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Arithmetic II

Mixtures & Alligations

ALL MODULES

CAT 2025 Lesson : Mixtures & Alligations - Examples 7 to 13

In the following example three liquids,

1

solution and

2

pure liquids, are mixed. We can combine the

2

pure liquids into a solution and use alligation.

Example 7

A vessel had

50

litres of milk solution (milk and water) that contained

20 \%

water. Kumaran then added equal quantities of pure milk and water to this vessel. If the resultant solution in the vessel had

60 \%

milk, then what was the quantity of pure milk (in litres) added to the vessel?

(1)

25

(2)

50

(3)

75

(4)

100

Solution

To the initial solution containing

20 \%

water (or

80 \%

milk), equal quantities of pure milk and water are added. The final solution contains

60 \%

milk.

Let the quantity of pure milk and water being added be

x

litres each. Total quantity of these two is

2x

and this mixture would contain

50 \%

water.

\therefore \dfrac{50}{2x} = \dfrac{10}{20}

⇒

x = 50

Answer: (2)

50

Example 8

24

litres of a solution that contains milk and water in the ratio of

5 : 3

is mixed with

27

litres of a solution that has milk and water in the ratio

7 : 2

. This solution is then mixed with

26

litres of a solution that has milk and water in the ratio of

8 : 5

, to form the final solution. What is the ratio of milk and water in the final solution?

Solution

The portion of milk in the three solutions are

\dfrac{5}{8}, \dfrac{7}{9}

and

\dfrac{8}{13}

.

The alligation method will be difficult as the LCM of denominators

(8, 9, 13) = 936

, is a very big number.

In this question, however, the litres of milk and water in each of the solutions have integral values. Therefore, we do not use alligation and directly find the portion of milk and water in each solution.

Qty (in Litres)	Total	Ratio	Milk	Water
Solution 1	$24$	$5 : 3$	$15$	$9$
Solution $2$	$27$	$7 : 2$	$21$	$6$
Solution $3$	$26$	$8 : 5$	$16$	$10$
Total			$\bm{52}$	$\bm{25}$

When the three solutions are mixed, we get a final solution containing

52

litres of milk and

25

litres of water.

∴ Ratio of milk to water in final solution

= 52 : 25

Answer:

52 : 25

In Examples 9 and 10, we deal with cases where a pure commodity is adulterated and the final mixture is sold at Cost Price. In these cases, the adulterant is the profit.

Example 9

2

kg of stones are mixed with

18

kg of rice and sold at the cost price of rice, then what is the profit percentage? [Assume the cost of stones is zero]

(1)

10 \%

(2)

11.1 \%

(3)

12.5 \%

(4) Cannot be determined

Solution

Let the CP of rice be Rs.

1

per kg.

CP of

18

kg of rice = Rs.

18

When

2

kg of stone is added, we get a total of

20

kg of adulterated rice, which is then sold at Rs.

x

per kg.

SP of

20

kg of rice = Rs.

20

Profit

\% = \dfrac{20 - 18}{18} \times 100 \% = 11.1 \%

Alternatively

For every

18

kg of rice purchased, extra

2

kg is sold. This additional

2

kg is the profit.

Profit

\% =\dfrac{2}{18} \times 100 \% = 11.1 \%

Answer:

(2) 11.1 \%

Example 10

A merchant buys and sells lemon concentrate. He wants to sell it at the cost price and achieve a profit of

10

%. He decides to achieve this by adding freely available water. How much water (in ml) must he add to each litre of lemon concentrate?

(1)

90

ml (2)

90.9

ml (3)

100

ml (4)

110

Solution

1

litre

= 1000

ml

As the lemon concentrate's SP = CP, the added water added will be the profit.

Let

x

ml of water be added to

1000

ml of pure lemon concentrate.

\dfrac{x}{1000} \times 100 \% = 10 \%

⇒

x = 100

ml

Answer:

(3) 100

3.1 Division of Parts

This is an extension of the alligation concept and an alternative to solve the alligation questions. Unlike the alligation method, division of parts can be used only when the ratio of weights is known. The following example details this concept. The solution may look long due to the detailed explanation. However, once you get a hang of it, this recommended approach saves time.

Example 11

The average marks of students of Classes A and B were

60

and

80

respectively. If there are

40

students in Class A and

60

students in Class B, then what is their combined average?

Solution

Ratio of number of students

= 40 : 60 = 2 : 3

The difference in the marks of Classes A and B is

80 - 60 =

$\bm{20}$ marks.

This difference is to be divided in the ratio of weights. As the ratio is

2 : 3

, we have a total of

2 + 3 =

$\bm{5}$ parts.

The difference of

20

marks can be divided into

5

parts with each part being $\bm{4}$ marks. This is shown in the diagram below.

As the ratio of A to B is

2 :3

,
- A has the strength to pull

2

parts from B
- B has the strength to pull

3

parts from A

Therefore, the average will be between $\bm{60}$ and $\bm{80}$ . So, to arrive at the weighted average, we either add to A's marks or subtract from B's marks, as shown below.

A pulls the average, from B, with it's strength of

2

parts (

= 8

marks).
∴ Weighted Average of Class = B's mark - Value of A's Parts

= 80 - 8 =

$\bm{72}$ marks

B pulls the average, from A, with it's strength of

3

parts (or

12

marks).
∴ Weighted Average of Class = A's mark + Value of B's Parts

= 60 + 12 =

$\bm{72}$ marks

This is shown in the diagram below.

Answer:

72

marks

Example 12

Ram purchased

600

Oranges at Rs.

27

each and

360

Apples at Rs.

39

each. What was the average price of a fruit?

Solution

Ratio of Oranges to Apples

= 600 : 360 = 5 : 3

Difference

= 39 - 27 =

Rs.

12

Total Parts

= 5 + 3 = 8

parts

1

Part

= \dfrac{12}{8} =

Rs.

1.5

Apples pull the average by

3

parts from the Price of Oranges.

Weighted average Price per fruit

= 27 + (3 \times 1.5) = 31.5

Answer: Rs.

31.5

Note: Weighted average computed from an Apple's Price will be

39 - (5 \times 1.5) = 31.5

Example 13

A class contains girls and boys in the ratio of

5 : 2

. In a recently conducted math test, the girls scored an average of

84 \%

and the class as a whole scored an average of

78 \%

. What was the average of the boys in the test?

(1)

63 \%

(2)

68 \%

(3)

70 \%

(4)

73 \%

Solution

Boys have pulled the weighted average by

2

parts from the girls.
∴

2

parts

= 84 \% - 78 \% = 6 \%

⇒

1

part

= 3 \%

Boys average will be

5

parts away from the weighted average ⇒

78 \% - (5 \times 3 \%) = 63 \%

Answer:

(1) 63 \%

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