#### Page No 7:

#### Question 1:

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

#### Answer:

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii)196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii)867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51,

Therefore, HCF of 867 and 255 is 51.

#### Page No 7:

#### Question 2:

Show that
any positive odd integer is of the form
,
or
,
or
,
where *q* is some integer.

#### Answer:

Let *a*
be any positive integer and *b* = 6. Then, by Euclid’s
algorithm,

*a* =
6*q* + *r*for some integer *q*** ≥ **0,
and *r* = 0, 1, 2, 3, 4, 5 because 0 **≤ ***r* <
6.

Therefore,
*a* = 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q + *3
or 6*q* + 4 or 6*q* + 5

Also, 6*q*
+ 1 = 2 × 3*q* + 1 = 2*k*_{1} + 1, where*
k*_{1} is a positive integer

6*q*
+ 3 = (6*q* + 2) + 1 = 2 (3*q* + 1) + 1 = 2*k*_{2}
+ 1, where* k*_{2} is an integer

6*q*
+ 5 = (6*q* + 4) + 1 = 2 (3*q* + 2) + 1 = 2*k*_{3}
+ 1, where* k*_{3} is an integer

Clearly,
6*q* + 1, 6*q* + 3, 6*q* + 5 are of the form 2*k*
+ 1, where *k* is an integer.

Therefore,
6*q* + 1, 6*q* + 3, 6*q* + 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 6*q* +
1, or 6*q* + 3,

or 6*q*
+ 5

#### Page No 7:

#### Question 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 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

#### Page No 7:

#### Question 4:

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

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

#### Answer:

Let *a*
be any positive integer and *b* = 3.

Then *a*
= 3*q* + *r* for some integer *q* ≥ 0

And *r*
= 0, 1, 2 because 0 ≤ *r* < 3

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

Or,

Where *k*_{1},
*k*_{2}, and *k*_{3} are some positive
integers

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

#### Page No 7:

#### Question 5:

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

#### Answer:

Let *a*
be any positive integer and *b* = 3

*a =*
3*q* + *r*, where *q *≥ 0 and 0 ≤ *r* <
3

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

**Case 1**:
When *a = 3q*,

Where *m*
is an integer such that *m* =

**Case 2**:
When *a* = 3*q* + 1,

*a*^{3
}*= (*3*q +*1*)*^{3}

*a*^{3}*
= *27*q*^{3 }*+ *27*q*^{2 }*+
*9*q + *1

*a*^{3}
= 9(3*q*^{3 }*+ *3*q*^{2 }*+
q*) + 1

*a*^{3
}*= *9*m* + 1

Where* m*
is an integer such that *m* = (3*q*^{3 }*+
*3*q*^{2 }*+ q)*

**Case 3**:
When *a* = 3*q* + 2,

*a*^{3
}*= (*3*q +*2*)*^{3}

*a*^{3}*
= *27*q*^{3 }*+ *54*q*^{2 }*+
*36*q + *8

*a*^{3}
= 9(3*q*^{3 }*+ *6*q*^{2 }*+
*4*q*) + 8

*a*^{3
}*= *9*m* + 8

Where *m*
is an integer such that *m* = (3*q*^{3 }*+
*6*q*^{2 }*+ *4*q)*

Therefore,
the cube of any positive integer is of the form 9*m*, 9*m *+
1,

or 9*m + *8.

#### Page No 11:

#### Question 1:

Express each number as product of its prime factors:

#### Answer:

#### Page No 11:

#### Question 2:

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

#### Answer:

Hence, product of two numbers = HCF × LCM

Hence, product of two numbers = HCF × LCM

Hence, product of two numbers = HCF × LCM

#### Page No 11:

#### Question 3:

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

#### Answer:

#### Page No 11:

#### Question 4:

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

#### Answer:

#### Page No 11:

#### Question 5:

Check whether 6^{n} can end with the digit 0 for any natural number *n*.

