QUESTIONS ON ALLEGATION OR MIXTURE (PART-I)
QUESTIONS ON ALLEGATION OR MIXTURE (PART-I)
VERY IMPORTANT: Remember that the weighted average/combined average/mean value is always written in the middle and the averages are written at the top. We write the smaller average/value in the left and greater one in the right. The mean value is subtracted from the value written in the right, and the value written in the left is subtracted from the mean value, figure thus got is considered positive (negative signs are ignored).
But in questions having multiple ratios as are given in queries 1, 2 and 3, we write the value of the thing that is first in the asked ratio in the left and the other value in the right.
QUERY 1
Two Vessels A & B contain milk and water in the ratio 7 : 3 and 3: 2 Find the ratio in which the quantities be taken from the two vessels so that the ratio of milk and water in the new mixture is 2 : 1
A) 1 : 2
B) 2 : 1
C) 1 : 3
D) 3 : 1
MAHA GUPTA
The resultant ratio will be the required ratio.
MILK IN A MILK IN B
| 7⁄10 | 3⁄5 | |||
| 2⁄3 | ||||
| 3⁄5 – 2⁄3 = 1⁄15 | 2⁄3 – 7⁄10 = 1⁄30 |
Hence, the ratio = 1⁄15 : 1⁄30 = 2 : 1 (option ‘B’)
ANOTHER METHOD
Let vessel A and B contain solutions 10x and 5y respectively.
Then milk in the resultant mixture = 7x + 3y
And water in the resultant mixture = 3x + 2y
The resultant ratio is 2 : 1 (given)
Therefore, (7x + 3y)⁄(3x + 2y) = 2⁄1
=> x : y = 1 : 1
Thus, ratio of two mixtures = 10x : 5y = 10*1 : 5*1 = 10 : 5 = 2 : 1 (option ‘B’)
RONNIE BANSAL
First understand how to solve these type of questions:
Two Vessels A & B contain milk and water in the ration p:q and r: s. Find the ratio in which the quantities be drawn or mixed from the two vessels so that the ratio of milk and water in the new mixture is m:n
Find p⁄(p+q); r⁄(r+s) & m⁄(m+n)`
In such a case (either drawn or mixed) m/m+n will be greater than either of the two ratios given just above; and smaller than the other. Now deduct smaller ratio from m⁄(m+n) & m⁄(m+n) from the greater.
We’ll find then a new ratio; the first term representing MILK (the first item) and the second term representing WATER (the second item). So if ratio of MILK & WATER is asked the answer will be the ratio thus found; otherwise reverse.
So solution to the above question
Here p⁄(p+q) = 7⁄(7+3) = 7⁄10
r⁄(r+s) = 3⁄(3+2) = 3⁄5
m⁄(m+n) = 2⁄(2+1) = 2⁄3
Here we see that 2⁄3 is greater than 3⁄5 and smaller than 7⁄10
Now, 2⁄3 – 3⁄5 = 1⁄15 & 7⁄10 – 2⁄3 = 1⁄30
=> 1⁄15 : 1⁄30 = 2 : 1
So as explained above 2 : 1 is the answer. (option ‘B’)
QUERY 2
In 2 type of brass, ratio of copper and zinc are 8 : 3 and 15 : 7. The ratio in which these two type of brass should be mixed so that the ratio of copper and zinc in this new type of brass becomes 5 : 2 is?
A) 5 : 2
B) 1 : 2
C) 4 : 3
D) 7 : 4
MAHA GUPTA
The resultant ratio will be the required ratio.
COPPER IN I COPPER IN II
| 8⁄11 | 15⁄22 | |||
| 5⁄7 | ||||
| 15⁄22 – 5⁄7 = 5⁄154 | 5⁄7 – 8⁄11 = 1⁄77 |
Hence, the ratio = 5⁄154 : 1⁄77 = 5 : 2 (option ‘A’)
ANOTHER METHOD
Let brass-I and II contain solutions 11x and 22y respectively.
Then milk in the resultant mixture = 8x + 15y
And water in the resultant mixture = 3x + 7y
The resultant ratio is 5 : 2 (given)
Therefore, (8x + 15y)⁄(3x + 7y) = 5⁄2
=> x : y = 5 : 1
Thus, ratio of two mixtures = 11x : 22y = 11*5 : 22*1 = 55 : 22 = 5 : 2 (option ‘B’)
RONNIE BANSAL
First of all see QUERY 1 here to know how to do such types
Here p⁄(p+q) = 8⁄11
r⁄(r+s) = 15⁄22
m⁄(m+n) = 5⁄7
Here we see that 5⁄7 is greater than 15⁄22 and smaller than 8⁄11
Now, 5⁄7 – 15⁄22 = 5⁄154 & 8⁄11 – 5⁄7 = 1⁄77
=> 5⁄154 : 1⁄77 = 5 : 2
So as explained above 5 : 2 is the answer (option ‘A’)
QUERY 3
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A) 4 litres, 8 litres
B) 6 litres, 6 litres
C) 5 litres, 7 litres
D) 7 litres, 5 litres
MAHA GUPTA
Milk in CAN-I = 100 – 25 = 75% = 3/4
Milk in CAN-II = 100 – 50 = 50% = 1/2
Milk in the new mixture = 5/8 — ratio of water to milk is given 3 : 5
CAN I CAN II
| 3⁄4 | 1⁄2 | |||
| 5⁄8 |
||||
| 1⁄2 – 5⁄8 = 1⁄8 | 5⁄8 – 3⁄4 = 1⁄8 |
Hence, the ratio of = 1⁄8 : 1⁄8 = 1 : 1
Therefore, quantity of milk taken from each can = 1⁄2 × 12 = 6 litres (option ‘B’)
ANOTHER METHOD
Let can-I and II contain solutions 4x and 2y respectively.
