Monday, December 29, 2014

Problems on Ages



To solve the problems based on ages, students are required the knowledge of linear equations. This method needs some basic concepts as well as some more time than it deserves. Sometimes it is easier to solve the problems by taking the given choices in account. But this hit-and-trial method proves costly sometimes, when we reach our solution much later. We have tried to evaluate some easier as well as quicker methods to solve this type of questions. Although we are not able to cover each type of questions in this section, our attempt is to minimize your difficulties.





Have a look at the following questions

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

Ex. 2. : At present the age of the father is five times that of the age of his son. Three years hence, the father’s age would be four times that of his son. Find the present ages of the father and the son.

Ex. 3. : Three years earlier the father was 7 times as old as his son. Three years hence the father’s age would be four times that of his son. What are the present ages of the father and the son?



5. If the current age is x, then1of the age isx.
nn






By the conventional method:

Sol. 1. : Let the present age of son = x years.
           Then, the present age of father = 5x years.
           3 years ago,
                 7 (x – 3) = 5x – 3
           or,   7x – 21 = 5x – 3
           or,           2= 18
                        x = 9 years
           Therefore, son’s age = 9 years.
           Father’s age = 45 years.

Sol. 2. : Let the present age of son = x years.
           Then, the present age of father = 5x years.
           3 years hence,
                  4(x + 3) = 5x + 3
           or,   4x + 12 = 5x + 3
                        x = 9 years.
           Therefore, son’s age = 9 years
           and father’s age = 45 years.

Sol. 3.Let the present age of son = x years.
           and the present age of father = y years.
           3 years earlier,
                  7(x – 3) = y – 3
           or,      7x – y = 18                    ...(1)
           3 years hence,
                  4(x + 3) = y + 3
           or,   4x + 12 = y + 3
           or,      4x – y = – 9                   ...(2)
           Solving (1) & (2) we get, x = 9 years & y = 45 years.

Quicker Method :

Sol. 1. :          
Son's age = 3 × (7 - 1) = 9 yrs.
7 - 5
and father’s age = 9 × 5 = 45 years.
Undoubtably you get confused with the above method, but it is very easy to understand and remember. See the following form of question :

Q. : tyears earlier the father’s age was x times that of his son. At present the father’s age is y times that of his son. What are the present ages of the son and the father?
Son's age = t1(x - 1)
x - y
Sol. 2. :         
Son' s age = (4 -1) × 3 = 9 yrs
5 - 4
and father’s age = 9 × 5 = 45 years.
To make more clear, see the following form :

Q. : The present age of the father is y times the age of his son. t2 years hence, the father’s age become z times the age of his son. What are the present ages of the father and his son?
Son's age =  (z - 1)t2
y - z
Sol. 3. :        
Son's age =  3(4 - 1) + 3(7 - 1) = 9 + 18 = 9 yrs.
7 - 43
To make the above formula clear see the following form of question :

Q. : tyears 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. What are the present ages of the son and the father?
of his son. What are the present ages of the son and the father?
Son's age =  t2(z - 1) + t1(x -1)
(x - z)
In the tabular form the above three types can be arranged as:
Unlimited Problems on ages question for Bank clerk, po, ssc
Ex. 4. : The age of a man is 4 times that of his son. 5 years ago, the man was nine times as old as his son was at that time. What is the present age of the man?
Sol. :  By the table, we see that formula (1) will beused.    
Son's age = 5(9 - 1) = 8 yrs.
(9 - 4)
∴   Father’s age = 4 × 8 = 32 years.
Note :  The relation between ‘earlier’ and ‘present’ ages are given; so we look for the formula derived from the two corresponding columns of the table. That gives the formula (1).
Ex. 5. : After 5 years the age of a father will be thrice the age of his son, whereas five years ago, he was 7 times as old as his son was. What are their present ages?

Sol. :  Formula (3) will be used in this case. So
Son's age =  5(7 - 1) + 5(3 - 1) = 10 yrs.
7 - 3
From the first relationship of ages, if F is the age of the father then
                   F + 5 = 3 (10 + 5)
                     F = 40 years.
Ex. 6. : 10 years ago, Sita’s mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. What is the present age of Sita?
Sol. :  In this case also, formula (3) will be used.
Daughter's age = 10(4 - 1) + 10(2 -1) = 20 yrs.
4 - 2
Ex. 7.: One year ago the ratio between Samir’s and Ashok’s age was 4 : 3. One year hence the ratio of their ages will be 5 : 4. What is the sum of their present ages in years?

