TCS New Verbal Elimination Round – 2014 – (1)

TCS New Verbal Elimination Round – 2014 – (1)

Directions:
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 
XXX

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
Regards
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
better.
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.
Regards
Project Lead
xxxx

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
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
(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
Explanation:
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
Explanation:
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
Explanation:
Let us assume we need “a” liters of 13% acid solution and “b” liters of 43% acid solution. Now
⇒18=a×13+b×43a+b⇒18=a×13+b×43a+b
⇒ab=51⇒ab=51
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
Explanation:
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

Answer:
Explanation:
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
Explanation:

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.
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
a+b+c=4024
b+c+d= 4087
a+c+d=4524
a+b+d=4573
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
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:

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
Answer:
Explanation:

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
d2−(r1−r2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾√d2−(r1−r2)2
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
Remember:
(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
A3=A2+1=A2+2(2)=8A3=A2+1=A2+2(2)=8
A4=A3+1=A3+2(3)=14A4=A3+1=A3+2(3)=14
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   
d. CIDA
Answer: a
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
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
Explanation:
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%
Explanation:

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
Explanation:
Let the tank capacity = vv liters.
Given, v7+22=v5v7+22=v5
v5−v7=22⇒v=385v5−v7=22⇒v=385

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

Answer: 5/36
Explanation:
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?

Answer:
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

Explanation:
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
⇒3‾√4(2)2:33‾√2(1)2⇒34(2)2:332(1)2
⇒2:3⇒2:3

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
Explanation:
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 n=ap.bq…n=ap.bq…
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
⇒531301=(56)K.51⇒531301=(56)K.51
5313019=(56)K.519=1K.59=55313019=(56)K.519=1K.59=5

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
Explanation:
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
Explanation:
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.
⇒12yx−2410+x=45⇒12yx−2410+x=45
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
Explanation:
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:
Explanation:
3450=23×34×51=a×b×c3450=23×34×51=a×b×c
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
Explanation:
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
Explanation:
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
Explanation:
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:
Explanation:
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
Explanation:
11,23,47,83,131
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
Answer:
Explanation:
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
⇒7n+147=13,524⇒7n+147=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
Explanation:
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
Explanation:
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
Explanation:
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
⇒1296−72x+x2=x2+144⇒1296−72x+x2=x2+144
⇒72x=1296−144⇒72x=1296−144
⇒x=16⇒x=16


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?

Explanation:
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
Shortcut:
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
Explanation:

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?

Explanation:
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
Explanation:
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
Explanation:
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

TCS latest Questions with Answers – 27

TCS latest Questions with Answers – 27

1. I bought a certain number of marbles at a rate of 27 marbles for rupees 2 times M,where M is an integer.  I divided these marbles into 2 parts of equal numbers,one part of which I sold at the rate of 13 marbles for Rs.M and the other at the rate of 14 marbles for Rs.M.I spent and received an integral no of rupees,but bought the least number of marbles.How many did I buy?
a. 870
b. 102660
c. 1770
d. 9828
Answer: d
Explanation:
Let he bought 2x marbles.
27 marbles costs = Rs.2M so 1 marble costs = Rs.2M/27
Therefore, x marbles costs = Rs. (2M × x) / 27
Now we calculate the selling prices.
He sold x marbles at the rate of 13 for Rs.M so 1 marble selling price = M/13
x marbles selling price = x × M/13
He sold another x marbles at the rate of 14 for Rs.M so 1 marble selling price = M/14
x marbles selling price = x × M/14
Now 2Mx27,xM13,xM142Mx27,xM13,xM14 are integers.
So x marbles must be divisible by 27, 13, 14.  LCM of 4914
So 2x = 9828

