# NCERT solution for class 10 maths real numbers ( Chapter 1) #### Solution for Exercise 1.1

Q1. Use Euclid’s division algorithm to find the HCF of :
1)135 and 225
2)196 and 38220
3).867 and 255

1). 135 and 225
We have 225 > 135,
So, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Here remainder $$90 \ne 0$$, we apply the division lemma again to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder $$45 \ne 0$$, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
The divisor at this stage is 45
Therefore, the HCF of 135 and 225 is 45.

2). 196 and 38220
We have 38220 > 196,
So, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
The divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.

3). 867 and 255
We have 867 > 225,
So, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Here remainder $$102 \ne 0$$, we apply the division lemma again to 255 and 102 to obtain
255 = 102 × 2 + 51
Here remainder $$51 \ne 0$$, we apply the division lemma again to 102 and 51 to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
The divisor at this stage is 51
Therefore, the HCF of 867 and 255 is 51.

Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer $$q \ne 0$$, and r = 0, 1, 2, 3, 4, 5 because $$0 \leq r < 6$$.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = $$2k_1 + 1$$, where $$k_1$$ is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = $$2k_2 + 1$$, where $$k_2$$ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = $$2k_3 + 1$$, where $$k_3$$ is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,
or 6q + 5

Q3.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?

We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.
To find the HCF, we can use Euclid’s algorithm.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Q4. 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.

Let a be any positive integer and b = 3.
Then a = 3q + r for some integer $$q \ne 0$$
And r = 0, 1, 2 because $$0 \leq r < 3$$
Therefore, a = 3q or 3q + 1 or 3q + 2
$$a^2 = (3q)^2$$ or $$(3q + 1)^2$$ or $$(3q + 2)^2$$
We know,
$$a^2 = 9q^2$$ or $$9q^2 + 6q + 1$$ or $$9q^2 + 12q + 4$$
=> $$a^2 = 3 × (3q^2)$$ or $$3 × (3q^2 + 2q)$$ or $$3 × (3q^2 + 4q + 1) + 1$$
=> $$3k_1$$ or $$3k_2 + 1$$ or $$3k_3 + 1$$
Where $$k_1,k_2,k_3$$ 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.

Q5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Let a be any positive integer and b = 3
a = 3q + r, where q $$\ne$$ 0 and 0 < r < 3
a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
We have three cases.
Case 1 : When a = 3q,
$$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$$
Where m is an integer such that $$m = 3q^3$$
Case 2 :When a = 3q + 1
=> $$a^3 = (3q + 1)^3$$
=> $$a^3 = 27q^3 + 27q^2 + 9q + 1$$
=> $$a^3 = 9( 3q^3 + 3q^2 + q) + 1$$
=> $$a^3 = 9m + 1$$
When m is an integer such that $$m = (3q^3 + 3q^2 + q)$$
Case 3 :When a = 3q + 2,
=> $$a^3 = (3q + 2)^3$$
=> $$a^3 = 27q^3 + 54q^2 + 36q + 8$$
=> $$a^3 = 9( 3q^3 + 6q^2 + 4q) + 8$$
=> $$a^3 = 9m + 8$$
When m is an integer such that $$m = (3q^3 + 6q^2 + 4q)$$
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

#### Solution for Exercise 1.2

Q1. Express each number as product of its prime factors:
1). 140
2). 156
3) 3825
4). 5005
5). 7429

140 = $$2 × 2 × 5 × 7$$ = $$2^2 × 5 × 7$$
156 = $$2 × 2 × 3 × 13$$ = $$2^2 × 3 × 13$$
3825 = $$3 × 3 × 5 × 5 × 17$$ = $$3^2 × 5^2 × 17$$
5005 = $$5 × 7 × 11 × 13$$
7429 = $$17 × 19 × 23$$

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
1) 26 and 91
2) 510 and 92
3) 336 and 54

1.26 and 91
$$26 = 2 × 13$$
$$91 = 7 × 13$$
HCF = 13
LCM = $$2 × 7 × 13 = 182$$
Product of two numbers 26 and 91 = $$26 × 91 = 2366$$
HCF × LCM = $$13 × 182 = 2366$$
Product of two numbers is HCF × LCM, so it is verified.
2.510 and 92
$$510 = 2 × 3 × 5 × 17$$
$$92 = 2 × 2 × 23$$
HCF = 2
LCM = $$2 × 2 × 3 × 5 × 17 × 23 = 23460$$
Product of two numbers 510 and 92 = $$510 × 92 = 46920$$
HCF × LCM = $$2 × 23460 = 46920$$
Product of two numbers is HCF × LCM, so it is verified
3.336 and 54
$$336 = 2 × 2 × 2 × 2 × 3 × 7 = 2^4 + 3 + 7$$
$$54 = 2 × 3 × 3 × 3 = 2 × 3^3$$
HCF = $$2 × 3 = 6$$
LCM = $$2^4 × 3^3 × 7 = 3024$$
Product of two numbers 336 and 54 = $$336 × 54 = 18144$$
HCF × LCM = $$6 × 3024 = 18144$$
Product of two numbers is HCF × LCM, so it is verified.

