RS Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 2 - Factors And Multiples

Question 1:

Question 2:

Factor: A factor of a number is an exact divisor of that number.
Multiple: A multiple of a number is a number obtained by multiplying it by a natural number.

Example 1: We know that 15 = 1 × 15 and 15 = 3 × 5

∴ 1, 3, 5 and 15 are the factors of 15
In other words, we can say that 15 is a multiple of 1, 3, 5 and 15.
 
Example 2: We know that 8 = 8 × 1, 8 = 2 × 4 and 8 =  4 × 2

∴ 1, 2, 4 and 8 are the factors of 8.
In other words, we can say that 8 is a multiple of 1, 2, 4 and 8.
 
Example 3: We know that 30 = 30 × 1, 30 = 5 × 6 and 30 = 6 × 5

∴ 1, 5, 6 and 30 are factors of 30.
In other words, we can say that 30 is a multiple of 1, 5, 6 and 30.
 
Example 4: We know that 20 = 20 × 1, 20 = 4 × 5 and 20 = 5 × 4

∴ 1, 4, 5 and 20 are factors of 20.
In other words, we can say that 20 is a multiple of 1, 4, 5 and 20.
 
Example 5: We know that 10 = 10 × 1, 10 = 2 × 5 and 10 = 5 × 2

∴ 1, 2, 5 and 10 are factors of 10.
In other words, we can say that 10 is a multiple of 1, 2, 5 and 10.

Question 3:

(i) 20
20 = 1 × 20; 20 = 10 × 2 and 20 = 4 × 5
The factors of 20 are 1, 2, 4, 5, 10 and 20.

(ii) 36
36 = 1 × 36; 36 = 2 × 18; 36 = 3 × 12 and 36 = 4 × 9
The factors of 36 are 1, 2, 3, 4, 6, 9, 12 and 36.

(iii) 60
60 = 1 × 60; 60 = 2 × 30; 60 = 3 × 20; 60 = 4 × 15 and 60 = 5 × 12
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15 and 60.

(iv) 75
75 = 1 × 75; 75 = 3 × 25 and 75 = 5 × 15
The factors of 75 are 1, 3, 5, 15, 25 and 75.

Question 4:

(i) 17
17 × 1 = 17; 17 × 2 = 34; 17 × 3 = 51; 17 × 4 = 68 and 17 × 5 = 85
∴ The first five multiples of 17 are 17, 34, 51, 68 and 85.

(ii) 23
23 × 1=23; 23 × 2 = 46; 23 × 3 = 69; 23 × 4 = 92 and 23 × 5 = 115
∴ The first five multiples of 23 are 23, 46, 69, 92 and 115.

(iii) 65
65 × 1 = 65; 65 × 2 = 130; 65 × 3 = 195; 65 × 4 = 260 and 65 × 5 = 325
∴ The first five multiples of  65 are 65, 130, 195, 260 and 325.

(iv) 70
70 × 1=70; 70 × 2 = 140; 70 × 3 = 210; 70 × 4 = 280 and 70 × 5 = 350
∴ The first five multiples of 70 are 70, 140, 210, 280 and 350.

Question 5:

(i) 32
Since 32 is a multiple of 2, it is an even number.
(ii) 37
Since 37 is not a multiple of 2, it is an odd number.
(iii) 50
Since 50 is a multiple of 2, it is an even number.
(iv) 58
Since 58 is a multiple of 2, it is an even number.
(v) 69
Since 69 is not a multiple of 2, it is an odd number.
(vi) 144
Since 144  is a multiple of 2, it is an even number.
(vii) 321
Since 321  is not a multiple of 2, it is an odd number.
(viii) 253
Since 253 is not a multiple of 2, it is an odd number.

Question 6:

Prime number: A number is called a prime number if it has only two factors, namely 1 and itself .

Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 are prime numbers.

Question 7:

(i) All prime numbers between 10 and 40 are 11, 13, 17, 19, 23, 29, 31 and 37.
(ii) All prime numbers between 80 and 100 are 83, 89 and 97.
(iii) All prime numbers between 40 and 80 are 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.
(iv) All prime numbers between 30 and 40 are 31 and 37.

Question 8:

(i) The smallest prime number is 2.
(ii) There is only one even prime number, i.e., 2.
(iii) The smallest odd prime number is 3.

Question 9:

(i) 87
The divisors of 87 are 1, 3, 29 and 87 i.e. 87 has more than 2 factors. Therefore 87 is not a prime number.

(ii) 89
The divisors of 89 are 1 and 89. Therefore 89 is a prime number.

(iii) 63
The divisors of 63 are 1, 3, 7, 9, 21 and 63 i.e. 63 has more than 2 factors. Therefore 63 is not a prime number.

(iv) 91
The divisors of 91 are 1, 7, 13 and 91 i.e. 91 has more than 2 factors. Therefore 91 is not a prime number.

Question 10:

90, 91, 92, 93, 94, 95 and 96 are seven consecutive numbers and none of them is a prime.

Question 11:

(i) No, there are no counting numbers with no factors at all because every number has at least two factors, i.e., 1 and itself.
(ii) There is only one number that has exactly one factor, i.e, 1.
(iii) The numbers between 1 and 100 that have exactly three factors are 4, 9, 25 and 49.

Question 12:

The numbers that have more than two factors are known as composite numbers.
Yes, a composite number can be odd.
The smallest odd composite number is 9.

Question 13:

Two consecutive odd prime numbers are called twin primes.
The pairs of twin primes between 50 to 100 are (59, 61) and (71, 73).

Question 14:

 If two numbers do not have a common factor other than 1, they are said to be co-primes.

Five pairs of co primes: (i) 2 and 3 (ii) 3 and 4 (iii) 4 and 5 (iv) 4 and 9 (v) 8 and 15

No, co–primes are not always primes.

