TCS placement solved question papers

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[Raman Vij]

Here is the TCS placement solved question paper:

TCS placement solved question paper

1. Ray writes a two digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the number.
a) 35
b) 42
c) 49
d) 57
Solution: Let the two digit number be xy.
4(x + y) +3 = 10x + y .......(1)
10x + y + 18 = 10 y + x ....(2)
Solving 1st equation we get 2x - y = 1 .....(3)
Solving 2nd equation we get y - x = 2 .....(4)
Solving 3 and 4, we get x = 3 and y = 5

2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ?
a) Greater than 14
b) less than or equal to 11
c) 13
d) 12
In a calender,
Number of months having 28 days = 1
Number of months having 30 days = 4
Number of months having 31 days = 7
28 x 1 + 30 x 4 + 31 x 7 = 365
Here, a = 1, b = 4, c = 7.

a+b+c = 12

3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete?
a) 11.30 am
b) 12 noon
c) 12.30 pm
d) 1 pm
Let the total work = 120 units.
As George completes this entire work in 8 hours, his capacity is 15 units /hour
Similarly, the capacity of paul is 12 units / hour
the capacity of Hari is 10 units / hour
All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74
Remaining work = 120 - 74 = 46
Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx)

So work gets completed at 1 pm

4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181)
a) 02
b) 82
c) 42
d) 22



5. J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in How many days?
a) 32
b) 48
c) 96
d) 24
Assume the total work = 48 units.
Capacity fo J = 48 / 16 = 3 units / day
Capacity of P = 48 / 24 = 2 units / day
Capacity of J, P, H = 48 / 8 = 6 units / day
From the above capacity of H = 6 - 2 - 3 = 1
So H takes 48 / 1 days = 48 days to dig the well

6. If a lemon and apple together costs Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a lemon. What is the cost of lemon?
L + A = 12 ...(1)
T + L = 4 .....(2)
L + 8 = A
Taking 1 and 3, we get A = 10 and L = 2

7. 3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs.144. What is the combined price of 1 apple, 1 peach, and 1 mango.
a) 37
b) 39
c) 35
d) 36
Note: It is 114 not 144.
3m + 4a = 85 ..(1)
5a + 6p = 122 ..(2)
6m + 2p = 114 ..(3)
(1) x 2 => 6m + 8a = 170
(3) => 6m + 2p = 114
Solving we get 8a - 2p = 56 ..(4)
(2) => 5a + 6p = 122
3 x (4) = 24a - 6p = 168
Solving we get a = 10, p = 12, m = 15
So a + p + m = 37


9. There are 5 sweets - Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday. A person eats one sweet a day, based on the following constraints.
(i) Ladu not eaten on monday
(ii) If Jamun is eaten on Monday, Ladu should be eaten on friday.
(iii) Peda is eaten the day following the day of eating Jilebi
(iv) If Ladu eaten on tuesday, kaju should be eaten on monday

based on above, peda can be eaten on any day except
a) tuesday
b) monday
c) wednesday
d) friday

From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on monday.

10. If YWVSQ is 25 - 23 - 21 - 19 - 17, Then MKIGF
a) 13 - 11 - 8 - 7 - 6
b) 1 - 2-3-5-7
c) 9 - 8 - 7 - 6 - 5
d) 7 - 8 - 5 - 3
MKIGF = 13 - 11 - 9 - 7 - 6
Note: this is a dummy question. Dont answer these questions

11. Addition of 641 + 852 + 973 = 2456 is incorrect. What is the largest digit that can be changed to make the addition correct?
a) 5
b) 6
c) 4
d) 7

641
852
963
------
2466

largest among tens place is 7, so 7 should be replaced by 6 to get 2456

12. Value of a scooter depriciates in such a way that its value at the end of each year is 3/4th of its value at the beginning of the same year. If the initial value of scooter is 40,000, what is the value of the scooter at the end of 3 years.
a) 23125
b) 19000
c) 13435
d) 16875
value of the scooter at the end of the year = 40000×(34)3 = 16875

13. At the end of 1994, R was half as old as his grandmother. The sum of the years in which they were born is 3844. How old R was at the end of 1999
a) 48
b) 55
c) 49
d) 53
In 1994, Assume the ages of GM and R = 2k, k
then their birth years are 1994 - 2k, 1994 - k.
But given that sum of these years is 3844.
So 1994 - 2k + 1994 - k = 3844
K = 48
In 1999, the age of R is 48 + 5 = 53