2. How many different integers can be expressed as the sum of three distinct numbers from the set {3, 10, 17, 24, 31, 38, 45, 52}?
option 
a) 8 
b) 56 
c) 16 
d) 15
Answer: c
Explanation:
Interesting question.  If you think that 8C38C3 = 56 is correct then it is wrong answer.  We are not asked how many ways we can select 3 numbers out of 8. But how many different numbers can be expressed as a sum of three numbers from the given set.  For example, 3 + 10 + 31 = 3 + 17 + 24 = 47.  So 47 can be expressed as a sum of 3 numbers in two different ways but 47 should be considered as only one number.
Now the minimum number that can be expressed as a sum of 3 numbers = 30. The next number is 37.  Similarly the largest number is 38 + 45 + 52 = 135.
So there exists many numbers in between, with common difference of 7.
Total numbers = l−ad+1=135−307+1l−ad+1=135−307+1 = 16.

3. How many different integers can be expressed as the sum of three distinct numbers from the set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?
option 
a) 8 
b) 56 
c) 120
d) 22
Answer: d
Explanation:
From the above discussion, minimum number = 24 and maximum number = 129.  So there exist many numbers in between these two numbers, with common difference of 5.  All these numbers can be expressed as a sum of 3 different integers from the given set.
Total numbers =  l−ad+1=129−245+1l−ad+1=129−245+1 = 22.



4. A owes B Rs.50. He agrees to pay B over a number of consecutive days starting on a Monday, paying single note of Rs.10 or Rs.20 on each day. In how many different ways can A repay B.
Explanation:
He can pay by all 10 rupee notes in 5 days = 1 way
3 Ten rupee + 1 twenty rupee = 4!3!×1!4!3!×1! = 4 ways
1 Ten rupee + 2 twenty rupee notes = 3!2!×1!3!2!×1! = 3 ways
Total ways = 1 + 4 + 3 = 8

5. HCF of 2472,1284 and a third number ‘n’is 12.If their LCM is 8*9*5*103*107.then the number ‘n’is..
a. 2^2*3^2*5^1
b. 2^2*3^2*7^1
c. 2^2*3^2*8103
d. None of the above.
Answer:
Explanation:
2472 = 23×3×10323×3×103
1284 = 22×3×10722×3×107
HCF = 22×322×3
LCM = 23×32×5×103×10723×32×5×103×107
HCF of the numbers is the highest number which divides all the numbers.  So N should be a multiple of 22×322×3
LCM is the largest number that is divided by the given numbers.  As LCM contains 32×532×5 these two are from N.
So N = 22×32×5122×32×51

6. What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3
a. 77! – 54!
b. 77! + 54!
c. 77!^2 – 54!^2
d. 77!
Answer: a
Explanation:
The above question can be written as 77!(77!−2∗54!)3(77!+54!)377!(77!−2∗54!)3(77!+54!)3+ 54!(2∗77!−54!)3(77!+54!)354!(2∗77!−54!)3(77!+54!)3
Let A = 77!, B = 54!
Then equation in the form
a(a−2b)3(a+b)3+b(2a−b)3(a+b)3a(a−2b)3(a+b)3+b(2a−b)3(a+b)3
a(a−2b)3a(a−2b)3 = a(a3−6a2b+12ab2−8b3)a(a3−6a2b+12ab2−8b3) = a4−6a3b+12a2b2−8ab3a4−6a3b+12a2b2−8ab3
b(2a−b)3b(2a−b)3 = b(8a3−12a2b+6ab2−b3)b(8a3−12a2b+6ab2−b3) = 8a3b−12a2b2+6ab3−b48a3b−12a2b2+6ab3−b4
Grouping similar terms, a4−6a3b+12a2b2−8ab3a4−6a3b+12a2b2−8ab3 + 8a3b−12a2b2+6ab3−b48a3b−12a2b2+6ab3−b4
= a4−b4+(−6a3b+8a3b)a4−b4+(−6a3b+8a3b) + (12a2b2−12a2b2)(12a2b2−12a2b2) + (−8ab3+6ab3)(−8ab3+6ab3)
= a4−b4+2a3b−2ab3a4−b4+2a3b−2ab3
= (a2−b2)(a2+b2)(a2−b2)(a2+b2)+2ab(a2−b2)2ab(a2−b2)
= (a2−b2)[(a2+b2)+2ab](a2−b2)[(a2+b2)+2ab]
= (a2−b2)(a+b)2(a2−b2)(a+b)2
= (a−b)(a+b)(a+b)2(a−b)(a+b)(a+b)2
= (a−b)(a+b)3(a−b)(a+b)3
Therefore, a(a−2b)3(a+b)3+b(2a−b)3(a+b)3a(a−2b)3(a+b)3+b(2a−b)3(a+b)3 = (a−b)(a+b)3(a+b)3=a−b(a−b)(a+b)3(a+b)3=a−b

