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Mr. and Mrs. Ahuja weigh $x$ kg and $y$ kg respectively. They both take a dieting course at the end of which Mr. Ahuja loses $5$ kg and weighs as much as the wife weighed before the course. Mrs. Ahuja loses $4$ kg and weighs $7$/$8$th of what her husband weighed before the course. From two equations in $x$ and $y$ and hence find their present weights.

I tried the following,

Mr. Ahuja's weight before the course$=$$x$ kg

Mrs. Ahuja's weight before the course$=$$y$ kg

                              After dieting course

Mr. Ahuja's weight: $E_1 =>x-5=y$

Mrs. Ahuja's weight: $E_2 =>y-4=\frac78 x$

Solving for $x$, I am getting $-27$ which is not possible.

Where did I go wrong?

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Hmm, I get sensible answers. Giving you a first step, did you get $x = \frac{7}{8}x + 9$? – Jason Knapp Jul 29 '14 at 19:24
up vote 4 down vote accepted

$$y = x-5 \;\text{and}\; y - 4 = \frac 78 x \implies y - 4 = \underbrace{(x-5)}_{\large y} - 4 = \frac 78 x$$

Multiplying both sides of the equation by $8$ gives us $$\begin{align} 8(x-5) - 8\cdot 4 = 7x & \iff 8x - 72 -7x = 0 \\ &\iff x = 72\text{ kg}.\end{align}$$

Now solve for $y = x - 5 = 72-5 = 67\;\text{ kg}$.

Recall that $x, y$ give the weights prior to losing weight. So we need to find current weights: Mr: $x- 5 = 72 - 5 = 67$, Mrs: $67 - 4 = 63$.

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Perhaps I am wrong, but plugging $x=44$ back into the original equations doesn't work out. – Vincent Jul 29 '14 at 19:30
But the value of x in the back of the book is 67 kg. – Abhishekstudent Jul 29 '14 at 19:31
But the question was to find their present weights, which are $x-5$ and $y-4$. – Per Manne Jul 29 '14 at 19:59
Indeed, @PerManne! – amWhy Jul 29 '14 at 20:03
@Abhishekstudent Just to clarify. You (we) set up the problem using $x, y$ to represent Mr and Mrs's weights, respectively, prior to losing weight. We are given that Mr. lost 5 kg, and Mrs lost 4 kg. Thus, since we are asked to find their present weights, Mr's present weight is $72 - 5 = 67$ kg. Mrs's present weight is $67 - 4 = 63$ kg. That should explain the discrepancy. – amWhy Jul 30 '14 at 12:14



$y=\frac78 x +4$

Set them equal to each other, solve for $x$. Then substitute $x$ into one of the original equations to find $y$.

Should find that $x=72$ kg and $y=67$ kg.

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Yes! That's exactly what I am getting. But the answer is not matching the answer at the back of the book. The answer according to the book should be 67 and 63. – Abhishekstudent Jul 29 '14 at 19:34
You can see that $67$ and $63$ obviously do not work out. Plug them into the first equation and you can see that $67-5=62$, not $63$. – Vincent Jul 29 '14 at 19:35
@Abhishekstudent Read the question more carefully. $67$ and $63$ kg are the present weights of mr. and mrs Ahuja, i.e., after the dieting course. – Per Manne Jul 29 '14 at 19:56

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