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
1.12, 15 and 21
2.17, 23 and 29
3.8, 9 and 25

1.12, 15 and 21
$$12 = 2^2 × 3$$
$$15 = 3 × 5$$
$$21 = 3 × 7$$
HCF = 3
LCM = $$2^2 × 3 × 5 × 7 = 420$$
2.17, 23 and 29
$$17 = 1 × 17$$
$$23 = 1 × 23$$
$$29 = 1 × 29$$
HCF = 1
LCM = $$17 × 23 × 29 = 11339$$
3.8, 9 and 25
$$8 = 2 × 2 × 2 = 2^3$$
$$9 = 3 × 3 = 3^2$$
$$25 = 5 × 5 = 5^2$$
HCF = 1
LCM = $$2^3 × 3^2 × 5^2 = 1800$$

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Since product of two numbers is equal to the product of HCF and LCM,
$$LCM × HCF = 306 × 657$$
$$LCM = {{306 × 657} \over HCF} = {{306 × 657} \over 9}$$
LCM = 22338

Q5. Check whether $$6^n$$ can end with the digit 0 for any natural number n.

A number ending with 0 is divisible by 10 according to the divisibility rule
And a number divisible by 10 is also divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of $$6^n = (2 × 3)^n$$
5 is not in the prime factorisation of $$6^n$$
So, for any value of n, $$6^n$$ will not be divisible by 5
Hence, $$6^n$$ cannot end with the digit 0 for any natural number n.

Q6. Explain why $$7 × 11 × 13 + 13$$ and $$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$$ are composite numbers.

There are two types of number prime and composite
Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
So, $$7 × 11 × 13 + 13$$
=>$$13 × (7 × 11 + 1)$$
=>$$13 × (77 + 1)$$
=>$$13 × 78 = 13 × 13 × 6$$
6 and 13 are the factors of $$7 × 11 × 13 + 13$$
And we know that numbers which have factors other than one and itself are composite number
$$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$$
=>$$5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)$$
=>$$5 × (1008 + 1)$$
=>$$5 × 1009$$
1009 is a prime number and cannot be factorized further
So, 5 and 1009 are the factors of $$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$$
And we know that numbers which have factors other than one and itself are composite number

Q7.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?

Ravi is taking less time than Sonia in completing one round. Since they are going in the same direction they will surely meet again but at that time Ravi will have completed one round with respect to Sonia.
To find the minutes after which Ravi and Sonia will meet again we have to find the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
$$18 = 2 × 3 × 3$$
$$12 = 2 × 2 × 3$$
LCM of 12 and 18 = $$2 × 2 × 3 × 3 =36$$
So, Ravi and Sonia will meet again at the starting point after 36 minutes.

#### Solution for Exercise 1.3

Q1. Prove that $$\sqrt{5}$$ is irrational.

Let us suppose that $$\sqrt{5}$$ is rational,
So there should be two co- prime integers (p and q, where q $$\ne$$ 0 ) such that:
$$\sqrt{5} = {{p}\over{q}}$$
$$=>\sqrt{5} × q = p$$
On squaring both sides we'll get:
$$=>5q^2 = p^2$$ ............................(i)
From equation (i) it can be clearly seen that 5 is a factor of $$p^2$$
Therefore, 5 is also factor of p.
Let p = 5r, where r is an integer.
On substituting the value of p in equation (i),
$$=>5q^2 = 25r^2$$
$$=>q^2 = 5r^2$$.............................(ii)
From equation (ii) it can be clearly seen that 5 is a factor of $$q^2$$
Therefore, 5 is also factor of q.
Which means 5 is a common factor of p and q which contradicts our assumption,
that p and q are co-prime and also that $$\sqrt{5}$$ is rational.
Hence it is proved that $$\sqrt{5}$$ is irrational.

Q2. Prove that $$3 + 2\sqrt{5}$$ is irrational.

