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#### Page No 1.10:

#### Question 1:

If *a* and *b* are two odd positive integers such that *a > b*, then prove that one of the two numbers $\frac{a+b}{2}\mathrm{and}\frac{a-b}{2}$is odd and the other is even.

#### Answer:

Given: If *a* and* b* are two odd positive integers such that *a* > *b*.

To Prove: That one of the two numbers and is odd and the other is even.

Proof: Let *a* and *b* be any odd odd positive integer such that *a* > *b*.

Since any positive integer is of the form *q*, 2*q* + 1

Let *a* = 2*q** + *1 and *b* = 2*m** + *1, where, *q* and *m* are some whole numbers

*$\Rightarrow \frac{a+b}{2}=\frac{(2q+1)+(2m+1)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+b}{2}=\frac{2(q+m)+1)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+b}{2}=(q+m+1)\phantom{\rule{0ex}{0ex}}$*

which is a positive integer.

Also,

*$\Rightarrow \frac{a-b}{2}=\frac{(2q+1)-(2m+1)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a-b}{2}=\frac{2(q-m)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a-b}{2}=(q-m)$*

Given, *a *> *b*

$\therefore $ 2*q** + *1 > 2*m** + *1

$\Rightarrow $ 2*q* > 2*m*

$\Rightarrow $ *q* > *m*

$\therefore $ $\frac{a-b}{2}=(q-m)0$

Thus, $\frac{(a-b)}{2}$ is a positive integer.

Now, we need to prove that one of the two numbers $\frac{(a+b)}{2}$ and$\frac{(a-b)}{2}$ is odd and other is even.

Consider, $\frac{(a+b)}{2}-\frac{(a-b)}{2}=\frac{(a+b)-(a-b)}{2}=\frac{2b}{2}=b$, which is odd positive integer.

Also, we know from the proof above that $\frac{(a+b)}{2}$ and$\frac{(a-b)}{2}$ are positive integers.

We know that the difference of two positive integers is an odd number if one of them is odd and another is even. (Also, difference between two odd and two even integers is even)

Hence it is proved that If *a* and *b* are two odd positive integers such that *a* > *b* then one of the two numbers and is odd and the other is even.

#### Page No 1.10:

#### Question 2:

Prove that the product of two consecutive positive integers is divisible by 2.

#### Answer:

To Prove: that the product of two consecutive integers is divisible by 2.

Proof: Let *n *− 1 and *n* be two consecutive positive integers.

Then their product is *n* (*n* − 1) = *n*^{2} − *n*

We know that every positive integer is of the form 2*q* or 2*q* + 1 for some integer *q*.

So let *n* = 2*q** *

So, *n*^{2} − *n* = (2*q*)^{2} − (2*q*)

Let *n* = 2*q* + 1

So, *n*^{2} − *n* = (2*q* + 1)^{2} − (2*q* + 1)

Hence it is proved that that the product of two consecutive integers is divisible by 2

#### Page No 1.10:

#### Question 3:

Prove that the product of three consecutive positive integer is divisible by 6.

#### Answer:

To Prove: the product of three consecutive positive integers is divisible by 6.

Proof: Let *n* be any positive integer.

Since any positive integer is of the form 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q* + 3 or 6*q* + 4, 6*q* + 5

If *n* = 6*q*

, which is divisible by 6

If *n* = 6*q* + 1

Which is divisible by 6

If *n* = 6*q* + 2

Which is divisible by 6

Similarly we can prove others.

Hence it is proved that the product of three consecutive positive integers is divisible by 6.

#### Page No 1.10:

#### Question 4:

For any positive integer *n*, prove that *n*^{3} – *n* divisible by 6.

#### Answer:

To Prove: For any positive integer *n*, *n*^{3} − *n* is divisible by 6.

Proof: Let *n* be any positive integer.

Since any positive integer is of the form 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q* + 3 or 6*q* + 4, 6*q* + 5

If *n* = 6*q*

If *n* = 6*q* + 1

If *n* = 6*q* + 2

Similarly we can prove others.

Hence it is proved that for any positive integer *n*, *n*^{3} − *n* is divisible by 6.

#### Page No 1.10:

#### Question 5:

Prove that if a positive integer is of the form 6*q** *+ 5, then it is of the form 3*q* + 2 for some integer* q*, but not conversely.

#### Answer:

To Prove: that if a positive integer is of the form 6*q* + 5 then it is of the form 3*q* + 2 for some integer *q*, but not conversely.

Proof: Let *n* = 6*q* + 5

Since any positive integer n is of the form of 3*k* or 3*k* + 1, 3*k* + 2

If *q* = 3*k*

If *q* = 3*k* + 1

If *q* = 3*k* + 2

Consider here 8 which is the form 3*q** *+ 2 *i.e.* 3 × 2 + 2 but it can’t be written in the form 6*q* + 5. Hence the converse is not true

#### Page No 1.10:

#### Question 6:

Prove that the square of any positive integer of the form 5*q* + 1 is of the same form.

#### Answer:

To Prove: that the square of a positive integer of the form 5*q* + 1 is of the same form

Proof: Since positive integer *n* is of the form 5*q* + 1

If *n* = 5*q* + 1

$\mathrm{Then}{n}^{2}={\left(5q+1\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}={\left(5q\right)}^{2}+{\left(1\right)}^{2}+2\left(5q\right)\left(1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}=25{q}^{2}+1+10q\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}=25{q}^{2}+10q+1\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}=5\left(5{q}^{2}+2q\right)+1$

Hence *n*^{2} integer is of the form 5*m* + 1.

#### Page No 1.10:

#### Question 7:

Prove that the square of any positive integer is of the form 3*m* or, 3*m* + 1 but not of the form 3*m* + 2.

#### Answer:

To Prove: that the square of an positive integer is of the form 3*m* or 3*m* + 1 but not of the form 3*m* + 2.

Proof: Since positive integer *n* is of the form of 3*q*, 3*q* + 1 and 3*q* + 2

If *n* = 3*q*

If *n* = 3*q* + 1

Then, *n*^{2} = (3*q* + 1)^{2}

If *n* = 3*q* + 2

Then, *n*^{2} = (3*q* + 2)^{2}

Hence *n*^{2} integer is of the form 3*m*, 3*m* + 1 but not of the form 3*m** *+ 2.

#### Page No 1.10:

#### Question 8:

Prove that the square of any positive integer is of the form 4*q** *or 4*q** +* 1 for some integer *q*.

#### Answer:

To Prove: that the square of any positive integer is of the form 4*q* or 4*q* + 1 for some integer *q*.

Proof: Since positive integer *n* is of the form of 2*q* or 2*q* + 1

If *n* = 2*q*

If *n* = 2*q* + 1

Hence it is proved that the square of any positive integer is of the form 4*q* or 4*q* + 1, for some integer *q*.

#### Page No 1.10:

#### Question 9:

Prove that the square of any positive integer is of the form 5*q*, 5*q* + 1, 5*q* + 4 for some integer *q*.

#### Answer:

To Prove: that the square of any positive integer is of the form 5*q* or 5*q* + 1, 5*q* + 4 for some integer *q*.

Proof: Since positive integer* n* is of the form of 5*q* or 5*q* + 1, 5*q* + 4

If *n* = 5*q*

If *n* = 5*q* + 1

If *n* = 5*q* + 2

If *n* = 5*q* + 4

Hence it is proved that the square of a positive integer is of the form 5*q* or 5*q* + 1, 5*q* + 4 for some integer *q*.

#### Page No 1.10:

#### Question 10:

Show that the square of an odd positive integer is of the form 8*q* + 1, for some integer *q*.

#### Answer:

To Prove: that the square of an odd positive integer is of the form 8*q* + 1, for some integer *q*.

Proof: Since any positive integer *n* is of the form 4*m* + 1 and 4*m* + 3

If *n* = 4*m* + 1

If *n* = 4*m* + 3

Hence *n*^{2} integer is of the form 8*q* + 1, for some integer *q*.

#### Page No 1.11:

#### Question 11:

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

#### Answer:

To Show: That any positive odd integer is of the form 6*q* + 1 or 6*q* + 3 or 6*q* + 5 where *q* is any some integer.

Proof: Let *a* be any odd positive integer and *b *= 6.

Then, there exists integers *q* and *r* such that

*a* = 6*q* + *r*, 0 ≤ *r* < 6 (by division algorithm)

⇒ *a* = 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q* + 3 or 6*q* + 4

But 6*q* or 6*q* + 2 or 6*q* + 4 are even positive integers.

