TCS email writing questions with answers – 3


1. Use all the phrases given
2. Minimum words should be 50 otherwise your email cannot be validated
3. Addressing and signing should be done as in the question given.
4. Common grammatical rules, punctuation should be according to standard english.
5. you can use your own phrases along with the phrases given.Question : 1
As a resident, write an email to the Municipal commissioner of your city, Mr.Ashok, reporting nuissance of a building under construction beside your place.  Sign the email as Kumar.
building – construction – long time – three years – water usage – mosquitoes – unhygienic – construction workers – bad behaviour – attention -request – action – immediately
Suggested Answer:
Dear Mr.Ashok
I am a resident of Indira Nagar.  I would like to bring to your kind notice that M/S abc constructions limited has started a building construction in our locality. For the last three years the work has been progressing very slow and lot of water is being used indiscriminately by the company.  Due to unhygienic conditions created by the construction, mosquitos, pigs are growing fast in this locality.  In addition to this, most of the ladies and young women are reporting the construction workers bad behaviour.  I request you to pay attention to this problem and take action immediately.Thanks and Regards

Question : 2
As a student representative of your department, write an email to your batch mates, suggesting a party for Head of Department Prof.Sunil who is retiring next month.  Sign the email as Sam.
inform – retire – plan – surprise – party – host – family – exceptional teacher – guide – mentor – groom – students – helpful – together – memorable
Suggested answer:
Hi all
I would like to inform you that our Head of Department Prof.Sunil is goint to retire next month.  We all know that Prof.Sunil is an an exceptional teacher and guided us in many typical situations during our project time.  In addition to that, he mentored and groomed many of our seniors  to grow into leadership positions.   For his helpful contributions, we will host a small party to surprise him.  We also invite his family.  Please inform all the students and together we make it successful and this will be memorable for us forever.
Thanks and regards

Question : 3
As a student representative of your college, write an email to the Principal of Professional Engineering College, Prof.Deb Chatterjee, inviting his institute to participate in the Technical symposium being organized in your college.  Sign the email as Sam.
Invite – technical Symposium – previous – success – expecting – huge participation – latest technology – stalls – demos – interaction – topics – complete – exchange ideas – exciting prizes.
Sample Answer:
Dear prof. Chatterjee,
I would like to inform you that we are going to organize a technical symposium on fourth sunday of this month.  Many eminent personalities form industry and academia are going to participate in this event. Looking at the previous year success of the event, we are expecting huge participation this year.  In this event the students shall get to know about the latest technologies.  Many eminent vendors are setting up their stalls to give demos about their products. In the symposium, there will be some topics for interaction.  Various competitions are being held as a part of the symposium.  It is wonderful opportunity to exchange ideas and win exciting prizes.
Thanks and regards

Question : 4
As a supplier, write an email to the manager of M/S Big wheel Manufacturing Company, Mr.Chopra, intimating of their payment that is due for the products delivered to them three months ago. Sign the email as Ramesh
On time – delivery of goods – three months – credit period – overdue – payment – of the earliest – longstanding – relationship
Sample Answer:
Hi Mr.Chopra
You are a valuable customer of our company for a very long time and we appreciate your business.  And you always make payments on time .  But recently, we observed that we have not received payment for the delivery of goods we made on 15th may this year.  Three months credit period was also over and payment is over due.  I request you to make payment for the above goods delivered of the earliest.  We are looking forward for a longstanding relationship with your company.
Thanks and regards

TCS Email writing latest Questions – 2


1. Use all the phrases given
2. Minimum words should be 50 otherwise your email cannot be validated
3. Addressing and signing should be done as in the question given.
4. Common grammatical rules, punctuation should be according to standard english.
5. you can use your own phrases along with the phrases given.Question : 1
As a member of your residential society, write an email to inspector of local Police station, Mr.Sharma, informing him about miscreants who ride their bikes rashly every evening outside your society.  Sign the email as william.
residential area – ride – rashly – children – play – elderly – walk – grocery shop – across the road – dangerous – accidents – nuisance – action – immediately.

Sample Answer:
Dear Mr.Sharma
We are the residents of Siddartha Nagar.  We would like to bring to your notice that a few guys are riding their bikes very rashly in the evening hours in the main road of the colony. As you know that this is the time when children play on the road and elderly go for an evening walk.  Also there is a grocery shop across the road and many housewifes used to cross the road to buy any groceries. In the recent times we observed that due this rash driving many accidents were happened and several injured.  This is creating a constant nuisance for all.  So we would like to request you to take necessary action to curb these activities
Thanking you
Yours sincerely

Question 2:
As a recent buyer of their car, write an email to the Manager of Smart Automative company, Mr.Ahmed, regarding the poor quality of service facility available in the city.  Sign the email as Chopra.
very few – service centers – complaints – pending problems – maintenance – cost – time – delivery – increase – customer satisfaction

Sample Answer:Dear Mr. Ahmed
I recently bought Fiat palio from “Sridhar Fiat show room” in Nagole.  Recently I faced small problem with car AC and bought the car for maintenance.  But to my utter surprise, the showroom staff told me that service is not available in their showroom and they asked me to take the car to near by service center.  I found that there are very few service centers available compared to sales showrooms, and there are many complaints regarding this.  This in turn is causing many pending problems and increased maintenance cost, time and delivery time.  I would like to suggest you that if more service centers are opened in the city, customer satisfaction also goes up which finally converts into more sales.
Thanks and Regards

Question 3: 
As a former student, write an email to your professor, Mr.Matt, thanking her for teaching and guidance that contributed to your overall development.  Sign the email as peter.
Successful – Placed – grateful – help – advice – grooming – values – shaping my future – sincere – professional Sample Answer:Dear Mr.Matt
I am very happy to tell you that I got successful in the recently conducted campus placement drive at my college.  I am placed with TCS.  I am extremely grateful for your help regarding my preparation.  More over your advice regarding personality development helped for my personal grooming.  In addition to that, your style of teaching inculcates not only those skills related to professional success but also for developing values which I believe helps for shaping my career. Once again I would like to thanks for your sincere and professional help.
with warm regards

Question 4:
As an intern at ABC consulting Pvt.Ltd, write an email to your internship Project Manager, Mr.Ramesh, informing about the progress that you are making and some difficulties that your are encountering.  Sign the email as Ben.
Thank – challenging – progress – tight schedule – support – report – analytics – guidance – access – doubt – requirements – design. 

Sample Answer:Dear Mr.Ramesh
Thank you for allotting a challenging project for my internship. I am making steady progress and learning many new things.  The project is due next month and we are on tight schedule.  I need some additional support with regard to the reporting of Analytics.  Your guidance helped me access the database with ease but I have several doubts regard to the requirements of the design.  But I am facing little problem in reporting.
Thanks and regards

TCS New Verbal Elimination Round – 2014 – (1)

TCS New Verbal Elimination Round – 2014 – (1)

1. Use all the phrases given 
2. Minimum words should be 50 otherwise your email cannot be validated
3. Addressing and signing should be done as in the question given. 
4. Common grammatical rules, punctuation should be according to standard english.
5. you can use your own phrases along with the phrases given. 

