Question

A B and C can do the piece of work in 4 days, A alone can do it in 10 days, B alone can it in 18 days, in how many days will C alone do the same work??​

Answer 1

Answer :

C alone can do the same work in 10.59 days.

Step-by-step explanation :

Given :

• A, B and C can do the piece of work in 4 days
• A alone can do it in 10 days
• B alone can it in 18 days

To find :

the number of days taken by C to do the work alone

Solution :

Let "x" be the number of days taken by C to do the work alone.

Work done by C alone in one day = 1/x

A alone can do the work in 10 days :

Work done by A alone in one day = 1/0

B alone can do the work in 18 days :

Work done by B alone in one day = 1/18

A, B and C can do the work together in 4 days :

Work done by them together in one day = 1/4

So,

Work done by them together in one day = Work done by A alone in one day + Work done by B alone in one day + Work done by C alone in one day

  $\sf \dfrac{1}{4}=\dfrac{1}{10}+\dfrac{1}{18}+\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\bigg(\dfrac{1}{10} \times \dfrac{9}{9}\bigg) + \bigg(\dfrac{1}{18} \times \dfrac{5}{5} \bigg) +\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\dfrac{9}{90}+\dfrac{5}{90}+\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\dfrac{9+5}{90}+\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\dfrac{14}{90}+\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\dfrac{2 \times 7}{2 \times 45}+\dfrac{1}{x} \\\\ \sf \dfrac{1}{4}=\dfrac{7}{45}+\dfrac{1}{x}$

  $\sf \dfrac{1}{x}=\dfrac{1}{4} -\dfrac{7}{45} \\\\ \sf \dfrac{1}{x}=\dfrac{1 \times 45}{4 \times 45} -\dfrac{7 \times 4}{45 \times 4} \\\\ \sf \dfrac{1}{x}=\dfrac{45}{180} -\dfrac{28}{180} \\\\ \sf \dfrac{1}{x}=\dfrac{45-28}{180} \\\\ \sf \dfrac{1}{x}=\dfrac{17}{180} \\\\ \sf x=\dfrac{180}{17} \\\\ \sf x \simeq 10.59$

Therefore, C alone can do the work in approximately 10.59 days

Answer 2

$\sf{Answer}$

Step by step explanation:-

Given :-

A,B,C Can do work in 4days

A alone can do in 10days

B alone can do in 18days

To find :-

How many days required for C alone to complete the work

Solution :-

For finding the C alone work We have to find first one day work

Let the work done by C is x

One day work of A is 1/10

One day of B is 1/18

One day work of C is 1/x

One day work of A,B,C is 1/4

So,

Work done by together in one day = Work done by A alone + Work done by B alone + Work done by C alone

So,

$\frac{1}{4}$ = $\frac{1}{10}$ + $\frac{1}{18}$ + $\frac{1}{x}$

$\frac{1}{4}$-$\frac{1}{x}$ = $\frac{1}{18}$+$\frac{1}{10}$

$\frac{x-4}{4x}$ = $\frac{ 9+5}{90}$

$\frac{x-4}{4x}$ = $\frac{14}{90}$

${90x-360}$ =${ 56x}$

${-360}$ = ${56x -90x}$

${-360}$ = ${-34x}$

${360}$ = ${34x}$

${x}$ = $\frac{360}{34}$

${x}$ = ${10.588}$

x = 10.599

So, work done by C alone is 10.59 days

Hope my answer helps to u

Thank u :)