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301. In how many ways can 5 different toys be packed in 3 identical boxes such that no. box is empty, if any of the boxes may hold all of the toys?
Answer: Option A
Explanation:Number of ways of achieving the second option 3 - 1 - 1
Three toys out of the 5 can be selected in 5C3ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case a + number of ways of achieving Case b
= 15 + 10 = 25 ways.
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302. Two cars starts from A and B and travel towards each other at a speed of 50 kmph and 60 kmph respectively. At the time of their meeting the second car has travelled 120 kmph more than the first. The distance between A and B is:
Answer: Option C
Explanation:Here is no explanation for this answer
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303. How many 4-bit digit numbers that do not contain the digits 3 or 6 are there?
Answer: Option D
Explanation:Let the number be _ _ _ _
Left digit can be any of the (1,2,4,5,7,8,9) --> 7 ways
Other digits can be (0,1,2,4,5,7,8,9) --> 8 ways
So the total ways = 7*8*8*8 = 3584 or 3584 numbers without any 3 or 6.
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304. A tourist wants to visit 3 or more of the 5 major cities in India: Chennai, Bangalore, Mumbai, Delhi and Kolkata. In how many ways can be plan his tour such that Chennai is always includes? Two plan of the tour are different if the cities in the tour or the order of the cities are different.
Answer: Option B
Explanation:As chennai is always there,
So tourist have to choose 2 cities out of 4 cities(excluding chennai) and this can be done by
Chennai must always be includes.
for 3 cities 4C2*3! => 6*3! => 36
for 4 cities 4C3*4! => 4*4! => 96
for 5 cities 4C4*5! => 1*5! => 120
So, Total = 5!+4*4!+6*3!
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305. 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:Initially sivaji had only one coin so he earns only 1000/-
to earn more he must convert one 100 note to 10 notes and then one ten note to ten coins
now he has Rs.990 and 10 coins
after converting 10 coins he has 10990
after converting another 10 coins he has 20980
after converting another 10 coins he has 30970
after converting another 10 coins he has 40960
after converting another 10 coins he has 50950
after converting another 10 rupee note to coins and using only 3
he has seven coins and 53940
so sum is 53940+7=53947
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306. George can do some work in 8 hours: Paul can do some work in 10 hours while Hari can do the same work in 12 hours. All the three of them start working at 9 a.m. while George stops work at 11 a.m. and the remaining two complete the work. Approximately at what time the work be finished?
Answer: Option B
Explanation:Chocolate method : Take Lcm of george,paul and Hilllari we get LCM(8,10,12)=120
Let the no of chocolates are 120 in that
George can eat 15 chocolates/hr
paul can eat 12cho/hr
Hillari can eat 10cho/hr
total choclates they can eat in 1 hr =Paul + George+hillari=15+12+10=37choclates
for 9:00 to 11:00 difference -2hrs
so for 2 hrs they can eat 37*2=74
George left after 11 remaing choclates are 120-74=46
Paul and hillari can eat=12+10=32/hr
=46/32 =2 hrs approx so ans is 11:00 + 2hrs =1:00 pm
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307. A number when successively divided by 5.3.2 gives remainders of 0, 2 and 1 respectively in that order. What will be the remainder when the same number is divided successively by 2, 3 and 5 in that order?
Answer: Option C
Explanation:Let the number = N
Now N = 5K
K = 3L + 2
L = 2M + 1
K = 3(2M + 1) + 2 = 6M + 5
N = 5(6M + 5) = 30 M + 25
For M = 0 we get the least number as 25.
Now when 25 is divided by 2, we get 12 as quotient and 1 as remainder.
When 12 is divided by 3 we get 4 as quotient, and 0 as remainder. When 4 is divided by 5 we gt 4 as remainder.
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308. Arun makes a popular brand of ice-cream in a rectangular shaped bar 6 cm long 5cm wide and 2 cm thick. To cut costs the company had decided to reduced the volume of bar by 19%. The thickness will remain the same, but the length and width will be decreased by the same %. The new width will be
Answer: Option D
Explanation:Volume = l*b*h
=> 6*5*2 = 60cm^3
Volume is reduced by 19%
New Volume = (100-19)100*60 = 486000 = 48.6
Thickness is same but length and breadth reduces by x%
So,new volume = (x/100*6)(x/100*5)2 = 48.6
=> x=90
So, Length and Breadth reduced by 10%
=> new width = 5-(10 * 5/100) = 4.5
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309. Ashok, Esha, Farookh and Gouri ran a race. Ashok said, "I did not finish 1ST or 4TH". Esha said, "I did not finish 4TH". Farookh said, "I finished 1ST". Gouri said, "I finished 4TH". There were no ties in the competition and exactly three of the children told the truth. Who finished 4TH?
Answer: Option D
Explanation:Here is no explanation for this answer
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310. In the polynomial f(x) =2*x^4 - 49*x^2+54, what is the product of the roots, and what is the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)?
Answer: Option A
Explanation:sum of roots= -(coeff of x^3/coeff of x^4)= -b/a = -0/2=0
product of roots=(-1)^n*(coeff of constant term/coeff of x^4)= e/a = 54/2 =27
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Here is the list of questions asked in TCS written test topics TCS aptitude questions. 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