Question

A milkman buys milk contained in $10$ vessels of equal size. If he sells his milk at ₹ 5 per litre, he loses ₹ 200; if he sells it at ₹ 6 per litre, he would gain ₹ 150 on the whole. Find the number of litres contained in each vessel.

• A. $20$ litres
• B. $30$ litres
• C. $25$ litres
• D. $35$ litres

Hint:

Translate "loss" as $C-R=\text{loss}$ and "gain" as $R-C=\text{gain}$. Setting up two linear equations in $c$ and $x$ makes such mixture/price problems straightforward.

Answer 1

The Correct Option is D

Step 1 (Variables).
Let each vessel contain $x$ litres $\Rightarrow$ total quantity $=10x$ litres.
Let cost price per litre be ₹ $c$ $\Rightarrow$ total cost $C=10x\,c$. Step 2 (Loss at ₹ 5 per litre).
Revenue at ₹ 5 per litre: $R_1=5\cdot 10x=50x$.
Loss ₹ 200 means $C-R_1=200$: \[ 10x\,c-50x=200 \Rightarrow x(c-5)=20 \Rightarrow \boxed{\,c-5=\dfrac{20}{x}\,}. \tag{1} \] Step 3 (Gain at ₹ 6 per litre).
Revenue at ₹ 6 per litre: $R_2=6\cdot 10x=60x$.
Gain ₹ 150 means $R_2-C=150$: \[ 60x-10x\,c=150 \Rightarrow 6-c=\frac{15}{x} \Rightarrow \boxed{\,c=6-\dfrac{15}{x}\,}. \tag{2} \] Step 4 (Solve for $x$).
From (1): $c=5+\dfrac{20}{x}$. Equate with (2): \[ 5+\frac{20}{x}=6-\frac{15}{x} \Rightarrow \frac{20}{x}+\frac{15}{x}=1 \Rightarrow \frac{35}{x}=1 \Rightarrow \boxed{x=35}. \] Step 5 (Answer).
Each vessel contains $\boxed{35\ \text{litres}}$.