NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIEEE 2003

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
A
$$7!\, \times 5!\,\,$$
B
$$6!\, \times 5!$$
C
$$30!$$
D
$$5!\, \times 4!$$

## Explanation

6 men can sit at the round table = $$\left( {6 - 1} \right)! = 5!$$ ways

Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.

So total no of ways 6 men and 5 women can dine at the round table

= $$5!\, \times {}^6{C_5} \times 5!$$

= $$5!\, \times 6 \times 5!$$

= $$5!\, \times 6!$$
2

### AIEEE 2002

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are
A
312
B
3125
C
120
D
216

## Explanation

Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.

Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.

Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)

So total no of arrangement = 1$$\times$$2$$\times$$3$$\times$$4$$\times$$5 = 5!

Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)

So total no of arrangement = 4$$\times$$4$$\times$$3$$\times$$2$$\times$$1 = 4.4!

$$\therefore$$ Total arrangement = 5! + 4.4! = 216
3

### AIEEE 2002

The sum of integers from 1 to 100 that are divisible by 2 or 5 is
A
3000
B
3050
C
3600
D
3250

## Explanation

According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$\times$$5 = 10.

Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100

This is an A.P where first term = 2, last term = 100 and total terms = 50.

$$\therefore$$ Sum of the numbers divisible by 2

= $${{50} \over 2}\left[ {2 + 100} \right]$$

= 25$$\times$$102

= 2550

Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100

$$\therefore$$ Sum of the numbers divisible by 5

= $${{20} \over 2}\left[ {5 + 100} \right]$$

= 10$$\times$$105

= 1050

And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100

$$\therefore$$ Sum of the numbers divisible by 10

= $${{10} \over 2}\left[ {10 + 100} \right]$$

= 5$$\times$$110

= 550

$$\therefore$$ Required sum = 2550 + 1050 - 550 = 3050
4

### AIEEE 2002

Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is
A
125
B
105
C
374
D
625

## Explanation

There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed

Case 1 : First digit is 1 = 1 _ _ _

Possible numbers starting with 1 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.

Case 2 : First digit is 2 = 2 _ _ _

Possible numbers starting with 2 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

Case 3 : First digit is 3 = 3 _ _ _

Possible numbers starting with 3 = 1$$\times$$5$$\times$$5$$\times$$5 = 125

Total possible numbers = 124 + 125 + 125 = 374

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12