Alligation or Mixture Quiz

Question 1

A mixture contains milk and water in the ratio 5:1. On adding 5 litres of water, the ratio of milk and water becomes 5:2. The quantity of milk in the mixture is:

Accepted Answer

Let the initial quantity of milk be 5x litres and water be x litres.
Total mixture = (5x + x) = 6x litres.

After adding 5 litres of water:
Quantity of milk remains = 5x litres
Quantity of water becomes = (x + 5) litres

The new ratio of milk to water is 5:2.
So, (5x) / (x + 5) = 5/2

Cross-multiplying:
5x * 2 = 5 * (x + 5)
10x = 5x + 25
10x - 5x = 25
5x = 25
x = 5

The quantity of milk in the mixture is 5x = 5 * 5 = 25 litres.

Question 2

A container contains 40 litres of milk. From this container, 4 litres of milk was taken out and replaced by water. This process was repeated two more times. How much milk is now contained in the container?

Accepted Answer

Initial quantity of milk in the container = 40 litres.
Quantity of milk taken out and replaced with water in each operation = 4 litres.
The process is repeated a total of 1 (first time) + 2 (two more times) = 3 times.

The formula for the remaining quantity of the original liquid after 'n' operations is:
Remaining Quantity = Initial Quantity * (1 - (Quantity removed / Initial Quantity))^n

Remaining milk = 40 * (1 - 4/40)^3
= 40 * (1 - 1/10)^3
= 40 * (9/10)^3
= 40 * (729/1000)
= (4 * 729) / 100
= 2916 / 100
= 29.16 litres.

Question 3

In what ratio must a grocer mix two varieties of tea worth ₹60 a kg and ₹65 a kg so that by selling the mixture at ₹68.20 a kg at a gain of 10%?

Accepted Answer

First, we need to find the Cost Price (CP) of the mixture.
Selling Price (SP) of the mixture = ₹68.20 per kg
Profit percentage = 10%

If CP is X, then SP = X + 10% of X = 1.10X
68.20 = 1.10X
X = 68.20 / 1.10 = 682 / 11 = ₹62 per kg.
So, the mean cost price of the mixture is ₹62 per kg.

Now, we use the Alligation Rule:
Cost of 1st variety = ₹60/kg
Cost of 2nd variety = ₹65/kg
Mean cost of mixture = ₹62/kg

Cost of 1st tea (₹60)     Cost of 2nd tea (₹65)
               \           /
                Mean Cost (₹62)
               /           \
(₹65 - ₹62) : (₹62 - ₹60)
    3       :     2

The ratio in which the two varieties of tea must be mixed is 3:2.

Question 4

Two vessels P and Q contain milk and water in the ratios 5:2 and 7:6 respectively. In what ratio should the contents of P and Q be mixed so as to obtain a mixture containing milk and water in the ratio 8:5?

Accepted Answer

Let's consider the fraction of milk in each vessel and in the final mixture.

Fraction of milk in vessel P = 5 / (5 + 2) = 5/7
Fraction of milk in vessel Q = 7 / (7 + 6) = 7/13
Fraction of milk in the final mixture = 8 / (8 + 5) = 8/13

Now, apply the Alligation Rule:

Milk in P (5/7)           Milk in Q (7/13)
                \             /
                   Mean Milk (8/13)
                  /             \
(8/13 - 7/13) : (5/7 - 8/13)
   (1/13)     : ( (5*13 - 7*8) / (7*13) )
   (1/13)     : ( (65 - 56) / 91 )
   (1/13)     : ( 9 / 91 )

To simplify the ratio, multiply both sides by the LCM of 13 and 91, which is 91.
(1/13 * 91) : (9/91 * 91)
7           : 9

Thus, the contents of P and Q should be mixed in the ratio 7:9.

Question 5

A mixture of 30 litres contains milk and water in the ratio 7:3. How much water should be added to it so that the ratio of milk and water becomes 3:7?

Accepted Answer

Initial quantity of mixture = 30 litres.
Ratio of milk to water = 7:3.

Quantity of milk = (7 / (7+3)) * 30 = (7/10) * 30 = 21 litres.
Quantity of water = (3 / (7+3)) * 30 = (3/10) * 30 = 9 litres.

Let 'x' litres of water be added to the mixture.

New quantity of milk = 21 litres (as milk is not added or removed).
New quantity of water = (9 + x) litres.

The new ratio of milk to water is 3:7.
So, 21 / (9 + x) = 3/7

Cross-multiplying:
21 * 7 = 3 * (9 + x)
147 = 27 + 3x
147 - 27 = 3x
120 = 3x
x = 120 / 3
x = 40 litres.

Therefore, 40 litres of water should be added.

Question 6

A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:

Accepted Answer

We can use the Alligation Rule to find the ratio of quantities sold at different profit percentages.

Profit on 1st part = 8%
Profit on 2nd part = 18%
Overall profit = 14%

1st part (8%)           2nd part (18%)
             \             /
                Overall (14%)
               /             \
(18 - 14)     : (14 - 8)
    4         :     6

Simplifying the ratio: 4:6 = 2:3

This ratio (2:3) represents the ratio of the quantity of sugar sold at 8% profit to the quantity sold at 18% profit.
Total quantity of sugar = 1000 kg.