14. When numbers are written in base b, we have 12 x 25 = 333, the value of b is?
a) 8
b) 6
c) None
d) 7
Let the base = b
So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3
2b2+9b+10=3b2+3b+3
b2−6b−7=0
Solving we get b = 7 or -1
So b = 7

15. How many polynomials of degree >=1 satisfy f(x2)=[f(x)]2=f(f(x)
a) more than 2
b) 2
c) 0
d) 1
Let f(x) = x2
f(x2)=[x2]2=x4
(f(x))2=[x2]2=x4
f(f(x))=f(x2)=[x2]2=x4

Only 1


17. In the question, A^B means, A raised to power B. If x*y^2*z < 0, then which one of the following statements must be true?
(i) xz < 0 (ii) z < 0 (iii) xyz < 0
a) (i) and (iii)
b) (iii) only
c) None
d) (i) only
As y^2 is always positive, x*y^2*z < 0 is possible only when xz < 0. Option d is correct.

18. The marked price of a coat was 40% less than the suggested retail price. Eesha purchased the coat for half the marked price at the fiftieth anniversary sale. What percentage less than the suggested retail price did Eesha pay?
a) 60
b) 20
c) 70
d) 30
Let the retail price is Rs.100. then market price is (100-40) % of 100 = 60. Eesha purchased the coat for half of this price. ie., 30 only. which is 70 less than the retail price. So Option C is correct.

2. A circle has 29 points arranged in a clock wise manner from o to 28. A bug moves clockwise manner from 0 to 28. A bug moves clockwise on the circle according to following rule. If it is at a point i on the circle, it moves clockwise in 1 sec by (1 + r) places, where r is the remainder (possibly 0) when i is divided by 11. If it starts in 23rd position, at what position will it be after 2012 sec.
Ans: After 1st second, it moves 1 + (23/11)r = 1 + 1 = 2, So 25th position
After 2nd second, it moves 1 + 25/11 = 1 + 3 = 4, So 29th position = 0
After 3rd second, it moves 1 + 0/11 = 1 + 0 = 1, So 1st position
After 4th second, it moves 1 + 1 = 3rd position
after 5th, 1 + 3/11 = 4 So 7th
After 6th, 1 + 7/11 = 8 so 15th
After 7th, 1 + 15/11 = 5 so 20th
After 8th, 1 + 20/11 = 10th, So 30th = 1st
So it is on 1st after every 3 + 5n seconds. So it is on 1st position after 2008 seconds (3 + 5 x 401) So on 20th after 2012 position.

3. In a city 100% votes are registered, in which 60% vote for congress and 40% vote for BJP. There is a person A, who gets 75% of congress votes and 8% of BJP votes. How many votes got by A?
Assume total votes are 100. So A got
75% of 60 = 45
8% of 40 = 3.2
A total of 48.2 %

4. Mean of 3 numbers is 10 more than the least of the numbers and 15 less than greatest of the 3. If the median of 3 numbers is 5, Find the sum of the 3 numbers?
Ans: Median is when the given numbers are arranged in ascending order, the middle one. Let the numbers are x, 5, y where x is the least and y is greatest.
Given that x+5+y3=x+10
and x+5+y3=y−15
Solving we get x = 0 and y = 25.
So sum of the numbers = 0 + 5 + 25 = 30

5. A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm.
Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am.
B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km

6. In a particular year, the month of january had exactly 4 thursdays, and 4 sundays. On which day of the week did january 1st occur in the year.
a) monday
b) tuesday
c) wednesday
d) thursday
Ans: If a month has 31 days, and it starts with sunday, Then Sundays, Mondays, tuesdays are 5 for that month. If this month starts with monday, then mondays, tuesdays, and wednesdays are 5 and remaining days are 4 each. so this month start with Monday.

