IBM Placement Paper 2014 (2)

This set of Placement paper is collected from on-campus recruitment drive for the session (2013-2014). Try to practice each question from this set as IBM repeats its questions. You will surely get some 4 to 5 questions common in your Online Test. If you are in the lucky side, even you can expect more common questions. We are posting the exact questions that are asked in IBM on-campus recruitment.

Here are some questions from number series.

1)  2, 1, 2, 3, 8, 0, 1, 9, 1, 2, ?

Ans. 1

Explanation: 2¹=2
2³=8
8⁰=1
1⁹=1
1²=1.
 
2) 7, 9, 11, 6, 11, 8, 5, 13, 5, 4, 15, ?

Ans. 2

Explanation: Divide the series in set of 3
1st set-  7  9 11
2nd set- 6 11 8
3rd set-  5 13 5
4th set-  4 15 ?
1st column- 7,6,5,4  (Subtract  1)
2st column- 9,11,13,15 ( Add 2 )
3st column- 11,8,5,? (Subtract 3)
Therefore, the missing term is 5 - 3 = 2


3) 2/3, 5/3, 19/6, 31/6, 23/3, 32/3, 84/6, ?

Ans. 108/6

Explanation: 2/3+1=5/3
19/6+2=31/6
23/3+3=32/3
84/6+4=108/6

4)  9, 12, 4, 8, 12, 15, 5, 9, 3, 6, 2, 6, 6, 9, ?

Ans. 3

Explanation: Divide the series into set of 4.
1st set-  9,12,4,8
2nd set- 12,15,5,9
3rd set- 3,6,2,6
4th set- 6,9, ?
Rules applied in each set
9+3=12, 12/3=4, 4+4=8
12+3=15, 15/3=5, 5+4=9
3+3=6, 6/3=2, 2+4=6
Therefore, 6+3=9, 9/3=3

5) 2, 3, 10, 15,26, ?

Ans.35

Explanation: 2*2-1=3
3*3+1=10
4*4-1=15
5*5+1=26
6*6-1=35

6) 507,169,248,62,36,12,168,42,168, ?

Ans. 56

Explanation: 507/3=169
248/4=62
36/3=12
168/4=42
168/3= 56

7)  4, 3, 144, 1.5, 2, 9, 5, 0.3, 2.25, 3.2, 2,?

Ans. 40.96

Explanation: 4 * 3= 12²= 144
1.5 * 2 =3²=9
5 * 0.3 = 1.5 ² = 2.25
Therefore, 3.2*2=6.4²=40.96

8) 49, 7, 98, 16, 4, 48, 9, 36, 25, 5,?

Ans.125

Explanation: 7²=49,49*2=98
4²=16,16*3=48
3²=9 ,9*4=36
5²=25,25*5=125

9) 96, 64, 128,192, 128, 256, 288, 192, ?

Ans.384

Explanation: Divide the series in set of 3.
1st set- 96,64,128
2nd set-192,128,256
3rd set- 288,192,?

Take 1st column-96 * 2=192, 96*3=288
2nd column- 64 * 2 =128, 64*3= 192
Therefore, 3rd column =128*2= 256, 128* 3= 384

10) 11 ,8 ,17 ,14 ,23 ,24 ,31 ,32,?

Ans. 39

Explanation: First divide the series
11 8 17- Difference bet. 2nd and 1st term is 3 and 3rd and 2nd term is 9
17 14 23 - Difference bet. 2nd and 1st term is 3 and 3rd and 2nd term is 9
23 24 31-Difference bet. 2nd and 1st term is  and 3rd and 2nd term is 7
Therefore, applying same rules we get 31 32 39

11) 3, 2, 1, 4, 5, 4, 5, 1, 0, 6, 3, ?

Ans. 2

Explanation:3+2-1=4
4+5-4=5
5+1-0=6
6+3-x=7 Therefore, x=2

12) 3, 7, 10, 8, 4, 12, 0, 5, 5, 3, 2, 5,?

