Question

An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal C is

• A. 70
• B. 96
• C. 48
• D. 84

Answer 1

The Correct Option is D

Given:

• Volume ratio of metals A : B : C = 3 : 4 : 7
• Weight per unit volume (density) ratio of A : B : C = 5 : 2 : 6
• Total weight of the alloy = 130 kg

Let the common volume unit be \( x \). Then:

Weight of A = \( 3x \times 5 = 15x \)

Weight of B = \( 4x \times 2 = 8x \)

Weight of C = \( 7x \times 6 = 42x \)

Total weight of alloy:

\[ 15x + 8x + 42x = 65x = 130 \]

Solve for \( x \):

\[ x = \frac{130}{65} = 2 \]

Weight of Metal C:

\[ \text{Weight of C} = 42x = 42 \times 2 = \boxed{84 \text{ kg}} \]

✅ Final Answer: 84 kg

Answer 2

Let the volumes of metals A, B, and C be:

\[ \text{A: } 3x,\quad \text{B: } 4x,\quad \text{C: } 7x \]

Let the weights per unit volume (density equivalents) be:

\[ \text{A: } 5y,\quad \text{B: } 2y,\quad \text{C: } 6y \]

Then, their weights will be:

\[ \text{Weight of A = } 3x \cdot 5y = 15xy \] \[ \text{Weight of B = } 4x \cdot 2y = 8xy \] \[ \text{Weight of C = } 7x \cdot 6y = 42xy \]

Total weight of the alloy is given as:

\[ 15xy + 8xy + 42xy = 65xy = 130 \]

Solving for \(xy\):

\[ xy = \frac{130}{65} = 2 \]

Now, the weight of metal C is:

\[ 42xy = 42 \cdot 2 = \boxed{84 \text{ kg}} \]

✅ Final Answer: \( \boxed{84 \text{ kg}} \)