# Placement Preparation Questions on Simplification & Approximation for Campus Recuritment

**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 quantitative 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 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.

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

**Tips and Tricks to solve the Simplification and Approximation 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 complete without questions based on Simplification & Approximations.
- Methods of learning in one chapter can be used in other chapters.
- Whether the Question is simple or difficult you can know the options.
- Focus on all areas in quantitative will help you crack quantitative section

Watch the below video to learn Tips & Tricks to crack Questions on Simplification and Approximation in Campus Recruitment Exams.

The Following questions are based on Simplification and Approximation:

**Simplification involving Large Numbers:**

**1. ****-87+1934-271-?=893**

A.679

B.683

C.687

D.689

**Answer: B**

**Explanation:**

From the given question Do you notice all the option are very close to each other

Do you notice except the first and last option remaining two options have a different unit digit at the end of the number

Now we have to take the unit digit of each number along with the sign

-7+4-1-?=3

Instead of taking the 4 we have to take 14 or 24 and so on

Because if we take the number 4 it will get negative digits. But our answer is in positive

So, we have to take 14 instead of 4

-7+14-1-?=3

6-?=3

In the place of the question mark, the number 3 has to come

According to the option, the unit digit which ends with 3 is 683.

**2. ****12738-2314-1781+9816=?**

A.18459

B.18461

C.19459

D.19461

**Answer: A**

**Explanation:**

**First part**

Look at the following options 459&461 are repeated two times

The unit digit could be 1 or 9

Based on that we have to eliminate two options

**Second part**

The unit digit of each number will be

8-4-1+6=9

we need the number ending with '9'

so, 2&4 options are eliminated

Now we are Looking at the thousand terms

12-2-1+9=18

It will be 18000

**Third part**

Suppose if we have two options starting with 18 and the units place is 9 we have to use the third part

From the above question, we have two 7 hundred numbers if they are canceled it get -50 or -40 difference

Then 816 & -314 it will get 500 difference(18000-50+500)

So, the answer is approximately equal to 18,450

so the answer is 18,459.

**Simplification involving Fractions**

**3. ****3 4/5+5 6/7=**

A.9 23/35

B.9 24/35

C.9 25/35

D.9 33/35

**Answer: A**

**Explanation:**

First, we have to consider

3+4/5+5+6/7=8+0.8+5+0.85≈9.65

Then 9 ?/?

we know that denominator should be 5 or 7 or 35

For the safe, you have to assume 35

then 9 ?/35

The numerator will be 23 or 24 or 25

Because of the fraction should be 0.65

But in the given options numerator has given 23,24,25

so we have to find given numerator

3+4/5+5+6/7=8+(28+30)/35=8+1+23/25=9 23/35

**4. ****3 4/5+? 3/8=9 ?/40**

A.5,6

B.4,8

C.5,7

D.4,7

**Answer: C**

**Explanation:**

Here 9 ?/40 in this '?' should be less than 40

Step 1 Isolate the whole & fraction

3+4/5+x+3/8+=9+?/40

Step 2 Estimate the whole number

3+4/5+x+3/8+=9+?/40

'x' should be equal to 5 due to '9'

Step 3 Simplifying the fraction

4/5+3/8=(32+15)/40=47/40=1+7/40

Here 7/40 is the fraction

so, the answer is 5,7

**5. ****4 3/7-8 2/5+6 1/5-7 4/7=?**

A.-4 15/35

B.-5 12/35

C.-5 15/35

D.-4 12/35

**Answer: B**

**Explanation:**

**whole numbers fractions**

** **4 +3/7

-8 -2/5

+6 +1/5

-7 -4/7

-5 -(1/7)-(1/5)

-[5+1/7+1/5]=-[5+12/35]=-5 12/35

# 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

**Answer: B**

**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

**Answer: D**

**Explanation:**

Multiplication Factor (M.F) of 60% is 1.6

Multiplication Factor (M.F) of 20% is 0.8

Cost Price = 837+ (1/5) + 837

= 837+ (6/5)

Marked Price = (837+ (6/5)) * 1.6

Selling Price = CP*MP*Discount

= CP * 1.6 * 0.8

= CP * 1.28 (Discount is Should always < 1)

Hence, There is a 28%of Profit if he sold at 20% of discount (1.28- 1= 0.28)