Quantity sold at 18% profit = (3 / (2 + 3)) * 1000
= (3/5) * 1000
= 3 * 200
= 600 kg.

Question 7

The ratio of milk and water in a 45-litre mixture is 4:1. How much water should be added to make the ratio 3:2?

Accepted Answer

Initial quantity of mixture = 45 litres.
Initial ratio of milk to water = 4:1.

Quantity of milk = (4 / (4+1)) * 45 = (4/5) * 45 = 36 litres.
Quantity of water = (1 / (4+1)) * 45 = (1/5) * 45 = 9 litres.

Let 'x' litres of water be added.

New quantity of milk = 36 litres (milk remains constant).
New quantity of water = (9 + x) litres.

The new ratio of milk to water is 3:2.
So, 36 / (9 + x) = 3/2

Cross-multiplying:
36 * 2 = 3 * (9 + x)
72 = 27 + 3x
72 - 27 = 3x
45 = 3x
x = 45 / 3
x = 15 litres.

Therefore, 15 litres of water should be added.

Question 8

In what ratio must water be mixed with milk to gain 20% on selling the mixture at the cost price of milk?

Accepted Answer

Let the Cost Price (CP) of 1 litre of pure milk be ₹1.
When water is mixed with milk, the cost of water is ₹0.

The problem states that the mixture is sold at the cost price of milk, so the Selling Price (SP) of 1 litre of mixture is ₹1.
There is a gain of 20% on this sale.

We know that SP = CP + Profit = CP + 0.20 * CP = 1.20 * CP (of the mixture).
So, 1 = 1.20 * CP (of mixture)
CP (of mixture) = 1 / 1.20 = 1 / (6/5) = ₹5/6 per litre.

Now, we use the Alligation Rule to find the ratio of water to milk:
Cost of Water (₹0/litre)    Cost of Milk (₹1/litre)
                 \           /
                   Mean Cost (₹5/6/litre)
                  /           \
(₹1 - ₹5/6)     : (₹5/6 - ₹0)
   (1/6)        : (5/6)

Multiplying by 6 to remove the denominators:
(1/6 * 6) : (5/6 * 6)
1         : 5

So, water and milk must be mixed in the ratio 1:5.

Question 9

A vessel contains 60 litres of milk. 12 litres of milk is taken out and replaced with water. This process is repeated once more. The amount of milk (in litres) in the vessel after the second operation is:

Accepted Answer

Initial quantity of milk = 60 litres.
Quantity of milk taken out and replaced with water = 12 litres.

The process is repeated once more, meaning it happens a total of 2 times.

Using the formula for the remaining quantity of the original liquid:
Remaining Quantity = Initial Quantity * (1 - (Quantity removed / Initial Quantity))^n

Remaining milk = 60 * (1 - 12/60)^2
= 60 * (1 - 1/5)^2
= 60 * (4/5)^2
= 60 * (16/25)
= (60 * 16) / 25
= (12 * 16) / 5 (dividing 60 and 25 by 5)
= 192 / 5
= 38.4 litres.

Question 10

A mixture of 70 litres of milk and water contains 10% water. How much water must be added to make the water 12.5% of the total mixture?

Accepted Answer

Initial quantity of mixture = 70 litres.
Initial percentage of water = 10%.

Quantity of water = 10% of 70 = (10/100) * 70 = 7 litres.
Quantity of milk = 70 - 7 = 63 litres.

Let 'x' litres of water be added to the mixture.

New total mixture = (70 + x) litres.
New quantity of water = (7 + x) litres.
Quantity of milk remains = 63 litres.

The new percentage of water is 12.5%.
So, (New quantity of water / New total mixture) * 100 = 12.5
(7 + x) / (70 + x) = 12.5 / 100
(7 + x) / (70 + x) = 1/8 (since 12.5 = 100/8)

Cross-multiplying:
8 * (7 + x) = 1 * (70 + x)
56 + 8x = 70 + x
8x - x = 70 - 56
7x = 14
x = 14 / 7
x = 2 litres.

Therefore, 2 litres of water must be added.

Question 11

In what ratio must a grocer mix two varieties of pulses costing ₹15 and ₹20 per kg respectively so as to get a mixture worth ₹16.50 per kg?

Accepted Answer

Using the rule of Alligation:
Cost of cheaper pulse: ₹15
Cost of dearer pulse: ₹20
Mean price (desired price): ₹16.50

Quantity of cheaper pulse (part 1) : Quantity of dearer pulse (part 2)
= (Cost of dearer - Mean price) : (Mean price - Cost of cheaper)
= (20 - 16.50) : (16.50 - 15)
= 3.50 : 1.50
To simplify the ratio, multiply by 100 to remove decimals: 350 : 150
Divide by 50: 7 : 3
Thus, the pulses must be mixed in the ratio 7:3.

Question 12

A container contains 40 liters of milk. From this container, 4 liters of milk was taken out and replaced by water. This process was repeated two more times. How much milk is now contained in the container?

