Free Playing with Numbers 02 Practice Test - 8th Grade 

Question 1

123 can be expressed as:

A.

10×1+10×2+10×3
 

B.

100×1+100×2+100×3
 

C.

100×1+10×2+1×3
 

D.

1×1+1×2+1×3

SOLUTION

Solution : C

Let the number be abc.

Then the general form of abc will be:

100×a+10×b+1×c

Thus, 123=(100×1)+(10×2)+(1×3)

Question 2

1000×2+100×3+10×2+1×4  is expressed as:

A.

4232

B.

1111

C.

2024

D.

2324

SOLUTION

Solution : D

1000×2+100×3+10×2+1×4=2000+300+20+4=2324.

Question 3

Aarush asked Soham to take any 2 or 3 digit number and reverse the digits, then subtract the larger number from the smaller number. Soham did the same and concluded that the result is exactly divisible by 9. What Soham concluded is:

A.

True

B.

False

SOLUTION

Solution : A

Let's consider a two digit number ab.Its general form will be 10×a+1×b.
If we reverse the digits, its general form will be 10×b+1×a. If we subtract both of the numbers, we get:
 (10×a+1×b)(10×b+1×a)
= (10a + b) -  (10b + a)
= 10a  + b  -10b - a
= 9a - 9b
= 9 (a-b)
Observe that the result is a multiple of 9 and hence is exactly divisible by 9.
Similarly, for 3 digit numbers, the difference of the number and number obtained by reversing the digits (if abc is the number) will be given by:
(100×a+10×b+c)(100×c+10×b+a)
=100a + 10b + c - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99 (a-c)
= 9×11(ac) 
which is also divisible by 9.

Note that this happens with any 2 or 3 number.

Question 4

When Rashi ordered pizza and coke, the bill had the digits A and B not visible clearly.

Pizza           A12Coke       + 31B                     525

Using the information visible on her bill, find the digits A and B respectively.

A.

A = 2, B = 1

B.

A = 3, B = 5

C.

A = 1, B = 3

D.

A = 2, B = 3

SOLUTION

Solution : D

In the tens place, the numbers being added are 1 and 1. The sum given in the tens place is 2, which means there is no carry over throughout the summation.

A + 3 = 5 i.e A = 5 - 3 = 2  and

2 + B = 5 i.e B = 5 - 2 = 3

 

Pizza           212Coke       + 313                     525

Question 5

Soham borrowed money from his mother for  buying pens. He bought 7 pens and the price of each pen is a two digit number, 1A. If the total amount spent by him is 10A, then the value of A is 

___

SOLUTION

Solution :

Price of one Pen =1×10+A

Number of Pens = 7

Total price =(10+A)×7

(10+A)×7=(100+A)

70 + 7A = 100 + A

6A = 30

A = 5

Question 6

By which of the following numbers is 392 divisible?

A.

9

B. 3
C. 2
D. 4

SOLUTION

Solution : C and D

A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9 respectively.

Here, sum of digits = 3 + 9 + 2 = 14 which is not divisible by either 3 or 9. Hence, the number is not divisible by 3 and 9.

Being an even number, the given number is divisible by 2.

A number is divisible by 4 if the number formed by the last two digits are divisible by 4.
In 392, 92 which is divisible by 4. Hence, the number is also divisible by 4.

Question 7

198 is divisible by:

A.

11

B. 5
C. 8
D. 6

SOLUTION

Solution : A and D

A number is divisible by 11 if the difference between the sum of digits at its odd places and that of the digits at the even places is divisible by 11.
(1 + 8) - 9 = 0 which is divisible by 11.

198 is even number and thus, divisible by 2.
The sum of digits 1 + 9 + 8 = 18 which is divisible by 3.
So, 198 is also divisible by 3.
Since, the number is divisible by both 2 and 3, it is also divisible by 6.

A number to be divisible by 5 must end in 0 or 5. Hence, 198 is not divisible by 5.

A number is divisible by 8 if the number formed by the last 3 digits of the number are divisible by 8, hence 198 is not divisible by 8.

Question 8

Soham took a 3 digit number and formed two other 3 digit numbers using the digits of the original number. He then added the three resulting numbers and divided their sum by 37 . He concluded that the result was exactly divisible by 37, no matter what number he chose. Is Soham's statement true or false?

A.

True

B.

False

SOLUTION

Solution : A

Let the number be abc.

abc = 100a + 10b + c. 

By rearranging the digits, let us say the two other numbers formed are cab and bca.

Expressing these two numbers in the expanded form,

cab = 100c + 10a + b

bca = 100b + 10c + a

abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10a + a + 10b + b + 100b + c + 100c + 10c
= 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3 × (a + b + c), since 111 is the product of 37 and 3.

Hence the sum is divisible by 37.

Question 9

Soham got a call from Manish who was not very clear with the concepts of divisibility. Manish then asked, "Is 244 divisible by 2?” . Soham replied yes. Is Soham's reply true or false?

A.

True

B.

False

SOLUTION

Solution : A

If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

For the number 244 in the ones place we find 4 which is divisible by 2. Therefore we can say that the number 244 is divisible by 2.

Question 10

Find Q in the addition
   3 1 Q+ 1 Q 3   5 0 1

A.

1

B.

4

C.

5

D.

8

SOLUTION

Solution : D

Study the addition in the ones column:
Q + 3 = 1, a number whose ones digit is 1.

If Q = 8, then 8 + 3 = 11.
So, the puzzle can be solved as shown below:

   3 1 8+ 1 8 3   5 0 1
Hence, Q = 8