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# Aptitude Questions & Answers

Updated on March 1, 2012

## Time and Work problems

Ex.1. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job. How long should it take both A and B, working together but independently, to do the same job?

Sol: Worker A can do a job in 8 days. That means Worker A can do 1/8th of work in single day.
Similarly, Worker B can do a job in 10 days. That means Worker B can do 1/10th of the work in single day.
The work both A and B combine can do in single day = (1/8) + (1/10) = (5+4)/40 = 9/40
So the time taken by A and B to complete the work is inverse of the work they can do in a single day. That is 40/9 days.

Ex.2. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together (42/5) (not 8) hours a day?

Sol. In this problem we have to solve by considering hours.
A can complete the work in (7 x 9) = 63 hours.
B can complete the work in (6 x 7) = 42 hours.
The work that A can do in 1 hour = (1/63)
The work that B can do in 1 hour = (1/42)
The work that both A and B can do in 1 hour = (1/63)+(1/42)=(5/126)
Both A and B will finish the work in (126/5) hrs.
Now we have to convert it into days.
Taking 42/5 hours work per day into consideration, then number of days to complete the work = (126 x 5)/(5 x 42)=3 days

Ex.3. A and B can do a piece of work in 18 days; Band C can do it in 24 days A and C can do it in 36 days. In how many days will A, B and C finish it together?

Sol. The amount of work both A and B together will do in 1 day = (1/18)
The amount of work both B and C together will do in 1 day = (1/24)
Similarly, The amount of work both A and C together will do in 1 day = (1/36)
Adding, we get: 2 (A + B + C)'s 1 day's work =¬(1/18 + 1/24 + 1/36)
=9/72 =1/8
To get (A +B + C)'s 1 day's work we have to divide with 2, that gives the work they will do in one day = 1/16
Thus, A, B and C together can finish the work in 16 days.

Ex.4. A is twice as good a workman as B and together they finish a piece in 18 days. In how many days will A alone finish the work?

Sol. In this problem A and B are not same type. They work efficiency is different.
The ratio of A's 1 day work to B's 1 day work = 2 : 1.
The work A and B will do in 1 day = 1/18
In 1/18 A and B divided the work in 2:1 ratio.
So A will do the 2/3 of the work in1/18.
A's 1 day work =(1/18*2/3)=1/27
Hence, A alone can finish the work in 27 days.

Ex.5. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?

Sol. The work done by A in 1 day = 1/80
Work done by A in 10 days = (1/80*10)=1/8
Remaining work = (1- 1/8) =7/ 8
Now, 7/ 8th work is done by B in 42 days.
Number of days require to do whole work by B alone = 42/ (8/7) = 48 days.
A's 1 day's work = 1/80 and B's 1 day's work = 1/48
The work both A and B together will do in 1 day = (1/80+1/48) = 8/240 =1/30
Hence, both will finish the work in 30 days.

Ex.6. A and B working separately can do a piece of work in 9 and 12 days respectively, If they work for a day alternately, A beginning, in how many days, the work will be completed?

Sol: This is a good problem.
A and B are doing the work in alternate days. So let us take the work done by A on 1st day and the work done by B on 2nd day as one cycle.
The amount of work done by A in 1 day = 1/9
The amount of work done by B in 1 day = 1/12
Now the work done by A and B in one cycle = (1/9+1/12)=7/36
Work done in 5 cycles = (5*7/36) =35/36
Now the remaining work = (1-35/36) =1/36
So in 5 cycles i.e., 10 days they will complete 35/36 work. The remaining work is 1/36 which will be done on 11th day. 5 cycles completed that means on 11th day A will work.
The time taken by A to complete 1/36 work = (1/36)/(1/9) = 9/36 = ¼
So it will take 10 days and extra ¼ day.

Ex.7. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?

Sol: This is a complicated problem.
Let us consider the amount of work a man can do in 1 day = x
Let us consider the amount of work a boy can do in 1 day = y
The amount of work done by 2 men and 3 boys in 1 day = 1/10
The amount of work done by 3 men and 2 boys in 1 day = 1/8
Then, 2x+3y = 1/10
and 3x+2y = 1/8
By solving above two equations we will get x = 7/200 and y = 1/100
The amount of work done by 2 men and 1 boy in one day = (2*7/200)+(1/100) = 8/100
= 2/25
Number of days will take to complete the work by 2 men and 1 boy = 25/2

## Clock Problems

Tips:
Angle traced by hour hand in 12 hours = 360 degrees
Angle traced by hour hand in 1 hour = 30 degrees
Angle traced by hour hand in 1 minute = 0.5 degrees
Angle traced by minute hand in 1 hour = 360 degrees
Angle traced by minute hand in 1 minute = 6 degrees

Ex 1: Find the angle between the hour hand and the minute hand of a clock when 3.25.

Sol: Angle traced by hour hand in three hours 25 min = (3*30+25*0.5)° =102*1/2°
Angle traced by it in 25 min. = (25*6)° = 150°
Required angle = 1500 - 102*1/2°= 47*1/2°

Ex 2: At what time between 2 and 3 o'clock will the hands of a clock be together?

