The ratio of the ages of A and B, 3 years ago, was 5 : 7. The ratio of their ages, 13 years from now, will be 9 : 11. If the present age of C is 11 years less than that of B, then what is the present age of C in years?

Question

Accepted Answer

Let the present ages of A and B be a and b respectively.

Given:
- 3 years ago, the ratio of ages of A and B was 5:7.  
  So, (a - 3) / (b - 3) = 5 / 7  => 7(a - 3) = 5(b - 3)  
  => 7a - 21 = 5b - 15  
  => 7a - 5b = 6  ...(1)

- 13 years from now, the ratio of ages of A and B will be 9:11.  
  So, (a + 13) / (b + 13) = 9 / 11  => 11(a + 13) = 9(b + 13)  
  => 11a + 143 = 9b + 117  
  => 11a - 9b = -26  ...(2)

Now, solve equations (1) and (2):  
From (1): 7a - 5b = 6  
From (2): 11a - 9b = -26

Multiply (1) by 9 and (2) by 5 to eliminate b:  
(1)*9: 63a - 45b = 54   
(2)*5: 55a - 45b = -130

Subtract (2)*5 from (1)*9:    
(63a - 45b) - (55a - 45b) = 54 - (-130)    
63a - 45b - 55a + 45b = 54 + 130    
8a = 184    
=> a = 23

Substitute a = 23 into (1):    
7(23) - 5b = 6    
161 - 5b = 6      
-5b = 6 - 161 = -155         
b = 31

Given that present age of C is 11 years less than B:        
C = b - 11 = 31 - 11 = 20

Therefore, the present age of C is 20 years.

Answer

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