Question

Please urgent !!! ( Best will be marked as brainliest )
A salt solution containing 60% salt and another salt solution containing 30% salt are mixed so as to get 20 litres of a 45% salt solution. Find how many litres of each type of solution should be mixed so as to achieve the desired result. 

Answer 1

Let 1 solution be x
2 solution be y
ATQ
Case1
60%of x +30% of y=20% of 45
60/100 *x+30/100*y=20/100*45
60x+30y=900_____eq(1)
Case2
x+y=20_____eq(2)
Solve two equations by elimination method
60x+30y= 900___(1)
x+y=20___(2)
Multiply eq(2) by 30
60x+30y=900
30x+30y=600
Now subtract it
60x+30y=900
-30x-30y=-600

30x=300

x=10 lit
Now substitute the value of x in eq (2)

x+y=20
10+y=20
y=10 lit
The two solutions are 10 and10 lit.

Answer 2

Answer:

Here percentage of salt in second solution is missing.

So let us take it to be 30%

Let quantity of 60% solution mixed be x l and that of 30% solution be y lNow total quantity of the mixture is given 20 lHence x+y = 20 .........(1)Also concentration of salt in 60% solution+ concentration of salt in 30% solution= concentration of salt in 40% solutionHence 60% of x + 30% of y = 45% of 20⇒60100x+30100y = 45100×20=60x+30y =900⇒2x+y=30...........(2)Now subtracting (1) from (2) we getx =10Putting x=10 in (1) we get,y = 10Hence quantity of 60% solution to be mixed is 10 l and quantity of 30% solution to be mixed is 10 l