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Time & Work

Expert replies
by sukhman » Tue Nov 05, 2013 4:26 am
To do a certain work , C alone takes thrice as long as A & B together. A alone would take as long as B& C together . All three together complete the work in 3 days. How long would each take separately ?
Please solve this question No choices are there !
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Source: — Problem Solving |

by ganeshrkamath » Tue Nov 05, 2013 5:52 am
sukhman wrote:To do a certain work , C alone takes thrice as long as A & B together. A alone would take as long as B& C together . All three together complete the work in 3 days. How long would each take separately ?
Please solve this question No choices are there !
A's rate of work = 1/a
B's rate of work = 1/b
Their combined rate of work = 1/a + 1/b
sukhman wrote:C alone takes thrice as long as A & B together
C's rate of work = 1/c = 1/3 * (1/a + 1/b)
sukhman wrote:All three together complete the work in 3 days
Combined rate of work of A,B, and C = 1/a + 1/b + 1/c
= 1/a + 1/b + 1/3 * (1/a + 1/b)
= 4/3 (1/a + 1/b) = 1/3
1/a + 1/b = 1/4_______________(1)
1/c = 1/12
sukhman wrote:A alone would take as long as B& C together
1/a = 1/b + 1/c
1/a = 1/b + 1/12
1/a - 1/b = 1/12_______________(2)

(1) + (2) => 2/a = 1/3
1/a = 1/6

1/b = 1/12

A alone takes 6 days.
B alone takes 12 days.
C alone takes 12 days.

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by GMATGuruNY » Tue Nov 05, 2013 6:22 am
sukhman wrote:To do a certain work , . A alone would take as long as B& C together . All three together complete the work in 3 days. How long would each take separately ?
Please solve this question No choices are there !
Let A = A's rate alone, B = B's rate alone, and C = C's rate alone.

A alone would take as long as B and C together.
In other words, A is as fast as B and C together:
A = B + C.

C alone takes thrice as long as A & B together.
In other words, A and B together are 3 times as fast as C:
A + B = 3C
A = 3C - B.

Since B+C = A and 3C-B = A, we get:
B+C = 3C-B
2B = 2C
B=C.

Let B=1 unit per hour and C=1 unit per hour.
Since A = B+C, A = 1+1 = 2 units per hour.
Combined rate for A+B+C = 2+1+1 = 4 units per hour.
Since A+B+C take 3 days to complete the job, the amount of work = r*t = 4*3 = 12 units.

Time for A alone to produce 12 units = w/r = 12/2 = 6 days.
Time for B alone to produce 12 units = w/r = 12/1 = 12 days.
Time for C alone to produce 12 units = w/r = 12/1 = 12 days.
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by Patrick_GMATFix » Sun Nov 10, 2013 2:49 pm
Tough question, but can be solved easily, without much math.

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by Mathsbuddy » Mon Nov 11, 2013 8:45 am
While I don't dispute the above answers, which are totally correct, I do wish to throw in a different (and wrong) answer to amuse:

If A and C individually takes 4.5 days alone, but together work inefficiently, taking turns to do the work, while the other sips tea and eats sandwiches, thereby taking 9 days together;
and if B's motivational presence with either or both of them knocks 1.5 days off their time, then the times for them individually are:

A = 4.5 days
C = 4.5 days
B = -1.5 days!!!

Check it out. The maths works!
As the given information about B is always with another worker, and the question does not state that A,B or C are constants, then the following is quite a realistic answer (relating to true people behaviour):

A alone = 4.5 days
C alone = 4.5 days
B alone = 0 days (as he doesn't work - he/she stands with a whip)
B with other(s) reduces production time by 1.5 days.
A and C together work at 50% power due to chatting too much and drinking tea!
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