TCS Placement Questions & Answers :: TCS
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The fourteen digits of a credit card are to be written in the boxes shown above. If the sum of every three consecutive digits is 18, then the value of x is:
A3
B2
C1
DCannot be determined
Answer: Option A
Explanation:Let us assume right most two squares are a , b
Then Sum of all the squares = 18 x 4 + a + b ------ (1)
Also Sum of the squares before 7 = 18
Sum of the squares between 7, x = 18 and
sum of the squares between x , 8 = 18
So Sum of the 14 squares = (18 + 7 + 18 + x + 18 + 8 + a + b) ------- (2)
By Solving equation 1 and 2
x = 3
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402 / 652
Given that 0 < a < b < c < d, which of the following the largest?
A(c+d) / (a+b)
B(a+d) / (b+c)
C(b+c) / (a+d)
D(b+d) / (a+c)
Answer: Option A
Explanation:Take a = 1, b = 2, c = 3, d = 4
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403 / 652
Thomas takes 7 days to paint a house completely whereas Raj would require 9 days to paint the same house completely. How many days will take to paint the house if both them work together. (Give answers to the nearest integer)?
A4
B2
C5
D3
Answer: Option A
Explanation:Here is no explanation for this answer
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404 / 652
In 2003, there are 28 days in February and there are 365 days in the year. In 2004, there are 29 days in February and there are 366 days in the year. If the date March 11, 2003 is Tuesday, then which on of the following would the date March 11, 2004 be?
AMonday
BThursday
CWednesday
DTuesday
Answer: Option B
Explanation:March 11, 2003 is Tuesday. So March 11, 2004 weekday will be 2 days after Tuesday. i.e., Thursday.
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405 / 652
X = 101102103104105106107......146147148149150 (From numbers 101-150). Find out the remainder when this number is divided by 9.
A7
B4
C2
D3
Answer: Option C
Explanation:The divisibility rule for 9 is sum of the digits is to be divisible by 9. So
We calculate separately, sum of the digits in hundreds place, tenths place, and units place.
Sum of the digits in hundreds place: 1 * 50 = 50
Sum of the digits in tenths place : 0 * 9 + 1 * 10 + 2 * 10 + 3 * 10 + 4 * 10 + 5 * 1 = 105
Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) * 5 = 225
So total = 380
So remainder = 380 / 9 = 2
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406 / 652
A number is 101102103104...150. As 101 102 103 103.... 150. What is reminder when divided by 3?
A5
B2
C7
D1
Answer: Option B
Explanation:Divisibility rule for 3 also same as 9.
so from the above discussion sum of the digits = 380
and remainder = 380/3 = 2
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407 / 652
7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?
A34
B51
C50
D56
Answer: Option B
Explanation:Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats.
So for every 4 terms we get one term with 3 in its unit digit.
So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.
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408 / 652
If two-third of a bucket is filled in one minute then the time taken to fill the bucket completely will be .
A90 seconds
B70 seconds
C60 seconds
D100 seconds
Answer: Option A
Explanation:If two-third of a bucket is filled in one minute
So, time taken to fill the bucket completely = 60*3/2= 90 seconds
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409 / 652
If a ladder is 100 m long and distance between bottom of ladder and wall is 60 m. What is the maximum size of cube that can be placed between the ladder and wall.
A34.28
B24.28
C21.42
D28.56
Answer: Option A
Explanation:8/6 = tan(y)
x/6*10 -x =tan(y)
8/6=x/60-x
x=34.28
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410 / 652
7,77,777,7777,77777,777777....
Find last 3 digits of sum of first 26 terms
A452
B456
C732
D672
Answer: Option C
Explanation:At unit digit 7 would be coming 26 timing
so, 26*7=182.
2 at unit place
Then 25*7=175+ (18 carry)=193
Then 24*7=187
So, Last digits will be 732.,
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Here is the list of questions asked in TCS Aptitude Test Question with Answers page 41. 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