Question

If A works alone, he would take 4 days more to complete the job than if both A and B worked together. If B worked alone, he would take 16 days more to complete the job than if A and B work together. How many days would they take to complete the work if both of them worked together?​

Answer 1

Answer: 8 days

Step-by-step explanation:

Let the work completed by both A and B together be x.

∴ (x + 4) days will be required to complete the work done by A alone.

∴ (x + 16) days will be required to complete the work done by B alone.

According to the question:

If both work together, work will be completed in x days.

⇒ $\frac{1}{x + 4} + \frac{1}{x + 16} = \frac{1}{x}$

⇒ $\frac{x + 16 + x + 4}{(x + 4) (x + 16)} = \frac{1}{x}$

⇒ $\frac{2x + 20}{x^{2} + 20x + 64} = \frac{1}{x}$

⇒ $2x^{2} + 20x = x^{2} + 20x + 64$

⇒ $x^{2} = 64$

∴ x = 8 days

∴ 8 days will be required to complete the work , if both A and B works together.

Answer 2

Answer:

Step-by-step explanation:

Given:

Working together, A and B do the job in $x$ days

A does the job in $x + 4$ days. A completes $\frac{1}{x + 4}$ of the job per day

B does the job in $x + 16$ days. B completes $\frac{1}{x + 16 }$ of the job per day

To Find:

We need to find the number of days would take if both of them worked together

Solution:

$\frac{1}{x+4} + \frac{1}{1+16} = \frac{1}{x}$

$\frac{(x+16)+(x+4)}{(x+4)(x+16)}=\frac{1}{x}$

$\frac{x+16+x+4}{(x+16)(x+4)}=\frac{1}{x}$

$x (2x + 20) = (x + 4) (x + 16)$

$2x^{2} + 20x = x^{2} + 20x + 64\\x^{2} = 64\\x = \sqrt{64} \\x = +8, -8$

$-8$ is an extraneous solution, so $x = 8$

If A and B are worked together, it would take $8$ days to completes the job.

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