1Book Back Questions66 questions
Q.1The number of prime numbers between 11 and 60 is _______v
Q.2The numbers 29 and _______ are twin primes.v
Q.33753 is divisible by 9 and hence divisible by _______v
Q.4The number of distinct prime factors of the smallest 4 digit number is______v
Q.5The sum of distinct prime factors of 30 is ________v
Q.6The sum of any number of odd numbers is always even.v
Q.7Every natural number is either prime or composite.v
Q.8If a number is divisible by 6, then it must be divisible by 3.v
Q.916254 is divisible by 2, 3, 6, and 9.v
Q.10The number of distinct prime factors of 105 is 3.v
Q.11Write the smallest and the biggest two-digit prime number.v
Answer:Smallest two-digit prime number – 11
Biggest two-digit prime number – 97.
Q.12Write the smallest and the biggest three-digit composite number.v
Answer:Smallest three-digit composite number – 100
Biggest three-digit composite number – 999
Q.13The sum of any three odd natural number is odd. Justify this statement with an example.v
Answer:True.
1 + 3 + 5 = 9 (Odd)
Q.14The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pair exists up to 100.v
Answer:(17, 71), (37, 73) and (79, 97)
Q.15Your friend says that every odd number is prime. Give an example to prove him/her wrong.v
Answer:False. 15 is an odd number. But not prime.
Q.16Each of the composite numbers has at least three factors. Justify this statement with an example.v
Answer:Factors of 4 are 1, 2, 4
Q.17Find the dates of any month in a calendar which are divisible by both 2 and 3.v
Answer:Every month the dates 6, 12, 18, 24, and 30 (excluding February) are divisible by both 2 and 3.
Q.18I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. Who am I?v
Answer:Two-digit prime numbers with a sum of digits 10 are 19, 37, 73.
of these factors of 57 is 19. the number is 19.
Q.19Find the prime factorisation of each number by factor tree method and division method. (i) 60 (ii) 128 (iii) 144 (iv) 198 (v) 420 (vi) 999v
Answer:60 = 2 × 30 = 2 × 2 × 15
= 2 × 2 × 3 × 5128 = 2 × 2 × 2 × 2 × 2 ×2 × 2
128 = 2 × 64 = 2 × 2 × 32
= 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8
128 = 2 × 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2 × 2144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 2 × 72 = 2 × 2 × 36
= 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3198 = 2 × 3 × 3 × 11
198 = 2 × 99 = 2 × 3 × 33 = 2 × 3 × 3 × 11420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7999 = 3 × 333 = 3 × 3 × 111
= 3 × 3 × 3 × 37
Q.20If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.v
Answer:143 = 11 × 13 (11, 13) or (13, 11)
143 = 11 × 13
Q.21The difference between two successive odd numbers isv
Q.22The only even prime number isv
Q.23Which of the following numbers is not a prime?v
Q.24The sum of the factors of 27 isv
Q.25The factors of the number are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. What is the number?v
Q.26The prime factorisation of 60 is 2 × 2 × 3 × 5. Any other number which has the same prime factorisation as 60 isv
- A. 30
- B. 120
- C. 90
- D. Impossible
Q.27If the number 6354*97 is divisible by 9 then the value of * isv
Q.28The number 87846 is divisible byv
- A. 2 only
- B. 3 only
- C. 11 only
- D. all of these
Q.29The HCF of 45 and 75 is …….v
Q.30The HCF of two successive even numbers is …….v
Q.31If the LCM of 3 and 9 is 9, then their HCF is ………v
Q.32The LCM of 26, 39 and 52 is ……..v
Q.33The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ………v
Q.34The numbers 57 and 69 are co-primes.v
Q.35The HCF of 17 and 18 is 1.v
Q.36The LCM of two successive numbers is the product of the numbers.v
Q.37The LCM of two co-primes is the sum of the numbers.v
Q.38The HCF of two numbers is always a factor of their LCM.v
Q.39Find the HCF of each set of numbers using the prime factorisation method. (i) 18, 24 (ii) 51, 85 (iii) 61, 76 (iv) 84, 120 (v) 27, 45, 81 (vi) 45, 55, 95v
Answer:(i) 18, 24HCF = 2 × 3 = 6
(ii) 51, 85HCF = 17(iii) 61, 76
61 = 1 × 61
For 61, 76, the common factor is 1
HCF = 1
(iv) 84, 120HCF = 2 × 2 × 3 = 12
(v) 27, 45, 81HCF = 3 × 3 = 9
(vi) 45, 55, 95Factors of 55 = 5 × 11
Factors of 45 = 3 × 3 × 5
Factors of 95 = 5 × 19
HCF = 5
Q.40Find the LCM of each set of numbers using the prime factorisation method. (i) 6, 9 (ii) 8, 12 (iii) 10, 15 (iv) 14, 42 (v) 30, 40, 60 (vi) 15, 25, 75v
Answer:(i) 6, 9Factors of 6 = 2 × 3
Factors of 9 = 3 × 3
LCM of 6 and 9 = 3 × 2 × 3 = 18
(ii) 8, 12(iii) 10, 15Factors of 10 = 2 × 5
Factors of 15 = 3 × 5
LCM of 10 and 15 = 5 × 2 × 3 = 30(iv) 14, 42Factors of 14 = 2 × 7
Factors of 42 = 2 × 3 × 7
LCM of 14 and 42 = 7 × 2 × 3 = 42
(v) 30, 40, 60Factors of 30 = 2 × 3 × 5
Factors of 40 = 2 × 2 × 2 × 5
Factors of 60 = 2 × 2 × 3 × 5
LCM of 30, 40, 60 = 2 × 5 × 3 × 2 × 2 = 120
(vi) 15, 25, 75Factors of 15 = 3 × 5
Factors of 25 =5 × 5
Factors of 75 = 3 × 5 × 5
LCM of 15, 25, 75 = 5 × 3 × 5 = 75
Q.41Find the HCF and the LCM of the numbers 154, 198, 286v
Answer:HCFFactors of 154 = 2 × 7 × 11
Factors of 198 = 2 × 3 × 3 × 11
Factors of 286 = 2 × 13 × 11
HCF of 154, 198, 286 = 11 × 2 = 22
LCMLCM = 2 × 11 × 7 × 9 × 13
= 22 × 63 × 13
= 18018
Q.42What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?v
Answer:This is an HCF related problem. So, we need to find the HCF of 80, 100, 12080 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres
Q.43The traffic lights at three different road junctions change after every 40 seconds, 60 seconds, and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?v
Answer:This is an LCM related problem. So, we need to find the LCM of 40, 60, 72LCM of 40, 60 and 72
= 2 × 2 × 3 × 5 × 2 × 1 × 1 × 3
= 360 seconds
= 6 minutes
The traffic lights will change simultane¬ously again at 8 : 06 am.