Hence it is proved that any positive odd integer is of the form 6*q* + 1 or 6*q* + 3 or 6*q* + 5, where *q* is any some integer.

#### Page No 1.11:

#### Question 12:

Prove that one of every three consecutive positive integers is divisible by 3.

#### Answer:

Let *n*, *n *+ 1, *n* + 2 be three consecutive positive integers, where *n *is any natural number.

By Euclid's division lemma,

*n* = *aq *+ *r*, where 0 ≤ *r *< *a*.

Now, if we divide *n* by 3, then* n* can be written in the form of 3*q*, 3*q*+1 or 3*q*+2.

This implies that we have three possible cases.

Case I:

If *n *= 3*q*, then *n* is divisible by 3.

However, *n *+ 1 and *n* + 2 are not divisible by 3.

Case II:

If *n* = 3*q* + 1, then *n* + 2 = 3*q* + 3 = 3(*q* + 1), which is divisible by 3.

However,* n* and *n* + 1 are not divisible by 3.

Case III:

If *n* = 3*q* + 2, then *n* + 1 = 3*q* + 3 = 3(*q* + 1), which is divisible by 3.

However, *n* and* n* + 2 are not divisible by 3.

Hence, we conclude that one of any three consecutive positive integers must be divisible by 3.

#### Page No 1.11:

#### Question 13:

Show that the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer *m*.

#### Answer:

Suppose *a* be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers *a* and 6, there exists non-negative integers *a* and *r* such that

*a* = 6*q *+* r*, where $0\le r<6$

$\Rightarrow {a}^{2}={\left(6q+r\right)}^{2}=36{q}^{2}+{r}^{2}+12qr\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+2qr\right)+{r}^{2}.....\left(1\right)\mathrm{where},0\le r6$

Case: 1

When* r *= 0.

Putting *r* = 0 in (1), we get

*${a}^{2}=6{\left(6q\right)}^{2}=6m$*

Where, *$m=6{q}^{2}\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 2

When* r* = 1.

Putting *r* = 1 in (1), we get

*${a}^{2}=6\left(6{q}^{2}+2q\right)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6m+1\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(6{q}^{2}+2q\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 3

When* r* = 2.

Putting *r* = 2 in (1), we get

*${a}^{2}=6\left(6{q}^{2}+4q\right)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6m+4\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(6{q}^{2}+4q\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 4

When* r* = 3.

Putting *r* = 3 in (1), we get

*${a}^{2}=6\left(6{q}^{2}+6q\right)+9\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+6q\right)+6+3\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+6q+1\right)+3\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6m+3\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(6{q}^{2}+6q+1\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 5

When* r* = 4

Putting *r* = 4 in (1), we get

*${a}^{2}=6\left(6{q}^{2}+8q\right)+16\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+8q\right)+12+4\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+8q+2\right)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6m+4\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(6{q}^{2}+8q+2\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 6

When* r* = 5.

Putting *r* = 5 in (1), we get

*${a}^{2}=6\left(6{q}^{2}+10q\right)+25\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+10q\right)+24+1\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6\left(6{q}^{2}+10q+4\right)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=6m+1\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(6{q}^{2}+10q+1\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Hence, the square of any positive integer cannot be of the form 6*m* + 2 or 6*m* + 5.

#### Page No 1.11:

#### Question 14:

Show that the cube of a positive integer is of the form $6q+r$ , where $q$ is ana integer and $r$ = 0, 1, 2, 3, 4, 5 .

#### Answer:

Suppose *a* be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers *a* and 6, there exists non-negative integers *a* and *r* such that

*a* = 6*q *+* r*, where $0\le r<6$

$\Rightarrow {a}^{3}={\left(6q+r\right)}^{3}=216{q}^{3}+{r}^{3}+3\times 6q\times r\left(6q+r\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(216{q}^{3}+108{q}^{2}r+18q{r}^{2}\right)+{r}^{3}.....\left(1\right)\mathrm{where},0\le r6$

Case: 1

When* r *= 0.

Putting *r* = 0 in (1), we get

*${a}^{3}=216{q}^{3}=6\left(36{q}^{3}\right)=6m$*

Where, *$m=36{q}^{3}$ *is an integer

Case: 2

When* r* = 1.

Putting *r* = 1 in (1), we get

*${a}^{3}=\left(216{q}^{3}+108{q}^{2}+18q\right)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(36{q}^{3}+18{q}^{2}+3q\right)+1\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6m+1\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(36{q}^{3}+18{q}^{2}+3q\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 3

When* r* = 2.

Putting *r* = 2 in (1), we get

*${a}^{3}=\left(216{q}^{3}+216{q}^{2}+72q\right)+8\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=\left(216{q}^{3}+216{q}^{2}+72q+6\right)+2\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(36{q}^{3}+36{q}^{2}+12q+1\right)+2\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6m+2\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(36{q}^{3}+36{q}^{2}+12q+1\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 4

When* r* = 3.

Putting *r* = 3 in (1), we get

*${a}^{3}=\left(216{q}^{3}+324{q}^{2}+162q\right)+27\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=\left(216{q}^{3}+324{q}^{2}+162q+24\right)+3\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(36{q}^{2}+54{q}^{2}+27q+4\right)+3\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6m+3\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(36{q}^{2}+54{q}^{2}+27q+4\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 5

When* r* = 4

Put *r* = 4 in (1), we get

*${a}^{3}=\left(216{q}^{3}+432{q}^{2}+288q\right)+64\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=\left(36{q}^{3}+72{q}^{2}+48q+60\right)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(36{q}^{3}+72{q}^{2}+48q+10\right)+4\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6m+4\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(36{q}^{3}+72{q}^{2}+48q+10\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Case: 6

When* r* = 5.

Putting *r* = 5 in (1), we get

*${a}^{3}=216{q}^{3}+540{q}^{2}+450q+125\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6\left(36{q}^{3}+90{q}^{2}+75q+20\right)+5\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=6m+5\phantom{\rule{0ex}{0ex}}\mathrm{Where},m=\left(36{q}^{3}+90{q}^{2}+75q+20\right)\mathrm{is}\mathrm{an}\mathrm{integer}$*

Hence, the cube of any positive integer of the form 6*q* + *r*, where *q* is an integer and *r *= 0, 1, 2, 3, 4, 5.

#### Page No 1.11:

#### Question 15:

Show that one and only one out of $n,n+4,n+8,n\hspace{0.17em}+12\mathrm{and}n+16$ is divisible by 5 , where *n* is any positive integer .

#### Answer:

Consider the numbers $n,\left(n+4\right),\left(n+8\right),\left(n+12\right)\mathrm{and}\left(n+16\right)$, where *n* is any positive integer.

$\mathrm{Suppose}n=5q+r,\mathrm{where}0\le r5\phantom{\rule{0ex}{0ex}}n=5q,5q+1,5q+2,5q+3,5q+4$

(By Euclid's division algorithm)

Case: 1

$\mathrm{When}n=5q.\phantom{\rule{0ex}{0ex}}n=5q\mathrm{is}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+4=5q+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+8=5q+5+5+3=5\left(q+1\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+12=5q+10+2=5\left(q+2\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+16=5q+15+1=5\left(q+3\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.$

Case: 2

$\mathrm{When}n=5q+1.\phantom{\rule{0ex}{0ex}}n=5q+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+4=5q+1+4=5\left(q+1\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+8=5q+1+5+3=5\left(q+1\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+12=5q+1+12=5\left(q+2\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+16=5q+1+16=5\left(q+3\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.$

Case: 3

$\mathrm{When}n=5q+2.\phantom{\rule{0ex}{0ex}}n=5q+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+4=5q+2+4=5\left(q+1\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+8=5q+2+8=5\left(q+2\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+12=5q+2+12=5\left(q+2\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+16=5q+2+16=5\left(q+3\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.$

Case: 4

$\mathrm{When}n=5q+3.\phantom{\rule{0ex}{0ex}}n=5q+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+4=5q+3+4=5\left(q+1\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+8=5q+3+8=5\left(q+2\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+12=5q+3+12=5\left(q+3\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+16=5q+3+16=5\left(q+3\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.$

Case: 5

$\mathrm{When}n=5q+4.\phantom{\rule{0ex}{0ex}}n=5q+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+4=5q+4+4=5\left(q+1\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+8=5q+4+8=5\left(q+2\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+12=5q+4+12=5\left(q+3\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5.\phantom{\rule{0ex}{0ex}}n+16=5q+4+16=5\left(q+4\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5.$

Hence, in each case, one and only one out of $n,n+4,n+8,n+12\mathrm{and}n+16$ is divisible by 5.

