Question

A can do a job in 16 days and B can do the same job in 12 days. With the help of C, they can finish the job in 6 days only. Then, C alone can finish it in​

Answer 1

Given

Let the A's One day work be 1 / 16

Let the B's One day work be 1/ 12

Let A,B and C's One day work be 1/ 6

To Find

how much time will be taken for C alone.

Solution

C's One day work will be the subtraction of A and B's work from the total ,

$\longrightarrow \: \dfrac{1}{6} - ( \frac{1}{12} + \frac{1}{16} )$

$\implies \: \dfrac{1}{6} - \dfrac{1}{16 } - \dfrac{1}{12}$

Here L.C.M = 48 and therefore ,

$\implies \: \: \dfrac{8 - 3 - 4}{48}$

$\implies \: \dfrac{1}{48}$

So , C person can complete the work within 48 days.

Answer 2

Answer:

Given :-

• A can do a job in 16 days and B can do the same job in 12 days.
• With the help of C, they finish the job in 6 days only.

To Find :-

• How many days need to finish the job by C alone.

Solution :-

Given :

$\bigstar$ A can do a job in 16 days.

Let,

$\mapsto \bf A\: can\: do\: a\: job\: in\: 1\: day = \dfrac{1}{16}$

Again,

$\bigstar$ B can do the same job in 12 days.

Let,

$\mapsto \bf B\: can\: do\: the\: job\: in\: 1\: day =\: \dfrac{1}{12}$

According to the question :

$\bigstar$ A, B and C can do the same job in 6 days.

So,

$\implies \sf\bold{\purple{A + B + C =\: \dfrac{1}{6}}}$

By putting those values we get,

$\implies \sf \dfrac{1}{16} + \dfrac{1}{12} + C =\: \dfrac{1}{6}$

$\implies \sf \dfrac{3 + 4}{48} + C =\: \dfrac{1}{6}$

$\implies \sf \dfrac{7}{48} + C =\: \dfrac{1}{6}$

$\implies \sf C =\: \dfrac{1}{6} - \dfrac{7}{48}$

$\implies \sf C =\: \dfrac{8 - 7}{48}$

$\implies \sf C =\: \dfrac{1}{48}$

$\implies \sf\bold{\red{C =\: 48}}$

$\therefore$ C can finish the job alone in 48 days .

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