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Bunuel
cleetus
A, B, C are three taps connected to a tank such that 6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank. 3 times the time taken by C to fill the tank is 10 times the time taken by A and B together to fill the tank. If A, B and C together fill the tank in 60/13 hours, then find the time taken by B alone to fill the tank?

A) 10 hrs
B) 15 hrs
C) 20 hrs
D) 25 hrs
E) 30 hrs

Check this: word-translations-rates-work-104208.html?hilit=time%20work#p812628

Let a, b and c be the time needed for A, B and C respectively to fill the tank alone.

Given:
\(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\) - combined rate of B and C is 7/6 of rate of A;

\(\frac{10}{3}*\frac{1}{c}=\frac{1}{a}+\frac{1}{b}\) - combined rate of A and B is 10/3 of rate of C;

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{13}{60}\) - combined rate of A, B and C is 13/60 tank/hour;

Solving:
\(\frac{1}{a}+\frac{7}{6a}=\frac{13}{60}\) --> \(a=10\);

\(\frac{1}{c}+\frac{10}{3c}=\frac{13}{60}\) --> \(c=20\);

\(\frac{1}{10}+\frac{1}{b}+\frac{1}{20}=\frac{13}{60}\) --> \(b=15\).

Answer: B.

Hi Bunuel,

I have a really stupid question, but I hope you can help me out. Why isn't it 6 (1/A) = 7(1/B + 1/C) leading to a final equation of 6/7 (1/A) = 1/B + 1/C?

Thanks in advanced - as always!!

6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank:

Time taken by A is a.
Time taken by B and C together to fill the tank is reciprocal of combined rate of B and C, thus it's bc/(b+c).

Given: \(6a=7*\frac{bc}{b+c}\) --> \(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\).

Hope it's clear.
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Bunuel

6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank:

Time taken by A is a.
Time taken by B and C together to fill the tank is reciprocal of combined rate of B and C, thus it's bc/(b+c).

Given: \(6a=7*\frac{bc}{b+c}\) --> \(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\).

Hope it's clear.
Isn't time taken by A supposed to be (1/a)?
And 6 time the time taken by A in 6*(1/a) as per the R*T=W?
I don't get how it's possible to say that a is the time taken by A, and then
when multiplying, you do the multiplication and only then take the reciprocal....
Can you explain this please?
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Bunuel

6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank:

Time taken by A is a.
Time taken by B and C together to fill the tank is reciprocal of combined rate of B and C, thus it's bc/(b+c).

Given: \(6a=7*\frac{bc}{b+c}\) --> \(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\).

Hope it's clear.
Isn't time taken by A supposed to be (1/a)?
And 6 time the time taken by A in 6*(1/a) as per the R*T=W?
I don't get how it's possible to say that a is the time taken by A, and then
when multiplying, you do the multiplication and only then take the reciprocal....
Can you explain this please?

a is the time taken by A, because I denoted it that way.
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Bunuel

6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank:

Time taken by A is a.
Time taken by B and C together to fill the tank is reciprocal of combined rate of B and C, thus it's bc/(b+c).

Given: \(6a=7*\frac{bc}{b+c}\) --> \(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\).

Hope it's clear.
Isn't time taken by A supposed to be (1/a)?
And 6 time the time taken by A in 6*(1/a) as per the R*T=W?
I don't get how it's possible to say that a is the time taken by A, and then
when multiplying, you do the multiplication and only then take the reciprocal....
Can you explain this please?

a is the time taken by A, because I denoted it that way.
If so, then why did you take the reciprocal of it for the equations?
Why not leave them with a,b, and c?
I am missing something, but I can't tell what....
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Bunuel
ronr34

Isn't time taken by A supposed to be (1/a)?
And 6 time the time taken by A in 6*(1/a) as per the R*T=W?
I don't get how it's possible to say that a is the time taken by A, and then
when multiplying, you do the multiplication and only then take the reciprocal....
Can you explain this please?

a is the time taken by A, because I denoted it that way.
If so, then why did you take the reciprocal of it for the equations?
Why not leave them with a,b, and c?
I am missing something, but I can't tell what....

