SSC CGL Tier1 Quantitative Aptitude Time and Work Test 5

$\left(A + B\right)$s 20 days work =$ \left(\dfrac{1}{30} \times 20\right) $=$ \dfrac{2}{3} $.

Remaining work =$ \left(1 -\dfrac{2}{3} \right) $=$ \dfrac{1}{3} $.

Now,$ \dfrac{1}{3} $work is done by A in 20 days.

Therefore, the whole work will be done by A in $\left(20 \times 3\right)$ = 60 days.

P can complete the work in $\left(12 \times 8\right)$ hrs. = 96 hrs.

Q can complete the work in $\left(8 \times 10\right)$ hrs. = 80 hrs.

$\therefore$ Ps1 hours work =$ \dfrac{1}{96} $and Qs 1 hours work =$ \dfrac{1}{80} $.

$\left(P + Q\right)$s 1 hours work =$ \left(\dfrac{1}{96} +\dfrac{1}{80} \right) $=$ \dfrac{11}{480} $.

So, both P and Q will finish the work in$ \left(\dfrac{480}{11} \right) $hrs.

$\therefore$ Number of days of 8 hours each =$ \left(\dfrac{480}{11} \times\dfrac{1}{8} \right) $=$ \dfrac{60}{11} $days = 5$ \dfrac{5}{11} $days.

$\left(20 \times 16\right)$ women can complete the work in 1 day.

$\therefore$ 1 womans 1 days work =$ \dfrac{1}{320} $.

$\left(16 \times 15\right)$ men can complete the work in 1 day.

$\therefore$ 1 mans 1 days work =$ \dfrac{1}{240} $

So, required ratio =$ \dfrac{1}{240} $:$ \dfrac{1}{320} $

=$ \dfrac{1}{3} $:$ \dfrac{1}{4} $

= 4 : 3 [cross multiplied]

$\left(A + B + C\right)$s 1 days work =$ \left(\dfrac{1}{24} +\dfrac{1}{6} +\dfrac{1}{12} \right) $=$ \dfrac{7}{24} $.

So, all the three together will complete the job in$ \left(\dfrac{24}{7} \right) $ days=3$ \dfrac{3}{7} $days.

$\left(A + B\right)$s 1 days work =$ \left(\dfrac{1}{15} +\dfrac{1}{10} \right) $=$ \dfrac{1}{6} $.

Work done by A and B in 2 days =$ \left(\dfrac{1}{6} \times 2\right) $=$ \dfrac{1}{3} $.

Remaining work =$ \left(1 -\dfrac{1}{3} \right) $=$ \dfrac{2}{3} $.

Now,$ \dfrac{1}{15} $work is done by A in 1 day.

$ \dfrac{2}{3} $work will be done by a in 15 $\times \dfrac{2}{3} $ = 10 days.

Hence, the total time taken = $\left(10 + 2\right)$ = 12 days.

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

Bs 1 days work =$ \dfrac{1}{12} $.

As 1 days work =$ \dfrac{1}{6} $; [2 times of Bs work]

$\left(A + B\right)$s 1 days work =$ \left(\dfrac{1}{6} +\dfrac{1}{12} \right) $=$ \dfrac{3}{12} $=$ \dfrac{1}{4} $.

So, A and B together can finish the work in 4 days.

As 1 days work =$ \dfrac{1}{15} $;

Bs 1 days work =$ \dfrac{1}{20} $;

$\left(A + B\right)$s 1 days work =$ \left(\dfrac{1}{15} +\dfrac{1}{20} \right) $=$ \dfrac{7}{60} $.

$\left(A + B\right)$s 4 days work =$ \left(\dfrac{7}{60} \times 4\right) $=$ \dfrac{7}{15} $.

Therefore, Remaining work =$ \left(1 -\dfrac{7}{15} \right) $=$ \dfrac{8}{15} $.

$\left(B + C\right)$s 1 days work =$ \left(\dfrac{1}{9} +\dfrac{1}{12} \right) $=$ \dfrac{7}{36} $.

Work done by B and C in 3 days =$ \left(\dfrac{7}{36} \times 3\right) $=$ \dfrac{7}{12} $.

Remaining work =$ \left(1 -\dfrac{7}{12} \right) $=$ \dfrac{5}{12} $.

Now,$ \dfrac{1}{24} $work is done by A in 1 day.

So,$ \dfrac{5}{12} $work is done by A in 24 $\times \dfrac{5}{12} $= 10 days.

$\left(A + B + C\right)$s 1 days work =$ \dfrac{1}{6} $;

$\left(A + B\right)$s 1 days work =$ \dfrac{1}{8} $;

$\left(B + C\right)$s 1 days work =$ \dfrac{1}{12} $.

$\left(A + C\right)$s 1 days work =$ \left(2 \times \dfrac{1}{6} \right) $-$ \left(\dfrac{1}{8} +\dfrac{1}{12} \right) $

=$ \left(\dfrac{1}{3} -\dfrac{5}{24} \right) $

=$\dfrac{3}{24}$

=$\dfrac{1}{8}$.

So, A and C together will do the work in 8 days.

Work done by X in 8 days =$ \left(\dfrac{1}{40} \times 8\right) $=$ \dfrac{1}{5} $.

Remaining work =$ \left(1 -\dfrac{1}{5} \right) $=$ \dfrac{4}{5} $.

Now,$ \dfrac{4}{5} $work is done by Y in 16 days.

Whole work will be done by Y in$ \left(16 \times\dfrac{5}{4} \right) $= 20 days.

Xs 1 days work =$ \dfrac{1}{40} $, Ys 1 days work =$ \dfrac{1}{20} $.

$\left(X + Y\right)$s 1 days work =$ \left(\dfrac{1}{40} +\dfrac{1}{20} \right) $=$ \dfrac{3}{40} $.

Hence, X and Y will together complete the work in$ \left(\dfrac{40}{3} \right) $= 13$ \dfrac{1}{3} $days.