1. Computer A takes 3 minutes to process an input while computer B takes 5 minutes. If computers A, B and C can process an average of 14 inputs in one hour, how many minutes does computer C alone take to process one input ?
a) 4 minutes
b) 6 minutes
c) 10 minitus
d) None of above
Explanation: Number of units processed by computer A in 1 minute
$$ = \frac{1}{3}$$
Number of units processed by computer B in 1 minute
$$ = \frac{1}{5}{\text{ }}$$
Number of units processed by computer A, B and C in 1 minute
$$\eqalign{ & = \frac{{14 \times 3}}{{60}} \cr & {\text{ = }}\frac{7}{{10}} \cr} $$
Number of units processed by computer C in 1 minute
$$\eqalign{ & = \frac{7}{{10}} - \left( {\frac{1}{3} + \frac{1}{5}} \right) \cr & = \frac{7}{{10}} - \frac{8}{{15}} \cr & = \frac{5}{{30}} \cr & = \frac{1}{6} \cr} $$
Computer C takes 6 minutes to process one input alone.
2. 5 men and 2 women working together can do four times as much work per hour as a men and a women together. The work done by a men and a women should be in the ratio ?
a) 1 : 2
b) 2 : 1
c) 4 : 1
d) 1 : 3
Explanation:
$$\frac{{{\text{5 men}} + {\text{2 women}}}}{{4{\text{work}}}}$$ = $$\left( {1{\text{ men}} + {\text{1 women}}} \right)$$
$$5{\text{ men}} + {\text{2 women}}$$ = $${\text{4 men}} + {\text{4 women}}$$
$$\eqalign{ & {\text{1 men}} = {\text{2 women}} \cr & \frac{{{\text{Men}}}}{{{\text{Women}}}} = \frac{2}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}:{\text{W}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2:1 \cr} $$
3. If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do the work in ?
a) $${\text{5}}\frac{6}{{13}}{\text{ months}}$$
b) $${\text{6 months}}$$
c) $${\text{5}}\frac{7}{{13}}{\text{ months}}$$
d) $${\text{11}}\frac{1}{{13}}{\text{ months}}$$
Explanation: 40 men = 60 women = 80 children
2 men = 3 women = 4 children
2 men = 3 women
1 women = $$\frac{2}{3}$$ men → 10 women
$$ \to \frac{2}{3} \times 10 = \frac{{20}}{3}{\text{ men}}$$
Similarly,
2 men = 4 children
1 children = $$\frac{1}{2}$$ men → 10 children
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{2} = {\text{5 men}}$$
10 men = 10 women = 10 children
$$\eqalign{ & {\text{10 men}} + \frac{{20}}{3} + 5 \cr & \Rightarrow \frac{{30 + 20 + 15}}{3} \cr} $$
10 men + 10 women + 10 children = $$\frac{{65}}{3}$$ men
40 men can do a piece of work in 6 months
1 men can do a piece of work in 6 × 40
$$\frac{{65}}{3}$$ men can do a piece of work in
$$\eqalign{ & = \frac{{6 \times 40}}{{\frac{{65}}{3}}} \cr & = 11\frac{1}{{13}}{\text{ months}} \cr} $$
4. Two workers A and B working together completed a job in 5 days. If A had worked twice as efficiently as he actually did, the work would have been completed in 3 days. To complete the job alone, A would require?
a) $${\text{5}}\frac{1}{5}{\text{ days}}$$
b) $${\text{6}}\frac{1}{4}{\text{ days}}$$
c) $${\text{7}}\frac{1}{2}{\text{ days}}$$
d) $${\text{8}}\frac{3}{4}{\text{ days}}$$
Explanation: L.C.M. of Total Work = 15
One day work of A + B = $$\frac{{15}}{5}$$ = 3 unit/day
One day work of (2A + B) = $$\frac{{15}}{3}$$ = 5 unit/day
Assume A's efficiency is 2 units, B's is 1 unit.
