Question

Metal A costs Rs. 8.40 per gm and metal B Rs. 0.21 per gm. In what proportion must these metals be mixed so that 1 gram of the mixture may be worth Rs. 5.67?

• A. 1 : 2
• B. 1 : 3
• C. 2 : 1
• D. 2 : 3

Hint:

When solving mixture problems, set up an equation for the cost of the mixture per unit weight, then solve for the proportion of the components using the given prices.

Answer 1

The Correct Option is C

Let the amount of metal A and metal B be \( x \) gm and \( y \) gm, respectively. The total cost of \( x \) gm of metal A and \( y \) gm of metal B will be: \[ \text{Total cost} = 8.40x + 0.21y \] The total weight of the mixture is \( x + y \) grams, and the price of the mixture is given as Rs. 5.67 per gram. So the total cost of 1 gram of the mixture is: \[ \frac{8.40x + 0.21y}{x + y} = 5.67 \] Multiplying both sides by \( x + y \), we get: \[ 8.40x + 0.21y = 5.67(x + y) \] Expanding the right side: \[ 8.40x + 0.21y = 5.67x + 5.67y \] Now, bring the terms involving \( x \) to one side and the terms involving \( y \) to the other side: \[ 8.40x - 5.67x = 5.67y - 0.21y \] Simplifying: \[ 2.73x = 5.46y \] Dividing both sides by \( 2.73 \): \[ \frac{x}{y} = \frac{5.46}{2.73} = 2 \] So, the ratio of metal A to metal B is \( 2 : 1 \). Final Answer: The correct answer is (c) 2 : 1.