**TCS sample and model placement paper questions for learn and practice 2013-2014 new batch asked questions some users contributions from different campus recruitments all over India **

**1) ** The wages of 24 men and 16 women amounts to Rs.11600/day. Half the number of men and 37 women earn the same amount /day. What is the daily wage of a man

Explanations

Let the wage of a man be m and woman be w

24m+16W=11600

12m+37w=11600

24m*37-12m*16=11600*(37-16)

12m(2*37-16)=11600*(37-16)

Solving m=11600*21/(12*58) =200*7/4=**350**

**2) ** The sum of three digits in a number is 17 The sum of square of the digits is 109 .If we subtract 495 from the number, the number is reversed. Find the number

Explanations

Let the digits abc

a+b+c=17-Eqn1

a^2+b^2+c^2 =109 -Eqn -2

Also 100a+10b+c-495=100c+10b+a

99(a-c)=495 Thus a-c =5 The possible combinations are (6,1)(7,2)(8,3)(9,4)

Case 1: (6,1)

From Eqn 1,b+17-7=10 which not possible as b is a single digit

Case 2:(7,2)

From eqn 1 b=17-9=8

Verifying with eqn 2,49+4+64=117and hence not valid

Case 3: (8,3)

From eqn 1,b=17-11=6

Verifying with eqn 2,64+9+36 = 109 and hence valid

**Thus the number is 863**

**3) ** Find the last two0 digits of (102^3921)+(3081^3921)

**Ans 02**

To find out the last 2 digits of a term where the number ends with 1, multiply the tens digit of the number with the last digit of the exponent This will give you the tens digit Unit digit will always be 1

Applying this shortcut

for 102^3921,last 2 digits is (2^1) 1=21

for 3081^3921,last 2 digits is (8^1)=81

Adding both we get 102

Hence last two digits of the expression is** 02**

**4) ** To find out the last 2 digits of a term where the number ends with 1, multiply the tens digit of the number with the last digit of the exponent This will give you the tens digit Unit digit will always be 1

Applying this shortcut

for 102^3921,last 2 digits is (2^1) 1=21

for 3081^3921,last 2 digits is (8^1)=81

Adding both we get 102

Hence last two digits of the expression is 02

**4** A series is given as 1,6,7,13,20,33,…….and so on Find the sum of first 52 terms

1st and 2nd terms are 1,6

3rd term is 7 which is sum of first+second

4the term is 13 which is sum of second +thirds

Expanding the series further by adding a few more terms it is 1,6,7,13,20,33,53,86,139,and so on

Now if we add the terms for upto 1,2,3,4,5,6 terms it is

1,2,14,27,47,80,133,219 and so on

**5) **

**Ans 26.25 sq units**

The Radius of the circle R circumscribing a triangle is given by the formula

R-ABC /(4*Area of Triangle)

Area of Triangle – 1/2*base*height – 1/2*BC*Ad-1.5BC (As Ad -3)

A,B,C are the side which are 17.5 9 and BC respectively

Hence radius R -17 .5*9*BC/(4*1.5BC) -17.5*9/6-26.25 squints

**6)** 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 round off or overflow errors) will be

Ever n 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 2^{n}. So after the 16 pressings the power becomes 2^{16}

So the final number will be 6^{216}=6^{65536}

**7) ** How many two digit numbers are there which when subtracted 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.

**8) ** 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.

**9)** 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

**10)** 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=a^{p}.b^{q}.c^{r..}. is (p+1)(q+1)(r+1)…

As n3 has 16 factors n3 can be one of the two formats given below

n^{3} =a^{15}

n3 = a^{3}.b^{3}

If n^{3} =a^{15} then n = a^{5} and number of factors of n^{4} = 21

n^{3} = a^{3}.b^{3} then n = ab and number of factors n^{4} = 25

**11)**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 traveling 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