For example, 3 and 4 are co-prime numbers, where 3 is a prime number and 4 is not a prime number.

Question 15:

(i) 36
36 as the sum of two odd prime numbers is (36 = 31 + 5).
(ii) 42
42 as the sum of two odd prime numbers is (42 = 31 + 11).
(iii) 84
84 as the sum of two odd prime numbers is (84 = 41 + 43).
(iv) 98
98 as the sum of two odd prime numbers is (98 = 31 + 67).

Question 16:

(i) 31
31 can be expressed as the sum of three odd prime numbers as (31 = 5 + 7 + 19).
(ii) ) 35
35 can be expressed as the sum of three odd prime numbers as (35 = 17 + 13 + 5).
(iii) 49
49 can be expressed as the sum of three odd prime numbers as (49 = 13 + 17 + 19).
(iv)  63
63 can be expressed as the sum of three odd prime numbers as (63 = 29 + 31 + 3).

Question 17:

(i) 36
36 can be expressed as the sum of twin primes as (36 = 17 + 19).
(ii) 84
84 can be expressed as the sum of twin primes as (84 = 41 + 43).
(iii) 120
120 can be expressed as the sum of twin primes as (120 = 59 + 61).
(iv) 144
144 can be expressed as the sum of twin primes as (144 = 71 + 73).

Question 1:

(i) False. 2 is the smallest prime number.
(ii) False. 2 is an even prime number.
(iii) False. 3 and 7 are two prime numbers and their sum is 10, which is even.
(iv) False. 4 and 9 are co-primes but neither of them is a prime number.

Question 2:

A number is divisible by 2 if its ones digit is 0, 2, 4, 6 or 8.
(i)   Since the digit in the ones place in 26250 is 0, it is divisible by 2
(ii)  Since the digit in the ones place in 69435 is not 0, 2, 4, 6 or 8, it is not divisible by 2.
(iii) Since the digit in the ones place in 59628 is 8, it is divisible by 2.
(iv) Since the digit in the ones place in 789403 is not 0, 2, 4, 6, or 8, it is not divisible by 2.
(v) Since the digit in the ones place in  357986 is 6, it is divisible by 2.
(vi) Since the digit in the ones place in 367314 is 4, it is divisible by 2.

Question 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.
(i)  733 is not divisible by 3 because the sum of its digits, 7 + 3 + 3, is 13, which is not divisible by 3.
(ii) 10038 is divisible by 3 because the sum of its digits, 1 + 0 + 0 + 3 + 8, is 12, which is divisible by 3.
 (iii) 20701 is not divisible by 3 because the sum of its digits, 2 + 0 + 7 + 0 + 1, is 10, which is not divisible by 3.
(iv) 524781 is divisible by 3 because the sum of its digits, 5 + 2 + 4 + 7 + 8 + 1, is 27, which is divisible by 3.
(v) 79124 is not divisible by 3 because the sum of its digits, 7 + 9 + 1 + 2 + 4, is 23, which is not divisible by 3.
(vi) 872645 is not divisible by 3 because the sum of its digits, 8 + 7 + 2 + 6 + 4 + 5, is 32, which is not divisible by 3.

Question 4:

A number is divisible by 4 if the number formed by the digits in its tens and units place is divisible by 4.

(i)  618 is not divisible by 4 because the number formed by its tens and ones digits is 18, which is not divisible by 4.
(ii) 2314 is not divisible by 4 because the number formed by its tens and ones digits is 14, which is not divisible by 4.
(iii) 63712 is divisible by 4 because the number formed by its tens and ones digits is 12, which is divisible by 4.
(iv) 35056 is divisible by 4 because the number formed by its tens and ones digits is 56, which is divisible by 4.
(v)  946126 is not divisible by 4 because the number formed by its tens and ones digits is 26, which is not divisible by 4.
(vi) 810524 is divisible by 4 because the number formed by its tens and ones digits is 24, which is divisible by 4.

Question 5:

A number is divisible by 5 if its ones digit is either 0 or 5.

(i) 4965 is divisible by 5, because the digit at its ones place is 5.
  
(ii) 23590 is divisible by 5, because the digit at its ones place is 0.
    
(iii) 35208 is not divisible by 5, because the digit at its ones place is 8.
     
 (iv) 723405 is divisible by 5, because the digit at its ones place is 5.
     
(v) 124684 is not divisible by 5, because the digit at its ones place is 4.
     
(vi) 438750 is divisible by 5, because the digit at its ones place is 0.
    

Question 6:

A number is divisible by 6 if it is divisible by both 2 and 3.

i)  Since 2070 is divisible by 2 and 3, it is divisible by 6.
     Checking the divisibility by 2: Since the number 2070 has 0 in its units place, it is divisible by 2.
     Checking the divisibility by 3: The sum of the digits of 2070, 2 + 0 + 7 + 0, is 9, which is divisible by 3. So, it is divisible by 3.

   
(ii) Since 46523 is not divisible by 2, it is not divisible by 6.

Checking the divisibility by 2: Since the number 46523 has 3 in its units place, it is not divisible by 2.
  
(iii) Since 71232 is divisible by both 2 and 3, it is divisible by 6.
     
Checking the divisibility by 2: Since the number has 2 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of the number, 7 + 1 + 2 + 3 + 2, is 15, which  is divisible by 3. So, the number is divisible by 3. 
   
(iv) Since 934706 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: Since the sum of the digits of the number, 9 + 3 + 4 + 7 + 0 + 6, is 29, which is not divisible by 3. So, the number is not divisible by 3.

(v) Since 251780 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 2 + 5 + 1 + 7 + 8 + 0, is 23, which is not divisible by 3. So, the number is not divisible by 3.

(vi)  Since 872536 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 8 + 7 + 2 + 5 + 3 + 6, is 31, which is not divisible by 3. So, the number is not divisible by 3.

