{"id":55108,"date":"2020-11-16T18:01:10","date_gmt":"2020-11-16T18:01:10","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55108"},"modified":"2023-09-19T02:10:37","modified_gmt":"2023-09-19T02:10:37","slug":"ncert-solutions-for-class-10th-mathematics-chapter-1-real-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-1-real-numbers\/","title":{"rendered":"NCERT Solutions for Class 10th Mathematics: Chapter 1 Real Numbers"},"content":{"rendered":"\n

Class 10: Mathematics Chapter 1 solutions. Complete Class 10 Mathematics Chapter 1 Notes.<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 1 Real Numbers<\/strong><\/h2>\n\n\n\n

NCERT 10th Mathematics Chapter 1, class 10 Mathematics Chapter 1 solutions<\/p>\n\n\n\n

Page No: 7<\/p>\n\n\n\n

Exercise 1.1<\/strong><\/p>\n\n\n\n

1. Use Euclid’s division algorithm to find the HCF of:<\/p>\n\n\n\n

(i) 135 and 225<\/p>\n\n\n\n

(ii) 196 and 38220<\/p>\n\n\n\n

(iii) 867 and 255<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 225 > 135 we always divide greater number with smaller one.<\/p>\n\n\n\n

Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 \u00d7 1 + 90<\/p>\n\n\n\n

Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 \u00d7 1 + 45<\/p>\n\n\n\n

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 \u00d7 45+ 0<\/p>\n\n\n\n

As there are no remainder so divisor 45 is our HCF.<\/p>\n\n\n\n

(ii) 38220 > 196 we always divide greater number with smaller one.<\/p>\n\n\n\n

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 \u00d7 195 + 0<\/p>\n\n\n\n

As there is no remainder so divisor 196 is our HCF.<\/p>\n\n\n\n

(iii) 867 > 255 we always divide greater number with smaller one.<\/p>\n\n\n\n

Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 \u00d7 3 + 102<\/p>\n\n\n\n

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 \u00d7 2 + 51<\/p>\n\n\n\n

Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 \u00d7 2 + 0<\/p>\n\n\n\n

As there is no remainder so divisor 51 is our HCF.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 1, class 10 Mathematics Chapter 1 solutions<\/p>\n\n\n\n

2. Show that any positive odd integer is of the form 6q<\/em> + 1, or 6q<\/em> + 3, or 6q<\/em> + 5, where q<\/em> is some integer.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let take a<\/em> as any positive integer and b<\/em> = 6.<\/p>\n\n\n\n

Then using Euclid\u2019s algorithm we get a = 6q<\/em> + r<\/em> here r<\/em> is remainder and value of q<\/em> is more than or equal to 0 and r<\/em> = 0, 1, 2, 3, 4, 5 because 0 \u2264 r<\/em> < b and the value of b<\/em> is 6 <\/p>\n\n\n\n

So total possible forms will 6q <\/em>+ 0 , 6q <\/em>+ 1 , 6q <\/em>+ 2,6q <\/em>+ 3, 6q<\/em> + 4, 6q<\/em> + 5<\/p>\n\n\n\n

6 is divisible by 2 so it is a even number <\/p>\n\n\n\n

6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number <\/p>\n\n\n\n

6 is divisible by 2 and 2 is also divisible by 2 so it is a even number <\/p>\n\n\n\n

6q<\/em>  +3 <\/p>\n\n\n\n

6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number <\/p>\n\n\n\n

6 is divisible by 2 and 4 is also divisible by 2 it is a even number<\/p>\n\n\n\n

6q<\/em> + 5 <\/p>\n\n\n\n

6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number<\/p>\n\n\n\n

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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 \u00d7 19 + 8
32 = 8 \u00d7 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.<\/p>\n\n\n\n

