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261. In how many ways can the letters of the english alphabet be arranged so that there are seven letter between the letters A and B, and no letter is repeated
Answer: Option E
Explanation:We can fix A and B in two ways with 7 letters in between them. Now 7 letters can be selected and arranged in between A and B in 24P7 ways. Now Consider these 9 letters as a string. So now we have 26 - 9 + 1 = 18 letters
These 18 letters are arranged in 18! ways.
So Answer is 2 x 24P7 x 18!
OR 2 x 24P7 x 18! = 36 x 24!.
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262. When Usha was thrice as old as Nisha, her sister Asha was 25, When Nisha was half as old as Asha, then sister Usha was 34. their ages add to 100. How old is Usha?
Answer: Option D
Explanation:Let the age of Usha is 3x then Nisha is x and Asha is 25
Also Usha 34, Nisha y, and Asha 2y.
We know that 3x - 34 = x - 2y = 25 - 2y
Solving above three equations we get x = 9, y = 16
Their ages are 34, 16, 32. whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40
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263. You have been given a physical balance and 7 weights of 52, 50, 48, 44, 45, 46 and 78 kgs. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 183 kgs.
Answer: Option A
Explanation:52+50+78 = 180
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264. Find the number of zeroes in the expression 15*32*25*22*40*75*98*112*125
Answer: Option B
Explanation:Maximum power of 5 in the above expression can be calculated like this. Count all the powers of 5 in the above expression. So number of zeroes are 9.
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265. Find the probability that a leap year chosen at random will have 53 Sundays.
Answer: Option B
Explanation:A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun). Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7
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266. An ant starts moving on the mesh shown below along the wires towards a food particle.If the ant is at the bottom-left corner of cell A and the food is at the top-right corner of cell F, then find the number of optimal routes for the ant.
Answer: Option B
Explanation:Total ways to move from A to the junction: There are 13 upward ways, 3 right side ways this Ant can move. Now these 16 ways may be in any order. So number of ways of arrangements = 16!/(13! * 3!) = 560
Similarly, from the junction to F, Total 12 upward ways and 5 right-side ways. These 17 ways can be in any order. So Total ways = 17!/(12! * 5!) = 6188.
Total ways to move from A to F = 560 * 6188 = 3465280
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267. You need a 18% acid solution for a certain test, but your supplier only ships a 13% solution and a 43% solution. You need 120 lts of the 18% acid solution. the 13% solution costs Rs 82 per ltr for the first 67 ltrs, and Rs 66 per ltr for any amount in access of 67 ltrs. What is the cost of the 13% solution you should buy?
Answer: Option C
Explanation:Let us assume we need "a" liters of 13% acid solution and "b" liters of 43% acid solution. Now
=>18 = (a * 13 + b * 43)/(a + b)
=>a/b = 5/1
So we need 100 liters of 13% acid solution, and 20 liters of 18% acid solution.
Final cost = 82 * 67 + 66 * 33 = 7672
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268. Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs. and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. After how many hours will B reach city Y from the time A overtook him for the first time?
Answer: Option C
Explanation:
Vehicle A overtaken B at 10.30 am and reached X at 12 pm. It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance 'n', So it should be at Q at 11 pm. It is clear that Vehicle A takes 0.5 hour to cover distance 'm'.
Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m. So Speeds ratio = 4.5 : 0.5 = 9 : 1.
Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 * 9 + 1 = 41.5 hours
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269. Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is:
Answer: Option A
Explanation:We have to find the area of the blue shaded one and double it to get the area common to the both.
Now this can be calculated as Area of the sector OAB - Area of the Triangle OAB.
As OA and OB are perpendicular, area of the sector OAB = 90/360 π (1)^2 = π/4
Area of the triangle OAB = 1/2 * 1 * 1 = 1/2
Area common to both = (π/4 - 1/2)/2 = π/2 - 1
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270. Two consecutive numbers are removed from the progression 1, 2, 3, ...n. The arithmetic mean of the remaining numbers is 26 1/4. The value of n is
Answer: Option C
Explanation:As the final average is 105/4, initial number of pages should be 2 more than a four multiple. So in the given options, we will check option C.
Total = n(n+1)/2=(50 * 51)/2 = 1275
Final total = 48 * 105/4 = 1260
So sum of the pages = 15. The page numbers are 7, 8
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Here is the list of questions asked in TCS written test papers for freshers TCS written test papers. 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