A man buys Rs.7 apples. When after a few days the price of the apple goes down by Rs.2, he buys 6 more apples.

Both the amounts were in whole numbers, were two digits and has the same digits.

Find how many apples could be bought for Rs.144.

The equations I could find was :

i)  10a + b = 7x
ii) 10b + a = 6(x-2)


  • x : cost/apple for trans a
  • x-2 : cost/apple for trans b
  • 10a+b : trans a total
  • 10b+a : trans b total

There are three variables, how do I derive the third equation ?


There isn't a third equation per se, but another constraint is obtained from the proviso that all quantities (and therefore $a$ and $b$) are integers. If you subtract Equation (ii) from Equation (i), you obtain

$$ 9a-9b = x+12 $$


$$ 9(a-b) = x+12 $$

We know from this that $x+12$ is divisible by $9$, so $x$ could be $6$, $15$, $24$, etc. At this point, we introduce the condition that the total costs in both cases were two-digit quantities. If $x$ were $15$ or more, the seven original apples would cost $105$ or more, violating the condition. Therefore, $x = 6$.


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