Question 7:

To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits. If their difference is a multiple of 7, the number is divisible by 7.

(i) 826 is divisible by 7.
 We have 82 − 2 × 6 = 70, which is a multiple of 7.

(ii) 117 is not divisible by 7.
      We have 11 − 2 × 7 = −3, which is not a multiple of 7.

(iii) 2345 is divisible by 7.
 We have  234 − 2 × 5 = 224, which is a multiple of 7.

(iv) 6021 is divisible by 7. 
 We have 602 − 2 × 1 = 600, which is not a multiple of  7.

(v)  14126 is divisible by 7.
    We have 1412 − 2 × 6 = 1400, which is a multiple of 7.

(vi)  25368 is divisible by 7.
      We have 2536 − 2 × 8 = 2520, which is a multiple of 7.

Question 8:

A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.

(i) 9364 is not divisible by 8.
  It is because the number formed by its hundreds, tens and ones digits, i.e., 364, is not divisible by 8.

(ii) 2138 is not divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 138, is not divisible by 8.

(iii) 36792 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 792, is divisible by 8.

(iv) 901674 is not divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 674, is not divisible by 8.

(v) 136976 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 976, is divisible by 8.

(vi) 1790184 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 184, is divisible by 8.

Question 9:

A number is divisible by 9 if the sum of its digits is divisible by 9.

(i)  2358 is divisible by 9, because the sum of its digits, 2 + 3 + 5 + 8, is 18, which is divisible by 9.

(ii)  3333 is not divisible by 9, because the sum of its digits, 3 + 3 + 3 + 3, is 12, which is not divisible by 9.

(iii)  98712 is divisible by 9, because the sum of its digits, 9 + 8 + 7 + 1 + 2, is 27, which is divisible by 9.

(iv)  257106 is not divisible by 9, because the sum of its digits, 2 + 5 + 1 0 + 6, is 21, which is not divisible by 9.

(v)  647514 is divisible by 9, because the sum of its digits, 6 + 4 + 7 + 5 + 1 + 4, is 27, which is divisible by 9.

(vi)  326999 is not divisible by 9, because the sum of its digits, 3 + 2 + 6 + 9 + 9 + 9, is 38, which is not divisible by 9. 

Question 10:

A number is divisible by 10 if its ones digit is 0.

(i) 5790 is divisible by 10, because its ones digit is 0.
(ii) 63215 is not divisible by 10, because its ones digit is 5, not 0.
(iii) 55555 is not divisible by 10, because its ones digit is 5, not 0.

Question 11:

A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

(i)  4334 is divisible by 11.
           
Sum of the digits at odd places = (4 + 3) = 7
Sum of the digits at even places = (3 + 4) = 7
    Difference of the two sums = (7 − 7) = 0, which is divisible by 11.

(ii)  83721 is divisible by 11.
Sum of the digits at odd places = (1 + 7 + 8) = 16
Sum of the digits at even places = (2 + 3) = 5
     Difference of the two sums = (16 − 5) = 11, which is divisible by 11.

(iii) 66311 is not divisible by 11.
          
   Sum of the digits at odd places = (1 + 3 + 6) = 10
   Sum of the digits at even places = (1 + 6) = 7
   Difference of the two sums = (10 − 7) = 3, which is not divisible by 11.
         
(iv) 137269 is divisible by 11.
           
   Sum of the digits at odd places = (9 + 2 + 3) = 14
   Sum of the digits at even places = (6 + 7 + 1) = 14
     Difference of the two sums = (14 − 14) = 0, which is a divisible by 11.

(v)  901351 is divisible by 11.
           
  Sum of the digits at odd places = (0 + 3 + 1) = 4
 Sum of the digits at even places = (9 + 1 + 5) = 15
 Difference of the two sums = (4 − 15) = −11, which is divisible by 11.
          
(vi) 8790322 is not divisible by 11.
           
    Sum of the digits at odd places = (2 + 3 + 9 + 8) = 22
    Sum of the digits at even places = (2 + 0 + 7) = 9
    Difference of the two sums = (22 − 9) = 13, which is not divisible by 11.

Question 12:

(i) 2724
 Here, 2 + 7 + * + 4 = 13 + * should be a multiple of 3.
 To be divisible by 3, the least value of * should be 2, i.e., 13 + 2 = 15, which is a multiple of 3.
∴ * = 2

(ii) 53046
Here, 5 + 3 + * + 4 + 6 = 18 + * should be a multiple of 3.
As 18 is divisible by 3, the least value of  * should be 0, i.e., 18 + 0 = 18.
 ∴ * = 0

 (iii) 81711
Here, 8+ * + 7 + 1 + 1 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18 , which is a multiple of 3.
∴ * = 1

 (iv) 62235
Here, 6 + 2 + * + 3 + 5 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
∴ * = 2
 
(v) 234117
Here, 2+ 3 +4 + * + 1 + 7 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18, which is a multiple of 3.
∴ * =1

(vi) 621054
Here, 6 + * +1 + 0 + 5 + 4 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
∴ * =2

Question 13:

(i) 6525
Here, 6 + 5+ * + 5 = 16 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 9.
∴ * =2

(ii) 27135
Here, 2 + * + 1 + 3 + 5 = 11 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 7, i.e., 11 + 7 = 18, which is a multiple of 9.
∴ * = 7

(iii) 67023
Here, 6 + * + 7 + 0 + 2 = 15 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 3, i.e., 15 + 3 = 18, which  is a multiple of 9.
∴ * = 3

(iv) 91467
Here, 9 + 1 * + 6 + 7 = 23 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 4, i.e., 23 + 4 = 27, which is a multiple of 9.
∴ * = 4

(v) 667881
Here,  6 + 6 + 7 + 8 + * + 1 = 28 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 8, i.e., 28 + 8 = 36, which is a multiple of 9.
∴ * = 8
 
(vi) 835686
Here, 8 + 3 + 5 + * + 8 + 6 = 30 + * should be a multiple of 9.
To be divisible of 9, the least value of * should be 6, i.e.,  30 + 6 = 36, which is a multiple of 9.
∴ * = 6

Question 14:

(i) 26*5
Sum of the digits at odd places = 5 + 6 = 11
Sum of the digits at even places = * + 2
Difference = sum of odd terms – sum of even terms
= 11 – (* + 2)  
= 11 – * – 2
 = 9 – *
Now, (9 – *) will be divisible by 11 if * = 9.
i.e., 9 – 9 = 0
0 is divisible by 11.
∴ * = 9
Hence, the number is 2695.