Accepted Answer

Initial quantity of milk = 40 liters.
Quantity taken out in each step = 4 liters.
Total number of times the process is repeated = 1 (first time) + 2 (two more times) = 3 times.

The formula for the quantity of milk left after 'n' repetitions is:
Quantity left = Initial Quantity * (1 - (Quantity removed / Initial Quantity))^n

Quantity of milk left = 40 * (1 - (4/40))^3
= 40 * (1 - 1/10)^3
= 40 * (9/10)^3
= 40 * (729/1000)
= 4 * (729/100)
= 2916/100
= 29.16 liters.

Question 13

A milkman sells milk at the cost price but he mixes water in it and thereby gains 25%. The percentage of water in the mixture is:

Accepted Answer

Let the cost price (CP) of 1 liter of pure milk be ₹1.
Since the milkman sells at the cost price, the selling price (SP) of 1 liter of the mixture is ₹1.
He gains 25%, which means his profit is 25% of his cost.
The profit comes entirely from the quantity of water mixed, as water costs nothing.

If the profit is 25%, it means for every ₹100 worth of milk, he makes ₹25 profit by adding water.
So, if the cost of milk is 100 parts, the profit (due to water) is 25 parts.

Ratio of Water to Milk = Profit percentage / 100% = 25/100 = 1/4.
This implies that for every 4 parts of milk, there is 1 part of water.

Total mixture = Milk + Water = 4 + 1 = 5 parts.
Percentage of water in the mixture = (Quantity of Water / Total Mixture) * 100
= (1 / 5) * 100 = 20%.

Question 14

Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. In what ratio should quantities be taken from both vessels to form a mixture containing spirit and water in the ratio 8:5?

Accepted Answer

We can use the rule of alligation by considering the concentration of spirit (or water) in each mixture.
Concentration of spirit in Vessel A = 5 / (5+2) = 5/7
Concentration of spirit in Vessel B = 7 / (7+6) = 7/13
Desired concentration of spirit in the new mixture = 8 / (8+5) = 8/13

Applying the rule of Alligation:

Concentration of Spirit in A (C1) = 5/7
Concentration of Spirit in B (C2) = 7/13
Mean Concentration (M) = 8/13

(C1)         (C2)
    5/7          7/13
        \      /
          8/13
        /      \
(M - C2) : (C1 - M)
(8/13 - 7/13) : (5/7 - 8/13)
(1/13)        : ( (5*13 - 8*7) / (7*13) )
(1/13)        : ( (65 - 56) / 91 )
(1/13)        : (9/91)

To simplify the ratio, multiply both sides by the LCM of 13 and 91, which is 91:
(1/13) * 91 : (9/91) * 91
7 : 9

So, the quantities from vessels A and B should be mixed in the ratio 7:9.

Question 15

Two vessels contain milk and water in the ratio 3:1 and 7:1 respectively. If equal quantities of both mixtures are mixed, what is the ratio of milk and water in the new mixture?

Accepted Answer

Let's assume we take 1 unit of mixture from each vessel for calculation convenience, as equal quantities are mixed.

From Vessel 1 (Milk:Water = 3:1, Total parts = 4):
Quantity of Milk = (3/4) * 1 = 3/4
Quantity of Water = (1/4) * 1 = 1/4

From Vessel 2 (Milk:Water = 7:1, Total parts = 8):
Quantity of Milk = (7/8) * 1 = 7/8
Quantity of Water = (1/8) * 1 = 1/8

When these equal quantities are mixed:
Total Milk in the new mixture = (3/4) + (7/8)
To add these fractions, find a common denominator (LCM of 4 and 8 is 8):
Total Milk = (6/8) + (7/8) = 13/8

Total Water in the new mixture = (1/4) + (1/8)
Total Water = (2/8) + (1/8) = 3/8

The ratio of Milk to Water in the new mixture is:
(13/8) : (3/8)
Since the denominators are the same, the ratio is 13:3.

Question 16

A shopkeeper had 2000 kg of sugar. He sold a part of it at 10% profit and the rest at 20% profit. If he gained 14% on the whole, the quantity sold at 20% profit is:

Accepted Answer

Let the two parts of sugar be sold at 10% profit and 20% profit respectively. The overall profit is 14%.
We can use the rule of Alligation to find the ratio of quantities sold at each profit percentage.

Profit on 1st part (P1) = 10%
Profit on 2nd part (P2) = 20%
Mean profit (M) = 14%

(P1)         (P2)
    10%          20%
        \      /
          14%
        /      \
(P2 - M) : (M - P1)
(20% - 14%) : (14% - 10%)
6%          : 4%

The ratio of the quantity sold at 10% profit to the quantity sold at 20% profit is 6:4, which simplifies to 3:2.

Total quantity of sugar = 2000 kg.
The quantity sold at 20% profit corresponds to 2 parts of this ratio.
Quantity sold at 20% profit = (2 / (3+2)) * 2000
= (2/5) * 2000
= 2 * 400 = 800 kg.

Question 17

A man travelled a distance of 80 km in 7 hours, partly on foot at 8 km/hr and partly on bicycle at 16 km/hr. The distance travelled on foot is:

Accepted Answer

Let the time travelled on foot be T_foot and time travelled on bicycle be T_bicycle.
Total time = T_foot + T_bicycle = 7 hours.

