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ncert solutions for class 10 maths chapter 1 real numbers

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NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4 – Real Numbers

1. Without actually performing the long division, state weather the following rational numbers will have a terminating decimal expansion or a not- terminating repeating decimal expansion:

Theorem: Let x = \cfrac { p }{ q }  be a rational number, such that the prime factorization of q is of the form 2n5m, where n and m are non-negative integers. Then x has a decimal expansion which terminates.

(i) \cfrac { 13 }{ 3125 }

Solution:

\cfrac { 13 }{ 3125 } = \cfrac { 13 }{ { 5 }^{ 5 } } = \cfrac { 13\times { 2 }^{ 5 } }{ { 5 }^{ 5 }\times { 2 }^{ 5 } }

Therefore, if we take m = n = 5, then \cfrac { 13 }{ 3125 }  can be written as \cfrac { p }{ q } , where q is of the form 2n5m. Hence this rational number has terminating decimal expansion.

(ii) \cfrac { 17 }{ 8 }

Solution:

\cfrac { 17 }{ 8 } = \cfrac { 17 }{ { 2 }^{ 3 } } = \cfrac { 17\times { 5 }^{ 3 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 } }

Therefore, if we take m = n = 3, then \cfrac { 17 }{ 8 } can be written as \cfrac { p }{ q } , where q is of the form 2n5m.

Hence this rational number has terminating decimal expansion.

(iii) \cfrac { 64 }{ 455 }

Solution:

\cfrac { 64 }{ 455 } = \cfrac { 64 }{ 5\times 7\times 13 }

Since here 455 is not the form of 2n5m, then \cfrac { 64 }{ 455 }  will not have a terminating decimal expansion.

(iv) \cfrac { 15 }{ 1600 }

Solution:

\cfrac { 15 }{ 1600 } = \cfrac { 15 }{ { 2 }^{ 6 }\times 5^{ 2 } } = \cfrac { 15\times { 5 }^{ 4 } }{ { 2 }^{ 6 }\times { 5 }^{ 6 } }

Therefore, if we take m = n = 6, then \cfrac { 15 }{ 1600 } can be written as \cfrac { p }{ q } , where q is of the form 2n5m.

Hence this rational number has terminating decimal expansion.

(v) \cfrac { 29 }{ 343 }

Solution:

\cfrac { 29 }{ 343 } = \cfrac { 29 }{ 7^{ 3 } }

Since here 343 is not the form of 2n5m, then \cfrac { 29 }{ 343 }  will not have a terminating decimal expansion.

(vi) \cfrac { 23 }{ 2^{ 3 }{ 5 }^{ 2 } }

Solution:

Since the denominator of \cfrac { 23 }{ 2^{ 3 }{ 5 }^{ 2 } } = \cfrac { 23\times 5 }{ 2^{ 3 }{ 5 }^{ 3 } }  is of the form of 2n5m, where n = m = 3. Hence this rational number has terminating decimal expansion.

(vii) \cfrac { 129 }{ 2^{ 2 }{ 5 }^{ 7 }{ 7 }^{ 5 } }

Solution:

Since the denominator of \cfrac { 129 }{ 2^{ 2 }{ 5 }^{ 7 }{ 7 }^{ 5 } } is not of the form of 2n5m. Hence this rational will not have terminating decimal expansion.

(viii) \cfrac { 6 }{ 15 }

Solution:

\cfrac { 6 }{ 15 } = \cfrac { 2\times 3 }{ 3\times 5 } = \cfrac { 2 }{ 5 } = \cfrac { 2\times 2 }{ 5\times 2 }

Therefore if we take n = m = 1, then \cfrac { 6 }{ 15 } can be written as \cfrac { p }{ q } , where q is of the form 2n5m Hence this rational number has terminating decimal expansion.

(ix) \cfrac { 35 }{ 50 }

Solution:

\cfrac { 35 }{ 50 } = \cfrac { 5\times 7 }{ 2\times { 5 }^{ 2 } } = \cfrac { 5\times 7\times 2 }{ { 2 }^{ 2 }\times { 5 }^{ 2 } }

Then \cfrac { 35 }{ 50 }  can be written as \cfrac { p }{ q } , where q is of the form 2n5m, where m = n= 2. Hence this rational number has terminating decimal expansion.