Ans.3

Explanation: Divide the series in set of 3
1st set- 1,9,0  (1+9+0=10 )
2nd set- 2,7,1 (2+7+1=10)
3rd set- 3,5,2  (3+5+2=10)
4th set  4,3,?   Therefore, 4+3+x =10 , x=3.

13) 49, 7, 98, 16, 4, 48, 9, 3, 36, 25, 5, ?

Ans. 125

Explanation: Divide the series in set of 3
1st set- 49,7,98     logic-7*7 = 49 , 49*2=98
2nd set- 16,4,48   logic- 4*4=16 ,16*3=48
3rd set- 9,3,36      logic-3*3 = 9 , 9*4=36
4th set- 25,5,?      Therefore, 5*5=25 and 25*5=125


14) 5, 6, 6, 5, 6.5, 8, 7, 9.5, ?

Ans. 7.5

Explanation: There sub series inside the series
Take the alternate numbers to make the series
1st series- 5, 6.5, 8, 9.5 (Add  1.5)
2nd series-6, 6.5,7, ?(Add 0.5)

Therefore, the next term is 7+.5= 7.5

15)  8,16,3,27,4,8,7,19,2,0,7, ?

Ans. 9

Explanation: Divide the series in a set of 4
1st set-8,16,3,27   Logic-8+16+3=27
2nd set- 4,8,7,19   Logic-4+8+7=19
3rd set- 2,0,7,?      Therefore, 2+0+7=9


Soon, we are coming up with more IBM latest placement papers. 


We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT WITH US.

On Friday, February 21, 2014

Continue Reading

IBM Placement paper Set-1

This set of questions are collected from IBM on-campus recruitment (2013-14). In this set of questions we are only covering the general aptitude part. We will be posting the number series questions soon.

Round 1: Online Aptitude Test
No. of Question:  36     (18 questions on number series + 18 Aptitude question)
Time: 81 minute  (2:15 minute for each question)
Time allocated to each question: 2:15 minute

Here are some questions that are asked in aptitude test.

1. A software company has 630 employees of whom 80 percent are software engineers and 20 percent are sales representatives. To obtain a 25 percent sales representative proportion, how many additional employees should be added to the sales staff?

A. 157
B. 125
C. 31
D. 42
E. 33

Ans. 42

Explanation: Total no. of employee = 630
No. of Software engineers = 80 % of 630 = 504
No. of sales staffs = 20% of 630 = 126
Now, we have to make the percent sales staff to 25 % .
Let X no of sales staff will be added to make it 25 % of the total employees.
Therefore, (126+ x)/(630+x)=25/100
504 + 4x= 630 +x
3x= 126
x=42

2. A company brought twice as many small envelopes as large envelopes. It used up 5/15 of the small envelopes and 4/6 of the large envelopes. What fraction of the total number was left unused?

A. 4/9
B. 1/3
C. 5/9
D. 4/3
E. 2/3

Ans. 5/9

Explanation: Let the no. of large envelopes = x
Therefore no. of small envelopes = 2x.
No. of unused small envelopes = (1- 5/15)*2x = (10/15)* 2x=(4/3)x.
No of unused large envelopes = (1-4/6)*x = (2/6)*x = (1/3)x.
Total fraction of unused envelopes is = (((1/3)x) + ((4/3)x) /3x ) = 5/9

3. A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective?

A. 4/19
B. 7/19
C. 12/19
D. 21/95

Ans. 7/19

Explanation: Probability that none is defective out of two bulbs = (¹⁶c₂ / ²⁰c₂)=12/19.
Therefore the prob. that at least one is  defective is = 1-(12/19)=7/19

4. In 1978, a kg of paper was sold at Rs25/-. If the paper rate increases at 1.5% more than inflation rate which is of 6.5% a year, then what will be the cost of a kg of paper after 2 years?

