TCS Questions-5

TCS Questions-5

 

1.    There are members formed into three groups A, B and C.

 

Two of the members belong to all the three groups

Three members belong to A & B

Three members belong to B & C

Three members belong to C & A.

 

What is the minimum possible number of members in any of the groups.

 

Answer: 4  ( can be worked out through a Venn diagram)

 

2.    There are three numbers X, Y, and Z. The median is 5 and the mean is M. M is 10 more than the smallest number and 15 less than the biggest number. What is X+Y+Z?

Answer: 30

Let Y be the median. Then Y = 5. Let X and Z be the smallest and biggest numbers. Then we have an equation (X+5+Z)/3 = M From the given information we can write this as  (M-10) + 5 + (M+ 15) = 3M.   M = 10   Hence X + Y + Z = 3M = 30

3.    Initial price of an article was 40,000. It’s value gets reduced every year by ¾ of the previous year’s value. What will be the value of the article after three years?

 

Answer: 675

Simple mathematics.          Initial price   - 40,000

After one year (less by ¾ of the value)  -  10,000

After Two years                                      -    2,500

After three years                                  -       675   

 

4.    60 men will take 40 days to complete a work. Suppose they start the work and after every 5 days, 5 men leave the job, then in how many days the work will be completed?

 

Answer: The work will not be completed.

60 men takes 40 days to complete the work. 60 men in one day will do 1/40. One man in one day will do 1/2400  of work.

First 5 days 60 men will complete (60/2400) x 5  = 1/8 ie 12/96 work is completed

Next 5 days 55 men will do (55/2400) x 5            =  11/96

Next 5 days  50 men will complete (50/2400) x 5 = 10/96  on this basis

When the last 5 men complete the work the total completed work will be 78/96 and there still will remain uncompleted work of 18/96.

5.    Three dice are rolled. What is the probability of getting 10 as the sum on the three dice?

 

Answer: 27/216  ( There are 27 possibilities of getting sum 10 against the total possibilities of 6^3)

 

6.    If all the letters in MASTER is alphabetically sorted out, then what would be the 49^th word?

 

Answer: AREMST

 

Arrange the letters of Master according to their alphabetical position  A E M R S T.

Take the first two letters AE. The remaining four letters can be arranged in 4! Ways

Next take AM. The other four letters can be arranged in 4! Ways. Thus we have covered 48 words. Hence the 49th word would start with A M and the other four letters would follow in their respective position.

The answer is AREMST.

 

7.    There are three trucks A, B and C.

Truck A can load 10 kg in one minute.

Truck B can load 13 1/3 kg in one minute.

Truck C can unload 5 kg in one minute.

If all the three trucks work simultaneously, then what is the time taken to load 2.4 tonnes.

 

Answer: 131 minutes (approx.)

The net load done in 1 minute is …..  10 + 13 1/3 – 5 = 18 1/3 kg

So, to load 2400 kg, the time taken will be

2400/18 1/3  ie 131 minutes (approx.)  

 

8.    In six years Raj’s father will be twice the age of Raj. Raj’s mother is aged twice the age of Raj two years ago. Raj is aged 25 years now. What will be the sum of the ages of Raj’s parents after three years?

 

Answer: 108 years.

At present Raj is aged 25 Years. Two years ago Raj’s age was 23 years.

Raj’s mother at present is twice Raj’s age two years ago. Is 2 x 23 = 46 years.

After six years Raj will be 25+6 = 31 years.

After six years Raj’s father will be twice Raj’s age … 2 x 31 = 62 years.

After three years, Raj’s father will be 62 – 3 = 59 years.

After three years, Raj’s Mother will be 46 + 3 = 49 years.

Hence the sum of Raj’s parents age after three years will be 59 + 49 = 108 years.

 

9.    N is an integer and N>2. At most how many integers among N+2, N+4, N+5, N+6 and N+7 are prime integers?

Answer: Two

Since N>2, putting the value of N as 3, 5 or 7 would at best can give only maximum two prime numbers.

 

10.  If N! has 13 zeroes, then what is the highest and lowest value of N!?

 

Answer: 55! And 59!

 

 The number of zeroes in any factorial is given by the formulae  X/5 + X/5^2

In other words, for every multiple of 5 we get One zero and for every multiple of 25 we get another zero. Hence for all values between 55! And 59! We will get only 13 zeroes.

 

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