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CAT 2019 Slot 1 — QA Question 11

BasicsEasy

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently 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?

Answer & solution

  • A

    36

  • B

    24

  • C

    12

  • 18

Solution

Easy

Work in "units of work per day". Take total work as the LCM of the two completion times so the daily rates come out clean. Two scenarios give two linear equations in A's and B's rates; solve for A's rate, then time == work//rate.

1

Set up the work. Let total work =lcm⁡(12,9)=36= \operatorname{lcm}(12,9)=36 units, with daily rates aa (A) and bb (B).

usual together: a+b=3612=3(1)\begin{aligned} &\text{usual together: } a+b = \frac{36}{12} = 3 \quad\text{(1)} \end{aligned}
2

Second scenario. A at half efficiency (a2)\big(\tfrac{a}{2}\big), B at thrice (3b)\big(3b\big), finishing in 99 days.

a2+3b=369=4(2)\begin{aligned} &\frac{a}{2} + 3b = \frac{36}{9} = 4 \quad\text{(2)} \end{aligned}
3

Solve the system. From (1), b=3−ab = 3-a; substitute into (2).

a2+3(3−a)=4(sub step 1)⇒ a2+9−3a=4⇒ −5a2=−5⇒a=2\begin{aligned} &\frac{a}{2} + 3(3-a) = 4 \quad\text{(sub step 1)}\\ &\Rightarrow\ \frac{a}{2} + 9 - 3a = 4\\ &\Rightarrow\ -\frac{5a}{2} = -5 \Rightarrow a = 2 \end{aligned}
4

Time for A alone. Work divided by A's usual rate.

tA=36a=362=18 days\begin{aligned} &t_A = \frac{36}{a} = \frac{36}{2} = 18 \text{ days} \end{aligned}
tA=18 dayst_A = 18 \text{ days}
CAT 2019 Slot 1 QA Q11: At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficient — Solution | TheCATExam