# Free Playing with Numbers 03 Practice Test - 8th Grade

### Question 1

If I divide 487 by 100 what is the remainder?

48

7

47

87

#### SOLUTION

Solution :D

On dividing 487 by 100, the remainder is the number formed by the last 2 digits of the given number. Hence it is 87 in this case.

### Question 2

The number 152875 is divisible by

3

9

5

4

#### SOLUTION

Solution :C

The sum of the digits of 152875 is 28 which is not a multiple of 3, 152875 is not divisible by 3.

28 is not a multiple of 9 so the number is also not divisible by 9.

Number formed by last two digits is 75 and it is not divisible by 4. Therefore, the number is not divisible by 4.

The last digit of given number is 5, hence 152875 is divisible by 5.

### Question 3

The number 456318 is divisible by:

3

2

6

11

#### SOLUTION

Solution :A, B, and C

The sum of the digits of the given number is 4 + 5 + 6 + 3 + 1 + 8 = 27 which is divisible by 3.

Hence, the given number is divisible by 3.

The last digit of the number is even. Hence, it is divisible by 2.

Since the number is divisible by both 3 and 2, it is divisible by 6.

A number is divisible by 11 , if the difference between the sum of its digits in even places and the sum of its digits in odd places is either 0 or divisible by 11.

Here, sum of digits in even places

= 1 + 6 + 4 = 11

Sum of digits in odd places

= 8 + 3 + 5 = 16

Difference = 16 -11 = 5 which is not divisible by 11.

∴ The given number is not divisible by 11.

### Question 4

The sum of a three digit number and the number formed by the reversal of its digits (if for both the numbers middle digit is 0) is always divisible by 101

True

False

#### SOLUTION

Solution :A

Let abc be the number.

Thus abc = 100a + 10b + c

The number obtained by reversing the digits is cba.

cba = 100c + 10b + a

Adding the two numbers we get, abc + cba = (100a + 10b + c) + (100c + 10b + a) = 101a + 20b + 101c

Since middle digit is 0, the sum will be 101(a+c), which is divisible by 101.

### Question 5

There are 2 two digit numbers such that one has been obtained by reversing the digits of the other. On subtracting the smaller number from the larger number, I get a number divisible by

3

4

6

9

#### SOLUTION

Solution :A and D

Let the number be ab.

Here we have 'a' in tens place and 'b' in units place. So we can write the number as 10a + b. If we reverse it we get 10b+a.(10a + b) – (10b + a) = 10a + b – 10b – a

= 9a – 9b = 9(a – b).

Hence the number is divisible by 9.

Additionally the resulting number will also be divisible by factors of 9 which is 1 and 3 in this case.

Hence the number obtained will be divisble by 3 and 9.

### Question 6

Which of the following is not divisible by 2?

123123

34567896

2345678234

18322

#### SOLUTION

Solution :A

All the even numbers and the numbers ending with 0 are divisible by 2.

Among the given options, 123123 has its last digit as 3 which makes it an odd number. Hence, this number is not divisible by 2.

### Question 7

Which of the following numbers is not divisible by 5?

12345

1234560

1232

1230

#### SOLUTION

Solution :C

The numbers 12345, 1230 and 1234560 have last digits as 5, 0 and 0 respectively. Hence, all three numbers are divisible by 5.

On the other hand, 1232 ends with a number other than 0 or 5. Hence, it is not divisible by 5.

### Question 8

Write 908 in expanded form.

(102×9)+(101×0)+(100×8)

(103×9)+(101×0)+(100×8)

(103×9)+(102×0)+(101×8)

(102×9)+(101×8)

#### SOLUTION

Solution :A

908 can be written as

(100×9)+(10×0)+(1×8)

= (102×9)+(101×0)+(100×8)

### Question 9

The usual form of 100×7+10×5+6 is

#### SOLUTION

Solution :100×7+10×5+6=700+50+6=756

### Question 10

Find the values of A and B 1 2 A+ 6 A B A 0 9

A = 8, B = 1

A = 1, B = 8

A = 9, B = 3

A = 3, B = 9

#### SOLUTION

Solution :A

We know that A + 2 is a two digit number with unit digit as 0. So, A = 8

A + B is a single digit number and is equal to 9.

A + B = 9

B = 9 - A

B = 9 - 8 = 1

Therefore, A = 8 and B = 1

1 2 8+ 6 8 1 8 0 9