Solving method for fractions having Prime numbers as Denominator.

Question

I have a question , like

$30 = 1.18x$ , and $30 = 0.82 y$ , find $x+y$.

when solving, $x=1500/59$, $y=1500/41$. Since the denominator is involving prime numbers I was wondering is there any method or some logic is there to solve this question under 30 seconds.?

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The actual question goes like this :

A man sold two gifts at $\$30$ each. On one gift he gained $18\%$ and on the other gift he lost $18\%$. What is his overall gain/loss(in $)

Answer 1

Continuing from an4s's answer:

$$\frac{60}{1 - 0.18^2} = \frac{60}{1 - \left(\frac{9}{50} \right)^2} = \frac{60}{1 - \frac{81}{2500}} = \frac{60}{\frac{2500 - 81}{2500}} = \frac{60 \cdot 2500}{2500 - 81} = \frac{150 \ 000}{2419}.$$

Then using the options in the multiple-choice question:

$$\frac{150 \ 000}{2500} < \frac{150 \ 000}{2419} < \frac{150 \ 000}{2400}$$ $$\frac{150}{25} < \frac{150 \ 000}{2419} < \frac{500}{8}$$ $$60 < \frac{150 \ 000}{2419} < 62.5$$

Since the gifts are between $60$ and $62.5$ dollars, he has therefore made a loss. Then the answer must be between options $B$ and $C$. However, since his loss is strictly less than $60 - 62.5 = -2.5$ dollars, then the answer is option $B$.

Answer 2

Hint:

$$\begin{align}30 &= 1.18x \implies x = \dfrac{30}{1.18} \\ 30 &= 0.82y \implies y = \dfrac{30}{0.82} \end{align}$$

Therefore, $$x + y = \dfrac{30}{1.18} + \dfrac{30}{0.82} = \dfrac{30(1.18 + 0.82)}{1.18\cdot 0.82}$$

Now, notice that

$$\begin{align}1.18&\equiv1 + 0.18\\ 0.82&\equiv 1 - 0.18\end{align}$$

and $$(a + b)(a - b) = (a^2 - b^2).$$

Answer 3

If that was the original question then: $30=1.18x$ and $30=0.82y$ what is $x+y$--- is not the right question. The right question is what is $0.18x - 0.18y$.

If $g$ is $0.18X$ then gain on the first gift and $l = 0.18Y$ the loss on the second gift then:

$X +g = X + 0.18X = 30$ so $X = \frac {30}{1.18}$ and $g=0.18\frac {30}{1.18}$

And $Y - l = Y-0.18X = 30$ so $Y=\frac {30}{0.82}$ and we need $g-l= 0.18\cdot 30 \cdot (\frac 1{1.18} -\frac 1{0.82})=$

$0.18\cdot 30\cdot (\frac {1-0.18}{(1+0.18)(1-0.18)}- \frac {1+0.18}{(1+0.18)(1-0.18)})=0.18\cdot 30\cdot(\frac {-2\cdot 0.18}{1-0.18^2})=$

$-60\cdot \frac {0.18^2}{1-0.18^2}=$

$-60\cdot \frac {18^2}{10000-18^2}=$

$-60\cdot \frac {9^2}{2500-9^2}=$

$-60\cdot \frac {81}{2419}=$

$-\frac {4860}{2419} \approx 2.01$.

..... or we could have do this from the start.

$X=\frac {30}{1.18} \approx 25.43$ so he made $30-25.43 = 4.57$

$Y = \frac {30}{0.82} \approx 36.59$ so he lost $36.59-30 = 6.59$ cents

So his loss was $-6.59 + 4.57 = -2.02$ (a one cent rounding error).

Is there a way to solve this 30 seconds by evaluating $\frac {30}{1.18} = {1500}{59}$ and $\frac {30}{0.82} = \frac {1500}{41}$?

I.e. $(\frac {1500}{59}-\frac {1500}{41}) = 1500\frac {41-59}{59*41}= 1500\frac {-18}{(50+9)(50-9)}= -18*1500*\frac 1{2500-81}$ and

.... I dunno....Seems about the same amount of work as my first method. The highlighted method is probably the fastest and easiest.