CAT 2021 Slot 3 · Quant · Question 17

CAT 2021 Slot 3 previous year question on Linear Equations, with the answer and a step-by-step solution.

Linear EquationsAlgebraType in the answer (TITA)Medium

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

Answer & solution

Answer: 34

Use the ratio to carry one unknown: let the small and medium prices be 2k and 5k, and the large price L.

Original product: 2k\cdot 5k\cdot L=10{k}^{2}L=800.

After the increase: \left( 2k+6 \right)\left( 5k+6 \right)L=3200.

Dividing the second by the first removes L entirely — the key step, since L is unchanged:

\frac{\left( 2k+6 \right)\left( 5k+6 \right)}{10{k}^{2}}=4,

so 10{k}^{2}+42k+36=40{k}^{2}, giving 30{k}^{2}-42k-36=0, or 5{k}^{2}-7k-6=0.

Factorising, \left( k-2 \right)\left( 5k+3 \right)=0, so k=2.

The small and medium prices are ₹4 and ₹10. From 10{k}^{2}L=800 we get 40L=800, so the largest size costs ₹20. The sum is 4+10+20=34.

CAT 2021 Slot 3 · Quant · Question 17

More Linear Equations questions