Question

Two solution of milk and water are kept in two vessels A and B.the ratio of quantity of milk in vessels A and B is 4:5 while the ratio of quantity of water in vessels A and B is 8:9. it is known that the concentration of milk in one of these vessels between 60%and 80%.what could be the concentration of milk in other vessels.​

Answer 1

Answer:

54%< concentration of milk in other vessel < 88.89%

Step-by-step explanation:

Ratio of milk = 4 x : 5x

Ratio of water = 8 y : 9

vessel A milk to water ratio = 4x /8y = x/2y

vessel B milk to water ratio = 5x /9y

vessel A milk concentration in percentage

$vessel \: a \: = \frac{x}{2y} \times 100$

vessel B milk concentration in percentage

$vessel \: b \: = \frac{5x}{9y} \times 100$

In any of the one vessel milk concentration is between 60 % and 80 %

from vessel A to B

$60 < \frac{x}{2y} \times 100 < 80$

Multiply by 10/9 {x/2y * 10/9 = 5x /9y}

Concentration In Vessel B

$66.67 < \frac{5x}{9y} \times 100 < 88.89$

from vessel B to A

$60 < \frac{5x}{9y} \times 100 < 80$

Multiply by 9/10 {5x/9y *9/10 = x/2y}

Concentration in vessel A

$54 < \frac{x}{2y} \times 100 < 72$

Since we don't know given concentration percentage of milk for whether vessel A or B , we write what could be the possible range satisfying both condition.

54%< concentration of milk in other vessel < 88.89%