Aptitude Discussion

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 |

**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.**

**Sravan Reddy**

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**Uttam Kumar**

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answer should be 9 not 19

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.

**Sujal Padhiyar**

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Here it should be 1 year hence not 11.

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

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