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Arithmetic II

Mixtures & Alligations

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CAT 2025 Lesson : Mixtures & Alligations - Successive Replacement & Examples 14 to 17

4. Successive Replacement

In these questions, we have a vessel with a solution (or pure liquid). A certain quantity of the solution is removed from this vessel and replaced with another liquid. This process is then repeated a number of times.

Note that the quantity of liquid in the vessel remains the same after each replacement. However, the concentration of the original solution/liquid changes with each replacement. Let's try to understand this with an example.

A vessel contains $\bm{50}$ litres of pure milk. Let's say a jug of volume $\bm{5}$ litres is used to remove milk and replace it with water, which means, the volume of the jug is $\bm{10\%}$ of the vessel. As milk is removed and replaced with water (a different liquid) the quantity of milk in the vessel reduces by

10 \%

for every replacement.

\bold{1^{\text{st}} Replacement}

: In the first replacement,

10 \%

of milk is removed and replaced with water.

Milk quantity

= 50 \times (1 - 10 \%) = 50 \times \left(1 - \dfrac{10}{100} \right) = 45

litres

\bold{2^{\text{nd}} Replacement}

10 \%

of the resultant solution is removed and replaced with water. So, the milk quantity further reduces by

10 \%

.

Milk Quantity

= 50 \times \left(1 - \dfrac{10}{100} \right) \left(1 - \dfrac{10}{100} \right) = 50 \times \left(1 - \dfrac{10}{100} \right) = 40.5

\bold{n^{\text{th}} Replacement}

: Milk quantity has been reduced by

10 \%, n

number of times.

Milk Quantity

= 50 \times \left(1 - \dfrac{10}{100} \right)^{n}

4.1 Conditions and Formulae

Where I is the initial quantity of non-replaced liquid,
$\bm{r\%}$ is the replacement percentage,
$\bm{n}$ is the number of replacements,
$\bm{F}$ is the final quantity of the non-replaced liquid,

F = I \times \left(1 - \dfrac{r}{100} \right)^{n}

(This is an extension of the formula for successive changes from the Percentages lesson)

If the quantity replaced is provided (for example, in litres) instead of percentage,

F = I \times \left(1 - \dfrac{\text{Replacement Quantity}}{\text{Total Quantity}} \right)^{n}

Note:

I

and

F

can both be in percentages (Example 14) or both be in litres (Example 15). Also note that, this formula can be applied when the initial quantity a pure liquid (Example 14) or a solution (Example 15).

Example 14

Ram removes

25 \%

of a jar filled with milk and replaces it with water. He once again removes

25 \%

of the resultant mixture and replaces it with water. He repeats this process a total of

5

times. What is the concentration of water in the final mixture?

Solution

Pure milk is the non-replaced liquid.

25 \%

is replaced a total of

5

times. Initially the jar contains

100 \%

milk.

Final

\%

of milk

= 100 \% \times \left(1 - \dfrac{25}{100} \right)^{5} = 100 \% \times \left(\dfrac{3}{4} \right)^{5} = 100 \% \times \dfrac{243}{1024}

24 \%

Final

\%

of water

= 100 \% - 24 \% = 76 \%

Answer:

76 \%

Example 15

A jar contains

10

litres of

75 \%

spirit solution. A

2

litre container is used to remove the solution from the jar and replace it with water. This process is repeated two more times. How many litres of spirit is left in the jar after the final replacement?

Solution

Initial quantity of Spirit

= 0.75 \times 10 = 7.5

litres

F = I \times \left(1 - \dfrac{\text{Replacement Quantity}}{\text{Total Quantity}} \right)^{n} = 7.5 \times \left(1 - \dfrac{2}{10} \right)^{3}

= 7.5 \times \left(\dfrac{4}{5} \right)^{3} = \dfrac{96}{25} = 3.84

litres

Answer:

3.84

litres

Example 16

60

-litre vessel is filled to the brim with milk and water in the ratio of

1 : 5

. Sonam replaces

15

litres of this liquid with milk. She then replaces

12

litres of the resulting liquid with milk, and then replaces

10

litres of the resulting liquid with milk. What percentage of the final solution is milk (rounded off to two decimal places)?

