Aptitude Time and Work Test Paper 5

Answer: D

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

Answer: B

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.

Answer: C

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.

Answer: D

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.

Answer: B

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.