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Emdadul28
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A and B can do a piece of work in 10 days, while B and C can do the sa Updated on: 26 Feb 2019, 04:39
Originally posted by Emdadul28 on 08 Nov 2016, 02:45.
Last edited by Bunuel on 26 Feb 2019, 04:39, edited 2 times in total.
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These
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D
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A and B can do a piece of work in 10 days, while B and C can do the sa 08 Nov 2016, 03:03
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Emdadul28
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?

A. 16
B. 32
C. 64
D. 96
E. None of These
Show more

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 = 1/10;
B and C can do the same work in 15 days: b + c = 1/15;
C and A can do the same work in 25 days: c + a = 1/25.

Sum the above 3 equations: 2(a + b + c) = 31/150 --> a + b + c = 31/300
Subtract a + b = 1/10 from above to get c = 1/300.

For 4 days all 3 worked and completed 4*(a + b + a) = 124/300 of the work.
For the next 4 days B and C worked and they completed 4(b + c) = 4/15 = 80/300 of the work.

So, by the time C is left alone 1 - (124/300 + 80/300) = 96/300 of the work is left to be completed by C alone.
Time = Job/Rate = (96/300)/(1/300) = 96 days.

Answer: D.
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A and B can do a piece of work in 10 days, while B and C can do the sa 08 Nov 2016, 11:01
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These
Show more

Let the total work be 150 units

Efficiency of A + B = 15 units/day
Efficiency of B + C = 10 units/day
Efficiency of A + C = 6 units/day

Efficiency of A + B + C = 15.5 units/day

So, Efficiency of A is = Efficiency of A + B + C - Efficiency of B + C

Or, Efficiency of A is = 15.5 - 10

Or, Efficiency of A is = 5.5

Similarly Efficiency of B is = 9.5 and, Efficiency of C is = 0.50


Quote:
they started working together, after 4 days A left.
Show more

Work done by A + B + C in 4 days is 15.5*4 = 62

Now, B + C works

Quote:
After another 4 days B left.
Show more

Work done by B + C in 4 days is 10*4 = 40 units

So, total work done is 102 units, and work left is 48 units..

Time taken by C to finish this work is \(\frac{48}{0.5}\) = 96 days...

Hence, answer will be (D) 96 days..
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A and B can do a piece of work in 10 days, while B and C can do the sa 08 Nov 2016, 05:01
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These

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 = 1/10;
B and C can do the same work in 15 days: b + c = 1/15;
C and A can do the same work in 25 days: c + a = 1/25.

Sum the above 3 equations: 2(a + b + c) = 31/150 --> a + b + c = 31/300
Subtract a + b = 1/10 from above to get c = 1/300.

For 4 days all 3 worked and completed 4*(a + b + a) = 124/300 of the work.
For the next 4 days B and C worked and they completed 4(b + c) = 4/15 = 80/300 of the work.

So, by the time C is left alone 1 - (124/300 + 80/300) = 96/300 of the work is left to be completed by C alone.
Time = Job/Rate = (96/300)/(1/300) = 96 days.

Answer: D.
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Quote:
Assume that total units of work =LCM (10,15,25) = 60 units

Since A and B finish the work in 10 days, work done by them in a day = 60/10 = 6 units/day

AND B and C finish the work in 15 days, work done by them in a day = 60/15 = 4 units/day

AND C and A finish the work in 20 days, work done by them in a day = 60/20 = 3 units/day
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@@@ Thanks for this. Can I do this math like this way? If so then help me to complete this math.
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A and B can do a piece of work in 10 days, while B and C can do the sa 08 Nov 2016, 06:35
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These

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 = 1/10;
B and C can do the same work in 15 days: b + c = 1/15;
C and A can do the same work in 25 days: c + a = 1/25.

Sum the above 3 equations: 2(a + b + c) = 31/150 --> a + b + c = 31/300
Subtract a + b = 1/10 from above to get c = 1/300.

For 4 days all 3 worked and completed 4*(a + b + a) = 124/300 of the work.
For the next 4 days B and C worked and they completed 4(b + c) = 4/15 = 80/300 of the work.

