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+1 vote
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$A$ can complete half work in $7$ days $B$ can do $\dfrac{1}{3}$ of work in $14$ days .while $C$ can complete the $20\%$ of the remaining work in $\dfrac{28}{5}$ days. In how many days $A,B,C$ will complete the work together ?
asked in Quantitative Aptitude by (38 points) 3 4
edited by | 147 views
0
$\dfrac{28}{5}$days OR $2\dfrac{8}{5}$ days ?

which one
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ACTUALLY THIS QUESTION IS FROM A VIDEO LECTURE WHICH SAYS ANSWER IS 84/11 BUT WHEN I SOLVED I GOT 168/17.......SO DONT KNOW ACTUAL ANSWER BUT HOW DID U GOT ????
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I mean to say C can complete $20\%$ of the remaining work in $\dfrac{28}{5}$ days or $2\dfrac{8}{5}$ days?
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it is  28/5.(28 upon 5)
+1
Yes, I also got $\dfrac{168}{17} = 9.88$ days
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@eyeamgj This is not your self doubt - so please add the actual source of your question.

3 Answers

+2 votes
$A$ does $\dfrac{1}{2}^{th}$ work in $7$ days

So, in $\dfrac{7}{\dfrac{1}{2}} = 7 \times 2 = 14 $ days $A$ can complete the whole work

$B$ does $\dfrac{1}{3}^{th}$ work in $14$ days

So, in  $\dfrac{14}{\dfrac{1}{3}} = 14 \times 3 = 42 $ days $B$ can complete the whole work

Now, remaining work = $\left ( 1 - \dfrac{1}{2} + \dfrac{1}{3} \right ) = \left ( 1 - \dfrac{5}{6} \right ) = \dfrac{6 - 5}{6} = \dfrac{1}{6}$

& $20\%$ of the remaining work = $\dfrac{1}{6} \times 20\% =\dfrac{1}{6} \times \dfrac{1}{5} =\dfrac{1}{30}$

∴ $C$ does $\dfrac{1}{30}^{th}$ work in $\dfrac{28}{5}$ days

So, in $\dfrac{\dfrac{28}{5}}{\dfrac{1}{30}} = \dfrac{28}{5} \times 30 = 28 \times 6 = 168$ days $C$ can complete the whole work

Now, LCM of $(14, 42, 168) = 168$

∴ Total units of work = $168 $ units

∴ $A$ can complete $\dfrac{168}{14} = 12$ units of work in $1$ day

$B$ can complete $\dfrac{168}{42} = 4$ units of work in $1$ day

$C$ can complete $\dfrac{168}{168} = 1$ units of work in $1$ day

When $A, B, C$ works together in $1$ day total work done = $12+4+1 = 17$ units

∴ To complete $168$ units $A, B, C $ have to work for $\dfrac{168}{17} = 9.88$ days

∴ To complete the whole work $A, B, C$ have to work together for $9.88$ days.
answered by (2.3k points) 5 8 14
edited by
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THANK U SO MUCH ......
0
great explanation
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A can do a full work in 14 days

B can do a full work in 42 days

C can do a full work in 28/5 is 5.6

What is 20 %

So full work done by c is 5.6*5/1 = 28

Then take lcm of (14,42,28) =84

So a do 14/84 = 6 work in a day

B do 42/84 = 2 work in a day

C do 28/84 = 3 work in a day

So A B and C do 11 work in a day.

So the answer is 84/11 days taken by A B C complete work together.
0 votes
84/17
answered by (18 points) 1
0 votes
$A$ can complete half work in $7$ days. So, speed of work of $A = \frac{x}{2\times 7} = \frac{x}{14}$ where $x$ is the amount of work to be done.

Similarly speed of work of $B = \frac{x}{42}$

Given that $A$ is doing half the work and $B$ is doing $1/3$ the work. So, remaining work $= 1 - 1/2 - 1/3 = 1/6.$ $C$ can complete $20\%$ of this in $28/5$ days. So,

speed of work of $C = \frac{0.2 \times 5x}{28 \times 6} = \frac{x}{168}$

Now, when $A, B$ and $C$ work together, their combined speed will be$ = \frac{x}{14} + \frac{x}{42} + \frac{x}{168} = \dfrac{12x+4x+x}{168} = \dfrac{17x}{168} $

So, amount of time required $=\dfrac{x}{\frac{17x}{168}} = \dfrac{168}{17}$
answered by (7.6k points) 51 68 112

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