Then water in the resultant mixture = x + y
And milk in the resultant mixture = 3x + y
The resultant ratio is 3 : 5 (given)
Therefore, (x + y)⁄(3x + y) = 3⁄5
=> x : y = 1 : 2
Thus, ratio of two mixtures = 4x : 2y = 4*1 : 2*2 = 4 : 4 = 1 : 1
But quantity of milk in the new mixture is 12 litres
Hence, milk to be taken from each of the cans = 12 × 1⁄2 = 6 litres (option ‘B’)
QUERY 4
The ratio of water and milk in a vessel is 1 : 2 and in other vessel is 3 : 4. 1-1 kg of mixture is drawn off from both vessels and poured into a third vessel. Then new ratio of water and milk?
A) 1 : 3
B) 8 : 13
C) 5 : 12
D) 2 : 5
MAHA GUPTA
1-1 kg of mixture is drawn from both the vessels
So quantities drawn from the 1st vessel:
WATER = 1 × 1⁄3 = 1⁄3 kg
MILK = 1 × 2⁄3 = 2⁄3 kg
And Quantities drawn from the 2nd vessel:
WATER = 1 × 3⁄7 = 3⁄7 kg
MILK = 1 × 4⁄7 = 4⁄7 kg
Therefore total of WATER drawn = 1⁄3 + 3⁄7 = 16⁄21 kg
And total of MILK drawn = 2⁄3 + 4⁄7 = 26⁄21 kg
Hence the new ratio of WATER and MILK = 16⁄21 : 26⁄21 = 8 : 13 (option ‘B’)
QUERY 5
A container contains 60 kg of milk. From this container 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is?
A) 42 kg
B) 58 kg
C) 43.74 kg
D) 42. 74 kg
MAHA GUPTA
Milk left in the container:
After 1st attempt……………………. 60 – 6 = 54 kg
After 2nd attempt…………………….54 – (54⁄60)*6 = 48.6 kg
After 3rd (last) attempt……………… 48.6 – (48.6⁄60)*6 = 43.74 kg (option ‘C’)
Another Method
This process is just like COMPOUND INTEREST that we see in calculating the amount (principal + interest) after certain number of years. The only difference is, in compound interest the sum keeps increasing whereas here the amount of milk will be decreasing every time, but the manner is the same.
Well, the formula to find amount in case of CI = P(1 + r⁄100)t, where P = principal and r= rate and t = term
As a quantity is to be taken out here, we just have to use sign of minus ( – ) in place of sign of addition ( + )
Therefore milk left in the container after 3 attempts = 60 (1 – 6⁄60)³ = 60*(9⁄10)³ = 43.74 kg (option ‘C’)
QUERY 6
4 liters are drawn from a container full of milk and is then filled with water. This operation is performed three more times. The ratio of the quantity of milk left in the container and that of water is 16 : 65. How much milk did the container hold initially?
A) 24 liters
B) 12 liters
C) 15 liters
D) 25 liters
MAHA GUPTA
This process of drawing of milk from the container is just like COMPOUND INTEREST that we see in calculating the amount (principal + interest) after certain number of years. The only difference is, in compound interest the sum keeps increasing whereas here the amount of milk will be decreasing every time, but the manner is same.
Well, the formula to find amount in case of CI = P (1 + r⁄100)t, where P = principal and r= rate and t = term
As the quantity is to be taken out here, we just have to use sign of minus ( – ) in place of sign of addition ( + )
As the operation of drawing of milk from the container is performed three more times after the first attempt already has been made, actual operation of drawn of milk is 4 times therefore.
Now, Let the quantity of milk in the container in the beginning be x liters
so rate of draw is 4⁄x; and the term is 4 —- 4 liters of milk is drawn every time
Thus quantity of milk left in the container after 4 attempts = x(1 – 4⁄x)4 liters
The ratio of milk to water now is 16 : 65, means milk is 16/81
Thus, [x(1 – 4⁄x)4]/x = 16⁄81
=> [x(1 – 4⁄x)4]/x = (2⁄3)4
=> (1 – 4⁄x)4 = (2⁄3)4
=> 1 – 4⁄x = 2⁄3
=> x = 12 liters (option ‘B’)
QUERY 7
9 liters are drawn from a cask full of water and it is filled with milk. Then 9 liters of mixture is drawn and the cask is again filled with milk. The quantity of water now left in the cask to that of milk is 16 : 9. How much does the cask hold?