Question 1: 
Using the following phrases, write an email with minimum of 70 words to the customer Mr. Gill Roy explaining delay to the project.
Payment processing system – Schedule – 10th May (Friday) – Unexpected power outage – 3 days – Overall delay – 7 days – includes recovery of lost work – will not recur
Sample Answer:
Dear Gill Roy
The project “Payment processing system” was scheduled to be delivered on 10th May (Friday). However, due to an unexpected power outage in our offshore site for the past 3 days, work did not progress as expected. Also we lost a few of our works as backup systems did not come online.  Hence we are expecting an overall delay includes recovery of lost work in the delivery of the project for a maximum of 7 days within which our team will work on the issues. Apologies for the delay and we will ensure that the mistake will not recur in future again. 
Thanks and Regards 

Question 2: 
You are a part of corporate communication team in your company.  The working time period is revised as 8:30 am to 5:00 pm. Using the following phrases, write an email with a minimum of 70 words and a maximum of 100 words to the employees in your company informing the same. 
by 30 minutes to avoid traffic – effect from next week – lunch duration-revised working time – reduced by 10 minutes-free breakfast-office will start earlier-till the end of rainy season-will be in effect.
Dear All
We hereby announce a change in the work timings as 8.30 AM to 5.00 PM, with effect from next week, till the end of rainy season Which means, office hours would commence 30 mins earlier to cover up minimum 30 Mins extra time being spent during peak hour traffic during monsoons. Also, additional changes include reduction of lunch duration by 10 minutes & timings of free breakfast are now applicable from 7.30 AM to 8.30 AM only. Since request to each one of you to adhere to the new timings.
Have a nice day
Lead – Corporate Communications

Question 3: As your company is doing good business and expanding, your company is relocating it’s office to a new address.  Using the following phrases, write an email with a minimum of 70 words and a maximum of 100 words to your customer informing the change in address.
near outer ring road-shifting to-bigger office space-November 10-change in telephone number-new address is provided below-fourth floor-Cesina Business Park.
Sample Answer:
Dear All
We are happy to announce that we are moving out to much spacious office from November 10th onwards.  It is indeed a great sign of our ever growing business & our increasing clientele.
Hence, for a better productivity results, our management has taken a decision of increasing the team size & allocate us a much more spacious facility with all the modern state of art amenities.
Our new abode will be Fourth Floor, Cesina Business Park, Near Outer Ring Road, Bangalore/Chennai/Hyderabad etc.,
Please also make a note of new board line number (reception number)-xxx-xxxxxxxx.  Lets us all make the most use of the resources available in the said new office to server our clients
Thanks & Regards 

Question 4: 
You are the project leader for a team of 20 members.  As the team members are not submitting the weekly time sheets regularly, you need to email them stressing the need to submit without fail.  Using the following phrases, write an email with a minimum of 70 words and a maximum of 100 words to your team members informing the same.  
can be accessed online-lead to loss of pay-every week-do not default-used to bill client-actual working hours-by friday-failure to adhere-time sheet filling application.
Sample Answer:
Dear All
It has been observed that many of you are not filling the timesheets on regular basis.  Let me tell you, filling up time sheets is the only way, to measure your hardwork, as long as you are working on this project. So please do not default on this.  More over, This is important for us to report it to our client, the actual amount of work done by each one of you in terms of number of hours per day, at the end of every week. 
Only on the basis of this, we can bill you all to the client, which is directly linked to you monthly salary.  In our words, it leads to loss of pay for any particular day, for which time sheet is not filled.
Please adhere to the company guidelines & fill the same on daily basis or atleast weekly basis.
Time sheet filling application is easily accessible in our intranet portal, which needs your login credentials.
Please do the needful on regular basis.
Project Lead

TCS Aptitude test syllabus and test pattern

TCS conducts written test and three rounds of interview to select freshers as assistant system engineer in their organisation.  In the first round they conduct one and half hour written test, followed by three interview rounds namely Technical round, Managerial round, and HR round. The written test has email writing component.

Test pattern:

Duration for test:
10 min + 80 min = 90 min.
You have to write an email using the given clues in around 70 words. You have to type the email in the space given. The most important thing is you have to use all the phrases given without missing even single one.
After the email writing, you have to take an aptitude test of 30 questions in 80 minutes.

Syllabus for written test:

According to the model questions given in the TCS recruitment portal we can assume the following chapters are very important. There is no verbal part in TCS aptitude test except for email writing.  1/3rd negative marking will be there.

Important topics:

Number system, Equations, Ratio and Proportion, Percentages, Profit and Loss, Time and Work, Time speed Distance, Areas and Mensuration, Averages, Permutations and Combinations, Probability, Plane geometry, Seating Arrangements, Sets, Progressions, Functions.

Interview process:

There is no hard and fast rule that you may be asked only Technical questions in technical interview. or HR questions in HR interview.  You have to prepare well before for all the types of interviews.
For technical interview, you have to focus on C, DBMS and JAVA. It is important you have to be through with at least a couple of your core subjects.
All the best.

TCS Open sesame 2016 questions and solutions

TCS Open sesame 2016 questions and solutions

1. The rupee/coin changing machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100 rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10 hundred rupee notes when you put a one rupee coin!Sivaji, after being ruined by his rivals in business is left with a one rupee coin and discovers the flaw in the machine by accident. By using the machine repeatedly, which of the following amounts is a valid amount that Sivaji can have when he gets tired and stops at some stage (assume that the machine has an infinite supply of notes and coins):

a. 26975 b. 53947
c. 18980 d. 33966

Answer: B
The process works like this:
Rs.1 Coin  ⇒ 10 × 100 = Rs.1000
Rs.100  ⇒ 10 × 10
Rs.10  ⇒ 1 × 10
Sivaji gets more money when he inserts a rupee coin only.  For each rupee coin he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000.  So he has 1999 with him.  Now if he inserts another coin, he has 1998 + 1000 = 2998.

Now each of these numbers are in the form of 999n + 1.  So option B can be written as 54 × 999 + 1.

2. Seven movie addicts- Guna, Isha, Leela, Madhu, Rinku, Viji and Yamini attend a film festival. Three films are shown, one directed by Rajkumar Hirani ,one by S.Shankar,and one by Mani Ratnam. Each of the film buffs sees only one of the three films. The films are shown only once, one film at a time. The following restrictions must apply :- Exactly twice as many of the film buffs sees the S.shankar film as see the Rajkumar Hirani film.- Guna and Rinku do not see the same film as each other.- Isha and Madhu do not see same film as each other.- Viji and Yamini see the same film as each other.- Leela sees the S.Shankar film.- Guna sees either the Rajkumar Hirani film or the Mani Ratnam film.Which one of the following could be an accurate matching of the film buffs to films ?(A) Guna: the S.Shankar film; Isha: the Mani Ratnam film; Madhu: the S.Shankar film(B) Guna: the Mani Ratnam film; Isha: the Rajkumar Hirani film; Viji: the Rajkumar Hirani film(C) Isha : the S.Shankar film; Rinku: the Mani Ratnam film; Viji: the Rajkumar Hirani film(D) Madhu: the Mani Ratnam film; Rinku: the Mani Ratnam film; Viji: the Mani Ratnam film

a. A b. C
c. D d. B

Answer: Option C
Guna × Rinku
Isha × Madhu
(Viji + Yamini)
Leela – Film: Shankar
Guna = RKH/Mani Ratnam

The following options are possible:

RKH Shankar Mani Ratnam
1 2 4
2 4 1

We will take options and check them.
Option A: Guna should not watch Shankar’s Film. So ruled out
Option B:

RKH Shankar Mani Ratnam
Isha _ Guna
Viji _ _

Now Yamini also watch RKH. Which is not possible.
Option C:

RKH Shankar Mani Ratnam
Viji Isha Rinku
Yamini Leela _

As Guna should not be watching Shankar’s movie she should watch Mani ratnam’s which is not possible.
Option D:

RKH Shankar Mani Ratnam
Guna Leela Madhu
_ Isha Rinku
_ _ Viji and Yamini

3. Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460?

a. 955 b. 980
c. 797 d. 618

Answer: C
5678 – 35 + (one of the answer option) should be divisible by 460.  Only option C satisfies.