(x) \cfrac { 77 }{ 210 }

Solution:

\cfrac { 77 }{ 210 } = \cfrac { 7\times 11 }{ 2\times 3\times 5\times 7 }

Since the denominator of \cfrac { 77 }{ 210 } is not of the form of 2n5m. Hence this rational will not have terminating decimal expansion.

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

Solution:  The numbers which have terminating decimal expansions are

(i) \cfrac { 13 }{ 3125 }

= \cfrac { 13\times { 2 }^{ 5 } }{ { 5 }^{ 5 }\times { 2 }^{ 5 } }

= \cfrac { 13\times { 2 }^{ 5 } }{ { 10 }^{ 5 } }

= \cfrac { 416 }{ { 10 }^{ 5 } }

= 0.00416

(ii) \cfrac { 17 }{ 8 }

= \cfrac { 17 }{ { 2 }^{ 3 } }

= \cfrac { 17\times { 5 }^{ 3 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 } }

= \cfrac { 17\times { 5 }^{ 3 } }{ { 10 }^{ 3 } }

= \cfrac { 2125 }{ { 10 }^{ 3 } }

= 2.125

(iii) \cfrac { 15 }{ 1600 }

\cfrac { 15 }{ { 2 }^{ 6 }\times 5^{ 2 } }

= \cfrac { 15\times { 5 }^{ 4 } }{ { 2 }^{ 6 }\times { 5 }^{ 6 } }

= \cfrac { 9375 }{ { 10 }^{ 6 } }

= 0.009375

(iv) \cfrac { 23 }{ 2^{ 3 }{ 5 }^{ 2 } }

= \cfrac { 23\times 5 }{ 2^{ 3 }{ 5 }^{ 3 } }

= \cfrac { 115 }{ { 10 }^{ 3 } }

= 0.115

(v) \cfrac { 6 }{ 15 }

\cfrac { 2\times 3 }{ 3\times 5 }

= \cfrac { 2 }{ 5 }

= \cfrac { 2\times 2 }{ 5\times 2 }

= \cfrac { 4 }{ 10 }

= 0.4

(vi) \cfrac { 35 }{ 50 }

= \cfrac { 5\times 7 }{ 2\times { 5 }^{ 2 } }

= \cfrac { 5\times 7\times 2 }{ { 2 }^{ 2 }\times { 5 }^{ 2 } }

= \cfrac { 70 }{ { 10 }^{ 2 } }

= 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 \cfrac { p }{ q } what can you say about the prime factors of q?

(i) 43.123456789
(ii) 0.120120012000120000. . .
(iii) 43.\overline { 123456789 }

Solution:  There are two types of decimal numbers. (1) terminating and (2) non-terminating. If a number is terminating then it is rational. If a number is non-terminating and repeating then it is rational and if non-terminating and non-repeating then it is irrational.

(i) 43.123456789

= \cfrac { 43123456789 }{ 1000000000 }

= \cfrac { 43123456789 }{ { 10 }^{ 9 } }

Since this has terminating decimal expansions, then it can be written as \cfrac { p }{ q }  and as q = 109. Then the prime factors of q are 2 and 5 since 109 = 2959.

(ii) Since the real number 0.120120012000120000. .. is non-terminating and non- repeating, So it is not a rational number. Hence it can’t expressed as \cfrac { p }{ q } .

(iii) In the real number 43.\overline { 123456789 } the term 123456789 is repeated and non-terminating, it is a rational number and can be written as \cfrac { p }{ q } . But the prime factor of q is not of the form 2n5m e, the prime factor of q will also have a factor other than 2 or 5.

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4 – Real Numbers

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NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 – Real Numbers

1. Prove that √5 is irrational.

Proof: Let √5 is not irrational, i.e., rational number. Then there exist two integers a and b, where (b ≠ 0) such that √5 = \cfrac { a }{ b } .

Let a and b have a common factor other than 1, then dividing a and b by that common factor we get √5 = \cfrac { { a }_{ 1 } }{ { b }_{ 1 } } , where a1 and b1 are co prime.

