TCS Placement Paper | MCQ 2
This is a TCS model placement paper for aptitude preparation. This placement paper will cover aptitude questions that are asked in TCS recruitment drives and also strictly follows the pattern of questions asked in TCS interviews. It is recommended to solve each one of the following questions to increase your chances of clearing the TCS interview.
1. There is a set of 30 numbers. The average of the first 10 numbers is equal to the average of last 20 numbers. What is the sum of last 20 numbers?
a) Twice the sum of the first ten numbers
b) Sum of first 10 numbers.
c) Twice the sum of the last ten numbers
d) Cannot be determined.
Answer: a) Twice the sum of the first ten numbers
Let the sum of the first 10 numbers is equal to ‘x’
Let the sum of the last 20 numbers is equal to ‘y’
According to the question:
x/10 = y/20
Therefore, y = 2x
2. There is a town called Metron, where the wheels of the front and rear of vehicles are of different sizes. The measurement unit followed in the town is the meter. The circumference of the front wheel of the car is 133 meters and that of rear wheels is 190 meters. So what is the distance travelled by the cart in meters when the front wheel has done nine more revolutions than the rear wheel?
Answer: d) 3990
At first, we calculate the LCM of 133 and 190 which is 1330. So, the front wheels take 10 rounds to cover 1330 meters and the rear wheels take 7 rounds to cover the same.
So, for 3 rounds extra, 1330 m distance has to be travelled.
To take 9 rounds extra, 1330 * 3 = 3990 m has to be traveled.
3. Let a number ‘x’ when divided by 406 leaves a remainder 115. What will be the number when the number is divided by 29?
According to the question, the number is equal to 406 * y +115
(406 * y) + 115 = Number
(406 * y) is divisible by 29.
but when 115 is divided by 29 leaves 28 as remainder.
So, 28 is the remainder, which is the answer.
4. A sequence of an alpha-numeric is to be formed. The sequence consisting of two alphabets followed by two numbers is to be formed with no repetitions. In how many ways can it be formed?
Answer: c) 58500
Total Number Of Alphabets = 26 (A -Z)
1st Alphabet can be selected from 26 available Alphabets
2nd Alphabet can be Selected from 25 ( as one Alphabet is already used and repetition is not allowed)
Numeric digits are 10 ( 0 to 9)
1st digit can be selected from 10 available numbers
2nd digit can be selected from 9 available numbers ( as one number is already used and repetition is not allowed)
So total Possible Combination = 26*25*10*9
= 58500 as the answer
5. According to a particular code language, A=0, B=1, C=2, …, Y=24, Z=25 then how can ONE+ONE (in the form of alphabets only) be coded?
Answer: c) BDAI
This is a 26 base question. Just like there is the Decimal system consisting of 10 digits from 0 to 9, the Base 26 system consist of 26 alphabets where A = 0, B = 1, Z = 25 and so on.
Let’s calculate, O N E + O N E
=> E + E
=> 4 + 4
=> 13 + 13
On converting 26 to Base 26 we get 1 0. Keeping 0(A) and taking 1 as carry
=> O + O + 1
Dividing 29 by 26 we get 1(B) 3(D)
So answer is BDAI
6. In the sequence of problemsolvingproblemsolvingproblemsolving… what is the 2015th alphabet?
Answer: d) n
‘problemsolving’ consist of 14 letters. On dividing 2015 by 14 we get 13. So the 13th letter is n and hence the answer.
7. What is the remainder when the number 101102103104105106107…148149150 is divided by 9?
The divisibility rule for 9 is that the sum of all digits of a number should be divisible by 9. Let’s calculate the sum of the digits:
There are 50 1’s (unit place) = 50
There are 10 1’s (tens place) = 10
There are 10 2’s (tens place) = 20
There are 10 3’s (tens place) = 30
There are 10 4’s (tens place) = 40
There is one 5 (tens place) = 5
For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225
=> So sum of all digits = 380
=> 380 / 9 = 2 (Answer)
8. 30 liters of 78% of a concentrated acid solution is to be prepared. How many litres of 90% concentrated acid needs to be mixed with 75% solution of concentrated acid to get the result?
Answer: b) 6
Let’s apply the weighted-average formula.
Let there be n1 litre of 90% acid solution and n2 litre of 75% solution
=> 78 = ((90 * n1) + (75 * n2))/(n1 + n2)
=> n1/n2 = 1 / 4
So 30 litres needed to be divided in the ratio of 1:4, which gives us 6 litre as the answer.
9. Ram will be 32 years old in eight years from now. In 4 years, Ram’s fathers age will be twic as Ram’s age and two years ago, his mother’s age will be twice as his age. What will be the present age of Ram’s father and mother?
Answer: Father's age = 52, Mother's age = 46
Ram will be 32 years old in next 8 years. So his present age is 32 – 8 = 24 years old
After 4 years Ram will be 28 years old. So his father will be 28 * 2 = 56 years old.
Therefore, fathers present age is 56 – 4 = 52 years old
Two years ago Ram was 22 years old. So his mothers age the was 22 * 2 = 44 years old
Therefore mothers present age is 44 + 2 = 46 years old.
10. In a class, the number of boys is equal to the number of girls. What was the total number of students if twice the number of boys as girls remain when 12 girls entered out?
Let ‘b’ be the number of boys and ‘g’ be the number of girls. According to the question:
=> b / (g – 12) = 2 / 1
Since b = g;
we get g = 24.
So the total number of students = 24 + 24 = 48