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?

Important Formulas on "Problems on Ages" :

1. If the current age is x, then n times the age is nx.

2. If the current age is x, then age n years later/hence = x + n.

3. If the current age is x, then age n years ago = x - n.

4. The ages in a ratio a : b will be ax and bx.

5. If the current age is x, then	1	of the age is	x	.
	n		n

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,           2x = 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. : t1 years 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 - 4		3

To make the above formula clear see the following form of question :

Q. : 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. 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:

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 ) 3 4	=	1  +  1 3 4	= 7 yrs.
	4  -  5 3 4		1 12

 
Now, by the first relation:

(s - 1)	=	4
7 - 1		3

∴   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 =	3	of 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 – x – 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 + 1		3

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.

Important Note : We can solve the above question without changing the form of ‘times’. When the question remains in its original form, we find the age of father directly as: Father's age =  56 - 4(1/3 - 1)  =  56 + (8/3)  =  176  = 44 yrs. (1/3 + 1) 4/3 4 And hence we may get the age of son. Thus this reveals an important fact that in each of the examples from 1 to 11, we may get the second age directly by inverting the ratio. For example, for 4 times we may use 1/4, for 4/3,we may use 3/4, and so on. Try to solve each of the above examples by inverting the ratio.

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 = x ×	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

Now the direct formula comes as :

           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 = x ×	T(a - b)
	difference of cross product

Note: 1. While evaluating the difference of cross-product always take the +ve sign. 2. Both the above direct formulas look similar. The only difference you can find is in the denominators. But it has been simplified as “difference of cross-products” to make it easier to remember. So with the help of one formula only you can solve both the questions. 3. The above two questions (Q. 14 and Q. 15) are similar to the questions discussed earlier (Q. 1 & Q. 2). Do you get the point? If you change ‘ratio’ in ‘times’ you will get the same thing. Now, you have two simple methods. Use the method which you feel easier to remember. Thus, you may solve Q. 14 as Q. 2 and vice versa. Q. 15 as Q. 1 and vice versa. 4. We suggest you to go through both the methods and choose the better of the two.

Ex. 16. : A man’s age is 125% of what it was 10 years ago,  

but 83	1	% 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 83	1	% of (x + 10) = x
	3

125% of (x – 10) = 83	1	% of (x + 10)
	3

          

or,	5	(x - 10) =	5	(x + 10)
	4		6

or,	5	x -	5	x =	50	+	50	or,	5x	=	250	∴   x = 50 yrs.
	4		6		6		4		12		12

Direct Method : With the help of the above detail method, we can define a general formula as:

=	125 × No. of years ago + 83 1 × No. of years after 3
	125 - 83 1 3

=	125 × 10 + 83 1 ×10 3
	125 - 83 1 3

=	10(625) / 3	=	10 × 625	= 50 yrs.
	(375 - 250) / 3		125

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