#### Page No 1.11:

#### Question 16:

Show that the square of an odd positive integer can be of the form 6*q *+ 1* *or 6*q *+ 3 for some integer *q.*

#### Answer:

It is known that any positive integer can be written in the form of $6m,6m+1,6m+2,6m+3,6m+4,6m+5$ for some integer *m*.

Thus, an odd positive integer can be of the form $6m+1,6m+3,6m+5$.

We have, ${\left(6m+1\right)}^{2}=36{m}^{2}+12m+1=6\left(6{m}^{2}+2m\right)+1=6q+1,\mathrm{where}q=6{m}^{2}+2m$ is an integer

${\left(6m+3\right)}^{2}=36{m}^{2}+36m+9=6\left(6{m}^{2}+6m+1\right)+3=6q+3,\mathrm{where}q=6{m}^{2}+6m+1$ is an integer

${\left(6m+5\right)}^{2}=36{m}^{2}+60m+25=6\left(6{m}^{2}+10m+4\right)+1=6q+1,\mathrm{where}q=6{m}^{2}+10m+4$ is an integer

Thus, the square of an odd positive integer can be of the form 6*q* + 1 or 6*q* + 3 for some integer *q.*

#### Page No 1.11:

#### Question 17:

A positive integer is of the form $3q+1,q$ being a anatural number. Can you write its square in any form other than $3m+1,3m$or $3m+2$ for some integer $m$ ? Justify your answer.

#### Answer:

By Euclid's lemma, $b=aq+r,0\le ra.$

Here, *b* is a positive integer and *a *= 3.

∴ *$b=3q+r,\mathrm{for}0\le r3$*

This must be in the form $3q,3q+1\mathrm{or}3q+2$.

Now,

${\left(3q\right)}^{2}=9{q}^{2}=3m,\mathrm{where}m=3{q}^{2}\phantom{\rule{0ex}{0ex}}{\left(3q+1\right)}^{2}=9{q}^{2}+6q+1=3\left(3{q}^{2}+2q\right)+1=3m+1,\mathrm{where}m=3{q}^{2}+2q\phantom{\rule{0ex}{0ex}}{\left(3q+2\right)}^{2}=9{q}^{2}+12q+4=3\left(3{q}^{2}+4q+1\right)+1=3m+1,\mathrm{where}m=3{q}^{2}+4q+1$

Therefore, the square of a positive integer 3*q* + 1 is always in the form of 3*m* or 3*m* + 1 for some integer *m*.

#### Page No 1.11:

#### Question 18:

Show that the square of any positive integer cannot be of the form $3m+2$, where $m$ is a natural number..

#### Answer:

By Euclid's lemma, $b=aq+r,0\le ra.$

Here, *b* is a positive integer and *a *= 3.

∴ *$b=3q+r,\mathrm{for}0\le r3$*

This must be in the form $3q,3q+1\mathrm{or}3q+2$.

Now,

${\left(3q\right)}^{2}=9{q}^{2}=3m,\mathrm{where}m=3{q}^{2}\phantom{\rule{0ex}{0ex}}{\left(3q+1\right)}^{2}=9{q}^{2}+6q+1=3\left(3{q}^{2}+2q\right)+1=3m+1,\mathrm{where}m=3{q}^{2}+2q\phantom{\rule{0ex}{0ex}}{\left(3q+2\right)}^{2}=9{q}^{2}+12q+4=3\left(3{q}^{2}+4q+1\right)+1=3m+1,\mathrm{where}m=3{q}^{2}+4q+1$

Therefore, the square of a positive integer cannot be of the form 3*m* + 2, where *m* is a natural number.

#### Page No 1.27:

#### Question 1:

Define HCF of two positive integers and find the HCF of the following pairs of numbers:

(i) 32 and 54

(ii) 18 and 24

(iii) 70 and 30

(iv) 56 and 88

(v) 475 and 495

(vi) 75 and 243.

(vii) 240 and 6552

(viii) 155 and 1385

(ix) 100 and 190

(x) 105 and 120

#### Answer:

(i) We need to find H.C.F. of 32 and 54.

By applying division lemma

54 = 32$\times $1 + 22

Since remainder, apply division lemma on 32 and remainder 22

32 = 22$\times $1 + 10

Since remainder, apply division lemma on 22 and remainder 10

22 = 10$\times $2 + 2

Since remainder, apply division lemma on 10 and remainder 2

10 = 2$\times $5 + 0

Therefore, H.C.F. of 32 and 54 is

(ii) We need to find H.C.F. of 18 and 24.

By applying division lemma

Since remainder, apply division lemma on divisor 18 and remainder 6

Therefore, H.C.F. of 18 and 24 is

(iii) We need to find H.C.F. of 70 and 30.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on divisor 30 and remainder 10

Therefore, H.C.F. of 70 and

(iv) We need to find H.C.F. of 56 and 88.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on 56 and remainder 32

Since remainder, apply division lemma on 32 and remainder 24

Since remainder, apply division lemma on 24 and remainder 8

Therefore, H.C.F. of 56 and.

(v) We need to find H.C.F. of 475 and 495.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on 475 and remainder 20

Since remainder, apply division lemma on 20 and remainder 15

Since remainder, apply division lemma on15 and remainder 5

Therefore, H.C.F. of 475 and.

(vi) We need to find H.C.F. of 75 and 243.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on 75 and remainder 18

Since remainder, apply division lemma on divisor 18 and remainder 3

Therefore, H.C.F. of 75 and.

(vii) We need to find H.C.F. of 240 and 6552.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on divisor 240 and remainder 72

Since remainder, apply division lemma on divisor 72 and remainder 24

Therefore, H.C.F. of 240 and.

(viii) We need to find H.C.F. of 155 and 1385.

By applying Euclid’s Division lemma

Since remainder, apply division lemma on divisor 155 and remainder 145

Since remainder, apply division lemma on divisor 145 and remainder 10

Since remainder, apply division lemma on divisor 10 and remainder 5

Therefore, H.C.F. of 155 and.

(ix) We need to find H.C.F. of 100 and 190.

By applying Euclid’s division lemma

Since remainder, apply division lemma on divisor 100 and remainder 90

Since remainder, apply division lemma on divisor 90 and remainder 10

Therefore, H.C.F. of 100 and.

(x) We need to find H.C.F. of 105 and 120.

By applying Euclid’s division lemma

Since remainder, apply division lemma on divisor 105 and remainder 15

Therefore, H.C.F. of 105 and.

#### Page No 1.27:

#### Question 2:

Use Euclid's division algorithm to find the HCF of

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(iv) 184, 230 and 276

(v) 136, 170 and 255

(vi) 1260 and 7344

(vii) 2048 and 960

#### Answer:

(i) Given integers are 225 and 135. Clearly 225 > 135. So we will apply Euclid’s division lemma to 225 and 135, we get,

Since the remainder. So we apply the division lemma to the divisor 135 and remainder 90. We get,

Now we apply the division lemma to the new divisor 90 and remainder 45. We get,

The remainder at this stage is 0. So the divisor at this stage is the H.C.F.

So the H.C.F of 225 and 135 is 45.

(ii) Given integers are 38220 and 196. Clearly 38220 > 196. So we will apply Euclid’s division lemma to 38220 and 196, we get,

The remainder at this stage is 0. So the divisor at this stage is the H.C.F.

So the H.C.F of 38220 and 196 is 196.

(iii) Given integers are 867 and 255. Clearly 867 > 225. So we will apply Euclid’s division lemma to 867 and 225, we get,

Since the remainder. So we apply the division lemma to the divisor 225 and remainder 192. We get,

Now we apply the division lemma to the new divisor 192 and remainder 33. We get,

Now we apply the division lemma to the new divisor 33 and remainder 27. We get,

Now we apply the division lemma to the new divisor 27 and remainder 6. We get,

Now we apply the division lemma to the new divisor 6 and remainder 3. We get,

The remainder at this stage is 0. So the divisor at this stage is the H.C.F.

So the H.C.F of 867 and 255 is 3.

(iv) Given integers are 184, 230 and 276.

Let us first find the HCF of 184 and 230 by Euclid lemma.

Clearly, 230 > 184. So, we will apply Euclid’s division lemma to 230 and 184.

230 = 184 × 1 + 46

Remainder is 46 which is a non-zero number. Now, apply Euclid’s division lemma to 184 and 46.