4. Find the probability that a leap year chosen at random will have 53 Sundays.

a. 1/7 b. 2/7
c. 1/49 d. 3/7

Answer: B
A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), …. (Sat + sun).  Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7

5. An ant starts moving on the mesh shown below along the wires towards a food particle.If the ant is at the bottom-left corner of cell A and the food is at the top-right corner of cell F, then find the number of optimal routes for the ant.

a. 13884156 b. 3465280
c. 4368 d. 6748

Answer: B
(Please read “Counting” to understand this question: Click here )
Total ways to move from A to the junction: There are 13 upward ways, 3 right side ways this Ant can move. Now these 16 ways may be in any order. So number of ways of arrangements = 16!13!×3!16!13!×3! = 560
Similarly, from the junction to F, Total 12 upward ways and 5 right-side ways. These 17 ways can be in any order. So Total ways = 17!12!×5!17!12!×5! = 6188
Total ways to move from A to F = 560 × 6188 = 3465280

6. You have been given a physical balance and 7 weights of 52, 50, 48, 44, 45, 46 and 78 kgs. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 183 kgs.

a. 180 b. 181
c. 182 d. 178

Answer: A
52+50+78 = 180

7. Two consecutive numbers are removed from the progression 1, 2, 3, …n.  The arithmetic mean of the remaining numbers is 26 1/4.  The value of n is

a. 60 b. 81
c. 50 d. Cannot be determined

Answer: C
As the final average is 105/4, initial number of pages should be 2 more than a four multiple. So in the given options, we will check option C.
Total = n(n+1)2=50×512=1275n(n+1)2=50×512=1275
Final total = 48×1054=126048×1054=1260
So sum of the pages = 15.  The page numbers are 7, 8

8. You need a 18% acid solution for a certain test, but your supplier only ships a 13% solution and a 43% solution. You need 120 lts of the 18% acid solution. the 13% solution costs Rs 82 per ltr for the first 67 ltrs, and Rs 66 per ltr for any amount in access of 67 ltrs. What is the cost of the 13% solution you should buy?

a. 8002 b. 7012
c. 7672 d. 7342

Answer: C
Let us assume we need “a” liters of 13% acid solution and “b” liters of 43% acid solution. Now
So we need 100 liters of 13% acid solution, and 20 liters of 18% acid solution.
Final cost = 82 × 67 + 66 × 33 = 7672

9. A spherical solid ball of radius 58 mm is to be divided into eight equal parts by cutting it four times longitudinally along the same axis.Find the surface area of each of the final pieces thus obtained( in mm^2) ? (where pi= 22/7)

a. 3365pi b. 5046pi
c. 1682pi d. 3346pi

Answer: B
If a sphere is cut into 8 parts longitudinally, It something looks like below

Now We have to find the surface area of one piece.   This is 18th18th of the initial sphere + 2 × area of the half circle 
= 18(4πr2)+π×r218(4πr2)+π×r2

= 18(4×π×582)+π×58218(4×π×582)+π×582
= 5046pi

10. There is a lot of speculation that the economy of a country depends on how fast people spend their money in addition to how much they save.  Auggie was very curious to test this theory.Auggie spent all of his money in 5 stores. In each store, he spent Rs.4 more than one-half of what he had when he went in. How many rupees did Auggie have when he entered the first store?

a. 248 b. 120
c. 252 d. 250

As he has spent all his money, He must spend Rs.8 in the final store. 
a simple equation works like this.  Amount left = 12x−412x−4
For fifth store this is zero. So x = 8. That means he entered fifth store with 8.
Now for fourth store, amount left = 8 so 12x−4=8⇒12x−4=8⇒ x = 24
For third store, amount left = 24 so 12x−4=24⇒12x−4=24⇒ x = 56
For Second store, amount left = 56 so 12x−4=56⇒12x−4=56⇒ x = 120
For first store, amount left = 120 so 12x−4=120⇒12x−4=120⇒ x = 248
So he entered first store with 248. 

11. A sudoku grid contains digits in such a manner that every row, every column, and every 3×3 box accommodates the digits 1 to 9, without repetition.  In the following Sudoku grid, find the values at the cells denoted by x and y and determine the value of 6x + 15y.

a. 87 b. 75
c. 66 d. 99

Answer: B

So x = 5, y = 3.
6x + 15y = 75

12. In how many ways can the letters of the english alphabet be arranged so that there are seven letter between the letters A and B, and no letter is repeated

a. 24P7 * 2 * 18! b. 36 * 24!
c. 24P7 * 2 * 20! d. 18 * 24!

Answer: B. Option A also correct.
We can fix A and B in two ways with 7 letters in between them. Now 7 letters can be selected and arranged in between A and B in 24P724P7 ways. Now Consider these 9 letters as a string. So now we have 26 – 9 + 1 = 18 letters
These 18 letters are arranged in 18! ways. So Answer is 2 x 24P724P7 x 18!
Infact, 2 x 24P724P7 x 18! = 36 x 24!. So go for Option B as it was given as OA.

13. A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x.  The value of f(1) is

a. 1 b. 2
c. 3 d. Cannot be determined

Answer: C
Put x =1 ⇒ f(1)+2*f(6-1) = 1 ⇒ f(1) + 2*f(5) = 1
Put x = 5 ⇒ f(5)+2*f(6-5) = 5 ⇒ f(5) + 2*f(1) = 5
Put f(5) = 5 – 2*f(1) in the first equation
⇒ f(1) + 2*(5 – 2*f(1)) = 1
⇒ f(1) + 10 – 4f(1) = 1
⇒ f(1) = 3

14.  Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?

A. 59 
B. 61 
C. 49 
D. 56
Answer: D
Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
(Explanation: See LCM formula 1 and 2: Click here)
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
⇒ 15x + 1 = 7c + 6  ⇒ c = 15x−5715x−57 ⇒ c = 2x+x−572x+x−57
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N – M = 56

15. The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573.  What is the largest of the numbers A, B, C, D?

a. 1712
b. 1650
c. 1164
d. 1211
Answer: a
b+c+d= 4087
Combining all we get 3(a+b+c+d) = 17208
⇒ a + b + c +d  = 3736
Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

16.  Anand packs 304 marbles into packets of 9 or 11 so that no marble is left.  Anand wants to maximize the number of bags with 9 marbles.  How many bags does he need if there should be atleast one bag with 11 marbles

a. 33
b. 32
c. 31
d. 30
Answer: B
Given 9x + 11y = 304.
x = 304−11y9304−11y9 = 33+7−2y9−y33+7−2y9−y
So y = – 1 satisfies. Now x = 35.  But y cannot be negative.
Now other solutions of this equation will be like this.  Increase or decrease x by 11, decrease or increase y by 9. So we have to maximise x. next solution is x = 24 and y = 8. So bags required are 32.