Sol: In 2 hours the angle traced by hours hand = 2*30 = 60 degrees
In 'x' minutes the angle traced by hours hand = x*0.5 = x/2 degrees
In 'x' minutes the angle traced by minutes hand = x*6 = 6x degrees
Required angle = 0 degrees
60+x/2-6x = 0
11x/2 = 60
Then, x = 120/11
So at 2 hours 120/11 minutes both hands will meet.

Ex. 3. A watch which gains uniformly, is 6 min. slow at 8 o'clock in the morning Sunday and it is 6 min. 48 sec. fast at 8 p.m. on following Sunday. When was it correct?

Sol. Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours.
The watch gains 6+6 minute 48 sec = 12min. 48 sec in 180 hrs.
48 sec means 0.8 minutes.
So 12.8 minutes gained in 180 hours.
For 1 minute it will take a time of = 180/12.8 = 14.0625 hours
To show the right time it has to cover initial 6 minutes late.
The time taken to gain 6 min = 14.0625*6 = 84.375 hours = 84 hours 22 minutes 30 sec.
= 3 days 12 hours 22 minutes 30 sec
That means on Wednesday night 8 hours 22 minutes and 30 sec it will show the exact time.

Ex. 4. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?

Sol. Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
In 24 hours the time it will gain = 10 minutes
In 1 hour the time it will gain = 10/24 minutes
In 29 hours the time it will gain = 29*10/24 minutes = 12 minutes 5 sec
So the time it will show is 1 hour and 12minutes 5 sec evening time.

## IMPORTANT FACTS AND FORMULAE

1. In water, the direction along the stream is called downstream and the direction against the stream is called upstream.

2. If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed downstream = (u+v) km/hr.

Speed upstream = (u-v) km/hr.

3. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = 1/2(a+b) km/hr

Rate of stream = 1/2(a-b) km/hr

SOLVED EXAMPLES

EX.1. A man can row upstream at 7 kmph and downstream at 10kmph.find man’s rate in still water and the rate of current.

Sol. Rate in still water=1/2(10+7) km/hr=8.5 km/hr.

Rate of current=1/2(10-7) km/hr=1.5 km/hr.

EX.2. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2hours30minutes to cover a distance of 5km upstream. Find the speed of the river current in km/hr.

Sol. Rate downstream = (15/3 ¾) km/hr = (15*4/15) km/hr = 4km/hr.

Rate upstream = (5/2 ½) km/hr = (5*2/5) km/hr = 2km/hr.

Speed of current = 1/2(4-2) km/hr =1km/hr

EX.3. A man can row 18 kmph in still water. It takes him thrice as long to row up as to row down the river. Find the rate of stream.

Sol. Let man’s rate upstream be x kmph. Then, his rate downstream = 3xkmph.

So, speed in still water = (3x+x)/2 = 2x = 18 or x = 9.

Rate upstream = 9 km/hr, rate downstream = 27 km/hr.

Hence, rate of stream = 1/2(27-9) km/hr = 9 km/hr.

EX.4. There is a road beside a river. Two friends started from a place A, moved to a temple situated at another place B and then returned to A again. One of them moves on a cycle at a speed of 12 km/hr, while the other sails on a boat at a speed of 10 km/hr. If the river flows at the speed of 4 km/hr, which of the two friends will return to place A first?

Sol. Clearly the cyclist moves both ways at a speed of 12 km/hr.

The boat sailor moves downstream @ (10+4) i.e., 14 km/hr and upstream @ (10-4) i.e., 6km/hr.

Let distance between A and B = x km

Time taken in downstream = x/14

Time taken in upstream = x/6

Average speed = 2x/((x/14)+(x/6))

= (2*14*6/14+6) km/hr

= 42/5 km/hr = 8.4 km/hr.

Since the average speed of the cyclist is greater, he will return to A first.

EX.5. A man can row 7 ½ kmph in still water. If in a river running at 1.5 km/hr it takes him 50 minutes to row to a place and back, how far off is the place?

Sol. Speed downstream = (7.5+1.5)km/hr=9 km/hr;

Speed upstream = (7.5-1.5) kmph=6kmph.

Let the required distance be x km then,

(x/9)+(x/6) = 50/60.

2x+3x = (5/6*18)

5x = 15

X = 3

Hence, the required distance is 3km.

EX.6. In a stream running at 2kmph, a motorboat goes 6km upstream and back again to the starting point in 33 minutes. Find the speed of the motorboat in still water.

Sol.let the speed of the motorboat in still water be x kmph then,

Time taken in upstream = 6/(x-2)

Time taken in downstream = 6/(x+2)

Total time taken = 6/(x+2) + 6/(x-2)

6/(x+2) +6/(x-2) = 33/60

11x2-240x-44=0

11x2-242x+2x-44=0

(x-22)(11x+2)=0

x=22.

EX.7. A man can row 40km upstream and 55km downstream in 13 hours. Also, he can row 30km upstream and 44km downstream in 10 hours. Find the speed of the man in still water and the speed of the current.

Sol.let rate upstream = x km/hr and rate downstream = y km/hr.

Then, 40/x +55/y =13………. (i)

And 30/x +44/y =10………... (ii)

Multiplying (ii) by 4 and (i) by 3 and subtracting, we get 11/y=1 or y=11.

Substituting y =11 in (i), we get x=5.

Rate in still water =1/2(11+5) kmph=8kmph.

Rate of current=1/2(11-5) kmph=3kmph

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