Let us suppose that $$3 + 2\sqrt{5}$$ is rational
So there should be two co- prime integers (p and q, where q $$\ne$$ 0 ) such that:
$$3 + 2\sqrt{5} = {{p}\over{q}}$$
$$=>2\sqrt{5} ={{p}\over{q}} - 3$$
$$=>2\sqrt{5} ={{p}\over{q}} - 3$$
$$=>\sqrt{5} =\frac{1}{2}(\frac{p}{q} - 3)$$ .................(i)
p and q are integers which means $$\frac{1}{2}(\frac{p}{q} - 3)$$ is a rational number.
But we know $$\sqrt{5}$$ is irrational.
This is not possible since LHS $$\ne$$ RHS.
That proves that our assumption that $$3 + 2\sqrt{5}$$ is rational is wrong.
Hence it is proved that $$3 + 2\sqrt{5}$$ is irrational.

Q3. Prove that the following are irrationals.
i.$${{1}\over \sqrt{2}}$$
ii.$$7 \sqrt{5}$$
iii.$$6 + \sqrt{2}$$

i.$${{1}\over \sqrt{2}}$$
Let us suppose that $${{1}\over \sqrt{2}}$$ is rational
So there should be two co- prime integers (p and q, where q $$\ne$$ 0 ) such that:
$${{1}\over \sqrt{2}} = {{p}\over{q}}$$
$$\sqrt{2} = {{q}\over{p}}$$
p and q are integers which means $${{q}\over{p}}$$ is a rational number.
But we know $$\sqrt{2}$$ is irrational.
This is not possible since LHS $$\ne$$ RHS.
This contradiction has risen because of our incorrect assumption that $${{1}\over \sqrt{2}}$$ is rational.
Hence it is proved that $${{1}\over \sqrt{2}}$$ is irrational.
ii.$$7 \sqrt{5}$$
Let us suppose that $$7 \sqrt{5}$$ is rational.
So there should be two co- prime integers (p and q, where q $$\ne$$ 0 ) such that:
$$7 \sqrt{5} = {{p}\over{q}}$$
$$\sqrt{5} = {{p}\over{7q}}$$
p and q are integers which means $${{p}\over{7q}}$$ is a rational number.
But we know $$\sqrt{5}$$ is irrational.
This is not possible since LHS $$\ne$$ RHS.
This contradiction has arisen because of our incorrect assumption that $$7 \sqrt{5}$$ is rational.
Hence it is proved that $$7 \sqrt{5}$$ is irrational.
iii. $$6 + \sqrt{2}$$
Let us suppose that $$6 + \sqrt{2}$$ is rational
So there should be two co- prime integers (p and q, where q $$\ne$$ 0 ) such that:
$$6 + \sqrt{2} = {{p}\over{q}}$$
$$\sqrt{2} = {{p}\over{q}} - 6$$
p and q are integers which means $${{p}\over{q}}-6$$ is a rational number.
But we know $$\sqrt{2}$$ is irrational.
This is not possible since LHS $$\ne$$ RHS.
That proves that our assumption that $$6 + \sqrt{2}$$ is rational is wrong.
Hence it is proved that $$6 + \sqrt{2}$$ is irrational.

#### Solution for Exercise 1.4

Q1. 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.
i.$${13 \over {3125}}$$
ii.$${17 \over {8}}$$
iii.$${64 \over {455}}$$
iv.$${15 \over {1600}}$$
v. $${29 \over {343}}$$
vi.$${23 \over {2^3 × 5^2}}$$
vii.$${{129} \over{2^2 × 5^7 × 7^5}}$$
viii.$${6 \over {15}}$$
ix.$${35 \over {50}}$$
x.$${77 \over {210}}$$

Ans(1)According to Theorem, any given rational number of the form $${{p}\over{q}}$$ where p and q are co-prime, has a terminating decimal expansion if q is of the form $$2^n × 5^m$$ , where m and n are non-negative integers
i. $${13 \over {3125}}$$

