Paliyal Solution For Class 10th CBSE MATHS Chapter 1 : Real Numbers

Paliyal Solution For Class 10th CBSE MATHS Chapter 1 : Real Numbers.

Mathematics Exercise 1.1     

Page No: 7

Exercise No :1.1

Question 1. Use Euclid 's division algorithm to Find HCF OF......

 (i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Answer (i) 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder ...

So that 225 = 135 × 1 + 90 So we apply division lemma again...

Divide 135 by 90 we get 1 quotient and 45 as remainder......

So that 135 = 90×1+45   So we apply division lemma again .....

Divide 90 by 45 we get quotient 2 and no remainder . So we can write it as-----

90 = 45×2+0  there are no remainder  So Divisor 45 is our HCF  So Complete this.  By

Kishori Lal -----

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(ii) 196 and 38220

Answer (ii) 38220>196 we always divide greater number with smaller one.

Divide 38220 by 196 we get 196×195 quotient and no remainder ...

Divide 38220 by 196 we get quotient 195 and no remainder . So we can write it as-----

38220 = 196×195+0  there are no remainder  So Divisor 196 is our HCF  So Complete this. 

By Kishori Lal -----------------

(iii) 867 and 255

Answer (i) 867 > 255 we always divide greater number with smaller one.

Divide 867 by 255 we get 1 quotient and 90 as remainder ...

So that 867 = 255 × 3 + 102 So we apply division lemma again...

Divide 255 by 102 we get 2 quotient and 51 as remainder......

So that 255 = 102×2+51   So we apply division lemma again .....

Divide 102 by 51 we get quotient 2 and no remainder . So we can write it as--

102 = 51×2+0  there are no remainder  So Divisor 51 is our HCF  So Complete this.  By

Kishori Lal --- So complete 1 Question Here

2. Show that any positive odd integer is of form 6q+ 1, or 6q+ 3, or 6q+ 5, where q is some integer.

Answer

Let take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will 6q + 0, 6q + 2, 6q + 3, 6q + 4, 6q + 5

6q + 0

6 is divisible by 2 so it is a even number

6q + 1

6 is divisible by 2 but 1 is not divisible by 2 so it is not odd number

6q + 2

6 is divisible by 2 and 2 is also divisible by 2 so it is a even number

6q + 3

6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6q + 4

6 is divisible by 2 and 4 is also divisible by 2 so it is a ,even number

6q + 5

6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns.

What is the maximum number of columns in which they can march?

Answer

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the HCF.

616 = 32 x 19 + 8

32 = 8 x 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hind: Let x be any positive integer then it is of the form 3q, 3q+1 or 3q+2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer

Let a be any positive integer and b = 3

Then a =3q+r for some integer q≥ 0

And r=0, 1, 2 because 0 ≤r ≤3

Therefore, a =3q or 3q + 1 or 3q+2

Or,

a²= (3q)² or( 3q+1)² or (3q=2)²

a²= (9q)² or 9q²+6q+1or 9q²+12q+4

= 3x(3q²) or 3(3q²+2q)+1 or 3(3q²+4q+1)+1

3k₁ or 3k₂+1 or 3k₃+1

Where k₁, k₂ and k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1or 9m+8.

Answer

Let a be any positive integer and b=3

a=3q = r, where q ≥ 0 and 0 ≤ r ≤3

a=3q or 3q + 1or 3q + 2

Therefore, every number can be represented as these three forms, There are three cases.

Case 1: When a=3q,

a³= (3q)³=27q³=9(3q)³=9m,

Where m is an integer such that m=3q³

Case 2: When a=3q+1,

a³= (3q + 1)³

a³=27q³+27q²+9q+1

a³=9(3q³+3q²+q) +1

a³=9m+1

Where m is an integer such that m=(3q³+3q²+q)

Case 3: When a =3q+2,

a³=(3q+2)³

a³=27q³+54q²+36q+8

a³=9(3q³+6q²+4q)+8

a³=9m+8

Where m is an integer such that m = (3q³+6q²+4q)

Therefore, the cube of any positive integer is of the form 9m, 9m+1, or 9m+8.

