Jar A has ?x? ml mixture of milk and water, of which 40% is water. Jar B also has mixture of milk and water, of which 20% is water. The quantity of mixture in Jar B is twice that of Jar A. If the content of Jar B is emptied into Jar A completely and the resultant quantity of milk in Jar A is 198 ml, what is the value of ?x? ?
Jar A has ?x? ml mixture of milk and water, of which 40% is water. Jar B also has mixture of milk and water, of which 20% is water. The quantity of mixture in Jar B is twice that of Jar A. If the content of Jar B is emptied into Jar A completely and the resultant quantity of milk in Jar A is 198 ml, what is the value of ?x? ?
1). 45
2). 90
3). 85
4). 95
2 answers
Total quantity of mixture in jar A = $x$ ml
% of milk in jar A = 100 - 40 = 60%
Quantity of milk in jar A = $\frac{60}{100} \times x = \frac{3 x}{5}$ ml
Total quantity of mixture in jar B = $2x$ ml
% of milk in jar B = 100 - 20 = 80%
Quantity of milk in jar B = $\frac{80}{100} \times 2x = \frac{8 x}{5}$ ml
=> Total quantity of milk in jar A and B = $\frac{3 x}{5} + \frac{8 x}{5} = 198$
=> $\frac{11 x}{5} = 198$
=> $x = 198 \times \frac{5}{11} = 18 \times 5$
=> $x = 90$ ml
Total quantity of mixture in jar A = $x$ ml
% of milk in jar A = 100 - 40 = 60%
Quantity of milk in jar A = $\frac{60}{100} \times x = \frac{3 x}{5}$ ml
Total quantity of mixture in jar B = $2x$ ml
% of milk in jar B = 100 - 20 = 80%
Quantity of milk in jar B = $\frac{80}{100} \times 2x = \frac{8 x}{5}$ ml
=> Total quantity of milk in jar A and B = $\frac{3 x}{5} + \frac{8 x}{5} = 198$
=> $\frac{11 x}{5} = 198$
=> $x = 198 \times \frac{5}{11} = 18 \times 5$
=> $x = 90$ ml