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A is thrice as good a workman as B. C alone takes 48 days to paint a house. All three A, B and C working together take 16 days to paint the house. It will take how many days for B alone to paint the house?​

Answer 1

If A is thrice as good as B and A,B & C can work for 16 days and C work for 48 days then, it will take 32 days for B alone to paint the house.

Step-by-step explanation:

C alone can paint the house in 48 days

So, C’s 1-day work = $\frac{1}{48}$

A, B & C together can paint the house in 16 days

So, (A+B+C)’s 1-day work = $\frac{1}{16}$

∴ (A+B)’s 1-day work is given by,

= [(A+B+C)’s 1-day work] – [C’s 1-day work]

= $\frac{1}{16}$ – $\frac{1}{48}$

= $\frac{3-1}{48}$

= $\frac{2}{48}$

= $\frac{1}{24}$ …… (i)

Now, we are given that A is 3 times as good a workman as B i.e., the ratio of efficiency of A & B = 3:1

Since Efficiency ∝ Work done

So, the ratio of the work done by A & B = 3:1

We can consider A alone can work for “3x” days and B alone can work for “x” days.  

So, A’s 1-day work = 1/3x & B’s 1-day work = 1/x ….. (ii)

Therefore, combining (i) & (ii), we can form an equation as,

$\frac{1}{3x}$ + $\frac{1}{x}$ = $\frac{1}{24}$

⇒ $\frac{x+3x}{3x*x}$ = $\frac{1}{24}$

⇒ $\frac{4}{3x}$ = $\frac{1}{24}$

⇒ x = 32 days

Thus, B alone can paint the house in 32 days.

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Answer 2

B alone can paint the house in 32 days.

Step-by-step explanation:

Since we have given that

Let the number of days B works be 'x'

Let the number of days A works be '3x'.

Number of days C works = 48

Work done by A = $\dfrac{1}{x}$

Work done by B = $\dfrac{1}{3x}$

Work done by  C = $\dfrac{1}{48}$

All three A, B, and C work in 16 days.

So, According to question, it becomes,

$\dfrac{1}{x}+\dfrac{1}{3x}+\dfrac{1}{48}=\dfrac{1}{16}\\\\\dfrac{3+1}{3x}=\dfrac{1}{16}-\dfrac{1}{48}\\\\\dfrac{4}{3x}=\dfrac{3-1}{48}\\\\\dfrac{4}{3x}=\dfrac{2}{48}\\\\x=\dfrac{48\times 4}{6}\\\\x=32$

Hence, B alone can paint the house in 32 days.

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