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An alloy A contains two elements, copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of the pure copper mixed in the alloy C?
  • a)
    3 kg
  • b)
    4 kg
  • c)
    5 kg
  • d)
    7 kg
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
An alloy A contains two elements, copper and tin in the ratio of 2 : 3...
Let the amount of pure copper added be ‘x’ kg
Amount of copper in 20 kg of alloy A = 2/ (2 + 3) × 20 = 8 kg
Amount of copper in 28 kg of alloy B = 3/ (3 + 4) × 28 = 12 kg
Total amount of copper in alloy C = 8 + 12 + x = 20 + x
Total amount of alloy C produced = 20 + 28 + x = 48 + x
20 + x = 6/(6 + 7) × (48 + x)
20 + x = 6/13 × (48 + x)
260 + 13x = 288 + 6x
7x = 28
∴ x = 4 kg
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Most Upvoted Answer
An alloy A contains two elements, copper and tin in the ratio of 2 : 3...
Given information:
- Alloy A contains copper and tin in the ratio of 2:3.
- Alloy B contains copper and tin in the ratio of 3:4.
- Alloy C contains copper and tin in the ratio of 6:7.
- 20 kg of alloy A is mixed with 28 kg of alloy B to form alloy C.

Calculating the quantities of copper and tin in alloys A, B, and C:
Let's assume that alloy A contains 2x kg of copper and 3x kg of tin.
Similarly, alloy B contains 3y kg of copper and 4y kg of tin.
And alloy C contains 6z kg of copper and 7z kg of tin.

Calculating the total quantity of copper and tin in alloy C:
The total quantity of copper in alloy C is the sum of the quantities of copper in alloy A, alloy B, and the pure copper added.
Total copper in alloy C = 2x + 3y + pure copper

The total quantity of tin in alloy C is the sum of the quantities of tin in alloy A and alloy B.
Total tin in alloy C = 3x + 4y

Given that the ratio of copper to tin in alloy C is 6:7, we can write the following equation:
(2x + 3y + pure copper) / (3x + 4y) = 6/7

Calculating the quantities of copper and tin in alloy A and B:
From the given ratios, we can write the following equations:
2x / 3x = 2/3
3y / 4y = 3/4

Simplifying these equations, we get:
x = (3/2)(2x) = (2/3)(3x)
y = (4/3)(3y) = (3/4)(4y)

This means that the quantities of copper and tin in alloy A and alloy B are in the same ratio.

Calculating the quantity of pure copper added:
We have the equation:
(2x + 3y + pure copper) / (3x + 4y) = 6/7

Since the quantities of copper and tin in alloy A and alloy B are in the same ratio, we can substitute their values:
(2x + 3y + pure copper) / (3x + 4y) = 6/7
(2x + 3y + pure copper) / (2x + 3y) = 6/7

Simplifying this equation, we get:
(2x + 3y + pure copper) = (6/7)(2x + 3y)

Expanding and simplifying further, we get:
7(2x + 3y + pure copper) = 6(2x + 3y)
14x + 21y + 7(pure copper) = 12x + 18y

Subtracting 12x + 18y from both sides, we get:
2x + 3y + 7(pure copper) = 0

Rearranging the equation, we get:
7(pure copper) = -(2x + 3y)

Dividing both sides by
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An alloy A contains two elements, copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of the pure copper mixed in the alloy C?a)3 kgb)4 kgc)5 kgd)7 kgCorrect answer is option 'B'. Can you explain this answer?
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An alloy A contains two elements, copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of the pure copper mixed in the alloy C?a)3 kgb)4 kgc)5 kgd)7 kgCorrect answer is option 'B'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about An alloy A contains two elements, copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of the pure copper mixed in the alloy C?a)3 kgb)4 kgc)5 kgd)7 kgCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alloy A contains two elements, copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of the pure copper mixed in the alloy C?a)3 kgb)4 kgc)5 kgd)7 kgCorrect answer is option 'B'. Can you explain this answer?.
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