A) 29.12
B. 29.72
C. 30.12
D. 32.65
E. none of these

Ans. 29.72

Explanation: Inflation Rate = 6.5%
Paper rate is 1.5% more than 6.5% = 1.5% +6.5% =8%
Therefore , the cost of paper after 2 year = 25 (1 +  8/100)²=25(1.08)²=29.16

5. On her first week on the job, a work station operator processed a batch of documents. On her second week, she processed 2 times as many, and on her third week she processed 15200 documents. If she processed a total of 34100 documents in the 3 weeks, how many documents did she process in her first week.

A. 4200
B. 10133
C. 5066
D. 9450
E. 6300

Ans. 6300

Explanation: Let the work station operator processed X documents in first week.
Then it processed 2x document in second week.
Third week it processed 15200 documents. (Given)
Therefore , X + 2X + 15200 = 34100
3X =18900
X= 6300

6. In a company, 60% are men and 40% are women.  Out of which 80% of men and 60% of women are in are senior citizen. How much percent of employees are senior citizen?

Ans. 72%

Explanation: 60% are men, of which 80% are senior citizen.
Therefore, percent of men senior citizen = (60/100)*(80/100)=48/100
Now 40% are women, of which 60% are senior citizen.
Therefore, percent of women senior citizen = (40/10)*(60/10)=24/100
Total percent  of senior citizen is (48/100) +(24/100)= (72/100)=72%

7. A television manufacture makes 7 different model sets in 5 screen sizes and 2 different model sets in 3 screen sizes. How many television units are required for a store to display each type of TV?

A. 39
B. 35
C. 13
D. 41
E. 9

Explanation: For the first  set - 7 x 5 =35
For the second set – 2 x 3 =6
Therefore 35 + 6 = 41

 8. Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?

Ans. 671

Explanation: Total no. of outcomes when 4 dices are thrown = 6⁴ =1296
Total no. of outcomes, where no dice result is 5 = 5⁴ = 625
Therefore, total no. of outcomes in which at least one dice result is 6 = 1296-635= 671.

In our upcoming post, we will cover the number series questions from recent IBM placement papers. Soon we are coming up with huge collection of latest Placement papers and Interview Experience.
   

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT WITH US.

On Tuesday, February 18, 2014

Continue Reading

TCS Latest Placement Paper 2014 Set-7

This set of placement paper  is collected from on-campus placement drive for the 2014 batch. TCS will follow the same pattern for the off–campus drive of 2014 batch.


1. There are 3 trucks A,B and C. A loads at the rate of 10kg/min and B loads at the rate of 13 1/3 kg/min. C unloads at the rate of 5kg/min. If all the 3 trucks are acting simultaneously, find the time taken to load 2.4 tonnes.

Ans.  130.91 min

2. The price of a book in four different shops and the successive discounts offered for the books is given below. Select the option in which the price of the book is the least.
(a) 10%, 5%, and 5% discount on Rs.195
(c) 12.5% and 12.5% discounts on Rs.205
(b) 25%, discount on Rs.200
(d) 10%,  and  15%  discounts  on  a  marked  price of Rs.190

Ans. D

3. When all possible six-letter arrangements of the letters of the word “MASTER” are sorted in alphabetical order, what will be the 49 th word?

Ans. AREMST

Explanation: First by arranging the given word in alphabetical order we get                                A,E,M,R,S,T
There are 24 words starts with AE
There are 24 words starts with AM
So the 49th word will be AREMST

4. A workman starts his work on Monday works for 8 days and takes every 9 th day as his holiday. His 12 th holiday will fall on?

Ans. Wednesday

5. Initial price of the scooter is 40000 and it reduces to 3/4th of the previous price every year. What will be the price after 3 years?

