## Quantitative Aptitude Questions on Percentages for Campus Recruitment Training

The Aptitude test consists of basic problems in quantitative aptitude, 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 is the problem-solving skills. It is essential to understand what the problem is and how the problem should be solved. So, the application of the learning is the Aptitude test.

## 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

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

## Tips and Tricks to solve the percentages 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. And, Quantitative Aptitude contributes a lot to increase your scores. No aptitude test is complete without questions based on 'percentages'.
• Methods used in the learning of one chapter can be used in the other chapters
• Calculate the percentages only to the extent where you can spot the answer to that question.
• Whether the question is simple or difficult you can know through the options.
• Mixers are also used to find percentages through a shortcut.
• Focus on all areas of quantitative aptitude that will help you crack the quantitative section

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

The following questions based on percentages:

### Calculation Based Questions:

1. What is 38.3% of  647 +  51.1% of  943?

A.687

B.729

C.773

D.813

Explanation:

Where the hint is: the options are not very close to each other. So, you can find the approximate value easily.

Now, 38.3% of  647 = 40% of  650  (take near values to the actual Values because options are not very close)

51.1% of  943 =  50% of  940

Where,  40% of  650 is 260  (10% of  650 is 65→ So, 40% of  650 is 260)

50% of  940 is 470 ( 10% of 940 is 94→ So, 50% of 940 is 470)

→   260+470 = 730

730 is near to 729. Hence option B) 729 is the answer

### Multiplication factors on Profit and Loss

2. A shopkeeper purchased a chair in the second-hand market for Rs.837. He then invested 1/5th of the amount in getting it painted and packaged. He priced it in such a way that he gets a 60% profit. Eventually, the article had to be sold at a discount of 20%. What is the overall profit or loss percentage?

A. 20% loss

B. No loss No profit

C. 20% profit

D. 28% Profit

E. 33.6% Profit

Explanation:

CP = Rs. 837

1/5 of Rs. 837 = Rs. 167.4 spent on painting and packaging.

Hence, total CP = 837 + 167.4 = Rs. 1004.4

P% = 60 % which is equivalent to SP = 1.6 when CP = 1.

But, d of 20 % provided — Hence, final SP  will be 0.8 x 1.6 or 0.8 x 1.6 x 1004 = Rs. 1285 approximately.

Then, the Profit = SP - CP = 1285 - 1004 = Rs. 281 and profit % = 28 % making option D as the answer.

### Difference between SI and CI

3. A person lent a sum of money at 12% per annum SI and received an interest of 2700 at the end of 2 years. What would be the interest if the same sum was invested at 12% per annum Compound Interest?

A. 2700

B. 2661.12

C. 2738.88

D. 2958.74

Explanation:

A = P(1.12)2 = P(1 + 0.12)2

= P(1 + 2*0.12 + 0.0144)

= P(1 + 0.24 + 0.0144)

= P + 0.24P + 0.0144P

CI should be greater than  SI. So, A.2700 and B. 2661.12 are not considered as CI. When we calculate above then we may get at most 100 extra then D.2958.74 is also not considered as CI.  Hence only C. 2738.88 is the suitable answer because it is greater than and near to SI

### Mix and Match - Find a rate

4. A salesperson sold 40% of his wares at a profit of 5%. At what profit % should he sell the remaining goods to get an overall profit of 11%?

A.15%

B.20%

C.25%

D.30%

Explanation:

Let the amount invested for 100C with the profit of 11C

Sold goods = 40C with the profit of 5C

= 40C *(1/20) = 2C

Remaining goods = 100C-40C= 60C

remaining goods sold profit is 11C - 2C = 9C

=(9C/60C)*100

= 15%

5. A person invested 40% of his money at 5% per annum. At what rate of interest should he invest the rest of his money to earn an overall interest of 11% per annum?

A.15%

B.20%

C.25%

D.30%

Explanation:

Here,

→  40% of the money invested at 5% profit

→ 11% is overall Profit

→ 6% is the difference between invested profit and overall Profit

→ Investment Ratio is 40% : 60% = 4% : 6%

→ Remaining Profit is 4% + 11% =15%

6. A person invested Rs.11000 in two investment schemes - one offering 7% p.a and another offering 12.5% p.a. How much should he be investing at 12.5% in order to get an overall return of 9.5%p.a?

A. One part is 5500 and Second part is 5500

B. One part is 8000 and Second part is 3000

C. One part is 7000 and Second part is 4000

D. One part is 6000 and Second part is 5000

Explanation:

One offering 7%p.a

Difference between 7% and 9.5% =2.5

= 2.5 *2 = 5

amount =5000

another offering 12.5%p.a

Difference between 12.5% and 9.5% =3

= 3*2 = 6

amount =6000

So, 1.6000 is invested for 7%p.a

2. 5000 is invested for 12.5%p.a

Total amount = 6000+5000= Rs.11,000

### Dealing with the multiple objects

7. A person sold a Sofa and Chair at 10% profit and 10% loss respectively to get an overall profit of 6%. If the Sofa was sold for Rs.700 more than the Chair, what is the selling price of the Chair?