Average speed for the entire journey = Total distance / Total time
Average speed = 80 km / 7 hours.

Using the rule of Alligation for speeds (or time taken):

Speed on foot (S1) = 8 km/hr
Speed on bicycle (S2) = 16 km/hr
Mean speed (M) = 80/7 km/hr

(S1)             (S2)
    8 km/hr          16 km/hr
        \          /
           80/7 km/hr
        /          \
(S2 - M) : (M - S1)
(16 - 80/7) : (80/7 - 8)
( (16*7 - 80)/7 ) : ( (80 - 8*7)/7 )
( (112 - 80)/7 ) : ( (80 - 56)/7 )
(32/7)        : (24/7)

The ratio of the time taken (Time on foot : Time on bicycle) is 32:24, which simplifies to 4:3.

Total time = 7 hours.
Time travelled on foot = (4 / (4+3)) * 7 = (4/7) * 7 = 4 hours.
Time travelled on bicycle = (3 / (4+3)) * 7 = (3/7) * 7 = 3 hours.

Distance travelled on foot = Speed on foot * Time on foot
= 8 km/hr * 4 hours = 32 km.

Question 18

A 20-liter mixture contains milk and water in the ratio 5:3. If 4 liters of this mixture is replaced by 4 liters of milk, the ratio of milk and water in the new mixture will be:

Accepted Answer

Initial total mixture = 20 liters.
Ratio of Milk : Water = 5:3.

Initial quantity of Milk = (5/8) * 20 = 100/8 = 12.5 liters
Initial quantity of Water = (3/8) * 20 = 60/8 = 7.5 liters

Step 1: 4 liters of the mixture is removed.
When 4 liters of mixture is removed, milk and water are removed in their original ratio (5:3).
Quantity of Milk removed = (5/8) * 4 = 2.5 liters
Quantity of Water removed = (3/8) * 4 = 1.5 liters

Remaining quantities after removal:
Milk remaining = 12.5 - 2.5 = 10 liters
Water remaining = 7.5 - 1.5 = 6 liters
Total mixture remaining = 10 + 6 = 16 liters.

Step 2: 4 liters of milk is added to the remaining mixture.
New quantity of Milk = 10 + 4 = 14 liters
New quantity of Water = 6 liters (as only milk was added)

New total mixture = 14 + 6 = 20 liters.

The new ratio of Milk : Water = 14 : 6.
Simplifying this ratio by dividing both by 2:
14 : 6 = 7 : 3.

Question 19

The average marks of boys in a class is 60 and the average marks of girls is 75. The average marks of the entire class is 66. What is the ratio of the number of boys to the number of girls in the class?

Accepted Answer

We can use the rule of Alligation to find the ratio of the number of boys to the number of girls.

Average marks of Boys (A_B) = 60
Average marks of Girls (A_G) = 75
Average marks of the entire class (Mean Average, M) = 66

(A_B)         (A_G)
    60            75
        \      /
          66
        /      \
(A_G - M) : (M - A_B)
(75 - 66) : (66 - 60)
9         : 6

The ratio of the number of boys to the number of girls is 9:6.
Simplifying this ratio by dividing both by 3:
9 : 6 = 3 : 2.

Question 20

A mixture contains milk and water in the ratio 4:3. If 2 liters of water is added to the mixture, the ratio of milk and water becomes 8:7. The quantity of milk in the given mixture is:

Accepted Answer

Let the initial quantity of milk in the mixture be 4x liters and the initial quantity of water be 3x liters.

Initial ratio of Milk : Water = 4x : 3x.

When 2 liters of water is added to the mixture:
Quantity of Milk = 4x liters (remains unchanged)
Quantity of Water = (3x + 2) liters

The new ratio of Milk : Water becomes 8:7.
So, we can set up the equation:
4x / (3x + 2) = 8 / 7

Now, cross-multiply to solve for x:
7 * (4x) = 8 * (3x + 2)
28x = 24x + 16

Subtract 24x from both sides:
28x - 24x = 16
4x = 16

Divide by 4:
x = 4

The question asks for the quantity of milk in the given (initial) mixture.
Quantity of milk = 4x = 4 * 4 = 16 liters.

Question 21

A mixture of 20 litres of milk and water contains 10% water. How much more water should be added so that the new mixture contains 25% water?

Accepted Answer

Initial mixture = 20 litres
Quantity of water = 10% of 20 = 2 litres
Quantity of milk = 20 - 2 = 18 litres
Let 'x' litres of water be added.
New total mixture = (20 + x) litres
New quantity of water = (2 + x) litres
The new mixture contains 25% water.
So, (2 + x) / (20 + x) = 25/100 = 1/4
4(2 + x) = 20 + x
8 + 4x = 20 + x
3x = 12
x = 4 litres.

Question 22

A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with liquid B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained in the can initially?

