**Campus Recruitment | Quantitative Aptitude Questions on Ratios & Averages**

**What is Campus Recruitment Process? **

Generally, a Campus Placement Process has four components:

* Pre Placement Talk

* Aptitude test

* Group Discussion

* Interview which can be various combinations of Technical and HR rounds** **

**Aptitude Test for Placements: **

The aptitude test consists of basic problems in numerical ability, logical reasoning, and verbal ability. The primary objective of this test is to check the problem-solving skills of the test-taker. Any job out there necessitates the presence of the candidate's problem-solving skills. It is essential to understand what the problem is and how the problem should be approached and solved. Hence, the aptitude test is an essential step towards identifying candidates with problem-solving skills to qualify the campus recruitment test.

**Tips and Tricks to solve Ratios & Averages based Questions for Campus Recruitment Test**

• If you want to stand out & beat the competition in your Campus Placements, the first step is to ace the Written Aptitude Test. Quantitative Aptitude contributes a lot to increase your scores. No aptitude test is deemed complete without questions based on Ratios & Averages

• Methods implemented in learning one chapter can be implemented to solve other chapters also.

• You can find the difficulty level of the questions by analyzing the options.

• Focusing on all major areas of quantitative aptitude will help you crack the quantitative section with ease

Watch the below video by Adhitya Lanka (IIM-Calcutta) having 10+ years of experience in Teaching Quantitative Aptitude and Reasoning to learn Tips & Tricks to crack Questions on Ratios & Averages in Campus Recruitment Exams

**Dilution- Finding Concentration (in Ratios)**

**1. 8 liters of water is added to 24 liters of a solution containing milk and water in the ratio 3: 1. What is the ratio of milk to water in the resulting solution?**

A. 18: 14

B. 12: 10

C. 16: 17

D. 20: 23

**Answer: A**

**Explanation:**

It can be solved in two ways:

**Method 1:**

the ratio of milk and water = 3:1

Volume of milk and water (in the same order) = 18L, 6L ( the value 6 is 1/4th of 24 liters. So, 3*6=18 & 1*6=6)

8 liters of water is added then it becomes, 6+8= 14

Now the ratio of Milk and water = 18: 14 or 9: 7

**Method 2:**

The initial volume of solution, V1=24,

The final volume of solution, V2 =24+8=32

Now, V2/V1 = 32/24 = 4/3

This means when we multiply the initial solution with 4/3, we get the final solution. The concentration is inversely proportional to volume so it should be 3/4.

The equation becomes like this - Final concentration = Cinint* (3/4) (→ V and C are inversely proportional to each other)

Cint = 3/4 (in total concentration 3/4 parts of milk - we take milk ratio as this time water is added to the solution)

Final Concentration Ratio for Milk =(3/4) * (3/4) = 9/16 (The 9/16 ratio means: if the total volume of solution 16L, then milk comprises 9L then the remaining 7L is water)

Hence Ratio of milk to water is 9: 7 or 18: 14.

**Check the aptitude test question for various companies:**

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**2. 8 liters of milk is added to 32 liters of a solution containing milk and water in the ratio 3: 1. What is the ratio of milk to water in the resulting solution?**

A. 1: 5

B. 3: 7

C. 4: 1

D. 5: 8

**Answer: C**

**Explanation:**

It can be solved in two ways:

**Method 1:**

the ratio of milk and water = 3:1

milk and water in liters = 24, 8 ( the value 8 comes from 1/4th of 32 liters. So, 3*8=24, 1*8=8)

8 liters of milk is added then 24+8= 32

Now the ratio of Milk and water = 32: 8 = 4:1

**Method 2:**

The initial volume of solution, V1=32,

The final volume of solution, V2 =32+8=40

Now, V2/V1 = 40/32 = 5/4

This means when we multiply the initial solution volume with 5/4, we get the final solution volume. The concentration is inversely proportional to volume so it should be 4/5.

The equation becomes like this - Final concentration = Cint* (4/5) (→ V and C are inversely proportional to each other)

Cint of water = 1/4 (in total concentration 1/4 parts is water - we take water ratio as this time milk is added to the solution)

Final Concentration Ratio for water =(4/5) * (1/4) = 1/5 (The 1/5 ratio means: if the total volume of solution 5L, then water comprises 1L then the remaining 4L is milk)

Hence Ratio of milk to water is 4:1.

**Dilution- Finding Quantity to be added**

**3. How many liters of water needs to be added to 30 liters of milk with 90% concentration to make a solution containing 85% milk?**

A. 2.5

B. 1.76

C. 4

D. 3

**Answer: B**

**Explanation:**

Cinitial = 90%, Cfinal = 85%

Cfinal/Cinitial = 85/90 = 17/18

As Concentration is inversely proportional to volume so,

vfinal/vinitial = 18/17 = (1+1/17) (we can write 18/17 as (1+1/17))

vfinal = 30 x (1+1/17)

= 30 + (30/17) = 30 + 1.76

Where 1.76 liters is the amount of water needed.

