Kseeb Class 6 Mathematics Chapter 3 Playing With Numbers Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 3 Playing With Numbers. Here students can easily find all the solutions for Playing With Numbers Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.
Playing With Numbers Exercise 3.1 Solutions :
(1) Write all the factors of the following numbers :
(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36
Solution :
(a) The factors of 24
24 = 1 × 24, 2 × 12, 3 × 8, 4 × 6
The factors of 24 is 1, 2, 3,4,6,12,24
(b) The factors of 15
15 = 1 × 15, 3 × 5
The factors of 15 is 1, 3,5, 15
(c) The factors of 21
21 = 1 × 21, 3 × 7
The factors of 21 is 1, 3, 7, 21
(d) The factors of 21
27 = 1 × 27, 3 × 9
The factors of 21 is 1, 3, 9, 27
(e) The factors of 12
12 = 1 × 12, 2 × 6, 3 × 4
The factors of 12 is 1, 2, 3, 4, 6, 12
(f) The factors of 20
20 = 1 × 20, 2 × 10, 4 × 5
The factors of 24 is 1, 2, 4, 5, 10, 20
(g) The factors of 18
18 = 1 × 18, 2 × 9, 3 × 6
The factors of 12 is 1, 2, 3, 6, 9,18
(h) The factors of 23
23 = 1 × 23
The factors of 23 is 1,23
(i) The factors of 36
36 = 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6
The factors of 24 is 1, 2, 3,4,6,9,12,36
(2) Write first five multiples of :
(a) 5
(b) 8
(c) 9
Solution :
(a) First five multiples of 5 = 5 × 1, 5 × 2, 5×3, 5× 4, 5×5
First five multiples of 5 is 5, 10, 15, 20, 25
(b) First five multiples of 8 = 8 × 1, 8 × 2, 8×3, 8× 4, 8×5
First five multiples of 8 is 8, 16, 24, 32, 40
(c) First five multiples of 9 = 9 × 1, 9 × 2, 9×3, 9× 4, 9×5
First five multiples of 9 is 9, 18, 27, 36, 45
(4) Find all the multiples of 9 upto 100.
Solution :
Multiples of 9 = 9 × 1, 9 × 2, 9×3, 9× 4, 9×5, 9×6 , 9 ×7 , 9 × 8, 9 × 9, 9 × 10, 9 × 11.
multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99
Playing With Numbers Exercise 3.2 Solutions :
(1) What is the sum of any two (a) Odd numbers? (b) Even numbers?
Solution :
(a) Sum of any two Odd numbers is even number.
(b) Sum of any two Even numbers is even number.
(2) State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution :
(a) False, The sum of three odd numbers is odd.
(b) True
(c) True
(d) False, If an even number is divided by 2, the quotient is always even.
(e) False
(f) False
(g) False, Sum of two prime numbers is always odd.
(h) True
(i) False, All even numbers are not composite numbers.
(j) True.
(3) The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution :
17 and 71, 37 and 73, 79 and 97 are the pairs of prime numbers.
(4) Write down separately the prime and composite numbers less than 20.
Solution :
Primer number mean the number which is divided by only 1 and itself whereas Composite number is the number which is divided by other number except 1 and itself.
Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18
(5) What is the greatest prime number between 1 and 10?
Solution :
Prime number between 1 and 10 are following
2, 3, 5, 7
So, 7 is the greatest prime number between 1 and 10
(6) Express the following as the sum of two odd primes.
(a) 44
(b) 36
(c) 24
(d) 18
Solution :
(a) 44 = 3 + 41 (13 + 31)
(b) 36 = 5 + 31 (17 + 19)
(c) 24 = 5 + 19 (11 + 13)
(d) 18 = 5 + 13 (7 + 11)
(7) Give three pairs of prime numbers whose difference is 2.
Solution :
3, 5; 5, 7 ; 11, 13 ; 41,43 are three pairs of prime numbers whose difference is 2.
(8) Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26
Solution :
(a) 23, (c) 37 are prime numbers.
(9) Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution :
The numbers are – 90, 91, 92, 93, 95, 96
(10) Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution :
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23 or 11+ 13 +7
(c) 53 = 11 + 13 + 29
(d) 61 = 7 + 13 + 41
(11) Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
Solution :
The five pairs of prime numbers are –
(i) 2, 13)
(ii) (3, 17)
(iii) (3, 7)
(iv) (2, 3)
(v) (13, 17)
(12) Fill in the blanks :
(a) A number which has only two factors is called a ___
(b) A number which has more than two factors is called a ___
(c) 1 is neither ___ nor ___
(d) The smallest prime number is ___
(e) The smallest composite number is ___
(f) The smallest even number is ___
Solution :
(a) Prime number
(b) Composite number
(c) Prime number nor Composite number
(d) 2
(e) 4
(f) 2
Playing With Numbers Exercise 3.3 Solutions :
(2) Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:
(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150
Solution :
(a) 572
divisibility tests of 4 =number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 =number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
72 is the last number which is divisible by 4
572 is divisible by 4.