Accepted Answer

Let the total initial quantity of the mixture be 'Q' litres.
Initial quantity of liquid A = (7/12)Q
Initial quantity of liquid B = (5/12)Q
When 9 litres of mixture are drawn off, the remaining mixture is (Q-9) litres. The ratio of A:B remains 7:5.
Remaining quantity of A = (7/12)(Q-9)
Remaining quantity of B = (5/12)(Q-9)
Now, 9 litres of liquid B are added.
New quantity of A = (7/12)(Q-9)
New quantity of B = (5/12)(Q-9) + 9
The new ratio of A:B is 7:9.
[(7/12)(Q-9)] / [(5/12)(Q-9) + 9] = 7/9
Cross-multiply: 9 * (7/12)(Q-9) = 7 * [(5/12)(Q-9) + 9]
Divide both sides by 7: (9/12)(Q-9) = (5/12)(Q-9) + 9
(3/4)(Q-9) - (5/12)(Q-9) = 9
Multiply by 12 to clear denominators: 9(Q-9) - 5(Q-9) = 108
4(Q-9) = 108
Q-9 = 27
Q = 36 litres
Initial quantity of liquid A = (7/12) * Q = (7/12) * 36 = 7 * 3 = 21 litres.

Question 23

In what ratio must water be mixed with milk costing ₹12 per litre so that the mixture costs ₹8 per litre?

Accepted Answer

Cost of water = ₹0 per litre
Cost of milk = ₹12 per litre
Mean price of mixture = ₹8 per litre
Using the Alligation Rule:
Cost of Milk (₹12)       Cost of Water (₹0)
         \           /
            Mean Price (₹8)
         /           \
(8 - 0)            (12 - 8)
   8                   4
The ratio of water to milk (quantity) is (12-8) : (8-0) = 4 : 8 = 1:2.

Question 24

A jar contains a mixture of liquid P and liquid Q in the ratio 4:1. When 10 litres of the mixture is taken out and 10 litres of liquid Q is poured into the jar, the ratio becomes 2:3. How many litres of liquid P was contained in the jar initially?

Accepted Answer

Let the initial quantity of liquid P be 4x and liquid Q be x. Total mixture = 5x.
When 10 litres of mixture is taken out:
Quantity of P removed = (4/5) * 10 = 8 litres
Quantity of Q removed = (1/5) * 10 = 2 litres
Remaining P = 4x - 8
Remaining Q = x - 2
Now, 10 litres of liquid Q is poured into the jar.
New quantity of P = 4x - 8
New quantity of Q = (x - 2) + 10 = x + 8
The new ratio P:Q is 2:3.
(4x - 8) / (x + 8) = 2/3
3(4x - 8) = 2(x + 8)
12x - 24 = 2x + 16
10x = 40
x = 4
Initial quantity of liquid P = 4x = 4 * 4 = 16 litres.

Question 25

Two vessels A and B contain milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio 9:4 is:

Accepted Answer

Consider the proportion of milk in each mixture.
Proportion of milk in Vessel A = 8/(8+5) = 8/13
Proportion of milk in Vessel B = 5/(5+2) = 5/7
Desired proportion of milk in the new mixture = 9/(9+4) = 9/13
Using the Alligation Rule:
Mixture A (Milk = 8/13)       Mixture B (Milk = 5/7)
            \           /
               Mean Mixture (Milk = 9/13)
            /           \
   |9/13 - 5/7|       |8/13 - 9/13|
   |(63-65)/91|       |-1/13|
   | -2/91 |          | -1/13 |
    2/91                 1/13
The ratio of the quantity of Mixture A to Mixture B is (Difference from B's side) : (Difference from A's side)
Ratio A:B = (2/91) : (1/13)
To simplify, multiply both sides by 91:
Ratio A:B = 2 : (91/13)
Ratio A:B = 2 : 7.

Question 26

A solution of salt and water contains 5% salt. If 20 litres of water are evaporated from the solution, the salt becomes 15%. Find the original quantity of the solution.

Accepted Answer

Let the original quantity of the solution be 'x' litres.
Initial quantity of salt = 5% of x = 0.05x litres.
When 20 litres of water are evaporated, the new total volume of the solution becomes (x - 20) litres.
The quantity of salt remains constant.
In the new solution, salt is 15%.
So, 0.05x = 15% of (x - 20)
0.05x = 0.15(x - 20)
Divide both sides by 0.05:
x = 3(x - 20)
x = 3x - 60
2x = 60
x = 30 litres.

Question 27

In 80 litres of mixture of milk and water, the ratio of milk to water is 7:3. In order to make the ratio 7:4, how much water must be added to the mixture?

Accepted Answer

Initial mixture = 80 litres.
Quantity of milk = (7 / (7+3)) * 80 = (7/10) * 80 = 56 litres
Quantity of water = (3 / (7+3)) * 80 = (3/10) * 80 = 24 litres
Let 'x' litres of water be added to the mixture.
New quantity of milk = 56 litres (as milk is not added or removed)
New quantity of water = (24 + x) litres
The new ratio of milk to water is 7:4.
56 / (24 + x) = 7/4
Cross-multiply: 56 * 4 = 7 * (24 + x)
Divide both sides by 7: 8 * 4 = 24 + x
32 = 24 + x
x = 32 - 24
x = 8 litres.

Question 28

How much water must be added to 100 ml of 80% acid solution to dilute it to 50% acid solution?

