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CAT 2021 Slot 3 — QA Question 18

Solving Quadratic EquationsEasy

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

Answer & solution

Answer: 34

Solution

Easy

Let the small and medium prices be 2x2x and 5x5x (ratio 2:52:5) and the large price be pp. The two product conditions — original =800=800 and increased =3200=3200 — give a ratio of 44, which removes pp and leaves a quadratic in xx.

1

Set up the two products.

2x⋅5x⋅p=800...(1)(2x+6)(5x+6) p=3200...(2)\begin{aligned} &2x\cdot 5x\cdot p = 800 \quad\text{...(1)}\\ &(2x+6)(5x+6)\,p = 3200 \quad\text{...(2)} \end{aligned}
2

Divide (2) by (1). pp cancels and the right side is 44.

(2x+6)(5x+6)10x2=4⇒ 10x2+42x+36=40x2⇒ 30x2−42x−36=0⇒ 5x2−7x−6=0\begin{aligned} &\frac{(2x+6)(5x+6)}{10x^2} = 4\\ &\Rightarrow\ 10x^2 + 42x + 36 = 40x^2\\ &\Rightarrow\ 30x^2 - 42x - 36 = 0\\ &\Rightarrow\ 5x^2 - 7x - 6 = 0 \end{aligned}
3

Solve for xx. Factor the quadratic.

(5x+3)(x−2)=0⇒ x=2(reject x=−35, prices>0)\begin{aligned} &(5x+3)(x-2)=0\\ &\Rightarrow\ x = 2 \quad\text{(reject }x=-\tfrac35,\ \text{prices}>0) \end{aligned}
4

Recover the three prices and sum. Use (1) for pp.

small=2x=4,medium=5x=10⇒ p=8004⋅10=20⇒ Sum=4+10+20=34\begin{aligned} &\text{small}=2x=4,\quad \text{medium}=5x=10\\ &\Rightarrow\ p = \frac{800}{4\cdot 10} = 20\\ &\Rightarrow\ \text{Sum} = 4+10+20 = 34 \end{aligned}
Sum=34\text{Sum} = 34
CAT 2021 Slot 3 QA Q18: A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different — Solution | TheCATExam