Solution

Water is the non-replaced liquid in this case.

Initial quantity of water

= \dfrac{5}{6} \times 60 = 50

litres

Final quantity of water

= 50 \times \left(1 - \dfrac{15}{60} \right) \times \left(1 - \dfrac{12}{60} \right) \times \left(1 - \dfrac{10}{60} \right)

= 50 \times \dfrac{3}{4} \times \dfrac{4}{5} \times \dfrac{5}{6} = 25

litres

∴ Milk Quantity in the final mixture

= 60 - 25 = 35

litres

\%

of Milk in final solution

= \dfrac{35}{60} \times 100 = \dfrac{175}{3} = 58.33 \%

Answer:

58.33 \%

5. Hybrid Replacement

5.1 Switching simultaneously and concentrations become equal

Where the quantities of

2

mixtures of different unknown concentrations are $\bm{a}$ and $\bm{b}$ , and a quantity of

x

is removed from each of the mixtures and placed in the other mixtures, and this results in both the mixtures having the same concentrations, then

x = \dfrac{ab}{a + b}

The following example provides two solutions – one using the formula and the other without.

Example 17

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs

8

kg and the second one weighs

16

kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?
[IIFT 2013]

(1)

3.33

kg (2)

4.67

kg (3)

5.33

kg (4) None of the above

Solution

The weights of the two alloys are

8

kg and

16

kg and the concentrations are unknown. Let

x

be the weight of the replaced portion from the two alloys.

x = \dfrac{8 \times 16}{8 + 16} = 5.33

kg

Alternatively

Let the portion of aluminium in the alloys weighing

8

kg and

16

kg be

m

and

n

respectively. Let

x

be the weight of the replaced portion from the two alloys.

Weight of aluminium in the two alloys are $\bm{8m}$ kg and $\bm{16n}$ kg respectively. Let weight of aluminium in

x

kg of alloy removed be $\bm{xm}$ kg and $\bm{xn}$ kg respectively.

Weight of aluminium after replacements are $\bm{8m - xm + xn}$ and $\bm{16n - xn + xm}$

Concentration of aluminium is the weight of aluminium divided by the total weight. These are equal for both the alloys.

Concentration

= \dfrac{8m - xm + xn}{8} = \dfrac{16n - xn + xm}{16}

⇒

16m - 2xm + 2xn = 16n - xn + xm

⇒

16m - 16n = 3xm - 3xn

⇒

16 (m - n) = 3x (m - n)

⇒

x = \dfrac{16}{3} = 5.33

Answer:

(3) 5.33

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Arithmetic II

Mixtures & Alligations

ALL MODULES

CAT 2025 Lesson : Mixtures & Alligations - Successive Replacement & Examples 14 to 17

4. Successive Replacement

10 \%

for every replacement.

\bold{1^{\text{st}} Replacement}

: In the first replacement,

10 \%

of milk is removed and replaced with water.

Milk quantity

= 50 \times (1 - 10 \%) = 50 \times \left(1 - \dfrac{10}{100} \right) = 45

litres

\bold{2^{\text{nd}} Replacement}

10 \%

of the resultant solution is removed and replaced with water. So, the milk quantity further reduces by

10 \%

.

Milk Quantity

= 50 \times \left(1 - \dfrac{10}{100} \right) \left(1 - \dfrac{10}{100} \right) = 50 \times \left(1 - \dfrac{10}{100} \right) = 40.5

\bold{n^{\text{th}} Replacement}

: Milk quantity has been reduced by

10 \%, n

number of times.