17. When Usha was thrice as old as Nisha, her sister Asha was 25, When Nisha was half as old as Asha, then sister Usha was 34.  their ages add to 100.  How old is Usha?

a. 37
b. 44
c. 45
d. 40
Answer: D
Let the age of Usha is 3x then Nisha is x and Asha is 25
Also Usha 34, Nisha y, and Asha 2y.
We know that 3x – 34 = x – 2y = 25 – 2y
Solving above three equations we get x = 9, y = 16
Their ages are 34, 16, 32. whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40

18. Find the number of zeroes in the expression 15*32*25*22*40*75*98*112*125

a. 12
b. 9
c. 14
d. 7
Answer: B
Maximum power of 5 in the above expression can be calculated like this.  Count all the powers of 5 in the above expression.  So number of zeroes are 9.  (Read this chapter)

19. Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. Afer how many hours will B reach city Y from the time A overtook him fro the first time?

a. 50 hrs
b. 49.5 hrs
c. 41.5 hrs
d. 37.5 hrs
Answer: C

Let us understand the diagram. Vehicle A overtaken B at 10.30 am and reached X at 12 pm.  It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance ‘n’, So it should be at Q at 11 pm.  It is clear that Vehicle A takes 0.5 hour to cover distance ‘m’.  Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m.  So Speeds ratio = 4.5 : 0.5 = 9 : 1.
Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 × 9 + 1 = 41.5 hours

20. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the twocircles is:

a. (π/2) – 1
b. 4
c. √2 – 1
d. √5

We have to find the area of the blue shaded one and double it to get the area common to the both. Now this can be calculated as Area of the sector OAB – Area of the Triangle OAB.
As OA and OB are perpendicular, area of the sector OAB = 90360π(1)290360π(1)2 = π4π4
Area of the triangle OAB = 12×1×112×1×1 = 1212
Area common to both = 2(π4−12)=π2−1

TCS Open sesame – 2014 with Solutions

TCS Open sesame – 2014 with Solutions 

1. What are the total number of divisors of 600(including 1 and 600)? 
a.  24 
b.  40 
c.  16 
d.  20 
Sol: Option a
If N=ap×bq×cr….N=ap×bq×cr…. then the number of factors of N = (p+1)(q+1)(r+1) ….
600 = 23×3×5223×3×52 
So number of factors of 600 = (3+1)(1+1)(2+1) = 24
2. What is the sum of the squares of the first 20 natural numbers (1 to 20)? 
a.  2870 
b.  2000 
c.  5650 
d.  44100 
Sol: Option a
Use formula n(n+1)(2n+1)6n(n+1)(2n+1)6
3. What  is ∑K=028K2(28KC)∑K=028K2(K28C) where 28KCK28C is the number of ways of choosing k items from 28 items? 
a.  406 *  227227  
b.  306 *  226226   
c.  28 *   227227
d.  56 *   227227
Sol: A
(1+x)n(1+x)n = nC0+nC1x+nC2x2nC0+nC1x+nC2x2 + . . . + nCnxnnCnxn – – – (1)
Differentiating w.r.t x we get
n(1+x)n−1n(1+x)n−1 = nC1+nC2(2x)nC1+nC2(2x) + nC3(3×2)nC3(3×2) + . . . + nCn(nxn−1)nCn(nxn−1)
Multiplying by x on both sides,
x.n(1+x)n−1x.n(1+x)n−1 = nC1.x+nC2(2×2)nC1.x+nC2(2×2) + nC3(3×3)nC3(3×3) + . . . + nCn(nxn)nCn(nxn)
Now again differentiating w.r.t to x
n[(1+x)n−1+x.(n−1)(1+x)n−2]n[(1+x)n−1+x.(n−1)(1+x)n−2] = nC1+nC2(4x)nC1+nC2(4x) + nC3(3×3)nC3(3×3) + . . . + nCn(n2xn−1)nCn(n2xn−1)
Putting x = 1, we get
n[(2)n−1+(n−1)(2)n−2]n[(2)n−1+(n−1)(2)n−2]= nC1+nC2(22)nC1+nC2(22) + nC3(32)nC3(32)+ . . . + nCn(n2)nCn(n2)
n(n+1)2n−2n(n+1)2n−2 = nC1+22nC2nC1+22nC2 + 32nC3+42nC432nC3+42nC4 + … + nCn(n2)nCn(n2)
n.2n−2(2+n−1)n.2n−2(2+n−1) = nC1+(22)nC2nC1+(22)nC2 + (32)nC3+(42)nC4(32)nC3+(42)nC4 + … + nCn(n2)nCn(n2)
n.2n−2(2+n−1)n.2n−2(2+n−1) = nC1+(22)nC2nC1+(22)nC2 + (32)nC3+(42)nC4(32)nC3+(42)nC4 + … + nCn(n2)nCn(n2)
n.2n−2(n+1)n.2n−2(n+1) = nC1+22nC2nC1+22nC2 + 32nC3+42nC432nC3+42nC4 + … + nCn(n2)nCn(n2)
Now substituting n = 28
28.226.2928.226.29 = 812.226812.226 = 406.227406.227

4. What is ∑K=0283K(28KC)∑K=0283K(K28C) where 28KCK28C  is the number of ways of choosing k items from 28 items? 
a. 256256
b.  3* 227227
c. 329329
d. 3* 427427 
Sol: Option A
We know that C0+3C1+32C2C0+3C1+32C2 + . . . + 3nCn=4n3nCn=4n
Substitute n = 28
We get ∑K=0283K(28KC)∑K=0283K(K28C) = 428428= 256256

