POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-AGE RELATED PROBLEMS

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-AGE RELATED PROBLEMS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

                                     AGES

Shortcut Methods for Age related Problems

1.    t 1 years earlier the fathers age was x times that of his son. At present the fathers age is y times that of his son. What are the present ages of his son and father

Sons age = t1 (x-1) / x-y

2.    The present age of the father is y times the age of his son. t 2 years hence, the fathers age becomes z times the age of his son. What are the present ages of the fathesr and son

Sons age = (z-1) t2 / y-z

3.    T1 years earlier the age of the father was x times the age of his son. t2 years hence, the age of the father becomes z times the age of his son. the present ages of the son and father

Sons age =  (z-1)t2 + t1(x-1)

                   ----------------------

                            (x-z)

4.    n years ago age = Total + Number of Years (Times-1)

                              --------------------------------------------------

                                             Times+1

5.    N years hence age = Total - Number of Years (Times-1)

                              --------------------------------------------------

                                             Times+1

6.    Present age = x:y, After T years  = a:b

Sons age =

                                       y * T(a-b)/ difference of cross products

Fathers age=

x * T(a-b)/ difference of cross products

7.    n years before the ages of Father and Son were in the ratio x:y. In another n years from now the ratio will be a:b. what are their Present ages

2n(a-b) / Difference of cross products

8.    If a is n years older than b. After x years y times of a equal to z times b. then their ages now

a age = zn/ z-y,  b age = yn/y-z                                          

  

                                                          Examples

1 The age of the father 3 years ago was 7 times the age of his son. At present the father's age is 5 times that of his son. What are the present ages of the father and the son?

Sol: Let the present age of son = x years

Then the present age of father = 5x yrs

3 years ago, 7(x-3) =5x - 3

2x = 18, x= 9 yrs, so father's age = 45 yrs

2. The ratio of the father's age to that of son's age is 4:1 the product of their ages is 196.What will be ratio of their ages after 5 yrs?

Sol:  let the ratio of proportionality be x yrs

4x*x = 196 or 4x2 = 196 or x= 7

Thus father's age = 28 yrs, Son's age = 7 yrs

After 5 yrs, father's age = 33 yrs, son's age =12 yrs

Ratio = 33:12= 11:4

3. The ratio of Vimals age and Arunas age is 3: 5 and the sum of their ages is 80 years. The ratio of their ages after 10 years will be

Sol: Let their ages be 3x and 5x years. 
Then,3x + 5x = 8O so x = 1O. 
Ratio of their ages after 10 years = (3x+ 10) : (5x+ 10) 
40 : 60 = 2 : 3.

4. One year ago, Mrs Promila was four times as old as her daughter Swati. Six years hence, Mrs Promilas age will exceed her daugther’s age by 9 years. The ratio of the present ages of Promila and her daughter is:

Sol: Let Swatis age 1 year ago = x. 
Then, Promilas age 1 year ago = 4x. 
(4x+6) - (x+6) = 9 or x = 3. 
Present age of Promila = (12 + 1) years = 13 years. 
Present age of Swati = (3+ 1) years =4 years. 
Ratio of their ages = 13 : 4.

5. The age of father 10 years ago was thrice the age of his son. Ten years hence, the father’s age will be twice that of his son. The ratio of their present ages is:

Sol: Let sons age 10 years ago = x. 
Then, fathers age 10 years ago = 3x. 
2(x+ 10+ 10) = (3x + 10 + 10)so x = 20. 
Ratio of their present ages = (3x + 10) : Cx + 10) 70: 30 = 7 :3.

6. Snehs age is1/6 th of her fathers age. Snehs fathers age will be twice of Vimals age after 10 years. If Vimals eighth birthday was celebrated two years before, then what is Snehs present age?.

Sol: Vimals age after 10 years = (8 + 2 + 10) years = 20 years. 
Snehs father’s age after 10 years = 40 years. 
Snehs father’s present age = 30 years. 
Snehs age= 1/6 * 30 years = 5 years.

7. Ten years ago, Chandrawati s mother was four times older than her daughter. After ten years, the mother will be twice older than daughter. The present age of Chandrawati is

Sol: Let Chandrawatis age 10 years ago is x years. 
Her mother’s age 10 years ago 4x. 
(4x+10)+10=2(x+10+10) or 2x=20 or x=10. 
Present age of Chandrawati (x + 10) = 20 years.

