mixture

1. In a mixture of milk and water the proportion of water by weight was 75%. If in 60 gm of mixture 15 gm water was added, what would be the percentage of water? (Weight in gm)
Solution:
Water in 60 gm mixture=60*75/100=45 gm. and Milk=15 gm.After adding 15 gm. of water in mixture,
total water=45+15=60 gm and weight of a mixture=60+15=75 gm.So % of water=100*60/75=80

2. Potatoes are made up of 99% water and 1% potato matter. Jack bought 100 pounds of potatoes and left them outside in the sun for a while. When he returned, he discovered that the potatoes had dehydrated and were now only made up of 98% water. How much did the potatoes now weigh?
Solution:
100 POUNDS = 99 WATER + 1 POTATO MATTER LETS SAY NOW ITS X POUNDS0.98X+1=X=50 POUNDS

(or)

initially water weigh 99 pounds & patato matter weigh 1 pound.
after dehydration water = 98%=> potato matter = 2% = 1 pound100% = (1/2) * 100 = 50 poundsSo, potato now weigh 50 pounds.

3. There are 7 dozen candles kept in a box. If there are 14 such boxes, how many candles are there in all the boxes together?
Solution:
7 dozen=7*12=84 in 1 boxin 14 box=14*84=1176

(or)
there are 7 dozen candles1 dozen contains= 12 candles
so 7*12 = 8414 such boxes so 14 * 84 = 1176


4. In a mixture of 90 litres, the ratio of milk and water is 4 : 2. How much water should be added to the mixture so that the ratio of milk and water becomes 6 : 4 ?
Solution:
Water in the 90 liter mixture= 90*2/6=30 liters and remaining 60 liters is milkIf 'x' liters of water is added and ratio of milk and water becomes 6:4 , then
60/(30+x) =6/4x=10

5. A mixture of milk and water contains 7% water. What quantity of pure milk should be added to 12 litres of mixture to reduce water to 4%?
Solution:
12 litre of mixture contain 7 % of water i.e quantity of water = 12*7/100 =0.84 litre
Now, 11.16 litre is milk in 12 litre of mix. So, 
from question,let x amount of milk is to be added 
(11.16+x)/0.84 = 96/4 
( It's saying the % water is reduced to 4% i.e new ratio of water:milk = 96/4)
Solving this equation you will get x=9.

6. Sugar worth 40 Rs per kg, 50 per kg and x Rs per kg are mixed in the ratio 2:3:4 to get a mixture worth 50 Rs per kg. Find the value of x.
Solution:
40*2+3*50+4*x=50*9
80+150+4x=450
4x=220x=55

7. Two solutions have milk & water in the ratio 7:5 and 6:11. Find the proportion in which these two solutions should Be mixed so that the resulting solution has 1 part milk and 2 parts water?Solution:
7x+6x/5y+11y=1/2
26x=16y
x/y=16/26
i.e 8/13

8. The amount of water (in ml) that should be added to reduce 9 ml lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is:
Solution:
50% of 9ml = 30% of x ml. so, x= 15ml. now,
 amount of water to be added= 15-9= 6ml, 
as alcohol remains same.

9. A vessel is filled to its capacity with pure milk. Ten litres are withdrawn from it and replaced by water. This procedure is repeated again. The vessel now has 32 litres of milk. Find the capacity of the vessel (in litres).
Solution:
by water. After n operations, the quantity of pure liquid=x(1−y/x)^n
quantity of pure liquid is 32 lit
i,e 32=x(1-10/x)^2
=>(x(x-10)^2)/x^2=32
=>(x-10)^2=32x
x^2+100-20x=32x
=>x^2+100-52x=0
by factorization 
(x-2)(x-50)=0
therefore x=50 or 2....since 2 is not possible practically bcaz present milk level is 32 
x=50

10. Pure Milk contains 89% water.How much water should be added to form a sample of 22 litre containing 90% water?
Solution:
a sample of 22 litre containing 90% water will have 10% milk i.e. 2.2 ltr milk.
2.2 ltr milk in a mixture of 11% milk means total mixture is 100*2.2/11= 20 ltr.,
so 2 ltr water should be added in 20 ltr mixture of 89% water to make it 22 ltr mixture having 90% water.

