ARITHMETICMATHSTIME AND WORK

QUESTIONS ON TIME & WORK (PART-II)

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QUESTIONS ON TIME & WORK (PART-II)

QUERY 11

A and B can do a work in 12 days, B and C can do the same work in 15 days, C and A can do the same work in 20 days. The time taken by A, B and C to do the same work is?

A) 12 days
B) 11 days
C) 8 days
D) 10 days

MAHA GUPTA
(A+B)’s 1 day work = 1/12
(B+C)’s 1 day work = 1/15
(C+A)’s 1 day work = 1/20

Adding above
2(A+B+C)’s 1 day work = 1/12 + 1/15 + 1/20 = 12/60
=> (A+B+C)’s 1 day work = (12/60)(1/2) = 1/10

Therefore the time taken by A, B and C to do the same work = 10 days (option ‘D’)


QUERY 12

A does a work in 6 days, B in 8 days, and C in 10 days. A worked alone for 2 days before leaving and then B and C worked together for 2 days and after that B left. The remaining work was finished by C alone. How long did it take to finish the work?

A) 37/6 days
B) 37/5 days
C) 6 days
D) 8 days

MAHA GUPTA
A’s 1 day work = 1/6
B’s 2 day work = 1/8
C’s 1 day work = 1/10

Therefore A’s 2 days work = 2*(1/6) = 1/3
(B+C)’s 2 days work= 2*(1/8 + 1/10)= 9/20
Work finished so far = 1/3 + 9/20 = 47/60

Remaining work = 1 – 47/60 = 13/60
It has to be done by C alone

So time taken by C to do this work = 10*(13/60) = 13/6 days
Hence total time needed to finish the work = Time taken by A alone + Time taken by B+C together + Time taken by C alone = 2 + 2 + 13/6
= 37/6 days (option ‘A’)


QUERY 13

Worker A is 50% as efficient as B. C does half of the work done by A & B together. If C alone does the work in 40 days, then A, B, C together can do the work in how many days?

A) 40 days
B) 20 days
C) 40/3 days
D) 40/7 days

MAHA GUPTA
A is 50% as efficient as B
Therefore A’s work : B’s work = 50 : 100 = 1 : 2

Now, let A’s and B’s 1 day work be x and 2x respectively
Therefore according to the question C’s 1 day work = (x + 2x)/2 = 3x/2
But C’s 1 day work = 1/40 (as he alone can do the total of work in 40 days)
Hence, 3x/2 = 1/40
=> x = 1/60

Therefore A/s 1 day work = x = 1/60
And B’s 1 day work = 2x = 2*1/60 = 1/30
C;s 1 day work = 1/40 (shown above)

Thus, (A+B+C)’s 1 day work = 1/60 + 1/30 + 1/40 = 9/120 = 3/40
So, the time in which they all together can do the work = 40/3 days (option ‘C’)

TRICK
As A is 50% efficient as B; so Work of A’s ratio to that of B = 1 : 2
But C is half efficient of A+B, means C = (1+2)/2 = 3/2 = 1.5

Therefore ratio of the work of A, B and C = 1 : 2 : 1.5
We see sum of the ratio of all i.e. 4.5 is thrice that of C
Hence the work will be finished by all working together in 1/3 (1/3 of thrice) of 40 days = 40/3 days (option ‘C’)


QUERY 14

If 30 men working 7 hours per day can do a work in 18 days in how many days will 21 men working 8 hours a day do the same work?

A) 20.5 days
B) 22.5 days
C) 22.8 days
D) 5 days

MAHA GUPTA
30 men’s total hours if the work is done 7 hours a day = 7*18 = 126 hours

Now 30 men are working 8 hours per day, days taken by them = 126/8 days
If 21 men work 8 hours per day, days taken by them = (126*30)/(8*21) = 22.5 days (option ‘B’)


QUERY 15

A does 2/3 part of a work in 4 day. B does 3/5 part of that work in 6 day. If they work together then in how many day work has to be done?

A) 3 days
B) 2 days
C) 15/4 days
D) 23/8 days

MAHA GUPTA
A’s 1 day work = (2/3)/4 = 1/6
B’s 1 day work = (3/5)/6 = 1/10

Therefore (A+B)’s 1 day wok = 1/6 + 1/10 = 4/15
Hence if they work together they’ll finish the work in 15/4 days (option ‘C’)


QUERY 16

A is 60% more than B, and B is 20% less than C; then what is the ratio between A : B : C.