(ii) 39*43
Sum of the digits at odd places = 3 + * + 3 = 6 + *
Sum of the digits at even places = 4 + 9 = 13
Difference = sum of odd terms – sum of even terms
= 6 + * – 13  
= * – 7
Now, (* – 7) will be divisible by 11 if * = 7.
i.e., 7 – 7 = 0
0 is divisible by 11.
∴ * = 7
Hence, the number is 39743.
 
(iii) 86*72
Sum of the digits at odd places  2 + * + 8 = 10 + *
Sum of the digits at even places  6 + 7 = 13
Difference = sum of odd terms – sum of even terms
= 10 + * – 13  
= * – 3
  Now, (* – 3) will be divisible by 11 if * = 3.
i.e., 3 – 3 = 0
0 is divisible by 11.
∴ * = 3
 Hence, the number is 86372.

(iv) 467*91
Sum of the digits at odd places 1  + * + 6 = 7 + *
Sum of the digits at even places  9 + 7 + 4 = 20
Difference = sum of odd terms – sum of even terms
= (7 + *) − 20  
   = * − 13
  Now, (* −13) will be divisible by 11 if * = 2.
i.e., 2− 13 = −11
−11 is divisible by 11.
∴ * = 2
 Hence, the number is 467291.

(v) 1723*4
Sum of the digits at odd places 4+ 3+ 7= 14
Sum of the digits at even places  *+2+1 = 3 + *
Difference = sum of odd terms – sum of even terms
= 14 – (3 + *)  
   = 11 − *
  Now, (11 − *) will be divisible by 11 if * = 0.
i.e., 11 − 0 = 11
11 is divisible by 11.
∴ * = 0
 Hence, the number is 172304.

(vi) 9*8071
Sum of the digits at odd places 1+0+*  = 1 + *
Sum of the digits at even places  7 + 8 + 9 = 24
Difference = sum of odd terms – sum of even terms
=1 + *  – 24  
   =   * − 23
  Now, (* − 23) will be divisible by 11 if * = 1.
i.e., 1 − 23 = −22
−22 is divisible by 11.
∴  * = 1
Hence, the number is 918071.

Question 15:

(i) 10000001 by 11
10000001 is divisible by 11.
Sum of digits at odd places = (1 + 0 + 0 + 0) = 1
Sum of digits at even places = (0 + 0 + 0 + 1) = 1
Difference of the two sums = (1 − 1) = 0, which is divisible by 11.
 
 
(ii) 19083625 by 11
19083625 is divisible by 11.
Sum of digits at odd places = (5 + 6 + 8 + 9) = 28
Sum of digits at even places = (2 + 3 + 0 + 1) = 6
 Difference of the two sums = (28 − 6) = 22, which is divisible by 11.

(iii) 2134563 by 9
    2134563 is not divisible by 9.
It is because the sum of its digits, 2 + 1 + 3 + 4 + 5 + 6 + 3, is 24, which is not divisible by 9.
          
(iv) 10001001 by 3
10001001 is divisible by 3. 
It is because the sum of its digits, 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1, is 3, which is divisible by 3.

(v) 10203574 by 4                  
  10203574 is not divisible by 4.
It is because the number formed by its tens and the ones digits is 74, which is not divisible by 4.
(vi) 12030624 by 8
     12030624 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits is 624, which is divisible by 8.

Question 16:

A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.

(i) 103 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(ii) 137 is a prime number, because it is not divisible by 2, 3, 5, 7 and 11.
(iii) 161 is a not prime number, because it is divisible by 7.
(iv) 179 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(v) 217 is a not prime number, because it is divisible by 7.
(vi) 277 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(vii) 331 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(viii) 397 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.

Question 17:

(i) 14 is divisible by 2, but not by 4.
(ii) 12 is divisible by 4, but not by 8.
(iii) 24 is divisible by both 2 and 8, but not by 16.
(iv) 30 is divisible by both 3 and 6, but not by 18.

Question 1:

(i) If a number is divisible by 4, it must be divisible by 8. False
Example: 28 is divisible by 4 but not divisible by 8.

(ii) If a number is divisible by 8, it must be divisible by 4. True
Example: 32 is divisible by both 8 and 4.

(iii) If a number divides the sum of two numbers exactly, it must exactly divide the numbers separately. False
Example: 91 (51 + 40) is exactly divisible by 13. However, 13 does not exactly divide 51 and 40.

(iv) If a number is divisible by both 9 and 10, it must be divisible by 90. True
Example: 900 is both divisible by 9 and 10. It is also divisible by 90.

(v)  A number is divisible by 18 if it is divisible by both 3 and 6. False
A number has to be divisible by 9 and 2 to be divisible by 18.
Example: 48 is divisible by 3 and 6, but not by 18.

 (vi) If a number is divisible by 3 and 7, it must be divisible by 21. True
Example: 42 is divisible by both 3 and 7. It is also divisible by 21.

(vii) The sum of consecutive odd numbers is always divisible by 4. True
 Example: 11 and 13 are consecutive odd numbers.
11 + 13 = 24, which is divisible by 4.  