11. How many kg of rice priced at $3.6 per kg should be added to 24 kg of rice priced at $2.7 per kg in order to make 5% profit on selling the mixture at $ 3.15 per kg?
Solution:
If 'x' kg of rice prices at 3.6$/kg is mixed with 24 kg. of rice priced at 2.7$/kg, then average price=
(x*3.6 + 24*2.7)/(24+x) =3.15*100/105
(3.6+64.8)/(24+x) =3

On solving x=12

12. 20 litres of mixture of acid and water contain 10% water. How much water should be added so that percentage of water becomes 20% in this mixture?
Solution:
total mixture=20 ltr
water=20*(10/100)=2
let x ltr water should be added,
so (2+x)=(20+x)*(20/100)
=>(2+x)=(20+x)*5
=>x=2.5 ltr

13. A milkman mixed 10 liters of water to 50 litres of milk of Rs.16 per liter, then cost price of mixture per liter is 
Solution:
Let price of water per liter be re. 1((10*1)+(50*16))/60 =13.33

(or)

Total mixture = 10w + 50m = 60
Total S.P= 16*60= Rs.960 @ Rs.16/lit
Now one can easily see in the statement that the milkman will get profit of 10 litres.
So, Gain = 10*60= Rs.160
Also, Gain = SP - CP 
160=960-CP
CP = 800 for 60litres
so CP for 1 litre = 800/60 = 13.33

14. The concentration of spirit in three different vessels A, B and C are 45%, 30% and 25% respectively. If 4 litres from vessel A, 5 litres from vessel B and 6 litres from vessel C are mixed, find the concentration of spirit in the resultant solution.
Solution:
=(4*45/100)+(5*30/100)+(5*25/100)
=1.8+1.5+1.5=4.8 so 100*4.8(4+5+6)
=32

(or)

(4*45/100)=1.8(5*30/100)=1.5(6*25/100)
=1.51.8+1.5+1.5=4.8(4.8*100)/(4+5+6)=32% 

15. An oil of Rs.42 per kg is mixed with 10 kg oil of Rs.69 per kg. If price of mixture oil is Rs.60 per kg then what is the quantity of first type of oil in the mixture?
 Solution:
Rule of Mixture
(Quantity of cheaper/Quantity of dearer)=(C.P fo dearer)-(mean price)/(mean price)-( C.P of cheaper)
let quantity of cheaper=x 
Quantity of dearer=10 kg
C.P fo dearer=69
mean price=60
C.P of cheaper=42
x/10 =(69-60)/(60-42)
x/10=9/18
x=90/18
so x=5kg

16. A seller mixed 4 dozen bananas costing Rs.12 per dozen with 6 dozen bananas at Rs.8 per dozen then what is the cost price of mixed bananas per dozen?
Solution:
4 dozens of bananas cost is rs12
then 4*12=486 dozens of bananas cost is rs8
then 6*8=4848+48/4+6=96/10=9.6

17. In what ratio must rice of Rs.16 per kg be mixed with rice of Rs.24 per kg so that cost of mixture is Rs.18 per kg? 
Solution:
(18-24)/(16-18)=6/2=3:1

18. If 10 litres of an oil of Rs.50 per litres be mixed with 5 litres of another oil of Rs.66 per litre then what is the rate of mixed oil per litre?
Solution:
50*10 = 50066*5 = 330830/15 = 55.33

19. In what ratio tea of Rs.80 per kg be mixed with 12kg tea of Rs.64 per kg, so thai cost price of mixture is Rs.74 per kg? 
Solution:
Unit Price of Cheapest Unit Price of dearer
64  80
74
80-74 | 74-64
Ratio = 6:10 =3:5

20. In what ratio must tea of Rs.42 per kg be mixed with tea of Rs.50 per kg so that cost of mixture is Rs.45 per kg?
Solution:
50-45:45-42=5:3

21. In what ratio mental A at Rs.68 per kg be mixed with another metal at Rs.96 per kg so that cost of alloy (mixture) is Rs.78 per kg?
Solution:96-78:78-68=9:5

22. A man buys eggs at 2 for Re 1 and an equal number at 3 for Rs 2 and sells the whole at 5 for Rs 3. His gain or loss percent is :
Solution:
Total cp of 2 eggs=1/2+2/3=7/6
Cp for 1 egg=7/12
Now Sp of 1 egg=3/5
Profit=3/5-7/12=1/60
gain %=(1/60)/(7/12)*100=20/7
So ans will be 2 6/7