=> $${13 \over {3125}} = {13 \over {5^5}}$$
Denominator of above expression is $$5^5$$, which is of the form $$2^n × 5^m$$ (n=0,m=5)
So, according to the above theorem $${13 \over {3125}}$$ has a terminating decimal expansion.
ii. $${17 \over {8}}$$
=> $${17 \over {8}} = {17 \over {2^3}}$$
Denominator of above expression is $$2^3$$, which is of the form $$2^n × 5^m$$ (n=3,m=0)
So, according to the above theorem $${17 \over {8}}$$ has a terminating decimal expansion.
iii.$${64 \over {455}}$$
=> $${64 \over {455}} = {64 \over {5 × 7 × 13}}$$
Denominator of above expression is $$5 × 7 × 13$$, which is not of the form $$2^n × 5^m$$
So, according to the above theorem $${64 \over {455}}$$ has a non terminating repeating decimal expansion.
iv. $${15 \over {1600}}$$
=> $${15 \over {1600}} = {3 × 5\over {2^6 × 5^2}} = {3\over {2^6 × 5}}$$
Denominator of above expression is $$2^6 × 5$$, which is of the form $$2^n × 5^m$$ (n=6,m=1)
So, according to the above theorem $${15 \over {1600}}$$ has a terminating decimal expansion.
v.$${29 \over {343}}$$
=> $${29 \over {343}} = {29 \over {7^3}}$$
Denominator of above expression is $$7^3$$, which is not of the form $$2^n × 5^m$$
So, according to the above theorem $${29 \over {343}}$$ has a non terminating repeating decimal expansion.
vi $${23 \over {2^3 × 5^2}}$$
Denominator of above expression is $$2^3 × 5^2$$, which is of the form $$2^n × 5^m$$ (n=3,m=2)
So, according to the above theorem $${29 \over {343}}$$ has a terminating decimal expansion.
vii.$${129 \over {2^2 × 5^7 × 7^5}}$$
Denominator of above expression is $$2^2 × 5^7 × 7^5$$, which is not of the form $$2^n × 5^m$$
So, according to the above theorem $${129 \over {2^2 × 5^7 × 7^5}}$$ has a non terminating repeating decimal expansion.
viii. $${6 \over {15}}$$
=> $${6 \over {15}} = {2 \over {5}}$$
Denominator of above expression is $$5^1$$, which is of the form $$2^n × 5^m$$ (n=0,m=1)
So, according to the above theorem $${6 \over {15}}$$ has a terminating decimal expansion.
ix. $${35 \over {50}}$$
=> $${35 \over {50}} = {5 × 7\over {2 × 5^2}} = {7\over {2 × 5}}$$
Denominator of above expression is $$2 × 5$$, which is of the form $$2^n × 5^m$$ (n=1,m=1)
So, according to the above theorem $${35 \over {50}}$$ has a terminating decimal expansion.
x.$${77 \over {210}}$$
=> $${77 \over {210}} = {7 × 11 \over {2 × 3 × 5 × 7}} = {11\over {2 × 3 × 5}}$$
Denominator of above expression is $$2 × 3 × 5$$, which is not of the form $$2^n × 5^m$$
So, according to the above theorem $${77 \over {210}}$$ has a non terminating repeating decimal expansion.

Q2. Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.

Ans(2).i.$${13 \over {3125}} = {13 \over {5^5}} = {13 × 2^5 \over {5^5 × 2^5}} = {13 × 2^5 \over {10^5}} = {416 \over {10^5}} = 0.00416$$

• ii.$${17 \over {8}} = {17 \over {2^3}} = {17 × 5^3 \over {2^3 × 5^3}} = {17 × 5^3 \over {10^3}} = {2125 \over {10^3}} = 2.125$$
iv.$${15 \over {1600}} = {15 \over {2^6 × 5^2}} = {15 × 5^4 \over {2^6 × 5^2 × 5^4}} = {15 × 5^4 \over {10^6}} = {9375 \over {10^6}} = 0.009375$$
vi.$${23 \over {2^3 × 5^2}} = {23 × 5^1 \over {2^3 × 5^2 × 5^1}} = {115 \over {10^3}} = 0.115$$
viii.$${6 \over {15}} = {2 × 3 \over {3 × 5}} = {2 \over {5}} = {2 × 2 \over {5 × 2}} = {4 \over {10}} = 0.4$$
ix.$${35 \over {50}} = {5 ×7 \over {2 ×5^2}} = {7 \over {2 × 5}} = {7 \over {10}} = 0.7$$

• Q3. 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 $$\frac{p}{q}$$, what can you say about the prime factors of q?
1) 43.123456789
2) 0.120120012000120000…
3) 43.123456789123456789...

=>$$43.123456789 = {43123456789 \over {10^9}} = {43123456789 \over {2^9 × 5^9}}$$
The above expression has denominator of the form $$2^n × 5^m$$ where n and m are non-negative integers.
It has denominator of the form $$2^n × 5^m$$ where n and m are non-negative integers.