Playing With Numbers(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

PLAYING WITH NUMBERS MCQs

Total Questions : 54 | Page 1 of 6 pages

Question 1. The usual form of

100 \times 7 + 10 \times 5 + 6

is
___

Discuss Question

100 \times 7 + 10 \times 5 + 6 = 700 + 50 + 6 = 756

Question 2. The number 152875 is divisible by

Discuss Question

Answer: Option C. -> 5
:
C
The sum of the digits of 152875 is 28 which is not a multiple of 3, 152875 is not divisible by 3.
28 isnot 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. 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?

True
False
8
6

Discuss Question

Answer: Option A. -> True
:
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

\times

\times

(a + b + c), since 111 is the product of 37 and 3.
Hence the sum is divisible by 37.

Question 4. 123 can be expressed as:

10×1+10×2+10×3
100×1+100×2+100×3
100×1+10×2+1×3
1×1+1×2+1×3

Discuss Question

Answer: Option C. -> 100×1+10×2+1×3
:
C

Let the number be abc.

Then the general form of abc will be:

100 \times a + 10 \times b + 1 \times c

Thus, 123 = (100 \times 1) + (10 \times 2) + (1 \times 3)

Question 5. How many numbers are divisible by 2 from 1 to 100?

Discuss Question

Answer: Option C. -> 50
:
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.

Question 6. 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
6
11

Discuss Question

Answer: Option A. -> True
:
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 7. If I divide 487 by 100 what is the remainder?

Discuss Question

Answer: Option D. -> 87
:
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.

If I Divide 487 By 100 What Is The Remainder?

Question 8. Find the values of A and B

\begin{matrix} 12 A + 6 A B A 09 \end{matrix}

A = 8, B = 1
A = 1, B = 8
A = 9, B = 3
A = 3, B = 9

Discuss Question

Answer: Option A. -> A = 8, B = 1
:
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

\begin{matrix} 128 + 681 809 \end{matrix}

Question 9. 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)

Discuss Question

Answer: Option A. -> (102×9)+(101×0)+(100×8)
:
A
908 can be written as

(100 \times 9) + (10 \times 0) + (1 \times 8)

(10^{2} \times 9) + (10^{1} \times 0) + (10^{0} \times 8)

Question 10. Which of the following numbers is not divisible by 5?

12345
1234560
1232
1230

Discuss Question

Answer: Option C. -> 1232
:
C
The numbers 12345, 1230 and 1234560 have last digits as 5, 0 and 0 respectively. Hence, all threenumbers 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.