"> Word Problem 1 | Job Exam Rare Mathematics (JEMrare)

Word Problem 1

Problem 1: In what proportion must a man mix milk at Rs 11 per liter with milk at Rs 6 per liter, so that the mixture may be worth Rs 8 per liter?

a) 2:3              b) 3:2               c) 5:7                 d) 5:6



Solution:

Let milk type 1=M1 = Rs 11 per liter

Let milk type 2 =M2= Rs 6 per liter

Let proportion quantities for desired mixture  weight / liter wise be;

‘y ‘ for milk type M1

‘z’  for milk type M2

Now we arrange data as under;

Now mixture price should be equale to the sum of prices of both types of milks separately;

Mixture price = Price of M1 + Price of M2

8(y+z)=11y+6z

\Rightarrow 8y+8z=11y+6z

\Rightarrow 8z-6z=11y-8y

\Rightarrow 2z=3y

\Rightarrow \frac{2}{3}=\frac{y}{z}

\Rightarrow 2:3 = y:z

So, both types to be mixed in 2:3 ratio

(entrance exam basic maths)