7. A, E, F, and G ran a race.
A said "I did not finish 1st /4th
E said "I did not finish 4th"
F said "I finished 1st"
G said "I finished 4th"
If there were no ties and exactly 3 children told the truth, when who finishes 4th?
a) A
b) E
c) F
d) G
Ans: Option D


9. In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. Secretary takes the top letter and types it. Boss delivers in the order 1, 2, 3, 4, 5 which cannot be the order in which secretary types?
a) 2, 4, 3, 5, 1
b) 4, 5, 2, 3, 1
c) 3, 2, 4, 1, 5
d) 1, 2, 3, 4, 5
Ans: Option B

10. At 12.00 hours, J starts to walk from his house at 6 kmph. At 13.30, P follows him from J's house on his bicycle at 8 kmph. When will J be 3 km behind P?
By the time P starts J is 1.5 hr x 6 = 9 km away from his house.
J is 3 km behind when P is 3 km ahead of him. ie., P has to cover 12 km. So he takes 12 / (8 - 6) = 6 hrs after 13.30. So the required time is 19.30Hrs

11. J is faster than P. J and P each walk 24 km. Sum of the speeds of J and P is 7 kmph. Sum of time taken by them is 14 hours. Then J speed is equal to
a) 7 kmph
b) 3 kmph
c) 5 kmph
d) 4 kmph
Given J > P
J + P = 7, only options are (6, 1), (5, 2), (4, 3)
From the given options, If J = 4 the P = 3. Times taken by them = 244+243=14

12. In a G6 summit held at london. A french, a german, an italian, a british, a spanish, a polish diplomat represent their respective countries.
(i) Polish sits immediately next to british
(ii) German sits immediately next to italian, British or both
(iii) French does not sit immediately next to italian
(iv) If spanish sits immediately next to polish, spanish does not sit immediately next to Italian
Which of the following does not violate the stated conditions?
a) FPBISG
b) FGIPBS
c) FGISPB
d) FSPBGI
e) FBGSIP
Ans: Option D

13. Raj drives slowly along the perimeter of a rectangular park at 24 kmph and completes one full round in 4 min. If the ratio of length to bredth of the park is 3 : 2, what are the dimansions?
a) 450 m x 300 m
b) 150 m x 100 m
c) 480 m x 320 m
d) 100 m x 100 m
24 kmph = 24×100060=400 m / min
In 4 minutes he covered 4 x 400 = 1600 m
This is equal to the perimeter 2 ( l + b) = 1600
But l : b = 3:2
Let l = 3k, b = 2k
Substituting, we get 2 ( 3k + 2k ) = 1600 => k = 180
So dimensions are 480 x 320

14. M is 30% of Q, Q is 20% of P and N is 50% of P. What is M / N
ans: Take P = 100, then N = 50, Q = 20, M = 6. So M/N = 3/25

15. At what time between 6 and 7 are the hands of the clock coincide?
Ans. Total = 3600
For hour = 360/12 = 300/hr
For Minute = full rotation = 3600/hr
Let the line is 't' , for 6 = 6*30=1800
then
30 t + 180=360 t
330t = 180
t = 180/330
t = 6/11 hr 6/11*60=360/11=32611
Ans. is 6:32

16. Series 1, 4, 2, 8, 6, 24, 22, 88 ?
Sol : The given series is in the format: x 4, -2, x4, -2, x4, -2, x4....
1x4 = 4
4-2=2
8-2=6
6x4=24
24-2=22
22x4=88
88-2=86
Ans: 86

17. 4 Women & 6 men have to be seated in a row given that no two women can sit together. How many different arrangements are there.
Sol : Let us first sit all the 6 men in 6 positions in 6! ways. Now there are 7 gaps between them in which 4 women can sit in 7P4 ways.
So total ways are 6! x 7P4

18. xy+yx=46 Find x & y values ?
Sol: 145+451=46
Hence x = 1, y = 45

19. In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B the present age of B is
Soln: A +10=2(B-10) ........(1)
A =B + 9 ......... (2)
from equations. 1 & 2
we get B = 39 A will be 39+9=48 years old.

