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Placement Questions & Answers :: TCS

Company: TCS

Total Qs: 670+

Website: http://www.tcs.com

Total View: 120.8K


Not Attempted

271. A sudoku grid contains digits in such a manner that every row, every column, and every 3x3 box accommodates the digits 1 to 9, without repetition. In the following Sudoku grid, find the values at the cells denoted by x and y and determine the value of 6x + 15y.

View Answer | Submit Your Solution | Topic: Logical Problems | Asked In |

Answer: Option B

Explanation:

So x = 5, y = 3.
6x + 15y = 75

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Not Attempted

272. How many different integers can be expressed as the sum of three distinct numbers from the set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?

View Answer | Submit Your Solution | Important Formulas | Topic: Permutations & Combinations | Asked In |

Answer: Option A

Explanation:

minimum possible number = 24 ( 3+8+13)

maximum possible number = 129 ( 38+43+48)

so if we assume a AP series now where common difference is 5 and the series will start with 24 and last term will be 129 then, using AP formula

Total numbers = [(l-a)/d + 1] = [(129 - 24)/7+1 ] = 22

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Not Attempted

273. There is a lot of speculation that the economy of a country depends on how fast people spend their money in addition to how much they save. Auggie was very curious to test this theory.Auggie spent all of his money in 5 stores. In each store, he spent Rs.4 more than one-half of what he had when he went in. How many rupees did Auggie have when he entered the first store?

View Answer | Submit Your Solution | Topic: Logical Problems | Asked In |

Answer: Option A

Explanation:

As he has spent all his money, He must spend Rs.8 in the final store.
a simple equation works like this. Amount left = 1/2 x?4
For fifth store this is zero. So x = 8. That means he entered fifth store with 8.
Now for fourth store, amount left = 8 so 1/2 x?4=8=> x = 24
For third store, amount left = 24 so 1/2 x?4=24=> x = 56
For Second store, amount left = 56 so 1/2 x?4=56=> x = 120
For first store, amount left = 120 so 1/2 x?4=120=> x = 248
So he entered first store with 248.

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Not Attempted

274. HCF of 2472,1284 and a third number 'n'is 12.If their LCM is 8*9*5*103*107.then the number 'n'is..

View Answer | Submit Your Solution | Important Formulas | Topic: H.C.F and L.C.M | Asked In |

Answer: Option A

Explanation:

2472 = 2^3󫢩03
1284 = 2^2󫢩07
HCF = 2^23
LCM = 2^33^2󬊁03107
HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 2^23
LCM is the largest number that is divided by the given numbers. As LCM contains 325 these two are from N.
So N = 2^23^25^1

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Not Attempted

275. What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3

View Answer | Submit Your Solution | Important Formulas | Topic: Simplification | Asked In |

Answer: Option A

Explanation:

take 77!=a
and 54!=b
now the given expression will look like
(a(a-2b)^3 )/(a+b)^3 + (b(2a-b)^3 )/(a+b)^3

( a (a^3 - 6a^2 b + 12ab^2 - 8b^3) + b (8a^3 - 12a^2 b + 6ab^2 - b^3) ) / (a+b)^3
(a^4 - 6a^3 b + 12 a^2 b^2 - 8ab^3 + 8a^3 b - 12a^2 b^2 + 6ab^3 - b^4) / (a+b)^3
(a^4 + 2a^3 b - 2ab^3 - b^4) / (a+b)^3
( a^4 - b^4 + 2ab(a^2 - b^2) ) / (a+b)^3
( (a^2 + b^2)(a^2 - b^2) + 2ab(a^2 - b^2) ) / (a+b)^3
( (a^2 - b^2) (a^2 + b^2 + 2ab) ) / (a+b)^3
( (a^2 - b^2)(a+b)^2 ) / (a+b)^3
( (a+b)(a-b) (a+b)^2 ) / (a+b)^3
( (a+b)^3 (a-b) ) / (a+b)^3
a-b
i.e 77! - 54!