4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m<\/em> or 3m<\/em> + 1 for some integer m.
[Hint: Let x<\/em> be any positive integer then it is of the form 3q<\/em>, 3q<\/em> + 1 or 3q<\/em> + 2. Now square each of these and show that they can be rewritten in form 3m<\/em> or 3m<\/em> + 1.]<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let a be any positive integer and b<\/em> = 3.
Then a = 3q<\/em> + r<\/em> for some integer q<\/em> \u2265 0
And r<\/em> = 0, 1, 2 because 0 \u2264 r<\/em> < 3
Therefore, a<\/em> = 3q<\/em> or 3q<\/em> + 1 or 3q<\/em> + 2
Or,
a<\/em>2<\/sup> = (3q<\/em>)2<\/sup> or (3q<\/em> + 1)2<\/sup> or (3q<\/em> + 2)2<\/sup>
a<\/em>2<\/sup> = (9q<\/em>)2<\/sup> or 9q<\/em>2<\/sup> + 6q<\/em> + 1 or 9q<\/em>2<\/sup> + 12q<\/em> + 4
= 3 \u00d7 (3q<\/em>2<\/sup>) or 3(3q<\/em>2<\/sup> + 2q<\/em>) + 1 or 3(3q<\/em>2<\/sup> + 4q<\/em> + 1) + 1
= 3k<\/em>1<\/sub> or 3k<\/em>2<\/sub> + 1 or 3k<\/em>3<\/sub> + 1<\/p>\n\n\n\n

Where k<\/em>1<\/sub>, k<\/em>2<\/sub>, and k<\/em>3<\/sub> are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m<\/em> or 3m<\/em> + 1.<\/p>\n\n\n\n

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m<\/em>, 9m <\/em>+ 1 or 9m<\/em> + 8.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let a be any positive integer and b = 3
a<\/em> = 3q<\/em> + r<\/em>, where q<\/em> \u2265 0 and 0 \u2264 r<\/em> < 3
\u2234 a<\/em> = 3q or 3q<\/em> + 1 or 3q<\/em> + 2
Therefore, every number can be represented as these three forms. There are three cases.<\/p>\n\n\n\n

Case 1: When a<\/em> = 3q<\/em>,<\/p>\n\n\n\n

a<\/em>3<\/sup> = (3q<\/em>)3<\/sup> = 27q<\/em>3<\/sup> = 9(3q<\/em>)3<\/sup> = 9m<\/em>,
Where m<\/em> is an integer such that m<\/em> = 3q<\/em>3<\/sup><\/p>\n\n\n\n

Case 2: When a <\/em>= 3q + 1,
a<\/em>3<\/sup> = (3q<\/em> +1)3<\/sup>
a<\/em>3<\/sup>= 27q<\/em>3<\/sup> + 27q<\/em>2<\/sup> + 9q<\/em> + 1
a<\/em>3<\/sup> = 9(3q<\/em>3<\/sup> + 3q<\/em>2<\/sup> + q<\/em>) + 1
a<\/em>3<\/sup> = 9m<\/em> + 1
Where m<\/em> is an integer such that m<\/em> = (3q<\/em>3<\/sup> + 3q<\/em>2<\/sup> + q<\/em>)<\/p>\n\n\n\n

Case 3: When a<\/em> = 3q<\/em> + 2,
a<\/em>3<\/sup> = (3q<\/em> +2)3<\/sup>
a<\/em>3<\/sup>= 27q<\/em>3<\/sup> + 54q<\/em>2<\/sup> + 36q<\/em> + 8
a<\/em>3<\/sup> = 9(3q<\/em>3<\/sup> + 6q<\/em>2<\/sup> + 4q) + 8
a<\/em>3<\/sup> = 9m<\/em> + 8
Where m<\/em> is an integer such that m<\/em> = (3q<\/em>3<\/sup> + 6q<\/em>2<\/sup> + 4q<\/em>)<\/p>\n\n\n\n

Therefore, the cube of any positive integer is of the form 9m<\/em>, 9m<\/em> + 1,
or 9m<\/em> + 8.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 1, class 10 Mathematics Chapter 1 solutions<\/p>\n\n\n\n

Exercise 1.2<\/strong><\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 1<\/p>\n\n\n\n

1. Express each number as product of its prime factors:<\/strong><\/p>\n\n\n\n

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 140 = 2 \u00d7 2 \u00d7 5 \u00d7 7 = 22<\/sup> \u00d7 5 \u00d7 7
(ii) 156 = 2 \u00d7 2 \u00d7 3 \u00d7 13 = 22<\/sup> \u00d7 3 \u00d7 13
(iii) 3825 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 17 = 32<\/sup> \u00d7 52<\/sup> \u00d7 17
(iv) 5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13
(v) 7429 = 17 \u00d7 19 \u00d7 23<\/p>\n\n\n\n

2. Find the LCM and HCF of the following pairs of integers and verify that LCM \u00d7 HCF = product of the two numbers.<\/strong>
(i) 26 and 91<\/p>\n\n\n\n

