A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is
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Official Correct Answer: D. First, the problem says there are two mixtures of coffee and cocoa. The first mixture is 16% coffee and costs Rs 240 per kg. The second mixture is 36% coffee and costs Rs 320 per kg. Then there's a new mixture that costs Rs 376 per kg, and We find out how much coffee is in 10 kg of this new mixture. The options are given, and the correct answer is D) 5, but I need to work through it.
Hmm, mixtures and alligation usually involve finding the ratio in which two or more ingredients are mixed. alligation is a method to find the ratio in which two or more ingredients at different prices must be mixed to produce a mixture at a given price.
So, Let the two mixtures as Mixture 1 and Mixture 2.
Mixture 1: 16% coffee, cost Rs 240/kg.
Mixture 2: 36% coffee, cost Rs 320/kg.
The new mixture costs Rs 376/kg. I think we need to find the ratio in which Mixture 1 and Mixture 2 are mixed to get this new price. Once we have that ratio, we can find out how much coffee is in the new mixture.
Wait, but the new mixture's percentage of coffee isn't given. Hmm, maybe We find the percentage of coffee in the new mixture first, and then calculate the quantity in 10 kg.
Alternatively, maybe I can use alligation directly on the coffee percentages. In alligation, we consider the differences in the percentages. So, the idea is to find the ratio in which the two mixtures are combined based on the cost prices and the percentage of coffee.
Wait, but in this case, the cost prices are given, and the percentage of coffee is given for each mixture. So, perhaps I need to set up equations based on the cost and the coffee content.
Let me denote:
Let’s say we mix x kg of Mixture 1 and y kg of Mixture 2 to get the new mixture. The total weight is x + y kg, and we know that the cost per kg is Rs 376. So, the total cost is 376(x + y).
The cost contributed by Mixture 1 is 240x, and by Mixture 2 is 320y. So, 240x + 320y = 376(x + y).
Let me write that equation:
240x + 320y = 376(x + y)
Simplify this:
240x + 320y = 376x + 376y
Bring like terms to one side:
240x - 376x + 320y - 376y = 0
-136x -56y = 0
Multiply both sides by -1:
136x + 56y = 0
Wait, that can't be right because x and y are positive quantities. Did I make a mistake in the signs?
Wait, let's go back. 240x + 320y = 376x + 376y
Subtract 240x and 320y from both sides:
0 = 136x + 56y
Which implies 136x + 56y = 0
But x and y can't be negative, so this suggests that the equation is not correct. Maybe I set up the equation wrong.
Wait, perhaps I should think in terms of the cost per kg. The cost of the new mixture is 376, which is a weighted average of 240 and 320 based on the quantities x and y.
So, the correct equation is:
(240x + 320y)/(x + y) = 376
Yes, that makes sense. So, cross-multiplying:
240x + 320y = 376x + 376y
Which simplifies to:
240x + 320y = 376x + 376y
Bring all terms to left:
240x - 376x + 320y - 376y = 0
-136x -56y = 0
Which is the same as before. Hmm, but this gives a negative equation. Maybe I should rearrange it differently.
Wait, perhaps I should express it as:
(240 - 376)x + (320 - 376)y = 0
Which is:
-136x -56y = 0
Divide both sides by -4:
34x + 14y = 0
Still, x and y can't be negative. So, maybe I need to approach this differently.
Alternatively, perhaps I should consider the percentage of coffee in the new mixture. Let the percentage of coffee in the new mixture as C%.
Then, the amount of coffee from Mixture 1 is 16% of x, which is 0.16x.
The amount of coffee from Mixture 2 is 36% of y, which is 0.36y.
The total coffee in the new mixture is 0.16x + 0.36y.
The total weight is x + y, so the percentage of coffee is (0.16x + 0.36y)/(x + y) = C%.
But I don't know C%. Hmm, maybe I can relate this to the cost.
Wait, perhaps I can use the cost equation and the coffee percentage equation together.
From the cost equation:
240x + 320y = 376(x + y)
Which simplifies to:
240x + 320y = 376x + 376y
Rearranged:
(240 - 376)x + (320 - 376)y = 0
-136x -56y = 0
Divide both sides by -4:
34x + 14y = 0
Wait, this still gives 34x = -14y, which would imply x and y are negative, which doesn't make sense. So, I must have made a mistake in setting up the equations.
Wait, maybe I should think about the cost per kg in terms of coffee and cocoa. Let the cost of coffee as C and cocoa as D.
But the problem doesn't give the individual costs of coffee and cocoa, only the costs of the mix