Showing posts with label Quantitative Aptitude. Show all posts

Sunday, 3 September 2017

TIME AND DISTANCE

IMPORTANT FACTS AND FORMULAE

Distance Distance

1. Speed = Time , Time= Speed , Distance = (Speed * Time)

2. x km / hr = x * 5

3. x m/sec = (x * 18/5) km /hr

4. If the ratio of the speeds of A and B is a:b , then the ratio of the times taken by them to cover the same distance is 1: 1 a b

or b:a.

5. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km / hr . Then , the average speed during the whole journey is 2xy km/ hr.

x+y

SOLVED EXAMPLES

Ex. 1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Sol. Aditya’s speed = 20 km/hr = {20 * 5} m/sec = 50 m/sec

18 9

\Time taken to cover 400 m= { 400 * 9 } sec =72 sec = 1 12 min 1 1 min.

50 60 5

Ex. 2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?

Sol. Speed = { 750 } m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr

150 5

Ex. 3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.

Sol. Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.

Then , 3x = 4y => x = 4 y => 4x = 16 y.

3 3

\ Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time

= 4x : 5y = 16 y : 5y =16 : 5 = 16:15

3 3

Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second?

Sol. Let the speed be x km/hr.

Then, distance covered in 1 hr. 40 min. i.e., 1 2 hrs = 5x km

3 3

Remaining distance = { 24 – 5x } km.

\ 5x = 5 { 24 - 5x } ó 5x = 5 { 72-5x } ó 7x =72 –5x

3 7 3 3 7 3

ó 12x = 72 ó x=6

Hence speed = 6 km/hr ={ 6 * 5 } m/sec = 5 m/sec = 1 2

18 3 3

Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.

Sol. Let the total distance be x km . Then,

2 x 1 x

3 + 3 = 7 ó x + x = 7 ó 7x = 42 ó x = 6

4 5 5 6 15 5

Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

Sol. Average speed = { 2xy } km/hr ={ 2*25*4 } km/hr = 200 km/hr

x+y 25+4 29

Distance traveled in 5 hours 48 minutes i.e., 5 4 hrs. = { 200 * 29 } km = 40 km

5 29 5

Distance of the post-office from the village ={ 40 } = 20 km

Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.

Sol. :

Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,

x/200+x/400+x/600+x/800=4x/yó25x/2500ó4x/yóy=(2400*4/25)=384

hence average speed =384 km/hr

Ex. 8.Walking at 5 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.

Sol. :New speed =5/6 of the usual speed

New time taken=6/5 of the usual time

So,( 6/5 of the usual time )-( usual time)=10 minutes.

=>1/5 of the usual time=10 minutes.

ð usual time=10 minutes

Ex. 9.If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Sol. Let the required distance be x km

Difference in the time taken at two speeds=1 min =1/2 hr

Hence x/5-x/6=1/5<=>6x-5x=6

óx=6

Hence, the required distance is 6 km

Ex. 10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?

Sol. Suppose they meet x hours after 10 a.m. Then,

(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.

65x + 35(x-1) = 390 => 100x = 425 => x = 17/4

So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.

Ex. 11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.

Sol. Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours

10x = 4 x 90 or x =36.

So, speed of goods train = 36kmph.

Ex. 12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?

Sol. Relative speed of the policeman = (10-8) km/hr =2 km/hr.

Time taken by police man to cover 100m 100 x 1 hr = 1 hr.

1000 2 20

In 1 hrs, the thief covers a distance of 8 x 1 km = 2 km = 400 m

20 20 5

Ex.13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

Sol. Let the distance be x km. Then,

( Time taken to walk x km) + (time taken to ride x km) =37 min.

( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.

But, the time taken to walk 2x km = 55 min.

Time taken to ride 2x km = (74-55)min =19 min.

PIPES AND CISTERNS

IMPORTANT FACTS AND FORMULAE

1. Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is

known as an outlet.

