# QUESTIONS ON TIME & WORK (PART-II)

## 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 = ^{1}⁄_{60}

B’s 1 day work = ^{1}⁄_{40}

(A+B)’s 1 day work = ^{1}⁄_{60} + ^{1}⁄_{40} = ^{1}⁄_{24}

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 = ^{1}⁄_{24}

So their 20 days’ work = (^{1}⁄_{24})*20 = ^{5}⁄_{6}

Amount of work left to be done = 1 – ^{5}⁄_{6} = ^{1}⁄_{6}

But this work has to be finished by A alone now

Time taken by A to do ^{1}⁄_{60} of the work = 1 day

Hence time to be taken by him to do ^{1}⁄_{6} of the work = ^{60}⁄_{6} = 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 = ^{1}⁄_{60}

Therefore his 4 days work = (^{1}⁄_{60})*4 = ^{1}⁄_{15}

Obviously the remaining work i.e. 1 – ^{1}⁄_{15} = ^{14}⁄_{15} 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 ^{14}⁄_{15} of the work = 24*(^{14}⁄_{15}) = ^{112}⁄_{5} = 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’)