# PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

How many liters of the mixture from a 90 liter bottle full of 90% concentrated milk has to be replaced with water to make the resultant 60% concentrated?

- (a)
60

- (b)
50

- (c)
30

- (d)
None of these

Answer: Option C

**Explanation** :

Let ‘r’ litres is removed and replaced with water.

∴ 60 = 90 × $\left(1-\frac{r}{90}\right)$

⇒ $\frac{2}{3}$ = 1 - $\frac{r}{90}$

⇒ r = 30 litres.

Hence, option (c).

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two jugs A and B contain 50 litres each of milk and water respectively. 10 litres of milk is taken from jug A and poured into jug B. Then, 10 litres of solution from jug B is removed and poured into jug A. Let the quantity (in litres) of water in jug B and milk in jug A be *v*_{b} & *v*_{a} respectively. Which of the following is true?

- (a)
*v*_{a}=*v*_{b} - (b)
*v*_{a}<*v*_{b} - (c)
*v*_{a}>*v*_{b} - (d)
Cannot be determined

Answer: Option A

**Explanation** :

**Step 1**: 10 litres is transferred from A to B.

Since A contains only milk initially, 10 litres of milk will be transferred from A to B.

Hence, quantity of milk and water in A and B will be

**Step 2**: 10 litres is transferred from B to A.

Now 10 litres out of 60 litres is transferred from B to A.

⇒ Fraction of solution being transferred = 10/60 = 1/6th

B has 10 litres of milk.

∴ Quantity of milk transferred from B to A = 10 × 1/6 = 5/3 litres

B has 50 litres of water.

∴ Quantity of water transferred from B to A = 50 × 1/6 = 25/3 litres

Hence, quantity of milk and water in A and B will be 3

∴ Final quantity of water in jug B (*v*_{b}) = 125/3 litres

and, final quantity of milk in jug A (*v*_{a}) = 125/3 litres

⇒ *v*_{a} = *v*_{b}

**Note**: If this entire process is repeated again, still the quantity of water in jug B will be same as quantity of milk in jug A, while quantity of milk in jug B will be same as quantity of water in jug A.

Hence, option (a).

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

From a barrel having 36 litres alcohol, 6 litres are drawn and replaced with water. Then, 12 litres of the mixture is replaced with water. Then, 18 litres of the mixture is replaced with water. Find the final quantity of alcohol in the barrel (in litres).

- (a)
30

- (b)
20

- (c)
10

- (d)
None of these

Answer: Option C

**Explanation** :

**Step 1:**

Since 6 litres out of 36 is removed, fraction of solution removed = 6/36 = 1/6^{th}

∴ Fraction of solution remaining = 5/6^{th}

⇒ Alcohol remaining after 1st replacement = 36 × 5/6 = 30 liters.

**Step 2:**

Now 12 litres out of 36 is removed, fraction of solution removed = 12/36 = 1/3^{rd}

∴ Fraction of solution remaining = 2/3^{rd}

⇒ Alcohol remaining after 1st replacement = 30 × 2/3 = 20 liters.

**Step 3:**

Now 18 litres out of 36 is removed, fraction of solution removed = 18/36 = 1/2

∴ Fraction of solution remaining = 1/2

⇒ Alcohol remaining after 1st replacement = 20 × 1/2 = 10 liters.

∴ Quantity of alcohol remaining in the barrel = 10 litres.

Hence, option (c).

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

Adish bought a 1000 ml bottle of Rum. On the 1st day, he drank 5% of the contents in the bottle and replaced that quantity with water. Next day, he drank 10% of the solution and replaced it with water. He continued this way. On the 19th day, he drank 95% of the solution and replaced it with water. On the 20^{th} i.e. the last day; he drank the entire contents of the bottle. Find the total quantity of water that he drank in the entire process. (Enter your answer as the nearest possible integer in ml).

Answer: 9500

**Explanation** :

Instead of focusing on how much water Adish drinks every day we shall focus on the amount of water added every day.

At the end of 20^{th} day Adish drinks the entire bottle, hence what ever amount of water is added in the bottle, he would drink it all.