8. Ten years ago A was half of B in age. If the ratio of their present ages is 3: 4, what will be the total of their present ages

Sol: Let As age 10 years ago = x. 
Then, Bs age 10 years ago = 2x.
(x + 10) / (2x+ lO) = 3/4 hence x = 5. Total of their present ages =(x + 10 + 2x + 10) 
(3x + 20) = 35 years.

9. Kamla got married 6 years ago. Today her age is 1.25 times her age at the time of marriage. Her son’s age is 1/10 times her age. Her sons age is:

Sol: Let Kamlas age 6 years ago = x years. Kamlas present age = (x + 6) years. 
x+6= (5/4)*x or x=24. 
Kamla’s present age= 30 years. 
Sons present age = (1/10)*30 years = 3 years.

10.Pushpa is twice as old as Rita was 2 years ago. If the difference of their ages be 2 years, how old is Pushpa today

Sol: Let Ritas age 2 years ago = x years. 
Pushpas age now = 2x years.
2x - (x + 2) = 2 so x = 4. 
Pushpas age now = 8 years.

11. Sushil was thrice as old as Snehal 6 years back. Sushil will be times as old as Snehal 6 years hence. How old is Snehal today?

Sol: Let Snehals age 6 years back = x. 
Then, Sushils age 6 years back = 3x. 
(5/3) * (X + 6 + 6) = (3X + 6 + 6) hence 5(x+ 12) = 3(3x+ 12) so x=6. 
Snehal today = (x+ 6) years = 12 years.

12. The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the father’s age at that time. The present age of father and son, respectively are: 

Sol: Let sons age = x years. Then fathers age = (45 - x)years. 
(x—5)(45—x—5) = 4(45- x - 5) hence (x—5) = 4 so x = 9 
their ages are 36 years and 9 years.

13. Two years ago a man was 6 times as old as his son. After 18 years, he will be twice as old as his son. Their present ages (in years) are

Sol: Let sons age 2 years ago be x. 
Mans age 2 years ago = 6x. 
2(x + 2 + 18) = (6x + 2 + 18) hece 4x=20 so x = 5. 
Their present ages are (6x +2) and (x +2) i.e. 32 years and 7 years.

14. A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father (in years) is:

Sol: Let sons age = x. Then, fathers age = 2x. 
12(x—2O) = (2x - 20) so x = 22 
Fathers present age = 44 years.

15. The ages of two persons differ by 20 years. If 5 years ago, the elder one be 5 times as old as the younger one, their present ages (in years) are respectively:

Sol: Let their ages be x and (x + 20) years. 
5 (x - 5) = (x + 20 - 5) or 4x = 40 or x = 10. 
Their present ages are 30 years and 10 years.

16. The ages of Ram and Mukta are in the ratio of 3: 5. After 9 years, the ratio of their ages will becomes 3:4. The present age of Mukta (in years) is.

Sol: Let Rams age = 3x & Muktas age = 5x. 
3x + 9 / 5x + 9 = 3/ 4 so 4*(3x+ 9) = 3 (5x+ 9) hence x= 3. 
Muktas age = 15 years.

17. If 6 years are subtracted from the present age of Gulzar and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Mahesh whose age is 5 years, then what is the present age of Guizar

Sol: Anups age =(5—2)years = 3 years.
Let Guizar’s age x years 
Then x - 16/18 = 3 so x - 6 = 54 so x 60

18. The average age of an adult class is 40 years. Twelve new students with an average age of 32 years join the class, thereby decreasing the average of the class by 4 years. The original strength of the class was

Sol: Let, original strength = x. 
40x + 12*32 / (x + 12) = 36 or x = 12

19. Sachin was twice as old as Ajay 10 years back. How old is Ajay today if Sachin will be 40 years old 10 years hence

Sol: Sachins age today = 30 years. 
Sachins age 10 years back = 20 years. 
Ajays age 10 years back = 10 years. 
Ajays age today =20 years.

20. Five years ago, the total of the ages of father and his son was 40 years. The ratio of their present ages is 4 : 1. What is the present age of the father?

Sol: Let sons age = x. Then, fathers age = 4x. 
(x - 5)+(4x - 5) = 40 ⇒ x = 10 
Present age of father = 40 years.

21.The ratio of the ages of Meena and Meera is 4 : 3. The sum of their ages is 2 years. The ratio Of their ages after 8 years will be

Sol: Let Meenas age = 4x & Meeras age = 3x. Then, 4x+3x=28 hence x=4. 
Meenas age = 16 years and Meeras age = 12 years. 
Ratio of their ages after 8 years = (16 + 8): (12 + 8) = 24 : 20 = 6 : 5.