184 = 46 × 4 + 0

The remainder at this stage is zero. Therefore, 46 is the HCF of 230 and 184.

Now, again use Euclid’s division lemma to find the HCF of 46 and 276.

276 = 46 × 6 + 0

The remainder at this stage is zero. Therefore, 46 is the HCF of 184, 230 and 276.

(v) Given integers are 136, 170 and 255.

Let us first find the HCF of 136, 170 by Euclid lemma.

Clearly, 170 > 136. So, we will apply Euclid’s division lemma to 136 and 170.

170 = 136 × 1 + 34

Remainder is 34 which is a non-zero number. Now, apply Euclid’s division lemma to 136 and 34.

136 = 34 × 4 + 0

The remainder at this stage is zero. Therefore, 34 is the HCF of 136 and 170.

Now, again use Euclid’s division lemma to find the HCF of 34 and 255.

255 = 34 × 7 +17

Remainder is 17 which is a non-zero number. Now, apply Euclid’s division lemma to 34 and 17.

34 = 17 × 2 + 0

The remainder at this stage is zero. Therefore, 17 is the HCF of 136, 170 and 255.

1260 < 7344

Thus, we divide 7344 by 1260 by using Euclid's division lemma

7344 = 1260 × 5 +1044

∵ Remainder is not zero,

∴ we divide 1260 by 1044 by using Euclid's division lemma

1260 = 1044 × 1 + 216

∵ Remainder is not zero,

∴ we divide 1044 by 216 by using Euclid's division lemma

1044 = 216 × 4 + 180

∵ Remainder is not zero,

∴ we divide 216 by 180 by using Euclid's division lemma

216 = 180 × 1 + 36

∵ Remainder is not zero,

∴ we divide 180 by 36 by using Euclid's division lemma

180 = 36 × 5 + 0

Since, Remainder is zero

Hence, HCF of 1260 and 7344 is 36.

(vii) 2048 and 960

2048 > 960

Thus, we divide 2048 by 960 by using Euclid's division lemma

2048 = 960 × 2 + 128

∵ Remainder is not zero,

∴ we divide 960 by 128 by using Euclid's division lemma

960 = 128 × 7 + 64

∵ Remainder is not zero,

∴ we divide 128 by 64 by using Euclid's division lemma

128 = 64 × 2 + 0

Since, Remainder is zero

Hence, HCF of 2048 and 960 is 64.

#### Page No 1.27:

#### Question 3:

Find the HCF of the following pairs of integers and express it as a linear combination of them.

(i) 963 and 657

(ii) 592 and 252

(iii) 506 and 1155

(iv) 1288 and 575

#### Answer:

(i) We need to find the H.C.F. of 963 and 657 and express it as a linear combination of 963 and 657.

By applying Euclid’s division lemma

Since remainder, apply division lemma on divisor 657 and remainder 306

Since remainder, apply division lemma on divisor 306 and remainder 45

Since remainder, apply division lemma on divisor 45 and remainder 36

Since remainder, apply division lemma on divisor 36 and remainder 9

Therefore, H.C.F. = 9.

Now,

(ii) We need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.

By applying Euclid’s division lemma

592 = 252×2+88

Since remainder, apply division lemma on divisor 252 and remainder 88

252 = 88×2+76

Since remainder, apply division lemma on divisor 88 and remainder 76

88 = 76×1+12

Since remainder, apply division lemma on divisor 76 and remainder 12

76 = 12×6+4

Since remainder, apply division lemma on divisor 12 and remainder 4

12 = 4×3+0.

Therefore, H.C.F. = 4.

Now,

$4=76-12\times 6\phantom{\rule{0ex}{0ex}}=76-\left[88-76\times 1\right]\times 6\phantom{\rule{0ex}{0ex}}=76-88\times 6+76\times 6\phantom{\rule{0ex}{0ex}}=76\times 7-88\times 6\phantom{\rule{0ex}{0ex}}=\left(252-88\times 2\right)\times 7-88\times 6\phantom{\rule{0ex}{0ex}}=252\times 7-88\times 14-88\times 6\phantom{\rule{0ex}{0ex}}=252\times 7-88\times 20\phantom{\rule{0ex}{0ex}}=252\times 7-\left[592-252\times 2\right]\times 20\phantom{\rule{0ex}{0ex}}=252\times 7-592\times 20+252\times 40\phantom{\rule{0ex}{0ex}}=252\times 47-592\times 20\phantom{\rule{0ex}{0ex}}=\overline{)252\times 47+592\times \left(-20\right)}$

(iii) We need to find the H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155.

By applying Euclid’s division lemma

Since remainder, apply division lemma on divisor 506 and remainder 143

Since remainder, apply division lemma on divisor 143 and remainder 77

Since remainder, apply division lemma on divisor 77 and remainder 66

Since remainder, apply division lemma on divisor 66 and remainder 11

Therefore, H.C.F. = 11.

Now,

(iv) We need to find the H.C.F. of 1288 and 575 and express it as a linear combination of 1288 and 575.

By applying Euclid’s division lemma

Since remainder, apply division lemma on divisor 575 and remainder 138

Since remainder, apply division lemma on divisor 138 and remainder 23

Therefore, H.C.F. = 23.

Now,

#### Page No 1.27:

#### Question 4:

Find the largest number which divides 615 and 963 leaving remainder 6 in each case.

#### Answer:

We need to find the largest number which divides 615 and 963 leaving remainder 6 in each case.

The required number when divides 615 and 963, leaves remainder 6, this means and are completely divisible by the number.

Therefore,

The required number = H.C.F. of 609 and 957.

By applying Euclid’s division lemma

Therefore, H.C.F. = 87.

Hence, the required number is.

#### Page No 1.27:

#### Question 5:

If the HCF of 408 and 1032 is expressible in the form 1032 *m* – 408 × 5, find *m*.

#### Answer:

We need to find *m* if the H.C.F of 408 and 1032 is expressible in the form.

Given integers are 408 and 1032 where.

By applying Euclid’s division lemma, we get

Since the remainder, so apply division lemma on divisor 408 and remainder 216

Since the remainder, so apply division lemma on divisor 216 and remainder 192

Since the remainder, so apply division lemma on divisor 192 and remainder 24

We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.

Therefore,

#### Page No 1.27:

#### Question 6:

If the HCF of 657 and 963 is expressible in the form 657*x* + 963 × – 15, find *x*.

#### Answer:

We need to find *x* if the H.C.F of 657 and 963 is expressible in the form.

Given integers are 657 and 963.

By applying Euclid’s division lemma, we get

Since the remainder, so apply division lemma on divisor 657 and remainder 306

Since the remainder, so apply division lemma on divisor 306 and remainder 45

Since the remainder, so apply division lemma on divisor 45 and remainder 36

Since the remainder, so apply division lemma on divisor 36 and remainder 9

Therefore, H.C.F. = 9.

Given H.C.F =.

Therefore,

#### Page No 1.27:

#### Question 7:

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:

We are given that 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. We need to find the maximum number of columns in which they can march.

Members in army = 616

Members in band = 32.

Therefore,

Maximum number of columns = H.C.F of 616 and 32.

By applying Euclid’s division lemma

Therefore, H.C.F. = 8

Hence, the maximum number of columns in which they can march is.

#### Page No 1.27:

#### Question 8:

A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

#### Answer:

The merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively.

So the greatest capacity of the tin for filling three different types of oil is given by the H.C.F. of 120,180 and 240.

So first we will calculate H.C.F of 120 and 180 by Euclid’s division lemma.

The divisor at the last step is 60. So the H.C.F of 120 and 180 is 60.

Now we will find the H.C.F. of 60 and 240,

The divisor at the last step is 60. So the H.C.F of 240 and 60 is 60.

Therefore, the tin should be of

#### Page No 1.27:

#### Question 9:

During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would you need to buy?

#### Answer:

We are given that during a sale, color pencils were being sold in packs of 24 each and crayons in packs of 32 each. If we want full packs of both and the same number of pencils and crayons, we need to find the number of each we need to buy.

Given that

Number of color pencils in one pack = 24

Number of crayons in pack = 32.

Therefore, the least number of both colors to be purchased

Hence, number of packs of pencils to bought

,

And number of packs of crayon to be bought

.

#### Page No 1.28:

#### Question 10:

144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have?

#### Answer:

Given that 144 cartons of coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and contains cartons of the same drink We need to find the greatest number of cartons, each stack would have

Given that

Number of cartons of coke cans = 144

Number of cartons of Pepsi cans = 90.

