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Mixtures & Alligations

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CAT 2025 Lesson : Mixtures & Alligations - Hybrid, Evaporation & Past Questions

5.2 Switching simultaneously and concentrations do not become equal

There is no shortcut/formula we recommend here. We apply the basics to solve this.

Example 18

Two alloys A and B weigh

10

kg each. The percentage of copper in A and B are

80 \%

and

60 \%

respectively. Two kg is removed from each of the alloys A and B. The portion removed from A is added to B and vice versa. What is the percentage point difference in the copper concentration of the two alloys?

Solution

	Alloy A	Alloy B
Copper in $10$ kg of	$0.8 \times 10 = 8$ kg	$0.6 \times 10 = 6$ kg
Copper in $2$ kg of	$0.8 \times 2 = 1.6$ kg	$0.6 \times 2 = 1.2$ kg
*After Replacement*
Copper Quantity	$8 - 1.6 + 1.2 = 7.6$ kg	$6 - 1.2 + 1.6 = 6.4$ kg
Copper $\%$	$76 \%$	$64 \%$

Percentage point difference

= 76 \% - 64 \% = 12 \%

Answer:

12 \%

5.3 Non-Simultaneous Transfer

Example 19

Vessel A contains

32

litres of milk solution where the ratio of milk to water is

3 : 1

. Vessel B contains

40

litres of milk solution where the ratio of milk to water is

1 : 3

8

litres of the solution is transferred from Vessel A to Vessel B.

12

litres of the solution in Vessel B is then transferred to Vessel A. What is the ratio of milk to water in Vessel A after the two transfers?

Solution

Applying the given ratios, milk and water in Vessel A are

24

litres and

8

litres respectively, while in Vessel B are

10

litres and

30

litres respectively.

Initially one-fourth of Vessel A (

8

out of

32

litres) is transferred to B. Therefore, Milk and Water in Vessel A will reduce by one-fourth.

Now, the total quantity in Vessel B is

40 + 8 = 48

litres. Now, one-fourth of Vessel B (

12

out of

48

litres) is transferred to A. Therefore, Milk and Water in Vessel B will reduce by one-fourth.

(in litres)	Vessel A			Vessel B
	Milk	Water	Total	Milk	Water	Total
Initial Quantity	$24$	$8$	$32$	$10$	$30$	$40$
Transferring $8$ litres to B	$24 - 6$ $= 18$	$8 - 2$ $= 6$	$32 - 8$ $= 24$	$10 + 6$ $= 16$	$30 + 2$ $= 32$	$40 + 8$ $= 48$
Transferring $12$ litres to A	$18 + 4$ $= 22$	$6 + 8$ $= 14$	$24 + 12$ $= 36$	$16 - 4$ $= 12$	$32 - 8$ $= 24$	$48 - 12$ $= 36$

Final ratio of milk and water in Vessel A

= 22 : 14 = 11 : 7

Answer:

11 : 7

6. Evaporation

These questions involve mixture of two items, where only one item is evaporating (or reducing), with the other remaining constant. Here, we first find the quantity of the item that does not reduce and then use this to calculate the total quantity of the mixture.

Example 20

Fresh grapes contain

90 \%

water by weight while dry grapes contain

20 \%

water by weight. What is the weight of dry grapes available from

20

kg of fresh grapes?
[CAT 2001]

(1)

2

kg (2)

2.4

kg (3)

2.5

kg (4) None of the above

Solution

When water evaporates from Fresh Grapes (

90 \%

water and

10 \%

pulp), we get Dry Grapes (

20 \%

water and

80 \%

pulp). This is because only the water quantity reduces, whereas the quantity of pulp remains the same.

Quantity of pulp in

20

kg of Fresh Grapes

= 10 \%

20

= 2

kg

Let the quantity of Dry Grapes be

x

kg. In these dry Grapes

2

kg should constitute the

80 \%

pulp.

\dfrac{80}{100} \times x = 2

⇒

x = 2.5

Answer: (3)

2.5

7. Past Questions

Question 1

A manufacturer has

200

litres of acid solution which has

15 \%

acid content. How many litres of acid solution with

30 \%

acid content may be added so that acid content in the resulting mixture will be more than

20 \%

but less than

25 \%

?
[XAT 2010]

		More than $100$ litres but less than $300$ litres
		More than $120$ litres but less than $400$ litres
		More than $100$ litres but less than $400$ litres
		More than $120$ litres but less than $300$ litres
		None of the above

Observation/Strategy

1

) As there are only

2

quantities involved, we can use alligations approach.

2

) As we need to find the range with the weighted average being between

20 \%

and

25 \%

, we can use two alligations to find the range of the quantities.

Let

x

and

y

be the litres of the

30 \%

acid solution that is required to form mixtures of

20 \%

and

25 \%

acid solutions respectively.

The following alligations provide the range of quantity of

30 \%

acid solutions to be between

100

litres and

400

litres.

Answer: (

3

) More than

100

litres but less than

400

litres

Question 2

Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a

10

litre container filled with orange juice, and replacing it with pineapple juice. If Gopal draws out another jug of the resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres, is
[XAT 2012]

		$2$
		$> 2$ and $< 2.5$
		$2.5$
		$> 2.5$ and $\le 3$
		$\ge 3$

Observation/Strategy

1

) This is

2

successive replacements.

2

) Pineapple juice is replacing the orange juice.

Let

x

be the size of a jug.

Final Orange juice quantity

= 10 \times \left(1 - \dfrac{x}{10} \right)^{2} = 5

⇒

\left(1 - \dfrac{x}{10} \right) = \sqrt{0.5}

\implies \dfrac{x}{10} = 1 - \sqrt{0.5}

(As

\sqrt{0.49} = 0.7,

\sqrt{0.5}

0.71

)

⇒

\dfrac{x}{10} = 1 - 0.71

⇒

x = 2.9

∴ The volume of the jug is between

2.5

and

3

litres.

Answer: (

4

)

> 2.5

and

\le 3

Question 3

There are two alloys P and Q made up of silver, copper and aluminium. Alloy P contains

45 \%

silver and rest aluminium. Alloy Q contains

30 \%

silver,

35 \%

copper and rest aluminium. Alloys P and Q are mixed in the ratio of

1 : 4.5

. The approximate percentages of silver and copper in the newly formed alloy is:
[IIFT 2015]

		$33 \%$ and $29 \%$
		$29 \%$ and $26 \%$
		$35 \%$ and $30 \%$
		None of the above

Observation/Strategy

1

) We can use alligations or simple weighted average here.

2

) The ratio of

1 : 4.5

can be simplified as

2 : 9

.

Percentage of Silver in final alloy

= \dfrac{(2 \times 45) + (9 \times 30)}{(2 + 9)} = \dfrac{360}{11}

32.72

Percentage of Copper in final alloy

= \dfrac{(2 \times 0 + 9 \times 35)}{2 + 9} = \dfrac{315}{11}

28.64

The answers are closest to

33 \%

And

29 \%

mentioned in option

1

.

Answer: (

1

)

33 \%

and

29 \%

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