# IBPS PO 2018 Mains: Quants Test Day 1

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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Given 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.**CorrectAnswer:

**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= (1-2/4)*8= 4 days

X will complete remaining work of D= (1-4/7)*6= 18/7 days

Required %= ((4-(18/7)/ (18/7))* 100= 500/9%IncorrectAnswer:

**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= (1-2/4)*8= 4 days

X will complete remaining work of D= (1-4/7)*6= 18/7 days

Required %= ((4-(18/7)/ (18/7))* 100= 500/9% - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Given 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.**CorrectAnswer:

**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= (1-13/21)= 8/21days

But M completes 2/7 of work D.

Remaining (8/21-2/7) = 2/21 is completed by N in 4 days

So, N alone will complete in work D= 4÷ 2/21= 42 daysIncorrectAnswer:

**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= (1-13/21)= 8/21days

But M completes 2/7 of work D.

Remaining (8/21-2/7) = 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 Aptitude**Given 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**CorrectAnswer:

**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 daysIncorrectAnswer:

**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 Aptitude

**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?**CorrectAnswer:

**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= (24-20)/9 = 4/9 days

Ratio of time taken by Y and Z in completing work B

= 4: 4/9

=9: 1

Ratio of efficiency= 1:9IncorrectAnswer:

**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= (24-20)/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 Aptitude

**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?**CorrectAnswer:

**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 daysIncorrectAnswer:

**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 Aptitude**The following pie-chart 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?**CorrectAnswer –

**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 = 384IncorrectAnswer –

**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 Aptitude**The following pie-chart 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 non-fresher candidates from cityX who passed the SBI exam in same year?**CorrectAnswer –

**4)280**

Explanation

Non fresger candidates who passed from city X

=1250 x 28/100 x 80/100

=280IncorrectAnswer –

**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 Aptitude**The following pie-chart 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 non-fresher candidates passed from city Y in year 2010 were 180, then how many total candidates passed the SBI exam from all cities together?**CorrectAnswer-

**3)1500**

Explanation

Toatl passed candidates from city Y

=180 x 100/75 = 240

A total candidates passed from all cities = 240×100/16 = 1500IncorrectAnswer-

**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 Aptitude

**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?**CorrectAnswer –

**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 = 2156-1344

=812IncorrectAnswer –

**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 = 2156-1344

=812 - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**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 non-fresher passed from city Z?**CorrectAnswer –

**1)112:187**

Explanation

Total candidates passed = 320 x 100/16 = 2000

Candidates passed from city Z = 11/100 x 2000 = 220

Non-fresher 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: 187IncorrectAnswer –

**1)112:187**

Explanation

Total candidates passed = 320 x 100/16 = 2000

Candidates passed from city Z = 11/100 x 2000 = 220

Non-fresher 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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