Accepted Answer

Initial volume of acid solution = 100 ml.
Quantity of acid = 80% of 100 ml = 80 ml.
Let 'x' ml of water be added.
New total volume of solution = (100 + x) ml.
The quantity of acid remains constant at 80 ml.
The new acid concentration is 50%.
So, 80 / (100 + x) = 50/100 = 1/2
Cross-multiply: 80 * 2 = 1 * (100 + x)
160 = 100 + x
x = 160 - 100
x = 60 ml.

Question 29

A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

Accepted Answer

Since the milkman sells at cost price, any gain he makes is solely due to the water added (assuming water is free).
Gain percentage = 25%
This means if the cost of milk is 100 units, the profit from water is 25 units.
So, for every 100 units of pure milk (cost-wise), there are 25 units of water.
Ratio of water to milk (by volume) = 25 : 100 = 1 : 4.
This means for every 1 part of water, there are 4 parts of milk.
Total mixture parts = 1 (water) + 4 (milk) = 5 parts.
Percentage of water in the mixture = (Quantity of Water / Total Quantity of Mixture) * 100
= (1 / 5) * 100 = 20%.

Question 30

A vessel contains 60 litres of milk. 12 litres of milk is taken out and replaced with water. This process is repeated one more time. What is the final quantity of milk in the vessel?

Accepted Answer

Initial quantity of milk = 60 litres.
Quantity of milk removed in each step = 12 litres.
Fraction of milk remaining after one operation = (Total Volume - Quantity Removed) / Total Volume
= (60 - 12) / 60 = 48 / 60 = 4/5.
This process is repeated two times.
Final quantity of milk = Initial quantity * (Fraction remaining after one operation)^Number of operations
Final quantity of milk = 60 * (4/5)^2
= 60 * (16/25)
= (60 * 16) / 25
= (12 * 16) / 5 (dividing 60 and 25 by 5)
= 192 / 5
= 38.4 litres.

Question 31

A grocer mixes two varieties of pulses costing ₹15 per kg and ₹20 per kg respectively. In what ratio should these two varieties be mixed so that the average cost of the mixture is ₹16.50 per kg?

Accepted Answer

Let the cost of the first variety be C1 = ₹15 per kg.
Let the cost of the second variety be C2 = ₹20 per kg.
Let the mean cost of the mixture be Cm = ₹16.50 per kg.

Using the rule of alligation:

Cost of cheaper (₹15)        Cost of dearer (₹20)
          \             /
           Mean Cost (₹16.50)
          /             \
(C2 - Cm)               (Cm - C1)
(20 - 16.50)            (16.50 - 15)
3.50                    1.50

So, the ratio of the quantities of the cheaper to the dearer variety is 3.50 : 1.50.
To simplify the ratio, multiply by 100: 350 : 150
Divide by 50: 7 : 3

Thus, the two varieties should be mixed in the ratio 7:3.

Question 32

In what ratio must water be mixed with milk costing ₹60 per litre so that the mixture costs ₹48 per litre?

Accepted Answer

Let the cost of milk be ₹60 per litre. The cost of water is ₹0 per litre.
The mean cost of the mixture is ₹48 per litre.

Using the rule of alligation:

Cost of water (₹0)         Cost of milk (₹60)
          \             /
           Mean Cost (₹48)
          /             \
(60 - 48)               (48 - 0)
12                      48

So, the ratio of the quantity of water to milk is 12 : 48.
Simplifying the ratio: 12/12 : 48/12 = 1 : 4.

Thus, water and milk must be mixed in the ratio 1:4.

Question 33

A shopkeeper has two types of sugar. One costs ₹45 per kg and the other costs ₹60 per kg. If he mixes them in the ratio 2:3, what will be the price per kg of the mixture?

Accepted Answer

Let the price of the first type of sugar be P1 = ₹45 per kg.
Let the price of the second type of sugar be P2 = ₹60 per kg.
The ratio in which they are mixed is R1:R2 = 2:3.

The price per kg of the mixture (P_mixture) can be found using the weighted average formula:
P_mixture = (P1 * R1 + P2 * R2) / (R1 + R2)
P_mixture = (45 * 2 + 60 * 3) / (2 + 3)
P_mixture = (90 + 180) / 5
P_mixture = 270 / 5
P_mixture = ₹54 per kg.

Alternatively, using alligation:
Cost of cheaper (₹45)     Cost of dearer (₹60)
         \              /
          Mean Cost (x)
         /              \
(60 - x)                (x - 45)

Given ratio is 2:3. So, (60 - x) / (x - 45) = 2/3
3 * (60 - x) = 2 * (x - 45)
180 - 3x = 2x - 90
180 + 90 = 2x + 3x
270 = 5x
x = 270 / 5 = ₹54 per kg.

Question 34

A container contains 40 litres of milk. From this container, 4 litres of milk was taken out and replaced by water. This process was repeated two more times. How much milk is now in the container?

Accepted Answer

Initial quantity of milk = 40 litres.
Quantity taken out in each step = 4 litres.
Number of times the process is repeated = 1 (first time) + 2 (two more times) = 3 times.

