Question 56

# A tea shop offers tea in cups of three different sizes. The product of the prices, in 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 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 INR, is

Solution

Let price of smallest cup be 2x and medium be 5x and large be y
Now by condition  1
we get $$2x\ \times\ \ 5x\ \times\ y\ =800$$
we get $$x^2y\ =80$$    (1)
Now as per second condition ;
$$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$$        (2)
Now dividing (2) and (1)
we get $$\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$$
we get $$10x^2+42x+36\ =\ 40x^2$$
we get $$\ 30x^2-42x-36=0$$
$$5x^2-7x-6=0$$
we get x=2
So 2x=4 and 5x=10
Now substituting in (1) we get y =20
Now therefore sum = 4+10+20 =34