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251. (p/q-q/p)=21/10. Then find 4p/q+4q/p?
Answer: Option B
Explanation:let us take p/q=x and q/p=1/x
x-1/x=21/10
then by solving we get a quadratic equation like 10x^2-10-21x=0
10x^2-25x+4x-10=0
then by taking factors we get two values for x,x=-2/5 and x=5/2
by substituting x=5/2
4(5/2)+4(2/5)=58/5
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252. 4 men can check exam papers in 8 days working 5 hours regularly. What is the total hours when 2 men will check the double of the papers in 20 days?
Answer: Option A
Explanation:Let a man can do 1 unit of work in 1 hour.
Total units of work = 4 * 8 * 5 = 160 units.
Now work = 2 * 160 = 320 units.
Now 2 men work for 20 days. Let in x hours they have to work per day.
Now total work = 2*x*20 = 40 x
40x = 320 So x = 320/40 = 8 hours.
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253. 7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?
Answer: Option A
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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254. In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?
Answer: Option A
Explanation:Find the percentages. Paper A = 18/70 * 100 = 25.7
Paper B = 14/30 * 100 = 46.6
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255. Find the total no of divisors of 1728 (including 1 and 1728)
Answer: Option C
Explanation:1728= 2^6 * 3^3
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
Imp: if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
=> (6+1)(3+1) = 28
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256. A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. The value of f(1) is
Answer: Option C
Explanation:Put x =1 => f(1)+2*f(6-1) = 1 => f(1) + 2*f(5) = 1
Put x = 5 => f(5)+2*f(6-5) = 5 => f(5) + 2*f(1) = 5
Put f(5) = 5 - 2*f(1) in the first equation
=> f(1) + 2*(5 - 2*f(1)) = 1
=> f(1) + 10 - 4f(1) = 1
=> f(1) = 3
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257. The rupee/coin changing machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100 rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10 hundred rupee notes when you put a one rupee coin! Sivaji, after being ruined by his rivals in business is left with a one rupee coin and discovers the flaw in the machine by accident. By using the machine repeatedly, which of the following amounts is a valid amount that Sivaji can have when he gets tired and stops at some stage (assume that the machine has an infinite supply of notes and coins):
Answer: Option B
Explanation:The process works like this:
Rs.1 Coin => 10 × 100 = Rs.1000
Rs.100 => 10 × 10
Rs.10 => 1 × 10
Sivaji gets more money when he inserts a rupee coin only. For each rupee coin, he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000. So he has 1999 with him. Now if he inserts another coin, he has 1998 + 1000 = 2998.
Now each of these numbers is in the form of 999n + 1. So option B can be written as 54 * 999 + 1.
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258. Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?
Answer: Option D
Explanation:Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
=> 15x + 1 = 7c + 6 => c = (15x?5)/7 => c = 2x+(x?5)/7
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N - M = 56
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259. The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573. What is the largest of the numbers A, B, C, D?
Answer: Option A
Explanation:a+b+c=4024
b+c+d= 4087
a+c+d=4524
a+b+d=4573
Combining all we get 3(a+b+c+d) = 17208
=> a + b + c +d = 3736
Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.
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260. Anand packs 304 marbles into packets of 9 or 11 so that no marble is left. Anand wants to maximize the number of bags with 9 marbles. How many bags does he need if there should be atleast one bag with 11 marbles
Answer: Option B
Explanation:Given 9x + 11y = 304.
x = (304?11y)/9 = 33 + (7?2y)/9 + y
So y = - 1 satisfies. Now x = 35,
Now other solutions of this equation will be like this. Increase or decrease x by 11, decrease or increase y by 9. So we have to maximize x. next solution is x = 24 and y = 8. So bags required are 32.
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Here is the list of questions asked in TCS placement papers TCS previous papers with solutions. 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