It satisfies the equation of both cases
Actual efficiency of A is 2 units/day
A alone can complete the work in
$$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{15}}{2} \cr & = 7\frac{1}{2}{\text{days}} \cr} $$
5. 3 men and 7 women can do a job in 5 days, while 4 men and 6 women can do it in 4 days. The number of days required for a group of 10 women working together, at the same rate as before, to finish the same job in ?
a) 30 days
b) 36 days
c) 20 days
d) 40 days
Explanation (3 men + 7 women) × 5 days = (4 men + 6 women) × 4 days
1 men = 11 women
∴ 3 men + 7 women
(3 × 11) women + 7 women
= 40 women
40 women can do a work in 5 days
1 women can do a work in (5 × 40) days
10 women can do a work in = $$\frac{{5 \times 40}}{{10}}$$ = 20 days
6. Tapas works twice as fast as Mihir. If both of them together complete a work in 12 days, Tapas alone can complete it in ?
a) 15 days
b) 18 days
c) 20 days
d) 24 days
Explanation:
| Tapas | : | Mihir | |
| Efficiency units/day |
2 | : | 1 |
Tapas alone complete the whole work in
$$\eqalign{ & = \frac{{36}}{2} \cr & = 18{\text{ days}} \cr} $$
7. 2 men and 3 women together or 4 men can complete a piece of work in 20 days. 3 men and 3 women will complete the same work in =
a) 12 day
b) 16 day
c) 18 days
d) 19 days
Explanation: According to the question,
$$\eqalign{ & {\text{2m}} + {\text{3w}} = {\text{4m}} \cr & {\text{3w}} = {\text{4m}} - {\text{2m }} \cr & {\text{3w}} = {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{3m}} + {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{5m}} \cr} $$
4 men can do work in 20 days
1 men can do work in 20 × 4 days
5 men can do work in $$\frac{{{\text{20}} \times {\text{4}}}}{5}$$ = 16 days
8. Twenty women together can complete a piece of work in 16 days, 16 men together can complete the same work in 15 days. The ratio of the working capacity of a man to that of a women is =
a) 3 : 4
b) 4 : 3
c) 5 : 3
d) 4 : 5
Explanation:
$$\eqalign{ & {\text{20 w}} \times {\text{16}} = {\text{16 m}} \times {\text{15}} \cr & {\text{20 w}} = {\text{15 m}} \cr & {\text{4 w}} = {\text{3 m}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{4}{3} \cr & {\text{Man}}:{\text{Women}} \cr & \,\,\,\,\,\,4:3 \cr} $$
9. A conveyor belt delivers baggage at the rate of 3 tons in 5 minutes and second conveyor belt delivers baggage at the rate of 1 ton in 2 minutes. How much time will it take to get
33 tons of baggage delivered using both the conveyor belts together ?
a) 25 minutes 30 seconds
b) 30 minutes
c) 35 minutes
d) 45 minutes
Explanation: Baggage delivered by first belt in 1 minute
$$ = \left( {\frac{3}{5}} \right){\text{tons}}$$
Baggage delivered by second belt in 1 minute
$$ = \left( {\frac{1}{2}} \right){\text{tons}}$$
Baggage delivered by both belt in 1 minute
$$\eqalign{ & = \left( {\frac{3}{5} + \frac{1}{2}} \right){\text{tons}} \cr & = \frac{{11}}{{10}}{\text{ tons}} \cr & {\text{Required time}} \cr & = \left( {33 \div \frac{{11}}{{10}}} \right){\text{ minutes}} \cr & = \left( {33 \times \frac{{10}}{{11}}} \right){\text{minutes}} \cr & = {\text{30 minutes}} \cr} $$
10. A manufacturer builds a machine which will address 500 envelopes in 8 minutes. He wishes to build another machine so that when both are operating together they will address 500 envelopes in 2 minutes. The equation used to find how many minutes x it would require the second machine to address 500 envelopes alone, is =
a) $$8 - x = 2$$
b) $$\frac{1}{8} + \frac{1}{x} = \frac{1}{2}$$
c) $$\frac{{500}}{8} + \frac{{500}}{x} = 500$$
d) $$\frac{x}{2} + \frac{x}{8} = 1$$
Explanation: Number of envelopes addressed by first machine in 1 minute
$$ = \frac{{500}}{8}$$
Number of envelopes addressed by second machine in 1 minute
$$ = \frac{{500}}{x}$$
Number of envelopes addressed by both machine in 1 minute
$$\eqalign{ & {\text{ = }}\frac{{500}}{2} \cr & \frac{{500}}{8} + \frac{{500}}{x} = \frac{{500}}{2} \cr & \frac{1}{8} + \frac{1}{x} = \frac{1}{2} \cr} $$