Practice MCQ Questions and Answer on Time and Work

Solution: Mango Packing rate of Ram Lal = $$\frac{{70}}{7}$$ = 10 per hour Guavas Packing rate of Ram Lal = $$\frac{{56}}{7}$$ = 8 per hour Mango Packing rate of Ram Lal's Wife = $$\frac{{30}}{6}$$ = 5 per hour Guavas Packing rate of Ram Lal's Wife = $$\frac{{24}}{6}$$ = 4 per hour On comparing, 8 Guavas = 10 Mangoes 1 Guavas = $$\frac{{10}}{8}$$ Mangoes 2400 Guavas = $$\frac{{10 \times 2400}}{8}$$   = 3000 Mangoes This means that they need to pack (6300 + 3000) Mangoes Combined Mango Packing rate of Both = 10 + 5 = 15 per hour We will take 2 days = 1 time unit. Working 10 hours a day they will pack 150 Mangoes in 1 time unit Total time taken to pack 6300 Mangoes = $$\frac{{6300}}{{150}}$$  = 42 time unit So, total time taken = 2 × 42 = 84 days

Solution:   (T.W.)60 2 units/day   ↓   30 days(A + B) $$\eqalign{ & \left( {{\text{A }} + {\text{ B}}} \right){\text{ worked for 20 days}} \cr & {\text{So, they completed}} \cr & 20 \times 2 = 40{\text{ units}} \cr & {\text{Work left}} = 60 - 40 = {\text{20 units}} \cr} $$   (T.W.) 2 0 1 units/day   ↓   20 days (A) A's efficiency 1 units/day A will complete the whole work in $$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{60}}{1} \cr & = 60{\text{ days}} \cr & \cr & {\bf{Alternate:}} \cr & \left( {{\text{A}} + {\text{B}}} \right) \times {\text{10}} = {\text{A}} \times {\text{20}} \cr & \frac{{{\text{A + B}}}}{{\text{A}}} = \frac{2}{1} \cr & {\text{Total work}} \cr & = 2 \times 30 \cr & = {\text{60 units}} \cr & {\text{Time taken by A alone}} \cr & = \frac{{60}}{1} \cr & = 60{\text{ days}} \cr} $$

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

Solution: Ratio of efficiencies of A, B and C, = 5x : 4x : 6x Number of days required by A and B = $$\frac{{100}}{{9{\text{x}}}}$$ ------ (1) Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ ------ (2) $$\eqalign{ & \frac{{100}}{{9{\text{x}}}} - \frac{{100}}{{15{\text{x}}}} = \frac{8}{3} \cr & \Rightarrow {\text{x}} = \frac{5}{3} \cr} $$ Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ = $$\frac{{100}}{{15 \times \frac{5}{3}}}$$ = 4 days

Solution: A = 20 days B = 30 days (A + B) × 9 + C × 6 = 60 (3 + 2) × 9 + C × 6 = 60 C = $$\frac{5}{2}$$ unit C = $$\frac{{60 \times 2}}{5}$$  = 24 days (B + C) × x = 60 × $$\frac{3}{4}$$ $$\left( {2 + \frac{5}{2}} \right){\text{x}} = 45$$ 9x = 90 x = 10 days

Solution: Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{1 men's 1 day's work}} \cr & = \frac{1}{{99 \times 2}} \cr & = \frac{1}{{198}} \cr & {\text{1 women's 1 day's work}} \cr & = \frac{1}{{99 \times 6}}. \cr & = \frac{1}{{594}} \cr & {\text{1 boy's 1 day's work}} \cr & = \frac{1}{{99 \times 4}} \cr & = \frac{1}{{396}} \cr} $$ (1 men + 1 women + boy)'s 1 day's work $$\eqalign{ & = \frac{1}{{198}} + \frac{1}{{594}} + \frac{1}{{396}} \cr & = \frac{{11}}{{1188}} \cr & = \frac{1}{{108}} \cr} $$ Hence, 1 mens , 1 women and 1 boys together take 108 days to finish the work.

Solution: $$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow 2{\text{A}} = {\text{3B}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{3}{2} \cr & \Rightarrow {\text{Then efficiency ratio }} \cr & {\text{A}}:{\text{B}} = 3:2 \cr} $$ ⇒ We know that time is inversely proportional to efficiency ⇒ Then time taken by them in ratio $$\eqalign{ & {\text{A}}:{\text{B}} = \mathop {\mathop {{\text{ }}2}\limits_{4 \times \downarrow } {\text{ }}}\limits_{8{\text{days}}} :\mathop {\mathop 3\limits_{{\text{ }} \downarrow \times 4} }\limits_{12{\text{days}}} \cr & \because {\text{A can do the work in 8 days}} \cr & \Rightarrow {\text{i}}{\text{.e, 2 units}} \to {\text{8}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1 unit}} \to {\text{4}} \cr & \Rightarrow {\text{Time taken by B}} \to {\text{3 units}} \cr & = 3 \times 4 \cr & = 12{\text{ days}} \cr} $$

Solution: Let total work = 40 units $$\eqalign{ & {\text{X's 1 day work}} = 1{\text{ unit}} \cr & {\text{X's 8 days work is}} \cr & = 8 \times 1 \cr & = 8{\text{ units}} \cr & {\text{Work left}} = 40 - 8 = 32 \cr & {\text{Y's 1 day work}} = 2{\text{ unit}} \cr & {\text{X's 8 days work is}} = 1{\text{ units}} \cr} $$ X + Y complete the whole work together in $$\eqalign{ & = \frac{{40}}{{2 + 1}} \cr & = 13\frac{1}{3}\,{\text{days}} \cr} $$

Solution: Let the number of pages typed in one hour by P, Q and R be x, y and z respectively Then, $$\eqalign{ & \Rightarrow x + y + z = \frac{{216}}{4} \cr & \Rightarrow x + y + z = 54z.....{\text{(i)}} \cr & {\text{ }}z - y = y - x \cr & \Rightarrow 2y = x + z.....{\text{(ii)}} \cr & {\text{ }}5z = 7x \cr & \Rightarrow x = \frac{5}{7}z......{\text{(iii)}} \cr} $$ Solving (i), (ii) and (iii), we get $$\eqalign{ & x = 15, \cr & y = 18,{\text{ }} \cr & z = 21 \cr} $$

Solution: Ratio of their daily wages = 5 : 6 : 4 Total wage = Rs. 1800 6 × 5x + 4 × 6x + 4 × 9x = 1800 [As A, B, C works for 6, 4, 9 days respectively] Or, 90x = 1800 Or, x = $$\frac{{1800}}{{90}}$$ Total amount of A = $$\frac{{1800 \times 30}}{{90}}$$   = Rs. 600