22. Ratio of Ashoks age to Sandeeps age is 4 : 3. Ashok will be 26 years old after 6 years. How old is Sandeep now?

Sol: Let their ages be 4x and 3x years. 
Now 4x + 6 = 26 hence x = 5 
. so Age of Sandeep = 3x = 15 years.

23. In 10 years, A will be twice as old5as B was 10 years ago. If A is now 9 years older than B, the present age of B is:

Sol: Let Bs age = x years. Then, As age = (x+ 9) years. 
(x+9+10)=2(x—10) hence x=39. 
Present age of B = 39 years.

24. The average age of 12 students is 20 years1f the age of one more student is included, the average decreases by 1. What is the age of the new student?

Sol: New average = (20 - 1) = 19. 
Age of new student = (13*19 - 12*20) years = 7 years

25. Average age of A and B is 24 years and average age of B, C and D is 22 years. The sum of the ages of A, B, C and D is:

Sol: A+B = (24X2) = 48 and B + C + D = (22*3) = 66 
With this data, we cannot find A + B + C + D.
So, the data is inadequate,

26. The ratio of the ages of Swati and Varun is 2: 5. After 8 years, their ages will be in the ratio of 1: 2. The difference in their present ages (in years) is:

Sol: Let Swati age = 2x & Varuns age = 5x. 
2x + 8 / 5x + 8 = 1/2 hence 2(2x+8) = (5x+8) so x=8. 
Swatis age = 16 years & Varun’s age =40 years. 
Difference of their ages = 24 years.

27. The Average age of a class of 22 subjects in 21 years. The average increased by 1 when the teacher’s age also included. What is the age of the teacher?

Sol: Avg x Number = Total

21 years x 22 nos = 462 years ... (1)

22 years x 23 nos = 506 years ... (2)

Teacher’s age = (2) - (1) = 506 – 462 = 44 Years.

28. Ten years ago, Kumar was thrice as old as Selva was but 10 years hence, he will be only twice as old. Find Kumar’s present age

Sol: Let Kumar’s present age be X years and Selva’s present age be Y years.

Then, according to the first condition,

X - 10 = 3(y - 10)

or, X - 3Y = -20 ..(1)

Now. Kumar's age after 10 years

=(X + 10) years

Selva's age after 10 years = (Y + 10)

(X+10) = 2(Y+10)

or, X - 2Y = 10 ..(2)

Solving (1) and (2), we get

X = 70 and Y = 30

Kumar's age = 70 years and Selva's age = 30 years.

29. A father is four times as old as his son today. After 20 years, he would be just twice as old. At the time of birth of his son, how much old must the father be?

Sol: Let the ages of son and father today by 'X' and '4x' years respectively.

4X + 20 = 2 (X+20) ---> As per information

X = 10 years ==> 4X = 40 years.

At the time of birth of his son, father must be 40 - 10 = 30 old

Sushil was thrice as old as Snehal 6 years back. Sushil will be 5  times as old as Snehal 3

6 years hence. How old is Snehal today?

Let Snehal's age 6 years back = X Then, Sushil's age 6 years back = 32. 

Then	5	(X + 6 + 6) = (3X + 6 + 6)
	3

5(X + 12) = 3 (3X + 12)

4X = 24

X = 6

Snehal's age today = (X + 6) years = 12 years.

30. The ages of Kamaraj and Deenan differ by 16 years, Six years ago, Mohan’s age was thrice as that of Kamaraj’s find their present ages.

Sol:

Let Kamaraj's age = X years

So, Mohan's age = (X + 16) years

Also, 3(X - 6) = X + 16 - 6 or, X = 14

Kamaraj's age = 14 years

and, Mohan's age = 14 + 16 = 30 years.

Courtesy : http://akulapraveen.blogspot.in/

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-TRAIN PROBLEMS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

TRAIN PROBLEMS

Train problems broadly center on the following three types.

1.      Time taken by a train to cross a stationary point

2.      Time taken by a train to cross a stationary length

3.      Time taken by a train to cross a moving length

* km/hr – m/s conversion:

X km/hr = [X * (5/18)] m/s.

* m/s – km/hr conversion:

X m/s = [X * (18/5)] km/hr.

§  Time taken by a train to cross a stationary point = train length/train speed

§  Time taken by a train to cross a stationary length = (train length + stationary length)/train speed

§  Time taken by a train to cross another train moving in the same direction = sum of length of the two trains/difference in their speeds

§  Time taken by a train to cross another train moving towards it = sum of length of the two trains/sum of their speeds

§   Time taken by a train of length 1 meter to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover 1 meter.