Sol. :  One year ago Samir’s age was  of Ashok’s age.
           One year hence Samir’s age will be  of Ashok’s age.
           ∴     Ashok’s age (by formula(3));
One year ago Samir's age was 4 of Ashok's age.
3

One year hence Samir's age will be 5 of Ashok's age.
4
∴  Ashok's age(By formula(3));
A =  
1(4 - 1) + 1(5 - 1)
34
  =  
1 + 1
34
 = 7 yrs.
4 - 5
34
1
12
 
Now, by the first relation:
(s - 1) = 4
7 - 13
∴   S = 8 + 1 = 9 years.
∴   Total ages = A + S = 9 + 7 = 16 years.
           where A = Ashok’s present age
           and S = Samir’s present age.

Ex. 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. :  10 years ago A was 1/2 of B’s age.
           At present A is 3/4 of B’s age.         
 B’s age [use formula (1)] = 10(1/2 - 1) = 20 yrs.
(1/2) - (3/4)

A's age = 3of 20 = 15 yrs.
4
Ex. 9. : The sum of the ages of a mother and her daughter is 50 years. Also 5 years ago, the mother’s age was 7 times the age of the daughter. What are the present ages of the mother and the daughter?

Sol. :  Let the age of the daughter be x years.
           Then, the age of the mother is (50 – x) years.
           5 years ago,
                 7 (x – 5) = 50 – – 5
           or,           8x = 50 – 5 + 35 = 80
                        x = 10
           Therefore, daughter’s age = 10 years
           and mother’s age = 40 years.

Quicker Method (Direct Formula) :                 
Daughter’s age = Total ages + No. of years ago(Times - 1)
Times + 1

50 + 5(7 - 1) = 10 yrs.
7 + 1
Thus, daughter's age = 10 yrs and Mother's age = 40 years.

Ex. 10. : The sum of the ages of a son and father is 56 years. After 4 years, the age of the father will be three times that of the son. What is the age of the son?

Sol. :  Let the age of the son be x years.
           Then, the age of the father is (56 – x) years.
           After 4 years,
                  3(x + 4) = 56 – x + 4
           or,           4x = 56 + 4 – 12 = 48
                        x = 12 years.
           Thus, son’s age = 12 years.

Quicker Method (Direct Formula):
Son’s age = Total ages - No. of years after(Times - 1)
Times + 1

56 - 4(3 -1) = 48 = 12 yrs.
3 + 13

Note:  Do you get the similarities between the above two direct methods? They differ only in signs in the numerator. When the question deals with ‘ago’ a +ve sign exists and when it deals with ‘after’ a –ve sign exists in the numerator.
Ex. 11.:  The sum of the present ages of the father and the son is 56 years. 4 years hence the son’s age will be  1/3 that of the father. What are the present ages of the father and the son?

Sol. :  Son’s age is 1/3 that of fahter.
         ⇒ Father’s age is 3 times that of son.
           Now we use the formula as in Ex. 10. Try it.
Father's age = 56 - 4(1/3 - 1) = 56 + (8/3) = 176 = 44 yrs.
(1/3 + 1)4/34
Ex. 12. : The ratio of the father’s age to the son’s age is 4 : 1. The product of their ages is 196. What will be the ratio of their ages after 5 years?

Sol. :  Let the ratio of proportionality be x, then
                 4x ×  x = 196
           or,      4x2 = 196
           or,          x = 7
           Thus, Father’s age = 28 years, Son’s age = 7 years.
           After 5 years,
           Father’s age = 33 years, Son’s age = 12 years.
           ∴   Ratio = 33 : 12 = 11 : 4

Ex. 13: The ratio of Rita’s age to the age of her mother is 3 : 11. The difference of their ages is 24 years. What will be the ratio of their ages after 3 years?

Sol. :  Difference in ratios = 8
           Then, 8 ≡ 24 ∴ 1 ≡ 3
           ie, value of 1 in ratio is equivalent to 3 years
ie, value of 1 in ratio is equivalent to 3 years
           Thus Rita’s age = 3 × 3 = 9 years.
           Mother’s age = 11 × 3 = 33 years.
           After 3 years, the ratio = 12 : 36 = 1 : 3

Ex. 14: The ratio of the ages of the father and the son at present is 6 : 1. After 5 years the ratio will become 7 : 2. What is the present age of the son?