**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

**Answer: C**

**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%

**Answer: A**

**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%

**Answer: A**

**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

**Answer: D**

**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

**Answer: A**

**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

# 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:**

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

**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:**

Infosys Campus Recruitment Process

Accenture Campus Recruitment Process

Amazon Campus Recruitment Process

Capgemini Campus Recruitment Process

Tech Mahindra Campus Recruitment Process

IBM Campus Recruitment Process

# 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

Interview round and group tasks help you in HR Interviews

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

**Answer/Explanation: **a) 29

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

**Answer/ Explanation:** b) 229

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

**Answer/ Explanation:** c) 12

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

**Answer/ Explanation:** 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.

a) had been moved

b) was moving

c) moved

d) has Moved

**Answer/ Explanation:** b) was moving

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’

Learn More: Infosys Verbal Ability Questions

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

IBM Campus Recruitment Process

# Amazon Aptitude Questions & Answers Online for Freshers

'Amazon' was started by the name Cadabra in the year 1994 by Jeff Bezos. It went online as 'amazon.com' in the year 1995. What started out as an online bookstore has today turned out to become the world's largest online retailer. The company’s name 'Amazon' can be correlated with the Amazon river from which it took inspiration.

The selection process of Amazon consists of three rounds.

a) Written Round

b) Technical Interview (Sometimes two technical interviews are conducted with the second one being a stress interview)

c) HR Interview

Read our popular technical blogs:

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JAVA Interview Questions and Answers for Freshers

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How to Ace Interviews & Group Tasks?

**Amazon Written Round:**

- The Written round consists of 3 sections,viz., Aptitude, Verbal Ability and Technical.
- There are 30 Questions in the Aptitude section. The important topics include Time, Speed and Work, Percentages, Profit and Loss, Ratio and Proportion, Probability, Triangles, and Functions.
- There are 30 questions in the Verbal Ability section. The important topics include Reading Comprehension, Sentence Correction, Identifying error in a sentence, Synonyms and Antonyms, etc.
- There are 44 questions in the Technical Section. The important topics include C, Data Structures, and DBMS.
- The allotted time to solve the questions in these sections is 120 minutes. There is no negative marking in this round.

A candidate is tested not only for general aptitude skills but also for programming/ coding abilities and communication skills is undoubtedly a must. Find the complete Amazon recruitment process here. Below are few sample questions based on previous Amazon placement papers. Read on to find out the rigour and subject topics of the questions generally expected in Amazon placement papers for freshers.

**Amazon Aptitude Sample Question 1: **The difference between 4/17 of 17/19 of a number and 3/38 of the same number is 17.5. Find the number.

a)17. 5

b) 18.5

c) 20

d) 133

**Answer: **(d)

**Explanation:**

Let the number be X.

Then, 4/17 of 17/19 of the number is (4/17) * (17/19) * X = (4/19) * X

3/38 of the number= (3/38) * X

Given:

the difference between these two values is 17.5.

Hence, [(4/19) * X - (3/38) * X ]= 17.5

Solving the equation, we have X=133

Therefore, the number is 133.

**Amazon Aptitude Sample Question 2:** a, b, c, and d are four numbers in arithmetic progression. Mean of these four numbers is 20. The common difference between the numbers is 6. Find the product of first and the last numbers.

a) 13

b) 34

c) 319

d) 124

**Answer:** (c)

**Explanation:**

Let the four numbers be a, a+b, a+2b, and a+3b

Mean of the four numbers = (a+a+b+a+2b+a+3b)/4 = (4a+6b)/4 = 20

Substituting b as 6, we get the value of a= 11

The product of the first and the last number is 11× 29 which is equal to 319.

Amazon Quantitative Aptitude Questions

**Amazon Aptitude Sample Question 3:** Find the correct group of signs for the equation to be in order.

**48_6_9_17**

a) /, +, =

b) +, /, =

c) =, +, /

d) +, =, /

**Answer:** (a)

**Explanation:**

One can solve these type of problems using trial and error method.

48 divided by 6 gives 8. 8 added to 9 gives 17.

**Amazon Aptitude Sample Question 4:** Today is Sunday. I will be attending an interview 4 days after the day before yesterday. On which day will I be attending the interview?

a) Tuesday

b) Wednesday

c) Monday

d) Thursday

**Answer:** (a)

**Explanation:**

If today is Sunday, Four days after Sunday will be Friday, the day before yesterday is Tuesday.

Also read: Amazon Logical Reasoning Questions

**Amazon Aptitude Sample Question 5:** Arrange the following phrases into a complete sentence.

A) of desks, and kids are working

B) on countless group assignments.

C) Nowadays, a typical classroom has pods

a) ABC

b) CAB

c) BCA

d) BAC

**Answer:** (b)

**Explanation:**

b) The complete sentence would look like this: Nowadays, a typical classroom has pods of desks, and kids are working on countless group assignments.

By looking at the set of sentences, one can understand that 'Nowadays' has to be the beginning of the sentence. From then on, it is simply putting the phrases together to form a coherent sentence.

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