√5 = \cfrac { { a }_{ 1 } }{ { b }_{ 1 } }  gives a1 = √5 b1

Squaring both side,

(a1)2= 5 (b1)2 …..(1).

 Therefore (a1)2 is divisible by 5 and hence a1 is divisible by 5

So we can write a1 = 5k, for some integer k. substituting this in (1) we get

25k2= 5 (b1)2

(b1)2 = 5k

(b1)2 is also divisible by 5 and hence b1 is divisible by 5.

Therefore a1 and b1 have least common factor 5, which is a contradiction to the fact that a1 and b1 are co prime. So, our assumption is wrong that √5 is rational.

2. Prove that 3+ 2√5 is irrational.

Proof: Let 3+2√5 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

3+2√5 = \cfrac { a }{ b }  

⇒ \cfrac { a }{ b }  – 3 = 2√5

⇒ \cfrac { a-3b }{ 2b } = √5

Since a and b are integers then \cfrac { a-3b }{ 2b } is rational and therefore √5 is rational, which is a contradiction to the fact that √5 is irrational.

Hence, our assumption is wrong that 3+ 2√5 is rational.

Therefore, 3+ 2√5 is irrational.

3. Prove that the following are irrational.

(i) \cfrac { 1 }{ \sqrt { 2 } }    
(ii) 7√5   
(iii) 6+√2

(i) Let \cfrac { 1 }{ \sqrt { 2 } } is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

\cfrac { 1 }{ \sqrt { 2 } } = \cfrac { a }{ b }  

⇒ \cfrac { b }{ a }  = √2.

Since, a and b are integers then \cfrac { b }{ a }  is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

Hence, our assumption is wrong and \cfrac { 1 }{ \sqrt { 2 } }  is irrational.

(ii) Let 7√5 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

7√5 = \cfrac { a }{ b }  

⇒ \cfrac { a }{ 7b }  = √5.

Since, a and b are integers, then \cfrac { a }{ 7b }  is rational

Therefore, √5 is rational, which is a contradiction to the fact that √5 is irrational.

Hence, our assumption is wrong and 7√5 is irrational.

(iii) Let  6+√2 is not irrational, i.e., rational number. Then there exist two co prime integers a and b, where (b ≠ 0) such that

6+√2 = \cfrac { a }{ b }

⇒ \cfrac { a }{ b }  – 6 = √2

⇒ \cfrac { a - 6b }{ b }  = √2

Since, a and b are integers, then \cfrac { a - 6b }{ b }  is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

Hence, our assumption is wrong and 6+√2 is irrational.

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 – Real Numbers

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NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers

1. Express each number as a product of its prime factors:

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

Solution:

(i) 140 = 2 × 2 × 5 × 7
= 22 × 5 × 7
Prime factors of 140 are 2, 5, 7

(ii) 156 = 2 × 2 × 3 × 13
= 22 × 3 × 13
Prime factors of 156 are 2, 3, 13.

(iii) 3825 = 5 × 5 × 3 × 3 ×17
= 52 × 32 ×17
Prime factors of 3825 are 5, 3, 17.

(iv) 5005 = 5 × 7 × 11 × 13
Prime factors of 5005 are 5, 7, 11, 13.

(v) 7429 = 17 × 19 × 23
Prime factors of 7429 are 17, 19, 23.

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

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

Solution:

We know that HCF of two numbers equal to the product of the smallest power of each common prime factor in the numbers.
And LCM of two numbers equal to product of the greatest power of each prime factor, involved in the numbers.

(i) 26 = 2 × 13
91 = 7 × 13
Here 13 is common in both the prime factors of 26 and 91.
HCF of 26 and 91 is 13
LCM of these two numbers = 2 × 7 × 13 =182.
HCF × LCM = 13 × 182 = 2366
Product of 26 and 91 = 26 × 91 = 2366.
Hence, LCM × HCF = Product of the given two numbers

(ii) 510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Here, 2 is common in both the prime factors of 510 and 92
HCF of 510 and 92 is 2
LCM = 2 × 2 × 3 × 5  17 × 23 = 23460
HCF × LCM = 2 × 23460 = 46920
Product of 510 and 92 is = 510 × 92 = 46920.
Hence, LCM × HCF = Product of the given two numbers

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
Here, 2 and 3 are common prime factors of 336 and 54.
HCF = 2×3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
HCF × LCM = 6 × 3024 = 18144
Product of336 and 54 = 336 × 54 = 18144
Hence, LCM × HCF = product of the given two numbers

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

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

Solution: We know that HCF of two numbers equal to the product of the smallest power of each common prime factor in the numbers.
And LCM of two numbers equal to product of the greatest power of each prime factor, involved in the numbers.

