A and B Can Do a Piece Of Work in 10 Days, While B And C Can Do The Same Work in 15 Days And C and A in 25 Days GMAT Problem Solving

Question: A and B can do a piece of work in 10 days, while B and C can do the same work in 15 days and C and A in 25 days. they started working together, after 4 days A left. After another 4 days B left. In how many days C can finish the remaining work?

1. 16
2. 32
3. 64
4. 96
5. None of These

“A and B can do a piece of work in 10 days, while B and C can do the same work in 15 days and C and A in 25 days.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantative Review 2022”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Let the rates of A, B, and C be a, b, and c respectively.
A and B can do a piece of work in 10 days: a + b = \(\frac{1}{10}\);
B and C can do the same work in 15 days: b + c = \(\frac{1}{15}\);
C and A can do the same work in 25 days: c + a = \(\frac{1}{25}\).
Sum the above 3 equations: 2(a + b + c) = \(\frac{31}{150}\)
That implies; a + b + c = \(\frac{31}{300}\)
Subtract a + b = \(\frac{1}{10}\)from above to get c = \(\frac{31}{300}\).
For 4 days all 3 worked and completed 4*(a + b + a) = \(\frac{124}{300}\)of the work.
For the next 4 days B and C worked and they completed 4(b + c) = \(\frac{4}{15}\)= \(\frac{80}{300}\)of the work.
So, by the time C is left alone 1 - (\(\frac{124}{300}\)+ \(\frac{80}{300}\)) = \(\frac{96}{300}\)of the work is left to be completed by C alone.
Time = Job/Rate = (\(\frac{96}{300}\))/(\(\frac{1}{300}\)) = 96 days.Hence the required answer is 96 days.

Correct Answer: D

Approach Solution 2:
Let us consider that A, B and C and finish the job in a, b and c days respectively. A and B can do the job in 10 days. So, \(\frac{1}{a}\) + \(\frac{1}{b}\)= \(\frac{1}{10}\)(equation 1).
B and C can do the job in 15 days. So, \(\frac{1}{b}\)+ \(\frac{1}{c}\)= \(\frac{1}{15}\)(equation 2).
C and A can do the job in 25 days. So, \(\frac{1}{c}\)+ \(\frac{1}{a}\)= \(\frac{1}{25}\)(equation 3).
So adding equation 1+ equation 2+ equation 3, we get-
2(\(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{c}\))=\(\frac{1}{10}\)+\(\frac{1}{15}\)+\(\frac{1}{25}\)
that implies; \(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{c}\)= \(\frac{31}{300}\)(equation 4).
B, C started working after 4 days of A left. So, in 4 work done= 4*\(\frac{31}{300}\)= \(\frac{124}{300}\)
Remaining work = 1 - \(\frac{124}{300}\)= \(\frac{44}{75}\)
The remaining work is done by B and C together in 4 days. After 4 days again B left. So, the work done by B and C in 4 days = 4 * \(\frac{1}{15}\)= \(\frac{4}{15}\)
Therefore the total remaining work = \(\frac{4}{75}\)- \(\frac{4}{15}\)= \(\frac{24}{75}\)= \(\frac{8}{25}\)
Hence, the remaining work done by C alone is
1C = \(\frac{31}{300}\)- \(\frac{1}{10}\)= \(\frac{1}{300}\)
The number of days taken by C = \(\frac{8}{25}\)/\(\frac{1}{300}\)=8*30025 = 96 days

Correct Answer: D

Approach Solution 3:

Let the total work be 300 units

A+B = 30 units/day
B+C = 20 units/day
A+C = 12 units/day

2(A+B+C) = 62 units/day
A+B+C = 31 units/day

Substituting for A+B,

30+C = 31 units/day
C=1 unit/day

In the first 4 days, work done = 31*4 = 124 units
In the next 4 days, work done = 20*4 = 80 units

This leaves [300-(124+80)] = 96 units of work to be done.

Since C works at only 1 unit/day, C will need 96 days to complete the remaining work

Correct Answer: D

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