23. In what ratio must rice at Rs 9.30 per Kg be mixed with rice at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:
(9.3x+10.8y)/(x+y)=10
9.3x+10.8y=10x+10y
7x=8y
x/y=8/7
8:7

24. Pure ghee costs Rs 100 per kg. After adulterating it with vegetable oil costing Rs 50 per kg, a shopkeeper sells the mixture at the rate of Rs 96 per kg, thereby making a profit of 20 %. In what ratio does he mix the two ?
Solution:
SP of mixture is 96 making profit 20%
therefor CP is 80

Now using alligation

100 50
/
/
80
/ 

(80-50)=30 (100-80)=20
so the ratio is= 30/20
3:2

25. 4 litres of a 20% solution of alcohol in water is mixed with 6 litres of an 80% solution of alcohol in water. What is the strength of alcohol in the resulting mixture?
Solution:
amount of alcohol in both solution = 4*20/100 + 6*80/100 = 4/5 + 24/5 = 28/5
total solution = 4+6 = 10 litre
percentage of alcohol in mixture = (28/5)/10 * 100 = 56%

26. 1 litre of water is added to 5 litres of a 20% solution of alcohol in water. The strength of alcohol is now?
Solution:
amount of alcohol in solution = 5*20/100 = 1litre
strength of alcohol in solution = 1/6 *100 = 16.67%

27. Two liquids are mixed in the proportion of 3:2 and the mixture is sold at $11 per litre at a 10% profit. If the first liquid costs $2 more per litre than the second, what does it cost per litre?Solution:
Given mixture ratio = 3:2
Lets assume second liquid = x,
So,first liquid = (x+2).
x-10/10-x-2 = 3/2
2x-20 = 24-3x
5x = 44
x=8.8
so first liquid cost is x+2 = 8.80+2 = 10.80

28. Two equal glasses filled with mixture of milk and water in the proportions of 2:1 and 1:1 respectively are emptied into a third glass. What is the proportion of milk and water in the third glass?
Solution:
first glass = 2:1
milk in first glass = 2/3 water in first glass = 1/3

second glass = 1:1
milk in second glass = 1/2 water in second glass = 1/2

third glass = (milk in first glass + milk in second glass)/(water in first glass + water in second glass)


29. 729ml of a mixture contains milk and water in the ratio 7:2. how much water is to be added to get a new mixture containing milk and water in the ratio 2:1?
Solution:
In 729ml milk amount is (7/9)*729=567
Water amount is (2/9)*729=162
our aim is water in the mixture half of milk.
Half of milk is 283.5
water to be added to get 283.5 is 162+121.5
So ans is 121.5ml

30. A mixture of milk and water measures 60 gallons. It contains 20% water. How many gallons of water should be added to it so that water may be 25%?
Solution:
milk+water=60 gallons
60 gallons=20% water+80%milk
20 % of 60=12 gallons
25 % of (60+x)=12+x;
x=4 gallons

31. Concentrations of three wines A, B and C are 10, 20 and 30 percent respectively. They are mixed in the ratio 2: 3 : x resulting in a 23% concentrated solution. Find x.
Solution:
Assume that we have 2a,3a,xa liters of A,B,C wines mixed respectively then

2a(0.10)+3a(0.20)+xa(o.3)=(2a+3a+xa)(0.23)

=0.8+(0.3x)=1.15+(0.23x)
then x=0.35/0.07=5

32. A vessel is full of a mixture of spirit and water in which there is found to be 17% of spirit by measure. Ten litres are drawn off and the vessel is filled up with water. The proportion of spirit is now found to be 15 1/9%. How much does the vessel hold?
Solution:
In 10L of water drawn amount of spirit present is = (17/100)*10 =1.7 L .
x is the capacity of the vessel
from 17% amount of spirit comes down to 15 1/9% .the difference between this percentage is (17% - 15 1/9%) = 17/9 % x 
17/9 % x =1.7 litres . from this equation we get the value of x as 90 L

33. A man buys oranges at Rs 5 a dozen and an equal number at Rs 4 a dozen. He sells them at Rs 5.50 a dozen and makes a profit of Rs 50. How many oranges does he buy ?
Solution:
50(avg of both is 4.5 so he gain 1 rupee and for profit of 50 is 50)

34. A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 21 per kg, then his gain percent is
Solution:
(20-x)/(x-18)=5:3
from here x=75/4
now the sell price is 21 and the cost price is 75/4 so profit is 12%

