Table of Contents
MCQ Questions for Class 8 Maths: Ch 16 Playing with Numbers
1. The generalised form of the number 33 is
(a) 10 × 3 + 3
(b) 10 × 3
(c) 3 + 3
(d) 3 × 3 + 3
► (a) 10 × 3 + 3
2. Which of the following statements is false?
(a) If a number is divisible by 8, it must be divisible by 4.
(b) If a number is divisible by both 9 and 10, it is divisible by 90.
(c) The sum of two consecutive odd numbers is always divisible by 4.
(d) If a number is not divisible by both 3 and 4, it is divisible by 12.
► (d) If a number is not divisible by both 3 and 4, it is divisible by 12.
3. Which of the given numbers is composite?
(a) 137
(b) 147
(c) 157
(d) 167
► (b) 147
4. If the division N ÷ 5 leaves a remainder of 0, what might be the one’s digit of N?
(a) 5
(b) Either 5 or 0
(c) 2
(d) 7
► (b) Either 5 or 0
5. Write in generalised form: 25
(a) 10 × 5 + 2
(b) 10 × 5 + 3
(c) 10 × 2 + 5
(d) 10 × 3 + 5
► (c) 10 × 2 + 5
6. Which of these numbers is divisible by 6?
(a) 5782
(b) 2666
(c) 6053
(d) 8964
► (d) 8964
7. By which of the following number 9042 is not divisible? 2, 3, 6, and 9
(a) 2
(b) 3
(c) 9
(d) 6
► (c) 9
8. Write in generalised form: 85
(a) 10 × 5 + 8
(b) 10 × 8 + 5
(c) 10 × 5 + 3
(d) 10 × 3 + 5
► (b) 10 × 8 + 5
9. What value should be given to * so that the number 653∗47 is divisible by 11?
(a) 1
(b) 6
(c) 2
(d) 9
► (a) 1
10. N is a 5-digit number divisible by 5. If N is bigger than 10000 and smaller than 10010, what is the value of N?
(a) 10000
(b) 10010
(c) 10005
(d) 10001
► (c) 10005
11. When is a number always divisible by 90?
(a) If it is divisible by both 2 and 45.
(b) If it is not divisible by both 5 and 18.
(c) If it is not divisible by both 9 and 10.
(d) If it is divisible by 3 and 20.
► (a) If it is divisible by both 2 and 45.
12. By which of the following numbers is 477 not divisible?
(a) 3
(b) 7
(c) 53
(d) 9
► (b) 7
13. Write in the usual form: 10×5 + 6
(a) 65
(b) 54
(c) 56
(d) 25
► (c) 56
14. Identify the missing digit in the number 234,4_6, if the number is divisible by 4.
(a) 2
(b) 6
(c) 4
(d) 5
► (d) 5
15. Write in the usual form: 100 × 7 + 10 × 1 + 8
(a) 871
(b) 718
(c) 178
(d) 781
► (b) 718
16. Which of the following is not prime?
(a) 107
(b) 127
(c) 153
(d) 197
► (c) 153
17. By which of the following number 168 is divisible?
(a) 5
(b) 10
(c) 9
(d) 2
► (d) 2
18. 32 + m is a prime number. What is the least value of ‘m’?
(a) 3
(b) 5
(c) 6
(d) 4
► (b) 5
19. Which of the following is divisible by 12?
(a) 284382
(b) 624876
(c) 926248
(d) 746174
► (b) 624876
20. If a number is divisible by 9, it is also divisible by which number?
(a) 3
(b) 6
(c) 2
(d) 4
► (a) 3
Playing with Numbers Class 8 Extra important Questions Very Short Answer Type
Question 1.
Write the following numbers in generalised form.
(a) ab
(b) 85
(c) 132
(d) 1000
Solution:
(a) ab = 10 × a + 1 × b = 10a + b
(b) 85 = 10 × 8 + 1 × 5 = 10 × 8 + 5
(c) 132 = 100 × 1 + 10 × 3 + 1 × 2 = 100 × 1 + 10 × 3 + 2
(d) 1000 = 1000 × 1
Question 2.
Write the following in usual form.
(a) 3 × 100 + 0 × 10 + 6
(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1
Solution:
(a) 3 × 100 + 0 × 10 + 6 = 300 + 0 + 6 = 306
(b) 5 × 1000 + 3 × 100 + 2 × 10 + 1 = 5000 + 300 + 20 + 1 = 5321
Question 3.
Which of the following numbers are divisible by 3?
(i) 106
(ii) 726
(iii) 915
(iv) 1008
Solution:
(i) Sum of the digits of 106 = 1 + 0 + 6 = 7 which is not divisible by 3.
Hence 106 is not divisible by 3.
(ii) Sum of the digits of 726 = 7 + 2 + 6 = 15 which is divisible by 3.
Hence 726 is divisible by 3.
(iii) Sum of the digits of 915 = 9 + 1 + 5 = 15 which is divisible by 3.
Hence 915 is divisible by 3.
(iv) Sum of the digits of 1008 = 1 + 0 + 0 + 8 = 9 which is divisible by 3.
Hence 1008 is divisible by 3.
Question 4.