A.180

B.800

C.880

D.700

Explanation:

Overall profit 6%

Profit 10% = 10-6 = 4

loss 10% = -10-6 = 16 (In alligation we don't consider Negative signs)

Ratio of cost= 4: 1

CP = 400x : 100x

profit and loss = 40x   :-10x

total = 440x : 90x

Difference = 440x-90x= 350x

=> 350x=700

=>x=2

Sofa cost is 880, and Chair cost is 180

There fore 880-180 = 700

Published in Test Patterns

## TCS Ninja Recruitment Process and Exam Pattern

TCS Ninja is a National Qualifier Test conducted by TCS (Tata Consultancy Services) in order to recruit candidates from a wide variety of areas. This exam provides jobs to almost 10,000 candidates, from a staggering number of 300,000 applicants. Thus, a candidate must have an in-depth knowledge about this exam in order to achieve their dream of working in TCS. This blog will contain all the necessary information regarding TCS Ninja exam such as TCS Ninja application process, exam pattern and date of application.

Published in Tips & Strategies

# Campus Placement Preparation for Freshers

As Colleges re-open for the new academic year, one phrase on every final year student’s mind is campus placements. Which companies would visit campus for recruitment this year? How many would they recruit? Are the recruitment processes going to change? What would be the level of difficulty of the Aptitude Tests? Questions, questions and more questions and hardly any answers! I shall attempt to answer some of these questions in the context of IT companies in engineering college campuses in this article.

Published in Tips & Strategies

## 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

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:

Capgemini Aptitude Test Sample Placement Question

Cognizant Aptitude Placement Questions 2018

Latest IBM Aptitude Practice Questions

Accenture Aptitude Test Questions for Freshers

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

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

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

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

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

Explanation:

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

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%

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

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

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:

Infosys Campus Recruitment Process

Accenture Campus Recruitment Process

Amazon Campus Recruitment Process

Capgemini Campus Recruitment Process

Tech Mahindra Campus Recruitment Process

Cognizant Placement Process

IBM Campus Recruitment Process

Deloitte Campus Recruitment Process

Wipro Recruitment Process

TCS Recruitment Process

Published in Test Patterns

# Latest Infosys Aptitude Placement Paper

Infosys is a multinational company providing information technology services and business consulting. The company has its headquarter in Bangalore, in the electronic city of India. Infosys is the 3rd largest IT Company in India and the 5th largest employer of H1-B visa as stated in the year 2013. Learn more about the company profile of Infosys here.

## Infosys Selection Process:

The company conducts the recruitment process every year to select new candidates. The selection process of the company consists of 2 rounds. These rounds are as follows:

• Written Exam
• HR Interview

Infosys Written exam consists of 3 sections - Quantitative Aptitude, Analytical & Logical Reasoning and, Verbal ability. There is no negative marking in the paper. Overall, the level of the paper is moderate. Only those candidates who clear the written exam will qualify for the next round. Find the Recruitment pattern here. Below are few sample questions based on previous Infosys placement papers. Read along to find out the rigor and subject topics of the questions generally expected in Infosys placement papers for freshers.

Infosys Aptitude Sample Question 1:

Students of St. Peter college are lined up in a row. Lata who is standing 16th from the left stands on the immediate right of Sita. When they exchange their places, Sita stands 14th from the right. Find out the total number of students in the row.

a) 29

b) 30

c) 28

d) 31

Number of students before Lata from the left (including Lata)= 16

When Sita and Lata exchange places, number of students from the right (excluding Lata as she has been counted once already)= 14-1= 13

Number of students in the row = 16+ 13 = 29

Infosys Aptitude Sample Question 2:

Coffee worth Rs. 183 per kg, 173 per kg and a third variety are mixed in the ratio of 2: 2: 3 and sold at Rs. 200 per kg at a price of no loss/ profit. What is the approximate cost per kg of the third variety of coffee?

a) 220

b) 229

c) 239

d) 240

Let the third variety of coffee be Rs. A Per Kg.

According to the question

((183×2) +(173×2)+(A×3)) / (2+2+3) = 200

Solving the above equation, we get the value of A as Rs.229.30

Must Practice Questions: Infosys Quantitative Aptitude Questions

Infosys Aptitude Sample Question 3:

Three friends went to a restaurant and found some pieces of cake in front of them. Shanti took one-fourth of them and returned 3. Sameera took one-half of them and returned one. John took one-seventh of them and returned none. What is the initial number of pieces of cake if there were 6 at the end?

a) 10

b) 24

c) 12

d) 08

The best way to solve this question is using the options to save time. Take options which are multiples of 1/4 and these are 24, 12 and 8. 8 is not possible given that there are 6 left at the end. Between 24 and 12, chose 12 as this doesn't lead to decimals (since the number of pieces is an integer). Now, check the data on initial number 12. Every condition holds true.

Read More: Infosys - Reasoning Questions

Infosys Aptitude Sample Question 4:

Ramesh was endowed______natural talent for music.

a) in

b) of

c) by

d) with

Endowed is always followed by the preposition ‘with’.

Infosys Aptitude Sample Question 5:

Ranjan jumped off the train while it_________.

Choose the right form of the verb to be filled in the blank from the options.

b) was moving

c) moved

d) has Moved

While is used for expressing an action in continuous tense. Since 'jumped off' is in past tense, past continuous form of the verb is ‘was moving’

Check the recruitment process for various companies:

Infosys Campus Recruitment Process

Accenture Campus Recruitment Process

Amazon Campus Recruitment Process

Capgemini Campus Recruitment Process

Tech Mahindra Campus Recruitment Process

Cognizant Placement Process

IBM Campus Recruitment Process

Deloitte Campus Recruitment Process

Published in Test Patterns
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