Therefore, the greatest number of cartons in one stack = H.C.F. of 144 and 90.

By applying Euclid’s division lemma

Hence, the greatest number cartons in one stack

#### Page No 1.28:

#### Question 11:

Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

#### Answer:

We need to find the greatest number which divides 285 and 1249 leaving remainder 9 and 7 respectively.

The required number when divides 285 and 1249, leaves remainder 9 and 7, this means are completely divisible by the number.

Therefore, the required number = H.C.F. of 276 and 1242.

By applying Euclid’s division lemma

Therefore, H.C.F. = 138

Hence, required number is.

#### Page No 1.28:

#### Question 12:

Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.

#### Answer:

_{We need to find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.}

_{The required number when divides 280 and 1245, leaves remainder 4 and 3, this means } are completely divisible by the number.

_{Therefore, the required number = H.C.F. of 276 and 1242.}

_{By applying Euclid’s division lemma}

_{Therefore, H.C.F. = 138.}

Hence, the required number is.

#### Page No 1.28:

#### Question 13:

What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.

#### Answer:

We need to find the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.

The required number when divides 626, 3127 and 15628 leaves remainders 1, 2 and 3 this means are completely divisible by the number.

Therefore, the required number = H.C.F. of 625, 3125 and 15625.

First we consider 625 and 3125.

By applying Euclid’s division lemma

H.C.F. of 625 and 3125 = 625

Now, consider 625 and 15625.

By applying Euclid’s division lemma

Therefore, H.C.F. of 625, 3125 and 15625 = 625

Hence, the required number is

#### Page No 1.28:

#### Question 14:

Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.

#### Answer:

Find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively.

The required number when divides 445, 572 and 699 leaves remainders 4, 5 and 6 this means are completely divisible by the number.

Therefore, the required number = H.C.F. of 441, 567 and 693.

First consider 441 and 567.

By applying Euclid’s division lemma

Therefore, H.C.F. of 441 and 567 = 63

Now, consider 63 and 693

By applying Euclid’s division lemma

Therefore, H.C.F. of 441, 567 and 693 = 63

Hence, the required number is.

#### Page No 1.28:

#### Question 15:

Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively

#### Answer:

Find the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively.

The required number when divides 2011 and 2623 leaves remainders 9 and 5 this means are completely divisible by the number.

Therefore, the required number = H.C.F. of 2002 and 2618

By applying Euclid’s division lemma

H.C.F. of 2002 and 2618 = 154

Hence, the required number is.

#### Page No 1.28:

#### Question 16:

Using Euclid's division algorithm , find the largest number that divides 1251, 9377, and 15628 leaving remainders 1, 2 and 3 respectively.

#### Answer:

It is given that 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively.

Subtracting these remainders from the respective numbers, we get

1251 − 1 = 1250

9377 − 2 = 9375

15628 − 3 = 15625

Now, 1250, 9375 and 15625 are divisible by the required number.

Required number = HCF of 1250, 9375 and 15625

By Euclid's division algorithm $a=bq+r,0\le rb$

For largest number, put *a* = 15625 and *b* = 9375

15625 = 9375 × 1 + 6250

$\Rightarrow 9375=6250\times 1+3125\phantom{\rule{0ex}{0ex}}\Rightarrow 6250=3125\times 2+0$

Since remainder is zero, therefore, HCF(15625 and 9375) = 3125

Further, take *c* = 1250 and *d* = 3125. Again using Euclid's division algorithm

$d=cq+r,0\le rc\phantom{\rule{0ex}{0ex}}\Rightarrow 3125=1250\times 2+625\left[\because r\ne 0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 1250=625\times 2+0$

Since remainder is zero, therefore, HCF(1250, 9375 and 15625) = 625

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.

#### Page No 1.28:

#### Question 17:

Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?

#### Answer:

We are given that two brands of chocolates are available in packs of 24 and 15 respectively. If he needs to buy an equal number of chocolates of both kinds, then find least number of boxes of each kind he would need to buy.

Given that

Number of chocolates of 1^{st} brand in one pack =24

Number of chocolates of 2^{nd} brand in one pack = 15.

Therefore, the least number of chocolates he need to purchase is

Therefore, the number of packet of 1^{st} brand is

And the number of packet of 2^{nd} brand is

.

#### Page No 1.28:

#### Question 18:

A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in cinches of the tile required that has to be cut and how many such tiles are required?

#### Answer:

A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft. by 8ft. We need to find the size in inches of the tile required that has to be cut and number of such tiles are required.

Size of bathroom = 10ft by 8ft

The largest size of tile required = H.C.F. of 120 and 96.

By applying Euclid’s division lemma

Therefore, H.C.F. = 24.

Thus, largest size of tile required = 24 inches.

Therefore,

#### Page No 1.28:

#### Question 19:

15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain?

#### Answer:

We are given that 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. We need to find the number of biscuit packets and number of pastries each box contain.

Given that

Number of pastries = 15

Number of biscuits packets = 12.

Therefore, required number of boxes to contain equal number = H.C.F. of 15 and 12.

By applying Euclid’s division lemma

Therefore, number of boxes required = 3.

Hence each box will contain pastries and biscuits packets.

#### Page No 1.28:

#### Question 20:

105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?

#### Answer:

We are given that, 105 goats, 140 donkeys and 175 cows. There is only one boat which will have to make many *y* trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. We need to tell the number of animals that went in each trip.

Given that

Number of goats = 105

Number of donkeys = 140

Number of cows = 175.

Therefore, the largest number of animals in 1 trip = H.C.F. of 105, 140 and 175.

First we consider 105 and 140.

By applying Euclid’s division lemma

Therefore, H.C.F. of 105 and 140 = 35

Now, we consider 35 and 175.

By applying Euclid’s division lemma

Therefore, H.C.F. of 105, 140 and 175 = 35

Hence, the number of animals went in each trip is.

#### Page No 1.28:

#### Question 21:

The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.

#### Answer:

We are given the length, breadth and height of a room as 8m 25cm, 6m 75cm and 4m 50cm, respectively. We need to determine the largest room which can measure the three dimensions of the room exactly.

We first convert each dimension in cm

Length of room = 8m 25cm = 825cm

Breadth of room = 6m 75cm = 675cm

Height of room = 4m 50cm = 450cm.

Therefore, the required longest rod = H.C.F. of 825, 675 and 450.

First we consider 675 and 450.

By applying Euclid’s division lemma

Therefore, H.C.F. of 675 and 450 = 225

Now, we consider 225 and 825.

By applying Euclid’s division lemma

Therefore, H.C.F. of 825, 675 and 450 = 75

Hence, the length of required longest rod is

#### Page No 1.28:

#### Question 22:

Express the HCF of 468 and 222 as 468*x* + 222*y* where* x, y *are integers in two different ways.

#### Answer:

We need to express the H.C.F. of 468 and 222 as

Where *x, y* are integers in two different ways.

Given integers are 468 and 222, where

By applying Euclid’s division lemma, we get

Since the remainder, so apply division lemma on divisor 222 and remainder 24

Since the remainder, so apply division lemma on divisor 24 and remainder 6

We observe that remainder is 0. So the last divisor 6 is the H.C.F. of 468 and 222 from we have

.

#### Page No 1.35:

#### Question 1:

Express each of the following integers as a product of its prime factors:

(i) 420

(ii) 468

(iii) 945

(iv) 7325

#### Answer:

TO EXPRESS: each of the following numbers as a product of their prime factors

(i) 420

(ii) 468

(iii) 945

(iv) 7325

#### Page No 1.35:

#### Question 2:

Determine the prime factorisation of each of the following positive integer:

(i) 20570

(ii) 58500

(iii) 45470971

#### Answer:

TO EXPRESS: each of the following numbers as a product of their prime factors

(i) 20570

$20570=2\times 5\times {11}^{2}\times 17$

(ii) 58500

(iii) 45470971

#### Page No 1.35:

#### Question 3:

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

#### Answer:

EXPLAIN: Why and are composite numbers

We can see that both the numbers have common factor 7 and 1.

And we know that composite numbers are those numbers which have at least one more factor other than 1.

Hence after simplification we see that both numbers are even and therefore the given two numbers are composite numbers

#### Page No 1.35:

#### Question 4:

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

#### Answer:

TO CHECK: Whether can end with the digit 0 for any natural number n.

We know that

Therefore, prime factorization of does not contain 5 and 2 as a factor together.

Hence can never end with the digit 0 for any natural number *n*

#### Page No 1.35:

#### Question 5:

Explain why $3\times 5\times 7+7$ is a composite number.

