वास्तविक संख्याएँ

कक्षा X

प्रश्नावली 1.1

प्रश्न 1. निम्नलिखित संख्याओं का HCF ज्ञात करने के लिए यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग कीजिए
(i) 135 और 225 (ii) 196 और 38220 (iii) 867 और 255

(i)
हलः
प्रश्न से स्पष्ट है कि
225>135
यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग करने पर
225 = 135x1 + 90
135 = 90x1 + 45
90 = 45x2 + 0
यहाँ शेषफल 0 प्राप्त हुआ।
अतः 135 और 225 का HCF = 45 Ans.

(ii)
हलः
प्रश्न से स्पष्ट है कि
38220>196
यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग करने पर
38220 = 196x195 + 0
यहाँ शेषफल 0 प्राप्त हुआ।
अतः 38220 और 196 का HCF = 196 Ans.

(iii)
हलः
प्रश्न से स्पष्ट है कि
867>255
यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग करने पर
867 = 255x3 + 102
255 = 102x2 + 51
102 = 51x2 + 0
यहाँ शेषफल 0 प्राप्त हुआ।
अतः 255 और 867 का HCF = 51 Ans.

प्रश्न 2. दर्शाइए कि कोई भी धनात्मक विषम पूर्णांक 6q + 1 या 6q + 3 या 6q + 5 के रुप का होता है, जहाँ q कोई पूर्णांक है।

हलः
माना कि a कोई धनात्माक पूर्णांक है तथा b = 6 है।
तब यूक्लिड विभाजन एल्गोरिथ्म से, किसी पूर्णांक q≥0 के लिए,
a = bq + r
जहाँ r = 0,1,2,3,4,5 है क्योंकि 0≤r<6 है।
इसलिए,
a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
हम जानते है कि कोई भी धनात्माक  विषम पूर्णांक  2 से विभाजित नही होता है।
अतः कोई भी धनात्माक  विषम पूर्णांक 

6q + 1 or 6q + 3 or 6q + 5 के रुप का होता है। Proved.

प्रश्न 3. किसी परेेड में 616 सदस्यों वाली एक सेना (आर्मी ) की टुकडी को 32 सदस्यों वाले एक आर्मी बैंड के पीछे मार्च करमा है। दोनों समूह को समान संख्या वाले स्तंभों की अधिकतम संख्या क्या है, जिसमें वे मार्च कर सकते हैं ?
हलः
प्रश्न से स्पष्ट है कि
a = 616 एवं b = 32
यूक्लिड विभाजन एल्गोरिथ्म से,
a = bq + r
616 = 32x19 + 8
32 = 8x4 + 0
यहाँ शेषफल 0 प्राप्त हुआ।
अतः HCF = 8
स्तंभों की अधिकत्तम संख्या = 8 Ans.

1. Real Numbers

Real Numbers

Class X

Chapter 1

NCERT Text Book Solution

Exercise 1.3

Q.No. 1. Prove that √5 is irrational.

Solution:

Let √5 is a rational number.

Therefore,

√5 = a/b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So,

b√5 = a

Now squaring both side

5b² = a²

a² divisible by 5, therefore, a is divisible by 5.

Let a = 5c

Therefore,

5b² = 25c²

b² = 5c²

Hence, b² is divisible by 5, therefore, b is divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √5 is a rational.

So, we conclude that √5 is irrational. Proved.

Q.No. 2. Prove that 3 + 2√5 is irrational.

Solution:

From the solution of question no. 1. It is proved that √5 is irrational.

We know that product of rational and irrational number is irrational.

So, 2√5 is irrational.

We also know that sum of rational and irrational number is irrational.

So, 3 + 2√5 is irrational. Proved.

Q.No. 3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Solution:

(i) Let √2 be rational.

Therefore,

√2 = a/b 

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√2 = a

Now squaring both side

2b² = a²

a² is divisible by 2, therefore a is divisible by 2.

Let a = 2c

Therefore,

2b² = 4c²

b² = 2c²

b² is divisible by 2, therefore, b is divisible by 2.

Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √2 is a rational.
So, we conclude that √2 is irrational.
We know that quotient of rational and irrational is irrational.
So, 1/√2 is irrational. Proved.

Solution:
(ii) 7√5
From the solution of question no. 1 it is proved that √5 is irrational.
We know that product of rational and irrational is irrational.
So, 7√5 is irrational. Proved.

Solution:
(iii) 6 + √2

From the solution of question no. 3 (i) it is proved that √2 is irrational.
We know that sum of rational and irrational is irrational.
So, 6 + √2 is irrational. Proved.

1. Real Numbers

Exercise 1.2

Class X

Q.No. 1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution:

(i)                 140

Prime factors of 140 = 2x2x5x7 Ans.

(ii)               156

Prime factors of 156 = 2x2x3x13 Ans.

(iii)             3825

Prime factors of 3825 = 3x3x5x5x17 Ans.

(iv)             5005

Prime factors of 5005 = 5x7x11x13 Ans.

(v)               7429

Prime factors of 7429 = 17x19x23 Ans.