#### Answer:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6^{n }*=* (2 ×3)^{n}

It can be observed that 5 is not in the prime factorisation of 6^{n}.

Hence, for any value of *n*, 6^{n} will not be divisible by 5.

Therefore, 6^{n} cannot end with the digit 0 for any natural number *n*.

#### Page No 11:

#### Question 6:

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

#### Answer:

Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)

= 13 × 78

= 13 ×13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

#### Page No 11:

#### Question 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:

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be 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

And, 12 = 2 **× **2 **× **3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes.

#### Page No 14:

#### Question 1:

Prove that is irrational.

#### Answer:

Let is a rational number.

Therefore,
we can find two integers *a*, *b* (*b* ≠ 0) such
that

Let *a*
and *b* have a common factor other than 1. Then we can divide
them by the common factor, and assume that *a* and *b* are
co-prime.

Therefore,
*a*^{2} is divisible by 5 and it can be said that *a*
is divisible by 5.

Let *a*
= 5*k*, where *k* is an integer

This
means that *b*^{2} is divisible by 5 and hence, *b*
is divisible by 5.

This
implies that *a* and *b* have 5 as a common factor.

And this
is a contradiction to the fact that *a* and *b* are
co-prime.

Hence,cannot be expressed as or it can be said that is irrational.

#### Page No 14:

#### Question 2:

Prove that is irrational.

#### Answer:

Let is rational.

Therefore, we can find two integers *a*, *b* (*b* ≠ 0) such that

Since *a* and *b* are integers, will also be rational and therefore,is rational.

This contradicts the fact that is irrational. Hence, our assumption that is rational is false. Therefore, is irrational.

#### Page No 14:

#### Question 3:

Prove that the following are irrationals:

#### Answer:

Let is rational.

Therefore, we can find two integers *a*, *b* (*b* ≠
0) such that

is rational as *a* and *b* are integers.

Therefore, is rational which contradicts to the fact that is irrational.

Hence, our assumption is false and is irrational.

Let is rational.

Therefore, we can find two integers *a*,* b* (*b* ≠
0) such that

for some integers *a* and *b*

is
rational as *a* and *b* are integers.

Therefore, should be rational.

This contradicts the fact thatis irrational. Therefore, our assumption that is rational is false. Hence, is irrational.

Let be rational.

Therefore, we can find two integers *a*, *b* (*b* ≠
0) such that

Since *a* and *b* are integers,
is also rational and hence,
should
be rational. This contradicts the fact that
is irrational. Therefore, our assumption is false and hence,
is irrational.

#### Page No 17:

#### Question 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:

#### Answer:

(i)** **

The denominator is of the form 5^{m}.

Hence, the decimal expansion ofis terminating.

(ii)** **

The denominator is of the form 2^{m}.

Hence, the decimal expansion of ** **is terminating.

(iii)

455 = 5 × 7 × 13

Since the denominator is not in the form 2^{m}^{ }× 5^{n}, and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.

(iv)** **

1600 = 2^{6} × 5^{2}

The denominator is of the form 2^{m}^{ }×^{ }5^{n}.

Hence, the decimal expansion of ** **is terminating.

(v)

Since the denominator is not in the form 2^{m}^{ }× 5^{n}, and it has 7 as its factor, the decimal expansion of is non-terminating repeating.

(vi)** **

The denominator is of the form 2^{m}^{ }× 5^{n}.

Hence, the decimal expansion of ** **is terminating.

(vii)

Since the denominator is not of the form 2^{m}^{ }× 5^{n}, and it also has 7 as its factor, the decimal expansion of is non-terminating repeating.

(viii)

The denominator is of the form 5^{n}.

Hence, the decimal expansion of ** **is terminating.

(ix)

The denominator is of the form 2^{m}^{ }× 5^{n}.

Hence, the decimal expansion of is terminating.