Shortcut:
If you try to solve this questions using above method, its almost impossible.  The best way is take a = 4, and b = 2. and substitute in the given equation. 0+ 2(8−2)363=22(8−2)363=2.  Now substitute a, b values in the given options and check where it is equal to 2.  Option a satisfies. If you like this shortcut, +1 this!!

7. The marked price of coat was 40% less than the suggested retail price. Eesha purchased the coat for half of the marked price at the 15th anniversary sale. What percent less than the suggested retail price did Eesha pay?
a) 60%
b) 20% 
c) 70% 
d) 30%
Answer:
Explanation:
Let the retail price = 100
So the market price will be = (100 – 40)% (100) = 60
Easha purchased price = 60/2 = 30
So she bought it for 70% less than retail price.

8. In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?
a. 200
b. 300
c. 450
d. 400
Answer: c
Explanation:
Let e = Number of engineering students, m = Number of MBA students and c = Number of CA students.
Given that,
4e + 3m + 5c = 3650- – – – (1)
3c = 2m , therefore c=2m3c=2m3
3e = 2c ⇒ e=2c3=23×2m3=4m9e=2c3=23×2m3=4m9
Substituting values of c and e in the given equation,
4×4m9+3m+5×2m3=36504×4m9+3m+5×2m3=3650
⇒16m+27m+30m9=3650⇒16m+27m+30m9=3650
⇒76m9=3650⇒76m9=3650
⇒m=450⇒m=450

9. A rectangle is divided into four rectangles with area 70, 36, 20, and x.  The value of x is

a. 350/9
b. 350/7
c. 350/11
d. 350/13
Answer: a
Explanation:
Areas are in proportion.
70x=3620⇒x=350970x=3620⇒x=3509

10. If a ladder is 10 m long and distance between bottom of ladder and wall is 6 m.  What is the maximum size of cube that can be placed between the ladder and wall.
a. 34.28
b. 24.28
c. 21.42
d. 28.56
Answer:
Explanation:
Here a = 6, and c = 10. b = 8 (∵∵ using Pythagorean theorem)

The maximum side of the square which can be inscribed in a right angle triangle = abca2+b2+ababca2+b2+ab (∵∵ see 7th questionhere for formula)
So side = 10×6×862+82+6×8=3.24310×6×862+82+6×8=3.243
Volume of the cube = 3.2433=34.0753.2433=34.075
Note:
The maximum side of a square is obtained when two sides of the square matches with a and b.  In this case side = ab/(a+b) = 3.428 which is higher than 3.243.

TCS latest Questions with Answers – 25


TCS latest Questions with Answers – 25


1. How many liters of a 90% of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid?

a. 3
b. 4
c. 6
d. 10

Answer: c
Explanation:
Let n1n1 liters of 90% concentration has to be mixed with n2n2 liters of 75% concentration to get 78% concentration solution. Using weighted average formula ax=n1×a1+n2×a2n1+n2ax=n1×a1+n2×a2n1+n2
78=n1×90+n2×75n1+n278=n1×90+n2×75n1+n2
n1n2=14n1n2=14
by dividing 30 in the ratio 1:4 we get 6 : 24. So we need 6 liters.

