TCS Placement Papers 2016-2017 With Solutions. TCS Off Campus Placement Papers 2016/2017. TCS Previous Year Placement Papers. Sample TCS Placement Papers with Soltions. TCS Placement Papers PDF.

1. The value of diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2:3. what is the loss percentage in the value?

**Sol:**Let weight be “x”

the cost of diamond in the original state is proportional to x2

when it is fallen it breaks into two pieces 2y and the 3y

x = 5y

Original value of diamond = (5y)2 = 25y2

Value of diamond after breakage = (2y)2+(3y)2=13y2

so the percentage loss will be = 25y2−13y225y2×100=48%

2. Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth?

a)David

b)Querishi

c)Chitra

d)Thara

**Sol:**As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.

Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.

3. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.

a) 2676

b) 2

c) 445

d) 86

**SOL:**This is a number series problem nothing to do with the data given.

1x 1+1=2

2 x 2+2=6

6 x 3+3=21

21 x 4+4=88 and not 86

88 x 5+5 = 445

445*6+6 = 2676

4. The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?

a) AOTDSP

b) AOTPDS

c) AOTDPS

d) AOSTPD

**SOL:**

In alphabetical order : A D O P S T

A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.

A D _ _ _ _ : empty places can be filled in 4!=24

A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.

A O D _ _ _ : 3!= 6

A O P _ _ _ : 3!=6

Till this 36 words are obtained, we need the 42nd word.

AOS _ _ _ : 3!= 6

Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.

So given word is AOSTPD

4. A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?

**SOL:**This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.

6C114C1×8C113C1=2491

5. There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?

Sol: At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.

case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21

case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21

case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21

Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)

= 312/16807

6. Total number of 4 digit number do not having the digit 3 or 6.

**Sol:**

consider 4 digits _ _ _ _

1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)

2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)

3st blank can be filled in 8C1 ways

4st blank can be filled in 8C1 ways

Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584

7. Find the missing in the series: 70, 54, 45, 41,____.

**Sol:**40

70-54 = 16 = 42

54-45 = 9 = 32

45-41 = 4 = 22

41-40 = 1 = 12

8. A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed?

**Sol:**We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.

Then rooms needed 12024+19224+14424 = 5 +8 + 6 = 19

9. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day)

a)7

b)4

c)30

d)37

**SOL:**

let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,................

finally the no. of roses left behind will be 4.

10. What is the 32nd word of "WAITING" in a dictionary?

**Sol:**Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W

Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways

so can't be arranged starting with A alone as it is asking for 32nd word so it is out of range

AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.

AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24

AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways

so,24+12 =36th word so out of range. So we should not consider all the words start with AGN

now AGNI_ _ _can be arranged in 3! ways =6 ways

so 24+6=30 within range

Now only two word left so, arrange in alphabetical order.

AGNTIIW - 31st word

AGNTIWI - 32nd word

11. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many “different” boxes of chocolates can be made?

**Sol:**

If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1

In this case x1+x2+x3......x6=10

in such a case the formula for non negative integral solutions is n+r−1Cr−1

Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003

12. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.

a. 1/3

b. 1/2

c. 5/9

d. 17/36

**Sol:**Their sum can be 3,4,6,8,9,12

For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.

Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.

So probability is (20/36)=(5/9)

13. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?

a. 2000

b. 4000

c. 5000

d. 3000

**Sol:**

Time taken by A and B is in the ratio of = 3:2

Ratio of the Work = 2 : 3 (since, time and work are inversely proportional)

Total money is divided in the ratio of 2 : 3 and B gets Rs.3000

14. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero.

a. 10

b. 11

c. 13

d. 12

**Sol:**

Let x ques were correct. Therefore, (26- x) were wrong

8x−5(26−x)=0

Solving we get x=10

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15. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,

a. 5.5

b. 4.5

c. 7.5

d. 6.5

**Sol:**

Volume =l×b×h = 6×5×2 = 60 cm3

Now volume is reduced by 19%.

Therefore, new volume = (100−19)100×60=48.6

Now, thickness remains same and let length and breadth be reduced to x%

so, new volume: (x100×6)(x100×5)2=48.6

Solving we get x =90

thus length and width is reduced by 10%

New width = 5-(10% of 5)=4.5

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

**Sol:**We have to consider the number of 4's in two digit numbers. _ _

If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)

So total 19 ways.

Alternatively:

There are total 9 4's in 14, 24, 34...,94

& total 10 4's in 40,41,42....49

thus, 9+10=19.

17. 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?

**Sol:**Let man daily wages and woman daily wages be M and W respectively

24M+16W=11600

12M+37W=11600

solving the above equations gives M=350 and W=200

18. 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?

**Sol:**

Profit = 4200

Profit =SP - CP

4200=SP - 300000 therefore SP=304200

x+y = 300000

1.2x + 0.9y = 304200

Solving for x = 114000 = CP of cow.

19. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4......

In the above sequence what is the number of the position 2888 of the sequence.

a) 1

b) 4

c) 3

d) 2

**Sol:**First if we count 1223334444. they are 10

In the next term they are 20

Next they are 30 and so on

So Using n(n+1)2×10≤2888

For n = 23 we get LHS as 2760. Remaining terms 128.

Now in the 24th term, we have 24 1's, and next 48 terms are 2's. So next 72 terms are 3's.

The 2888 term will be “3”.

20. How many 4-digit numbers contain no.2?

**Sol:**Total number of four digit numbers =9000 (i.e 1000 to 9999 )

We try to find the number of numbers not having digit 2 in them.

Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken

Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832

Total number of numbers having digit 2 in it = 9000-5832 =3168

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