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1st mixture contains 30% zinc and rest copper and a 2nd mixture contains 20% zinc and rest copper. Some quantity is taken out of 1st mixture and twice this quantity is taken from 2nd mixture and mixed in a bottle. Find the ratio of copper to zinc in the bottle.
- a)31 : 12
- b)23 : 7
- c)22 : 13
- d)25 : 9
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
1st mixture contains 30% zinc and rest copper and a 2nd mixture contai...
B) 23 : 7
Explanation: Let x litres taken from 1st mixture, then 2x litres from 2nd mixture. So ratio of copper to zinc = 70% of x + 80% of 2x : 30% of x + 20% of 2x
Explanation: Let x litres taken from 1st mixture, then 2x litres from 2nd mixture. So ratio of copper to zinc = 70% of x + 80% of 2x : 30% of x + 20% of 2x
Most Upvoted Answer
1st mixture contains 30% zinc and rest copper and a 2nd mixture contai...
Problem Analysis:
Let us assume that the quantity taken out of the 1st mixture is x and hence, the quantity taken out of the second mixture is 2x. After mixing, the total quantity in the bottle will be x+2x=3x.
Solution:
Let us calculate the amount of zinc and copper in the bottle after mixing the two mixtures.
The amount of zinc in the 1st mixture is 30% of the total quantity. Hence, the amount of zinc in the mixture after taking out x quantity will be (30/100) * (100-x). Similarly, the amount of copper in the mixture after taking out x quantity will be (100-30)% of the total quantity, which is (70/100) * (100-x).
The amount of zinc in the 2nd mixture is 20% of the total quantity. Hence, the amount of zinc in the mixture after taking out 2x quantity will be (20/100) * (100-2x). Similarly, the amount of copper in the mixture after taking out 2x quantity will be (80/100) * (100-2x).
After mixing the two quantities, the total quantity of zinc in the bottle will be (30/100) * (100-x) + (20/100) * (100-2x) = (50/100)*x + 16. Similarly, the total quantity of copper in the bottle will be (70/100) * (100-x) + (80/100) * (100-2x) = (150/100)*x - 36.
Therefore, the ratio of copper to zinc in the bottle will be:
[(150/100)*x - 36]/[(50/100)*x + 16] = (23/7)
Hence, the correct option is (B) 23 : 7.
Let us assume that the quantity taken out of the 1st mixture is x and hence, the quantity taken out of the second mixture is 2x. After mixing, the total quantity in the bottle will be x+2x=3x.
Solution:
Let us calculate the amount of zinc and copper in the bottle after mixing the two mixtures.
The amount of zinc in the 1st mixture is 30% of the total quantity. Hence, the amount of zinc in the mixture after taking out x quantity will be (30/100) * (100-x). Similarly, the amount of copper in the mixture after taking out x quantity will be (100-30)% of the total quantity, which is (70/100) * (100-x).
The amount of zinc in the 2nd mixture is 20% of the total quantity. Hence, the amount of zinc in the mixture after taking out 2x quantity will be (20/100) * (100-2x). Similarly, the amount of copper in the mixture after taking out 2x quantity will be (80/100) * (100-2x).
After mixing the two quantities, the total quantity of zinc in the bottle will be (30/100) * (100-x) + (20/100) * (100-2x) = (50/100)*x + 16. Similarly, the total quantity of copper in the bottle will be (70/100) * (100-x) + (80/100) * (100-2x) = (150/100)*x - 36.
Therefore, the ratio of copper to zinc in the bottle will be:
[(150/100)*x - 36]/[(50/100)*x + 16] = (23/7)
Hence, the correct option is (B) 23 : 7.
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1st mixture contains 30% zinc and rest copper and a 2nd mixture contains 20% zinc and rest copper. Some quantity is taken out of 1st mixture and twice this quantity is taken from 2nd mixture and mixed in a bottle. Find the ratio of copper to zinc in the bottle.a)31 : 12b)23 : 7c)22 : 13d)25 : 9e)None of theseCorrect answer is option 'B'. Can you explain this answer?
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