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471. How many 2's are there between the terms 112 to 375?
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
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472. 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?
Answer: Option B
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473. 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?
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.
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474. 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.
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
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475. In a group of five families, every family is expected to have a certain number of children, such that the number of children forms an arithmetic progression with a common difference of one, starting with two children in the first family. Despite the objection of their parents, every child in a family has as many pets to look after as the number of offsprings in the family. What is the total number of pets in the entire group of five families.
Answer: Option D
Explanation:As the number of children are in arithmetic progression starting with 2, the five families have 2, 3, 4, 5, 6 kids respectively.
As each children has kept the pets equal to the number of kids in the family, Each family has n^2 pets.
So total = 2^2+3^2+4^2 +5^2 +6^2 = 90
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476. Radius of the bigger circle is 1. Which area will be greater?
Answer: Option B
Explanation:Here is no explanation for this answer
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477. An old man and a young man are working together in an office and staying together in a near by apartment. The old man takes 30 minutes and the young 20 minutes to walk from appartment to office. If one day the old man started at 10.00 AM and the young man at 10:05AM from the apartment to office, when will they meet?
Answer: Option A
Explanation:Ratio of old man speed to young man speed = 2:3
The distance covered by old man in 5 min = 10
The 10 unit is covered with relative speed=10/(3-2)=10 min
so, they will meet at 10:15 am.
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478. In a potato race, 20 potatoes are placed in a line of intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Answer: Option C
Explanation:Given, total number of potatos = 20.
First potato 24 metres from the starting point. There are 4 meters in the intervals. A contestant is required to bring the potatoes back to the starting place one at a time. So for the first potato he has to travel 48 meters, for second 56 meters ...
48,56,64............20 terms.
a = 48, d= 8, n = 20.
Important Point:
Sum of n terms in A.P = Sn=n2[2a+(n?1)d]
S20 = 202[2*48+(20?1)8]
S20 = 202[96+152]
S20 = 10*248 = 2480
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479. Let a, b, c, d and e be distinct integers in ascending order such that(76-a)(76-b)(76-c)(76-d)(76-e) = 1127. What is a + b + c + d
Answer: Option B
Explanation: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.
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480. In this question, A^B refers to A raised to the power B.Ten tickets numbered 1, 2, 3, ..., 10. Six tickets are selected at random one of a time with replacement. The probability of the largest number appearing on the selected ticket is 7 is
Answer: Option B
Explanation:Let's first find out probability of that maximum number being any number between 1 to 7.
P(1 to 7) = (7/10) * (7/10) * (7/10) ... 6 times
Now find out probability of that maximum number being any number between 1 to 6.
P(1 to 6) = (6/10) * (6/10) * (6/10) ... 6 times
Now, probability that maximum number is exactly 7
= P(1 to 7) - P (1 to 6)
= (7^6 - 6^6) / 10^6
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Here is the list of questions asked in TCS Aptitude Test Question with Answers TCS Mock Test page 48. 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