2. Find the ratio of the area of square to area of triangle. 

a. 1:2
b. 2:1
c. 2:3
d. 3:2

Answer: b
Explanation:
Have a look at the diagram below.

Let the side of the square = 2 units.
Now the area of the square = 22 = 4.
Area of the triangle = 12×2×2=212×2×2=2
Ratio = 4 : 2 = 2 : 1.

3. In this question A^B means A raised to the power B.  If f(x) = ax^4 – bx^2 + x + 5 and f(-3) = 2, then f(3) =

a. 1
b. – 2
c. 3
d. 8

Answer: d
Explanation:
f(-3) = a(-3)4 – b(-3)2 + (-3) + 5 = 81a – 9b + 2 = 2 So 81a – 9b = 0
f(3) = a(3)4 – b(3)2 + (3) + 5 = 81a – 9b + 8
Substituting the value of 81a – 9b = 0 in the above we get f(3) = 8



4. Of a set of 30 numbers, average of first 10 numbers = average of last 20 numbers.  Then the sum of the last 20 numbers is?

a. Cannot be determined.
b. 2 x sum of last ten numbers
c. 2 x sum of first ten numbers
d. sum of first ten numbers

Answer: c
Explanation:
We know that sum = average x number of observations.
Let the common average = x
Now sum of first 10 numbers = 10x
Sum of the last 20 numbers = 20x.
So sum of the last 20 numbers = 2 × sum of the first ten numbers.

5. A play school has chocolates which can supply 50 students for 30 days. For the first ten days only 20 students were present. How many more students can be accommodated into the earlier group such that the entire chocolates get consumed in 30 days.  Assume each student takes the same number of chocolates.

a. 45
b. 60
c. 55
d. 70

Answer: a
Explanation:
Let each students gets 1 chocolate. Now total chocolates = 50 x 30 = 1500
If first 10 days only 20 students were present, then total chocolates consumed = 10 × 20 = 200
Now we are left with 1500 – 200 = 1300 chololates. These were to be consumed in 20 days.
So each day 1300/20 = 65 chocolates were to be distributed.
So we can add 65 – 20 = 45 students.

6. In the town of Unevenville, it is a tradition to have the size of the front wheels of every cart different from that of the rear wheels.  They also have special units to measure cart wheels which is called uneve.  The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled by the cart in uneves, when the front wheel has done nine more revolutions than the rear wheel?

a. 570
b. 1330
c. 3990
d. 399

Answer: c
Explanation:
LCM of 133 and 190 is 1330. So to cover this distance, front wheel takes 10 rounds, and back wheel takes 7 rounds.
So for 3 rounds extra, 1330 uneves distance has to be travelled. To take 9 rounds extra, 1330 × 3 = 3990 uneves has to be traveled.

7. There are 20 persons sitting in a circle.  In that there are 18 men and 2 sisters.  How many arrangements are possible in which the two sisters are always separated by a man?

a. 18!x2
b. 17!
c. 17×2!
d. 12

Answer: a
Explanation:
Let the first sister name is A. Now she can sit any where in the 20 places (Symmetrical). Now her sister B can sit to her left or right in 2 ways. Now the remaining 18 persons can be sit in 18 places in 18! ways. Total = 18! × 2

8. A number plate can be formed with two alphabets followed by two digits, with no repetition.  Then how many possible combinations can we get?

a. 58500
b. 67600
c. 65000
d. 64320

Answer: a
Explanation:
Easy.  26 × 25 × 10 × 9 = 58500

9. A alone can do 1/4th of the work in 2 days.  B alone can do 2/3th of the work in 4 days.  If all the three work together, they can complete it in 3 days so what part of the work will be completed by C in 2 days? 

a. 1/12
b. 1/8
c. 1/16
d. 1/20

Answer: a
Explanation:
A can do the total work in 8 days, and B can do it in 6 days.
Let the total work be 24 units. Now capacities are
A = 24/8 = 3,
B = 24/6 = 4,
A + B + C = 24/3 = 8
So Capacity of C = 1 unit.
In two days C will do 2 units which is 2/24th part of the total work. So 1/12th part.