35. There are two vessels A and B in which the ratio of milk and water are as 5:2 and 8:7 respectively. Two gallons are drawn from vessel A and 3 gallons from vessel B, and are mixed in another empty vessel. What is the ratio of milk and water in it?
Solution:
If 2 gallons are taken from A, the quantity of milk & water=2*5/7:2*2/7=10/7:4/7
If 3 gallons are taken from B,the quantity of milk & water=3*8/15:3*7/15=8/5:7/5
Now in the new mixture qtantity of milk 10/7+8/5=106/35 & water 4/7+7/5=69/35
Therefore ratio of milk:water=106/35:69/35=106:69

36. Gopal purchased 35 kg of rice at the rate of Rs 9.50 per kg and 30 kg at the rate of Rs 10.50 per kg. He mixed the two. Approximately, at what price per kg should he sell the mixture to make 35 % profit in the transaction ?
Solution:
35*9.50+30.10.5=647.5
647.5*1.30=874.125
874.125/65=13.44

37. Ajay bought 15 kg of dal at the rate of Rs 14.50 per kg and 10 kg at the rate of Rs 13 per kg. He mixed the two and sold the mixture at the rate of Rs 15 per kg. What was his total gain in this transaction ?
Solution:
15*14.50=217.5
10*13 =130
(217.5+130)=347.50
15+10=25 so,25*15=375
375-347.50=27.50

38. How much water must be added to 60 liters of milk at 11/2 liters for Rs 20 so as to have a mixture worth Rs 10 2/3 a liter?
Solution:
C.P. of 1 litre of milk = Rs. 20×2/3 = Rs. 40/3.
C.P. of 1 litre of water =Rs 0.
Mean price = Rs. 32/3.
By the rule of alligation, we have :
C.P of 1 litre        C.P of 1 litre
of water             of milk
   (0)                  (Rs. 40/3)
          \                /          Mean Price
          (Rs. 32/3)
         /               \
  40/3−32/3    32/(3−0)
      8/3              32/3
The ratio of water and milk =8/3:32/3.=>8:32=1:4.
Thus, Quantity of water to be added to 60 litres of milk:=>(1/4)×60 litres.=>15 litres.

39. By mixing two qualities of pulses in the ratio 2: 3 and selling the mixture at the rate of Rs 22 per kilogram, a shopkeeper makes a profit of 10 %. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is 
Solution:
Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x). 
Sell price of 5 kg = Rs. (22x5) = Rs. 110. 
(110 - (28 + 3x)/(28 + 3x)) * 100 = 82-3x/28 + 3x = 1 / 10 
820 - 30x = 28 +3x ; 33x = 792 ; 
x = 24

40. A certain amount of solution contains 18% alcohol. 8 litres of the solution is taken out and replaced with water. The resultant solution contains 15% alcohol. Find the volume(in litres) of the solution.
Solution:
8 liters=3%
? =remaing 15%
(15*8)/3=40

41. 6 kgs of tea at $6 per kg and 4 kgs of tea at $7 per kg are mixed together and the mixture is sold at a 10% profit. What is the selling price per kg of mixture?
Solution: 
6kg->$36,4kg->$28, then the mixture will be 10kg->$36+$28=$64.
if 10% profit->$6.4 for 10kg. now 10kg->$64+$6.4=$70.4. Hence per kg=$7.04

42. A 240 gallons solution contains milk and water in the ratio 5:4. How many gallons of water must be added so that the ratio of milk and water in the resulting solution is 4:5?
Solution:
Let X be the number of gallons of water added so that the ratio of milk and water changes to 4:5 from 5:4.
Amount of milk in the initial solution= 5/9 × 240 gallons
Amount of milk in the final solution= 4/9 x (240 + X)gallons
Since there is no change in the content of the milk we have
5/9 × 240 = 4/9 x (240 + X)
i.e. X = 60 gallons

43. A milk vendor sells milk at Cost Price but still gains 20%. Find the ratio of milk and water in every gallon that he sells
Solution:
Let Rs 100 be cost price and selling price
To get 20% profit by selling at Rs 100, quantity to be actual sold =100*100/120=250/3, 
In case of milk to compensate for the quantity water to be mixed=100-250/3=50/3
Therefore the ratio of milk:water=250/3:50/3=5:1