(i) 12 = 2 × 2 × 3, 15 = 3 × 5 and 21 = 3 × 7
Therefore HCF (12, 15, 21) = 3, since 3 is the only common factor.
And LCM (12, 15, 21) = 2 × 2 × 3 × 5  × 7 = 420

(ii) Here the numbers 17, 23, 29 are all primes
So no common prime factors are there.
Hence HCF (17, 23, 29) = 1
And LCM (17, 23, 29) = 17 × 23 × 29 = 11339.

(iii) 8 = 23 , 9 = 32  and 25 = 52
Since no common prime factors are there so HCF (8, 9, 25) = 1
And LCM (8, 9, 25) = 8 × 9 × 25 = 1800.

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

Solution:

We know that,
HCF × LCM = Product of the given two numbers.
Therefore 9 × LCM (306, 657) = 306 × 657
i.e., LCM (306, 657) = \cfrac { 306\times 657 }{ 9 }
LCM (306, 657)  = 22338

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

Solution: 

Let 6n can end with the digit 0. Then 5 must be a prime factor of it.
Now 6n = (2 × 3)n, i.e., 2 and 3 are the only prime factors, which is a contradiction to the fact that 5 is a prime factor of 6n. Hence our assumption is wrong.
Therefore, 6n can not end with the digit 0 for any natural number n.

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

Solution:

There are two types of numbers – Prime numbers and Composite Numbers.
Prime number is a natural number greater than 1 that has no positive divisors other than 1 and the number itself.
Composite numbers are numbers divisible by one or more numbers other than 1 and itself.
7 × 11 × 13 + 13 = 13 ( 7 × 11 + 1 ) = 13 × ( 77 + 1 ) = 13 × 78 = 13 × 13 × 6
Hence, given expression is divisible by 13 and 6
Therefore, 7 × 11 × 13 + 13 is a composite numbers.
Similarly, 7 × 6 × 5 × 4 × 3 × 2 × 1+5 = 5 ( 7 × 6 × 4 × 3 × 2 + 1) = 5 × (1008 + 1 ) = 5 × 1009
Hence, given expression is divisible by 5 and 1009.
Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1+5 is a composite numbers.

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 in the same direction. After how many minutes will they meet again at the starting point?

Solution:

To find the required time we have to find the LCM of these two time which they takes to drive one round.
18 = 2 × 3 × 3
12 = 2 × 2 × 3.
LCM (18, 12) = 2 × 2 × 3 × 3 = 36
Hence, after 36 minutes they will meet again.

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers

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NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 – Real Numbers

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

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

Solution: Euclid’s division lemma state that “Given positive integers  and , there exist unique integers and satisfying  a = bq + r, 0≤r<b”
If , r = 0 then b, is called HCF of a and b.

(i) Here 225 > 135. Applying division lemma to 225 and 135, we get
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, again apply the division lemma to 90 and 135
135 = 90 x 1 + 45
Since now remainder is 45 ≠ 0, apply division algorithm to 45 and 90
90 = 45 x 2 + 0
The remainder is now become zero and the divisor is 45 here.
Hence the HCF of 135 and 225 is 45

(ii) Since 38220 > 196, Applying division lemma to 38220 and 196, we get
38220 = 196 x 195 + 0
The remainder is become zero and the divisor is 196.
Hence the HCF of 38220 and 196 is 196.