(ii) 510 and 92 <\/p>\n\n\n\n

(i) 26 = 2 \u00d7 13
91 =7 \u00d7 13
HCF = 13
LCM =2 \u00d7 7 \u00d7 13 =182
Product of two numbers 26 \u00d7 91 = 2366
Product of HCF and LCM 13 \u00d7 182 = 2366
Hence, product of two numbers = product of HCF \u00d7 LCM<\/p>\n\n\n\n

(ii) 510 = 2 \u00d7 3 \u00d7 5 \u00d7 17
92 =2 \u00d7 2 \u00d7 23
HCF = 2
LCM =2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 17 \u00d7 23 = 23460
Product of two numbers 510 \u00d7 92 = 46920
Product of HCF and LCM 2 \u00d7 23460 = 46920
Hence, product of two numbers = product of HCF \u00d7 LCM<\/p>\n\n\n\n

(iii) 336 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 7
54 = 2 \u00d7 3 \u00d7 3 \u00d7 3
HCF = 2 \u00d7 3 = 6
LCM = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 7 =3024
Product of two numbers 336 \u00d7 54 =18144
Product of HCF and LCM 6 \u00d7 3024 = 18144
Hence, product of two numbers = product of HCF \u00d7 LCM.<\/p>\n\n\n\n

3. Find the LCM and HCF of the following integers by applying the prime factorization method.<\/strong><\/p>\n\n\n\n

(i) 12, 15 and 21 <\/p>\n\n\n\n

(ii) 17, 23 and 29 <\/p>\n\n\n\n

(iii) 8, 9 and 25<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 12 = 2 \u00d7 2 \u00d7 3
15 =3 \u00d7 5
21 =3 \u00d7 7
HCF = 3
LCM = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 7 = 420<\/p>\n\n\n\n

(ii) 17 = 1 \u00d7 17
23 = 1 \u00d7 23
29 = 1 \u00d7 29
HCF = 1
LCM = 1 \u00d7 17 \u00d7 19 \u00d7 23 = 11339<\/p>\n\n\n\n

(iii) 8 =1 \u00d7 2 \u00d7 2 \u00d7 2
9 =1 \u00d7 3 \u00d7 3
25 =1 \u00d7 5 \u00d7 5
HCF =1
LCM = 1 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5 = 1800<\/p>\n\n\n\n

4. Given that HCF (306, 657) = 9, find LCM (306, 657).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We have the formula that
Product of LCM and HCF = product of number
LCM \u00d7 9 = 306 \u00d7 657
Divide both side by 9 we get
LCM = (306 \u00d7 657) \/ 9 = 22338<\/p>\n\n\n\n

5. Check whether 6n<\/em> can end with the digit 0 for any natural number n<\/em>.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 \u00d7 5.<\/p>\n\n\n\n

So value 6n<\/em> should be divisible by 2 and 5 both 6n<\/em> is divisible by 2 but not divisible by 5 So it can not end with 0.<\/p>\n\n\n\n

6. Explain why 7 \u00d7 11 \u00d7 13 + 13 and 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 are composite numbers.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

7 \u00d7 11 \u00d7 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.<\/p>\n\n\n\n

7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5
Taking 5 common, we get
5(7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 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.<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

They will be meet again after LCM of both values at the starting point.
18 = 2 \u00d7 3 \u00d7 3
12 = 2 \u00d7 2 \u00d7 3
LCM = 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.<\/p>\n\n\n\n

Exercise 1.3<\/h5>\n\n\n\n

NCERT 10th Mathematics Chapter 1, class 10 Mathematics Chapter 1 solutions<\/p>\n\n\n\n

1. Prove that \u221a5 is irrational.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let take \u221a5 as rational number
If a<\/em> and b<\/em> are two co prime number and b<\/em> is not equal to 0.
We can write \u221a5 = a<\/em>\/b<\/em>
Multiply by b both side we get
b<\/em>\u221a5 = a<\/em>
To remove root, Squaring on both sides, we get
5b<\/em>2<\/sup> = a<\/em>2<\/sup> \u2026  (i)<\/strong><\/p>\n\n\n\n

Therefore, 5 divides a<\/em>2<\/sup> and according to theorem of rational number, for any prime number p<\/em> which is divides a<\/em>2<\/sup> then it will divide a<\/em> also.
That means 5 will divide a<\/em>. So we can write
a<\/em> = 5c<\/em>
Putting value of a<\/em> in equation (i)<\/strong> we get
5b<\/em>2<\/sup> = (5c<\/em>)2<\/sup>
5b<\/em>2<\/sup> = 25c<\/em>2<\/sup>
Divide by 25 we get<\/p>\n\n\n\n

b<\/em>2\/5 = c<\/em>2<\/sup><\/p>\n\n\n\n

Similarly, we get that b<\/em> will divide by 5
and we have already get that a<\/em> is divide by 5
but a<\/em> and b<\/em> are co prime number. so it contradicts.
Hence \u221a5 is not a rational number, it is irrational.<\/p>\n\n\n\n