 Time taken by a train of length 1 meters to pass a stationary object of length b meters is the time taken by the train to cover (1 + b) meters.

 Suppose two trains or two bodies are moving in the same direction at u m / s and v m/s, where u > v, then their relatives speed = (u - v) m / s.

 Suppose two trains or two bodies are moving in opposite directions at u m / s and v m/s, then their relative speed is = (u + v) m/s.

 If two trains of length a meters and b meters are moving in opposite directions at u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v)sec.

If two trains of length a meters and b meters are moving in the same direction at u m / s and v m / s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec.

 If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then
(A's speet) : (B’s speed) = (b1/2: a1/2).

Example:

1.If a train going at 90 kmph takes 28 seconds to go past a lamp post, but 80seconds to cross a platform, what is the length of (a) the train and (b) the platform?

90 kmph= (90 × 5/18) = 25 m/s.

Solution for (a):

Given the time to cross the lamp post as 28 seconds,

Formula:

(train length/25 )= 28 or train length = 700 m. 

Solution for (b):

Given the time to cross the platform as 80 seconds,

Formula:

{(train length + platform length)/25 } = 80

or (train length + platform length) = 2000 m.

Since train length is 700 m, platform length = 1300 m or 1.3 km. 

2. How long does it take for a train of length 800 m moving at 80 kmph to cross a train of length 1200 m coming at a speed of 100 kmph from the opposite direction?

The relative speed = 80 + 100 (Moving opposite direction)

= 180 kmph = 50 m/s.

The distance to be covered = 800 + 1200 ( Sum of the train length)

= 2000 m.

Formula:

Time taken for crossing = 2000/50 (Time = Sum of the train length / Relative Speed)         = 40 seconds.

Above two example problems are basic type. Here i will give three more examples with little bit different.

Example 3:

Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains meet?

In this problem, we need to find the time taken by trains to meet each other. So no need to consider the train length.

Distance between the trains = 100m

Relative speed = 54+72 kmph (Trains are in opposite direction)

= 126 kmph = 35mps

Time taken = Distance/ Relative speed

= 100/35 = 20/7 seconds

Example 4:

Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains cross each other?

In this problem, we need to find the time taken by trains to cross each other. So we need to consider the train length and also distance between the trains.

Distance to be covered = 100+200 +100 (Train length 1 + Train length 2 +distance between them)

Relative Speed = 35 mps

Time taken = Distance/ Relative speed

= 400/35 = 80/7 seconds

Example 5:

Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains be 100m apart again ?

In this problem, Trains have to cross each other and then be at 100m apart.

For distance , we need to consider initial distance , train length and their final distance between them.

Relative Speed = 35 mps

Distance to be covered = 100+100+200+100 = 500m

Time taken = 500/35 = 100/7 seconds.

The Questions will involve one train or two trains; therefore read the question whether the question involves one train or two trains, because you can decide what formula to use and solve quickly. The entire aptitude question on trains will contain any of these 5 questions. Note that sometimes trains are replaced by some other thing like person, man, car … etc

Single Train Questions (2 Types of questions)

Question 1:

What is the time taken by a train of length ‘L’ metres and travelling at a speed of ‘S’ to pass a pole or a standing man or a signal post or some stationary object?

Explanation

·         The Question will contain length of the train ‘L’

·         Train’s speed ‘S’ will be given in terms of either km/hr or m/sec.

If speed is given in terms of Km/hr, always remember to convert  it to m/sec which can be done               by this formula: =S x (5/18) m/sec

·         You have to find the time taken by the train to pass the stationary object. Use this formula

             Time taken to pass the object= L/S sec

Question 2

What is the time taken by a train of length ‘L’ metres and travelling at speed of ‘S’ to pass a pole or a standing man or a signal post or some stationary object of Length ‘M’?

Explanation

·         The Question will contain length of the train ‘L’

·         Length of the object ‘M’

·         Train’s speed ‘S’ will be given in terms of either km/hr or m/sec.

If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done                  by this formula: =S x (5/18) m/sec

·         You have to find the time taken by the train to pass the object of length ‘M’. Follow this procedure

               1. You know this: Time taken to pass the object= L/S sec

               2. Make changes to the above formula in this way…

                   Time taken to pass the object of length ‘M’ = (L+M)/S sec

Two Train Questions (3 Types)

Question 3:

What is the taken by the two trains that are of length ‘a’ and ‘b’ respectively and travelling opposite to each other at a speed of ‘u’ and ‘v’ respectively to cross each other?

