# Free Playing with Numbers 01 Practice Test - 8th Grade

990 + 9999 is divisible by

A.

3

B.

9

C.

11

D.

7

#### SOLUTION

Solution : A, B, and C

990 and 9999 are both divisible by 99. Hence, their sum is also divisible by 99.

99 is divisible by 9 and 11. 9 is divisible by 3. Hence, the sum of 990 and 9999 is divisible by 9, 11 as well as 3.

(f × 100) + (d × 10) + e is written as

A.

def

B.

efd

C.

fde

D.

edf

#### SOLUTION

Solution : C

(f × 100) + (d × 10) + e is written as fde which is a three digit number.

For example, (2 × 100) + (3 × 10) + 1 is the expanded form of  231.

Find A in the addition, given that B is a natural number.

A+A+AB A

A.

1

B.

5

C.

3

D.

2

#### SOLUTION

Solution : B

The sum of three A's is such a number that has A in its ones place.

Therefore, the sum of two A's must be a number whose ones digit is 0.

This happens only for A = 0 and A = 5

If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0.

Given that B is a natural number, therefore B cannot be equal to 0. So we don't consider this possibility.

Hence, A = 5   5+5+51 5

Therefore, A= 5 and B= 1

Find the digit A
B A×B 3   57A

A.

2

B.

3

C.

5

D.

4

#### SOLUTION

Solution : C

Ones digit of 3 × A is A. So, it must be either A = 0 or A = 5.

Now, if B = 1, then BA × B3 would at most be equal to 15 × 13; (Taking A= 5 to find the maximum value of the product) that is, it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.

If B = 3, then BA × B3 would be at least 30 × 33; (Taking A = 0 to find the minimum value of the product) that is 990. But 57A is less than 990. So, B cannot be equal to 3.

Putting these two facts together, we get B = 2.
So, the multiplication is either 20 × 23, or
25 × 23.

The first possibility fails, since
20 × 23 = 460. But, the second one works out correctly, since
25 × 23 = 575. So the answer is A = 5 and B = 2.
2 5×2 3   575

How many numbers are divisible by 2 from 1 to 100?

A.

23

B.

49

C.

50

D.

51

#### SOLUTION

Solution : C

From 1 to 4 there are 2 numbers divisible by 2, which are 2 and 4. From 1 to 10, there are 5 numbers, which are 2, 4, 6, 8, 10. Hence, the number of numbers divisible by 2 is half the number up to which we count. Half of 100 is 50. Hence 50 is the number of numbers divisible by 2 from 1 to 100.

I take a 3 digit number with distinct digits. I can get 6 different 3 digit numbers by rearranging the digits of this number.

A.

True

B.

False

#### SOLUTION

Solution : A

Let the 3 digit number be abc. On rearranging the digits I get, abc, bac, cba, cab, acb, bca. Hence I will get 6  three digit numbers.

If I add a 2 digit number and the number formed by reversing the digits, then the sum is surely divisible by

___
A.

2

B.

23

C.

11

D.

3

#### SOLUTION

Solution : C

Let a be the tens place digit and b be the units place digit, then  the original number is 10 x a + 1 x b.The new number after reversing the digits is 10 x b + 1 x b. Therefore the sum will be:
(10a + b) + (10b + a) = 11a + 11b

= 11 (a + b)
Which is divisible by 11.

If dividing the 3 digit number "3BA" by 10 gives 34 without leaving any remainder, the values of B and A are:

A.

B = 4

B.

B = 3

C.

A = 5

D.

A = 0

#### SOLUTION

Solution : A and D

Given: 3BA10=34
Since there is no remainder, 3BA is completely divisible by 10.
Now we know that all numbers exactly divisible by 10 will have 0 in the one's place.
A = 0
Now 3B010=34

So, 3B = 34
B = 4

Therefore, the number is 340.

Rashi asked Aarush to take a two digit number and then reverse the digits. After that, she asked him to add the first number and the new number, then  divide it by 11. She asked him do you get a  remainder ? Aarush's reply will be -

A.

Yes

B. No

#### SOLUTION

Solution : B

Let's take a two digit number 'ab'. Its general form is 10 × a + 1 × b. When we  reverse the digits,  the new number will be of the form  10 × b + 1 × a.

If we add both the numbers we get
(10
× a + 1 × b) + (10 × b + 1 × a)
=10a + b + 10b + a
= 11(a+b)

Hence when divided by 11 it will not leave any remainder.

What is the remainder when 1591 is divided by 1000?

A.

91

B.

0

C.

591

D.

1

#### SOLUTION

Solution : C

On dividing 1591 by 1000, the remainder is the last 3 digits. Hence it is 591.