Sol. :  Father : Son
           Present age = 6 : 1
           After 5 years = 7 : 2          
Son’s age = 1 × 5(7 - 2) = 5 yrs.
6 × 2 - 7 ×1

Father’s age = 6 × 5(7 - 2) = 30 yrs.
6 × 2 - 7 ×1
Then what direct formula comes?
           Father : Son
           Present age = x : y
           After T years = a : b
Then Son's age = y × T(a - b)
difference of cross product

and, Father’s age = × T(a - b)
difference of cross product
          
Ex. 15. : The ratio of the ages of the father and the son at present is 3 : 1. 4 years earlier, the ratio was 4 : 1. What are the present ages of the son and the father?

Sol. :  Father : Son
           Present age = 3 : 1
           4 years before = 4 : 1
Son’s age = 1 × 4(4 -1) = 12 yrs.
4 × 1 - 3 × 1     

Father’s age = 3 ×4(4 -1)= 36 yrs.
4 × 1 - 3 × 1
           Father : Son
           Present age = x : y
           T years before = a : b
Then Son's age = y × T(a - b)
difference of cross product

and, Father’s age = × T(a - b)
difference of cross product
Ex. 16. : A man’s age is 125% of what it was 10 years ago,  
but 831% of what it will be after 10 years. What is his present age?
3
Sol. :  Detailed Method : Let the present age be x years. Then
125% of (x – 10) = x; and 831% of (x + 10) = x
3

125% of (x – 10) = 831% of (x + 10)
3
          
or, 5(x - 10) = 5(x + 10)
46

or, 5x - 5x = 50 + 50 or, 5x = 250           ∴   x = 50 yrs.
46641212
Direct Method : With the help of the above detail method, we can define a general formula as:
=    
125 × No. of years ago + 831× No. of years after
3
125 - 831
3

=    
125 × 10 + 831×10
3
125 - 831
3

=    10(625) / 3 = 10 × 625 = 50 yrs.
(375 - 250) / 3125

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  1. a km/hr =a x5m/s.
    18
  2. a m/s =a x18km/hr.
    5
  3. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
  4. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
  5. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.
  6. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
  7. If two trains of length a metres and b metres are moving in opposite directions at um/s and v m/s, then:
    The time taken by the trains to cross each other =(a + b)sec.
    (u + v)
  8. If two trains of length a metres and b metres are moving in the same direction at um/s and v m/s, then:
    The time taken by the faster train to cross the slower train =(a + b)sec.
    (u - v)
  9. 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 speed) : (B's speed) = (b : a)




A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
A.120 metresB.180 metres
C.324 metresD.150 metres

2. 
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
A.45 km/hrB.50 km/hr
C.54 km/hrD.55 km/hr

3. 
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A.200 mB.225 m
C.245 mD.250 m

4. 
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A.1 : 3B.3 : 2
C.3 : 4D.None of these

5. 
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?
A.120 mB.240 m
C.300 mD.None of these
6. 
A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
A.65 secB.89 sec
C.100 secD.150 sec

7. 
Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
A.50 mB.72 m
C.80 mD.82 m

8. 
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
A.40 secB.42 sec
C.45 secD.48 sec

9. 
Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
A.36B.45
C.48D.49

10. 
A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
A.3.6 secB.18 sec
C.36 secD.72 sec


11. 
A 270 metres 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?
A.230 mB.240 m
C.260 mD.320 m
E.None of these

12. 
A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
A.230 mB.240 m
C.260 mD.270 m


13. 
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:
A.30 km/hrB.45 km/hr
C.60 km/hrD.75 km/hr


14. 
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:
A.9B.9.6
C.10D.10.8

15. 
A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
A.5 secB.6 sec
C.7 secD.10 sec

 View Answer


16. 
A train travelling at a speed of 75 mph enters a tunnel 31/2 miles long. The train is 1/4 mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
A.2.5 minB.3 min
C.3.2 minD.3.5 min

17. 
A train 800 metres 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:
A.130B.360
C.500D.540

18. 
A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?
A.320 mB.350 m
C.650 mD.Data inadequate

19. 
A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
A.50 mB.150 m
C.200 mD.Data inadequate

20. 
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?
A.69.5 km/hrB.70 km/hr
C.79 km/hrD.79.2 km/hr