Explanation

·         The Question will contain two trains and its lengths as; ‘a’ of the first train and ‘b’ of the second train.

·         The speed of the trains will also be included  as ‘u’ as speed of the first train and ‘v’ as the speed of the second train

If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done by             this formula

For train 1:      =u x (5/18) m/sec an

For train 2:     = v x (5/18) m/sec

Or

= (u+v) x (5/18) m/sec

·         Now you have find out the time taken by the trains to cross each other; use this formula

= (a+b)/ (u+v)

u+v is called as relative speed of trains travelling in opposite direction

Question 4

What is the taken by the two trains that are of length ‘a’ and ‘b’ respectively and travelling in the same direction or parallel to each other at a speed of ‘u’ and ‘v’ respectively to cross each other?

Explanation

·         The Question will contain two trains and its lengths as: ‘a’ of the first train and ‘b’ of the second train.

·         The speed of the trains will also be included  as ‘u’ as speed of the first train and ‘v’ as the speed of the second train

           If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done by             this formula

            For train 1: =u x (5/18) m/sec and

            For train 2: = v x (5/18) m/sec

           Or

           = (u+v) x (5/18) m/sec

·         Now you have find out the time taken by the trains to cross each other;

·         use this formula

            = (a+b)/ (u-v)

            u-v is called as relative speed of trains travelling in opposite direction

Question 5

What is the time taken by the two trains that start at their points: A for the first train and B for the second train that travels at a speed of ‘u’ and ‘v’ respectively to reach their destination after crossing each other?

Explanation

·         The Question will contain two trains and its’ Speed as: ‘u’ of the first train and ‘v’ of the second train.

·         You have to find out the time taken by the trains to reach their destination after crossing each other.

·         Use this formula             (√v: √u)

                           EXAMPLES

1.      A train 280 m long, running with a speed of 63 km/hr will pass a tree in:

Sol:Speed = 63 x 5 18  m/sec = 35 2  m/sec.

Time taken = 280 x 2 35  sec = 16 sec.

2.A train moves with a speed of 108 kmph. Its speed in meters per second is:

108 kmph =		108 x	5 18		m/sec = 30 m/sec.
3. A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train.
Speed = 132 6  m/sec = 22 x 18 5  km/hr = 79.2 km/hr.
3. How long does a train 110 meters long running at the speed of 72 km/hr take to cross a bridge 132 meters in length? Sol: Speed = 72 x 5 18  m/sec = 20 m/sec. Total distance covered = (110 + 132) m = 242 m. Required time = 242 20  sec = 12.1 sec. 4. The length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is: Sol: Speed = 45 x 5 18  m/sec = 25 2  m /sec; Time = 30 sec. Let the length of bridge be x meters. Then, 130 + x 30 = 25 2 2(130 + x) = 750 x = 245 m. 5. A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is: Sol:  Let the length of the train be x meters and its speed be y m/sec. They, x y  = 15 y = x 15 x + 100 25 = x 15  x = 150 m.
6.A train crosses a platform 100 m long in 60 seconds at a speed of 45 km/hr. The time taken by the train to cross an electric pole is :  Sol: Speed = 45 x 5 18  m/sec = 25 2  m/sec. Let the length of the train be x metres. Then, x + 100 25 2  = 60   or   x = 650 m Time taken by the train to cross an electric pole = 650 x 2 25  sec = 52 sec.  7.A train 240 m long passed a pole in 24 seconds. How long will it take to pass a platform 650 m long? Sol: Speed = 240 24  m/sec = 10 m/sec. Required time = 240 + 650 10  sec = 89 sec.  8.A train takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is: Sol:  Let the length of the train be x meters. x + 162 18 = x + 120 15  15 (x + 162) = 18 (x + 120) x = 90 m. 9.A goods train runes at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?  Sol: Speed = 72 x 5 18  m/sec = 20 m/sec; Time = 26 sec. Let the length of the train be x meters. Then, x + 250 26  = 20 x + 250 = 520 x = 270. 10.Two trains are running in opposite directions with the same speed. If the length of each train is 120 meters and they cross each in 12 seconds, then the speed of each train ( in km/hr) is Sol:  Let the speed of each train be x m/sec. Then, relative speed of the two trains = 2x m/sec. So, 2x = (120 + 120) 12 2x = 20 x = 10. Speed of each train = 10 m/sec = 10 x 18 5  km/hr = 36 km/hr. 11.Two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. The length of the faster train is:  Sol:
Relative speed = (36 + 45) km/hr = 81 x 5 18  m/sec = 45 2  m/sec.
Length of train = 45 2  x 8  m = 180 m.