2. Prove that 3 + 2\u221a5 is irrational.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let take that 3 + 2\u221a5 is a rational number.
So we can write this number as
3 + 2\u221a5 = a<\/em>\/b<\/em>
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2\u221a5 = a<\/em>\/b<\/em> \u2013 3
2\u221a5 = (a<\/em>-3b<\/em>)\/b<\/em>
Now divide by 2, we get
\u221a5 = (a<\/em>-3b<\/em>)\/2b<\/em>
Here a<\/em> and b<\/em> are integer so (a<\/em>-3b<\/em>)\/2b<\/em> is a rational number so \u221a5 should be a rational number But \u221a5 is a irrational number so it contradicts.
Hence, 3 + 2\u221a5 is a irrational number.<\/p>\n\n\n\n

3. Prove that the following are irrationals:
<\/strong>(i) 1\/\u221a2 (ii) 7\u221a5 (iii) 6 + \u221a2<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Let take that 1\/\u221a2 is a rational number.
So we can write this number as<\/p>\n\n\n\n

1\/\u221a2 = a<\/em>\/b<\/em>
Here a <\/em>and b<\/em> are two co prime number and b<\/em> is not equal to 0
Multiply by \u221a2 both sides we get
1 = (a<\/em>\u221a2)\/b<\/em>
Now multiply by b<\/em>
b<\/em> = a<\/em>\u221a2
divide by a we get
b<\/em>\/a<\/em> = \u221a2
Here a<\/em> and b<\/em> are integer so b\/a is a rational number so \u221a2 should be a rational number But \u221a2 is a irrational number so it contradicts.
Hence, 1\/\u221a2 is a irrational number<\/p>\n\n\n\n

(ii) Let take that 7\u221a5 is a rational number.
So we can write this number as
7\u221a5 = a<\/em>\/b<\/em>
Here a<\/em> and b<\/em> are two co prime number and b<\/em> is not equal to 0
Divide by 7 we get
\u221a5 = a<\/em>\/(7b<\/em>)
Here a<\/em> and b<\/em> are integer so a<\/em>\/7b<\/em> is a rational number so \u221a5 should be a rational number but \u221a5 is a irrational number so it contradicts.
Hence, 7\u221a5 is a irrational number.<\/p>\n\n\n\n

(iii) Let take that 6 + \u221a2 is a rational number.
So we can write this number as
6 + \u221a2 = a<\/em>\/b<\/em>
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
\u221a2 = a<\/em>\/b<\/em> \u2013 6
\u221a2 = (a<\/em>-6b<\/em>)\/b<\/em>
Here a<\/em> and b<\/em> are integer so (a-6b<\/em>)\/b<\/em> is a rational number so \u221a2 should be a rational number.<\/p>\n\n\n\n

But \u221a2 is a irrational number so it contradicts.
Hence, 6 + \u221a2 is a irrational number.<\/p>\n\n\n\n

Exercise 1.4<\/h2>\n\n\n\n

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
<\/strong>(i) 13\/3125<\/p>\n\n\n\n

(ii) 17\/8 <\/p>\n\n\n\n

(iii) 64\/455 <\/p>\n\n\n\n

(iv) 15\/1600 <\/p>\n\n\n\n

(v) 29\/343 <\/p>\n\n\n\n

(vi) 23\/23 <\/sup>\u00d7 52<\/sup> <\/p>\n\n\n\n

(vii) 129\/22 <\/sup>\u00d7 57 <\/sup>\u00d7 75<\/sup> <\/p>\n\n\n\n

(viii) 6\/15 <\/p>\n\n\n\n

(ix) 35\/50 <\/p>\n\n\n\n

(x) 77\/210<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 13\/3125
Factorize the denominator we get
3125 =5 \u00d7 5 \u00d7 5 \u00d7 5 \u00d7 5 = 55<\/sup><\/p>\n\n\n\n

So denominator is in form of 5m<\/sup> so it is terminating .<\/p>\n\n\n\n

(ii) 17\/8
Factorize the denominator we get
8 =2 \u00d7 2 \u00d7 2 = 23<\/sup>
So denominator is in form of 2m<\/sup> so it is terminating .<\/p>\n\n\n\n