The formula for the remaining quantity of the original liquid after 'n' operations is:
Remaining quantity = Initial quantity * (1 - (Quantity removed / Initial quantity))^n

Remaining milk = 40 * (1 - (4 / 40))^3
Remaining milk = 40 * (1 - 1/10)^3
Remaining milk = 40 * (9/10)^3
Remaining milk = 40 * (729 / 1000)
Remaining milk = (40 * 729) / 1000
Remaining milk = 29160 / 1000
Remaining milk = 29.16 litres.

So, 29.16 litres of milk is now in the container.

Question 35

Two vessels A and B contain mixtures of milk and water in the ratios 4:3 and 2:3 respectively. In what ratio should the contents of vessels A and B be mixed to obtain a new mixture containing milk and water in the ratio 1:1?

Accepted Answer

Let's consider the concentration of milk in each vessel.

Concentration of milk in vessel A (C_A) = 4 / (4+3) = 4/7
Concentration of milk in vessel B (C_B) = 2 / (2+3) = 2/5

Desired concentration of milk in the new mixture (C_M) = 1 / (1+1) = 1/2

Using the rule of alligation:

Concentration A (4/7)        Concentration B (2/5)
          \             /
           Mean Conc. (1/2)
          /             \
(C_B - C_M)               (C_M - C_A) - This is incorrect, it should be |C_M - C_B| and |C_M - C_A|

Correct application of alligation:
Quantity of A : Quantity of B = |C_M - C_B| : |C_M - C_A|

|C_M - C_B| = |1/2 - 2/5| = |(5-4)/10| = 1/10
|C_M - C_A| = |1/2 - 4/7| = |(7-8)/14| = |-1/14| = 1/14

Ratio of A to B = (1/10) : (1/14)
To remove fractions, multiply by the LCM of 10 and 14, which is 70.
Ratio = (1/10 * 70) : (1/14 * 70)
Ratio = 7 : 5

Thus, the contents of vessels A and B should be mixed in the ratio 7:5.

Question 36

A mixture of 70 litres contains milk and water in the ratio 3:4. How much water must be added to the mixture so that the ratio of milk to water becomes 2:5?

Accepted Answer

Initial total mixture = 70 litres.
Initial ratio of milk to water = 3:4.

Initial quantity of milk = (3 / (3+4)) * 70 = (3/7) * 70 = 30 litres.
Initial quantity of water = (4 / (3+4)) * 70 = (4/7) * 70 = 40 litres.

Let 'x' litres of water be added to the mixture.
New quantity of milk = 30 litres (milk quantity remains unchanged).
New quantity of water = (40 + x) litres.

New ratio of milk to water = 2:5.
So, 30 / (40 + x) = 2 / 5
Cross-multiply:
30 * 5 = 2 * (40 + x)
150 = 80 + 2x
150 - 80 = 2x
70 = 2x
x = 70 / 2 = 35 litres.

Therefore, 35 litres of water must be added.

Question 37

A container has a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the container is filled with 9 litres of liquid B, the ratio of A and B becomes 7:9. How many litres of liquid A was in the container initially?

Accepted Answer

Let the initial quantities of liquid A and B be 7x and 5x litres respectively. Total mixture = 12x litres.

When 9 litres of mixture are drawn off:
Quantity of A removed = (7 / 12) * 9 = 63/12 = 21/4 litres.
Quantity of B removed = (5 / 12) * 9 = 45/12 = 15/4 litres.

Remaining quantity of A = 7x - 21/4
Remaining quantity of B = 5x - 15/4

Now, 9 litres of liquid B are added.
New quantity of A = 7x - 21/4
New quantity of B = 5x - 15/4 + 9 = 5x + ((-15 + 36)/4) = 5x + 21/4

The new ratio of A to B is 7:9.
(7x - 21/4) / (5x + 21/4) = 7/9

Multiply both sides by 4 to clear denominators:
(28x - 21) / (20x + 21) = 7/9

Cross-multiply:
9 * (28x - 21) = 7 * (20x + 21)
252x - 189 = 140x + 147
252x - 140x = 147 + 189
112x = 336
x = 336 / 112
x = 3

Initial quantity of liquid A = 7x = 7 * 3 = 21 litres.
Initial quantity of liquid B = 5x = 5 * 3 = 15 litres.

Initial total mixture = 21 + 15 = 36 litres. This is consistent with the calculations.

So, there were 21 litres of liquid A in the container initially.

Question 38

How many kilograms of sugar costing ₹9 per kg must be mixed with 27 kg of sugar costing ₹7 per kg so that selling the mixture at ₹9.24 per kg may gain 10%?

Accepted Answer

Selling price of the mixture = ₹9.24 per kg.
Profit percentage = 10%.

Let the cost price of the mixture be CP_mixture.
SP = CP * (1 + Profit% / 100)
9.24 = CP_mixture * (1 + 10/100)
9.24 = CP_mixture * (110/100)
9.24 = CP_mixture * (11/10)
CP_mixture = 9.24 * (10/11)
CP_mixture = 92.4 / 11
CP_mixture = ₹8.40 per kg.

