# Solve this : 2. Two types of coffee C1 and C2 are blended in the ratio 3 : 13 by weight to form a standard blend X. A second standard blend, Y, is formed by blending C1, C2 & C3 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 C1 coffee in 800g of Z.

Hi,
Taking P and Q in place of C1 and C2 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