Consider a set A {1, 2, 3,……… 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5?
A. | 533 |
B. | 599 |
C. | 467 |
D. | 66 |
Option: C Explanation :
599 is the correct answer
1000/3 = 333.33 = 333
1000/5 = 200
1000/(3*5) = 1000/15=66
333+200-66=467
Click on Discuss to view users comments. sarathbabu said: (4:13am on Sunday 28th December 2014)
There are 333 numbers that are divisible by 3 since 333*3=999.Among those 333 numbers there are 66 numbers that are also divisble by 5 since 333/5 =66.6So 333-66 = 267 numbers that are divisble only by 3.There are 200 numbers that are divisible by 5 since 200*5 = 1000.Among those 200 numbers there are 66 numbers that are also divisble by 3.So 200-66 = 134 numbers that are divisible only by 5.267 - numbers that are divisible only by 3.134 - numbers that are divisible only by 5.66 - numbers that are divisble both by 3 and 5.So we have 267 134 66 = 467 numbers that are divisble either by 3 or 5.Answer : c
|
You have to sort a list L, consisting of a sorted list followed by a few 'random' elements. Which of the following sorting method would be most suitable for such a task?
A. | Bubble sort |
B. | Selection sort |
C. | Quick sort |
D. | Insertion sort |
Option: D Explanation : Click on Discuss to view users comments. |
An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element?
A. | 0.336 bits baud |
B. | 3 bits baud |
C. | 120,00,000 bits baud |
D. | None of the above |
Option: B Explanation : Click on Discuss to view users comments. asha said: (1:24pm on Wednesday 15th April 2015)
r=N/S, where N - bit rate, S -Baud rate. 6000/2000=3
|