A) 35 liters
B) 25 (liters
C) 53 liters
D) 45 liters
RONNIE BANSAL
Remember the following: Suppose a vessel contains some kind of liquid (let it be x units) from which y units are taken out and replaced by another liquid, after n operations the quantity of the first liquid left in the vessel = x(1 – y⁄x)n ,and the equation for the capacity of the vessel is
[x(1 – y⁄x)n]/x = first term of the ratio divided by the addition of both the term of the ratio.
Here we have to find the capacity of the cask, therefore [x(1 – 9⁄x)n]/x = 16⁄25
=> (1 – 9⁄x)² = 16⁄25
=> 1 – 9⁄x = 4⁄5
=> 5x – 45 = 4x
=> x = 45
So the cask can hold 45 liters in all. (option ‘D’)
QUERY 8
Ramesh earns Rs 4000 in a month. In the next month his income increases by 5%. Due to rise in prices his monthly expenses increase by 7% and his savings decrease by 1%. Find his increased expenditure.
A) Rs 2800
B) Rs 3000
C) Rs 3210
D) Rs 3600
MAHA GUPTA
INCOME = EXPENDITURE + SAVINGS
We see that INCOME is mixture of EXPENDITURE and SAVINGS, so we can solve this problem using Allegation Method. As income is the mixture, we’ll take 5 as the mean value.
Expenditure Savings
| 7 | -1 | |||
| 5 | ||||
| (-1) – 5 = 6 | 5 – 7 = 2 |
Therefore, initial EXPENDITURE : SAVINGS = 6 : 2 = 3 : 1
Hence, Ramesh’s initial expenditure = 4000 × 3⁄4
= 3000
And his increased expenditure = 3000 + 7% of 3000 = Rs 3210 (option ‘C’)
ANOTHER METHOD
Let Ramesh’s initial monthly expenditure = Rs x
And his initial monthly savings = Rs y
Hence, x + y = 4000 —- (i)
His monthly expenditure after increase in prices = x + 7% of x = 107x⁄100
And his monthly savings now = y – 1% of y = 99y⁄100
But his income also increases, so his income now = 4000 + 5% of 4000 = 4200
Therefore, 107x⁄100 + 99y⁄100 = 4200
=> 107x + 99y = 420000 —- (ii)
Solving (i) and (ii)
y = 3000
Therefore, Ramesh’s increased expenditure = 3000 + 7% of 3000 = 3210 (option ‘C’)
QUERY 9
Three glasses of capacity 2 litres, 5 litres and 9 litres contain mixture of milk and water with milk concentrations 90%, 80% and 70% respectively. The contents of three glasses are emptied into a large vessel. Find the ratio of milk to water in the resultant mixture?
A) 7 : 3
B) 121 : 39
C) 11 : 3
D) 11 : 39
SUCHI SHARMA
When the concentration of milk is 90%; the ratio of milk to water = 9 : 1
When the concentration of milk is 80%; the ratio of milk to water = 8 : 2
When the concentration of milk is 70%; the ratio of milk to water = 7 : 3
Therefore the resultant concentration ratio of milk to water = sum of milk concentration in different vessels : sum of water concentration in different vessels
= [2*(9⁄10) + 5*(8⁄10) + 9*(7⁄10)] : [2*(1⁄10) + 5*(2⁄10) + 9*(3⁄10)]
= 121⁄10 : 39⁄10
= 121 : 39 (option ‘B’)
QUERY 10
How much water must be added to 60 litres of milk at 3⁄2 litres for Rs 20 so as to have a mixture worth Rs 32⁄3 a litre?
A) 60
B) 18
C) 15
D) 30
MAHA GUPTA
The cost price (CP) of 60 liters of milk without adding water = 60*20*(2⁄3) = 800
But the cost price has to be same after mixing of water in the milk
Now let the quantity of water to be added = x liters.
Therefore the total mixture = (60 + x) liters
Now the cost price of the mixture per liter = Rs 32⁄3.
Therefore, (60 + x)*(32⁄3) = 800
=> x = 15
Hence 15 liters of water will be needed to mix in the milk (option ‘C’)
TRICK (Allegation Method)
Cost of milk per litre = 20*(2⁄3) = Rs 40⁄3, and water is available free of cost, so its value = 0
COST OF WATER COST OF MILK
| 0 | 40⁄3 | |||
| 32⁄3 | ||||
| 40⁄3 – 2⁄3 = 8⁄3 | 32⁄3 – 0 = 32⁄3 |
Hence, WATER : MILK = 8⁄3 : 32⁄3 = 8 : 32 = 1 : 4
Milk = 4 ratios = 60⁄4 = 15 per ratio
Water = 1 ratio = 1*15
Therefore, water to be added = 15 litres (option ‘C’)
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