Milk Quantity

= 50 \times \left(1 - \dfrac{10}{100} \right)^{n}

4.1 Conditions and Formulae

F = I \times \left(1 - \dfrac{r}{100} \right)^{n}

(This is an extension of the formula for successive changes from the Percentages lesson)

If the quantity replaced is provided (for example, in litres) instead of percentage,

F = I \times \left(1 - \dfrac{\text{Replacement Quantity}}{\text{Total Quantity}} \right)^{n}

Note:

I

and

F

Example 14

Ram removes

25 \%

of a jar filled with milk and replaces it with water. He once again removes

25 \%

of the resultant mixture and replaces it with water. He repeats this process a total of

5

times. What is the concentration of water in the final mixture?

Solution

Pure milk is the non-replaced liquid.

25 \%

is replaced a total of

5

times. Initially the jar contains

100 \%

milk.

Final

\%

of milk

= 100 \% \times \left(1 - \dfrac{25}{100} \right)^{5} = 100 \% \times \left(\dfrac{3}{4} \right)^{5} = 100 \% \times \dfrac{243}{1024}

24 \%

Final

\%

of water

= 100 \% - 24 \% = 76 \%

Answer:

76 \%

Example 15

A jar contains

10

litres of

75 \%

spirit solution. A

2

Solution

Initial quantity of Spirit

= 0.75 \times 10 = 7.5

litres

F = I \times \left(1 - \dfrac{\text{Replacement Quantity}}{\text{Total Quantity}} \right)^{n} = 7.5 \times \left(1 - \dfrac{2}{10} \right)^{3}

= 7.5 \times \left(\dfrac{4}{5} \right)^{3} = \dfrac{96}{25} = 3.84

litres

Answer:

3.84

litres

Example 16

60

-litre vessel is filled to the brim with milk and water in the ratio of

1 : 5

. Sonam replaces

15

litres of this liquid with milk. She then replaces

12

litres of the resulting liquid with milk, and then replaces

10

litres of the resulting liquid with milk. What percentage of the final solution is milk (rounded off to two decimal places)?

Solution

Water is the non-replaced liquid in this case.

Initial quantity of water

= \dfrac{5}{6} \times 60 = 50

litres

Final quantity of water

= 50 \times \left(1 - \dfrac{15}{60} \right) \times \left(1 - \dfrac{12}{60} \right) \times \left(1 - \dfrac{10}{60} \right)

= 50 \times \dfrac{3}{4} \times \dfrac{4}{5} \times \dfrac{5}{6} = 25

litres

∴ Milk Quantity in the final mixture

= 60 - 25 = 35

litres

\%

of Milk in final solution

= \dfrac{35}{60} \times 100 = \dfrac{175}{3} = 58.33 \%

Answer:

58.33 \%

5. Hybrid Replacement

5.1 Switching simultaneously and concentrations become equal

Where the quantities of

2

mixtures of different unknown concentrations are $\bm{a}$ and $\bm{b}$ , and a quantity of

x

is removed from each of the mixtures and placed in the other mixtures, and this results in both the mixtures having the same concentrations, then

x = \dfrac{ab}{a + b}

The following example provides two solutions – one using the formula and the other without.

Example 17

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs

8

kg and the second one weighs

16

3.33

kg (2)

4.67

kg (3)

5.33

kg (4) None of the above

Solution

The weights of the two alloys are

8

kg and

16

kg and the concentrations are unknown. Let

x

be the weight of the replaced portion from the two alloys.

x = \dfrac{8 \times 16}{8 + 16} = 5.33

kg

Alternatively

Let the portion of aluminium in the alloys weighing

8

kg and

16

kg be

m

and

n

respectively. Let

x

be the weight of the replaced portion from the two alloys.

Weight of aluminium in the two alloys are $\bm{8m}$ kg and $\bm{16n}$ kg respectively. Let weight of aluminium in

x

= \dfrac{8m - xm + xn}{8} = \dfrac{16n - xn + xm}{16}

⇒

16m - 2xm + 2xn = 16n - xn + xm

⇒

16m - 16n = 3xm - 3xn

⇒

16 (m - n) = 3x (m - n)

⇒

x = \dfrac{16}{3} = 5.33

Answer:

(3) 5.33

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