(iii) 64\/455
Factorize the denominator we get
455 =5 \u00d7 7 \u00d7 13
There are 7 and 13 also in denominator so denominator is not in form of 2m<\/sup> \u00d7 5n<\/sup> . so it is not terminating.<\/p>\n\n\n\n

(iv) 15\/1600
Factorize the denominator we get
1600 =2 \u00d7 2 \u00d7 2 \u00d72 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5 = 26<\/sup> \u00d7 52<\/sup>
so denominator is in form of 2m<\/sup> \u00d7 5n<\/sup>
Hence it is terminating.<\/p>\n\n\n\n

(v) 29\/343
Factorize the denominator we get
343 = 7 \u00d7 7 \u00d7 7 = 73<\/sup>
There are 7 also in denominator so denominator is not in form of 2m<\/sup> \u00d7 5n<\/sup>
Hence it is non-terminating.<\/p>\n\n\n\n

(vi) 23\/(23<\/sup> \u00d7 52<\/sup>)
Denominator is in form of 2m<\/sup> \u00d7 5n<\/sup>
Hence it is terminating.<\/p>\n\n\n\n

(vii) 129\/(22<\/sup> \u00d7 57<\/sup> \u00d7 75<\/sup> )
Denominator has 7 in denominator so denominator is not in form of 2m<\/sup> \u00d7 5n<\/sup>
Hence it is none terminating.<\/p>\n\n\n\n

(viii) 6\/15
divide nominator and denominator both by 3 we get 2\/5
Denominator is in form of 5m<\/sup> so it is terminating.<\/p>\n\n\n\n

(ix) 35\/50 divide denominator and nominator both by 5 we get 7\/10
Factorize the denominator we get
10=2 \u00d7 5
So denominator is in form of 2m<\/sup> \u00d7 5n<\/sup> so it is terminating.<\/p>\n\n\n\n

(x) 77\/210
simplify it by dividing nominator and denominator both by 7 we get 11\/30
Factorize the denominator we get
30=2 \u00d7 3 \u00d7 5
Denominator has 3 also in denominator so denominator is not in form of 2m<\/sup> \u00d7 5n<\/sup><\/p>\n\n\n\n

Hence it is none terminating.<\/p>\n\n\n\n

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 13\/3125 = 13\/55<\/sup> = 13\u00d725<\/sup>\/55<\/sup>\u00d725<\/sup> = 416\/105<\/sup> = 0.00416 (ii) 17\/8 = 17\/23<\/sup> = 17\u00d753<\/sup>\/23<\/sup>\u00d753<\/sup> = 17\u00d753<\/sup>\/103<\/sup> = 2125\/103<\/sup> = 2.125<\/p>\n\n\n\n

(iv) 15\/1600 = 15\/24<\/sup>\u00d7102<\/sup> = 15\u00d754<\/sup>\/24<\/sup>\u00d754<\/sup>\u00d7102<\/sup> = 9375\/106<\/sup> = 0.009375<\/p>\n\n\n\n

(vi) 23\/23<\/sup>52<\/sup> = 23\u00d753<\/sup>\u00d722<\/sup>\/23<\/sup> 52<\/sup>\u00d753<\/sup>\u00d722<\/sup> = 11500\/105<\/sup> = 0.115<\/p>\n\n\n\n

(viii) 6\/15 = 2\/5 = 2\u00d72\/5\u00d72 = 4\/10 = 0.4<\/p>\n\n\n\n

(ix) 35\/50 = 7\/10 = 0.7.<\/p>\n\n\n\n

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<\/em> , q<\/em> you say about the prime factors of q<\/em>?<\/strong><\/p>\n\n\n\n

(i) 43.123456789<\/p>\n\n\n\n

(ii) 0.120120012000120000…<\/p>\n\n\n\n

(iii) 43.123456789<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Since this number has a terminating decimal expansion, it is a rational number of the form p<\/em>\/q, and q<\/em> is of the form 2m<\/sup> \u00d7 5n<\/sup>.<\/p>\n\n\n\n

(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<\/em>\/q, and q<\/em> is not of the form 2m<\/sup> \u00d7 5n<\/sup>.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 1, class 10 Mathematics Chapter 1 solutions<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 1: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 1 Real Numbers<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for Class 10th Mathematics: Chapter 1 Real Numbers PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h4>\n\n\n\n