Time And Work - Aptitude MCQ Questions and Solutions with Explanations

Work done by X in 8 days = $$ {\frac{1}{{40}} \times 8} $$   = $$\frac{1}{5}$$
Remaining work = $$ {1 - \frac{1}{5}} $$   = $$\frac{4}{5}$$
Now, $$\frac{4}{5}$$ work is done by Y in 16 days
Whole work will be done by Y in = $$ {16 \times \frac{5}{4}} $$   = 20 days
∴ X's 1 day's work = $$\frac{1}{{40}}$$
∴ Y's 1 day's work = $$\frac{1}{{20}}$$
(X + Y)'s 1 day's work
$$\eqalign{ & = {\frac{1}{{40}} + \frac{1}{{20}}} \cr & = \frac{3}{{40}} \cr} $$
Hence, X and Y will together complete the work in
$$\eqalign{ & = {\frac{{{\text{40}}}}{{\text{3}}}} \cr & {\text{ = 13}}\frac{{\text{1}}}{{\text{3}}}\,\,{\text{days}} \cr} $$

$$\eqalign{ & \left( {{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right){\text{:}}\left( {{\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right) \cr & = \frac{7}{4}:1 = 7:4 \cr & {\text{Let}}\,{\text{A's}}\,{\text{and}}\,{\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\,{\text{be}} \cr & 7x\,{\text{and}}\,4x\,{\text{respectively}} \cr & {\text{Then}},\,7x + 4x = \frac{1}{7} \cr & \Rightarrow 11x = \frac{1}{7} \cr & \Rightarrow x = \frac{1}{{77}} \cr & \therefore {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{77}} \times {\text{7}}} = \frac{1}{{11}} \cr} $$
So, A will do the work in 11 days

$$\eqalign{ & {\text{Let}}\,{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = x\,{\text{and}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = y \cr & {\text{Then}},\,x + y = \frac{1}{{30}} \cr & 16x + 44y = 1 \cr & {\text{Solving}}\,{\text{these}}\,{\text{two}}\,{\text{equations,}}\,{\text{we}}\,{\text{get}} \cr & x = \frac{1}{{60}}\,{\text{and}}\,y = \frac{1}{{60}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{60}} \cr & {\text{Hence,}}\,{\text{B}}\,{\text{alone}}\,{\text{shall}}\,{\text{finish}}\,{\text{the}} \cr & {\text{whole}}\,{\text{work}}\,{\text{in}}\,{\text{60}}\,{\text{days}} \cr} $$

So let's solve now,
A's 1 day work = $$\frac{1}{{18}}$$
B's 1 day work = $$\frac{1}{9}$$ [because B take half time than A]
(A + B)'s one day work
$$\eqalign{ & = {\frac{1}{{18}} + \frac{1}{9}} \cr & = {\frac{{1 + 2}}{{18}}} \cr & = \frac{1}{6} \cr} $$

In these kind of questions, always take total work as L.C.M. of number of days.
Here L.C.M. of Total Work = 360
One day work of A + B = $$\frac{{360}}{{72}}$$  = 5 unit/day
One day work of B + C = $$\frac{{360}}{{120}}$$  = 3 unit/day
One day work of C + A = $$\frac{{360}}{{90}}$$  = 4 unit/day
∴ Total units/day = 5 + 3 + 4 = 12
(Here , 12 unit represents twice of the work done by A, B and C. So we will divide it by 2)
Work done by (A + B + C) per day = $$\frac{{12}}{2}$$ = 6 units/day
∴Total time taken by (A + B + C)
$$\eqalign{ & = \frac{{360}}{6} \cr & = 60{\text{ days}} \cr} $$

Ayesha's 1 day's work $$ = \frac{1}{{16}}$$
Amita's 1 day's work $$ = \frac{1}{{8}}$$
(Ayesha + Amitha)'s 1 day's work
$$\eqalign{ & = {\frac{1}{{16}} + \frac{1}{8}} \cr & = \frac{3}{{16}} \cr} $$
∴ Both together can complete the work in
$$\eqalign{ & = \frac{{16}}{3} \cr & = 5\frac{1}{3}{\text{ days}} \cr} $$

$$\eqalign{ & {\text{Required average,}} \cr & = \frac{{ {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{12}}} }}{5} \cr & = {\frac{{48}}{{60}} \times \frac{1}{5}} \cr & = \frac{4}{{25}} \cr & = 0.16 \cr} $$

Try to solve these kind of question by option
Because of his being absent he was paid Rs. 750 less
Now check with option
Since maximum possible daily wages is asked so it will be Rs. 250
OR :
Maximum possible daily wage
= HCF of Rs. 5750 and Rs. 5000
= Rs. 250

In these kind of questions, always take total work as L.C.M. of number of days
Here L.C.M. of Total Work = 120
One day work of A + B = $$\frac{{120}}{8}$$ = 15 unit/day
One day work of B + C = $$\frac{{120}}{10}$$ = 12 unit/day
One day work of C + A = $$\frac{{120}}{6}$$ = 20 unit/day
∴ Total units per day = 15 + 12 + 20 = 47
∴ Efficiency of :
$$\eqalign{ & 2\left( {{\text{A}} + {\text{B}} + {\text{C}}} \right) = 47 \cr & {\text{A}} + {\text{B}} + {\text{C}} = \frac{{47}}{2} \cr} $$
(A + B + C) will complete the whole work in
$$\eqalign{ & = \frac{{120}}{{\frac{{47}}{2}}} \cr & = \frac{{240}}{{47}} \cr & = 5\frac{5}{{47}}{\text{days}} \cr} $$

L.C.M. of total work = 24
One day work of A + B = $$\frac{{24}}{8}$$ = 3 unit/day
B's 1 day work = $$\frac{{24}}{12}$$   = 2 units/days
A's 1 day work = 3 - 2 = 1 units/days
4 days work of B = 4 × 2 = 8 units/days
Work left = 24 - 8 = 16 units
A will complete the remaining work in
$$\eqalign{ & = \frac{{{\text{16 units}}}}{{{\text{1 unit/day}}}} \cr & = {\text{16 days}} \cr} $$