A big list of TCS campus papers with Explainations

This post is for all freshers who are going to sit in the TCS campus drive. You may be knowing that TCS sets the questions from the old question papers. So here is the big list of all the questions asked in TCS campus drive.

This post is for all freshers who are going to sit in the TCS campus drive. You may be knowing that TCS sets the questions from the old question papers. So here is the big list of all the questions asked in TCS campus drive.
SET 1
1. 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, \${x_1} + {x_2} + {x_3}......{x_r} = n\$ each person is eligible to take any number of articles then the total ways are \${}^{n + r - 1}{C_{r - 1}}\$
In this case \${x_1} + {x_2} + {x_3}......{x_6} = 10\$
in such a case the formula for non negative integral solutions is \${}^{n + r - 1}{C_{r - 1}}\$
Here n =6 and r=10. So total ways are \${}^{10 + 6 - 1}{C_{6 - 1}}\$ = 3003
2. 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)
3. 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
4. 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
5. 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 times b times h\$ = \$6 times 5 times 2\$ = 60 \$c{m^3}\$
Now volume is reduced by 19%.
Therefore, new volume = \$displaystylefrac{{(100 - 19)}}{{100}} times 60 = 48.6\$
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: \$left( {displaystylefrac{x}{{100}} times 6} right)left( {displaystylefrac{x}{{100}} times 5} right)2 = 48.6\$
Solving we get x =90
thus length and width is reduced by 10%
New width = 5-(10% of 5)=4.5
6. 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.
7. 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
8. 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.
9. 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 \$displaystylefrac{{n(n + 1)}}{2} times 10 le 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”.
10. 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
 Tata Consultancy service campus drive
SET 2
1. In a staircase, there ar 10 steps. A child is attempting to climb the staircase. Each time she can either make 1 step or 2 steps. In how many different ways can she climb the staricase?
a) 10
b) 21
c) 36
d) None of these
Ans: d
Use fibonnacci series, with starting two terms as 1, 2. So next terms are 3, 5, 8, 13, 21, 34, 55, 89
2. A boy buys 18 sharpners, (Brown/white) for Rs.100. For every white sharpener, he pays one rupee more than the brown sharpener. What is the cost of white sharpener and how much did he buy?
a) 5, 13
b) 5, 10
c) 6, 10
d) None of these
Ans: C
Assume that he bought b, brown sharpeners and w, white sharpeners and the cost of brown sharpener is x and white sharpener is x + 1
So w(x+1) + bx = 100
w + b = 18
b = 18 - w
Substituting in equation 1, we get w(x+1) + (18 -w)x = 100 so w + 18 x = 100
Take option 1: If white sharpners are 13, x = (100 - 13) /18 = 4.833
Option 2, If white sharpeners are 10, x = (100 - 10)/18 = 5 So white sharpeners cost is 6.
Option 3 Satisfies this condition.
3. Letters of alphabets no from 1 to 26 are consecutively with 1 assigned to A and 26 to Z. By 27th letter we mean A, 28th B. In general 26m+n, m and n negative intezers is same as the letters numbered n.
Let P = 6, strange country military general sends this secret message according ot the following codification scheme. In codifying a sentence, the 1st time a letter occurs it is replaced by the pth letter from it. 2nd time if occurred it is replaced by P^2 letter from it. 3rd time it occurred it is replaced by p^3 letter from it. What is the code word for ABBATIAL
a) GHNNZOOR
b) GHKJZOHR
c) GHHGZOGR
d) GHLKZOIR
Ans: D
A should be coded as 1+6 = G (it occurred for first time)
B should be coded as 2+6 = H (it occurred for first time)
B Should be coded as 2 + 36 = 38 - 26 = 12 = L (it occurred for second time)
Option D is correct
4. Of a set of 30 numbers, average of 1st 10 numbers is equal to average of last 20 numbers. The sum of last 20 numbers is?
a) 2 x sum of last 10 numbers
b) 2 x sum of 1st 10 numbers
c) sum of 1st 10 numbers
d) Cannot be determined
Ans: B
Let average of first 10 numbers is a. Then sum = 10a
Average of last 10 nmbers also a. Then their sum = 20a
From the options B correct
5. In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.
a) 1565
b) 2256
c) 2456
d) 1243
Ans: B
Maximum 3 men can be played which means there can be 0, 1, 2, 3 men in the team.
(5C0×11C11)+(5C1×11C10)+(5C2×11C9)+(5C3×11C8)=2256
6. The wages of 24 men and 16 women amount to 11600 per day. Half the number of men and 37 women has same money. The daily wages paid to each man is
a) 375
b) 400
c) 350
d) 325
Ans: C
24m + 16w = 11600
12m + 37 w = 11600
Solving we get 12 m = 21w
Substituting in the first equation we get, 42w + 16 w = 11600 ⇒ w = 200
M = 350
7. A number when successively divided by 5, 3, 2 gives remainder 0, 2, 1 respectively in that order. What will be the remainder when the same number is divided successively by 2, 3, 5 in that order
a) 4, 3, 2
b) 1, 0,4
c) 2, 1, 3
d) 4, 1, 2
Ans: B
use this simple technique.[ (1 x 3) + 2] = 5
[(5 x 5) + 0] = 25
Procedure:
Let the number = N
Now N = 5K
K = 3L + 2
L = 2M + 1
K = 3(2M + 1) + 2 = 6M + 5
N = 5(6M + 5) = 30 M + 25
For M = 0 we get the least number as 25. Now when 25 is divided by 2, we get 12 as quotient and 1 as remainder. When 12 is divided by 3 we get 4 as quotient, and 0 as remainder. When 4 is divided by 5 we gt 4 as remainder.
8. a,b,c,d,e are distinct numbers. if (75-a)(75-b)(75-c)(75-d)(75-e)=2299 then a+b+c+d= ?
Hint:2299 is divisible by 11.
2299 = 11×11×19×1×1=11×−11×19×−1×1=
Two of the terms in the given expression should equal to 1. As all the digits are distinct, two of the terms should be negative.
One possible solution = (75 - 64)(75 - 56)(75 - 86)(75 - 74)(75 - 76)
Then a + b + c + d + e = 64 + 56 + 86 + 74 + 76 = 356
But as the sum of only 4 terms was asked, we have to subtract one term.
So given answer can be one of 292, 306, 270, 282, 280
9. If A ^B means A raised to the power of B, in which of the following choices must P be greater than Q
a) 0.9^P=0.9^Q
b) 0.9^P=0.92^Q
c) 0.9^P>0.9^q
Option A is wrong as P = Q
Option B is wrong as PQ=Log0.92Log0.9=0.79139
Option C is also wrong as aP>aQ then P>Q if a > 1
10. 2 gears one with 12 teeth and other one with 14 teeth are engaged with each other. One teeth in smaller and one tooth in bigger are marked and initially those 2 marked teeth are in contact with each other. After how many rotations of the smaller gear with the marked teeth in the other gear will again come into contact for the first time?
a)7
b) 12
c) Data insufficient
d) 84
Correct Option : A
Assume the distance between the teeth is 1 cm. Then the circumference of first gear is 12 cm and the second is 14 cm.
Now LCM (12, 14) = 84. So to cover 84 cm, the first gear has to rotate 8412 = 7 rounds (the second gear rotates 84 / 14 = 6 rounds as it is bigger)
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MindxMaster: A big list of TCS campus papers with Explainations
A big list of TCS campus papers with Explainations
This post is for all freshers who are going to sit in the TCS campus drive. You may be knowing that TCS sets the questions from the old question papers. So here is the big list of all the questions asked in TCS campus drive.
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MindxMaster
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