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** **__Solve this :__

2. Two types of coffee C_{1} and C_{2} are blended in the ratio 3 : 13 by weight to form a standard blend X. A second standard blend, Y, is formed by blending C_{1}, C_{2} & C_{3} coffee in the ratio 1 : 7 : 12 by weight. Calculate :

(i) The ratio of the three types of the coffee in a third standard blend Z, formed by mixing equal weights of X and Y.

(ii) The weight of C_{1} coffee in 800g of Z.

__Solve this :__

Taking P and Q in place of C

_{1}and C

_{2}for simplicity,

For X, P : Q = 3 : 13, total 3 + 13 = 16 units

For Y, P : Q : R = 1 : 7 : 12, total 1 + 7 + 12 = 20 units

Since blend Z is formed by mixing equal weights of X and Y, the total number of units for X and Y must be the same in order for reasonable comparison (units must be of the same size!)

Lowest common multiple of 16 and 20 = 80

(16 x 5 = 80, 20 x 4 = 80)

Therefore, 5 'sets' of X and 4 'sets' of Y (are blended together to get Z. (Only then will there be equal weight of X and Y blended. Though of course, 10 and 8 sets and other integer multiples are possible too. However, since this is about ratio, eventually, you will still need to simplify; might as well work with the lowest common multiple!)

3 : 13 = 15 : 65

1 : 7 : 12 = 4 : 28 : 48

Add up the P and the Q...

Final answer is 15 + 4 : 65 + 28 : 48 = 19 : 93 : 48

Once you get your answer for (i), (ii) is not difficult...

From (i), we know that for blend Z, P : Q : R = 19 : 93 : 48

Total of 19 + 93 + 48 = 160 units

(notice how it is twice of 80, the LCM of 16 and 20)

Therefore, weight of type P coffee in 800g of blend Z is simply

=$\frac{19}{160}\times 800=95g$

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