Ans. 16875

Explanation: Therefore price of scooter in 3 years = 40000*(3/4)*(3/4)*(3/4) =16875

6. A starts riding his bike at 10am with a speed of 20kmph and B also starts at 10am with a speed of 40 kmph from the same point in the same direction. A turns south at 12 o’clock and B turns north at 11 am. What will be the distance between A and B at 2 pm?

Ans 160 kmph

7. If there are six periods in each working day of a school. In how many ways can one set up the time table for a day such that each subject is allowed at least one period?

Ans. 3600

Explanation: The five subjects can be done in 5! ways. The remaining 1 period can be any of the 5 subjects and it can come in at any of the 6 different periods. So 5 * 6 = 30 ways.
The total ways is 5! * 30 = 3600 ways.

8. 1!+2!+3!...+50! when divided by 5!, the remainder is?

Ans. 33

Explanation: 5! is 120 and all numbers from 5! to 50! are divisible by 5!. We have to check for the first 4 numbers i.e, from 1! to 4!. The addition is 1+2+6+24 = 33. Therefore, the remainder is 33.

9. In month of 31 days, there are exactly 4 Thursdays and 4 Sundays. What is the day of  the week on the first of that month?

Ans. Monday

Explanation: If Thursdays and Sundays occur 4 times, then the days between them – Friday and Saturday – also will  occur 4 times. The remaining days Mon, Tue and Wed will occur 5 times. Hence the month starts on  Monday.

10. My name is PREET. But my son accidentally types the by interchanging a pair of letters in my name. What  is the probability that despite this interchange, the name remains unchanged?

Ans. 10%

Explanation: Using 5 letters one can form 10 combinations and the only way that the name remains unchanged is when both E’s are getting interchanged. That is one of 10, which is 10%.

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT.  

On Monday, February 10, 2014

Continue Reading

TCS sample placement paper uploaded by TCS

TCS had uploaded a sample paper in their Portal to give you the idea of type of question will be asked and to become familiar with their User Interface.I am providing the sample question with their solution. You will clearly get an idea about the difficulty level.

1. If ‘m’ is an odd integer and ‘n’ an even integer, which of the following is definitely odd?

Ans. m+n

Explanation: You just remember the following odd    odd = even; even    even = even; even    odd = odd. Also odd x odd = odd; even x even = even; even x odd = even.

2. If 3y + x > 2 and x + 2y  3, What can be said about the value of y?

Ans. y>-1

Explanation: Multiply the second equation with -1 then it will become - x - 2y >= - 3.  Add the equations.  You
will get y > -1.

3. 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?

Ans. 5

Explanation: Given R + B + G = 20; G = 7; and R + G < 13.  Substituting G = 7 in the last equation, We get R < 6.  So maximum value of R = 5.

4. 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?

Ans: 48

Explanation: If there are n cubes lie on an edge, then total number of cubes with one side painting is given by 6 X  (n-2)²  edge. Hence answer = 24.

5. 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 in the 15 the lap,at the same spot where they began to cycle?

Ans. 165

Explanation: L.C.M for 4  & 6 is 12 so first meet is 12.and 15 th lap is 15 * 12 =165

6. Arun, Akash, Amir and Aswanth go for a picnic.   When Arun stands on a weighing machine, Akash also climbs  on,  and  the  weight  shown  was  132  kg.   When  Akash  stands,  Amir  also climbs  on,  and  the machine shows 130 kg.  Similarly the weight of Amir and Aswanth is found as 102 kg and that of Akash and Aswanth is 116 kg.  What is Aswanth’s weight?

Ans. 44 kg

Explanation: 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.

7. 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?

Ans. 48

Explanation: R  =4+ E
R =2 + I
R+2 = 2(E+2 )
Solving all the above equations we get
R= 6,E= 2,I =4 so after 2 yrs it is R=8 and I =6,which is 48.

8.The telephone company wants to add an area code composed of 2 letters to every phone number. In order  to  do  so,  the  company  chose  a  special  sign  language  containing  124  different  signs.  If  the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?