(viii) If a number divides two numbers exactly, it must divide their sum exactly.  True
Example: 42 and 56 are exactly divisible by 7.
42+56 = 98, which is exactly divisible by 7. 

Question 2:

We use the division method as shown below:

$\begin{array}{l}2\overline{)12}\\ 2\overline{)6}\\ 3\overline{)\text{3 }}\\ 1\end{array}$
∴ 12 = 2 × 2 × 3
            = 2² × 3

Question 3:

We will use the division method as shown below:
$\begin{array}{l}2\overline{)18}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ 1\end{array}$

∴ 18 = 2 × 3 × 3
         = 2 × 3²

Question 4:

We will use the division method as shown below:

$\begin{array}{l}2\overline{)48}\\ 2\overline{)\text{24 }}\\ 2\overline{)\text{12 }}\\ 2\overline{)\text{6 }}\\ 3\overline{)\text{3 }}\\ \text{ 1 }\end{array}$

∴ 48 = 2 × 2 × 2 × 2 × 3
         =2⁴×3

Question 5:

We will use the division method as shown below:

  $\begin{array}{l}2\overline{)\text{56 }}\\ 2\overline{)\text{28 }}\\ 2\overline{)\text{14 }}\\ \text{ 7}\end{array}$
∴ 56 = 2 × 2 × 2 × 7
           = 2³×7

Question 6:

We will use the division method as shown below:

$\begin{array}{l} 2\overline{)\text{90 }}\\ 3\overline{)\text{45 }}\\ 3\overline{)\text{15 }}\\ \begin{array}{l} 5\overline{)\text{5 }}\\ \overline{)\text{1 }}\\ \end{array}\end{array}$

∴ 90 = 2 × 3 × 3 × 5
         = 2 × 3² × 5

Question 7:

 We will use the division method as shown below:
  
    $\begin{array}{l} 2\overline{)136}\\ 2\overline{)\text{68 }}\\ 2\overline{)\text{34 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\\ \end{array}$
∴ 136 = 2 × 2 × 2 × 17
              = 2³ × 17

Question 8:

We will use the division method as shown below:

$\begin{array}{l}2\overline{)252}\\ 2\overline{)\text{126 }}\\ 3\overline{)\text{63 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$

∴ 252 = 2 × 2 × 3 × 3 × 7 × 1
           = 2² × 3² × 7 × 1                         

Question 9:

We will use the division method as shown below:

$\begin{array}{l}2\overline{)420}\\ 2\overline{)\text{210 }}\\ 3\overline{)\text{105 }}\\ 7\overline{)\text{35 }}\\ 5\overline{)\text{5 }}\\ \text{ 1}\end{array}$
∴ 420 = 2 × 2 × 3 × 7 ×5 × 1
            =2²×3×5×7

Question 10:

We will use the division method as shown below:

$\begin{array}{l}7\overline{)\text{637 }}\\ 7\overline{)\text{91 }}\\ 13\overline{)\text{13 }}\\ \text{ 1}\end{array}$

∴ 637 = 7 ×7 × 13
              = 7² × 13

Question 11:

We will use the division method as shown below:

$\begin{array}{l}3\overline{)\text{945 }}\\ 3\overline{)\text{315 }}\\ 3\overline{)\text{105 }}\\ 5\overline{)\text{35 }}\\ 7\overline{)\text{7 }}\\ \text{ 1 }\end{array}$

∴ 945 = 3 × 3 × 3 × 5 × 7 × 1
            = 3³ × 5 × 7

Question 12:

We will use the division method as shown below:

$\begin{array}{l} 2\overline{)\text{1224 }}\\ 2\overline{)\text{612 }}\\ 2\overline{)\text{306 }}\\ 3\overline{)\text{153 }}\\ 3\overline{)\text{51 }}\\ \text{ 17 }\overline{)\text{17 }}\\ \text{ 1}\end{array}$
∴ 1224 = 2 × 2 × 2 × 3 ×3 × 17
                 = 2³ × 3² × 17

Question 13:

We will use the division method as shown below:

$\begin{array}{l}3\overline{)\text{1323 }}\\ 3\overline{)\text{441 }}\\ 3\overline{)\text{147 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$
∴ 1323 = 3 × 3 × 3 ×7 × 7 × 1
                  =3³ × 7²

Question 14:

We will use the division method as shown below:

$\begin{array}{l} 2\overline{)\text{8712 }}\\ 2\overline{)\text{4356 }}\\ 2\overline{)\text{2178 }}\\ 3\overline{)\text{1089 }}\\ 3\overline{)\text{363 }}\\ 11\overline{)\text{121 }}\\ 11\overline{)\text{11 }}\\ \text{ 1}\end{array}$
∴ 8712 = 2 × 2 × 2 × 3 × 3 × 11 × 11
                 = 2³ × 3² × 11²

Question 15:

 We will use the division method as shown below:

$\begin{array}{l} 7\overline{)\text{9317 }}\\ 11\overline{)\text{1331 }}\\ 11\overline{)\text{121 }}\\ 11\overline{)\text{11 }}\\ \text{ 1}\end{array}$
∴ 9317 = 7 × 11 × 11 × 11
             = 7 × 11³

Question 16:

  We will use the division method as shown below:

$\begin{array}{l} 3\overline{)\text{1035 }}\\ 3\overline{)\text{345 }}\\ 5\overline{)\text{115 }}\\ 23\overline{)\text{23 }}\\ \text{ 1}\end{array}$
∴ 1035 =3 × 3 × 5 × 23
                  = 3² × 5 × 23

Question 17:

We will use the division method as shown below:

$\begin{array}{l} 3\overline{)\text{1197 }}\\ 3\overline{)\text{399 }}\\ 7\overline{)\text{133 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\end{array}$
∴ 1197 = 3 × 3 × 7 × 19
                 = 3² × 7 × 19

