Question

A alone and B alone can do a work in respectively 18 and 8 days more than both working together. Find the number of days required if both work together.

• A. 18
• B. 12
• C. 16
• D. 36

Hint:

To solve this type of problem instantly, memorize the shortcut:
$\text{Time } (T) = \sqrt{\text{Extra days}_A \times \text{Extra days}_B}$
Here: $T = \sqrt{18 \times 8} = \sqrt{144} = 12$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a work-and-time algebraic word problem.
We can solve it by setting up rates of work or by using a direct mathematical shortcut.
Key Formula or Approach:
If A takes $a$ days more than (A+B) together, and B takes $b$ days more than (A+B) together, then the time $T$ taken by both working together is: \[ T = \sqrt{a \times b} \]

Step 2: Detailed Explanation:
Let the time taken by A and B together be $T$ days.
- A alone takes: $T + 18$ days.
- B alone takes: $T + 8$ days.
Using the algebraic relation: \[ \frac{1}{T + 18} + \frac{1}{T + 8} = \frac{1}{T} \] Cross-multiplying and simplifying: \[ T(T + 8 + T + 18) = (T + 18)(T + 8) \] \[ T(2T + 26) = T^2 + 26T + 144 \] \[ 2T^2 + 26T = T^2 + 26T + 144 \] \[ T^2 = 144 \implies T = 12 \text{ days} \] Thus, both working together require 12 days.

Step 3: Final Answer:
The required number of days is 12, matching Option (B).