Ans: 492

Explanation: The phone company already created 122*122 area codes, now it can create 124*124. 1242-1222 =(124+122)(124-122) = 246*2 = 492 additional codes.

9. A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday?

Ans. 4

Explanation: 4 rolls were not sold.Because half of them were sold by noon.So 20 rolls remain, and then 20% were not sold. So 20% of the 20 rolls is 4.

10. A necklace is made by stringing N individual beads together in the repeating pattern red bead, green bead, white bead, blue bead and yellow bead. If the necklace begins with a red bead and ends with a white bead, then N could be:

Ans: 68

Explanation: R G W B Y is the bead pattern and it repeats.
Bead want to end with White.
So, the 3rd, 8th, 13th, 18th... beads will be W.
this can be expressed as 5n+3,where n is an integer.(counting no of white bead)
Test each of the answer choices to determine which is multiple of 5 plus a value of 3.Of the options ,only 68=5(13)+3 can be written in the form 5n+3.
So,the answer is 68.

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT.   

On Monday, February 10, 2014

Continue Reading

TCS Latest Placement Paper 2014 Set-5

This year TCS had changed their question pattern and have added new questions to their database. There are few questions that are being repeated from the previous year's paper. The type of question this year are more difficult compared to previous year. Try to give more time and solve more questions from permutation, combination, probability and Geometry. Many students have said that they have got 15 to 16 question from probability, permutation and combination and other set of student got same no. of question from geometry.

1. 3 dies are thrown. What is the probability that sum ten appears?

Ans. 27/216

Explanation: When 3 dice are thrown, total number of possibilities are n(s) = 6^3 = 216

Sum is 10 : (1,3,6),(1,4,5),(1,5,4),(1,6,3)

(2,2,6),(2,3,5),(2,4,4),(2,5,3),(2,6,2)

(3,1,6),(3,2,5),(3,3,4),(3,4,3),(3,5,2),(3,1,6)

(4,1,5),(4,2,4),(4,3,3),(4,4,2),(4,5,1)

(5,1,4),(5,2,3),(5,3,2),(5,4,1)

(6,1,3),(6,2,2),(6,3,1)

Probability of getting a sum of ten is 27/216 = 1/8

2. In an examination 80% of student passed in mathematics, 55% in English, 29% failed in both subject. Find the percent of student passed in both subject ?

Ans. 64%

Explanation: Passed in math = 80%

Passed in English = 55

Failed in both = 29%

Total percent of student passed in atleast one subject =100 -29 = 71%

Total no pass student in both Subject = 80+55-71 = 64%

3. What is the next number of the following sequence.

18,24,5,21,27,8,24,30,...

Ans 11

Explanation: There are there series

18,21,24  ------------ +3

24,27,20 -------------+3

5,8,....     -------------+3

8+3=11.

4.  A man spends 1/3rd of his salary on food, 1/4th on rent and 1/5th on cloths. If he is left with 1800, then who much salary does he earn.

Ans. 3692

Explanation: Let total Salary = x

1/3 x+ 1/4 x+ 1/5 x =47 /60 x.

Remaining salary = x- 47/60 x= 13/60 x

13/60 x = 1800

x=3692 (approx)

5. For a car there are 5 tires including one spare tire. All tires are equally used. If the total distance travelled by the car is 40000 km then what is the average distance travelled by the each tire?

Ans. 32000

Explanation:  Total distance travelled by the car=4000 0km

Total distance travelled by 4 wheels=4*40000=160000

as all tires are equally used

So distance travelled by the each tire=160000/5=32000

6. If A speaks the truth 60% of the times, B speaks the truth 50% of the times. What is the probability that at least one will tell the truth ?

Ans. 8/10

Explanation: The prob. that both of them telling the lie = (40/100* 50/100)=2/10

                       Therefore prob. that atleast one will tell the truth = 1- 2/10 = 8/10= .8

7. If the letters in the word ACTORS are permuted in all possible ways and arranged in alphabetical order.Find the 40th word in permuted alphabetical order.

