Question

1. 300 grams of type a raisin bran is mixed with 500 grams of type b raisin bran to produce a mixture which is 11% raisins. type a raisin bran has twice as many raisins per kilogram as type b. what percentage of raisins are in each type of raisin bran?

Explanation:

Step1: Find raisins in Type A

Type A is 300 grams and 11% raisins. So raisins in A: $300\times0.11 = 33$ grams.

Step2: Let raisins in B be $x$ grams

Type A has twice as many raisins per kg as Type B. First, find raisins per kg in A: $\frac{33}{0.3}$ (since 300g = 0.3kg) = 110 grams per kg. So raisins per kg in B is $\frac{110}{2}=55$ grams per kg. Type B is 500g = 0.5kg, so raisins in B: $55\times0.5 = 27.5$ grams? Wait, no, wait. Wait, the problem says "Type A Raisin Bran has twice as many raisins per kilogram as Type B". Wait, maybe better to let percentage in B be $p$, so percentage in A is $2p$? Wait, no, original Type A is 11%? Wait, no, wait the problem: "300 grams of Type A Raisin Bran is mixed with 500 grams of Type B Raisin Bran to produce a mixture which is 11% raisins. Type A Raisin Bran has twice as many raisins per kilogram as Type B. What percentage of raisins are in each type of Raisin Bran?"

Oh, I misread. Let's redefine. Let percentage of raisins in Type B be $x$ (so per kg, it's $10x$ grams? Wait, no, percentage: if percentage is $x\%$, then per kg, it's $x$ grams per 100 grams? No, percentage is (mass of raisins / mass of bran) * 100. So let’s let:

Let percentage of raisins in Type B be $p$ (so in 1 kg of Type B, raisins are $p\%$ of 1000g = $10p$ grams? Wait, no: $p\%$ means $\frac{p}{100}$ of the mass is raisins. So in $m$ grams of Type B, raisins are $\frac{p}{100} \times m$ grams.

Type A has twice as many raisins per kilogram as Type B, so percentage of raisins in Type A is $2p$ (since per kg, raisins in A is $2p\%$ of 1000g = $20p$ grams, and in B is $10p$ grams, so twice).

Now, total mass of mixture: 300g + 500g = 800g.

Total raisins in mixture: 11% of 800g = $0.11 \times 800 = 88$ grams.

Raisins in Type A: $\frac{2p}{100} \times 300$ grams (since Type A is 300g, percentage $2p$).

Raisins in Type B: $\frac{p}{100} \times 500$ grams (Type B is 500g, percentage $p$).

So equation: $\frac{2p}{100} \times 300 + \frac{p}{100} \times 500 = 88$

Simplify:

$6p + 5p = 88$ (because $\frac{2p \times 300}{100} = 6p$, $\frac{p \times 500}{100} = 5p$)

$11p = 88$

$p = 8$

So percentage in Type B is 8%, percentage in Type A is $2p = 16\%$? Wait, but wait, original Type A was said to be 300g, but when we calculated, we assumed percentage in A is $2p$. Wait, let's check:

Raisins in A: 16% of 300g = $0.16 \times 300 = 48$ grams.

Raisins in B: 8% of 500g = $0.08 \times 500 = 40$ grams.

Total raisins: 48 + 40 = 88 grams. 88 grams is 11% of 800g (since 0.11*800=88). Correct. And Type A has 16% per kg, Type B has 8% per kg, so 16% is twice 8%. Correct.

Wait, but initially I misread the problem. So the steps:

1. Let percentage in B be $p$, so percentage in A is $2p$ (since twice per kg).

1. Total mass of mixture: 300 + 500 = 800g.

1. Total raisins in mixture: 11% of 800 = 88g.

1. Raisins in A: $2p\%$ of 300g = $\frac{2p}{100} \times 300 = 6p$ grams.

1. Raisins in B: $p\%$ of 500g = $\frac{p}{100} \times 500 = 5p$ grams.

1. Equation: 6p + 5p = 88 => 11p = 88 => p = 8.

So Type A: 16%, Type B: 8%.

Answer:

Type A: 16%, Type B: 8%