Q.No. 2. Find the LCM and HCF of the following pairs of integers and verify that LCMxHCF = Product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution:

(i)                 26 and 91

Prime factors of 26 = 2x13

Prime factors of 91 = 13x7

LCM = 2x7x13 = 182

HCF = 13

LCM x HCF = 182x13=2366

Product of number = 26x91 = 2366

LCMxHCF = Product of number is verified.

(ii)               510 and 92

Prime factors of 510 = 2x3x5x17

Prime factors of 92 = 2x2x23

LCM = 2x2x3x5x17x23 = 23460

HCF = 2

LCMxHCF = 23460x2 = 46920

Product of number = 46920

Hence, LCMxHCF = Product of number is verified.

(iii)             336 and 54

Prime factors of 336 = 2x2x2x2x3x7

Prime factors of 54 = 2x3x3x3

LCM = 2x2x2x2x3x3x3x7 = 3024

HCF = 2x3 = 6

LCMxHCF = 18144

Product of number = 336x54 = 18144

Hence, LCMxHCF = Product of number is verified.

Q.No. 3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15and 21 (ii) 17, 23 and 29 (iii) 8,9 and 25

Solution:

(i)                 12, 15 and 21

Prime factorisation of 12 = 2x2x3

Prime factorisation of 15 = 3x5

Prime factorisation of 21 = 3x7

LCM = 2x2x3x5x7 = 420

HCF = 3

(ii)               17, 23 and 29

Prime factorisation of 17 = 17

Prime factorisation of 23 = 23

Prime factorisation of 29 = 29

LCM = 17x23x29 = 11339

HCF = 1

(iii)             8, 9 and 25

Prime factorisation of 8 = 2x2x2 = 8

Prime factorisation of 9 = 3x3

Prime factorisation of 25 = 5x5

LCM = 2x2x2x3x3x5x5 = 1800

HCF = 1

Q.No. 4.Given that HCF (306, 657) = 9 , find LCM (306, 657)

Solution:

LCMxHCF = Product of numbers

or, LCM = Product of numbers/HCF

or, LCM = 306x657/9

or, LCM = 22338 Ans.

Q.No. 5. Check whether  can end with the digit 0 for any natural number n.

Solution:

Prime factorisation of 6 = 2x3

So, the prime factorisation of 6 is not divisible by 5.

Hence, cannot end with 0 for any natural number n according to Fundamental Theorem of Arithmetic.

Q.No. 6. Explain why 7x11x13 + 13 and 7x6x5x4x3x2x1 + 5 are composite numbers.

Solution:

These numbers can be expressed as the form of the factorisation of prime numbers.

So, these numbers are composite numbers.

Q.No. 7. There is 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 the same direction. After how many minutes will they meet again at the starting point.

Solution:

Prime factorisation of 18 = 2x3x3

Prime factorisation of 18 = 2x2x3x3

LCM of 12 and 18 = 2x2x3x3 = 36

So, they will meet again after 36 minutes. Ans.

1. Real Numbers

Class X

NCERT Text Book Solution

Exercise 1.1

1. Use Euclid's division algorithm to find the HCF of 

(i)135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution:

(i)

From question it is clear that

225 > 135.

Apply Euclid's division lemma, to 135 and 225.

225 = 135x1 + 90

135 = 90x1 + 45

90 = 45x2 + 0

The remainder has now become zero.

The HCF of 225 and 135 is 45 Ans.

(ii)

From question it is clear that 

38220>196

Apply Euclid's division lemma, to 38220 and 196.

38220 = 196x195 + 0

The remainder has now become zero.

The HCF of 38220 and 196 is 195 Ans.

(iii)

From question it is clear that

867>255

Apply Euclid's division lemma, to 867 and 255.

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0 

The remainder has now become zero.

The HCF of 867 and 255 is 51 Ans

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

Solution:

Let a be any positive integer and b = 6.

Now, by Euclid's division algorithm,

a = 6q + r for some integer q ≥ 0,

r = 0, 1, 2, 3, 4, 5

Because 0 ≤ r < b

So, a can be

6q,or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5.

Since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 ( since they are all divisible by 2)

Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Proved.

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 ?

Solution:

For this, it will be find out the HCF of 616 and 32.

Apply Euclid's algorithm, to 616 and 32.

616 = 32x19 + 8

32 = 8x4 + 0

So, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which they can march = 8 Ans.

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

Solution:

Let a be any positive integer and b = 3.

Now, Euclid's division lemma,

a = 3q + r for some integer q ≥ 0,

Since, 0 ≤ r < 3

a can be 3q, or 3q + 1, or 3q + 2

Therefore,

a² = 9q², or 9q² + 6q + 1, or 9q² + 12q + 1.

or, a² = 3x3q², or 3( 3q² + 2q) + 1, or 3(3q² + 4q) + 1

So, the square of any positive integer is either of the form 3m, or 3m + 1. Proved

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

Solution:

Let a be any positive integer and b = 9.

Now, Euclid's division lemma,

a = 9q + r for some integer q ≥ 0,

Since, 0 ≤ r < 9

a can be 9q, or 9q + 1, or 9q + 2, 9q + 3, or ...... 9q + 8.

Therefore,

a³ = 9x81q³, or 9(81q³ + 27q² + 3q)+ 1, or 9(81q³ + 54q² + 6q) + 8 ..................

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