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At their usual efficiency levels, A and B together finish a task in $12$ days. If A had worked half as efficiency as she usually does, and B had worked thrice as efficiency as he usually does, the task would have been completed in $9$ days. How many days would A take to finish the task if she works alone at her usual efficiency?

- $24$
- $18$
- $12$
- $36$

1 vote

Given that, $\text{A}$ and $\text{B}$ together finish a task in $12 \; \text{days}$.

- Total work $ = (\text{A} + \text{B}) 12 \; \text{days} \quad \longrightarrow (1)$

And, if $\text{A}$ had worked half as efficiently as she usually does, and $\text{B}$ had worked thrice as efficiently as he usually does, the task would have been completed in $ 9 \; \text{days}$. Then,

- Total work $ = \left(\frac{\text{A}}{2} + 3\text{B} \right) 9 \; \text{days} \quad \longrightarrow (2)$

Equate the equation $(1)$ and $(2),$ we get

$ (\text{A} + \text{B}) 12 = \left(\frac{\text{A}}{2} + 3\text{B} \right) 9 $

$ \Rightarrow 12\text{A} + 12\text{B} = \left(\frac{9}{2}\right)\text{A} + 27\text{B} $

$ \Rightarrow \left(12- \frac{9}{2} \right) \text{A} = (27-12) \text{B}$

$ \Rightarrow \left( \frac{24-9}{2} \right) \text{A} = (27-12) \text{B}$

$ \Rightarrow \frac{15}{2} \text{A} = 15 \text{B}$

$ \Rightarrow 15\text{A} = 30\text{B}$

$ \Rightarrow \frac{\text{A}}{\text{B}} = \frac{2}{1}$

$\Rightarrow \boxed{ \text{A} = 2\text{B}}$

Using formula $: \boxed{\text{Total Work = Efficiency} \times \text{Time}}$

$ \text{Total work} = ( \text{A+B}) \times 12 $

$ \Rightarrow \text{Total work} = (2\text{B} + \text{B}) \times 12 $

$ \Rightarrow \text{Total work} = 3\text{B} \times 12$

$ \Rightarrow \text{Total work} = 36\text{B}$ units.

$\therefore$ Number of days in which $\text{A}$ take to finish the task if she works alone at her usual efficiency $ = \frac {\text{Total Work}}{\text{Efficiency of A}} = \frac{36\text{B}}{\text{A}} = \frac{36\text{B}}{2\text{B}} = 18 \; \text{days}.$

Correct Answer $: \text{B}$