Question 18:

 We will use the division method as shown below:

$\begin{array}{l} 3\overline{)\text{4641 }}\\ 7\overline{)\text{1547 }}\\ 13\overline{)\text{221 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\end{array}$
∴ 4614 = 3 × 7 × 13 × 17

Question 19:

 We will use the division method as shown below:

$\begin{array}{l} 3\overline{)\text{4335 }}\\ 5\overline{)\text{1445 }}\\ 17\overline{)\text{289 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\end{array}$
∴ 4335 = 3 × 5 × 17 × 17
                  = 3 × 5 × 17²

Question 20:

We will use the division method as shown below:

$\begin{array}{l} 3\overline{)\text{2907 }}\\ 3\overline{)\text{969 }}\\ 17\overline{)\text{323 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\end{array}$
∴ 2907 = 3 × 3 × 17 × 19
              = 3² × 17 × 19

Question 1:

  We will use the division method as shown below:

$\begin{array}{l} 5\overline{)\text{13915 }}\\ 11\overline{)\text{2783 }}\\ 11\overline{)\text{253 }}\\ \begin{array}{l}23 \overline{)\text{23 }}\\ \overline{)\text{1 }}\end{array}\end{array}$
∴ 13915 = 5 × 11 × 11 × 23
                    = 3 × 11² × 23

Question 2:

The given numbers are 84 and 98.

We have:

$\begin{array}{l}2\overline{)\text{84 }}\\ 2\overline{)\text{42 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$$\begin{array}{l}2\overline{)\text{98 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$
 
84 = 2 × 2 × 3 × 7 = 2² × 3 × 7
98 = 2 × 7 × 7 = 2 × 7²
  
∴ HCF of the given numbers = 2 × 7 = 14

Question 3:

The given numbers are 170 and 238.

We have:

$\begin{array}{l} 2\overline{)\text{170 }}\\ 5\overline{)\text{85 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\end{array}$$$\begin{gathered} \begin{array}{l} 2\overline{)\text{238 }}\\ 7\overline{)\text{119 }}\\ 17\overline{)\text{17 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$

170 = 2 × 5 × 17
238 = 2 × 7 × 17

∴ H.C.F. of the given numbers = 2 × 17 = 34

Question 4:

The given numbers are 504 and 980.

We have:

$\begin{array}{l}2\overline{)\text{504 }}\\ 2\overline{)\text{252 }}\\ 2\overline{)\text{126 }}\\ 3\overline{)\text{63 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$       $\begin{array}{l}2\overline{)\text{980 }}\\ 2\overline{)\text{490 }}\\ 5\overline{)\text{245 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$

    504 = 2 × 2 ×2 × 3 × 3 × 7 = 2³ × 3² × 7
    980 = 2 × 2 × 5× 7 × 7 = 2² × 5 × 7²
∴ HCF of the given numbers = 2² × 7 = 28

Question 5:

The given numbers are 72, 108 and 180

We have:

$\begin{array}{l}2\overline{)\text{72 }}\\ 2\overline{)\text{36 }}\\ 2\overline{)\text{18 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$   $\begin{array}{l}2\overline{)\text{108 }}\\ 2\overline{)\text{54 }}\\ 3\overline{)\text{27 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$    $\begin{array}{l}2\overline{)\text{180 }}\\ 2\overline{)\text{ 90 }}\\ 3\overline{)\text{45 }}\\ 3\overline{)\text{15 }}\\ 5\overline{)\text{ 5 }}\\ \text{ 1}\\ \end{array}$
Now, 72=2 × 2 × 2 × 3 × 3 = 2³ × 3²
108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³
180 = 2 × 2 × 3 ×3 × 5 = 2² × 3² × 5
∴ HCF =2²×3² =36

Question 6:

The given numbers are 84, 120 and 138.

We have:

$\begin{array}{l}2\overline{)\text{84 }}\\ 2\overline{)\text{42 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1 }\end{array}$       $\begin{array}{l}2\overline{)\text{120 }}\\ 2\overline{)\text{60 }}\\ 2\overline{)\text{30 }}\\ 3\overline{)\text{15 }}\\ 5\overline{)\text{5 }}\\ \text{ 1}\\ \end{array}$        $\begin{array}{l} 2\overline{)\text{138 }}\\ 3\overline{)\text{69 }}\\ 23\overline{)\text{23 }}\\ \text{ 1}\end{array}$

Now, 84 = 2 × 2 ×3 × 7
120 = 2 × 2× 2 ×3 × 5
138 = 2 × 3 × 23
∴ HCF = 2 × 3 = 6

Question 7:

The given numbers are 106, 159 and 371.
We have:

$\begin{array}{l} 2\overline{)\text{106 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$$\begin{array}{l} 3\overline{)\text{159 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$$\begin{array}{l} 7\overline{)\text{371 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$
Now, 106 = 2 × 53
159 = 3 × 53
371 = 7 × 53
∴ HCF = 53

Question 8:

Given numbers are 272 and 425.

We have:

$\begin{array}{l} 2\overline{)\text{272 }}\\ 2\overline{)\text{136 }}\\ 2\overline{)\text{68 }}\\ 2\overline{)\text{34 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\\ \end{array}$        $\begin{array}{l} 5\overline{)\text{425 }}\\ 5\overline{)\text{85 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\\ \end{array}$
Now, 272 = 2 × 2× 2 × 2 × 17
425 = 5 × 5 × 17
∴ The required HCF is 17.