I think you should go through the basics once more. For example, check here: work-word-problems-made-easy-87357.html or here: two-consultants-can-type-up-a-report-in-12-5-hours-and-edit-126155.html#p1030079

(time)*(rate)=)(job done), thus time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).
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A, B, C are three taps connected to a tank such that 6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank. 3 times the time taken by C to fill the tank is 10 times the time taken by A and B together to fill the tank. If A, B and C together fill the tank in 60/13 hours, then find the time taken by B alone to fill the tank?

A) 10 hrs
B) 15 hrs
C) 20 hrs
D) 25 hrs
E) 30 hrs

Check this: word-translations-rates-work-104208.html?hilit=time%20work#p812628

Let a, b and c be the time needed for A, B and C respectively to fill the tank alone.

Given:
\(\frac{7}{6}*\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\) - combined rate of B and C is 7/6 of rate of A;

\(\frac{10}{3}*\frac{1}{c}=\frac{1}{a}+\frac{1}{b}\) - combined rate of A and B is 10/3 of rate of C;

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{13}{60}\) - combined rate of A, B and C is 13/60 tank/hour;

Solving:
\(\frac{1}{a}+\frac{7}{6a}=\frac{13}{60}\) --> \(a=10\);

\(\frac{1}{c}+\frac{10}{3c}=\frac{13}{60}\) --> \(c=20\);

\(\frac{1}{10}+\frac{1}{b}+\frac{1}{20}=\frac{13}{60}\) --> \(b=15\).

Answer: B.

Hi Bunuel, how good one should be to finish this problem in 2 min? Because I spent 1:30 to set up all the equations, and even if the solving part is not that complicated it is not even that quick.
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Given: A, B, C are three taps connected to a tank such that 6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank. 3 times the time taken by C to fill the tank is 10 times the time taken by A and B together to fill the tank.

Asked: If A, B and C together fill the tank in 60/13 hours, then find the time taken by B alone to fill the tank?

Let the time taken by taps A, B & C be x, y & z hours respectively

6 times the time taken by A to fill the tank is 7 times the time taken by B and C together to fill the tank.
6x = 7yz/(y+z)
7/x = 6(1/y + 1/z)

3 times the time taken by C to fill the tank is 10 times the time taken by A and B together to fill the tank.
3z = 10xy/(x+y)
10/z = 3(1/x + 1/y)

A, B and C together fill the tank in 60/13 hours
1/x + 1/y + 1/z = 13/60

7/x = 6(1/y + 1/z) = 6(13/60 - 1/x) = 13/10 - 6/x
13/x = 13/10
x = 10 hours

10/z = 3(1/x + 1/y) = 3(13/60 - 1/z) = 13/20 - 3/z
13/z = 13/20
z = 20 hours

1/10 + 1/y + 1/20 = 13/60 = 3/20 + 1/y
1/y = 13/60 - 3/20 = 4/60 = 1/15
y = 15 hours

IMO B
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Assume Ta, Tb, Tc are time taken by A, B and C to fill the full tank individually.
Given 6*Ta = 7* T(a & b)
3 * Tc = 10 * T(a & b)
Ta + Tb + Tc = 60/13

Ta/ T(a & b) = 7/6 and Tc/ T(a &b) = 10/3

Ta : T(a & b)
7 : 6
As work done is constant, time and efficiency are inversely proportional
Ea : E(b & c)
6 : 7 ----- (1)

Similarly,
Tc : T(a & b)
10 : 3

Ec : E(a & b)
3 : 10 ----- (2)


Adding (1) and (2)
Ea + Ec = 6 + 3 = 9 ---- (3)
Ea + 2Eb + Ec = 17 ---- (4)

(4)-(3)
2Eb = 8
Eb =4

Therefore, Ea=6, Eb=4 and Ec = 3

TW = (Ta + Tb + Tc) * (Ea + Eb + Ec)
= ( 60/13) * ( 6+4+3) [Given total time taken by A,B,C together to fill the tank = 60/13)
= (60/13) * (13)
= 60
TW = 60

Total time taken by B to fill the tank = 60/Eb = 60/4 = 15 hours
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