(b) 726352
Divisibility tests of 4 =number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 =number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
52 is the last number which is divisible by 4
352 is the last number which is divisible by 8.
726352 is divisible by 4 and 8
(c) 5500
Divisibility tests of 4 =number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 = number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
00 is the last number which is divisible by 4
5500 is divisible by 4
(d) 6000
Divisibility tests of 4 = number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 = number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
00 is the last number which is divisible by 4
000 the last number which is divisible by 8
6000 is divisible by 4 and 8
(e) 12159
12159 is not divisible by 4 or 8
(f) 14560
Divisibility tests of 4 = number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 =number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
60 is the last number which is divisible by 4
560 is the last number which is divisible by 8.
14560 is divisible by 4 and 8
(g) 21084
Divisibility tests of 4 = number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 = number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
84 is the last number which is divisible by 4
21084 is divisible by 4
(h) 31795072
Divisibility tests of 4 = number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 = number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
72 is the last number which is divisible by 4
072 is the last number which is divisible by 8
31795072 is divisible by 4 and 8
(i) 1700
Divisibility tests of 4 = number contain last two digit 0 or last two number are divisible by 4 then the number is divisible by 4.
Divisibility tests of 8 = number contain last Three digit 0 or last three number are divisible by 8 then the number is divisible by 8.
00 is the last number which is divisible by 4
1700 is divisible by 4
(j) 2150
2150 is not divisible by 4 or 8
(3) Using divisibility tests, determine which of following numbers are divisible by 6:
(a) 297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852
Solution :
(a) 297144
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
297144 = this is even number and divisible by 3.
297144 is divisible by 6
(b) 1258
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
1258 = this is even number but not divisible by 3.
1258 is not divisible by 6
(c) 4335
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
4335 = this is not even number.
4335 is not divisible by 6
(d) 61233
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
61233 = this is not even number
61233 is not divisible by 6
(e) 901352
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
901352 = this is even number but not divisible by 3.
901352 is not divisible by 6
(f) 438750
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
438750 = this is even number and divisible by 3.
438750 is divisible by 6
(g) 1790184
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
1790184 = this is even number and divisible by 3.
1790184 is divisible by 6
(h) 12583
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
12583 = this is not even number
12583 is not divisible by 6
(i) 639210
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
639210 = this is even number and divisible by 3.
639210 is divisible by 6
(j) 17852
Divisibility test 6 = the numbers which are even and divisible by 3 are divisible by 6.
17852 = this is even number but not divisible by 3.
17852 is not divisible by 6
(4) Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153
Solution :
(a) 5445
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 5 + 4 = 9 and 4 + 5 = 9
Subtracting sum = 9 -9 = 0
5445 is divisible by 11.
(b) 10824
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 4 + 8 +1 = 13 and 2 + 0 = 2
Subtracting sum = 13 -2 = 11
10824 is divisible by 11.
(c) 7138965
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 5 + 9 +3 +7 = 24 and 6 + 8 + 1 = 15
Subtracting sum = 24 -15 = 9
7138965 is not divisible by 11.
(d) 70169308
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 8 + 3 +6 +0 = 17 and 0 + 9 + 1 + 7 = 17
Subtracting sum = 17 -17 = 0
70169308 is divisible by 11.
(e) 10000001
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 1+ 0 +0+0 = 1 and 0 + 0 + 0 + 1 = 1
Subtracting sum = 1 -1 = 0
10000001 is divisible by 11.
(f) 901153
Divisibility test of 11 = we are adding alternate number and subtract the sum which we get and this result are 0 or multiple of 11 then the number is divisible by 11.
By adding alternate numbers 3+ 1 +0 = 4 and 5 + 1 + 9 = 15
Subtracting sum = 15 -4 = 11
901153 is divisible by 11.
(5) Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :
(a) __ 6724
(b) 4765 __ 2
Solution :
(a) __ 6724
Divisible by 3 = adding all numbers and sum is divisible by 3 then the number is divisible by 3.
Smallest digit = 2 + 6 + 7 + 2 + 4 =21 is divisible by 3.
Greatest digit = 8 + 6 + 7 + 2 + 4 =27 is divisible by 3
Smallest digit = 2
Greatest digit = 8
(b) 4765 __ 2
Divisible by 3 = adding all numbers and sum is divisible by 3 then the number is divisible by 3.