5. A call center agent has a list of 305 phone numbers of people in alphabetic order of names (but she does not have any of the names).  She needs to quickly contact Deepak Sharma to convey a message to him.  If each call takes 2 minutes to  complete, and every call is answered, what is the minimum amount of time in which she can guarantee to deliver the message to Mr Sharma. 
a.  18 minutes 
b.  610 minutes 
c.  206 minutes 
d.  34 minutes 
Sol: Option A
6. The times taken by a phone operator to complete a call are 2,9,3,1,5 minutes respectively.  What is the average time per call? 
a.  4 minutes 
b.  7 minutes 
c.  1 minutes 
d.  5 minutes 
Sol: Option A
7. The times taken by a phone operator to complete a call are 2,9,3,1,5 minutes respectively.  What is the median time per call? 
a.  5 minutes 
b.  7 minutes 
c.  1 minutes 
d.  4 minutes 
 Sol: NO option is correct. Median is 3
8. Eric throws two dice, and his score is the sum of the values shown.  Sandra throws one die, and her score is the square of the value shown.  What is the probability that Sandra’s score will be strictly higher than Eric’s score? 
a.  137/216 
b.  17/36 
c.  173/216 
d.  5/6 
Sol: A
9. What is the largest integer  that divides  all three numbers 23400,272304,205248 without leaving a remainder? 
a.  48 
b.  24 
c.  96 
d.  72 
Sol: Option B
Find GCD
10. Of the 38 people in my office, 10 like to drink chocolate, 15 are cricket fans, and 20 neither like chocolate nor like cricket.  How many people like both cricket and chocolate? 
a.  7 
b.  10 
c.  15 
d.  18 
Sol: Option A
11. If f(x) = 2x+2 what is f(f(3))? 
a.  18 
b.  8 
c.  64 
d.  16 
 Sol: Option A
12. If   f(x) = 7 x +12, what is f-1(x) (the inverse function)? 
a.  (x-12)/7 
b.  7x+12 
c.  1/(7x+12) 
d.  No inverse exists 
Sol: Option A
13. A permutation is often represented by the cycles it has.  For example, if we permute the numbers in the natural order to 2 3 1 5 4, this is represented as (1  3 2) (5 4).  In this the (132) says that the first number has gone to the position 3, the third number has gone to the position 2, and the second number  has gone to position 1, and (5 4) means that the fifth number  has gone to position 4 and the fourth number  has gone to position 5.  The numbers with brackets are to be read cyclically.  If a number  has not changed position, it is kept as a single cycle.  Thus 5 2 1 3 4 is represented as (1345)(2). We may apply permutations on itself If we apply the permutation (132)(54) once, we get 2 3 1 5 4.  If we apply it again, we get 3 1 2 4 5 , or (1 2 3)(4) (5) If we consider the permutation of 7 numbers (1457)(263), what is its order (how many 
times must it be applied before the numbers appear in their original order)? 
a.  12 
b.  7 
c.  7! (factorial of 7) 
d.  14 
Sol: Not yet solved
14. What is the maximum value of x3y3 + 3 x*y when x+y = 8? 
a.  4144 
b.  256 
c.  8192 
d.  102 
Sol: Option A
The question probably be x3.y3+3x∗yx3.y3+3x∗y
Sustitute x = 4 and y = 4
15. Two circles of radii 5 cm and 3 cm touch each other at A and also touch a line at B and C. The distance BC in cms is? 
a.  60‾‾‾√60 
b.  62‾‾‾√62 
c.  68‾‾‾√68 
d.  64‾‾‾√64 
Sol: Option A
d = distance between centers
16. In Goa beach, there are three small picnic tables. Tables 1 and 2 each seat three people.   Table 3 seats  only one person, since two of its seats are broken. Akash, Babu, Chitra, David, Eesha, Farooq, and Govind all sit at seats at these picnic tables. Who sits with whom and at which table are determined by the following constraints: 
a.  Chitra does not sit at the same table as Govind. 
b.  Eesha does not sit at the same table as David. 
c.  Farooq does not sit at the same table as Chitra.  
d.  Akash does not sit at the same table as Babu. 
e.  Govind does not sit at the same table as Farooq. 
Which of the following is a list of people who could sit together at table 2? 
a.  Govind, Eesha, Akash 
b.  Babu, Farooq, Chitra 
c.  Chitra, Govind, David. 
d.  Farooq, David, Eesha. 
Sol: Option A
17. There are a number of chocolates in a bag.  If they were to be equally divided among 14 children, there are 10 chocolates left.  If  they were to be equally divided among 15 children, there are 8 chocolates left.  Obviously, this can be satisfied if any multiple of 210 chocolates are added to the bag.  What is the remainder when the minimum feasible number of chocolates in the bag is divided by 9?   
a.  2 
b.  5 
c.  4 
d.  6 
Sol: Option B
18. Let f(m,n) =45*m + 36*n, where m and n are integers (positive or negative)  What is the minimum positive value  for f(m,n) for all values of m,n (this may be achieved for various values of m and n)? 
a.  9 
b.  6 
c.  5 
d.  18 
Sol: Option A
19. What is the largest number that will divide 90207, 232585 and 127986 without leaving a remainder? 
a.  257 
b.  905 
c.  351 
d.  498 
Sol: Option A
20. We have an equal arms two pan balance and need to weigh objects with integral weights in the range 1 to 40 kilo grams. We have a set of standard weights and can  place the weights in any pan. . (i.e) some weights can be in a pan with objects and some weights can be in the other pan. The minimum number of standard weights required is: 
a.  4 
b.  10 
c.  5 
d.  6 
 Sol: Option A
21. A white cube(with six faces) is painted red on two different faces.  How many different ways can this be done (two paintings are considered same if on a suitable rotation of the cube one painting can be carried to the other)? 
a.  2 
b.  15 
c.  4 
d.  30  
 Sol: Option A
22. How many divisors (including 1, but excluding 1000) are there for the number 1000? 
a.  15 
b.  16 
c.  31 
d.  10 
 Sol: Option A
23. In the polynomial   f(x) =2*x^4 – 49*x^2 +54, what is the product of the roots, and what is the sum of the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)? 
a.  27,0 
b.  54,2 
c.  49/2,54 
d.  49,27 
 Sol: Option A
24. In the polynomial f(x) = x^5 + a*x^3 + b*x^4 +c*x + d, all coefficients a, b, c, d are integers. If 3 + sqrt(7) is a root, which of the following must be also a root?(Note that x^n denotes the x raised to the power n, or x multiplied by itself n times. Also sqrt(u) denotes  the square root of u, or the number which when multiplied by itself, gives the number u)? 
a.  3-sqrt(7) 
b.  3+sqrt(21) 
c.  5 
d.  sqrt(7) + sqrt(3) 
Sol: Option A

TCS Opensesame 2013 with solutions and Explainations

TCS Opensesame 2013 with solutions and Explainations

1.  If 3y + x > 2 and x + 2y ≤ 3, What can be said about the value of y?
A. y = -1
B. y >-1
C. y <-1
D. y = 1
Answer: B
Multiply the second equation with -1 then it will become – x – 2y ≥ – 3.  Add the equations.  You will get y > -1.

2. If the price of an item is decreased by 10% and then increased by 10%, the net effect on the price of the item is
A. A decrease of 99%
B. No change
C. A decrease of 1%
D. An increase of 1%
Answer: C
If a certain number is increased by x% then decreased by x% or vice versa, the net change is always decrease.  This change is given by a simple formula − (x⁄10)2 = − (10⁄10)2 = −1 %.  Negitive sign indicates decrease.

3. If m is an odd integer and n an even integer, which of the following is definitely odd?
A. (2m+n)(m-n)
B. (m + n2) + (m – n2)
C. m2 + mn + n2
D. m +n
Answer: C and D (Original Answer given as D)
You just remember the following odd ± odd = even; even ± even = even; even ± odd = odd
Also odd × odd = odd; even × even = even; even × odd = even.

4.  What is the sum of all even integers between 99 and 301?
A. 40000
B. 20000
C. 40400
D. 20200
Answer: D
The first even number after 99 is 100 and last even number below 301 is 300.  We have to find the sum of even numbers from 100 to 300.  i.e., 100 + 102 + 104 + …………… 300.
Take 2 Common.  2 x ( 50 + 51 + ………..150)
There are total 101 terms in this series.  So formula for the sum of n terms when first term and last term is known is n2(a+l)n2(a+l)
So 50 + 51 + ………..150 = 1012(50+150)1012(50+150)
So 2 x 1012(50+150)1012(50+150) = 20200

5. There are 20 balls which are red, blue or green.  If 7 balls are green and the sum of red balls and green balls is less than 13, at most how many red balls are there?
A. 4
B. 5
C. 6
D. 7
Answer: B
Given R + B + G = 17; G = 7; and R + G < 13.  Substituting G = 7 in the last equation, We get R < 6. So maximum value of R = 6

6.  If n is the sum of two consecutive odd integers and less than 100, what is greatest possibility of n?
A. 98
B. 94
C. 96
D. 99
Answer : C
We take two odd numbers as (2n + 1) and (2n – 1).
Their sum should be less than 100. So (2n + 1) + (2n – 1) < 100 ⇒⇒ 4n < 100.
The largest 4 multiple which is less than 100 is 96

7. x2x2 < 1/100, and x < 0 what is the highest range in which x can lie?
A. -1/10 < x < 0
B. -1 < x < 0
C. -1/10 < x < 1/10
D. -1/10 < x
Answer: A
(x – a)(x – b) < 0 then value of x lies in between a and b.
(x – a)(x – b) > 0 then value of x does not lie inbetween a and b. or ( −∞−∞, a) and (b, −∞−∞) if a < b
x2x2 < 1/100 ⇒⇒
(x2−1/100)<0(x2−1/100)<0 ⇒⇒ (x2−(1/10)2)<0(x2−(1/10)2)<0 ⇒⇒ (x−1/10)(x+1/10)<0(x−1/10)(x+1/10)<0
So x should lie in between – 1/10 and 1/10.  But it was given that x is -ve. So x lies in -1/10 to 0

8.  There are 4 boxes colored red, yellow, green and blue.  If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?
A. 1
B. 6
C. 9
D. 5
Answer: 5
Total ways of selecting two boxes out of 4 is 4C24C2 = 6.  Now, the number of ways of selecting two boxes where none of the green or red box included is only 1 way.  (we select yellow and blue in only one way).  If we substract this number from total ways we get 5 ways.

