Question

Pot 1 contains alcohol and spirit in 4:3. Pot 2 contains spirit and alcohol in 4:3. How much of 2nd mixure must be mixed with 7 litre of first to have alcohol and spirit in 6:5

Answer 1

Answer:

Explanation:

by alligation method

4/4+3 -6/6+5   :   3/3+4-6/6+5

4/7-6/6/11= 9/77

3/7-6/11= 2/77

9/77 :2/77

so the ratio is 9 : 2 ans

fly away

Answer 2

Answer:

The amount of mixture that is mixed in mixture 1 from mixture 2 to have alcohol and spirit in the ratio of 6:5 is 14/9 liters.

Explanation:

Given,

Pot 1 contains alcohol and spirit in a 4:3 ratio:

i.e. The alcohol has a quantity of 4x, and the spirit has a quantity of 3x.

Pot 2 contains spirit and alcohol in a 4:3 ratio:

i.e. The spirit has a quantity of 4x, and the alcohol has a quantity of 3x.

To find,

The amount of 2nd mixture must be mixed with 7 liters of first to have alcohol and spirit in 6:5.

Calculation,

Let 'y' amount 2nd mixture is mixed with the 7 liters of the 1st mixture.

Now as the amount of 1st mixture given is 7 liters, then the amount of alcohol is 4 liters and the amount of spirit is 3 liters.

Similarly, for mixture 2, the amount of alcohol is 3 liters and the amount of spirit is 4 liters.

Now, from mixture 2 if the 'y' amount is mixed with the 7 liters of the 1st mixture, then the amount of alcohol, and spirit mixed will be 3y/7 and 4y/7.

So, the amount of alcohol, and spirit present in mixture 1 is 4 + (3y/7), and 3 + (4y/7) respectively.

But the ratio of alcohol to the spirit in the final mixture 1 is 6: 5

Then,

$\frac{4+ \frac{3y}{7} }{3 + \frac{4y}{7} } = \frac{6}{5}$

Solving for x :

y = 14/9

Hence, the amount of alcohol mixed in mixture 1 is (3/7)(14/9) = 6/9, and the amount of spirit mixed in mixture 1 is (4/7)(14/9)= 8/9.

Therefore, the amount of mixture that is mixed in mixture 1 from mixture 2 to have alcohol and spirit in the ratio of 6:5 is 14/9 liters.

#SPJ2