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A, B, and C worked together to complete a job in $10$ days. However, C only worked for the first three days when $37/100$ of the job was done. Also, the work done by A in $5$ days is equal to the work done by B in $4$ days. How many days would be required by the fastest worker to complete the entire job?

  1. $30$
  2. $28$
  3. $20$
  4. $45$
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Let the speed of working of $A$ be $x$ that of $B$ be $y$ and that of $C$ be $z.$

Let the total amount of work be $X$

$\dfrac{0.37X}{x+y+z} = 3\to (1) \qquad\dfrac{0.63X}{x+y}=7 \to (2)$

$y = \frac{5}{4}x (B$ is faster than $A)$

Substituting $y$ in $(2)$ we get $\dfrac{0.63X}{\frac{9}{4}x} = 7 \implies x = 0.04X \implies y = 0.05X $

From $(1)$ we get $\dfrac{0.37X}{0.09X+z} = 3 \implies 0.37X =0.27X+ 3z \implies 3z = 0.1 X \implies z = 0.033 X $

So, $z < x < y$ and $B$ is the fastest worker with working speed of $0.05X$ per day.

So, total number of days required by $B$ = $\dfrac{X}{0.05X} =20.$
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Let us assume work done in one day B = x then A = 4x/5 and c worked only for 3 days.

Remaining work is 1- 37/100 = 63/100

Remaining days work =>

7((5/4x) +(1/x)) =63/100 => x= 25

i.e B can complete in 25 days and A can complete in 20 days
Answer:

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