9. All faces of a cube with an eight – meter edge are painted red.  If the cube is cut into smaller cubes with a two – meter edge, how many of the two meter cubes have paint on exactly one face?
A. 24
B. 36
C. 60
D. 48
Answer : A
If there are n cubes lie on an edge, then total number of cubes with one side painting is given by 6×(n−2)26×(n−2)2.  Here side of the bigger cube is 8, and small cube is 2.  So there are 4 cubes lie on an edge. Hence answer = 24

10. Two cyclists begin training on an oval racecourse at the same time.  The professional cyclist completes each lap in 4 minutes; the novice takes 6 minutes to complete each lap.  How many minutes after the start will both cyclists pass at exactly the same spot where they began to cycle?
A. 10
B. 8
C. 14
D. 12
Answer: D
The faster cyclist comes to the starting point for every 4 min so his times are 4, 8, 12, ………  The slower cyclist comes to the starting point for every 6 min so his times are 6, 12, 18, ………  So both comes at the end of the 12th min.

11. M, N, O and P are all different individuals; M is the daughter of N; N is the son of O; O is the father of P; Among the following statements, which one is true?
A. M is the daughter of P
B. If B is the daughter of N, then M and B are sisters
C. If C is the granddaughter of O, then C and M are sisters
D. P and N are bothers. 
Answer: B

From the diagram it is clear that If B is the daughter of N, then M and B are sisters.  Rectangle indicates Male, and Oval indicates Female.

12. In the adjoining diagram, ABCD and EFGH are squares of side 1 unit such that they intersect in a square of diagonal length (CE) = 1/2.  The total area covered by the squares is 

A. Cannot be found from the information
B. 1 1/2
C. 1 7/8
D. None of these
Answer:  C

Let CG = x then using pythogerous theorem CG2+GE2=CE2CG2+GE2=CE2
⇒⇒  x2+x2=(1/2)2⇒2×2=1/4⇒x2=1/8×2+x2=(1/2)2⇒2×2=1/4⇒x2=1/8
Total area covered by two bigger squares = ABCD + EFGE – Area of small square = 2 – 1/8 = 15/8

13. There are 10 stepping stones numbered 1 to 10 as shown at the side.  A fly jumps from the first stone as follows; Every minute it jumps to the 4th stone from where it started – that is from 1st it would go to 5th and from 5th it would go to 9th and from 9th it would go to 3rd etc.  Where would the fly be at the 60th minute if it starts at 1?

A. 1
B. 5
C. 4
D. 9
Answer : A

Assume these steps are in circular fashion. 
Then the fly jumps are denoted in the diagram.  It is clear that fly came to the 1st position after 5th minute.  So again it will be at 1st position after 10th 15th …..60th. min.
So the fly will be at 1st stone after 60th min. 

14. What is the remainder when 617+1176617+1176  is divided by 7?
A. 1
B. 6
C. 0
D. 3
Answer: C
617617 = (7−1)17(7−1)17 =
17C0.717−17C1.716.1117C0.717−17C1.716.11+. . . + 17C16.71.11617C16.71.116 – 17C17.11717C17.117
If we divide this expansion except the last term each term gives a remainder 0.  Last term gives a remainder of – 1.
Now From Fermat little theorem, [ap−1p]Rem=1[ap−1p]Rem=1
So [1767]Rem=1[1767]Rem=1
Adding these two remainders we get the final remainder = 0

15. In base 7, a number is written only using the digits 0, 1, 2, …..6.  The number 135 in base 7 is 1 x 7272 + 3 x 7 + 5 = 75 in base 10.  What is the sum of the base 7 numbers 1234 and 6543 in base 7. 
A. 11101
B. 11110
C. 10111
D. 11011
Answer: B

In base 7 there is no 7.  So to write 7 we use 10.  for 8 we use 11…… for 13 we use 16, for 14 we use 20 and so on.
So from the column d, 4 + 3 = 7 = 10, we write 0 and 1 carried over.  now 1 + 3 + 4 = 8 = 11, then we write 1 and 1 carried over.  again 1 + 2 + 5 = 8 = 11 and so on

16. The sequence {An}{An} is defined by A1A1 = 2 and An+1=An+2nAn+1=An+2n what is the value of A100A100
A. 9902
B. 9900
C. 10100
D. 9904
Answer: A
We know that A1A1 = 2 so A2=A1+1=A1+2(1)=4A2=A1+1=A1+2(1)=4
So the first few terms are 2, 4, 8, 14, 22, ……
The differences of the above terms are 2, 4, 6, 8, 10…
and the differences of differences are 2, 2, 2, 2.  all are equal.  so this series represents a quadratic equation.
Assume  AnAn = an2+bn+can2+bn+c
Now A1A1 = a + b + c = 2
A2A2 = 4a + 2b + c = 4
A3A3 = 9a + 3b + c = 8
Solving above equations we get a = 1, b = – 1 and C = 2
So substituting in AnAn = n2+bn+cn2+bn+c = n2−n+2n2−n+2
Substitute 100 in the above equation we get 9902.

17.Find the number of rectangles from the adjoining figure (A square is also considered a rectangle)

A. 864
B. 3276
C. 1638
D. None
Answer: C
To form a rectangle we need two horizontal lines and two vertical lines.  Here there are 13 vertical lines and 7 horizontal lines.  The number of ways of selecting 2 lines from 13 vertical lines is 13C213C2 and the number of ways of selecting 2 lines from 7 horizontals is 7C27C2. So total rectangles = 7C2x13C27C2x13C2

18. A, B, C and D go for a picnic.  When A stands on a weighing machine, B also climbs on, and the weight shown was 132 kg.  When B stands, C also climbs on, and the machine shows 130 kg.  Similarly the weight of C and D is found as 102 kg and that of B and D is 116 kg.  What is D’s weight
A. 58kg
B. 78 kg
C. 44 kg
D. None
Answer : C
Given A + B = 132; B + C = 130; C + D = 102, B + D = 116
Eliminate B from 2nd and 4th equation and solving this equation and 3rd we get D value as 44.

19.  Roy is now 4 years older than Erik and half of that amount older than Iris.  If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy’s age multiplied by Iris’s age?
A. 28
B. 48
C. 50
D. 52
Answer: 48

20. X, Y, X and W are integers.  The expression X – Y – Z is even and the expression Y – Z – W is odd.  If X is even what must be true?
A. W must be odd
B. Y – Z must be odd
C. W must be odd
D. Z must be odd
Answer: A or C (But go for C)

21.  Mr and Mrs Smith have invited 9 of their friends and their spouses for a party at the Waikiki Beach resort.  They stand for a group photograph.  If Mr Smith never stands next to Mrs Smith (as he says they are always together otherwise). How many ways the group can be arranged in a row for the photograph?
A. 20!
B. 19! + 18!
C. 18 x 19!
D. 2 x 19!
Answer: C

22.  In a rectangular coordinate system, what is the area of a triangle whose vertices whose vertices have the coordinates (4,0), (6, 3) adn (6 , -3)
A. 6
B. 7
C. 7.5
D. 6.5
Answer: A

23. A drawer holds 4 red hats and 4 blue hats.  What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
A. 1/2
B. 1/8
C. 1/4
D. 3/8
Answer: B

24. In how many ways can we distribute 10 identical looking pencils to 4 students so that each student gets at least one pencil?
A. 5040
B. 210
C. 84
D. None of these
Answer: C

