Rs Aggarwal 2018 for Class 10 Math Chapter 1 - Real Numbers

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Page No 9:

Question 1:

What do you mean by Euclid's division algorithm.

ANSWER:

Euclid's division algorithm states that for any two positive integers a and b, there exist unique integers q and r, such that a = bq + r, where 0 ≤ r < b.

Page No 9:

Question 2:

A number when divided by 61 gives 27 as quotient and 32 as remainder.
Find the number.

ANSWER:

We know, Dividend = Divisor ×$\times$ Quotient + Remainder
 Given: Divisor = 61, Quotient = 27, Remainder = 32
Let the Dividend be x.
∴ x = 61 × 27 + 32$61 \times 27 + 32$
        = 1679

Hence, the required number is 1679.

Page No 9:

Question 3:

By what number should 1365 be divided to get 31 as quotient and 32 as remainder?

ANSWER:

Given: Dividend = 1365, Quotient = 31, Remainder = 32
Let the divisor be x.
Dividend = Divisor ×$\times$ Quotient + Remainder
                       1365 =  x ×$\times$ 31 + 32
   ⇒                1365 − 32 = 31x
   ⇒                      1333 = 31x

   ⇒                        x = 133331$\frac{1333}{31}$ = 43
Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

Page No 9:

Question 4:

Using Euclid's algortihm, find the HCF of

(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575

ANSWER:

(i)

On applying Euclid's algorithm, i.e. dividing 2520 by 405, we get:
         Quotient = 6, Remainder = 90
         ∴ 2520 = 405 ×$\times$ 6 + 90
   Again on applying Euclid's algorithm, i.e. dividing 405 by 90, we get:
          Quotient = 4, Remainder = 45
          ∴ 405 = 90 ×$\times$ 4 + 45
      Again on applying Euclid's algorithm, i.e. dividing 90 by 45, we get:
         ∴ 90 = 45 ×$\times$ 2 + 0                                                                  
                                                                                                                 
Hence, the HCF of 2520 and 405 is 45.

(ii)

On applying Euclid's algorithm, i.e. dividing 1188 by 504, we get:
       Quotient = 2, Remainder = 180
         ∴ 1188 = 504 ×$\times$ 2 +180                            
    Again on applying Euclid's algorithm, i.e. dividing 504 by 180, we get:                      
        Quotient = 2, Remainder = 144                        
            ∴ 504 = 180×$\times$ 2 + 144                                  
   Again on applying Euclid's algorithm, i.e. dividing 180 by 144, we get:                    
        Quotient = 1, Remainder = 36                                    
              ∴ 180 = 144 ×$\times$ 1 + 36                                      
 Again on applying Euclid's algorithm, i.e. dividing 144 by 36, we get:                    
             ∴ 144 = 36 ×$\times$ 4 + 0                                          

Hence, the HCF of 1188 and 504 is 36.
                                                                                             
(iii)

On applying Euclid's algorithm, i.e. dividing 1575 by 960, we get:
         Quotient = 1, Remainder = 615      
        ∴ 1575 = 960 ×$\times$ 1 + 615                      
    Again on applying Euclid's algorithm, i.e. dividing 960 by 615, we get:                    
       Quotient = 1, Remainder = 345              
        ∴ 960 = 615 ×$\times$ 1 + 345                    
  Again on applying Euclid's algorithm, i.e. dividing 615 by 345, we get:                
     Quotient = 1, Remainder = 270                                  
  l;\ xv
\
     ∴ 615 = 345 ×$\times$ 1 + 270                     4
 Again on applying Euclid's algorithm, i.e. dividing 345 by 270, we get:              
    Quotient = 1, Remainder = 75    
      ∴ 345 = 270 ×$\times$ 1 + 75                    
Again on applying Euclid's algorithm, i.e. dividing 270 by 75, we get:                                  
   Quotient = 3, Remainder = 45                                                                              
      ∴270 = 75 ×$\times$ 3 + 45                                                            
 Again on applying Euclid's algorithm, i.e. dividing 75 by 45, we get:                      
   Quotient = 1,  Remainder = 30                                                                                            
     ∴ 75  = 45 ×$\times$ 1 + 30                                                                                        
Again on applying Euclid's algorithm, i.e. dividing 45 by 30, we get:
  Quotient = 1, Remainder = 15
     ∴ 45  = 30 ×$\times$ 1 + 15
Again on applying Euclid's algorithm, i.e. dividing 30 by 15, we get:
 Quotient = 2, Remainder = 0
   ∴ 30 = 15 ×$\times$ 2 + 0

