Alligation or Mixture - Aptitude Questions and Answers

Ratio of milk and water in jar A = 180 L : 36 L = 5 : 1
Let, x L of mixture taken out from jar A.

So, Quantity of milk taken out from jar A = x*5/6 = 5x/6
and Quantity of water taken out from jar A = x*1/6 = x/6

Now, according to the question,
(5x/6) / (x/6 + 6) = 5/2
(5x/6) / (x + 36)/6 = 5/2
5x / (x + 36) = 5/2
10x = 5x + 180
5x = 180
x = 36
Hence, 36 L of mixture was taken out from Jar A.

Let X is the cost of pure wine. Then,
Cost of resultant mixture, X + 25% of X = 50
100X + 25X = 50*100
125X = 50 *100
X = 5000/125 = Rs. 40

So, wine of Rs. 40(CP) sold at Rs. 50(SP)

Now, in Rs. 60 = 1 litre wine.
So, in Rs.40 = 40/60 = 2/3 part wine, remaining part 1/3 water.
Therefore ratio Water : Wine = 1/3 : 2/3
= 1:2

By the rule of alligation: If two ingredients are mixed, then
Quantity of cheaper/Quantity of dearer =[(C.P. of dearer) - (Mean price)] / (Mean price) - [(C.P. of cheaper)]

i.e. (Cheaper Quantity):(Dearer quantity)= (d-m):(m-c).

C.P. of cheaper(in paise) ,Water c=0/lit
C.P. of dearer(in paise),Wine d=60/lit
Mean price(in paise) or resultant mixture costs, m= 40/lit

Thus ,(Cheaper Quantity):(Dearer quantity)= (60-40) / (40- 0)
=> 20:40 =>1:2

Average price of the resultant mixture= (16×12+5×33) / (16+5)
=357 / 21
=17

There is 10% water in 100 liters mixture of milk and water. How much more water is to be added to the mixture so that the resultant solution contains 25% water?

Initial Solution = 100
Initial water =10% of 100 = 10

Let x is the water to be added to the mixture so that the new mixture must have 25% water

new Solution = 100+x
water in new Solution = 10+x

10+x = 25% of (100+x)
10+x = (25/100)*(100+x)
10+x = (1/4)*(100+x)
40+4x=100+x
3x = 60
x = 20 L

By the rule of alligation: If two ingredients are mixed, then
Quantity of cheaper/Quantity of dearer =[(C.P. of dearer) - (Mean price)] / (Mean price) - [(C.P. of cheaper)]

i.e. (Cheaper Quantity):(Dearer quantity)= (d-m):(m-c).

C.P. of cheaper(in paise) , c=750
C.P. of dearer(in paise), d=1100
Mean price(in paise) or resultant mixture costs, m= 825

Thus ,(Cheaper Quantity):(Dearer quantity)= (1100-825) / (825- 750)
= (1100-825) / (825- 750)
=275 / 75
=11 / 3

Initial Solution = 20 liters(water+sugar)
Initial sugar=20%
So, Initial Water=80% of 20 = 16

Let x is the water to be added to the mixture so that the new mixture must have ontains 8% sugar i.e. 92% water

Thus,new Solution = 20+x
water in new Solution = 16+x

16+x = 92% of (20+x)
16+x = (92/100)*(20+x)
16+x = (23/25)*(20+x)

400+25x=460+23x
2x = 60
x = 30 L

Let there is 100 kg ore.
25% of ore contains 90% copper,which means 25 kg contains; 25*90/100 = 22.5 kg of copper.

22.5 kg copper included in 100 kg of ore.
Then, 1 kg of copper = 25/100 kg ore contains
Hence, 60 kg copper = 100*60/22.5 = 266.66 kg of ore contains

By the rule of alligation: If two ingredients are mixed, then
Quantity of cheaper/Quantity of dearer =[(C.P. of dearer) - (Mean price)] / (Mean price) - [(C.P. of cheaper)]

i.e. (Cheaper Quantity):(Dearer quantity)= (d-m):(m-c).

ratio of a liquid1 in mixture1, c=3/8
ratio of a liquid1 in mixture2, d=6/7
Mean ratio of a liquid1 or ratio of a liquid1 in resultant mixture, m= 7/10

Thus ,Required ratio = (6/7 - 7/10) / (7/10 - 3/8)
= ((60-49)/70) / ((28-15)/40)
=(11/70) / (13/40)
=(11*40) / (70*13)
=44 / 91

Initial Solution = 40
Initial water =10% of 40 = 4

Let x water be added then the new mixture must have 20% water

Final Solution = 40+x
Final water = 4+x

4+x = 20% of (40+x )
4+x = (20/100)*(40+x )
4+x = (1/5)*(40+x )
20+5x=40+x
4x=20
x = 5
water to be added to the mixture, x = 5

Alloy X
Total 25
Ratio 1:2
lead 25* (1/3)=25/3
tin 25* (2/3)=50/3

Alloy Y
Total 125
Ratio 2:3
lead 125* (2/5)=50
tin 125* (3/5)=75

When X and Y are mixed, then
Amount of lead = (25/3) + 50 = 175/3 kg
Amount of tin = (50/3) + 75 = 275/3 kg

Ratio of lead to tin in the mixture = (175/3) / (275/3) = 175 : 275 = 7 : 11

The Required ratio = cp of biggest - mean price / mean price- cp of lowest
=(80 – 60)/(60- 50)
=20:10
=2:1