TCS Placement Questions & Answers :: TCS
451 / 652
The letters in word ADOPTS are permuted in all possible ways and arranged in the alphabetical order. Find the word at position 42 in the permuted alphabetical order.
AAOTDSP
BAOSTPD
CAOTDPS
DAOTPDS
Answer: Option B
Explanation:No. of words starting with AD=4!=24, AO=4!
Thus the word has to start with AO because the last word with AO will be the 48th word.
Now no. of words formed with AOD=3!=6, AOP=3!=6, AOS=3!=6 Thus the last word formed with AOS is=4!+3!+3!+3!=24+6+6+6=42
Thus the last word formed with AOS is AOSTPD
Submit Your Solution
452 / 652
8 year old Eesha visited her grandpa. He gave her this riddle. I started working at 13. I spent 1/6 of my working life in a factory. I spent 1/4 of my working life in an office, and I spent 1/4 of my working life as a school caretaker. For the last 32 years of my working life I've been doing social service. How old am I?
A109
B102
C105
D113
Answer: Option A
Explanation:Let x be the number of years he worked.
=> x/6 + x/4 + x/4 + 32 = x
=> x = 96
So, his age = 96 + 13 = 109
Submit Your Solution
453 / 652
There is a set of 36 distinct points on a plane with the following characteristics:
* There is a subset A consisting of fourteen collinear points.
* Any subset of three or more collinear points from the 36 are a subset of A.
How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)
A7140
B4774
C1540
D6776
Answer: Option D
Explanation:The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22
OR
2 points from 14 and 1 points from 22
OR
3 points from 22
=> 14C1*22C2+14C2*22C1+22C3 = 6776
Submit Your Solution
454 / 652
What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct).
For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.
A441
B540
C84
D2100
Answer: Option B
Explanation:3600 = 2^4 * 3^2 * 5^2
Let abc = 2^4 * 3^2 * 5^2
We have to distribute four 2's to three numbers a, b, c in (4+3-1)C(3-1)=6C2 = 15 ways.
Now two 3's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Now two 5's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Total ways = 15 * 6 * 6 = 540
Submit Your Solution
455 / 652
How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?
A6480
B320
C2160
D720
Answer: Option D
Explanation:Submit Your Solution
456 / 652
In the town of Unevenville, it is a tradition to have the size of the front wheels of every cart different from that of the rear wheels. They also have special units to measure cart wheels which is called uneve. The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled by the cart in uneves, when the front wheel has done nine more revolutions than the rear wheel?
A570
B1330
C3990
D389
Answer: Option C
Explanation:Submit Your Solution
457 / 652
How many 2's are there between the terms 112 to 375?
A313
B159
C156
D315
Answer: Option C
Explanation:Let us calculate total 2's in the units place. (122, 132, 142 ... 192), (201, 212, 222, ... 292), (302, 312, ... 372) = 8 + 10 + 8 = 26
Total 2's in tenth's place, (120, 121, 122, ..., 129) + (220, 221, ..., 229) + (320, 321, ..., 329) = 30
Total 2's in hundred's place = (200, 201, ... 299) = 100.
Total 2's between 112 and 375 = 26 + 30 + 100 = 156
Submit Your Solution
458 / 652
Consider the sequence of numbers 0, 2, 2, 4,... Where for n > 2 the nth term of the sequence is the unit digit of the sum of the previous two terms. Let sn denote the sum of the first n terms of this sequence. What is the smallest value of n for which sn >2771?
A692
B693
C694
D700
Answer: Option B
Explanation:Submit Your Solution
459 / 652
Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?
A105
B60
C88
D71
Answer: Option D
Explanation:Let Sandra, David and Mary each has s, d and 11 respectively.
After the first distribution,
David has d + d = 2d, Mary has 11 + 11 = 22 and Sandra has s - d - 11.
After the second distribution,
Sandra has 2*(s - d - 11) , mary has 2*22 = 44 and david has 2d - (s - d - 11) - 22=3d - s -11.
After the third distribution,
Sandra has 2*2(s - d - 11), david has 2*(3d - s - 11) and mary has 44 - 2(s - d - 11) - (3d - s - 11) = 77 - s - d
It is given that finally Mary has Rs.17. So, 77 - s - d=17
=> s + d = 60
=> s + d + 11 = 60 + 11 = 71.
Submit Your Solution
460 / 652
Babla alone can do a piece of work in 10 days. Ashu alone can do it in 15 days. The total wages for the work is Rs.5000. How much should be Babla be paid if they work together for an entire duration of work.
A2000
B4000
C5000
D3000
Answer: Option D
Explanation:Bablu work = 1/10
Ashu work = 1/15
Total work = 1/10+1/15 = 3+2/30 = 5/30
Total work = 6 days
Bablu paid = 6/10*5000 = 3000
Submit Your Solution
Here is the list of questions asked in TCS Aptitude Test Question with Answers page 46. Practice TCS Written Test Papers with Solutions and take Q4Interview TCS Online Test Questions to crack TCS written round test. Overall the level of the TCS Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of TCS