12.How many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 lm/hr?

Sol: Speed of train relative to man = (63 – 3) km/hr = 60 km/hr

=  60 x 5 18  m/sec = 50 3  m/sec.

Time taken to pass the man = 500 x 3 50  sec = 30 sec.

13.A train of length 150 meters takes 40.5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr?

Sol: Speed = 150 + 300 40.5  m/sec = 450 40.5 x 18 5  km/hr = 40 km/hr.

14.A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, then the length of the train is:

Sol:

Speed = 12 10  x 60  km/hr = 72 x 5 18  m/sec = 20 m/sec.

Length of the train = (Speed x Time) = (20 x 6) m = 120 m.

15.A train 360m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

Sol:

Speed = 45 x 5 18  m/sec = 25 2  m/sec.

Total distance covered = (360 + 140) m = 500 m.

Required time = 500 x 2 25  sec = 40 sec.

16.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:

Sol:

Relative speed = (60 + 40 ) km/hr = 100 x 5 18  m/sec = 250 9  m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m

Required time = 300 x 9 250  sec = 54 5  sec = 10.8 sec.

17.A train 800 meters long is running at a speed of 78 km / hr. If it crosses a tunnel in 1 minute, then the length of the tunnel (in meters) is:

Sol:

Speed = 78 x 5 18  m/sec = 65 3  m /sec;

Time = 1 minute = 60 sec.

Let the length of the tunnel be x meters.

Then, 800 + x 60 = 65 3 3(800 + x) = 3900 x = 500.

18.A 270 meters long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

Sol:

Relative speed = (120 + 80) km/hr = 200 x 5 18  m/sec = 500 9  m/sec.

Let the length of the other train be x metres.

Then, x + 270 9 = 500 9 x + 270 = 500 x = 230.

19. Two trains, one from Warangal to Hyderabad and the other from Hyderabad to Warangal, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Sol:

Let us name the trains as A and B. Then,

(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3

20.In what time will a train 100 meters long cross an electric pole, if tis speed be 144 km/hr?

Sol:

Speed = 144 x 5 18  m/sec = 40 m/sec.

Time taken = 100 40  sec = 2.5 sec.

21.A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?

Sol:

Let the length of the train be x meters and its speed by y m/sec.

They, x y  = 8  x = 8y

Now, x + 264 20 = y 8y + 264 = 20y  y = 22.

Speed = 22 m/sec = 22 x 18 5  km/hr = 79.2 km/hr.

22. Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

Sol:

Let the speed of the slower train be x m/sec.

Then, speed of the faster train = 2x m/sec.

Relative speed = (x + 2x) m/sec = 3x m/sec.

(100 + 100) 8  = 3x 24x = 200 x = 25 3

So, speed of the faster train = 50 3  m/sec = 50 3 x 18 5  km/hr = 60 km/hr.

23. A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

Sol:

Speed = 54 x 5 18  m/sec = 15 m/sec.

Length of the train = (15 x 20) m = 300 m.

Let the length of the platform be x meters.

Then, x + 300 36  = 15 x + 300 = 540 x = 240 m.

24.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

Sol:

Speed = 60 x 5 18  m/sec = 50 3  m/sec.

Length of the train = (Speed x Time) = 50 3  x 9  m = 150 m.

25.A train 110 m long passes a man, running at 6 kmph in the direction opposite to that of the train, in 6 seconds. The speed of the train is:

Sol:

Speed of the train relative to man = 110 6  m/sec = 110 6 x 18 5  km/hr = 66 km/hr.

Let the speed of the train be x kmph. Then, relative speed = (x + 6) = kmph.

x + 6 = 66  or  x = 60 kmph.

26. A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

Sol:

Speed = 300 18  m/sec = 50 3  m/sec.

Let the length of the platform be x meters

Then, x + 300 39 = 50 3  3 (x + 300) = 1950 x = 350 m.

27. A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is:

Speed of the train relative to man = 125 10  m/sec = 25 2  m/sec.

=  25 2 x 18 5  km/hr = 45 km/hr.

Let the speed of the train be x kmph. Then, relative speed = (x – 5) kmph.

x - 5 = 45  or   x = 50 kmph.

28. Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

Sol : Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x -1) hours = 25 (x – 1) km.

20x + 25 (x - 1) = 110   45x = 135   x = 3.

So, they meet at 10 a.m.

29. Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.

Relative speed = (45 + 30) km/hr = 75 x 5 18  m/sec = 125 6  m/sec.