Smallest digit = 4 + 7 + 6 + 5 + 0 + 2 =24 is divisible by 3.
Greatest digit =4 + 7 + 6 + 5 + 9 + 2 = 33 is divisible by 3
Smallest digit = 0
Greatest digit = 9
(6) Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :
(a) 92 __ 389
(b) 8 __ 9484
Solution :
As we know that, A number is divisible by 11 if the sum of odd place and the sum of even place. And then their difference is divisible by 11.
Now, (a) 1___31
= 1 3 31
(b) 701___9308
= 701 6 9308
Playing With Numbers Exercise 3.4 Solutions :
(1) Find the common factors of :
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120
Solution :
(a) Factors of 20 = 1, 2,4,5,10,20
Factors of 28 = 1, 2,4,7,14,28
Common factors of 20 and 28 are 1, 2, and 4
(b) Factors of 15 = 1, 3, 5,15
Factors of 25 = 1,5,25
Common factors of 15 and 25 are 1,5.
(c) Factors of 35 = 1, 5,7,35
Factors of 40 = 1,2,4,5,10,20,40
Common factors of 35 and 40 are 1,5.
(d) Factors of 56 = 1, 2,4,7,8,14,28,56
Factors of 120 = 1,2,3,4,5,6,10,12,20,24,30,40
Common factors of 56 and 120 are 1, 2, 4, 8
(2) Find the common factors of :
(a) 4, 8 and 12
(b) 5, 15 and 25
Solution :
(a) Factors of 4 = 1, 2, 4
Factors of 8 = 1, 2,, 4,8
Factors of 12 = 1, 2,3,4,6
Common factors of 4, 8 and 12 are 1, 2, 4
(b) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5
Factors of 25 = 1, 5
Common factors of 5, 15 and 25 are 1,5.
(3) Find first three common multiples of :
(a) 6 and 8
(b) 12 and 18
Solution :
(a) 6 and 8
Multiples of 6 = 6, 12,18,24,30,36,42,48,54,60,66,72
Multiples of 8 = 8, 16, 24,32,40,48,56,64,72
First three common multiples of 6 and 8 are 24, 48, and 72
(b) 12 and 18
Multiples of 12 =12, 24, 36,48,60,72,84,96,18,120
Multiples of 18 =18, 36,54,72,90,108,126,144
First three common multiples of 12 and 18 are 36, 72, and 108
(4) Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution :
All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84, 96
(5) Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16
Solution :
(a) 18 and 35
Co-prime numbers are the numbers having only 1 as a common factor.
In 18 and 35 only 1 as a common factor.
18 and 35 are Co-prime numbers.
(b) 15 and 37
Co-prime numbers are the numbers having only 1 as a common factor.
In 15 and 37 only 1 as a common factor.
15 and 37 are Co-prime numbers.
(c) 30 and 415
Co-prime numbers are the numbers having only 1 as a common factor.
In 30 and 415 not a 1 as a common factor.
i.e. 30 and 415 also divisible by 5
30 and 415 are not Co-prime numbers.
(d) 17 and 68
Co-prime numbers are the numbers having only 1 as a common factor.
In 17 and 68 not a 1 as a common factor.
i.e. 17 and 68 also divisible by 17
17 and 68 are not Co-prime numbers.
(e) 216 and 215
Co-prime numbers are the numbers having only 1 as a common factor.
In 216 and 215 only 1 as a common factor.
216 and 215 are Co-prime numbers.
(f) 81 and 16
Co-prime numbers are the numbers having only 1 as a common factor.
In 81 and 16 only 1 as a common factor.
81 and 16 are Co-prime numbers.
(6) A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution :
A number is divisible by both 5 and 12. That number be always divisible by 60
12 × 5 = 60
(7) A number is divisible by 12. By what other numbers will that number be divisible?
Solution :
A number is divisible by 12
12 = 1 x 12 ;
12 = 2 x 6 ;
12 = 3 x 4
Therefore, 1, 2, 3, 4, 6 are the other number.
Playing With Numbers Exercise 3.5 Solutions :
(1) Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.a
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution :
(a) Given statement is – True
(b) Given statement is – True
(c) Given statement is – True
(d) Given statement is – True
(e) False, 9 and 10 are co-primes, but they both are composite.
(f) False, 12 is divisible by 4 but not by 8
(g) Given statement is- True
(I) False, if a number exactly divides the sum of two numbers, it not exactly divide the two numbers separately.
(2) Here are two different factor trees for 60. Write the missing numbers.
