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What is the next number?
11, 23, 47, 83, 131, ....
A145
B178
C176
D191
Answer: Option D
Explanation:23–11 = 12
47–23 = 24
83–47 = 36
131–83 = 48
Therefore, 131+60=191
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222 / 652
A sum of Rs.3000 is distributed among A, B, and C. A gets 2/3 of what B and C got together and c gets 1/3 of what A and B got together, C's share is?
A750
B740
C870
D787
Answer: Option A
Explanation:As A+B+C=3000. And A=2/3(B+C).
Also C=1/3(A+B).
So, 3C=A+B.
Hence now we can substitute A+B by 3C.
So final equation will be 3C+C=3000.
Hence , 4C=3000 => C=750.
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223 / 652
On door A - It leads to freedom
On door B - It leads to Ghost house
On door C - door B leads to Ghost house
The statement written on one of the doors is wrong.
Identify which door leads to freedom.
AA
BB
CC
DNone of these
Answer: Option C
Explanation:Case 1: A, B are true. In this case, Statement C also correct. So contradiction.
Case 2: B, C are true. In this case, B leads to ghost house and C confirms it. Now A is wrong. So door A does not lead to freedom. So Door C leads to freedom.
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224 / 652
What is the probability of getting sum 3 or 4 when 2 dice are rolled
A1/36
B5/36
C1/2
D1
Answer: Option B
Explanation:Required number of ways of sum 3 or 4 = (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5
Total ways = 6^2 = 36
Probability = 5/36
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225 / 652
a bb ccc dddd eeeee .........What is the 120th letter?
A14 O's.
B15 N's.
C15 P's.
D15 O's.
Answer: Option D
Explanation:Number of letters in each term are 1, 2, 3, 4, 5,….
The above number are in series of A.P.
So n*(n+1)/2 ? 120
For n = 15 above eq. condition will be true.
15th term contains 15 O's.
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226 / 652
Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300 (Hint 44^2=1936)
A1
B2
C3
D4
ECan't be determined
Answer: Option A
Explanation:The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 - 1) = 2116
47^2 = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.
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227 / 652
What is in the 200th position of 1234 12344 123444 1234444....?
A8
B3
C4
D1
Answer: Option C
Explanation:Here is no explanation for this answer
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228 / 652
What is the next number in the given series?
70, 54, 45, 41,.....
A35
B36
C38
D40
Answer: Option D
Explanation:Consecutive squares are subtracted from the numbers.
70 - 54 = 16
54 - 45 = 9
45 - 41 = 4
So next we have to subtract 1.
So answer = 41 - 1 = 40
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229 / 652
The sticks of the same length are used to form a triangle as shown below. If 87 such sticks are used then how many triangles can be formed?
A44
B43
C34
D45
Answer: Option B
Explanation:First triangle is formed by using 3 sticks, but any subsequent triangle may be formed by using 2 sticks.
Therefore, If 1st triangles uses 3 sticks,
Remaining sticks = 87 - 3 = 84. With these 84, we can form 42 triangles. So total = 42 + 1 = 43
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230 / 652
The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?
A7
B8
C5
D4
Answer: Option A
Explanation:From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.
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Here is the list of questions asked in Practice aptitude test for TCS. 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
