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A tea shop offers tea in cups of three different sizes. The product of the prices, in $\text{INR},$ of three different sizes is equal to $800.$ The prices of the smallest size and the medium size are in the ratio $2:5.$ If the shop owner decides to increase the prices of the smallest and the medium ones by $\text{INR} \; 6$ keeping the price of the largest size unchanged, the product then changes to $3200.$ The sum of the original prices of three different sizes, in $\text{INR},$ is

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Let the price of the smallest cup be $2a,$ and medium be $5a,$ and large be $b.$

Now, $2a \times 5a \times b = 800$

$\Rightarrow a^{2}b = 80 \quad \longrightarrow (1)$

And, $(2a+6) (5a+6)b = 3200 \quad \longrightarrow (2)$

Divide equations $(2)$ by equation $(1).$

$\Rightarrow \frac{(2a+6)(5a+6)b}{a^{2}b} = \frac{3200}{80}$

$\Rightarrow (2a+6)(5a+6) = 40 a^{2}$

$\Rightarrow 10a^{2} + 12a + 30a + 36 = 40a^{2}$

$\Rightarrow -30a^{2} + 42a + 36 = 0$

$\Rightarrow -5a^{2} + 7a + 6 = 0$

$\Rightarrow 5a^{2} - 7a - 6 = 0$

$\Rightarrow 5a^{2} - 10a + 3a - 6 = 0$

$\Rightarrow 5a(a – 2) – 3 (a - 2) = 0$

$\Rightarrow (a-2)(5a-3) = 0$

$\Rightarrow \boxed{a = 2}$

So,

- $2a = 4$
- $5a = 10$

Put the value of $a$ in equation $(1).$

$\Rightarrow a^{2}b = 80$

$\Rightarrow 2^{2}b = 80$

$\Rightarrow \boxed{b = 20}$

$\therefore$ The sum of the original prices of three different sizes $ = 2a + 5a + b = 4 + 10 + 20= 34.$

Correct Answer $:34$