# Aptitude Time and Work Test Paper 5

21) A can finish the job at the same time in which B and C together do it. If A and B together can finish the work in 10 days and C alone can do the work in 50 days, how many days B will take to complete the same job?

1. 20 days
2. 22 days
3. 22(1/2) days
4. 25 days

Explanation:

ATQ,

The efficiency of A = B+C
(A+B) can finish the work in 10 days, and C can finish the work in 50 days.

Now, A+B = 10 days
C = 50 days

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (10 and 50) = 50
Let the total work = 50

Note: One day work = (total work/ days)

Now,

C's one day work = 50/50 = 1
(A+B)'s one day work = 5

i.e., (A+B+C)'s one day work = 5+1=6

i.e., A+A = 6
Or, A's one day work efficiency = 3

A+B = 5, i.e., 3+B = 5
Or, B's one day efficiency=2
Hence, B alone can work in 50/2 = 25 days

22) A and B working together can finish a work in 12 days, B and C working together can finish the work in 16 days. If A works for 5 days, B works for 7 days, and C completes the remaining work in 13 days, C alone can complete the work in how many days?

1. 22 days
2. 24 days
3. 26 days
4. 28 days

Explanation:

ATQ,

A+B = 12 days
B+C = 16 days

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (12 and 16) = 48
Let the total work = 48

Note: One day work = (total work/ days)

Now,

(A+B)'s one day work = 48/12 = 4 unit
(B+C)'s one day work = 48/16 = 3 unit

As per the question:

A works for 5 days
B works for 7 days or (5 + 2) days, that means B works 5 days with A and remaining 2 days with C.
C works 13 days or (2+11) days, that means C works 2 days with B and remaining 11 days alone.

That means total work done by (A+B) in 5 days
A+B = 5 days * 4 unit = 20 units
And, total work done by (B+C) = 2 days * 3 unit = 6 units
So, A+B+C finish the 26 units of work.

Remaining work = 48-26 = 22 unit work
And C completes the remaining work in 11 days.
i.e., C's one day's works = 22/11 = 2 units.

C alone can finish total work in [total work/ C's one day work] = [48/2] = 24 days.

23) A can finish a work in 10 days and B can finish the same work in 15 days. If they work alternatively, find the time taken to finish the job.

1. 14 days
2. 15 days
3. 12 days
4. 14.5 days

Explanation:

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (10 and 15) = 30
Let the total work = 30

Note: One day work = (total work/ days)

Now,

A's one day work = 30/10 = 3 unit
B's one day work = 30/15 = 2 unit

That means A and B works 5 unit works in 2 days
i.e., 2 days = 5 unit work

To complete the 30 work, multiply both sides with 6.
i.e., 2 days * 6 = 5 unit work * 6

Or, 12 days = 30 works
So, A and B require 12 days to finish the work.

24) A, B, and C individually can finish a job in 10, 15, and 30 days respectively. If A starts the work and continues until the end, B and C work alternatively, in how many days work will be done?

1. 6 days
2. 4 days
3. 4[1/3] days
4. 6[3/5] days

Explanation:

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (10, 15, and 30) = 30
Let the total work = 30

Note: One day work = (total work/ days)

Now,

A's one day work = 30/10 = 3 unit
B's one day work = 30/15 = 2 unit
C's one day work = 30/30= 1 unit

ATQ, A work continuously and B and C works alternatively

i.e., (A+B)'s one day work = 3+2 = 5 unit
And (A+C)'s one day work = 3+1 = 4 unit

Or, 2 days work = 9 unit
To complete the 27 units work, multiply both sides with 3.

i.e., 2 days * 3 = 9 unit * 3
Or, 6 days = 27 units of work
Now, on the 7th day, the remaining work is done by A + B.
If (A+B)'s one day work = 5 unit work
i.e.,
1 work = 1/5 days
Or, 3 work = 3/5 days

Hence, the work will be finished in 6+3/5 days or 6[3/5] days.

25) A can finish a work in 10 days, B can finish the same work in 15 days, and C can finish it in 30 days. All three start working together, but after some days A leaves the job, then after one day B also left the job. C completes the remaining job in 3 days. Find the number of working days of B.

1. 4 days
2. 5 days
3. 6 days
4. 7 days

Explanation:

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (10, 15, and 30) = 30
Let the total work = 30

Note: One day work = (total work/ days)

Now,

A's one day work = 30/10 = 3 unit
B's one day work = 30/15 = 2 unit
C's one day work = 30/30= 1 unit
Let A works for days = D

Note: man * days = total work

ATQ, (A+B+C) works 6 units work per day till D days, so their total work = 6 * D
(B+C) works 3 unit work per day till 1 day, so their total work = 3 * 1 = 3
C works 1 unit work per day till 3 days, so his total work = 1 * 3 = 3

Or, 6*D + 3*1 + 1*3 = 30 (total work)
6D = 30-6
D = 24/6
D= 4

So, A works for 4 days, B works for 4+1=5 days, and C works for 4+1+3= 8 days.

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