### Playing with Numbers - Worksheets

CBSE Worksheet-1
Class 08 - Mathematics (Playing with Numbers)

General Instructions: All questions are compulsory. Q.1 to Q.2 carries one mark each. Q.3 to Q.7 carries two marks each. Q.8 and Q.9 carries three marks each. Q.10 to Q.12 carries four marks each.

1. Find the values of letters:
$\begin{array}{c}\underset{_}{\begin{array}{ccc}\begin{array}{c}\\ +\end{array}& \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}A\\ 5\end{array}\end{array}}\\ \begin{array}{ccc}& 5& 7\end{array}\end{array}$
2. Write 129 in generalized form.
3. State True and False:
1. A two digit number is always divisible by 2 if any of the two digits are even.
2. A two digit number ab is always divisible by 2 if b is even.
3. A three digit number abc is divisible by 5 if it is even
4. A three digit number abc is divisible by 5 if c is 0 or 5
4. Fill in the blanks :
1. If N ÷ 5 leaves remainder 0 then unit digit of N is ............
2. If N ÷ 2 leaves no remainder then N is an ......... number
3. If N ÷ 2 leaves remainder 1 then N is an .......... number
4. If N ÷3 leaves remainder 0 then least possible sum of digits of N is ..........
5. Match the following general form with usual form :
General formUsual form
1. 1000 + 4a. 1040
2. 1000 + 40b. 421
3. $3×100+3$c. 1004
4. 400 + 20 + 1d. 303
6. Take a 3 digit number 234. Reverse the order subtract smaller number from the larger number divide the answer by 99. Is there any remainder.
7. If 24x is a multiple of 3, where x is a digit, what is the value of x ?
8. Give five example of numbers, each one of which is divisible by 2 but not divisible by 4.
9. Give five example of number, each one of which is divisible by 4 but not divisible by 8.
10. Without performing actual division, find remainder when 430346 is divided by 9.
11. In a two digit number the digit in the one's place is three times the digit in the ten's place and the sum of the digits is equal to 8. What is the number?
12. A and B play a game with numbers. A asks B to choose a two digit number then to reverse the digits of the number. A askes B to add them up and divide. The number by 11. A guarantees there will be no remainder. What was logic behind it?
CBSE Worksheet-1
Class 08 - Mathematics (Playing with Numbers)
Solution

1. Since A + 5 = 7
A = 7 – 5 = 2,
2. 129 = 100 + 20 + 9
= 100 × 1 + 10 × 2 + 9.
1. False
2. True
3. False
4. True
1. 0 or 5
2. Even
3. Odd
4. 3
1. -c
2. -a
3. -d
4. -b
3. Chosen number = 234
Revised number = 432
On subtracting the smaller number from the larger one, we get = 432 – 234 = 198
Finally, 198 ÷ 99 = 2, with no remainder.
4. Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers; 0, 3, 6, 9, 12, 15, 18, ......... But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.
1. 31202
2. 16310
3. 13618
4. 6798
5. 2314
1. 43012
2. 23116
3. 63604
4. 313140
5. Here sum of digits 4 + 3 + 0 + 3 + 4 + 6 is 20
When 20 is divide by 9 then remainder will be 2.
6. Let ten's digit be 'a' then unit digit be '3a'
Number = $10×a+3a$
= 10a + 3a
= 13a.
According to the question
a + 3a = 8
4a = 8
a = 2
$\therefore$ Number $=13×2$
= 26
7. Let the number be ab
reverse number be ba
According to question
ab + ba = 11(a + b)
Hence by guarantees there will be no remainder when it is divisible by 11