Mixture problems

Question: Some 80% acid solutions is to be mixed with 35% acid solution to make 300 liters of 50% solution. How much of each acid should be used?

Solution:

Then the Equation as follows
0.80x+0.35(300-x)=300(.50)
0.80x+105-0.35x=150
0.45x=45
x=100 of 80% of acid solution
300-x=200 of 35% of Acid solution. When mixed together we get 300 liters of solution.
check: 
0.80(100) =80 liters
0.35(300-x)=70 liters
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300(.50) =150 liters
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Question: A total of 180 tickets are sold, Some at $3,some at $5, and some at $10 each. The total value of the tickets was $1100. The number of $5 tickets was 20 more than $3 tickets. How many of each of tickets were sold.

Solution: 

       Let x be no. of tickets sold at $3.
Then x+20 are the no. of tickets sold at $5.
For $10 ticket value, need to add $3 and $5 value then subtract it to 180 i.e., total tickets 
Let 180-2x-20 be the no. of the tickets sold at $10.

Amount of Tickets Sold	Rate	Values
x	3	3*x
x+20	5	5*(x+20)
180-(2x+20)	10	10*(160-2x)
180		1100

Then the equation as follows.

3x+5*(x+20)+10*(160-2x)=1100

3x+5x+100+1600-20x=1100
-12x+1700=1100
-12x=-600
x=50

Therefore the tickets sold at
$3 is 50,at $5 is (x+20)=>50+20=70, and at $10(180-2x-20)=>160-100=60.

Add all the above value.
50+70+60=180.
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Question: Fifty tickets were sold to chicken barbeque for total of $219.Children's ticket were sold at $2.50, Youth's ticket were sold at $3.50 and Adult's ticket were sold at $5.00.There were 10 more than youth tickets than the children's ticket. How many of each ticket was sold?

Solution:

Let x be the no.of Children's ticket sold at $2.50
then x+10 be the no. of youth tickets sold at $3.50

	Amount of Tickets Sold	Rate	Values
Children’s Ticket	x	2.5	2.5*x
Youth Tickets	x+10	3.5	3.5*(x+10)
Adult Tickets	50-(2x+10)	5	5*(40-2x)
	50		219

Then the equation as follows

2.5x+3.5*(x+10)+5*(40-2x)=219
2.5x+3.5x+35+200-10x=219
6x+235-10x=219
-4x=-16
x=4

Therefore the tickets sold 
Children's ticket is 4
youth Ticket is x+10=14
Adult tickets is 50-2x-10=32

Children's tickets+Youth Ticket+Adult Ticket=4+14+32=50.
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Question: How much water must be added to 50% of alcohol solution to obtain 100 liters of 10% of it solution?

Solution: 
Let x be the amount of the water to be added at its percentage is 0.00%
then let 100-x be the amount of alcohol solution at 50%

	Amount of solution	percent	Pure Stuff
Water	x	0.00	0*x
Alcohol Solution	100-x	0.50	0.5*(100-x)
Total	100	0.10	10

Then the equation as follows

0*x+0.5*(100-x)=10
50-0.5x=10
-0.5x=-40
x=80

The amount of water used is 80 liters

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Question: How much pure alcohol must be added to 100 liters of 10% of alcohol solution to create 80% of the solution?

Solution: 
Let x the be the amount of pure alcohol and it's percentage is 100%(pure)

	Amount of solution	percent	Pure Stuff
Pure alcohol	x	1.00	1*x
Alcohol Solution	100	0.10	0.1*100
Total	100+x	0.80	0.80(100+x)

then the equation as follows:

x+0.10*100=0.80(100+x)
x+10=0.8x+80
0.2x=70
x=350

Therefore 350 liters of pure alcohol is used to mix with alcohol solution.

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Question: How much liquid must be drained from a 24 liters radiator that is 25% antifreeze and replaced with pure antifreeze in order to bring the strength upto 50%?

Solution: 

	Amount of solution	percent	Pure Stuff
Begin	24	0.25	0.25*24
Drain	-x	0.25	0.25*(-x)
Add	+x	1.00	1.00(+x)
End	24	0.50	24*0.50

Let -x be the amount of the liquid which is drained at 25% of anti freeze

let +x be the amount of the liquid which is added after replaced with pure antifreeze which of 100%

Then the equation as follows

0.25*24-0.25*x+x=24*(0.50)

6-0.25x+x=12

0.75x=6

x=8.

Therefore the liquid which is drained from a 24 liters radiator is 8 liters.

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Question: How much liquid must be drained from a 18 liters radiator that is 10% antifreeze and replaced with pure antifreeze in order to bring the strength upto 50%?

Solution: 

	Amount of solution	percent	Pure Stuff
Begin	18	0.10	0.10*18
Drain	-x	0.10	0.10*(-x)
Add	+x	1.00	1.00(+x)
End	18	0.50	18*0.50

Let -x be the amount of the liquid which is drained at 10% of anti freeze

let +x be the amount of the liquid which is added after replaced with pure antifreeze which of 100%

Then the equation as follows

0.10*18-0.10*x+x=18*(0.50)

1.8-0.10x+x=9

0.90x=7.2

x=8.

Therefore the liquid which is drained from a 18 liters radiator is 8 liters.