#### Answer:

$\mathrm{Let}a=3\times 5\times 7+7=7\left(3\times 5+1\right)=7\times 16$

It can be seen that *a* has two more factors 7 and 16 other than 1 and the number itself.

Therefore, it is a composite number.

#### Page No 1.39:

#### Question 1:

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers:

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

(iv) 404 and 96

#### Answer:

TO FIND: LCM and HCF of following pairs of integers

TO VERIFY:

(i) 26 and 91

Let us first find the factors of 26 and 91

We know that,

$\mathrm{L}.\mathrm{C}.\mathrm{M}\times \mathrm{H}.\mathrm{C}.\mathrm{F}=\mathrm{First}\mathrm{number}\times \mathrm{Second}\mathrm{number}\phantom{\rule{0ex}{0ex}}\Rightarrow 182\times 13=26\times 91\phantom{\rule{0ex}{0ex}}\Rightarrow 2366=2366$

Hence verified

(ii) 510 and 92

Let us first find the factors of 510 and 92

We know that,

Hence verified

(iii) 336 and 54

Let us first find the factors of 336 and 54

We know that,

Hence verified

(iv)

$404=2\times 2\times 101\phantom{\rule{0ex}{0ex}}96=2\times 2\times 2\times 2\times 2\times 3\phantom{\rule{0ex}{0ex}}\mathrm{LCM}=2\times 2\times 2\times 2\times 2\times 3\times 101=9696\phantom{\rule{0ex}{0ex}}\mathrm{LCM}=9696\phantom{\rule{0ex}{0ex}}\mathrm{HCF}=2\times 2=4\phantom{\rule{0ex}{0ex}}\mathrm{Product}\mathrm{of}\mathrm{numbers}=96\times 404=38784\phantom{\rule{0ex}{0ex}}\mathrm{LCM}\times \mathrm{HCF}=4\times 9696=38784\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{LCM}\times \mathrm{HCF}=\mathrm{Product}\mathrm{of}\mathrm{numbers}$

#### Page No 1.39:

#### Question 2:

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

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(iv) 40, 36 and 126

(v) 84, 90 and 120

(vi) 24, 15 and 36

#### Answer:

TO FIND: LCM and HCF of following pairs of integers

(i) 15, 12 and 21

Let us first find the factors of 15, 12 and 21

(ii) 17, 23 and 29

Let us first find the factors of 17, 23 and 29

(iii) 8, 9 and 25

Let us first find the factors of 8,9 and 25

(iv) 40, 36 and 126

Let us first find the factors of 40, 36 and 126

(v) 84, 90 and 120

Let us first find the factors of 84, 90 and 120

(vi) 24, 15 and 36

Let us first find the factors of 24, 15 and 36.

#### Page No 1.39:

#### Question 3:

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

(ii) Write the smallest number which is divisible by both 306 and 657.

#### Answer:

(i)

GIVEN: HCF of two numbers 306 and 657 is 9.

TO FIND: L.C.M of number

We know that,

(ii) The smallest number divisible by both 306 and 657 is the LCM of 306 and 657.306 = 2 × 3 × 3 × 17

657 = 3 × 3 × 73

Therefore, LCM of 306 and 657 is 2 × 3 × 3 × 17 × 73 = 22,338

Hence, the smallest number which is divisible by both 306 and 657 is 22,338.

#### Page No 1.40:

#### Question 4:

Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.

#### Answer:

TO FIND: can two numbers have 16 as their H.C.F and 380 as their L.C.M

On dividing 380 by 16 we get 23 as the quotient and 12 as the remainder,

Since L.C.M is not exactly divisible by the H.C.F, two numbers cannot have 16 as their H.C.F and 380 as their L.C.M

#### Page No 1.40:

#### Question 5:

The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.

#### Answer:

GIVEN: LCM and HCF of two numbers are 2175 and 145 respectively. If one number is 725

TO FIND: Other number

We know that,

#### Page No 1.40:

#### Question 6:

The HCF to two numbers is 16 and their product is 3072. Find their LCM.

#### Answer:

GIVEN: HCF of two numbers is 16. If product of numbers is 3072

TO FIND: L.C.M of numbers

We know that,

#### Page No 1.40:

#### Question 7:

The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.

#### Answer:

GIVEN: LCM and HCF of two numbers are 180 and 6 respectively. If one number is 30

TO FIND: Other number

We know that,

#### Page No 1.40:

#### Question 8:

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

#### Answer:

TO FIND: Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

L.C.M OF 520 and 468

$520={2}^{3}\times 5\times 13$

$\mathrm{LCM}\mathrm{of}520\mathrm{and}468={2}^{3}\times {3}^{2}\times 5\times 13\phantom{\rule{0ex}{0ex}}=4680$

Hence 4680 is the least number which exactly divides 520 and 468 i.e. we will get a remainder of 0 in this case. But we need the Smallest number which when increased by 17 is exactly divided by 520 and 468.

Therefore

Hence is Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

#### Page No 1.40:

#### Question 9:

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

#### Answer:

TO FIND: The smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

L.C.M of 28 and 32.

Hence 224 is the least number which exactly divides 28 and 32 i.e. we will get a remainder of 0 in this case. But we need the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively

Therefore

Hence is the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively

#### Page No 1.40:

#### Question 10:

What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

#### Answer:

TO FIND: Smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case

L.C.M OF 35, 56 and 91

Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case

Therefore

Hence is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.

#### Page No 1.40:

#### Question 11:

A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. it is to be paved with square tiles of the same size. Find the least possible number of such tiles.

#### Answer:

GIVEN: A rectangular yard is 18 m 72 cm long and 13 m 20 cm broad .It is to be paved with square tiles of the same size.

TO FIND: Least possible number of such tiles.

Length of the yard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (∵ 1 m = 100 cm)

Breadth of the yard =13 m 20 cm = 1300 cm + 20 cm = 1320 cm

The size of the square tile of same size needed to the pave the rectangular yard is equal the HCF of the length and breadth of the rectangular yard.

Prime factorisation of 1872 = ${2}^{4}\times {3}^{2}\times 13$

Prime factorisation of 1320 = ${2}^{3}\times 3\times 5\times 11$

HCF of 1872 and 1320 = ${2}^{3}\times 3=24$

∴ Length of side of the square tile = 24 cm

Number of tiles required = $\frac{\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{courtyard}}{\mathrm{Area}\mathrm{of}\mathrm{each}\mathrm{tile}}=\frac{\mathrm{Lenght}\times \mathrm{Breadth}}{{\left(\mathrm{Side}\right)}^{2}}=\frac{1872\mathrm{cm}\times 1320\mathrm{cm}}{{\left(24\mathrm{cm}\right)}^{2}}=4290$

Thus, the least possible number of tiles required is 4290.

#### Page No 1.40:

#### Question 12:

Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

#### Answer:

TO FIND: Greatest number of 6 digits exactly divisible by 24, 15 and 36

The greatest 6 digit number be 999999

24, 15 and 36

Since

Therefore, the remainder is 279.

Hence the desired number is equal to

Hence is the greatest number of 6 digits exactly divisible by 24, 15 and 36.

#### Page No 1.40:

#### Question 13:

Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

#### Answer:

TO FIND: The number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

L.C.M Of 8, 15 and 21.

When 110000 is divided by 840, the remainder is obtained as 800.

Now, 110000 − 800 = 109200 is divisible by each of 8, 15 and 21.

Also, 110000 + 40 = 110040 is divisible by each of 8, 15 and 21.

109200 and 110040 are greater than 100000.

Hence, 110040 is the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

#### Page No 1.40:

#### Question 14:

Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

#### Answer:

TO FIND: Least number that is divisible by all the numbers between 1 and 10 (both inclusive)

Let us first find the L.C.M of all the numbers between 1 and 10 (both inclusive)

1 = 1

2 = 2

3 = 3

4 = 2^{2}

5 = 5

6 = 2 × 3

7 = 7

8 = 2^{3}

9 = 3^{2}

10 = 2 × 5

Hence is the least number that is divisible by all the numbers between 1 and 10 (both inclusive)

#### Page No 1.40:

#### Question 15:

A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again?

#### Answer:

GIVEN: A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60, and 72 km a day, round the field.

TO FIND: When they meet again.

In order to calculate the time when they meet, we first find out the time taken by each cyclist in covering the distance.