Solution :
Missing numbers for Tree (i) are –
Missing numbers for Tree (ii) are –
(3) Which factors are not included in the prime factorisation of a composite number?
Solution :
1 and the number itself are not included in the prime factorisation of a composite number.
(4) Write the greatest 4-digit number and express it in terms of its prime factors.
Solution :
Greatest 4-digit number = 9999
9999 = 3 × 3 × 11 × 101
(5) Write the smallest 5-digit number and express it in the form of its prime factors.
Solution :
Smallest 5-digit number =10000
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
(6) Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.
Solution :
All the prime factors of 1729
1729 = 7 × 13 × 19
The difference of two consecutive prime factors is 6.
(7) The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution :
Lets take 1, 2, 3 then 1 × 2 × 3 = 6, divisible by 6
2, 3, 4 then 2 × 3 × 4 = 24 divisible by 6
3, 4, 5 then 3 × 4 × 5 = 60, divisible by 6
4, 5, 6 then 4 × 5 × 6 = 120, divisible by 6
So, in every example we can see that the given statement is True.
(8) The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Solution :
Two consecutive odd numbers = 3, 5
Sum of two consecutive odd numbers = 3 + 5 = 8
8 is divisible by 4.
The sum of two consecutive odd numbers is divisible by 4. Verifies the statement
(9) In which of the following expressions, prime factorisation has been done?
(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9
Solution :
In (b) 56 = 7 × 2 × 2 × 2 and c) 70 = 2 × 5 × 7 prime factorisation has done.
(10) Determine if 25110 is divisible by 45.
Solution :
Divisibility test of 45 = the number follows divisibility test of 5 and 9 are divisible by 45.
Divisibility test of 5 = the number contain 0 or 5 at unit place.
Divisibility test of 9 = the sum of all the numbers present in number divisible by 9.
25110 contain 0 at unit place hence Divisible by 5.
Sum of numbers = 2 + 5+1+1+ 0 =9 divisible by 9.
25110 is divisible by 45
(11) 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.
Solution :
No.
This is Because Number 12 is divisible by both 4 and 6; but 12 is not divisible by 24.
(12) I am the smallest number, having four different prime factors. Can you find me?
Solution :
The smallest prime number is 2. Next is 3, 5, and 7.
Four consecutive prime numbers are 2,3,5,7
Four different prime factors = 2 × 3 × 5 × 7 = 210
The smallest number is 210.
Playing With Numbers Exercise 3.6 Solutions :
(1) Find the HCF of the following numbers :
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Solution :
(a) 18, 48
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
Factors of 18 = 1,2,3,6,9,18
Factors of 48 =1, 2, 3,4,6,8,12,24,48
∴ HCF of 18, 48 is 6
(b) 30, 42
Factors of 30 = 1, 2,3,5,6,10,15,30
Factors of 42 =1, 2, 3, 6,14,21,42
∴ HCF of 30, 42 is 6
(c) 18, 60
Factors of 18 = 1, 2, 3, 6,9,18
Factors of 60 =1, 2, 3,4,5, 6,10,15,30,60
∴ HCF of 18, 60 is 6
(d) 27, 63
Factors of 27 = 1, 3, 9, 27
Factors of 63 =1, 2, 3, 7, 9,
∴ HCF of 27, 63 is 9
(e) 36, 84
Factors of 36 = 1, 2, 3, 9, 12, 36
Factors of 84 =1, 2, 3, 4, 6, 12, 42
∴ HCF of 36, 84 is 12
(f) 34, 102
Factors of 34 = 1, 2, 17
Factors of 102 =1, 2, 3, 6, 17
∴ HCF of 34, 102 is 17
(g) 70, 105, 175
Factors of 70 = 1, 2, 5,7, 10,35,70
Factors of 105 =1, 3, 5, 35,105
Factors of 175 =1,5,35
∴ HCF of 70, 105, and 175 is 35
(h) 91, 112, 49
Factors of 91 = 1, 7, 13
Factors of 112 =1, 7, 14
Factors of 49 =1,7
∴ HCF of 91, 112, and 49 is 7
(i) 18, 54, 81
Factors of 18 = 1, 2, 9.
Factors of 54 =1, 2, 3, 6, 9.
Factors of 81 =1, 3,9.
∴ HCF of 18, 54, and 81 is 9
(j) 12, 45, 75
Factors of 12 = 1, 2, 3,4,6,12
Factors of 45 =1, 3, 5, 9, 45
Factors of 75 =1, 5,15,25.
∴ HCF of 12, 45 and 75 is 1
(2) What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) HCF of two consecutive numbers is 1.