25. The prime factorization of integer N is A x A x B x C, where A, B and C are all distinct prime integers.  How many factors does N have?
A. 12
B. 24
C. 4
D. 6
Answer: A

26. Tim and Elan are 90 km from each other.they start to move each other simultaneously Tim at speed 10 and elan 5 kmph. If every hour they double their speed what is the distance that Tim will pass until he meet Elan 
A. 45
B. 60
C. 20
D. 80
Answer: B

27. A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress.
A.  2/3
B.  1/3
C.  1/6
D.  1/9 
Answer: B

28. N is an integer and N>2, at most how many integers among N + 2, N + 3, N + 4, N + 5, N + 6,  and N + 7 are prime integers?
A. 1
B. 3
C. 2
D. 4
Answer: C

29. A turtle is crossing a field.  What is the total distance (in meters) passed by turtle? Consider the following two statements
(X) The average speed of the turtle is 2 meters per minute
(Y) Had the turtle walked 1 meter per minute faster than his average speed it would have finished 40 minutes earlier
A. Statement X alone is enough to get the answer
B. Both statements X and Y are needed to get the answer
C. Statement Y alone is enough to get the answer
D. Data inadequate
Answer: B

30. Given the following information, who is youngest?
C is younger than A; A is taller than B
C is older than B; C is younger than D
B is taller than C; A is older than D
A. D
B. B
C. C
D. A
Answer: B

31. If P(x) = ax4+bx3+cx2+dx+eax4+bx3+cx2+dx+e has roots at x = 1, 2, 3, 4 and P(0) = 48, what is P(5)
A. 48
B. 24
C. 0
D. 50 
Answer: A

TCS latest Questions with Answers – 30

TCS latest Questions with Answers – 30

Star mark question:

1. In particular language if A=0, B=1, C=2,…….. ..     , Y=24, Z=25 then what is the value of  ONE+ONE (in the form of alphabets only)

a. BDAI   
b. ABDI   
c. DABI   
Answer: a
This problem is based on Base 26 rather than regular base 10 (decimal system) that we normally use.  In base 10 there are 10 digits 0 to 9 exist.  In base 26 there are 26 digits 0 to 25 exist.  To convert any number into base 26, we have to divide the number with 26 and find the remainder. (Study this Base system chapter).
Here, ONE + ONE =
E has value of 4. So E + E = 8 which is equal to I.
Now N + N = 13 + 13 = 26.  But in base 26, there is no 26.  So (26)10=(10)26(26)10=(10)26

So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29

Therefore, (29)10=(13)26(29)10=(13)26
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI

2. Find the number of perfect squares in the given series 2013, 2020, 2027,……………., 2300  (Hint 44^2=1936)

a. 1  
b. 2   
c. 3  
d. Can’t be determined
Answer: a
The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k.  We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 – 1) = 2116
47^2 = 2116 + (2 x 47 – 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k.  One number satisfies.

3.  What is in the 200th position of 1234 12344 123444 1234444….?

Answer: 4
The given series is 1234, 12344, 123444, 1234444, …..
So the number of digits in each term are 4, 5, 6, … or (3 + 1), (3 + 2), (3 + 3), …..upto n terms = 3n+n(n+1)23n+n(n+1)2
So 3n+n(n+1)2≤2003n+n(n+1)2≤200
For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184.  And 16 them contains 16 + 3 = 19 digits.
Now 17 term contains 20 digits and 123444……417times123444……4⏟17times.  So last digit is 4 and last two digits are 44.

4. 2345 23455 234555 234555……….. what was last 2 numbers at 200th digit?

Answer: 55
Proceed as above.  The last two digits in the 200th place is 55.

5. There are equal number of boys and girls in a class. If 12 girls entered out, twice the boys as girls remain. What was the total number of students in a class?

Answer: 48
Let the boys = b and girls = g
Given bg−12=21bg−12=21
Substitute b = g in the above equation. g = 24. So total students = 24 + 24 = 48

6. a bb ccc dddd eeeee ………What is the 120th letter?

Answer: O
Number of letters in each term are in AP. 1, 2, 3, …
So n(n+1)2≤120n(n+1)2≤120
For n = 15, we get LHS = 120. So 15th letter in the alphabet is O. So 15th term contains 15 O’s.

7. There are 120 male and 100 female in a society. Out of 25% male and 20% female are rural. 20% of male and 25% of female rural people passed in the exam. What % of rural students have passed the exam?

Answer: 22%

From the above data, Rural male = 25%(120) = 30, Rural female = 20%(100) = 20.
Passed students from rural: male = 20%(30) = 6, female = 25%(20) = 5
Required percentage = 1150×100=22%1150×100=22%

8. 1/7 th of the tank contains fuel. If 22 litres of fuel is poured into the tank the indicator rests at 1/5th mark. What is the quantity of the tank?

Answer: 385
Let the tank capacity = vv liters.
Given, v7+22=v5v7+22=v5

9. What is the probability of getting sum 3 or 4 when 2 dice are rolled

Answer: 5/36
Required number of ways = (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5
Total ways = 62=3662=36
Probability = 536536

10. On the fabled Island of Knights and Knaves, we meet three people, A, B, and C, one of whom is a knight, one a knave, and one a spy. The knight always tells the truth, the knave always lies, and the spy can either lie or tell the truth.A says: “C is a knave.”B says: “A is a knight.”C says: “I am the spy.”Who is the knight, who the knave, and who the spy?

Explanation: A= Knight, B= Spy, C = Knave
Let us say A is Knight and speaks truth.  So C is Knave and B is spy. So C’s statement is false and B’s statement is true.  This case is possible.
Let us say B is Knight. This is not possible as A also becomes Knight as B speaks truth.
Let us say C is Knight. This is clearly contradicted by C’s statement itself.

TCS latest Questions with Answers – 29

TCS latest Questions with Answers – 29

1. The perimeter of a equilateral triangle and regular hexagon are equal.  Find out the ratio of their areas?

a. 3:2

b. 2:3
c. 1:6
d. 6:1
Correct Option: b

Let the side of the equilateral triangle = aa units and side of the regular hexagon is bb units.
Given that,  3a=6b3a=6b ⇒ab=21⇒ab=21
Now ratio of the areas of equilateral triangle and hexagon = 3‾√4a2:33‾√2b234a2:332b2

2. What is the remainder of (32^31^301) when it is divided by 9?

a. 3
b. 5
c. 2
d. 1

Correct option: b
See solved example 6 here
3231301932313019 = 53130195313019
Euler totient theorem says that [aϕ(n)n]Rem=1[aϕ(n)n]Rem=1
ϕ(n)=n(1−1a)(1−1b)…ϕ(n)=n(1−1a)(1−1b)… here……
Now ϕ(9)=9(1−13)=6ϕ(9)=9(1−13)=6
Therefore, 5656 when divided by 9 remainder 1.
Now 313016=1301=1313016=1301=1
So 3130131301 can be written as 6k + 1

3. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?

a. 980
b. 797
c. 955
d. 618
Correct option: b
Let xx be the number to be added to 5678.
When you divide 5678 + xx by 460 the remainder = 35.
Therefore, 5678 + xx = 460k + 35 here kk is some quotient.
⇒⇒ 5643 + xx should exactly divisible by 460.
Now from the given options x = 797.