Ans. AOSRTC

Explanation: Arranging in alphabetical order as ACORST.

No of words forming with  A=5!=120

No of words forming with AC=4!=24

No of words forming with AOC=3!=6

No of words forming with AOR=3!=6(36 words).

AOSCRT = 37th word

AOSCTR =38th word

AOSRCT =39th word

AOSRTC =40th word

8. What is the probability that a particular selected leap year has 53 Sundays?

Ans. 2/7

Explanation: A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.

The remaining 2 days may be any of the following :

Sunday and Monday

Monday and Tuesday

Tuesday and Wednesday

Wednesday and Thursday

Thursday and Friday

Friday and Saturday

Saturday and Sunday

For getting 53 Sundays in a year, one of the remaining 2 days must be a Sunday.

n(S) = 7

n(E) = 2

P(E) = n(E) / n(S) = 2 / 7

9. How many number plates with 1,2,3,4,5 digits as numbers without repetition.

Ans. 325

Explanation: No of number plates with 1 digit = 5

 No of number plates with 2 digit=5*4=20

 No of number plates with 3 digit=5*4*3=60

 No of number plates with 4 digit=5*4*3*2=120

 No of number plates with 5 digit=5*4*3*2*1=120

Therefore total no of plate=120+120+60+20+5=325

10. What is the sum of 1-2+3-4+5.......-98+99 ?

Ans. 50

Explanation: 1-2+3-4+5.......-98+99

=(1-2)+(3-4)+(5-6)+....+(97-98)+99

=(-1-1-1....49 times)+99

=-49+99

=50

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT.

On Monday, February 10, 2014

Continue Reading

TCS Latest Placement paper 2014 Set-3

These questions are from TCS new test pattern for 2014 batch. There are few repetitions from the previous year's question paper .Here are some of the TCS questions and solution with Explanation.

 1. P is thirty percentage of Q, Q  is twenty percentage of N. M is fifty percentage of N, Find the value of P/N ?

Ans. .06

Explanation: P = .3Q

Q = .2N

M = .5N

P = (.3*.2)N

P/N = .06

2. If twenty four  men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve  men and thirty seven women  work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?

Ans. 350

Explanation: Since the wages paid are equal, total work done by both the groups should also be equal.

Equating the total work done in terms of men days and women days.

24m+16w=12m+37w  -  (1)

 12m=21w (or) 4m=7w

Substituting w=4m/7 in eqn 1 we get

24m + 16w = 24w + 16(4m/7)

= (168m+64m)/7

= The total amount paid for 232m/7 = 11600

For each men = (11600*7)/232=350.

3. A takes 2 hours to make a publication. B takes 10 hours to make a publication. Find the time

taken by them to make two publications, working independently?

Ans. 10

Explanation: A can complete a publication in 2 hours and B in 10 hours and so the maximum time taken by both working independently to complete 2 publications will be 10 hours.

4. If all the numbers between 11 and 100 are written on a piece of paper. How many times will

the number 4 be used?

Ans. 19

Explanation: 14,24,34,44,54,64,74,84,94,40,41,42,43,44,45,46,47,48,49

Therefor there are 19 times the number 4 can be used between 11-100.

5. In a school , sixty percent of the students are girls and thirty five percent of the girl s are poor. If a student is randomly selected, what is the probability of selecting a poor girl student?

Ans. 21%

Explanation: Let the number of the students be X.

Therefore poor girl students = .35*.65*X

The probability of selecting poor girl is given by

= (.35*.65)*X

(100/100)*X =21%

6. Two beakers are on the table. The capacity of the first beaker is x litres and that of the second beaker is 2x litres. Two thirds of the first beaker and one fourth of the second beaker is filled with  wine. The remaining  space  is  filled  with  water.  If  the  content  in  both  the  beakers  are mixed in a large beaker of volume 3x litres, what is the proportion of wine in the beaker?

