. of fresher passed from city X and that of nonfresher passed from city Z?
Correct
Answer – 1)112:187
Explanation
Total candidates passed = 320 x 100/16 = 2000
Candidates passed from city Z = 11/100 x 2000 = 220
Nonfresher candidates passed from Z
=85/100 x 220 = 187
Candidates passed from city X
=28/100 x 2000 = 560
Fresher candidates passed from X = 20/100 x 560 = 112
Required ratio = 112: 187Incorrect
Answer – 1)112:187
Explanation
Total candidates passed = 320 x 100/16 = 2000
Candidates passed from city Z = 11/100 x 2000 = 220
Nonfresher candidates passed from Z
=85/100 x 220 = 187
Candidates passed from city X
=28/100 x 2000 = 560
Fresher candidates passed from X = 20/100 x 560 = 112
Required ratio = 112: 187
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