(iii) Here 867 > 255. Applying division lemma to 867 and 255, we get
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, again apply the division lemma to 102 and 255
255 = 102 x 2 + 51
Since now remainder is 51 ≠ 0, apply division algorithm to 102 and 51
102 = 51 x 2 + 0
The remainder is now become zero and the divisor is 51 here.
Hence the HCF of 867 and 255 is 51

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 positive odd integer which gives q as quotient and r as remainder when it is divide by 6.
Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,
a = 6q + r, 0 ≤ r < 6,
So possible values of r are 0, 1, 2, 3, 4 and 5.
Now, when r = 0, a = 6q = 2 x 3q, which is divisible by 2 so 6q is even positive number.
when, r = 1, a = 6q + 1, not divisible by 2, hence 6q + 1 is odd positive number.
when, r = 2, a = 6q + 2 = 2(3q + 1 ) i.e., divisible by 2, hence 6q + 2 is even positive number.
when, r = 3, a = 6q + 3 = 3(2q + 1), not divisible by 2, so 6q + 3 is odd positive number.
when, r = 4, a = 6q + 4 = 2(3q + 2),  divisible by 2, hence 6q + 4 is even positive number.
when, r = 5, a = 6q + 5, not divisible by 2, so it is an odd positive integer.
Hence only odd positive integers are 6q + 1, 6q + 3, 6q + 5.
Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

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

Solution: To find the maximum number of columns we have to find the HCF of 616 and 32.
Use the Euclid’s division algorithm to 616, 32 we have
616 = 32 x 19 + 8
32 = 8 x 4 + 0
Since remainder is 0 and divisor is 8 hence HCF of  is 616, 32 is 8
Therefore, the maximum number of columns in which the army can march is 8.

4. Use Euclid’s division lemma two show that the square of any positive integer is either of the form or  for some integer .

Solution: Let x be any positive integer, which gives q as quotient and r as remainder when it is divided by 3.
Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,
a = 3q + r, 0 ≤ r < 3,
So possible values of r are 0, 1, 2.
Now, when r = 0, x = 3q
Therefore, x2  = (3q)2  = 9q2 = 3 x 3q2 = 3m, where m= 3q2
When, r = 1, x = 3q + 1
x2 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q
And when r = 2, x = 3q + 2
Hence x2 = (3q + 2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m= 3q2 + 4q + 1) + 1
Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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

Solution: Let x be any positive integer, which gives q as quotient and r as remainder when it is divided by 9.
Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,
a = 9q + r, 0 ≤ r < 9 we get,
So possible values of r are 0, 1, 2, 3, 4, 5, 6, 7, 8.

Now, when r = 0, x = 9q
Therefore, x3 = (9q)3 = 9 x 81q3 = 9m, where m = 81q3
When r = 1, x = 9q + 1
x3 = (9q + 1)3 = (9q)3 + 3.(9q)2.1 + 3.9q.12 + 13
9(81q3 + 3 x 9q2 + 3q) + 1
= 9m + 1, where m = 81q3 + 3 x 9q2 + 3q

When r = 2, x = 9q + 2
Hence x3 = (9q + 2)3
= (9q)3 + 3.(9q)2.2 + 3 x 9q.22 + 23
= 9(81q3 + 3.9.2q2 + 3q.22) + 8
= 9m + 8, where m = 81q3 + 3.9.2q2 + 3q.22

When r = 3, x = 9q + 3
Hence x3 = (9q + 3)3
= (9q)3 + 3.(9q)2.3 + 3.9q.32 + 33
= 9(81q3 + 3.9.q2.3 + 3q.32) + 27
= 9(81q3 + 3.9.q2.3 + 3q.32 + 3)
= 9m, where m = 81q3 + 3.9.q2.3 + 3q.32 + 3

When r = 4, x = 9q + 4
Hence x3 = (9q + 4)3
= (9q)3 + 3.(9q)2.4 + 3.9q.42 + 43
= 9(81q3 + 3.9.q2.4 + 3q.42) + 64
= 9(81q3 + 3.9.q2.4 + 3q.42 + 7) + 1
= 9m + 1, where m = 81q3 + 3.9.q2.4 + 3q.42 + 7

Continuing this process for r = 5, 6, 7 and 8, we get that x3 is of the form 9m or 9m + 1 or 9m + 8.
Therefore, the cube of any positive integer is either of the form 9m or 9m + 1 or 9m + 8, for some integer m.

Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 – Real Numbers

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