A) 20 : 32 : 25
B) 30 : 32 : 45
C) 32 : 20 : 25
D) 25 : 32 : 20

MAHA GUPTA
Let C = 100
Then B = 100 – 20%100 = 100 – 20 = 80
And A = 80 + 60%of 80 = 80 + 48 = 128

Therefore A : B : C = 128 : 80 : 100
= 32 : 20 : 25 (option ‘C’)


QUERY 17

A and B can complete a work individually in 60 and 40 days. Both start working simultaneously, but 4 days before the work is scheduled to get over B leaves. Find the total number of days taken for the work to be completed.

A) 26 days
B) 26.4 days
C) 26. 8 days
D) 30 days

MAHA GUPTA
A’s 1 day work = 160
B’s 1 day work = 140

(A+B)’s 1 day work = 160 + 140 = 124

Means if both of them do the work till the end time taken will be 24 days

But B leaves 4 days before. Therefore they simultaneously do the work for 24 – 4 = 20 days

Their 1 day work = 124
So their 20 days’ work = (124)*20 = 56

Amount of work left to be done = 1 – 56 = 16
But this work has to be finished by A alone now
Time taken by A to do 160 of the work = 1 day
Hence time to be taken by him to do 16 of the work = 606 = 10 days

Hence total time to do whole of the work in actual = 20 + 10 = 30 days (option ‘C’)

Now look this question.
A and B can complete a work individually in 60 and 40 days. Both start working simultaneously, but B did not work on the last 4 days. Find the total number of days taken for the work to be completed.

If you see carefully both are different. In the first question B leaves 4 days before the scheduled time. But in the second one B leaves before the last 4 days when the work is finished. Means in the last 4 days A alone does the work.

Now A’s 1 day work = 160
Therefore his 4 days work = (160)*4 = 115

Obviously the remaining work i.e. 1 – 115 = 1415 is done by A and B together

From above they do whole of the work in 24 days
Hence time taken by them to do 1415 of the work = 24*(1415) = 1125 = 22.4 days

Therefore total time to finish the whole work = 4 + 22.4 = 26.4 days (option ‘B’)


QUERY 18

A is twice as efficient as B and B is thrice as efficient as C. If all 3 together can complete a work in 10 days, find the difference in the number of days taken to complete the work when A and B work together and when B and C work together.

A) 25
B) 125/6
C) 125/8
D) 125/9

MAHA GUPTA
A = 2B ==> A : B = 2 : 1
B = 3C ==> B : C = 3 : 1
==> A : B : C = 6 : 3 : 1

(A+B+C)’s 1 day work = 1/10
So A/s 1 day work = (1/10)*(6/10) = 6/100
And B’s 1 day work = (1/10)*(3/10) = 3/100
And C’s 1 day work = (1/10)*(1/10) = 1/100

Hence (A+B)’s 1 day work = 6/100 + 3/100 = 9/100
Therefore time to be taken by A+B to do whole of the work = 100/9 days

Now (B+C)’s 1 day work = 3/100 + 1/100 = 4/100
Therefore time to be taken by B+C to do whole of the work = 100/4 days

You see (B+C)’s 1 day work is greater
So the required difference = 100/4 – 100/9 = 125/9 days (option ‘D’)


QUERY 19

16 men do a piece of work in 16 days. 12 more men come after 4 day’s work. In how many days the remaining work will be completed?

A) 7 days
B) 9 days
C) 48/7 days
D) 47/4 days

MAHA GUPTA
Number of men in beginning = 16
Number of men after = 16+12 = 28

The remaining work to be finished by 16 men in = 16 – 4 = 12 days
The remaining work to be finished by 28 men in = (12*16)/28 = 48/7 days (option ‘C’)


QUERY 20

A group of men decided to do a job in 8 days. But since 10 men dropped out everyday, the job got completed at the end of 12th day. How many men were there at beginning?

A) 55
B) 165
C) 80
D) 90

MAHA GUPTA
Let the number of men in the beginning = a
Therefore time taken by 1 man to do whole of the work = 8a days

Because 10 men are dropping everyday and the work finishes in 12 day, time taken by 1 man to do whole of the work = [a + (a – 10) + (1 – 20) + (a – 30) + (a – 40) + (a – 50) + (a – 60) + (a – 70) + (a – 80) + (a – 90) + (a – 100) + (a – 110)] days

Therefore 8a = a + (a – 10) + (a – 20) + (a – 30) + (a – 40) + (a – 50) + (a – 60) + (a – 70) + (a – 80) + (a – 90) + (a – 100) + (a – 110)

=> 8a = 12a – 660
=> a = 660/4 = 165 (option ‘B’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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