Distance covered = (500 + 500) m = 1000 m.

Required time = 1000 x 6 125  sec = 48 sec.

30.A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is

Sol:

Let the speed of the second train be x km/hr.

Relative speed = (x + 50) km/hr = (x + 50) x 5 18  m/sec = 250 + 5x 18  m/sec.

Distance covered = (108 + 112) = 220 m.

220 250 + 5x 18  =  6 250 + 5x = 660 x = 82 km/hr.

31. The length of a train and that of a platform are equal. If with a speed of 90 km/hr, the train crosses the platform in one minute, then the length of the train (in metres) is:

Sol:

Speed = 90 x 5 18  m/sec = 25 m/sec; Time = 1 min. = 60 sec.

Let the length of the train and that of the platform be x metres.

Then, 2x 60  = 25 x = 25 x 60 2   = 750.

32.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:

Sol:

Relative speed = (60 + 40 ) km/hr = 100 x 5 18  m/sec = 250 9  m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m

Time taken 300 x 9 250  sec = 54 5  sec = 10.8 sec.

33.A train 150 m long passes a km stone in 15 seconds and another train of the same length travelling in opposite direction in 8 seconds. The speed of the second train is:

Sol:

Speed of first train = 150 15  m/sec = 10 m/sec.

Let the speed of second train be x m/sec.

Relative speed = (10 + x) m/sec.

300 10 + x  = 8 300 = 80 + 8x x = 220 8 = 55 2  m/sec.

So, speed of second train = 55 2 x 18 5  kmph = 99 kmph.

34. Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. First train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?

Sol:

Relative speed = (40 – 20) km/hr = 20 x 5 18  m/sec = 50 9  m/sec.

Length of faster train = 50 9  x 5  m = 250 9  m = 27 7 9  m.

35.Two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively. In how much time will they cross each other, if they are sunning the same direction?

Sol:

Relative speed = (45 – 40 ) kmph = 5 kmph = 5 x 5 18  m/sec = 25 18  m/sec.

Total distance covered = Sum of lengths of trains = 350 m.

Time taken 350 x 18 25  sec = 252 sec.

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-PIPES & CISTERNS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.

Pipes and Cisterns:

It’s based on the Time and work model.

Terms which are used in these problems are

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.

Properties:

§  If a pipe can fill a tank in x hours, then:

              part filled in 1 hour = 1/x

§  If a pipe can empty a tank in y hours, then:

part filled in 1 hour = 1/y

§  If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y>x), there on opening both the pipes, the net part filled in 1 hour = { 1/x – 1/y }

§  If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y

                E represents Total time taken to empty the tank

                F represents Total time taken to fill the tank

               L represents the L.C.M

               e represents time taken to empty the tank

               f represents time taken to fill the tank

§  Time for emptying, (emptying pipe is bigger in size.)

E = (f * e)/(f – e)

§  Time for filling, (Filling pipe is bigger in size.)

F = (e * f)/(e – f)

§  Pipes ‘A’ & ‘B’ can fill a tank in f1 hrs & f2 hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the three pipes are opened simultaneously then the tank is filled in ,

F = L/[(L/f1) + (L/f2) - (L/e)]

§  Two taps ‘A’ & ‘B’ can fill a tank in ‘t1′ & ‘t2′ hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the tank is full & all the three pipes are opened simultaneously. Then the tank will be emptied in,

E = L/[(L/e) - (L/f1) - (L/f2)]

§  Capacity of the tank is , F = (f * e)/(e – f)

§  A filling tap can fill a tank in ‘f’ hrs. But it takes ‘e’ hrs longer due to a leak at the bottom. The leak will empty the full tank in ,

E = [ t(f + e) * tf ] / [ t(f + e) – tf ]

                                                      Example:

1.If a pipe can fill the tank in 6 hrs but unfortunately there was a leak in the tank due to which it took 30 more minutes .Now if the tank was full how much time will it take to get emptied through the leak?

Sol: By last property,

t(f+e) = 6+0.5 =6.5hrs

tf = 6 hrs

E = 6.5*6 / (6.5 – 6)

= 78 hrs .

2.A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is

Sol: Work done by the inlet in 1 hour = (1/6) - (1/8) = 1/24
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 4 liters.
Volume of whole = (1440 * 4) litres = 5760 litres.

3.12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?

Sol: Capacity of the tank = (12 * 13.5) litres = 162 litres.
Capacity of each bucket = 9 litres
Number of buckets needed = (162/9) = 18.

