Question : A can do a certain piece of work in 2.4 times the number of days in which B and C together can do it. If A and B together can do the same piece of work in 27 days and C alone can do it in 75 days, then how many days will B take to do this piece of work alone?
Option 1: 54
Option 2: 48
Option 3: 45
Option 4: 42
Correct Answer: 45
Solution :
(A + B)'s 1 day's work = $\frac{1}{27}$
C's 1 day's work = $\frac{1}{75}$
(A + B + C)'s 1 day's work = $\frac{1}{27}+\frac{1}{75} = \frac{34}{675}$ ---(1)
A's 1 day's work = $\frac{1}{2.4}$(B + C)'s 1 day's work ---(2)
$\frac{3.4}{2.4}$(B + C)'s 1 day's work = $\frac{34}{675}$
⇒ (B + C)'s 1 day's work = $\frac{24}{675}$
A's 1 day work =$ \frac{34}{675}- \frac{24}{675}=\frac{10}{675}$
we know (A + B)'s 1 day's work = $\frac{1}{27}$
$\therefore$ B's 1 day's work will be = $\frac{1}{27} - \frac{10}{675}$
$\therefore$ B can do the whole task in $=\frac{{15}}{{675}}= 45$ days
Hence, the correct answer is 45.
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