Number of days 1^{st} cyclist took to cover 360 km = $\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Distance}\mathrm{covered}\mathrm{in}1\mathrm{day}}=\frac{360}{48}=7.5=\frac{75}{10}=\frac{15}{2}\mathrm{days}$

Similarly, number of days taken by 2^{nd} cyclist to cover same distance = $\frac{360}{60}=6\mathrm{days}$

Also, number of days taken by 3^{rd} cyclist to cover this distance = $\frac{360}{72}=5\mathrm{days}$

Now, LCM of $\left(\frac{15}{2},6\mathrm{and}5\right)=\frac{\mathrm{LCM}\mathrm{of}\mathrm{numerators}}{\mathrm{HCF}\hspace{0.17em}\mathrm{of}\mathrm{denominators}}=\frac{30}{1}=30\mathrm{days}$

Thus, all of them will take 30 days to meet again.

#### Page No 1.40:

#### Question 16:

In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

#### Answer:

GIVEN: In a morning walk, three persons step off together. Their steps measure 80 cm, 85 cm and 90 cm.

TO FIND: minimum distance each should walk so that all can cover the same distance in complete steps.

The distance covered by each of them is required to be same as well as minimum. The required distance each should walk would be the L.C.M of the measures of their steps i.e. 80 cm, 85 cm, and 90 cm,

So we have to find the L.C.M of 80 cm, 85 cm, and 90 cm.

Hence minimum distance each should walk so that all can cove the same distance in complete steps

#### Page No 1.40:

#### Question 17:

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

#### Answer:

It is given that, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively.

The minimum distance each should walk so that each can cover the same distance in complete steps is the LCM of 30, 36 and 40.

30 = 2 × 3 × 5

36 = 2 × 2 × 3 × 3

40 = 2 × 2 × 2 × 5

Therefore, LCM of 30, 36 and 40 is 2 × 2 × 2 × 3 × 3 × 5 = 360

Hence, the minimum distance each should walk so that each can cover the same distance in complete steps is 360 cm.

#### Page No 1.40:

#### Question 18:

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

#### Answer:

On subtracting 1, 2 and 3 from 1251, 9377 and 15628 respectively, we get 1250, 9375 and 15625.

Now, we find the HCF of 1250 and 9375 by using Euclid's division lemma

9375 = 1250 × 7 + 625

∵ Remainder is not zero,

∴ we divide 1250 by 625 by using Euclid's division lemma

1250 = 625 × 2 + 0

Since, remainder is zero,

Therefore, HCF of 1250 and 9375 is 625.

Now, we find the HCF of 15625 and 625 by using Euclid's division lemma

15625 = 625 × 25 + 0

Since, remainder is zero,

Therefore, HCF of 15625 and 625 is 625.

Thus, HCF of 1250, 9375 and 15625 is 625.

Hence, the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively is 625.

#### Page No 1.49:

#### Question 1:

Show that the following numbers are irrational.

(i) $\frac{1}{\sqrt{2}}$

(ii) $7\sqrt{5}$

(iii) $6+\sqrt{2}$

(iv) $3-\sqrt{5}$

#### Answer:

(i) Let us assume that is rational .Then , there exist positive co primes *a* and *b* such that

(ii) Let us assume that is rational .Then , there exist positive co primes *a* and *b* such that

We know that is an irrational number

Here we see that is a rational number which is a contradiction

(iii) Let us assume that is rational. Then , there exist positive co primes *a* and *b* such that

Here we see that is a rational number which is a contradiction as we know that is an irrational number

(iv) Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

Here we see that is a rational number which is a contradiction as we know that is an irrational number

#### Page No 1.49:

#### Question 2:

Prove that following numbers are irrationals:

(i) $\frac{2}{\sqrt{7}}$

(ii) $\frac{3}{2\sqrt{5}}$

(iii) $4+\sqrt{2}$

(iv) $5\sqrt{2}$

#### Answer:

(i) Let us assume that is rational .Then , there exist positive co primes a and b such that

(ii) Let us assume that $\frac{3}{2\sqrt{5}}$ is rational .Then , there exist positive co primes a and b such that

(iii) Let us assume that is rational .Then , there exist positive co primes *a* and *b* such that

(iv) Let us assume that is rational .Then , there exist positive co primes *a* and *b* such that

#### Page No 1.49:

#### Question 3:

Show that $2-\sqrt{3}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b *such that

$2-\sqrt{3}=\frac{a}{b}\phantom{\rule{0ex}{0ex}}\sqrt{3}=2-\frac{a}{b}$

This implies, $\sqrt{3}$ is a rational number, which is a contradiction.

Hence, is irrational number.

#### Page No 1.49:

#### Question 4:

Show that $3+\sqrt{2}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

This implies,

#### Page No 1.49:

#### Question 5:

Prove that $4-5\sqrt{2}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

#### Page No 1.49:

#### Question 6:

Show that $5-2\sqrt{3}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

#### Page No 1.49:

#### Question 7:

Prove that $2\sqrt{3}-1$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

#### Page No 1.49:

#### Question 8:

Prove that $2-3\sqrt{5}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes *a* and *b* such that

#### Page No 1.49:

#### Question 9:

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

#### Answer:

Let us assume that is rational .Then, there exist positive co primes* a* and* b *such that

Here we see that is a rational number which is a contradiction as we know that is an irrational number.

#### Page No 1.49:

#### Question 10:

Prove that $\sqrt{2}+\sqrt{3}$ is an irrational number .

#### Answer:

Let us assume that $\sqrt{2}+\sqrt{3}$ is rational . Then, there exist co-prime positive integers *a* and *b *such that

$\sqrt{2}+\sqrt{3}=\frac{a}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{a}{b}-\sqrt{2}$

Squaring on both sides, we get

$3=\frac{{a}^{2}}{{b}^{2}}-\frac{2a}{b}\sqrt{2}+2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}}{{b}^{2}}-1=\frac{2a}{b}\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}-{b}^{2}}{2ab}=\sqrt{2}$

$\Rightarrow \sqrt{2}$ is a rational number $\left(a,b\mathrm{are}\mathrm{integers},\mathrm{so}\frac{{a}^{2}-{b}^{2}}{2ab}\mathrm{is}\mathrm{rational}\right)$

This contradicts the fact that $\sqrt{2}$ is a rational number. So our assumption was incorrect.

Hence, $\sqrt{2}+\sqrt{3}$ is an irrational number.

#### Page No 1.49:

#### Question 11:

Given that $\sqrt{2}$ is irrational, prove that $\left(5+3\sqrt{2}\right)$ is an irrational number.

#### Answer:

Let us assume, to the contrary that $5+3\sqrt{2}$ is rational

That is, we can find coprime a and b $(b\ne 0)$ such that $5+3\sqrt{2}=\frac{a}{b}$

Therefore,

$5-\frac{a}{b}=-3\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{3}-\frac{a}{3b}=-\sqrt{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a-5b}{3b}=\sqrt{2}$

Since, a and b are integers, we get $\frac{a-5b}{3b}$ is rational, and so $\sqrt{2}$ is rational.

which contradicts the fact that $\sqrt{2}$ is irrational.

So, our assumption was wrong that $5+3\sqrt{2}$ is rational.

Hence, we conclude that $5+3\sqrt{2}$ is irrational.

#### Page No 1.49:

#### Question 12:

Prove that $\frac{2+\sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

#### Answer:

Let us assume that $\frac{2+\sqrt{3}}{5}$ is a rational number.

Thus, $\frac{2+\sqrt{3}}{5}$ can be represented in the form of $\frac{p}{q}$, where *p* and *q* both are integers, *q* ≠ 0, *p* and *q* are co-prime numbers.

$\frac{2+\sqrt{3}}{5}=\frac{p}{q}\phantom{\rule{0ex}{0ex}}\Rightarrow 2+\sqrt{3}=5\left(\frac{p}{q}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{5p}{q}-2\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{5p-2q}{q}$

since, $\frac{5p-2q}{q}$ is rational ⇒ $\sqrt{3}$ is rational.

But, it is given that $\sqrt{3}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $\frac{2+\sqrt{3}}{5}$ is an irrational number.

#### Page No 1.49:

#### Question 13:

Prove that $2+5\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

#### Answer:

Let us assume that $2+5\sqrt{3}$ is a rational number.

Thus, $2+5\sqrt{3}$ can be represented in the form of $\frac{p}{q}$, where *p* and *q* both are integers, *q* ≠ 0, *p* and *q* are co-prime numbers.