Prove that the sum of the given numbers and the numbers obtained by reversing their digits is divisible by 11.
(a) 89
(b) ab
(c) 69
(d) 54
Solution:
(a) Given number = 89
Number obtained by reversing the order of digits = 98
Sum = 89 + 98 = 187 ÷ 11 = 17
Hence, the required number is 11.
(b) Given number = ab = 10a + b
Number obtained by reversing the digits = 10b + a
Sum = (10a + b) + (10b + a)
= 10a + b + 10b + a
= 11a + 11b
= 11(a + b) ÷ 11
= a + b
(c) Given number = 69
Number obtained by reversing the digits = 96
Sum = 69 + 96 = 165 ÷ 11 = 15
Hence, the required number is 11.
(d) Given number = 54
Number obtained by reversing the digits = 45
Sum = 54 + 45 = 99 ÷ 11 = 9
Hence, the required number is 11.
Question 5.
Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9.
(i) 59
(ii) xy
(iii) xyz
(iv) 203
Solution:
(i) Given number = 59
Number obtained by reversing the digits = 95
Difference = 95 – 59 = 36 ÷ 9 = 4
Hence, the required number is 9.
(ii) Given number = xy = 10x + y
Number obtained by reversing the digits = 10y + x
Difference = (10x + y) – (10y + x)
= 10x + y – 10y – x
= 9x – 9y
= 9(x – y) ÷ 9
= x – y
Hence, the required number is 9.
(iii) Given number = xyz = 100x + 10y + z
Number obtained by reversing the digits = 100z + 10y + x
Difference = (100x + 10y + z) – (100z + 10y + x)
= 100x + 10y + z – 100z – 10y – x
= 99x – 99z
= 99(x – z)
= 99(x – 2) ÷ 9
= 11 (x – z)
Hence, the required number is 9.
(iv) Given number = 203
Number obtained by reversing the digits = 302
Difference = 302 – 203 = 99 ÷ 9 = 11
Hence, the required number is 9.
Question 6.
If a, b, c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37.
Solution:
We have abc + cab + bca
abc = 100a + 10b + c
cab = 100c + 10a + b
bca = 100b + 10c + a
Adding abc + cab + bca = 111a +111b + 111c
= 111 (a + b+ c)
= 37 × 3 (a + b + c) which is a multiple of 37.
Hence proved.
Playing with Numbers Class 8 Extra Questions Short Answer Type
Question 7.
Complete the magic square given below so that the sum of the numbers in each row or in each column or along each diagonal is 15.
Solution:
(i) A = 15 – (8 + 1) = 15 – 9 = 6
(ii) F = 15 – (8 + 5) = 15 – 13 = 2
(iii) C = 15 – (A + F) = 15 – (6 + 2) = 15 – 8 = 7
(iv) E = 15 – (1 + 5) = 15 – 6 = 9
(v) D = 15 – (E + F) = 15 – (9 + 2) = 15 – 11 = 4
(vi) B = 15 – (8 + 4) = 15 – 12 = 3
Hence the required square is
Question 8.
Find the values of P and Q from the given addition problem
Solution:
Here, 3 + Q = 7
⇒ Q = 7 – 3 = 4
Now taking second column, we get
4 + 7 = 11 i.e. 1 is carried over to third column
⇒ 1 + P + 2 = 9
⇒ 3 + P = 9
P = 9 – 3 = 6
Hence the value of P = 6 and Q = 4
Question 9.
Find the values of p, q and r in the following multiplication problem.
Solution:
6 × 4 = 24, Here 2 is carried over second column
⇒ 6 × p + 2 – 3 × 10 = 2 [∵ 21 – 3 × 6 = 3]
⇒ 6p – 30 = 0
⇒ p = 5
Now the multiplication problem becomes,
Here 2 + r = 4
⇒ r = 2
q × 354 = 1062
⇒ q = 3
Hence, p = 5, q = 3, r = 2
Question 10.
Observe the following patterns:
1 × 9 – 1 = 8
21 × 9 – 1 = 188
321 × 9 – 1 = 2888
4321 × 9 – 1 = 38888
Find the value of 87654321 × 9 – 1
Solution:
From the pattern, we observe that there are as many eights in the result as the first digit from the right which is to be multiplied by 9 and reduced by 1.
87654321 × 9 – 1 = 788888888
Playing with Numbers Class 8 Extra Questions High Order Thinking Skills (HOTS) Type
Question 11.
Complete the cross number puzzle with the given column.
Solution:
Hence the complete square is
Question 12.
The product of two 2-digit numbers is 1431. The product of their tens digits is 10 and the product of their units digits is 21. Find the numbers. Solution:
Let the required two 2-digit numbers be 10a + b and 10p + q as per the condition, we have
a × p = 10 and b × q = 21
a = 2 and p = 5 or a = 5 and p = 2
Similarly b × q = 21
b = 3 and q = 7 or b = 7 and q = 3
10p + q = 57 or 10p + q = 53
and 10a + b = 23 or 10a + b = 27
Since the units digit of product 1431 is 1.
Numbers are 57 and 23 or 53 and 27.
Now 57 × 23 = 1311 and 53 × 27 = 1431 which is given.
Hence, the required numbers are 53 and 27.
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