This set of placement paper is collected from on-campus recruitment for the session (2013-14) at NIT, Surathkal. Try to solve each question as all these questions are getting repeated.
1. There are 3 people A, B, C having a total share of Rs 3,000. A’s share is 2/3rd of the sum of B and C’s shares. C’s share is ½ of the sum of A and B’s shares. What is the value of C’s share?
Ans: 1000
Explanation: A + B +C = 3000 …………….(1)
½(A+B)+A+B=3000
3A +3B = 6000
A+B=2000 …………………….(2)
Now, substituting the value of equation (2) in equation (1)
A + B + C = 3000
2000 + C =3000
C =1000
2. The total income of Amit in 2003, 2004, 2005 is 364000. Every year there is an increase in his salary by 20%. What is the salary in 2005?
Ans. 144000
Explanation: Let Amit salary in 2003 is x.
Therefore, salary in 2004 was 120/100 x and salary in 2005 is 120/100(120/100)x.
x+ 120/100x + 120/100(120/100)x= 364000
x + 120/100 x + 144/100 x = 364000
(364/100)x = 364000
x= 1,00,000
Therefore, Salary in 2005 is (364000-120000-100000) = 144000
3. In how many ways a team of 11 members can be formed from 5 men and 11 women such that the team have at most 3 men?
Ans. 2256
Explanation: 1st Case: The team has no men member = 11C11 = 1
2nd Case: The team has 1 men member = 11C10 * 5C1 =11 * 5 = 55
3rd Case: The team has 2 men members = 11C9 * 5C2 = 55 *10 = 550
4th Case: The team has 3 men members= 11C8 *5C3 = 165 * 10 = 1650
Therefore, Total no of ways = 1+55+550+1650= 2256
4. In a set of 30 numbers, average of 1st 10 numbers is equal to the average of last 20 numbers .Find the sum of last 20 numbers?
Ans. Twice the sum of first 10 numbers
Explanation: Average = (Total sum of Numbers/ No. of numbers )
(Sum of first 10 numbers/10) = ( sum of last 20 numbers /20)
Therefore, Sum of last 20 numbers = 2 * sum of first 10 numbers .
5. In a class, there are 9 students in which there are 5 boys and 4 girls. In a row, there will always a boy between any two girls. Find out how many such arrangement possible?
Ans. 43200
Explanation: 5 boys can be arranged in 5! ways.
There are six place for girls. The girls can choose any 4 out of 6 .Therefore 6C4.
Now 4 girls will have 4! ways of arrangement between themselves.
Total no. of such arrangement = 120 * 15 * 24 = 43200
6. What no. should be added to 5678 so that the remainder is 35 when divided by 460?
Ans. 337
Explanation: 5678%460=158
(460-158)+35=337
7. In 2003 there are 28 days in February and 365 days in that year, in 2004 there are 29 days in February and 366 days in that year. If the date 11 march 2003 is Tuesday, then which would be the date for 11 march 2004?
Ans. Thursday
Explanation: It’s given that 11 march 2003 is Tuesday. We have to find which day is 11 march 2004
So, there is difference of 366 days between the two dates.
Therefore, no of odd days = (366%7)= 2. So, it will be Thursday.
We will soon post some more TCS latest placement papers.
1. There are 3 people A, B, C having a total share of Rs 3,000. A’s share is 2/3rd of the sum of B and C’s shares. C’s share is ½ of the sum of A and B’s shares. What is the value of C’s share?
Ans: 1000
Explanation: A + B +C = 3000 …………….(1)
½(A+B)+A+B=3000
3A +3B = 6000
A+B=2000 …………………….(2)
Now, substituting the value of equation (2) in equation (1)
A + B + C = 3000
2000 + C =3000
C =1000
2. The total income of Amit in 2003, 2004, 2005 is 364000. Every year there is an increase in his salary by 20%. What is the salary in 2005?
Ans. 144000
Explanation: Let Amit salary in 2003 is x.
Therefore, salary in 2004 was 120/100 x and salary in 2005 is 120/100(120/100)x.
x+ 120/100x + 120/100(120/100)x= 364000
x + 120/100 x + 144/100 x = 364000
(364/100)x = 364000
x= 1,00,000
Therefore, Salary in 2005 is (364000-120000-100000) = 144000
3. In how many ways a team of 11 members can be formed from 5 men and 11 women such that the team have at most 3 men?
Ans. 2256
Explanation: 1st Case: The team has no men member = 11C11 = 1
2nd Case: The team has 1 men member = 11C10 * 5C1 =11 * 5 = 55
3rd Case: The team has 2 men members = 11C9 * 5C2 = 55 *10 = 550
4th Case: The team has 3 men members= 11C8 *5C3 = 165 * 10 = 1650
Therefore, Total no of ways = 1+55+550+1650= 2256
4. In a set of 30 numbers, average of 1st 10 numbers is equal to the average of last 20 numbers .Find the sum of last 20 numbers?
Ans. Twice the sum of first 10 numbers
Explanation: Average = (Total sum of Numbers/ No. of numbers )
(Sum of first 10 numbers/10) = ( sum of last 20 numbers /20)
Therefore, Sum of last 20 numbers = 2 * sum of first 10 numbers .
5. In a class, there are 9 students in which there are 5 boys and 4 girls. In a row, there will always a boy between any two girls. Find out how many such arrangement possible?
Ans. 43200
Explanation: 5 boys can be arranged in 5! ways.
There are six place for girls. The girls can choose any 4 out of 6 .Therefore 6C4.
Now 4 girls will have 4! ways of arrangement between themselves.
Total no. of such arrangement = 120 * 15 * 24 = 43200
6. What no. should be added to 5678 so that the remainder is 35 when divided by 460?
Ans. 337
Explanation: 5678%460=158
(460-158)+35=337
7. In 2003 there are 28 days in February and 365 days in that year, in 2004 there are 29 days in February and 366 days in that year. If the date 11 march 2003 is Tuesday, then which would be the date for 11 march 2004?
Ans. Thursday
Explanation: It’s given that 11 march 2003 is Tuesday. We have to find which day is 11 march 2004
So, there is difference of 366 days between the two dates.
Therefore, no of odd days = (366%7)= 2. So, it will be Thursday.
We will soon post some more TCS latest placement papers.
Soon
we are coming up with huge collection of latest Placement papers and
Interview Experience. Like us on Facebook to get the updates.
Like & Share
Written by Interview Time Group
We provide latest placement papers, tricks to crack interview and Study materials. We are trying to create All-in-One placement hub all you need to access to catch up a job.
0 comments:
Post a Comment