Some money (dollars) is divided among friends A,B and C in the ratio of 5:6:9

After A gives fifty dollars to his mother, the ratio becomes 3:4:6

Find the amount of money A has after giving fifty dollars to his mother.


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  • $\begingroup$ What have you tried? $\endgroup$ – Tito Eliatron Nov 29 '18 at 18:37
  • $\begingroup$ I couldn't find a logic to solve itt. $\endgroup$ – Wardah Nov 29 '18 at 18:38

The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).

So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:

$$ \begin{align} \text{Before} \quad 5:6:9 &\iff 10:12:18 \\ \text{After} \quad 3:4:6 &\iff 9:12:18 \end{align} $$

Now, the problem states that A has given $\$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).

This indicates that one unit of the ratio corresponds to $\$50$.

Since now he has 9 units, he has $9 \times \$50 = \$450$.

  • $\begingroup$ Is there any technique to solve it with simple algebra? $\endgroup$ – Wardah Nov 29 '18 at 18:57
  • $\begingroup$ Yes, I believe the other answer solves it in an algebraic way. $\endgroup$ – Sean Lee Nov 29 '18 at 18:59
  • $\begingroup$ I am just a beginer. Found it very hard to understand. As it is not solved completly. $\endgroup$ – Wardah Nov 29 '18 at 19:02

Hint: Let $x$ be the amount $A$ has to begin with. Then: $\dfrac{x}{5} = \dfrac{b}{6}, \dfrac{x-50}{3} = \dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $\dfrac{6x}{5} = \dfrac{4(x-50)}{3}$. Can you take it from here ?.

  • $\begingroup$ I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth. $\endgroup$ – Wardah Nov 29 '18 at 19:03

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