Page No.11

Exercise 1.2

Express each number as product of its prime factors:

140

156

3825

5005

7429

Answer

(1) 140=2 x 2 x 5 x 7=2² x 5 x 7

(2) 156= 2 x 2 x 3 x 13 = 2² x 3 x 13

(3) 3825= 3 x 3 x 5 x 5 x 17= 3² x 5² x 17

(4) 5005= 5 x 7 x 11 x 13

(5) 7429= 17 x 19 x 23

2. Find the LCM and HCF of the following pairs of integers and verify the LCM X HCF = product of the two numbers

 (i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer

(i) 26=2 x 13

91=7 x 13

HCF=13

LCM=2 x 7 x 13=182

Product of HCF and LCM 13 x 182=2366

Hence, product of two numbers= product of HCF x LMC

(ii) 510 = 2 x 3 x 5 x 17

92 = 2 x 2 x 23

HCF = 2

LCM = 2 X 2 X 3 X 5 X 17 X 23 = 23460

Product of two numbers 510 x 92 = 46920

Product of HCF and LCM 2 x 23460 = 46920

Hence, Product of two numbers = product of HCF x LCM

(iii) 336 = 2 x 2 x 2 x 2 x 3 x 7

54 = 2 x 3 x 3 x 3

HCF = 2 X 3 = 6

LCM = 2 X 2 X 2 X 2 X 3 X 3 X 7 = 3024

Product of two numbers 336 x 54 = 18144

Product of HCF and LCM 6 x 3024 = 18144

Hence, product of two numbers = product of HCF x LCM.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Answer

(i) 12 = 2 x 2 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 X 2 X 3 X 5 X 7 = 420

(ii) 17 = 1 X 17

23 = 1 X 23

29 = 1 X 29

HCF = 1

LCM = 1 X 17 X 19 X 23 = 11339

(iii) 8 = 1 x 2 x 2 x 2

       9 = 1 x 3 x 3

      25= 1 x 5 x 5

HCF = 1

LCM = 1 X 2 X 2 X 2 X 3 X 5 X 5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer

We have the formula that

Product of LCM and HCF = product of number

LCM x 9 = 306 x 657

Divide both sides by 9 we get

LCM = (306 x 657) / 9 = 22338

5. Check whether 6n can end with the digit 0 for any natural number n.

Answer

 If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 x 5.

So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.

6. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Answer

7 x 11 x 13 + 13

Taking 13 common, we get

13 (7 x 11 + 1)

13(77 + 1)

13(78)

It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

Taking 5 common, we get

5(7 x 6 x 4 x 3 x 2 x 1 + 1)

5(1008 + 1)

5(1009)

It is product of two numbers and both numbers are more than 1 so it is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Answer

They will be meet again after LCM of both values at the starting point.

18 = 2 x 3 x 3

12= 2 x 2 x 3

LCM= 2 x 2 x 3 x 3 = 36

Therefore, they will meet together at the starting point after 36 minutes.

Page No: 14

Exercise 1.3

1. Prove that √5 is irrational.

Answer

Let take √5 as rational number

If a and b are two co prime number and b is not equal to 0.

We can write √5 = a/b

Multiply by b both side we get

B √5 = a

To remove root, Squaring on both sides, we get

5b² = a² … (i)

Therefore, 5 divides a² and according to theorem of rational number, for any prime number p which is divides a² then it will divide a also.

That means 5 will divide a. So we can write 

A = 5c

Putting value of a in equation (i) we get

5b²=(5c)²

5b²=25c²

Divide by 25 we get

b2/5=c²

Similarly, we get that b will divide by 5

and we have already get that a is divide by 5

but a and b are co prime number. So it contradicts. Hence √5 is not a rational number, it is irrational.

2. Prove that 3+2√5 is irrational

Answer

Let take that 3+2√5 is a rational number.

So we can write this number as

3+2√5=a/b

Here a and b are two co prime number and b is not equal to 0

Subtract 3 both side we get

2√5 = a/b -3

2√5 = (a-3b)/b

Now divide by 2, we get

√5= (a-3b)/2b

Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.

Hence, 3+2√5 is a irrational number

3. Prove that the following are irrationals:

(i) 1√2 (ii) 7√5 (iii) 6+√2

Answer

(i) Let take that 1√2 is a rational number.

