CAT 2023

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Profit and Loss — Relating Cost Price, Selling Price, and Profit/Loss Percentages
• Formula: $$ \text{Selling Price} = \text{Cost Price} \times \left(1 + \frac{\text{Profit%}}{100}\right), \quad \text{Profit%} = \frac{\text{SP} - \text{CP}}{\text{CP}} \times 100 $$

2. Calculation

Let the initial common selling price of both objects A and B be $p$.

Finding Cost Price of A: Gita makes a 20% profit on A, so $\text{SP}_A = 1.2 \times \text{CP}_A = p$, which gives $\text{CP}_A = \dfrac{p}{1.2} = \dfrac{5p}{6}$.

Finding Cost Price of B: Gita makes a 10% loss on B, so $\text{SP}_B = 0.9 \times \text{CP}_B = p$, which gives $\text{CP}_B = \dfrac{p}{0.9} = \dfrac{10p}{9}$.

Finding the New Selling Price: Now she increases the selling price so that a 10% profit is made on B. The new common selling price $s$ is: $$s = 1.10 \times \text{CP}_B = \frac{11}{10} \times \frac{10p}{9} = \frac{11p}{9}$$

Finding Profit % on A at new selling price: $$\text{Profit% on A} = \frac{s - \text{CP}_A}{\text{CP}_A} \times 100 = \frac{\dfrac{11p}{9} - \dfrac{5p}{6}}{\dfrac{5p}{6}} \times 100$$

First, compute the numerator: $\dfrac{11p}{9} - \dfrac{5p}{6} = p\left(\dfrac{11}{9} - \dfrac{5}{6}\right) = p\left(\dfrac{22}{18} - \dfrac{15}{18}\right) = \dfrac{7p}{18}$

Now divide by $\text{CP}_A = \dfrac{5p}{6}$: $$\text{Profit%} = \frac{\dfrac{7p}{18}}{\dfrac{5p}{6}} \times 100 = \frac{7p}{18} \times \frac{6}{5p} \times 100 = \frac{7 \times 6}{18 \times 5} \times 100 = \frac{42}{90} \times 100 = \frac{140}{3} \approx 46.67%$$

This is nearest to 47%.

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3. Solution

Answer = Option C (47%) ✅

The profit made on object A at the new selling price is approximately $\dfrac{140}{3}% \approx 46.67%$, which is nearest to 47%.