Ans. 7/6

7.Three non negative numbers, X, Y and Z are such that the mean is M and the median is 5. If M is 10 more than the smallest number and 15 less than the biggest number, find the va lue of X+Y+Z.

Ans. 30

Explanation: (X+Y+Z)3 = M (or) X +Y +Z +3M

Let Y be the middle value, then Y=5

X+Z=3M-5

X=M-10;

Z=M+15;

M-10+M+15=3M-5

M=10

X=0; Y=5; Z=25

8. From  5  men  and 11  women,  in  how  many ways  can  a  panel  of  11  be  formed  such that  the number of men is not more than 3?

Ans: 2266

Explanation: (5C3*11C8)+(5C2*11C9)+(5C1*11C10)+(11C11) = 2266

9. After 6 years Raju’s father will be twice that of his age and two years ago, his mothers age was twice of that of Raju’s age. What is the sum of Raju’s parent’s age?

Ans. 4 more than four times Raju’s age

Explanation: F+6=2(R+6)

F= 2R+6

M-2=2(R-2)

M= 2R-2

Therefore the sum of Raju’s Parent’s age is

F+M=2R+6+2R-2

F+M=4R+4

 4 more than four times Raju’s age

10. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?

Ans. 240000

Explanation: C+H=300000  -----eqution 1

1.20C+.90H=304200  --------equation 2

On equating 1 and 2 we get

C= Rs. 240000 and H= Rs.60000

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT.

 

On Monday, February 10, 2014

Continue Reading

TCS Latest Placement paper 2014 Set-1

This set of questions are asked at on-Campus placement drive for the session 2013-14. There are few questions that are repeated from the previous year's paper (2012-13) as told by many students. We are able to get some questions from them.

1. Ahmed, Babu,  Chitra,  David  and  Eesha  each  choose  a  large  different  number.  Ahmed says, “ My number is not the largest and not the smallest”. Babu says, “My number is not the largest and not the smallest”. Chitra says, “My number is the largest”. David says, “My number is the smallest”. Eesha says, “ My number is not the smallest”.

Exactly  one  of  the  five  children  is  lying.  The  others  are  telling  the  truth.  Who  has  the largest number?

Ans. Eesha

Explanation: Ahmed and Babu cannot lie because each of them say two facts (not the largest,Not the smallest) and there is no chance for both the facts to be wrong. David says “My number is smallest”. If David lies, one of the remaining four should lie. But  exactly one  person  lies  in  this  problem.  So  David  says  the  truth.If  David’s statement is true, Eesha’s statement is also true. The one who lies is Chitra and Eesha has the largest number.

2. A cow and a horse are bought for Rs.200000. The cow  is sold at a profit of 20% and the horse is sold at a loss of 10%. The overall gain is Rs.4000. The cost price of the cow is

Ans. 80,000

Explanation: Let the cost price of cow and horse is C and H Respectively

C + H = 200000  -  (1)

1.2C + .9H = 204000  -  (2)

Solving equation (1) & (2)

C = 80000.

3. If X^Y denotes X raised to the power Y, Find the last two digits of ( 1941 ^ 3843 ) + ( 1961^4181).

Ans.  82

Explanation: 1941^2 ends in 81. 1941^3 ends in 21, 1941^4 ends in 61, 1941^5 ends in 01 and 1941^6 ends in 41 and this cycle keeps repeating. Similarly the cycle for 1961 powers is 61, 21, 81, 41, 01 and the cycle repeats. After adding up the final two digits of these numbers for their respective powers, we find that the sum is 82.

4. George can do some work in 8 hours, Paul can do the same work in 10  hours while Hari can  do the  same  work  in  12  hours. All the  three  of them  start  working  at  9  a.m  while George stops work  at 11 a.m and remaining two complete the work. Approximately at what time will the work be finished?