Question 9:

The given numbers are 144, 252 and 630.
We have:

$\begin{array}{l}2\overline{)\text{144 }}\\ 2\overline{)\text{72 }}\\ 2\overline{)\text{36 }}\\ 2\overline{)\text{18 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$          $\begin{array}{l}2\overline{)\text{252 }}\\ 2\overline{)\text{126 }}\\ 3\overline{)\text{63 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$        $\begin{array}{l}2\overline{)\text{630 }}\\ 3\overline{)\text{315 }}\\ 3\overline{)\text{105 }}\\ 5\overline{)\text{35 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$   

Now, 144 = 2 × 2× 2 × 2× 3 × 3
252 = 2 ×2 × 3× 3 × 7
630 = 2 ×3 × 3× 5 × 7
∴ HCF = 2 × 3 × 3 =18

Question 10:

The given numbers are 1197, 5320 and 4389.

We have:

$\begin{array}{l} 3\overline{)\text{1197 }}\\ 3\overline{)\text{399 }}\\ 7\overline{)\text{133 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\\ \end{array}$      $\begin{array}{l} 2\overline{)\text{5320 }}\\ 2\overline{)\text{2660 }}\\ 2\overline{)\text{1330 }}\\ 5\overline{)\text{665 }}\\ 7\overline{)\text{133 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\\ \end{array}$        $\begin{array}{l} 3\overline{)\text{4389 }}\\ 7\overline{)\text{1463 }}\\ 19\overline{)\text{209 }}\\ 11\overline{)\text{11 }}\\ \text{ 1 }\end{array}$
Now, 1197 = 3 × 3× 7× 19 = 3² × 7 × 19
5320 = 2 × 2 × 2× 5 × 7 × 19 = 2³ × 5 × 7 × 19
4389 = 3 ×7 × 19 × 11
∴ Required HCF = 19 × 7 = 133

Question 11:

We have:


∴ The HCF of 58 and 70 is 2.

Question 12:

The given numbers are 399 and 437.

We have:

∴ The HCF is 19.

Question 13:

The given numbers are 1045 and 1520.
We have:


∴ The HCF of 1045 and 1520 is 95.

Question 14:

The given numbers are 1965 and 2096.

We have:
 

∴ The HCF is 131.

Question 15:

The given numbers are 2241and 2341.
We have:


∴ HCF = 83

Question 16:

The given numbers are 658, 940 and 1128.

First we will find the HCF of 658 and 940.


Thus, the HCF of 658 and 940 is 94.

Now, we will find the HCF of 94 and 1128.


Thus, the HCF of 94 and 1128 is 94.

∴ The HCF of 658, 940 and 1128 is 94.

​

Question 17:

The given numbers are 754, 1508 and 1972.
 
First, we will find the HCF of 754 and 1508.

So, the HCF of 754 and 1508 is 754.

Now, we will find the HCF of 754 and 1972.


So, the HCF of 754 and 1972 is 58.

∴ The HCF of 754, 1058 and 1972 is 58.

Question 18:

The given numbers are 391, 425 and 527.
First, we will find the HCF of 391 and 425.

So, the HCF of 391 and 425 is 17.
Now, we will find the HCF of 17 and 527.


So, the HCF of 17 and 527 is 17.
∴ The HCF of 391, 425 and 527 is 17.

Question 19:

The given numbers are 1794, 2346 and 4761.
First, we will find the HCF of 1794 and 2346.

So, the HCF of 1794 and 2346 is 138.
Now, we will find the HCF of 138 and 4761.

So, the HCF of 138 and 4761 is 69.

∴ The HCF of 1794, 2346 and 4761 is 69.

Question 20:

The given numbers are 59 and 97.

59=59×1
97=97×1

∴ HCF = 1

Since 59 and 97 does not have any common factor other than 1, the two numbers are co-primes.

Question 21:

The given numbers are 161 and 192.
We have:
$$\begin{gathered} \begin{array}{l} 7\overline{)161}\\ 23\overline{)23}\end{array} \\ \overline{)1} \end{gathered}$$$\begin{array}{l}2\overline{)192}\\ 2\overline{)\text{96 }}\\ 2\overline{)\text{48 }}\\ 2\overline{)\text{24 }}\\ 2\overline{)12}\\ 2\overline{)\text{6 }}\\ 3\\ \end{array}$

Now, 161 = 7 × 23 × 1
192 = 2 × 2× 2 ×2 × 2×  2 × 3 = 2⁶  × 3 × 1
∴ HCF = 1
Hence, 161 and 192 are co-primes.

Question 22:

The given numbers are 343 and 432.
We have:
$\begin{array}{l}7\overline{)\text{343 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$$\begin{array}{l}2\overline{)432}\\ 2\overline{)\text{216 }}\\ 2\overline{)\text{108 }}\\ 2\overline{)\text{54 }}\\ 3\overline{)27}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \overline{)\text{1 }}\end{array}$

Now, 343 = 7 × 7× 7 × 1 = 7³ × 1
432 = 2 × 2× 2 ×2 × 3× 3 ×3 = 2⁴ × 3³ × 1
∴ HCF =1
Hence, 343 and 432 are co-primes.

Question 23:

Given numbers are 512 and 945.
We have:
$\begin{array}{l}2\overline{)512}\\ 2\overline{)\text{256 }}\\ 2\overline{)\text{128 }}\\ 2\overline{)\text{64 }}\\ 2\overline{)32}\\ 2\overline{)\text{16 }}\\ 2\overline{)\text{8 }}\\ 2\overline{)\text{4 }}\\ 2\overline{)\text{2 }}\\ \text{ 1 }\\ \end{array}$
$\begin{array}{l}\mathrm{<br><br>\; 3}\overline{)\text{945 }}\\ 3\overline{)\text{315 }}\\ 3\overline{)\text{105 }}\\ 5\overline{)\text{35 }}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$
$\begin{array}{l}7\overline{)\text{7 }}\end{array}$
$\begin{array}{l}\end{array}$   $\begin{array}{l}\overline{)\text{1 }}\\ \mathrm{}\end{array}$$\begin{array}{l}\end{array}$

512 = 2 × 2 ×2 × 2 × 2× 2 × 2× 2 × 2 = 2⁹
 945 = 3 × 3 × 3 × 5 × 7 = 3³ × 5 × 7
Thus, the HCF of 512 and 945 is 1.
∴ 512 and 945 are co-primes.