Hence, the HCF of 960 and 1575 is 15

Page No 9:

Question 5:

Show that every positive integer is either even or odd.

ANSWER:

Let us assume that there exist a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n − 1 must be either odd or even.

Case 1: If n − 1 is even, n − 1 = 2k for some k.
             But this implies n = 2k + 1
             this implies n is odd.

Case 2: If n − 1 is odd, n − 1 = 2k + 1 for some k.
             But this implies n = 2k + 2 = 2(k + 1)
             this implies n is even.

In both ways we have a contradiction.
Thus, every positive integer is either even or odd.

Page No 9:

Question 6:

Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

ANSWER:

Let n be any arbitrary positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid's division lemma, we have

n = 6m + r, where 0 ≤ r < 6.

As 0 ≤ r < 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.
⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5

But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 (∵ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even interger whereas n is an odd integer)

⇒ n = 6m + 1 or n = 6m + 3 or n = 6m + 5

Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Page No 9:

Question 7:

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

ANSWER:

Let n be any arbitrary positive odd integer.
On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid's division lemma, we have

n = 4m + r, where 0 ≤ r < 4.

As 0 ≤ r < 4 and r is an integer, r can take values 0, 1, 2, 3.
⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3

But n ≠ 4m or n ≠ 4m + 2 (∵ 4m, 4m + 2 are multiples of 2, so an even interger whereas n is an odd integer)

⇒ n = 4m + 1 or n = 4m + 3

Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Page No 16:

Question 1:

Using prime factorization, find the HCF and LCM of

(i) 36, 84
(ii) 23, 31
(iii) 96, 404
(iv) 144, 198
(v) 396, 1080
(vi) 1152, 1664

ANSWER:

(i) 36, 84
   Prime factorisation:
   36 = 22 ⨯ 32
   84 = 22 ⨯ 3 ⨯ 7
​ HCF = product of smallest power of each common prime factor in the numbers = 22 ⨯ 3 = 12
 LCM = product of greatest power of each prime factor involved in the numbers = 22 ⨯ 32 ⨯ 7 = 252

(ii) 23, 31
   Prime factorisation:
   23 = 23
   31 = 31
​ HCF = product of smallest power of each common prime factor in the numbers = 1
 LCM = product of greatest power of each prime factor involved in the numbers = 23 ⨯ 31 = 713

(iii) 96, 404
   Prime factorisation:
   96 = 25 ⨯ 3
   404 = 22 ⨯ 101
​ HCF = product of smallest power of each common prime factor in the numbers = 22 = 4
 LCM = product of greatest power of each prime factor involved in the numbers = 25 ⨯ 3 ⨯ 101 = 9696

(iv) 144, 198
   Prime factorisation:
   144 = 24 × 32$2^4 \times 3^2$
  198 = 2 × 32 × 11$2 \times 3^2 \times 11$
​ HCF = product of smallest power of each common prime factor in the numbers = 2 × 32$2 \times 3^2$ = 18
 LCM = product of greatest power of each prime factor involved in the numbers = 24 × 32 × 11$2^{4 }\times 3^{2 }\times 11$ = 1584

(v) 396, 1080
    Prime factorisation:
   396 = 22 × 32  × 11$2^2 \times 3^{2 } \times 11$
  1080 = 23 × 33 × 5$2^3 \times 3^3 \times 5$
 ​ HCF = product of smallest power of each common prime factor in the numbers = 22 × 32$2^2 \times 3^2$ = 36
LCM = product of greatest power of each prime factor involved in the numbers = 23 × 33 × 5 ×11$2^3 \times 3^{3 }\times 5 \times 11$ = 11880

