CAT 2023Slot 2QAQuestion & Solution
Question
Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs.51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is
Solution
1. Concept Used
- Topic: Profit and Loss combined with Linear Diophantine Equations
- Formula: $$\text{Total Profit} = \text{Selling Price} - \text{Cost Price} = \frac{1}{8} \times (1000m + 1125n) = 51000$$
2. Calculation
Let the number of white shirts be $m$ and the number of blue shirts be $n$, where both $m, n$ are positive integers (since he bought both colors).
The total cost of all shirts is $1000m + 1125n$, and the total number of shirts is $m + n$.
The average cost per shirt is: $$\text{Avg Cost} = \frac{1000m + 1125n}{m + n}$$
The market price (MP) is set at 25% above the average cost: $$\text{MP} = \frac{5}{4} \times \frac{1000m + 1125n}{m + n}$$
He then sells at a 10% discount on MP, so the selling price per shirt is: $$\text{SP} = \frac{9}{10} \times \frac{5}{4} \times \frac{1000m + 1125n}{m + n} = \frac{9}{8} \times \frac{1000m + 1125n}{m + n}$$
The profit per shirt is: $$\text{Profit per shirt} = \text{SP} - \text{Avg Cost} = \frac{9}{8} \cdot \frac{1000m+1125n}{m+n} - \frac{1000m+1125n}{m+n} = \frac{1}{8} \cdot \frac{1000m+1125n}{m+n}$$
The total profit is: $$\text{Total Profit} = \frac{1}{8} \cdot \frac{1000m+1125n}{m+n} \times (m+n) = \frac{1000m + 1125n}{8}$$
Setting this equal to Rs. 51000: $$\frac{1000m + 1125n}{8} = 51000$$ $$1000m + 1125n = 408000$$
Dividing throughout by 125: $$8m + 9n = 3264$$
Solving for $m$: $$m = \frac{3264 - 9n}{8}$$
For $m$ to be a positive integer, $(3264 - 9n)$ must be divisible by 8. Since $3264 = 8 \times 408$ is already divisible by 8, we need $9n$ to be divisible by 8, which means $n$ must be divisible by 8.
So let $n = 8k$ where $k \geq 1$ (since $n \geq 1$ and $n$ must be a multiple of 8).
Then: $$m = \frac{3264 - 9(8k)}{8} = \frac{3264 - 72k}{8} = 408 - 9k$$
For $m \geq 1$: $408 - 9k \geq 1 \Rightarrow k \leq 45.2$, so $k \leq 45$.
The total number of shirts is: $$m + n = (408 - 9k) + 8k = 408 - k$$
To maximize $m + n$, we need to minimize $k$. The minimum value of $k$ is $1$ (since $n \geq 1$ and must be a multiple of 8, so $n_{\min} = 8$).
With $k = 1$: $n = 8$, $m = 408 - 9 = 399$
$$\text{Maximum total shirts} = 408 - 1 = 407$$
Verification: $1000(399) + 1125(8) = 399000 + 9000 = 408000$ ✓ and $408000 / 8 = 51000$ ✓
3. Solution
Answer = 407 ✅
The maximum possible total number of shirts Jayant could have bought is 407 (399 white shirts + 8 blue shirts).