So we can write this number as

1√2=a/b

Here a and b are two co prime number and b is not equal to 0

Multiply by √2 both sides we get

1=(a√2)/b

Now multiply by b

b=a√2

divide by a we get

b/a=√2

Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.

Hence, 1√2 is a irrational number.

(ii) Let take that 7√5 is a rational number.

So we can write this number as

7√5=a/b

Here a and b are two co prime number and b is not equal to 0

Divide by 7 we get

√5=a/(7b)

Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 should be a rational number but √5 is a irrational number so it contradicts.

Hence, 7√5 is a irrational number.

 (iii) Let take that 6 + √2 is a rational number.

So we can write this number as

6 + √2 = a/b

Here a and b are two co prime number and a and b is not equal to 0

Subtract 6 both side we get

√2 = a/b – 6

√2 = (a-6b)/b

Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number.

But √2 is a irrational number so it contradicts.

Hence, 6 + √2 is a irrational number.

Page No: 17

ALSO READ:

Exercise 1.4

1. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal

Expansion:

(i) 13/3125

(ii) 17/8

(iii) 64/455

(iv) 15/1600

(v) 29/343

(vi) 23/2³ x 5²

(vii) 129/2² x 5⁷ x 7⁵

(viii) 6/15

(ix)35/50

(x) 77/210

Answer

(i) 13/3125

Factorize the denominator we get

3125 = 5 x 5 x 5 x 5 x 5 = 5⁵

So denominator is in form of 5^m so it is terminating.

(ii) 17/8

Factorize the denominator we get

8 = 2 x 2 x 2 = 2³

So denominator is in form of 2^m so it is terminating.

(iii) 64/455

Factorize the denominator we get

455 = 5 x 7 x 13

There are 7 and 13 also in denominator so denominator is not in form of 2^m x 5ⁿ . so it is not terminating.

(iv) 15/1600

Factorize the denominator we get

1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 = 2⁶ x 5² 

So denominator is in form of 2^m x 5²

Hence it is terminating.

(v) 29/343

Factorize the denominator we get

343 = 7 x 7 x 7 = 7³

There are 7 also in denominator so denominator is not in form of 2^m x 5ⁿ

Hence it is non-terminating.

(vi) 23/(2³ x 5²)

Denominator is in form of 2^m x 5ⁿ

Hence it is terminating.

(vii) 129/(2² x 5⁷ x 7⁵)

Denominator has 7 in denominator so denominator is not in form of 2^m x 5ⁿ

Hence it is none terminating.

(viii) 6/15

Divide nominator and denominator both by 3 we get 2/5

Denominator is in form of 5^m so it is terminating.

(ix) 35/50 divide denominator and nominator both by 5 we get 7/10

Factorize the denominator we get

10 = 2 x 5

So denominator is in form of 2^m x 5ⁿ so it is terminating.

(x) 77/210

Simplify it by dividing nominator and denominator both by 7 we get 11/30

30 = 2 x 3 x 5

Denominator has 3 also in denominator so denominator is not in form of 2^m x 5ⁿ

Hence it is none terminating.

Page No. 18

2. Write down the decima lexpansions of those rational numbers in question 1 above which have terminating decimal expansions.

Answer

(i) 13/3125 = 13/5⁵ = 13x2⁵/5⁵x2⁵ = 416/10⁵ = 0.00416

(ii) 17/8 = 17/2³ = 17x5³/2³x5³ = 17x5³/10³ = 2125/10³ = 2.125

(iv) 15/1600 = 15/2⁴x10 = 15x5⁴/2⁴x5⁴x10² = 9375/10⁶ = 0.009375

(vi) 23/2³5² = 23x5³x2²/2³ 5²x5³x2² = 11500/10⁵ = 0.115

(viii) 6/15 = 2/5 = 2x2/5x2 = 4/10 = 0.4

(ix) 35/50 = 7/10 = 0.7

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p, q you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000….

(iii) 43.123456789

Answer

(i) Since this number has a terminating decimal expansion, it is a rational number of the form p/q, and q is of the form 2^m x 5ⁿ .

(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q, and q not of the form 2^m x 5ⁿ.

By Kishori Lal -----