Ans. 1 pm

Explanation: Total number of work to be done= 120 Units (LCM of 8,10,12)

George’s one hour work = 120/8 = 14 Units

Paul’s one hour work = 120/10 = 12 Units

Hari’s one hour work = 120/12 = 10 Units

Units of work finished at 11 AM = (14+12+10)*2 = 74

Remaining work to be done = 120-74 = 46 units

One hour Paul + Hari work = 22 units

Approximately they will take two hours to finish the work

So the work will get finished at 1 PM

5. If M is 30% of Q, Q is 20% of P and N is 50% of P then M/N =

Ans. 3/25

Explanation: M is 30% of Q

Q is 20% of P

Nis 50 % of P

Then M/N = ?

Let P= 100

N = 50

Q = 20

M = 6

M/N = 6/ 50 =3/25.

6. In a office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on the top of the pile in the secretary’s inbox. When there is time, the secretary takes the top letter off the pile and type’s it. If there are five letter inall , and the boss delivers in the order of 1 2 3 4 5, which of the following could NOT be the order in which secretary types them.

Ans. 4 5 2 3 1

Explanation: Going by the options and checking logically which order is  possible.

7. There  are  5  sweets  –  Jumun,  Kulfi,  Peda,  Laddu  and  Jilabi  that   I  wish  to  eat  on  5 consecutive days –  Monday through Friday, one sweet a day, based on the following self imposed constraints:

1)  Laddu is not eaten on Monday

2)  If Jamun is eaten on Monday, then Laddu must be eaten on Friday

3)  If Laddu is eaten on Tuesday, Kulfi should be eaten on Monday

4)  Peda is eaten the day following the day of eating Jilabi

 Based on the above, peda can be eaten on any day except?

Ans. Monday

Explanation: Peda  can  be  had  only  after  having  Jilabi.  So  Peda  can  never  be  had  on  the starting day, which is Monday.

8. At 12.00 hours Jake starts to walk from his house at 6 kms an hour. At 13.30 hours, Paul follows him from Jake’s  house on his bicycle at 8 kms per hour. When will Jake be 3 kms behind Paul?

Ans. 19:30 hours

Explanation: Jake starts at 12.00 and covers 6 km/h.

Paul starts at 1.30 and covers 8 km/h.

Relative speed between Jake & paul is 2 kmph, where Paul stating Jake is 9 km ahead  of Paul.  From  13.30  hours  paul  takes  4.30  hrs  to  meet  Jake.  Again  he  needs  1.30  hrs  to  lead  Jake  by  3  km  Relative  speed.  Totally he  takes  6  hrs.  so 13.30+6 = 19.30 hrs.

9. Jake can dig a well in 16 days. Paul can dig the same well  in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in ?

Ans. 48 days.

Explanation: Given that speed of Jake is greater than Paul.

Distance = 24 km

Sum of their speed is 7 km/h = J+P

So possible speed ratio between J & P is

Go by Option

6:1 Not in option

5:2 = (24/5)+(24/2) ≠ 14 Hours

4:3 = (24/4)+(24/3) = 14 Hours

So Jake’s speed is 4 km/h.

10. If  a lemon and an apple together cost Rs. 12.00, a tomato and a lemon  cost Rs. 4.00 and an apple cost Rs.8.00 more than a tomato or a lemon then which of the following can be Page  2/4 the price of a lemon?

Ans. 2

Explanation: Let cost of a Lemon is L

Let cost of a Apple is A

Let cost of a Tomato is T

L+A = 12 – (1)

T+L = 4  -  (2)

A = 8+L  -  (3)

A = 8+T  -  (4)

Sub (3) in (1)

L+8+L = 12

L = 2, A = 10, T = 2.

We launched our Facebook Page.  JUST HIT LIKE BUTTON TO CONNECT.  
 

On Monday, February 10, 2014

Continue Reading