**Replacement- Finding Final concentration**

**4. Ten liters of water is added to 40 liters of a mixture containing 70% milk. What is the concentration of the resulting mixture?**

A. 18: 21

B. 13: 16

C. 17:15

D. 14:11

**Answer:D**

**Explanation:**

Given: Vinitial=40, Vfinal= 40+10= 50, Cinitial = 70%, Cfinal= ?

Vfinal/Vinitial= 50/40 = 5/4

Cfinal/Cinitial = 4/5

Cfinal= Cinitial x (4/5)

= 70% x (4/5)

= 56% ( it implies that out of 100 parts of the solution 56 parts is milk and the remaining 44 parts is water)

So, the ratio = 56:44 = 14: 11

**Finding Final concentration**

**5. Ten liters of water is added to 40 liters of a mixture containing 70% milk. What is the concentration of the resulting mixture?**

A. 76: 24

B. 74:26

C. 20: 80

D. 60:40

**Answer:A**

**Explanation:**

Given: Vinitial=40, Vfinal= 40+10= 50, Cinitial= 100%-70% = 30%, Cfinal= ?

Vfinal/Vinitial= 50/40 = 5/4

Cfinal/Cinitial = 4/5

Cfinal= Cinitial x (4/5)

= 30% x (4/5)

= 24% ( it implies that out of 100 parts of the solution 24 parts is water and the remaining 76 parts is milk)

So, the ratio = 76:24

**Replacement- Finding the quantity**

**6. A cask contains 72 liters of a solution containing 90% milk. How many liters of the solution needs to be replaced with the water in order to decrease the concentration of the milk to 80%?**

A. 5

B. 2.5

C. 8

D. 1/7

**Answer:C**

**Explanation: **

Two steps 1: removing stuff( removing some part of the solution)

2: adding water

initial concentration milk, C1 = 90% (milk is being multiplied with some fraction)

amount of milk in liters = 90% x 72

After removal of a fraction = {90% x 72} (1-f) = 80% x 72

90% x (1-f) = 80% ( canceling 72 on both the sides)

(1-f) = 80%/ 90%

(1-f) = 8/ 9

f = 1/9

Quantity of water added = (1/9)* 72 = 8 liters

**Repeated Dilution- Find the concentration**

**7. A cask contains 80 liters of milk with 96% concentration. Every day a boy removes 20 liters of the solution from the cask replaces it with water. What is the ratio of milk to water in the cask after 3 days?**

A. 30.5%

B. 40.5%

C. 50.5%

D. 60.5%

**Answer: B**

**Explanation:**

Fraction f = 20/80 =1/4

Initial Concentration, C0 = 96%

Concentration after Day 1, C1 = 96% x (1-f)

Concentration after Day 2, C2 = [96% x (1-f)] x (1-f) = 96% x (1-f)2

Concentration after Day 3, C3 = [ 96% x (1-f)2] x (1-f) = 96% x (1-f)3 (use => Cfinal= Cinitial * (1-f)n)

=96% x (1-(1/4))3

= 96% x (3/4)3

= (96 x 27)/64 = 40.5%

**Repeated Dilution- Different Fractions**

**8. A cask contains 80 liters of milk with 96% concentration. On the first day, a boy removes 20 liters of the solution from the cask and replaces with water. On the second day, he removes 16 liters and replaces with water. Finally, on the Third day, he removes 24 liters and replaces with water. What is the ratio of milk to water in the cask after 3 days?**

A. 425: 625

B. 504: 746

C. 200: 550

D. 325: 525

**Answer: B**

**Explanation: **

Cfinal= 96% x [(1-(20/80)) x (1-(16/80)) x (1- (24/80))]

= (24/25) x (3/4) x (4/5) x (7/10)

= (504/1250) (where the total amount of the solution is 1250 out of which the amount of milk is 504 and the amount of water is 746)

Ratio = 504: 746

**Repeated replacement**

**9. A cask contains 80 liters of a solution containing milk and water in the ratio 3: 1. Each day, a boy takes out 'x' liters of the solution and replaces it with water. After two days, the ratio of milk to water became 12: 13. Find x?**

A. 20 liters

B. 26 liters

C. 36 liters

D. 16 liters

**Answer: D**

**Explanation:**

Cinitial = 3/4 x (milk ratio)

Cfinal = 12/25

Cfinal = Cinitial x (1-(x/80)) x (1-(x/80))

12/25 = 3/4 x (1-(x/80)) x (1-(x/80))

(12 x 4) /(3 x 25) = (1-(x/80))2

16/25 = (1-(x/80))2

(4/5)2 = (1-(x/80))2

(4/5) = 1-(x/80)

x/80 = 1- (4/5) = 1/5

x= 80/5 = 16 liters

**Please read the blogs below to know the detailed campus recruitment process of respective companies:**

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IBM Campus Recruitment Process