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?

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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.
Updated On: Mar 10, 2025
  • 1 : 2
  • 1 : 3
  • 2 : 1
  • 2 : 3
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The Correct Option is C

Solution and Explanation

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.
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