Question

A milkman has \(250\) litres of milk containing \(5\%\) fat. How many litres of milk containing \(15\%\) fat should he add to his stock so that the fat content in the mixture would be more than \(7\%\) but less than \(10\%\)?

• A. More than \(62.5\) litres but less than \(250\) litres
• B. More than \(100\) litres but less than \(200\) litres
• C. More than \(62.5\) litres but less than \(200\) litres
• D. More than \(100\) litres but less than \(250\) litres

Hint:

In mixture problems: \[ \text{New Percentage} = \frac{\text{Total amount of pure substance}} {\text{Total mixture}} \times 100 \] Form inequalities directly when the percentage is required to lie within a range.

Answer 1

The Correct Option is A

Step 1: Set up the unknown.
Let $x$ litres of $15\%$ fat milk be added to the $250$ litres of $5\%$ fat milk.

Step 2: Write the total fat.
Fat from the original milk is $250\times\tfrac{5}{100}=12.5$ litres. Fat from the added milk is $\tfrac{15}{100}x=0.15x$. Total fat is $12.5+0.15x$.

Step 3: Write the fat percentage of the mixture.
Total volume is $250+x$, so the fat percentage is $\dfrac{12.5+0.15x}{250+x}\times100$.

Step 4: Apply the lower bound (more than 7%).
$\dfrac{12.5+0.15x}{250+x}\times100>7$ gives $1250+15x>7(250+x)=1750+7x$, so $8x>500$ and $x>62.5$.

Step 5: Apply the upper bound (less than 10%).
$\dfrac{12.5+0.15x}{250+x}\times100<10$ gives $1250+15x<2500+10x$, so $5x<1250$ and $x<250$.

Step 6: Combine the two bounds.
Hence $62.5<x<250$: more than $62.5$ litres but less than $250$ litres. \[ \boxed{62.5<x<250\ \text{litres}} \]