4.One tap can fill a cistern in 2 hours and another tap can empty the cistern in 3 hours. How long will they take to fill the cistern if both the taps are opened ?

Sol: Net part filled in 1 hour = (1/2) - (1/3) = 1/6 Cistern will be full in 6 hours

5.A cistern can be filled in 9 hours but it takes 10 hours due to in its bottom. If the cistern is full, then the time that the leak will take to empty it, is:

Sol: Work done by the leak in 1 hour = (1/9 - 1/10) = 1/90.
Leak will empty the full cistern in 90 hrs

6.A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 litres a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litres does the cistern hod?

Sol: Work done by the inlet in 1 hour = (1/8) - (1/12) = 1/24
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 6 litres
Volume of whole = (1440 x 6) litres 8640 litres

7.An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. The leak can drain out all the water of the tank in :.

Sol: Work done by the leak in 1 hour = (1/3) - (2/7) = 1/21 .
Leak will mpty the tank in 21 hours.

8.TapsA and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket

Sol: Part filled in 3min = 3[(1/12) + (1/15)] = 3 * (9/60) = 9/20
Remaining part = 1 - (9/20) = 11/20
Part filled by B in 1 min = 1/15
(1/15) : (11/20) = 1 : x or x = (11/20) * 1 * (15/1) = 8 min 15 sec
Remaining part is filled by B in 8 ruin. 15 sec.

9. Two pipes A and B can fill a cistern in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time B should be closed so that the tank is full in 9 minutes ?

Sol: Let B be closed after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (9 — x) min, = 1
x[(1/12) + (1/16)] + (9 - x)(1/12) = 1 or (7x/48) + (9-x)/12 = 1
or7x + 36 — 4x = 48 or x=4.
So, B must be closed after 4 minutes

10.Bucket P has thrice the capacity as bucket Q It takes 60 turns for bucket P to fill the empty drum. How many turns it wifi take for both the buckets P and Q, having each turn together to fill the empty drum

Sol: Let capacity of P be y litres. Then, capacity of Q = y/3 litres. Capacity of the drum = 60y litres.
Required number of turns = 60y /(y + (y/3)) = (60 * (3/4y) = 45

11.To fill a cistern, pipes A, B and C take 20 minutes, 15 minutes and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern, is : .

Sol; Part filled by (A +B+ c) in 1 min. = (1/20) +(1/15) + (1/12) = 12/60 = 1/5
All the three pipes together will fill the tank in 5 min.

12. A tap can fill a tank in 16 minutes and another can empty it in8 minutes. If the tank is already half full and both the tanks are oped together, the tank will be:

Sol; Rate of waste pipe being more, the tank will be emptied when both the pipes are opened.
Net emptying work done by both in 1 min = (1/8) - (1/16) = 1/16
Now, full tank will be emptied by them in 16 min.
Half full tank will be emptied in 8 min.

13.A tank can be filled by a tap in 20 minutes and by another tap in 6O minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in

Sol; Part filled in 10 min = 10[(1/20) + (1/60)] = 10 * (4/60) = 2/3
Remaining part = (1 - (2/3)) = 1/3
Part filled by second tap in 1 min = 1/60
(1/60) : (1/3) ∷ 1 : x
Hence, the remaining part will be filled in 20 min

14.Two taps A and B can fill a tank in 10 hours and 15 hours respectively. If both the taps are opened together, the tank will be full in:

Sol: As hours work=1/10, Bs 1 hours work = 1/15,
(A+B)s 1 hours work = (1/10) + (1/15) = 5/30 = 1/6
Both the taps can fill the tank in 6 hours.

15.Two pipes can fill a tank in 10 hours 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 the tank will 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.

16.Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?

Part filled by A in 1 min. = 1 20  ; Part filled by B in 1 min. = 1 30

Part filled by (A + B) in 1 min. 1 20 + 1 30 = 1 12

Both the pipes can fill the tank in 12 minutes.

17.Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

Sol: Net part filled in 1 hour = 1 5 + 1 6 - 1 12 = 17 60

The Tank will be full in 60 17  hrs i,e., 3 9 17  hrs.

18.One pipe ca fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

Sol:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in x 3  minutes.

1 x + 3 x = 1 36 4 x = 1 36 x = 144 min.

19.A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour = 1 4 - 1 9 = 5 36

The cistern will be filled in 36 5  hrs i.e., 7.2 hrs.

20.A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x 2  and x 4  hours respectively to fill the tank

1 x + 2 x + 4 x = 1 5 7 x = 1 5 x = 35 hrs.

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