(x)** **

Since the denominator is not of the form 2^{m}^{ }× 5^{n}, and it also has 3 as its factors, the decimal expansion of ** **is non-terminating repeating.

#### Page No 18:

#### Question 2:

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

#### Answer:

(viii)

#### Page No 18:

#### Question 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 , what can you say about the prime factor of *q*?

(i) 43.123456789 (ii) 0.120120012000120000… (iii)

#### Answer:

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational number of the form and *q* is of the form

i.e., the prime factors of *q* will be either 2 or 5 or both.

(ii) 0.120120012000120000 …

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 and *q* is not of the form i.e., the prime factors of *q* will also have a factor other than 2 or 5.

#### Page No 29:

#### Question 17:

What is the least number that divisible by all the natural numbers from 1 to 10 (both inclusive)?

(a) 100

(b) 1260

(c) 2520

(d) 5040

#### Answer:

(c) 2520

We have to find the least number that is divisible by all numbers from 1 to 10.

∴ LCM (1 to 10) = ${2}^{3}\times {3}^{2}\times 5\times 7=2520$

Thus, 2520 is the least number that is divisible by every element and is equal to the least common multiple.

#### Page No 29:

#### Question 21:

*a* and *b* are two positive integers such that the least prime factor of *a* is 3 and the least prime factor of *b* is 5. Then, the least prime factor of (*a* + *b*) is

(a) 2

(b) 3

(c) 5

(d) 8

#### Answer:

(a) 2

Since 5 + 3 = 8, the least prime factor of *a* + *b* has to be 2, unless *a* + *b* is a prime number greater than 2.

If* a* + *b* is a prime number greater than 2, then *a* + *b *must be an odd number. So, either *a* or *b* must be an even number. If *a* is even, then the least prime factor of *a *is 2, which is not 3 or 5. So, neither *a* nor* b* can be an even number. Hence, *a* + *b* cannot be a prime number greater than 2 if the least prime factor of *a* is 3 or 5.

#### Page No 29:

#### Question 24:

Which of the following rational numbers is expressible as a terminating decimal?

(a) $\frac{124}{165}$

(b) $\frac{131}{30}$

(c) $\frac{2027}{625}$

(d) $\frac{1625}{462}$

#### Answer:

(c) $\frac{2027}{625}$

$\frac{124}{165}=\frac{124}{5\times 33}$; we know 5 and 33 are not the factors of 124. It is in its simplest form and it cannot be expressed as the product of ^{$({2}^{m}\times {5}^{n})$} for some non-negative integers $m,n$.

So, it cannot be expressed as a terminating decimal.

$\frac{131}{30}$ = $\frac{131}{5\times 6}$; we know 5 and 6 are not the factors of 131. Its is in its simplest form and it cannot be expressed as the product of ( ${2}^{m}\times {5}^{n}$) for some non-negative integers *$m,n$.*

So, it cannot be expressed as a terminating decimal.

$\frac{2027}{625}=\frac{2027\times {2}^{4}}{{5}^{4}\times {2}^{4}}=\frac{32432}{10000}=3.2432$; as it is of the form $({2}^{m}\times {5}^{n})$, where $m,n$ are non-negative integers.

So, it is a terminating decimal.

$\frac{1625}{462}=\frac{1625}{2\times 7\times 33}$; we know 2, 7 and 33 are not the factors of 1625. It is in its simplest form and cannot be expressed as the product of $({2}^{m}\times {5}^{n})$ for some non-negative integers $m,n$.

So, it cannot be expressed as a terminating decimal.

#### Page No 29:

#### Question 35:

$\left(2+\sqrt{2}\right)$ is

(a) an integer

(b) a rational number

(c) an irrational number

(d) none of these

#### Answer:

(c) an irrational number

$2+\sqrt{2}$ is an irrational number.

if it is rational, then the difference of two rational is rational

∴ $(2+\sqrt{2})-2$ = $\sqrt{2}$ = irrational

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