The rule of alligation help to find the ratio in which two or more variety of ingredients of a given price must be mixed to produce a mixture of desired price.It is a rule for the solution of problems concerning the compounding or mixing of ingredients
A mixture contains two or more commodities of certain quantity mixed together to get the desired quantity.
The cost of a unit quantity of the mixture is called the mean price.
$\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)$
Therefore, (Cheaper Quantity) : (Dearer Quantity) = (d – m) : (m-c)
1: Calculate the average price of the resulting mixture when two variety of sugar at ₹15 per Kg and ₹20 per Kg are mixed in the ratio of 2:3.
Options:
A. ₹17
B. ₹19
C. ₹18
D. ₹20
Solution:
Let average price be x
$\frac{2}{3} = \frac{(20 - x)}{ (x - 15)}$
$2x - 30 = 60 - 3x$
$5x = 90$
$x = \frac{90}{5} = ₹18$ per Kg
Correct option: C
2.Two varieties of rice are in the ratio of 3:2 such that the average price of the resulting mixture is ₹15 per Kg. The price of one of the varieties is ₹10 per Kg. Find the price of the other variety of rice.
Options:
A. ₹23 per Kg
B. ₹22.5 per Kg
C. ₹20.5 per Kg
D. ₹21 per Kg
Solution:
Let Price of another variety is $A_{2}$
$\frac{3}{2}= \frac{(A_{2}-15)}{(15-10)}$
Average age of the family =21
$\implies \frac{(31+29+A_{2})}{3} = 21$
$15 = 2A_{2}-30$
$\frac{45}{2} = A_{2}$
A2 = ₹22.5 per Kg
Correct option: B
3. Two varieties of tea are mixed in some ratio. The cost of the first variety is ₹20 per Kg and that of second variety is ₹30 per Kg. If the average cost of the resulting mixture is ₹25 per Kg, find the ratio.
Options:
A. 1:1
B. 2:1
C. 1:2
D. 1:3
Solution:
$\frac{x}{y}= \frac{(30-25)}{(25-20)}yx=(25−20)(30−25)\frac{x}{y} = \frac{5}{5} = 1:1$
Correct option: A
If a container has x unit of liquid A from which y units are taken out and replaced by water. This process is repeated n number of time, then the quantity of pure liquid will be given by:
After n operations, the quantity of pure liquid = $ \left ( x \left( 1 – \frac{y}{x} \right)n \right)$
1: A mixture of $300l$ contains pure milk and water in the ratio of $2:5$. Now $100 l$ of water is added to the mixture. Calculate the ratio of milk and water in the resulting mixture.
Options:
A. 2:5
B. 1:4
C. 2:3
D. 3:2
Solution:
The quantity of milk in initial mixture = $\frac{2}{5} \times 300 = 120 \text{l}$
Quantity of water $= 300 - 120 = 180l$
Water added $= 100l$
Concentration of water in the resulting mixture $= 300 + 100 = 400l$
The ratio of the resulting mixture $\frac{120}{280} = 1 : 4$
Correct option: B
2.A milkman mixes $30l$ of water in $90l$ of milk. He then sells $\frac{1}{4}th$ of this mixture. Now he adds water to replenish the quantity of milk sold. Find the current proportion of milk and water.
Options:
A. 5:6
B. 7:9
C. 4:5
D. 9:7
Solution:
The initial ratio of milk and water = 90:30 = 3:1
Now, $\frac{1}{4}th$ of the mixture is sold, that is the total volume of the mixture is reduced by $25\%$.
In other words, both water and milk are reduced by $25\%$.
So, the volume of milk and water is $67.5l$ and $22.5$ respectively.
Now, $30l$ ($25\%$ of the total mixture volume) of water is added to the mixture.
The volume of milk $= 67.5l$
Volume of water $= 22.5+30= 52.5l$
Current ratio $= \frac{67.5}{52.5} = 9:7$
Correct option: D
3.A 20 litres mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?
Options:
A. 3:9
B. 9:1
C. 6:5
D. 9:4
Solution:
In 20 liters of mixture
Milk =$\frac{3}{5} \times 20 = 12 l$
Water $= 8 l$
Milk $= 6 l$
Water $= 4 l$
On adding 10 liters of milk,
Milk $= 12 – 6 + 10 = 16 l$
Water $= 8 – 4 = 4 l$
Again, in 10 liters of mixture,
Milk $= \frac{4}{5} \times 10 = 8 l$
Water = 2 litres
On adding 10 litres of milk,
Milk $= 16 – 8 + 10 = 18$ litres
Water $= 2$ litres
Therefore, Required ratio $= 18 : 2 = 9 : 1$
Correct option: B