Question 24:

The given numbers are 385 and 621.
$$\begin{gathered} \begin{array}{l} 5\overline{)\text{385 }}\\ 7\overline{)\text{77 }}\\ 11\overline{)\text{11 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$$$\begin{gathered} \begin{array}{l} 3\overline{)\text{621 }}\\ \text{ 3}\overline{)\text{207 }}\\ 3\overline{)\text{69 }}\\ 23\overline{)\text{23 }}\end{array} \\ \overline{)\text{ 1 }} \end{gathered}$$

385 = 5 × 7 × 11 × 1
621 = 3 × 3 × 3 × 23 = 3³ × 23 × 1
∴ HCF = 1

Hence, they are co-primes.

Question 25:

The given numbers are 847 and 1014.

$$\begin{gathered} \begin{array}{l}7\overline{)\text{847 }}\\ 11\overline{)\text{121 }}\\ 11\overline{)\text{11 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$$$\begin{gathered} \begin{array}{l}2\overline{)\text{1014 }}\\ 3\overline{)\text{507 }}\\ 13\overline{)\text{169 }}\\ 13\overline{)\text{13 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$

847 = 7 × 11 × 11 × 1 = 7 × 11² × 1
1014 = 2 × 3 × 13 × 13 × 1
∴ HCF = 1
Hence, 847 and 1014 are co-primes.

Question 26:

Because the remainder is 6, we have to find the number that exactly divides (615 - 6) and (963 - 6).

Required number = HCF of 609 and 957
$\begin{array}{l}609\begin{array}{c}1\\ \overline{)\begin{array}{l}\text{ 957}\\ \underset{}{-609}\end{array}}\end{array}\\ \text{ 348}\overline{)609}(1\\ \underset{}{-348}\\ \text{ 261}\overline{)348}(1\\ -\underset{\_}{261}\\ \text{ 87}\overline{)261}(3\\ -\underset{\_}{261}\\ \text{ 0}\end{array}$

Therefore, the required number is 87. 

Question 27:

Clearly, we have to find the number which exactly divides (2011 − 9) and (2623 − 5).
So, the required number is the HCF of 2002 and 2618.
$\begin{array}{l}2002\begin{array}{c}1\\ \overline{)\begin{array}{l}\text{ 2618}\\ \underset{}{-2002}\end{array}}\end{array}\\ \text{ 616}\overline{)2002}(3\\ \underset{}{-1848}\\ \text{ 154}\overline{)616}(4\\ -\underset{}{616}\\ \text{ 0}\end{array}$

∴ The required number is 154.

Question 28:

Since the respective remainders of 445, 572 and 699 are 4, 5 and 6, we have to find the number which exactly divides (445-4), (572-5) and (696-6).

So, the required number is the HCF of 441, 567 and 693.
Firstly, we will find the HCF of 441 and 567.

∴ HCF = 63

Now, we will find the HCF of 63 and 693.

∴ HCF = 63

Hence, the required number is 63.

Question 29:

(i) $\frac{161}{207}$
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
 
Now, we will find the HCF of 161 and 207.

∴ HCF = 23

Dividing the numerator and the denominator by the HCF, we get:

$\frac{161÷23}{207÷23}=\frac{7}{9}$


(ii) $\frac{517}{799}$
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 517 and 799.

∴ HCF = 47

Dividing the numerator and the denominator by the HCF, we get:

$\frac{517÷47}{799÷47}=\frac{11}{17}$

(iii) $\frac{296}{481}$
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 296 and 481.



∴ HCF = 37

Dividing the numerator and the denominator by the HCF, we get:

$\frac{296÷37}{481÷37}=\frac{8}{13}$

Question 30:

The lengths of the three pieces of timber are 42 m, 49 m and 63 m.
The greatest possible length of each plank will be given by the HCF of 42, 49 and 63.

Firstly, we will  find the HCF of 42 and 49 by division method.

∴ The HCF of 42 and 49 is 7.
Now, we will find the HCF of 7 and 63.
​​

∴ The HCF of 7 and 63 is 7.
Therefore, HCF of all three numbers is 7
Hence, the greatest possible length of each plank is 7 m.

Question 31:

Three different containers contain 403 L, 434 L and 465 L of milk.

The capacity of the container that can measure the milk in an exact number of times will be given by the HCF of 403, 434 and 465.


∴ HCF = 31

Now, we will find the HCF of 31 and 465.


∴ HCF = 31

Hence, the capacity of the required container is 31 L.

Question 32:

Number of apples = 527
Number of pears = 646
Number of oranges = 748
The fruits are to be arranged in heaps containing the same number of fruits.
The greatest number of fruits possible in each heap will be given by the HCF of 527, 646 and 748.

Firstly, we will find the HCF of 527 and 646.


∴ HCF of 527, 646 and 748 = 17

So, the greatest number of fruits in each heap will be 17.

Question 33:

7 m = 700 cm
3 m 85 cm = 385 cm
12 m 95 cm = 1295 cm

The required length of the tape that can measure the lengths 700 cm, 385 cm and 1295 cm will be given bu the HCF of 700 cm, 385 cm and 1295 cm.

Evaluating the HCF of 700, 385 and 1295 using prime factorisation method, we have:
 
700 = 2 × 2 × 5 × 5 × 7 = 2² × 5² × 7

385 = 5 × 11 × 7

1295 = 5 × 7 × 37

∴ HCF = 5 ×7 = 35

Hence, the longest tape which can measure the lengths 7 m, 3 m 85 cm and 12 m 95 cm exactly is of 35 cm.