Now we use the rule of alligation with the cost prices:
Cost of cheaper sugar (C1) = ₹7 per kg.
Cost of dearer sugar (C2) = ₹9 per kg.
Mean cost of mixture (Cm) = ₹8.40 per kg.

Cost of cheaper (₹7)        Cost of dearer (₹9)
          \             /
           Mean Cost (₹8.40)
          /             \
(9 - 8.40)              (8.40 - 7)
0.60                    1.40

Ratio of cheaper sugar to dearer sugar = 0.60 : 1.40 = 6 : 14 = 3 : 7.

We are given that the quantity of cheaper sugar (costing ₹7 per kg) is 27 kg.
Let the quantity of dearer sugar (costing ₹9 per kg) be 'x' kg.

From the ratio, we have:
Quantity of cheaper / Quantity of dearer = 3 / 7
27 / x = 3 / 7
3x = 27 * 7
3x = 189
x = 189 / 3
x = 63 kg.

Therefore, 63 kg of sugar costing ₹9 per kg must be mixed.

Question 39

A milkman sells milk at the cost price, but he mixes water in it and thereby gains 25%. The ratio of milk and water in the mixture is:

Accepted Answer

The milkman gains 25% by mixing water and selling at cost price. This means the profit comes entirely from the water added.

Let the cost price of 1 litre of milk be ₹1.
Since he sells at cost price, the selling price of 1 litre of mixture is also ₹1.
He makes a profit of 25%.

If the cost price of the pure milk is CP_M, and the cost of water is 0.
Let the quantity of milk be 'M' and water be 'W'.

The profit is 25% on the cost price of milk (assuming water has no cost).
This means for every 100 parts of milk, he adds 25 parts of water to get a 25% profit.

Alternatively, consider the value.
If the milkman gains 25%, it means that the quantity of water added is 25% of the quantity of milk.
If milk quantity is 100 units, then water quantity is 25 units.
Ratio of water to milk = 25:100 = 1:4.
Since the question asks for the ratio of milk and water, it is 4:1.

Using alligation:
Cost of water = ₹0 per unit.
Cost of milk = Let it be ₹1 per unit (or any 'x' value, it will cancel out).
Selling price of mixture = Cost price of milk = ₹1 per unit.
Since there's a 25% profit, the actual cost price of the mixture must be:
CP_mixture = SP / (1 + Profit% / 100) = 1 / (1 + 25/100) = 1 / (1.25) = 1 / (5/4) = ₹4/5 per unit.

Now, apply alligation:
Cost of water (0)       Cost of milk (1)
          \           /
           Mean Cost (4/5)
          /           \
(1 - 4/5)             (4/5 - 0)
1/5                   4/5

Ratio of water : milk = 1/5 : 4/5 = 1 : 4.
So, the ratio of milk : water = 4 : 1.

Question 40

A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and thereby gains 16 2/3%. Find the percentage of water in the mixture.

Accepted Answer

The gain percentage is 16 2/3%.
16 2/3% can be written as 50/3%.
As seen in the previous question, the profit comes from the water added. So, the quantity of water is 16 2/3% of the quantity of milk.

Let the quantity of milk be M and water be W.
Profit = (Water / Milk) * 100
50/3 = (W / M) * 100
W/M = (50/3) / 100 = 50 / 300 = 1/6

So, the ratio of Water : Milk = 1 : 6.

Total mixture = Milk + Water = 6 + 1 = 7 parts.
Percentage of water in the mixture = (Water / Total mixture) * 100
Percentage of water = (1 / 7) * 100
Percentage of water = 100/7 % = 14.2857...
Approximately 14.28%.

Alternatively, using alligation (similar to Q9):
Let cost of milk = ₹1. Cost of water = ₹0.
SP of mixture = CP of milk = ₹1.
Profit = 16 2/3% = 1/6.
CP of mixture = SP / (1 + Profit) = 1 / (1 + 1/6) = 1 / (7/6) = 6/7.

Using alligation for ratio of water to milk:
Cost of water (0)        Cost of milk (1)
          \             /
           Mean Cost (6/7)
          /             \
(1 - 6/7)               (6/7 - 0)
1/7                     6/7

Ratio of Water : Milk = 1/7 : 6/7 = 1 : 6.

Percentage of water in the mixture = (1 / (1+6)) * 100 = (1/7) * 100 = 14.28%.

Question 41

In a mixture of 60 litres, the ratio of milk and water is 2:1. If the ratio is to be 1:2, then the amount of water (in litres) to be further added is:

Accepted Answer

Initial total mixture = 60 litres.
Initial ratio of milk to water = 2:1.

Initial quantity of milk = (2 / (2+1)) * 60 = (2/3) * 60 = 40 litres.
Initial quantity of water = (1 / (2+1)) * 60 = (1/3) * 60 = 20 litres.

Let 'x' litres of water be added to the mixture.
New quantity of milk = 40 litres (milk quantity remains unchanged).
New quantity of water = (20 + x) litres.

New ratio of milk to water = 1:2.
So, 40 / (20 + x) = 1 / 2
Cross-multiply:
40 * 2 = 1 * (20 + x)
80 = 20 + x
x = 80 - 20
x = 60 litres.

Therefore, 60 litres of water must be further added.