Discussion

Explanation:

Method 1:
(A + B)’s 4 days work = $4\left(\frac{1}{15}+\frac{1}{20}\right)$ = $\frac{7}{15}$

Remaining work = (1 - 7/15) = 8/15;

Now, $\frac{1}{20}$th work is finished by B in 1 day.

∴ $\frac{8}{15}$th work will be finished by B in $\frac{8/15}{1/20}$ = $\frac{8\times 20}{15}$ = $\frac{32}{3}$ = $10\frac{2}{3}$

Method 2:
Let the total work to be done = LCM(15, 20) = 60 units.
Efficiency of A = 60/15 = 4 units/day
Efficiency of B = 60/20 = 3 units/day

A and B work for 4 days and then let B finish the remaining work in x days.

∴ Total work done = Work done by A and B in 4 days + Work done by B in x days

⇒ 60 = (4 + 3) × 4 + 3 × x
⇒ 60 = 28 + 3x
⇒ 3x = 32
⇒ x = $\frac{32}{3}$ = $10\frac{2}{3}$

Method 3:
Let the total work to be done = LCM(15, 20) = 60 units.
Efficiency of A = 60/15 = 4 units/day
Efficiency of B = 60/20 = 3 units/day

A works for 4 days and B work for 4 + x days

∴ Total work done = Work done by A in 4 days + Work done by B in 4 + x days

⇒ 60 = 4 × 4 + (4 + x) × 3
⇒ 60 = 28 + 3x
⇒ 3x = 32
⇒ x = $\frac{32}{3}$ = $10\frac{2}{3}$

Hence, option (c).

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