So, by the time C is left alone 1 - (124/300 + 80/300) = 96/300 of the work is left to be completed by C alone.
Time = Job/Rate = (96/300)/(1/300) = 96 days.

Answer: D.

Quote:
Assume that total units of work =LCM (10,15,25) = 60 units

Since A and B finish the work in 10 days, work done by them in a day = 60/10 = 6 units/day

AND B and C finish the work in 15 days, work done by them in a day = 60/15 = 4 units/day

AND C and A finish the work in 20 days, work done by them in a day = 60/20 = 3 units/day

@@@ Thanks for this. Can I do this math like this way? If so then help me to complete this math.
Show more

You can continue basically exactly the same way with this approach.
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A and B can do a piece of work in 10 days, while B and C can do the sa 08 Nov 2016, 07:06
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These

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 = 1/10;
B and C can do the same work in 15 days: b + c = 1/15;
C and A can do the same work in 25 days: c + a = 1/25.

Sum the above 3 equations: 2(a + b + c) = 31/150 --> a + b + c = 31/300
Subtract a + b = 1/10 from above to get c = 1/300.

For 4 days all 3 worked and completed 4*(a + b + a) = 124/300 of the work.
For the next 4 days B and C worked and they completed 4(b + c) = 4/15 = 80/300 of the work.

So, by the time C is left alone 1 - (124/300 + 80/300) = 96/300 of the work is left to be completed by C alone.
Time = Job/Rate = (96/300)/(1/300) = 96 days.

Answer: D.

Quote:
Assume that total units of work =LCM (10,15,25) = 60 units LCM(10,15,25)=300

Since A and B finish the work in 10 days, work done by them in a day = 60/10 = 6 units/day 300/10= 30 units

AND B and C finish the work in 15 days, work done by them in a day = 60/15 = 4 units/day 300/15= 20 units

AND C and A finish the work in 20 days, work done by them in a day = 60/20 = 3 units/day
AND C and A finish the work in 25 days, work done by them in a day = 300/25 = 12 units/day

Certain data have been wrongly noted down

@@@ Thanks for this. Can I do this math like this way? If so then help me to complete this math.
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A and B can do a piece of work in 10 days, while B and C can do the sa 12 Aug 2019, 22:22
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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?

A. 16
B. 32
C. 64
D. 96
E. None of These
Show more

Given: 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.

Asked: In how many days C can finish the remaining work?

Let us assume that A, B & C can finish the job in a, b & c days respectively

A and B can do a piece of work in 10 days
\(\frac{1}{a} + \frac{1}{b} = \frac{1}{10}\) (1)
B and C can do the same work in 15 days
\(\frac{1}{b} + \frac{1}{c} = \frac{1}{15}\) (2)
C and A can do the same work in 25 days
\(\frac{1}{c} + \frac{1}{a} = \frac{1}{25}\) (3)

Adding (1) + (2) + (3)
\(2 (\frac{1}{a} + \frac{1}{b}+ \frac{1}{c}) = \frac{1}{10} + \frac{1}{15} + \frac{1}{25} = \frac{31}{150}\)
\frac{1}{a} + \frac{1}{b}+ \frac{1}{c} = \frac{31}{300}[/m] (4)


they started working together, after 4 days A left
In 4 days, work completed = \(4 * \frac{31}{300} = \frac{124}{300}\)
After 4 days, work balance = 1 - \frac{124}{300} = \frac{176}{300} = \frac{44}{75}

Balance work is done by B & C together for 4 days
After another 4 days B left.
Work done by B & C in 4 days\(= 4 * \frac{1}{15} = \frac{4}{15}\)

Work balance = \(\frac{44}{75} - \frac{4}{15} = \frac{24}{75} = \frac{8}{25}\)

Balance work is done by C alone
\(\frac{1}{c} = \frac{31}{300} - \frac{1}{10} = \frac{1}{300}\)
No of days taken by C \(= \frac{8}{25}/\frac{1}{300} = \frac{8*300}{25} = 96\)

IMO D
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A and B can do a piece of work in 10 days, while B and C can do the sa 12 Aug 2019, 22:35
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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

Answer is (D)
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A and B can do a piece of work in 10 days, while B and C can do the sa 15 Aug 2020, 08:01
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A + B = 10 days.
B + C = 15 days.
C + A = 25 days.

