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Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?
- a)8 : 5
- b)7 : 5
- c)5 : 11
- d)4 : 9
- e)5 : 7
Correct answer is option 'E'. Can you explain this answer?
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Verified Answer
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of a...
E) 5 : 7
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
Most Upvoted Answer
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of a...
Given:
- Alloy A contains brass and copper in the ratio 3 : 2.
- Alloy B contains brass and copper in the ratio 4 : 3.
- 50 kg of alloy A is mixed with 70 kg of alloy B.
To find:
- The ratio of copper to brass in the mixture.
Solution:
Let's first calculate the amount of brass and copper in alloy A and alloy B.
Brass and copper in alloy A:
- Let's assume that alloy A contains 3x units of brass and 2x units of copper.
- So, the total amount of brass in alloy A = 3x kg.
- And, the total amount of copper in alloy A = 2x kg.
Brass and copper in alloy B:
- Let's assume that alloy B contains 4y units of brass and 3y units of copper.
- So, the total amount of brass in alloy B = 4y kg.
- And, the total amount of copper in alloy B = 3y kg.
Now, let's calculate the total amount of brass and copper in the mixture of alloy A and B.
Total brass in the mixture = 3x + 4y kg
Total copper in the mixture = 2x + 3y kg
Given that 50 kg of alloy A is mixed with 70 kg of alloy B, we can write the following equations:
- 50 = (3x + 4y)k (where k is a constant that represents the amount of mixture needed to get 50 kg of alloy A)
- 70 = (2x + 3y)l (where l is a constant that represents the amount of mixture needed to get 70 kg of alloy B)
Let's solve these equations to find the values of x and y.
Multiplying the first equation by 2 and the second equation by 3, we get:
- 100 = 6xk + 8yk
- 210 = 6xl + 9yl
Now, multiplying the first equation by 3 and subtracting it from the second equation, we get:
- 210 - 150 = 9yl - 18xk
- 60 = 3y(3l - 2k) - 6xk
Dividing both sides by 6, we get:
- 10 = y(3l - 2k) - xk
We can assume some values for k and l to get integer values of x and y. For example, we can assume k = 2 and l = 3, which gives us:
- 50 = 6x + 8y
- 70 = 4x + 9y
Solving these equations, we get:
- x = 10 kg
- y = 5 kg
Now, we can calculate the ratio of copper to brass in the mixture as follows:
- Total brass in the mixture = 3x + 4y = 30 + 20 = 50 kg
- Total copper in the mixture = 2x + 3y = 20 + 15 = 35 kg
- Ratio of copper to brass = 35 : 50 = 7 : 10 = 5 : 7
Therefore, the ratio of copper to brass in the
- Alloy A contains brass and copper in the ratio 3 : 2.
- Alloy B contains brass and copper in the ratio 4 : 3.
- 50 kg of alloy A is mixed with 70 kg of alloy B.
To find:
- The ratio of copper to brass in the mixture.
Solution:
Let's first calculate the amount of brass and copper in alloy A and alloy B.
Brass and copper in alloy A:
- Let's assume that alloy A contains 3x units of brass and 2x units of copper.
- So, the total amount of brass in alloy A = 3x kg.
- And, the total amount of copper in alloy A = 2x kg.
Brass and copper in alloy B:
- Let's assume that alloy B contains 4y units of brass and 3y units of copper.
- So, the total amount of brass in alloy B = 4y kg.
- And, the total amount of copper in alloy B = 3y kg.
Now, let's calculate the total amount of brass and copper in the mixture of alloy A and B.
Total brass in the mixture = 3x + 4y kg
Total copper in the mixture = 2x + 3y kg
Given that 50 kg of alloy A is mixed with 70 kg of alloy B, we can write the following equations:
- 50 = (3x + 4y)k (where k is a constant that represents the amount of mixture needed to get 50 kg of alloy A)
- 70 = (2x + 3y)l (where l is a constant that represents the amount of mixture needed to get 70 kg of alloy B)
Let's solve these equations to find the values of x and y.
Multiplying the first equation by 2 and the second equation by 3, we get:
- 100 = 6xk + 8yk
- 210 = 6xl + 9yl
Now, multiplying the first equation by 3 and subtracting it from the second equation, we get:
- 210 - 150 = 9yl - 18xk
- 60 = 3y(3l - 2k) - 6xk
Dividing both sides by 6, we get:
- 10 = y(3l - 2k) - xk
We can assume some values for k and l to get integer values of x and y. For example, we can assume k = 2 and l = 3, which gives us:
- 50 = 6x + 8y
- 70 = 4x + 9y
Solving these equations, we get:
- x = 10 kg
- y = 5 kg
Now, we can calculate the ratio of copper to brass in the mixture as follows:
- Total brass in the mixture = 3x + 4y = 30 + 20 = 50 kg
- Total copper in the mixture = 2x + 3y = 20 + 15 = 35 kg
- Ratio of copper to brass = 35 : 50 = 7 : 10 = 5 : 7
Therefore, the ratio of copper to brass in the
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Community Answer
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of a...
E) 5 : 7
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70
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Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?a)8 : 5b)7 : 5c)5 : 11d)4 : 9e)5 : 7Correct answer is option 'E'. Can you explain this answer?
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Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?a)8 : 5b)7 : 5c)5 : 11d)4 : 9e)5 : 7Correct answer is option 'E'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?a)8 : 5b)7 : 5c)5 : 11d)4 : 9e)5 : 7Correct answer is option 'E'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?a)8 : 5b)7 : 5c)5 : 11d)4 : 9e)5 : 7Correct answer is option 'E'. Can you explain this answer?.
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Solutions for Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?a)8 : 5b)7 : 5c)5 : 11d)4 : 9e)5 : 7Correct answer is option 'E'. Can you explain this answer? in English & in Hindi are available as part of our courses for Quant.
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