100 points please solve it.
Answers
Step-by-step explanation:
To make the solution 10 litres containing 40 percent acids is we have to added 6 litres of solution from 1st and 4 litres of solution from second mixture.
Answer:
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58.
Let say Equal Distance = D km Total Distance = 20 km
Time with x km/h speed = D/x hr
Time with y km/h speed = D/y hr Total Time = D/x + D/y D(y + x)/xy hr
Average speed = Total Distance / Total Time
=> Total Distance = Average speed *
Total Time
=> 2D = 8 Y + x)/xy => xy = 4(x + y)
Let say Equal time = T hr
Total time = 2T hr
Total distance = xT + yT = (x + y)T
Average speed = (x + y)T/2T = (x + y)/2 =>
9 = (x + y)/2 x + y = 18.
=> xy = 4(18)
=> xy = 72
x+y = 18
=> x + 72/x = 18
=> x? + 72 = 18
=> x? - 18x + 72 = 0
=> x? 6x - 12x + 72 = 0
=> x(x-6) -12(x 6) = 0
=> (x-12)(x-6) = 0
=> x = 12 & 6
&=6 & 12
x>y
Solution: x = 12 & y = 6
59.
Let X be the amount of first solution and Y be the amount of second to be taken.
The amount of acid in x=0.5x
But another 50% has to be water.
The amount of water =0.5x.
The amount of acid in y=0.25y.
The amount of water in y=0.75y
The total amount of acid required in final solution =0.4×10=4 litres and remaining 6 lit would be water.
Therefore,
0.5x + 0.25y = 4..................(1).
Or,0.5x=4−0.25y
Similarly for water,
0.5x+0.75y=6 .................(2)
Substituting the value of 0.5x from Eq. (1) into Eq. (2) we have,
4−0.25y+0.75y=6
4+0.5y=6
Solution: y=4,x=6.