(vi) 1152 , 1664
    Prime factorisation:
   1152 = 27 × 32$2^7 \times 3^2$
  1664 = 27 × 13$2^7 \times 13$
HCF = product of smallest power of each common prime factor involved in the numbers = 27$2^7$ = 128
 LCM = product of greatest power of each prime factor involved in the numbers = 27 × 32 × 13$2^7 \times 3^2 \times 13$ = 14976

Page No 16:

Question 2:

Using prime factorization, find the HCF and LCM of

(i) 8, 9, 25
(ii) 12, 15, 21
(iii) 17, 23, 29
(iv) 24, 36, 40
(v) 30, 72, 432
(vi) 21, 28, 36, 45

ANSWER:

(i) 8 = 2 ⨯ 2 ⨯ 2 = 23
     9 = 3 ⨯ 3 = 32
     25 = 5 ⨯ 5 = 52
HCF = Product of smallest power of each common prime factor in the numbers = 1
LCM = Product of the greatest power of each prime factor involved in the numbers = 23 ⨯ 32 ⨯ 52 = 1800

(ii) 12 = 2 ⨯ 2 ⨯ 3 = 22 ⨯ 3
      15 = 3 ⨯ 5
      21 = 3 ⨯ 7
HCF = Product of smallest power of each common prime factor in the numbers = 3
LCM = Product of the greatest power of each prime factor involved in the numbers = 22 ⨯ 3 ⨯ 5 ⨯ 7 = 420

(iii) 17 = 17
       23 = 23
       29 = 29
HCF = Product of smallest power of each common prime factor in the numbers = 1
LCM = Product of the greatest power of each prime factor involved in the numbers = 17 ⨯ 23 ⨯ 29 = 11339

(iv) 24 = 2× 2 × 2 ×3$\times 2 \times 2 \times 3$ = 23 × 3$2^3 \times 3$
    36 = 2 × 2 ×3 × 3$2 \times 2 \times 3 \times 3$ = 22 × $2^2 \times$32$3^2$
    40 =  2 × 2 ×2 × 5$\times 2 \times 2 \times 5$ = 23$2^3$ × 5$\times 5$
  ∴ ​HCF = Product of smallest power of each common prime factor in the numbers = 22$2^2$ = 4
 ∴​ LCM = Product of the greatest power of each prime factor involved in the numbers =  23×32×5 = 360$2^3\times 3^2\times 5 = 360$

(v) 30 = 2 × 3 × 5$30 = 2 \times 3 \times 5$
     72 = 2 × 2 × 2 × 3 × 3$72 = 2 \times 2 \times 2 \times 3 \times 3$ = 23 × 32$2^{3 }\times 3^2$
    432 = 2 × 2 × 2 × 2 × 3 × 3× 3$432 = 2 \times 2 \times 2 \times 2 \times 3 \times 3\times 3$ = 24 × 33$2^4 \times 3^3$
  ∴ HCF = Product of smallest power of each common prime factor in the numbers = 2 × 3 = 6$2 \times 3 = 6$
  ∴ ​LCM = Product of the greatest power of each prime factor involved in the numbers = 24 × 33 × 5 = 2160$2^{4 }\times 3^3 \times 5 = 2160$

(vi) 21 = 3 × 7$21 = 3 \times 7$
     28  = 28 = 2 × 2 × 7$28 = 2 \times 2 \times 7$ = 22 × 7$2^2 \times 7$
     36 = 2 × 2 × 3 × 3$36 = 2 \times 2 \times 3 \times 3$ = 22 × 32$2^2 \times 3^2$
     45 = 5 × 3 × 3$45 = 5 \times 3 \times 3$ = 5 × 32$5 \times 3^2$
   ∴​ HCF = Product of smallest power of each common prime factor in the numbers = 1
   ∴​ LCM = Product of the greatest power of each prime factor involved in the numbers = 22 × 32 × 5 × 7 = 1260