10. How many prime numbers are there which are less than 100 and greater than 3 such that they are of the following forms

  • 4x + 1
  • 5y – 1
a. 11
b. 12
c. 7
d. None of the above

Answer: d
Explanation:
Let the number be N.
So N = 4x + 1 = 5y – 1
⇒x=5y−24⇒x=5y−24
y = 2 satisfies the equation. So minimum number satisfies both the equations is 9 and general format of the numbers which satisfies the equation = k. LCM (4, 5) + 9 = 20k + 9.
Now by putting values 1, 2, 3 . . . . for k, we get 29, 49, 69, 89. Of which only 29, 89 are primes.

11. Babla alone can do a piece of work in 10 days. Ashu alone can do it in 15 days.  The total wages for the work is Rs.5000.  How much should be Babla be paid if they work together for an entire duration of work.

a. 2000
b. 4000
c. 5000
d. 3000

Answer: d
Explanation:
Money should always be divided in the inversely proportion way.  So Babla will get 1515+10×5000=30001515+10×5000=3000

12. The shopkeeper charged 12 rupees for a bunch of chocolate. but i bargained to shopkeeper and got two extra ones, and that made them cost one rupee for dozen less then first asking price.  How many chocolates I received in 12 rupees ?

a. 10
b. 16
c. 14
d. 18

Answer: b
Explanation:
Let the number of chocolates bought = n or n/12 dozens
Assume this would cost x rupees.
Now given that (n+2)/12 dozens cost x – 1 rupee.
So 12n/12=x12n/12=x – – – (1)
and 12(n+2)/12=x−112(n+2)/12=x−1 – – – (2)
(1) – (2)  = 144n−144n+2=1144n−144n+2=1
From the options, 16 satisfies.

TCS latest Questions with Answers – 24

TCS latest Questions with Answers – 24


1. How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?

a. 6480
b. 320
c. 2160
d. 720

Answer: d
Explanation:
If the we have to form even numbers, units digit must be 2, 4, 6. i.e., 3 ways.  Also 5th digit should be even. So it can be filled in 2 ways.  Now remaining 5 digits can be filled in 5! ways. So total 5! × 3 × 2 = 720 ways.

2. The five tyres of a car (four road tyres and one spare) were used equally in a journey of 40,000 kms.  The number of kms of use of each tyre was

a. 40000
b. 10000
c. 32000
d. 8000

Answer: c
Explanation:
Total kilometers travelled by 4 tyre = 40000 × 4 = 1,60,000.   This has to be share by 5 tyres. So each tyre capacity = 1,60,000 / 5 = 32, 000.  You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre.  But If the tyres are rotated properly after each 8000 km, all the tyres are equally used.

3. In a group of five families, every family is expected to have a certain number of children, such that the number of children forms an arithmetic progression with a common difference of one, starting with two children in the first family. Despite the objection of their parents, every child in a family has as many pets to look after as the number of offsprings in the family. What is the total number of pets in the entire group of five families.

a. 99
b. 9
c. 55
d. 90

Answer: d
Explanation:
As the number of children are in arithmetic progression starting with 2, the five families have 2, 3, 4, 5, 6 kids respectively. As each children has kept the pets equal to the number of kids in the family, each family has n2 pets. So total = 22+32+42 +52 +62 = 90

4. According to the stock policy of a company, each employee in the technical division is given 15 shares of the company and each employee in the recruitment division is given 10 shares.  Employees belonging to both committees get 25 shares each.  There are 20 employees in the company, and each one belongs to at least one division.  The cost of each share is $10.  If the technical division has 15 employees and the recruitment division has 10 employees, then what is the total cost of the shares given by the company?

a. 2650
b. 3180
c. 3250
d. 3120

Answer: c
Explanation:

We have to use addition formula n(A∪B)n(A∪B) = n(A)+n(B)−n(A∩B)n(A)+n(B)−n(A∩B)
20 = 15 + 10 – x
x = 5
So total shares given to only technical = 10 × 15 = 150
Shares given to only Recruitment = 5 × 10 = 50
Share given to Technical as well as recruitment people = 5 × 25 = 125
Total shares = 150 + 50 + 125 = 325.
Total value = 325 × 10 = 3250



5. The average marks of 3 students A, B and C is 60.  When another student D joins the group, the new average becomes 56 marks.  If another student E, who has 3 marks more than D, joins the group, the average of the 4 students B, C, D and E becomes 55 marks.  How many marks did A get in the exam?

a. 50
b. 54
c. 51
d. 53

Answer: c
Explanation:
Given that A + B + C = 60 × 3 = 180
A + B + C + D = 56 × 4 = 224
Therefore, D = 44
E = 44 + 3 = 47
Given, B + C + D + E = 55 × 4 = 220
B + C + 44 + 47 = 220
⇒ B + C = 220 – 91 = 129
So A + 129 = 180 ⇒ A = 51

6. What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct).  For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.

a. 441
b. 540
c. 84
d. 2100

Answer: b
3600 = 24 × 32 × 52
Let abc = 24 × 32 × 52
We have to distribute four 2’s to three numbers a, b, c in 4+3−1C3−1=6C24+3−1C3−1=6C2 = 15 ways.
Now two 3’s has to be distributed to three numbers in 2+3−1C3−1=4C22+3−1C3−1=4C2 = 6 ways
Now two 5’s has to be distributed to three numbers in 2+3−1C3−1=4C22+3−1C3−1=4C2 = 6 ways
Total ways = 15 × 6 × 6 = 540

7. There is a 7-digit telephone number with all different digits.  If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?

a. 120
b. 30240
c. None of these
d. 6720

Answer: d
Explanation:
If left and right digits are fixed with 5 and 6, then the remaining 5 places can be filled by remaining 8 digits in 8P58P5 = 6720 ways.

8. A certain sum of money is sufficient to pay either George’s wages for 15 days or Mark’s wages for 10 days.  For how long will it suffice if both George and Mark work together?

a. 8
b. 6
c. 9
d. 5

Answer: b
Explanation:
Let the money to be paid = 30 rupees. Then George daily wage = 30/15 = 2, and Mark daily wage = 30/10 = 3.
If both are working, then 5 rupees to be paid. So given sum is sufficient for 30 / 5 = 6 days.

9. The remainder when m + n is divided by 12 is 8, and the remainder when m – n is divided by 12 is 6.  If m > n, then what is the remainder when mn divided by 6?

a. 3
b. 4
c. 2
d. 1

Answer: b
Explanation:
m + n = 12a + 8 ⇒ (m+n)2=144a2+192a+64(m+n)2=144a2+192a+64 – – – (1)
m – n = 12b + 6 ⇒ (m−n)2=144b2+144b+36(m−n)2=144b2+144b+36 – – – (2)
(1) – (2) ⇒ 4mn = 144a2+192a−144b2−144b+28144a2+192a−144b2−144b+28
mn = 36a2+48a−36b2−36b+736a2+48a−36b2−36b+7
Now mn is divided by 6, all the terms except 7 gives 0. So 7 divided by 6, remainder = 1

10. There is a set of 36 distinct points on a plane with the following characteristics:
* There is a subset A consisting of fourteen collinear points.
* Any subset of three or more collinear points from the 36 are a subset of A.
How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)

a. 7140
b. 4774
c. 1540
d. 6776

Answer: d
Explanation:
The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22 (or) 2 points from 14 and 1 points from 22 (or) 3 points from 22
⇒ 14C1×22C2+14C2×22C1+22C314C1×22C2+14C2×22C1+22C3=  6776