TCS Placement Questions & Answers :: TCS

Answer: Option A

As there are 4 full weeks i.e 28 days.
So, every day occurs min 4 times.

Then remaining 3 days (as jan has 31 days) will be Monday Tuesday Wednesday. so on 31st Jan comes Wednesday.
So, 1st Jan will be MONDAY

Answer: Option C

Total Initial Salary : 17*45000 = 765000
Average Salary After removal of 3 Teachers = 45000-2500 = 42500
Total Final Salary : 14*42500 = 595000
Sum of Salaries of 3 teachers who left : 765000 - 595000 = 170000

Answer: Option A

Lets assume the two digit number be 10x+y
Adding 2 after doubling the number is reverse of the number.
So 2(10x+y)+2 = 10y+x
19x -8y +2=0 ----- (i)
The above equation (i) satisfies for the value, x=2, y=5
So, the number = 10*2+5= 25

Answer: Option B

Product of 5 terms equal to 1127. As all the five terms are integers, given product should be a product of 5 numbers. Now factorize 1127.
1127 = 72 * 23 = 7 * 7 * 23
But given that all the a, b, c, d, e are distinct. And we are getting only 3 terms with 7 repeats.
Now the logic is, integers means positive and negative, 7 and - 7 possible and 1, - 1 also possible . As a,b, c, d, e are in ascending order, the factors should be in decreasing order. So (23, 7, 1, -1, -7)
Now a = 53; b = 69; c = 75; d = 77
a + b + c + d = 274.

Answer: Option C

1+3+5+7+9+11+13.........+199 i.e 100 terms and
sum of n odd natural numbers is n^2 i.e 100^2 =10,000

-(2+4+6+8+10+12...........+200) i.e 100 terms and
sum of n even natural numbers is n(n+1) i.e -(100(100+1))= -10,100
average of first 200 terms = (10,000+(-10,100))/200
=-100/200
=-1/2 Ans.

Answer: Option B

43 time the no. And 34 time the reverse no.
43(10x+ y)=34(10y+ x)
430x+ 43y= 340y+34x
430x+43y-340y-34x=0
To solve this equation become
4x-3y=0 -------------- (1)

Sum of no. And reverse no. Is 14
X+ Y= 14 ------------- (2)
To solve both equation value of x and y
X= 6 and y= 8
So, number will be 68

Answer: Option D

Here is no explanation for this answer

Answer: Option C

A's 1 day work = \(\frac{1}{14}\)
B' 1 day work = \(\frac{1}{21}\)

A left 3 day before the completion of work.
B's 3 days' work = \(\frac{1}{21} \times 3 = \frac{1}{7}\)
Remaining work = \(1 \times \frac{1}{7} = \frac{6}{7}\)

(A + B)'s 1 day's work = \(\left(\frac{1}{14} + \frac{1}{21}\right) = \frac{5}{42}\)

Now, \(\frac{5}{42}\) work is done by A and B = 1 day.
or
whole work done by A&B = \(\frac{42}{5}\)
\(\frac{6}{7}\) part of work done by A and B = \(\frac{42}{5} \times \frac{6}{7} = \frac{36}{5}\) days.

Hence, total time taken = (3+36/5)days = \(10\frac{1}{5}\) days.

Answer: Option D

First find the 1 day work of both (A & B)
A 1 day's work = 1/10
and
B 1 day's work = 1/15

So (A + B) 1 day's work = (1/10+1/15)
= (3/30+2/30) = 5/30 = 1/6

So Both (A & B) together can finish work in 6 days

Answer: Option A

Here is no explanation for this answer