2. (i) If a pipe can fill a tank in x hours, then : part filled in 1 hour = 1/x

(ii) If a pipe can empty a full tank in y hours, then : part emptied in 1 hour = 1/y

(iii) If a pipe can .fill a tank in x hours and another pipe can empty the full tank in y hours (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)

(iv) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)

SOLVED EXAMPLES

Ex. 1:Two pipes A and B can fill a tank in 36 bours and 46 bours respectively. If both the pipes are opened simultaneously, bow mucb time will be taken to fill the

tank?

Sol: Part filled by A in 1 hour = (1/36);

Part filled by B in 1 hour = (1/45);

Part filled by (A + B) In 1 hour =(1/36)+(1/45)=(9/180)=(1/20)

Hence, both the pipes together will fill the tank in 20 hours.

Ex. 2: Two pipes can fill a tank in 10hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

Sol: Net part filled In 1 hour =(1/10)+(1/12)-(1/20)=(8/60)=(2/15).

The tank will be full in 15/2 hrs = 7 hrs 30 min.

Ex. 3: If two pipes function simultaneously, tbe reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than tbe otber. How many hours does it take the second pipe to fill the reservoir?

Sol:let the reservoir be filled by first pipe in x hours.

Then ,second pipe fill it in (x+10)hrs.

Therefore (1/x)+(1/x+10)=(1/12) ó(x+10+x)/(x(x+10))=(1/12).

ó x^2 –14x-120=0 ó (x-20)(x+6)=0

óx=20 [neglecting the negative value of x]

so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir

Ex. 4: A cistern has two taps which fill it in 12 minutes and 15minutes respectively. There is also a waste pipe in the cistern. When all the 3 are opened , the empty cistern is full in 20 minutes. How long will the waste pipe take to empty the full cistern?

Sol: Workdone by the waste pipe in 1min

=(1/20)-(1/12)+(1/15) = -1/10 [negative sign means emptying]

therefore the waste pipe will empty the full cistern in 10min

Ex. 5: An electric pump can fill a tank in 3 hours. Because of a leak in ,the tank it took 3(1/2) hours to fill the tank. If the tank is full, how much time will the leak take

to empty it ?

Sol: work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21).

The leak will empty .the tank in 21 hours.

Ex. 6. Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes

are opened simultaneously and it is found that due to leakage in the bottom it tooki 32 minutes more to fill the cistern.When the cistern is full, in what time will the leak empty it?

Sol: Work done by the two pipes in 1 hour =(1/14)+(1/16)=(15/112).

Time taken by these pipes to fill the tank = (112/15) hrs = 7 hrs 28 min.

Due to leakage, time taken = 7 hrs 28 min + 32 min = 8 hrs

Work done by (two pipes + leak) in 1 hour = (1/8).

Work done by the leak m 1 hour =(15/112)-(1/8)=(1/112).

Leak will empty the full cistern in 112 hours.

Ex. 7: Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. after 7 min,C is also opened. In how much time, the tank is full?

Sol:Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20).

Remaining part=(1-(7/20))=(13/20).

Net part filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).

Now,(1/60) part is filled in one minute.

(13/20) part is filled in (60*(13/20))=39 minutes.

Ex.8: Two pipes A,B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.?

Sol: let B be closed after x min. then ,

Part filled by (A+B) in x min. +part filled by A in (18-x)min.=1

Therefore x*((1/24)+(1/32))+(18-x)*(1/24)=1 ó (7x/96) + ((18-x)/24)=1.

ó 7x +4*(18-x)=96.

Hence, be must be closed after 8 min.

TIME AND WORK

IIMPORTANT FACTS AND FORMULAE

1. If A can do a piece of work in n days, then A's 1 day's work = (1/n).

2. If A’s 1 day's work = (1/n),then A can finish the work in n days.

3. A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3.

SOLVED EXAMPLES

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? (IGNOU, 2003)

Sol. A’s 1 hour's work = 1/8

B's 1 hour's work = 1/10

(A + B)'s 1 hour's work = (1/8) +(1/10)=9/40

Both A and B will finish the work in 40/9 days.

Ex. 2. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work? (Bank P.O. 2003)

Sol. (A + B)'s 1 day's work = (1/4). A's 1 day's work = (1/12).

B's 1 day's work =((1/4)-(1/12))=(1/6)

Hence, B alone can complete the work in 6 days.