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Not Attempted

276. The marked price of coat was 40% less than the suggested retail price. Eesha purchased the coat for half of the marked price at the 15th anniversary sale. What percent less than the suggested retail price did Eesha pay?

View Answer | Submit Your Solution | Important Formulas | Topic: Profit and Loss | Asked In |

Answer: Option C

Explanation:

Let the retail price = 100
So the market price will be = (100 - 40)% (100) = 60
Easha purchased price = 60/2 = 30
So she bought it for 70% less than retail price.

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Not Attempted

277. A spherical solid ball of radius 58 mm is to be divided into eight equal parts by cutting it four times longitudinally along the same axis.Find the surface area of each of the final pieces thus obtained( in mm^2) ? (where pi= 22/7)

View Answer | Submit Your Solution | Important Formulas | Topic: Volume and Surface Area | Asked In |

Answer: Option B

Explanation:

If a sphere is cut into 8 parts longitudinally, It something looks like below Now We have to find the surface area of one piece. This is 1/8th of the initial sphere + 2 area of the half circle
= 1/8(4?r^2)+?r2
= 1/8(4 * ? * (58)^2) + ? *(58)^2
= 5046pi

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Not Attempted

278. In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?

View Answer | Submit Your Solution | Important Formulas | Topic: Problems on Numbers | Asked In |

Answer: Option C

Explanation:

Lets assume number of engg = e
number of Mba = m
and Number of CA = c
4e+3m+5c=3650
3c=2m m=3c/2
3e=2c e=2c/3 ------------- (1)
put the value in the above eq. (1)
4*2c/3+3*3c/2+5c=3650
c=300
m=3c/2
m=450

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Not Attempted

279. A rectangle is divided into four rectangles with area 70, 36, 20, and x. The value of x is

View Answer | Submit Your Solution | Important Formulas | Topic: Area | Asked In |

Answer: Option A

Explanation:

Areas are in proportion.
70/x = 36/20
=> x = 350/9

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Not Attempted

280. Seven movie addicts- Guna, Isha, Leela, Madhu, Rinku, Viji and Yamini attend a film festival. Three films are shown, one directed by Rajkumar Hirani ,one by S.Shankar,and one by Mani Ratnam. Each of the film buffs sees only one of the three films. The films are shown only once, one film at a time. The following restrictions must apply :
- Exactly twice as many of the film buffs sees the S.shankar film as see the Rajkumar Hirani film.
- Guna and Rinku do not see the same film as each other.
- Isha and Madhu do not see same film as each other.
- Viji and Yamini see the same film as each other.
- Leela sees the S.Shankar film.
- Guna sees either the Rajkumar Hirani film or the Mani Ratnam film.

Which one of the following could be an accurate matching of the film buffs to films?

(A) Guna: the S.Shankar film; Isha: the Mani Ratnam film; Madhu: the S.Shankar film

(B) Guna: the Mani Ratnam film; Isha: the Rajkumar Hirani film; Viji: the Rajkumar Hirani film

(C) Isha : the S.Shankar film; Rinku: the Mani Ratnam film; Viji: the Rajkumar Hirani film

(D) Madhu: the Mani Ratnam film; Rinku: the Mani Ratnam film; Viji: the Mani Ratnam film

View Answer | Submit Your Solution | Topic: Unknown| Asked In |

Answer: Option C

Explanation:

Guna Rinku
Isha Madhu
(Viji + Yamini)
Leela - Film: Shankar
Guna = RKH/Mani Ratnam

The following options are possible:
RKH Shankar Mani Ratnam
1 2 4
2 4 1
We will take options and check them.
Option A: Guna should not watch Shankar's Film. So ruled out
Option B:
RKH Shankar Mani Ratnam
Isha _ Guna
Viji _ _
Now Yamini also watch RKH. Which is not possible.
Option C:
RKH Shankar Mani Ratnam
Viji Isha Rinku
Yamini Leela _
As Guna should not be watching Shankar's movie she should watch Mani ratnam's which is not possible.
Option D:
RKH Shankar Mani Ratnam
Guna Leela Madhu
_ Isha Rinku
_ _ Viji and Yamini

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