4. A girl entered a store and bought x flowers for y dollars (x and y are integers). When she was about to leave, the clerk said, “If you buy 10 more flowers I will give you all for $$2, and you will save 80 cents a dozen”. The values of x and y are:

a. (15,1)
b. (10,1)
c. (5,1)
d. Cannot be determined from the given information.
Correct option: c
Given she bought xx flowers for yy dollars.
So 1 flower cost = yxyx
12 flowers or 1 dozen cost = 12yx12yx
Again, xx+10 cost = 2 dollars
1 flower cost = 210+x210+x
12 flowers or 1 dozen cost = 2×1210+x=2410+x2×1210+x=2410+x
Given that this new dozen cost is 80 cents or 4/5 dollar less than original cost.
From the given options, c satisfies this.

5. If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?

a. 9
b. 3
c. 5
d. 7
Correct option: c
Let ′N′′N′ be the given number.
N=357k+5N=357k+5 = 17×21k+517×21k+5
If this number is divided by 17 remainder is 5 as 357k is exactly divided by 17.

6. In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.

a. 450
b. 420
c. 350
d. 320
Correct option:
We have to distribute three 2’s to a, b, c in 3+3−1C3−1=5C2=103+3−1C3−1=5C2=10 ways
We have to distribute four 3’s to a, b, c in 3+4−1C3−1=6C2=153+4−1C3−1=6C2=15 ways
We have to distribute one 5 to a, b, c in 3 ways.
Total ways = 10×15×3=45010×15×3=450 ways.

7. On door A – It leads to freedomOn door B – It leads to Ghost houseOn door C – door B leads to Ghost houseThe statement written on one of the doors is wrong.Identify which door leads to freedom.

a. A
b. B
c. C
d. None
Correct option: c
Case 1: A, B are true. In this case, Statement C also correct. So contradiction.
Case 2: B, C are true. In this case, B leads to ghost house and C confirms it. Now A is wrong. So door A does not lead to freedom. So Door C leads to freedom.

8. In the given figure, If the sum of the values along each side is equal. Find the possible values a, b, c, d, e, and f.

a. 9, 7, 20, 16, 6, 38
b. 4, 9, 10, 13, 16, 38
c. 4, 7, 20, 13, 6, 38
d. 4, 7, 20, 16, 6, 33
Correct option: c
From the above table, 42 + a + b = 47 + e.  Therefore,  a + b = 5 + e.  Option a, b ruled out.
47 + e = 15 + f.   Therefore, 32 + e = f. Option d ruled out.
4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number
a. 5/18
b. 13/18
c. 1/36
d. 1/2

9. 70, 54, 45, 41……. What is the next number in the given series?

a. 35
b. 36
c. 38
d. 40
Correct option: d
Consecutive squares are subtracted from the numbers.
70 – 54 = 16
54 – 45 = 9
45 – 41 = 4
So next we have to subtract 1. So answer = 41 – 1 = 40

10. How many positive integers less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each digit being used only once.

a. 52
b. 68
c. 66
d. 34
Correct option:
Single digit number = 4
Double digit number = 4××3 = 12
Three digit numbers = 3××3××2= 18 (∵∵ If Hundred’s place is 5, then the number is greater than 500)
Total = 34.

TCS latest Questions with Answers – 28

TCS latest Questions with Answers – 28

1. 11, 23, 47, 83, 131, . What is the next number?

a. 145
b. 178
c. 176
d. 191
23–11 = 12
47–23 = 24
83–47 = 36
131–83 = 48
Therefore, 131+60=191

2. A series of book was published at seven year intervals.  When the seventh book was published the total sum of publication year was 13, 524.  First book was published in?

a. 1911
b. 1910
c. 2002
d. 1932
Let the years be n, n+7, n+14, …., n+42.  (∵∵ use formula Tn=a+(n−1)dTn=a+(n−1)d to find nth term)
Sum = Sn=n2(2a+(n−1)d)Sn=n2(2a+(n−1)d) = 72(2n+(7−1)7)72(2n+(7−1)7) = 13,524
⇒⇒ n = 1911

3. Crusoe hatched from a mysterious egg discovered by Angus, was growing at a fast pace that Angus had to move it from home to the lake. Given the weights of Crusoe in its first weeks of birth as 5, 15, 30,135, 405, 1215, 3645. Find the odd weight out.

a) 3645 
b) 135 
c) 30 
d) 15
Answer: c
5×3 = 15
15×3 = 45 ⇒⇒ Given as 30
45×3 = 135
135×3 = 405
405×3 = 1215
1215×3 = 3645

4. A can complete a piece of work in 8 hours, B can complete in 10 hours and C in 12 hours. If A,B, C start the work together but A laves after 2 hours. Find the time taken by B and C to complete the remaining work.

1) 2 (1/11) hours
2) 4 (1/11) hours
3) 2 (6/11) hours
4) 2 hours
A,B,C’s 1 hour work is = 18+110+11218+110+112 =  15+12+10120=3712015+12+10120=37120
A,B,C worked together for 2 hours, Therefore, 2 hours work is = 37120×2=376037120×2=3760
Remaining work = 1−3760=23601−3760=2360
(23/60 work is done by B and C together)
B, C’s 1 hour work = 110+112=6+560=1160110+112=6+560=1160
(2360)th(2360)th part of the work done by B, C in = (2360)1160(2360)1160 = 21112111 hours.

5. A tree of height 36m is on one edge of a road broke at a certain height.  It fell in such a way that the top of the tree touches the other edge of the road. If the breadth of the road is 12m, then what is the height at which the tree broke?

a. 16
b. 24
c. 12
d. 18
Let the tree was broken at x meters height from the ground and 36 – x be the length of other part of the tree.

From the diagram, (36−x)2=x2+122(36−x)2=x2+122

6. The sticks of same length are used to form a triangle as shown below.If 87 such sticks are used then how many triangles can be formed?

First triangle is formed by using 3 sticks, but any subsequent triangle may be formed by using 2 sticks.  Therefore, If 1st triangles uses 3 sticks, Remaining sticks = 87 – 3 = 84.  With these 84, we can form 42 triangles. So total = 42 + 1 = 43
To solve questions like these, use formula, 2n + 1 = k.  Here n = triangles, k = sticks
2n+1 = 87 ⇒⇒ n = 43.

7. 17 × 8 m rectangular ground is surrounded by 1.5 m width path. Depth of the path is 12 cm. Gravel is filled and find the quantity of gravel required.

a. 5.5
b. 7.5
c. 6.05
d. 10.08

Area of the rectangular ground = 17×8=136m217×8=136m2
Area of the big rectangle considering the path width = (17+2×1.5)×(8+2×1.5)=220m2(17+2×1.5)×(8+2×1.5)=220m2
Area of the path = 220−136=84m2220−136=84m2
Gravel required = 84m2×12100m=10.08m384m2×12100m=10.08m3

8. A sum of Rs.3000 is distributed among A, B, and C. A gets 2/3 of what B and C got together and c gets 1/3 of what A and B got together, C’s share is?

Let B+C together got 3 units, then A get 2 units. or B+CA=32B+CA=32 – – – (1)
Let A+B together got 3 units, then B get 1 units. or A+BC=31A+BC=31 – – – (2)
By using Componendo and Dividendo, we can re-write equations (1) and (2), A+B+CA=3+22=52=208A+B+CA=3+22=52=208 and A+B+CC=3+11=41=205A+B+CC=3+11=41=205
So A = 8, B = 7, C = 5
C’s share = 5(8+5+7)×3000=7505(8+5+7)×3000=750

9. The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?

a. 7
b. 8
c. 5
d. 4
From the given information, (272738 – 13, 232342 – 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.

10.  Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1).  For all natural numbers (Integers>0)m and n.  What is the value of f(17)? 

a. 5436
b. 4831
c. 5508
d. 4832
f(1) = 0
f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32
f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204
f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832