20. A student can select one of 6 different math book, one of 3 different chemistry book & one of 4 different science book.In how many different ways students can select book of math, chemistry & science.
Sol: 6C1×3C1×4C1 = 6x3x4=72 ways

21. Sum of two number is 50 & sum of three reciprocal is 1/12 so find these two numbers
Sol : x+y = 50 .....(1) x=50-y ....(2)
1x+1y=112 ⇒y+xxy=112⇒12(y+x)=xy ...(3)
put (2) in (4)
⇒ 12(y+50-y)=(50-y)y
⇒ 12y+600-12y=50y-y2
⇒ y2-50y+600=0
⇒ y2-30y-20y+600=0
⇒ y(y-30)-20(y-30)=0
⇒ (y-20) (y-30)=0
y=20 or y=30
if y=20 then x = 30
or y=30 then x = 20
two numbers are 30 & 20

1. Dinalal divides his property among his four sons after donating Rs.20,000 and 10% of his remaining property. The amounts received by the last three sons are in arithmetic progression and the amount received by the fourth son is equal to the total amount donated. The first son receives as his share RS.20,000 more than the share of the second son. The last son received RS.1 lakh less than the eldest son. 10. Find the share of the third son.
a) Rs.80,000
b) Rs.1,00,000
c) Rs.1,20,000
d) Rs.1,50,000
Ans: Assume the amounts received by the 2nd, 3rd, and 4th sons are a+d, a, a-d (as they are in AP)
Now Eldest son received Rs.20,000 more than the 2nd son. So He gets a+d+20,000
Last son received 1 lakh less than the eldest son. So (a+d+20,000) - (a-d) = 1,00,000 ⇒ 2d = 80,000 ⇒ d = 40,000
So Amounts received by the 4 sons are a + 60,000, a+40,000, a, a - 40,000
It was given that the youngest son's share is equal to 20,000 + 12(His property)
Assume His property = K rupees.
Then 20,000 + 12(K) = a - 40,000 ...........(1)
and the Remaining property = Sum of the properties received by all the four son's together.
Remaining property = 910(K-20,000)
⇒910(K-20,000) = ( a + 60,000 ) + (a+40,000) + a +( a - 40,000) ..(2)
Solving We get K = 40,000 and a = 1,20,000
So third son got Rs.1,20,000

In a quadratic equation, (whose coefficients are not necessarily real) the constant term is not 0. The cube of the sum of the squares of its roots is equal to the square of the sum of the cubes of its roots. Which of the following is true?
a)Both roots are real
b) Neither of the roots is real
c) At least one root is non-real
d)At least one root is real
Ans: Assume the given quadratic equation is ax2+bx+c=0 whose roots are p, q.
Now given that (α2+β2)3=(α3+β3)2
By expanding we get, α6+3.α4.β2+3.α2.β4+β6=α6+β6+2.α3.β3
3.α2.β2(α2+β2)=2.α3.β3
3.(α2+β2)=2.α.β
3.(α2+β2)+6.α.β−6.α.β=2.α.β
3.(α+β)2=8.α.β ...(1)
We know that sum of the roots = α+β=−ba
product of the roots = α.β=ca
Substituting in the equation (1) we get 3.(−ba)2=8.ca⇒3.b2=8.a.c

The nature of the roots can be determined by finding the magnitude of the determinant = b2−4ac
But we know that ac = 3b28
So b2−4ac = b2−4.3b28=−b88<0
So the roots are imaginary.

3. A man sold 12 candies in 10$ had loss of b% then again sold 12 candies at 12$ had profit of b% find the value of b.
Ans: Here 12 candies is immaterial.
Loss % = CP−SPCP×100
So Here SP = 10 and loss% = b%

CP−10CP×100=b⇒CP−10CP=b100

In the second case he got a profit of b%

So Profit % = SP−CPCP×100

So Here SP = 12 and profit% = b%

12−CPCP×100=b⇒12−CPCP=b100

Solving 1 and 2 we get b = 1/11 or 9.09%

4. find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?
Ans: 1 letter can be chosen in 2 ways. a or b
2 letters can be chosen in 3 way. aa, ab, bb
3 letters can be chosen in 3 ways. bbb, aab, bba
4 letters can be chosen in 2 ways. aabb, bbba
5 letters can be chosen in 1 way.
So total ways are 11

5. what is the sum of all the 4 digit numbers that can be formed using all of the digits 2,3,5 and 7?
Ans: use formula (n-1)! x (111..n times) x (Sum of the digits)
here n is number of different letters
So answer is 3 ! x 1111 x 17

6. 30^72^87 divided by 11 gives remainder
Ans: Fermat little theorem says, ap−1p remainder is 1.
ie., 3010 or 810when divided by 11 remainder is 1.
The unit digit of 7287 is 8 (using cyclicity of unit digits) Click here
So 7287 = 10K + 8
30(10K+8)11=(3010)K.30811=1k.30811
8811=22411=(25)4.2411=1611=5