∴ Water added on 1^{st} day = 5% of 1000 ml

Water added on 2^{nd} day = 10% of 1000 ml

Water added on 3^{rd} day = 15% of 1000 ml

…

Water added on 19^{th} day = 95% of 1000 ml

⇒ Total amount of water added = 5% of 1000 + 10% of 1000 + 15% of 1000 + … + 95% of 1000

= $\frac{1000}{100}$ × (5 + 10 + 15 + … + 95)

= 10 × 5 × (1 + 2 + 3 + … + 19)

= 10 × 5 × $\frac{19\times 20}{2}$

= 9500 ml

∴ Total amount of water Adish drinks = 9500 ml.

Hence, 9500.

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

From a container full of pure milk, ‘r’ litres of milk was drawn & replaced with water. ‘r’ litres of the solution was again drawn & replaced with water. The ratio of milk & water was then 64 : 225. What fraction of the container’s capacity is ‘r’?

- (a)
9/17

- (b)
8/17

- (c)
15/17

- (d)
None of these

Answer: Option A

**Explanation** :

Let the initial quantity of container = V liters.

∴ Initial quantity of milk = V liters [initially only milk is present]

∴ Final quantity of milk = $\frac{64}{64+225}$ × V = $\frac{64}{289}V$

Fraction of milk removed = $\frac{r}{V}$

∴ Fraction of milk remaining = $\left(1-\frac{r}{V}\right)$

This process is repeated twice.

∴ Final quantity of milk = (Initial quantity of milk) × ${\left(1-\frac{r}{V}\right)}^{2}$

⇒ $\frac{64}{289}V$ = $V{\left(1-\frac{r}{V}\right)}^{2}$

⇒ $\frac{64}{289}$ = ${\left(1-\frac{r}{V}\right)}^{2}$

⇒ $\frac{8}{17}$ = $\left(1-\frac{r}{V}\right)$

⇒ $\frac{r}{V}$ = $\frac{9}{17}$

Hence, option (a).

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

A bottle full of rum has 65% alcohol. A part of rum is removed & replaced by another variety having 44% alcohol & now the alcohol in the bottle is 56 %. Find the fraction of rum replaced.

- (a)
1/3

- (b)
2/3

- (c)
1/2

- (d)
3/7

Answer: Option D

**Explanation** :

Let the initial quantity of bottle = 1 liter and the amount of rum removed is ‘s’ liter.

Now ‘s’ liter of 44% alcohol solution is added to ‘1 – s’ liter of 65% alcohol solution to get a 56% alcohol solution.

∴ $\frac{\mathrm{Quantity}\mathrm{of}64\%\mathrm{solution}}{\mathrm{Quantity}\mathrm{of}44\%\mathrm{solution}}$ = $\frac{12}{9}$

⇒ $\frac{1-s}{s}$ = $\frac{4}{3}$

⇒ s = $\frac{3}{7}$ liter

∴ 3/7 liter was removed from 1 liter of rum bottle.

⇒ Fraction of bottle removed = $\frac{3/7}{1}$ = $\frac{3}{7}$ ^{th}

Hence, option (d).

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

In fresh grapes, 80% of the weight is water while in dry grapes only 20% of the weight is water. How many kg of dry grapes can be obtained from 20 kg of fresh grapes?

[Enter your answer as nearest possible integer in grams]

Answer: 5000

**Explanation** :

Dry grapes are obtained when water is evaporated from fresh grapes. Though the amount of pulp remains the same.

Let 'x' kgs of dry grapes are obtained from 20 kgs of fresh grapes.

∴ Amount of pulp in 20 kgs of fresh grapes = 20 × 20% = 4 kg.

Also, the amount of pulp in 'x' kg of dry grapes = x × 80% = 0.8x

⇒ 4 = 0.8x

⇒ x = 5 kgs = 5000 gms

Hence, 5000.

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**PE 2 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement**

A shopkeeper mixes three varieties of rice costing Rs. 15, Rs. 17 and Rs. 22 per kg. Which of the following can represent a possible ratio in which the three varieties are mixed, if the trader makes a profit of 20% by selling the mixture at Rs. 21.60 per kg?

- (a)
2 : 5 : 3

- (b)
2 : 1 : 3

- (c)
2 : 6 : 3

- (d)
1 : 2 : 3

Answer: Option C

**Explanation** :

20% profit is earned when the mixture is sold at Rs. 21.60/kg.

∴ Cost price of the mixture = 21.60/1.2 = Rs. 18/kg

Now, only option (c) will satisfy the given condition

$\frac{15\times 2+17\times 6+22\times 3}{2+6+3}$ = $\frac{198}{11}$ = 18

Hence, option (c).

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