Time and Work Aptitude Questions and Answers:

Answer: (b) \(5.2 \ days\)

Solution: Given, \(x = 10\), \(y = 20\), \(z = 25\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{10 \times 20 \times 25}{10 \times 20 + 20 \times 25 + 25 \times 10} $$ $$ = \frac{5000}{200 + 500 + 250} $$ $$ = \frac{5000}{950} $$ $$ = 5.2 \ days $$

Answer: (a) \(1.05 \ days\)

Solution: Given, \(x = 2\), \(y = 4\), \(z = 5\), then $$ = \frac{xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 4 \times 5}{2 \times 4 + 4 \times 5 + 5 \times 2} $$ $$ = \frac{40}{38} $$ $$ = 1.05 \ days $$

Answer: (b) \(5.45 \ days\)

Solution: Given, \(x = 5\), \(y = 10\) and \(z = 15\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 5 \times 10 \times 15}{50 + 150 + 75} $$ $$ = \frac{1500}{275} $$ $$ = 5.45 \ days $$

Answer: (a) \(6.4 \ days\)

Solution: Given, \(x = 8\), \(y = 10\) and \(z = 12\) then $$ = \frac{2xyz}{xy + yz + zx} $$ $$ = \frac{2 \times 8 \times 10 \times 12}{80 + 120 + 96} $$ $$ = \frac{1920}{296} $$ $$ = 6.4 \ days $$

Answer: (d) \(1 \ day\)

Solution: Both can finish the part of work in \(3\) days $$ = \left(\frac{1}{5} + \frac{1}{10}\right) \times 3 $$ $$ = \frac{9}{10} $$ remaining work $$ = 1 - \frac{9}{10} $$ $$ = \frac{1}{10} $$ then B will finish the remaining work in days $$ = \frac{1/10}{1/10} $$ $$ = 1 \ day $$

Answer: (c) \(2.5 \ days\)

Solution: Both can finish the part of work in \(5\) days $$ = \left(\frac{1}{10} + \frac{1}{15}\right) \times 5 $$ $$ = \frac{5}{6} $$ remaining work $$ = 1 - \frac{5}{6} $$ $$ = \frac{1}{6} $$ then Q will finish the remaining work in days $$ = \frac{1/6}{1/15} $$ $$ = 2.5 \ days $$

Answer: (d) \(10.2 \ days\)

Solution: one man can finish the work in days = \(10 \times 3\) = \(30\) days

one women can finish the work in days = \(8 \times 4\) = \(32\) days

one girl can finish the work in days = \(6 \times 5\) = \(30\) days

then one man, one women and one girl together can finish the part of work in one day $$ = \frac{1}{30} + \frac{1}{32} + \frac{1}{30} $$ $$ = \frac{94}{960} $$ Hence, one man, one women and one girl together can finish the work $$ = \frac{960}{94} $$ $$ = 10.2 \ days $$

Answer: (a) \(39.9 \ days\)

Solution: Let P requires \(k\) days to finish the work

and Q requires \(3k\) days to finish the work, then $$ \frac{1}{k} + \frac{1}{3k} = \frac{1}{10} $$ $$ k = 13.3 \ days $$ Hence, Q requires \(3k\) = \(39.9\) days to finish the work.

Answer: (a) \(3.4 \ days\)

Solution: Given, \(x = 6 \ days\), \(y = 8 \ days\)

If both P and Q start working together then $$ = \frac{xy}{x + y} $$ $$ = \frac{6 \times 8}{6 + 8} $$ $$ = \frac{48}{14} $$ $$ = 3.4 \ days $$ Hence, together they can finish the work in \(3.4 \ days\).

Answer: (c) \(20 \ days\)

Solution: together, they can finish the part of work in one day = \(\frac{1}{10}\)

the women alone can finish the part of work in one day = \(\frac{1}{20}\)

then, her son alone can finish the part of work in one day $$ = \frac{1}{10} - \frac{1}{20} $$ $$ = \frac{1}{20} $$ Hence, her son alone can finish the work in \(20\) days.