Algebra
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Q.

The sum of $A$ and $B$'s age is 43 years. 11 year hence, $A$'s age will be 7/6 times $B$'s age then. Find $B$'s present age.

 A.

22 years

 B.

20 years

 C.

24 years

 D.

19 years

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Solution:
Option(D) is correct

Let $A$'s age be $x$ and $B$'s age be $y$.

$x+y=43$ -------- (i)

\((x+11)=\dfrac{7}{6}(y+11)\)

$⇒ 6x−7y=11$ -------- (ii)

On solving both the equations, we get:

$y = 19 \textbf{ years}$

Edit: Based on comments from Harish and Sujal Padhiyar, updated the solution.

Edit 2: For an alternative solution, check comment by Sravan Reddy.


(7) Comment(s)


Akash Sarda
 ()

Add 11 to each option, and check which one is divisible by 6. Since A's age is integer. There is only one such option!

D. 19.



Sravan Reddy
 ()

Fast mind based calculation thought process:

=> 11 year hence - so add 22 to 43 = 65 years (sum of both ages after 11 years)

=> 7:6 ratio so dividing by 13 and multiplying by 7 and 6

=> Ages are 35 and 30 - 11 years earlier it was 19



Uttam Kumar
 ()

answer should be 9 not 19


Deepti
 ()

Can you show how 9 years is the answer?

Because my calculations suggest otherwise. I solved the given equations and got the result to be correct as 19 years.

Daphne Ritz
 ()

Could you please show your work? I did it using different methods and still got 19.


Sujal Padhiyar
 ()

Here it should be 1 year hence not 11.


Harish
 ()

I guess this should be '11 year hence'. And solution needs to be changed.