# Placement Questions & Answers :: TCS

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**TCS Practice Q&A**

461. Tim and Elan are 90 km from each other, they start to move each other simultaneously, Tim at speed 10 and Elan 5kmph, if every hour they double their speed what is the distance that Tim will pass until he meet Elan?

**Answer:**** Option B **

**Explanation:**

1st hour=tim 10kmph elan 5 the dist will be 90-15=75

2nd hr=tim->20kmph & elan->10 dist=75-30=45

3rd hr =tim->40 & elan->20..the totl dist for 3rd hr is 60.but we need only 45km to meet ..so we consider only 45mins instead of 3rd hr..

In 45mins =>tim ->30 & elan->15

totl dist travld by tim is 10+20+30=60

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462. A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X?

**Answer:**** Option C **

**Explanation:**

Cost price= 24*X,

spoiled= 24*1/3=8 kg, now total amount= (24-8)=16 kg

selling price= 16*200 =3200

profit= (3200-24*X)

profit=2* cost , (3200-24*X)= 2*24*X, x=400/9 =44 4/9

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463. A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?

**Answer:**** Option B **

**Explanation:**

Let the base be X.

Therefore (X*X + X*0 + 0) = (2*X +4) + (3*X + 2),

=> X * X = 5 * X + 6,

=> X * X - 5 * X -6 = 0,

=> (X- 6) (X+1) = 0

Therefore base is 6.

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464. On a certain island, 5% of the 10000 inhabitants are one-legged and half of the others go barefooted. What is the least number of Shoes needed in the island?

**Answer:**** Option A **

**Explanation:**

One-legged = 5% of 10000 = 500

Remaining = (10000-500) = 9500

Barefooted = 9500/2 = 4750

Remaining people = 9500-4750 = 4750

Hence required number of shoes = 4750*2+500*1 = 10000

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465. The letters in word ADOPTS are permuted in all possible ways and arranged in the alphabetical order. Find the word at position 42 in the permuted alphabetical order.

**Answer:**** Option B **

**Explanation:**

No. of words starting with AD=4!=24, AO=4!

Thus the word has to start with AO because the last word with AO will be the 48th word.

Now no. of words formed with AOD=3!=6, AOP=3!=6, AOS=3!=6 Thus the last word formed with AOS is=4!+3!+3!+3!=24+6+6+6=42

Thus the last word formed with AOS is AOSTPD

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466. 8 year old Eesha visited her grandpa. He gave her this riddle. I started working at 13. I spent 1/6 of my working life in a factory. I spent 1/4 of my working life in an office, and I spent 1/4 of my working life as a school caretaker. For the last 32 years of my working life I've been doing social service. How old am I?

**Answer:**** Option A **

**Explanation:**

Let x be the number of years he worked.

=> x/6 + x/4 + x/4 + 32 = x

=> x = 96

So, his age = 96 + 13 = 109

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467. There is a set of 36 distinct points on a plane with the following characteristics:

* There is a subset A consisting of fourteen collinear points.

* Any subset of three or more collinear points from the 36 are a subset of A.

How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)

**Answer:**** Option D **

**Explanation:**

The given data indicates that 14 points are collinear and remaining 22 points are non collinear.

A triangle can be formed by taking 1 points from 14 and 2 points from 22

OR

2 points from 14 and 1 points from 22

OR

3 points from 22

=> 14C1*22C2+14C2*22C1+22C3 = 6776

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468. What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct).

For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.

**Answer:**** Option B **

**Explanation:**

3600 = 2^4 * 3^2 * 5^2

Let abc = 2^4 * 3^2 * 5^2

We have to distribute four 2's to three numbers a, b, c in (4+3-1)C(3-1)=6C2 = 15 ways.

Now two 3's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways

Now two 5's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways

Total ways = 15 * 6 * 6 = 540

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469. How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?

**Answer:**** Option D **

**Explanation:**

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470.

**Answer:**** Option C **

**Explanation:**

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Here is the list of questions asked in TCS Aptitude Test Question with Answers TCS Mock Test page 47. 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