Ex. 3. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days

of 7 bours each. How long will they take to do it, working together 8 hours a day?

Sol. A can complete the work in (7 x 9) = 63 hours.

B can complete the work in (6 x 7) = 42 hours.

A’s 1 hour's work = (1/63) and B's 1 hour's work =(1/42)

(A + B)'s 1 hour's work =(1/63)+(1/42)=(5/126)

Both will finish the work in (126/5) hrs.

Number of days. of (42/5) hrs each =(126 x 5)/(5 x 42)=3 days

Ex. 4. 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, Band C finish it together and separately?

Sol. (A + B)'s 1 day's work = (1/18) (B + C)'s 1 day's work = (1/24)

and (A + C)'s 1 day's work = (1/36)

Adding, we get: 2 (A + B + C)'s 1 day's work =(1/18 + 1/24 + 1/36)

=9/72 =1/8

(A +B + C)'s 1 day's work =1/16

Thus, A, Band C together can finish the work in 16 days.

Now, A’s 1 day's work = [(A + B + C)'s 1 day's work] - [(B + C)'s 1 day work:

=(1/16 – 1/24)= 1/48

A alone can finish the work in 48 days.

Similarly, B's 1 day's work =(1/16 – 1/36)=5/144

B alone can finish the work in 144/5=28 4/5 days

And C’s 1 day work =(1/16-1/18)=1/144

Hence C alone can finish the work in 144 days.

Ex. 6. 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. (A’s 1 day’s work):)(B’s 1 days work) = 2 : 1.

(A + B)'s 1 day's work = 1/18

Divide 1/18 in the ratio 2 : 1.

:. A’s 1 day's work =(1/18*2/3)=1/27

Hence, A alone can finish the work in 27 days.

Ex. 6. A can do a certain job in 12 days. B is 60% more efficient than A. How many

days does B alone take to do the same job?

Sol. Ratio of times taken by A and B = 160 : 100 = 8 : 5.

Suppose B alone takes x days to do the job.

Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days.

Ex. 7. 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. Work done by A in 10 days =(1/80*10)=1/8

Remaining work = (1- 1/8) =7/ 8

Now,7/ 8 work is done by B in 42 days.

Whole work will be done by B in (42 x 8/7) = 48 days.

A’s 1 day's work = 1/80 and B's 1 day's work = 1/48

(A+B)'s 1 day's work = (1/80+1/48)=8/240=1/30

Hence, both will finish the work in 30 days.

Ex. 8. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. !find the share of each.

Sol :C's 1 day's work = 1/3-(1/6+1/8)=24

A : B : C = Ratio of their 1 day's work = 1/6:1/8:1/24= 4 : 3 : 1.

A’s share = Rs. (600 *4/8) = Rs.300, B's share = Rs. (600 *3/8) = Rs. 225.

C's share = Rs. [600 - (300 + 225») = Rs. 75.

Ex. 9. 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?

(A + B)'s 2 days' work =(1/9+1/12)=7/36

Work done in 5 pairs of days =(5*7/36)=35/36

Remaining work =(1-35/36)=1/36

On 11th day, it is A’s turn. 1/9 work is done by him in 1 day.

1/36 work is done by him in(9*1/36)=1/4 day

Total time taken = (10 + 1/4) days = 10 1/4days.

Ex 10 .45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?

(45 x 16) men can complete the work in 1 day.

1 man's 1 day's work = 1/720

45 men's 6 days' work =(1/16*6)=3/8

Remaining work =(1-3/8)=5/8

75 men's 1 day's work = 75/720=5/48

Now,5 work is done by them in 1 day.

5work is done by them in (48 x 5)=6 days.

8 5 8

Ex:11. 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?

Soln: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.

Then, 2x+3y = 1 and 3x+2y = 1

10 8

Solving,we get: x = 7 and y = 1

200 100

(2 men + 1 boy)’s 1 day’s work = (2 x 7 + 1 x 1 ) = 16 = 2

200 100 200 25

So, 2 men and 1 boy together can finish the work in 25 =12 1 days

2 2