Let the amount of work be LCM of (10,15,20) = 150 units.

A + B = 15 units per day
B + C = 10 units per day
C + A = 6 units per day

A + B + C = \(\frac{31 }{ 2}\) = 15.5 units per day

Alone C = 15.5 - ( A + B ) = 15.5 - 15 = 0.5 units per day

First 4 days: All worked together and hence units completed are: 15. 5 * 4 = 62 units.

Next 4 days, Only B and C worked together and hence units completed are: 10 * 4 = 40 units

The total units completed are 102.
Remaining : 150 - 102 = 48.

48 units will be completed by C alone at a rate of 0.5 units per day.

Hence, C = \(\frac{48 }{ 0.5}\) = 96 days.

Answer D
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A and B can do a piece of work in 10 days, while B and C can do the sa 12 Oct 2021, 19:37
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Let the total work be 150 units

Efficiency of A + B = 15 units/day
Efficiency of B + C = 10 units/day
Efficiency of A + C = 6 units/day

Efficiency of A + B + C = 15.5 units/day

So, Efficiency of A is = Efficiency of A + B + C - Efficiency of B + C

Or, Efficiency of A is = 15.5 - 10

Or, Efficiency of A is = 5.5

Similarly Efficiency of B is = 9.5 and, Efficiency of C is = 0.50

Now a worked for 4 days, B, 8 days and C lets assume (8+ x) days
now
4*A + 8*B + C(8+x)= 150
4*5.4+ 8*9.5 +.5(8+x)= 150
Solving for x = 48/.5 = 96

Or

Now a worked for 4 days, B, 8 days and C lets assume (x) days
now
4*A + 8*B + C(x)= 150
4*5.4+ 8*9.5 +.5(x)= 150
Solving for x = 104
This 104 already includes the 8 days that C worked together so we reduce that and find the number of days he has to work more
i.e. 104-8= 96 .
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A and B can do a piece of work in 10 days, while B and C can do the sa Updated on: 07 Sep 2025, 02:30
Originally posted by Ngocanhvy on 07 Sep 2025, 02:29.
Last edited by Ngocanhvy on 07 Sep 2025, 02:30, edited 1 time in total.
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Inspired by the solution of Karishma using convenient value, here is my solution:

Assump the work needed to be done is 150 (LCM(10,15,25) = 150)
A,B, C is the number of pieces of work done by A,B,C in one day. We have:
A + B = 15
B + C = 10
A+ C = 6

2(A+B+C) = 31 <=> A+B+C = 31/2
So in 4 days, three people can complete: 31*2 = 62 pieces
In 4 days (after A left), B & C can complete: 10*4 = 40 pieces
So the total pieces done is: 62+40 = 102
The remaining work: 150 - 102 = 48

We also have: (A+C) + (B+C) = (A+B) + 2C = 6 + 10 = 16
2C = 16 - (A+B) = 16 - 15 =1
So C can done 1/2 piece of work in one day.

So the number of days that C need to complete the remaining work is: 48/(1/2) = 48* 2 = 96
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A and B can do a piece of work in 10 days, while B and C can do the sa 07 Sep 2025, 02:29
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Inspired by the solution of Karishima using convenient value, here is my solution:

Assump the work needed to be done is 150 (LCM(10,15,25) = 150)
A,B, C is the number of pieces of work done by A,B,C in one day. We have:
A + B = 15
B + C = 10
A+ C = 6

2(A+B+C) = 31 <=> A+B+C = 31/2
So in 4 days, three people can complete: 31*2 = 62 pieces
In 4 days (after A left), B & C can complete: 10*4 = 40 pieces
So the total pieces done is: 62+40 = 102
The remaining work: 150 - 102 = 48

We also have: (A+C) + (B+C) = (A+B) + 2C = 6 + 10 = 16
2C = 16 - (A+B) = 16 - 15 =1
So C can done 1/2 piece of work in one day.

So the number of days that C need to complete the remaining work is: 48/(1/2) = 48* 2 = 96
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