Question

Two boxes a and b were filled with coffee and chicory, mixed in a in the ratio 5 : 3 and in b in the ratio 7 : 3. what quantity (in kg) must be taken from each to form a mixture which shall contain 6 kg of coffee and 3 kg of chicory, respectively?

Answer 1

let x amount is taken from box a and y amount taken from b
the amount of coffee is
5/8x+ 7/10y= 6
the amount of chicory is
3/8x + 3/10y = 3
by solving equations
x= 4 and y=5

Answer 2

Answer:

So the quantity should be taken is 4:5

Solution:

Let us assume that x and y quantity is taken from the mixture.

Mixture should contain 6kg of coffee, so

$\frac{5 x}{8}+\frac{7 y}{10}=6$

$50 x+56 y=6 \times 80$

$25x + 28y = 240$---------- (i)

Mixture should contain 3kg of chicory, so,

$\frac{3 x}{8}+\frac{3 y}{10}=3$

$10x +8y = 80$

$30x +24y =240$ ----- (ii) (// multiplying both side by 3)

Hence from (i) and (ii),

$25x +28y =30x + 24y$

$30x -25x = 28y -24y$

$5x = 4y$

$\frac{x}{y}=\frac{4}{5}$

$\therefore x: y=4: 5$