(b) HCF of two consecutive Even numbers is 2.
(c) HCF of two consecutive Odd numbers is 1.
(3) HCF of co-prime numbers 4 and 15 was found as follows by factorisation : 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Solution :
No, because we know 1 is common factor of all numbers.
Thus HCF of 4 and 15 is 1
Playing With Numbers Exercise 3.7 Solutions :
(1) Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
Solution :
Renu purchases two bags of fertiliser of weights 75 kg and 69 kg.
Maximum value of weight which can measure the weight of the fertiliser exact number of times =
We have to find HCF of 75 kg and 69 kg
Factor of 75 = 1, 3, 5,15,25,75
Factor of 69 =1, 3,13,23,69
HCF of 75 kg and 69 kg is 3
The maximum value of weight is 3 kg.
(2) Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Solution :
Three boys step off together from the same spot
Their steps measure 63 cm, 70 cm and 77 cm respectively.
The minimum distance each should cover so that all can cover the distance in complete steps
We have to find LCM of 63 cm, 70 cm and 77 cm respectively.
Factor of 63 = 1, 3, 7, 9, 21, 63
Factor of 70 =1, 2, 5,10,35,70
Factor of 77 =1, 7,11,77
LCM of 63 cm, 70 cm and 77 cm is 6930 cm.
The minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.
(3) The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly
Solution :
The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively.
The longest tape which can measure the three dimensions of the room exactly =
We have to find HCF of 825 cm, 675 cm and 450 cm.
Factor of 825 = 1, 75
Factor of 675 =1, 75
Factor of 450 =1, 75
HCF of 825 cm, 675 cm and 450 cm is 75
The longest tape is 75 cm.
(4) Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution :
The smallest 3-digit number which is exactly divisible by 6, 8 and 12.
We have to find LCM of 6, 8 and 12.
Factor of 6 = 1, 2, 3, 6
Factor of 8 =1, 2, 4, 8
Factor of 12 =1, 2, 3, 4,6,12
LCM of 6, 8 and 12 is 120
(5) Determine the greatest 3-digit number exactly divisible by 8, 10 and 12
Solution :
The greatest 3-digit number exactly divisible by 8, 10 and 12.
We have to find LCM of 8, 10 and 12.
Factor of 8 = 1, 2, 4, 8
Factor of 10 = 1, 2,5,10
Factor of 12 = 1, 2, 3, 4,6,12
LCM of 8, 10 and 12 is 960
(6) The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
Solution :
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively.
They change simultaneously at 7 a.m.
Time will they change simultaneously again = we have to find LCM of 48 seconds, 72 seconds and 108 seconds.
Factor of 48 =1, 2, 3, 4, 6, 8, 12,16,24,48
Factor of 72 =1, 2,3,4,6,8,9,12,18,24,36,72
Factor of 108 =1, 2, 3, 4,6,12,18,36,54,108
LCM of 48 seconds, 72 seconds and 108 seconds is 432 seconds.
432 seconds = 7 minutes 12 second.
Time will they change simultaneously again is 7 minutes 12 seconds past 7 a.m.
(7) Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
Solution :
Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively.
Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
We have to find HCF of 403 litres, 434 litres and 465 litres.
Factor of 403 = 1, 31
Factor of 434 =1, 31
Factor of 465 =1, 31
HCF of 403 litres, 434 litres and 465 litres is 31 litres.
Maximum capacity of a container is 31 litres.
(8) Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution :
The least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
We have to find LCM of 6, 15 and 18 and adding remainder 5
Factor of 6 = 1 * 2*3
Factor of 15 = 1*3*5
Factor of 18 = 1*3*6
LCM of 6, 15 and 18 is 90.
Adding remainder 5 = 90 + 5 = 95
The least number which when divided by 6, 15 and 18 leave remainder 5 in each case is 95.
(11) Find the LCM of the following numbers in which one number is the factor of the other
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45
Solution :
(a) 5, 20
LCM of 5, 20
Factor of 5 = 1 × 5
Factor of 20 = 1 × 5 × 4
LCM of 5, 20 = 20
(b) 6, 18
LCM of 6, 18
Factor of 6 = 1 × 6
Factor of 18 = 1 × 6 × 3
LCM of 6, 18 = 18
(c) 12, 48
LCM of 12, 48 =
Factor of 12 = 1 × 12
Factor of 48 = 1 × 12 × 4
LCM of 12, 48 = 48
(d) 9, 45 what do you observe in the results obtained?
LCM of 9, 45
Factor of 9 = 1 × 9
Factor of 45 = 1 × 9 × 5
LCM of 9, 45 = 45
LCM is the highest number.
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