Q.44The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible.v
Answer:Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs are possible.
Q.45The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.v
Answer:36x = 6 × 12 × 12
⇒ \(x=\frac{6 \times 12 \times 12}{36}=24\)Objective Type Questions
Q.46Which of the following pairs is co-prime?v
- A. 51, 63
- B. 52, 91
- C. 71, 81
- D. 81, 99
Q.47The greatest four-digit number which is exactly divisible by 8, 9, and 12 isv
- A. 9999
- B. 9996
- C. 9696
- D. 9936
Q.48The HCF of two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum isv
Q.49Which of the following cannot be the HCF of two numbers whose LCM is 120?v
Q.50Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number upto 16.v
Answer:Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13
Q.51Is 173 a prime? Why?v
Answer:Yes, 173 is a prime. Because it has only 1 and itself as factors.
Q.52For which of the numbers, from n = 2 to 8. Is 2n – 1 a prime?v
Answer:n = 2 ⇒ 2n – 1 = 2 × 2 – 1
= 4 – 1
= 3 (prime)
n = 3 ⇒ 2n – 1 = 2 × 3 – 1
= 6 – 1
= 5 (prime)
n = 4 ⇒ 2n – 1 = 2 × 4 – 1
= 8 – 1
= 7 (prime)
n = 5 ⇒ 2n – 1 = 2 × 5 – 1
= 10 – 1
= 9 (Not prime)
n = 6 ⇒ 2n – 1 = 2 × 6 – 1
= 12 – 1
= 11 (prime)
n = 7 ⇒ 2n – 1 = 2 × 7 – 1
= 14 – 1
= 13
n = 8 ⇒ 2n – 1 = 2 × 8 – 1
= 16 – 1
= 15 (Not prime)
Q.53Explain your answer with the reason for the following statements.v
- A. A number is divisible by 9 if it is divisible by 3.
- B. A number is divisible by 6 if it is divisible by 12.
Answer:(i) False, 42 is divisible by 3 but it is not divisible by 9
(ii) True, 36 is divisible by 12. Also divisible by 6.
Q.54Find A as required (i) The greatest 2 digit number 9 A is divisible by 2. (ii) The least number 567A is divisible by 3. (iii) The greatest 3 digit number 9A6 is divisible by 6. (iv) The number A08 is divisible by 4 and 9. (v) The number 225A85 is divisible by 11.v
Answer:(i) 98 A = 8
(ii) 5670 A = 0
(iii) 996 A = 9
(iv) 108 A = 1
(v) 225885 A = 8
Q.55Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.v
Answer:False 12 is divisible by both 4 and 6. But not divisible by 24
Q.56The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.v
Answer:True 3 + 5 = 8 is divisible by 4.
Q.57Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20cm, 3m 60 cm and 4 m.v
Answer:1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem. So, we need to find the HCF of 120,360 and 400.120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cmChallenge Problems
Q.58The sum of three prime numbers is 80. The difference between the two of them is 4. Find the numbers.v
Answer:Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 [∵ one number must be even]
Also their difference = 4 (given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41
Q.59Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.v
Answer:Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.
Q.60Find the smallest number which is exactly divisible by all the numbers from 1 to 9.v
Q.61The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.v
Answer:Yes. Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7
Q.62Malarvizhi, Karthiga, and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively, Assuming that they met each other on the 1st of October, when will all the three meet again?v
Answer:This is an LCM related problem. So, we need to find the LCM of 5, 6, and 10.LCM = 5 × 2 × 1 × 3 × 1
= 30
All the three will meet again once in 30 days.
Q.63In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?v
Answer:LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90, and 105.
Q.64The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?v
Answer:15 × 20 = 300
HCF of 15 and 20 is 5
The numbers are 15 and 20
Q.65Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help)v
Answer:564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.
Q.66Wilson, Mathan, and Guna can complete one round of a circular track in 10, 15, and 20 minutes respectively. If they start together from at the starting point of 7 am, at what time will they meet together again at the same starting point?v
Answer:This is an LCM related problem. So, we need to find the LCM of 10, 15, and 20.LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7.am + 60 minutes = 8 am.