7. 1234567891011121314151617181920......424344 what is remainder when divided by 45?
Ans: Let N = 1234567891011121314151617181920......424344
Remainder when N is divided by 5 is 4. So N = 5K + 4 .....(1)
Remainder when N is divided by 9 is Sum of the digits of N divided by 9. We know that 1+2+3+...44 = 990 Which gives digit sum as 9. So remainder when N is divided by 9 is 0.
So N = 9L .....(2)
Equation (1) and (2) we 9L = 5K + 4
For K = 1 this equation gets satisfied. So least possible number satisfies the condition is 9
So The general format of N = w(LCM of (9, 5)) + Least number satisfies the condition.
So N = w.45 + 9
When N is divided by 45, we get 9 as remainder.

1. The wages of 24 men and 16 women amounts to Rs.11600 per day. Half the number of men and 37 women earn the same amount per day. What is the daily wage of a man?
Let the wage of a man is m and woman be w.
24m+16w=11600
12m+37w = 11600
Solving we get m = 350

2. The sum of three digits a number is 17. The sum of square of the digits is 109. If we substract 495 from the number, the number is reversed. Find the number.
Let the number be abc.
Then a + b + c= 17 .....(1)
a2+b2+c2=109 .....(2)
100a+10b+c -495 = 100c+10b+a ......(3)
From 3, we get a - c = 5
So the possibilities for (a, c, b) are (6,1,10), (7,2,8), (8,3,6), (9,4,4)
From the above, (8,3,6) satisfies the condition.

3. A calculator has a key for squaring and another key for inverting. So if x is the displayed number, then pressing the square key will replace x by x^2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternatively presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be
Evern number of inverse key has no effect on the number.
By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of 2n. So after the 16 pressings the power becomes 216
So the final number will be 6216=665536

4. How many two digit numbers are there which when substracted from the number formed by reversing it's digits as well as when added to the number formed by reversing its digits, result in a perfect square.
Let the number xy = 10x + y
Given that, 10x+y - (10y - x) = 9(x-y) is a perfect square
So x-y can be 1, 4, 9. -------- (1)
So given that 10x+y +(10y +x) = 11(x+y) is a perfect square.
So x+y be 11. Possible options are (9,2), (8,3),(7,4),(6,5) ---------(2)
From the above two conditions only (6,5) satisfies the condition
Only 1 number 56 satisfies.

5. Find the 55th word of SHUVANK in dictionary
Sol: Arranging the letters in alphabetical order we get : A H K N S U V
Now Total words start with A are 6!
Total words start with AH are 5! = 120
Now
Total words start with AHK are 4! = 24
Total words start with AHN are 4! = 24
Total words start with AHSK are 3! = 6
Now AHSNKUV will be the last word required.

6. Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph
Relative speed = 60 - 40 = 20 kmph
Initial gap as car B leaves after 2 hours = 40 x 2 = 80 kms
Car B should be 9 km ahead of the A at a required time so it must be 89 km away
Time = 89 / 20 = 4.45 hrs or 267 mins

7. Find the average of the terms in the series 1-2+3-4+5....+199-200
Sol1-2) +(3-4) + (5-6) +........(199-200) = -100
Average = 100 / 200 = -0.5

8. n is a natural number and n^3 has 16 factors. Then how many factors can n^4 have?
Total factors of a number N=ap.bq.cr... is (p+1)(q+1)(r+1)...
As n3 has 16 factors n3 can be one of the two formats given below
n3 =a15
n3 = a3.b3
If n3 =a15 then n = a5 and number of factors of n4 = 21
n3 = a3.b3 then n = ab and number of factors n4 = 25

9. Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged?
Let the total time taken by the cars be a and b
Let the time after which the speed is interchanged be t
For car A, 60t+90(a-t) = 420, 90a - 30t = 420 .......(1)
For car B, 90t + 60(b-t) = 420, 60b + 30t = 420 ....(2)
Using both (1) and (2), we get 90a + 60b = 840
But as a - b =1, 90a + 60(a-1) = 840.
Solving a = 6.
Substituting in equation 1, we get t = 4