$2+5\sqrt{3}=\frac{p}{q}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\sqrt{3}=\frac{p}{q}-2\phantom{\rule{0ex}{0ex}}\Rightarrow 5\sqrt{3}=\frac{p-2q}{q}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{p-2q}{5q}$

since, $\frac{p-2q}{5q}$ is rational ⇒ $\sqrt{3}$ is rational.

But, it is given that $\sqrt{3}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $2+5\sqrt{3}$ is an irrational number.

#### Page No 1.49:

#### Question 14:

Prove that for any prime positive integer* p*, $\sqrt{p}$ is an irrational number.

#### Answer:

Let us assume that is rational .Then, there exist positive co primes a and b such that

#### Page No 1.50:

#### Question 15:

If *p, q *are prime positive integers, prove that $\sqrt{p}+\sqrt{q}$ is an irrational number.

#### Answer:

Let us assume that is rational. Then, there exist positive co primes *a* and *b* such that

Here we see that is a rational number which is a contradiction as we know that is an irrational number

#### Page No 1.56:

#### Question 1:

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

(i) $\frac{23}{8}$

(ii) $\frac{125}{441}$

(iii) $\frac{35}{50}$ [NCERT]

(iv) $\frac{77}{210}$ [NCERT]

(v) $\frac{129}{{2}^{2}\times {5}^{7}\times {7}^{17}}$

(vi) $\frac{987}{10500}$

#### Answer:

(i) The given number is $\frac{23}{8}$.

Here, $8={2}^{3}$ and 2 is not a factor of 23.

So, the given number is in its simplest form.

Now, $8={2}^{3}$ is of the form ${2}^{m}\times {5}^{n}$, where *m* = 3 and *n* = 0.

So, the given number has a terminating decimal expansion.

(ii) The given number is $\frac{125}{441}$.

Here, $441={3}^{2}\times {7}^{2}$ and none of 3 and 7 is a factor of 125.

So, the given number is in its simplest form.

Now, $441={3}^{2}\times {7}^{2}$ is not of the form ${2}^{m}\times {5}^{n}$.

So, the given number has a non-terminating repeating decimal expansion.

(iii) The given number is $\frac{35}{50}$ and HCF(35, 50) = 5.

∴ $\frac{35}{50}=\frac{35\xf75}{50\xf75}=\frac{7}{10}$

Here, $\frac{7}{10}$ is in its simplest form.

Now, $10=2\times 5$ is of the form ${2}^{m}\times {5}^{n}$, where *m* = 1 and *n* = 1.

So, the given number has a terminating decimal expansion.

(iv) The given number is $\frac{77}{210}$ and HCF(77, 210) = 7.

∴ $\frac{77}{210}=\frac{77\xf77}{210\xf77}=\frac{11}{30}$

Here, $\frac{11}{30}$ is in its simplest form.

Now, $30=2\times 3\times 5$ is not of the form ${2}^{m}\times {5}^{n}$.

So, the given number has a non-terminating repeating decimal expansion.

(v) The given number is $\frac{129}{{2}^{2}\times {5}^{7}\times {7}^{17}}$.

Clearly, none of 2, 5 and 7 is a factor of 129.

So, the given number is in its simplest form.

Now, ${2}^{2}\times {5}^{7}\times {7}^{17}$is not of the form ${2}^{m}\times {5}^{n}$.

So, the given number has a non-terminating repeating decimal expansion.

(vi) The given number is $\frac{987}{10500}$.

$\frac{987}{10500}=\frac{987\xf721}{10500\xf721}=\frac{47}{500}$

Now, 500 = 2^{2} × 5^{3}

The denominator can be written in the form of 2* ^{m}* × 5

*.*

^{n}So, the given number has a terminating decimal expansion.

$\frac{987}{10500}=\frac{47}{500}=\frac{47}{{5}^{3}\times {2}^{2}}\times \frac{2}{2}=\frac{94}{{5}^{3}\times {2}^{3}}=\frac{94}{1000}=0.094$

#### Page No 1.56:

#### Question 2:

Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2^{m} × 5^{n}, where, *m*, *n* are non-negative integers.

(i) $\frac{3}{8}$

(ii) $\frac{13}{125}$

(iii) $\frac{7}{80}$

(iv) $\frac{14588}{625}$

(v) $\frac{129}{{2}^{2}\times {5}^{7}}$ [NCERT]

#### Answer:

(i) The given number is $\frac{3}{8}$.

Clearly, 8 = 2^{3} is of the form 2^{m} × 5^{n}, where *m* = 3 and *n* = 0.

So, the given number has terminating decimal expansion.

$\therefore \frac{3}{8}=\frac{3\times {5}^{3}}{{2}^{3}\times {5}^{3}}=\frac{3\times 125}{{\left(2\times 5\right)}^{3}}=\frac{375}{{\left(10\right)}^{3}}=\frac{375}{1000}=0.375$

(ii) The given number is $\frac{13}{125}$.

Clearly, 125 = 5^{3} is of the form 2^{m} × 5^{n}, where *m* = 0 and *n* = 3.

So, the given number has terminating decimal expansion.

$\therefore \frac{13}{125}=\frac{13\times {2}^{3}}{{2}^{3}\times {5}^{3}}=\frac{13\times 8}{{\left(2\times 5\right)}^{3}}=\frac{104}{{\left(10\right)}^{3}}=\frac{104}{1000}=0.104$

(iii) The given number is $\frac{7}{80}$.

Clearly, 80 = 2^{4} × 5 is of the form 2^{m} × 5^{n}, where *m* = 4 and *n* = 1.

So, the given number has terminating decimal expansion.

$\therefore \frac{7}{80}=\frac{7\times {5}^{3}}{{2}^{4}\times 5\times {5}^{3}}=\frac{7\times 125}{{\left(2\times 5\right)}^{4}}=\frac{875}{{\left(10\right)}^{4}}=\frac{875}{10000}=0.0875$

(iv) The given number is $\frac{14588}{625}$.

Clearly, 625 = 5^{4} is of the form 2^{m} × 5^{n}, where *m* = 0 and *n* = 4.

So, the given number has terminating decimal expansion.

$\therefore \frac{14588}{625}=\frac{14588\times {2}^{4}}{{2}^{4}\times {5}^{4}}=\frac{14588\times 16}{{\left(2\times 5\right)}^{4}}=\frac{233408}{{\left(10\right)}^{4}}=\frac{233408}{10000}=23.3408$

(v) The given number is $\frac{129}{{2}^{2}\times {5}^{7}}$.

Clearly, 2^{2} × 5^{7 }is of the form 2^{m} × 5^{n}, where *m* = 2 and *n* = 7.

So, the given number has terminating decimal expansion.

$\therefore \frac{129}{{2}^{2}\times {5}^{7}}=\frac{129\times {2}^{5}}{{2}^{2}\times {5}^{7}\times {2}^{5}}=\frac{129\times 32}{{\left(2\times 5\right)}^{7}}=\frac{4182}{{\left(10\right)}^{7}}=\frac{4182}{10000000}=0.0004182$

#### Page No 1.57:

#### Question 3:

Write the denominator of the rational number $\frac{257}{5000}$ in the form ${2}^{m}\times {5}^{n}$, where $m,n$ are non-negative integers. Hence, write the decimal expansion, without actual division.

#### Answer:

The given number is $\frac{257}{5000}$.

Now, 5000 = 2^{3} × 5^{4}

Clearly, 2 and 5 are factors of 5000.

The denominator can be written in the form of 2* ^{m}* × 5

*.*

^{n}So, the given number has a terminating decimal expansion.

$\frac{257}{5000}=\frac{257}{{5}^{4}\times {2}^{3}}\times \frac{2}{2}=\frac{514}{{10}^{4}}=\frac{514}{10000}=0.0514$

#### Page No 1.57:

#### Question 4:

What can you say about the prime factorisations of the denominators of the following rationals:

(i) 43.123456789

(ii) $43.\overline{)123456789}$

(iii) $27.\overline{142857}$

(iv) 0.120120012000120000 ...

#### Answer:

(i) Since 43.123456789 has terminating decimal expansion.

So, its denominator is of the form 2* ^{m} *× 5

*, where*

^{n}*m*,

*n*are non-negative integers.

(ii) Since $43.\overline{)123456789}$ has non-terminating decimal expansion.

So, its denominator has factors other than 2 or 5.

(iii) Since $27.\overline{142857}$ has non-terminating decimal expansion.

So, its denominator has factors other than 2 or 5.

(iv) Since 0.120120012000120000 ... has non-terminating decimal expansion.

So, its denominator has factors other than 2 or 5.

#### Page No 1.57:

#### Question 5:

A rational number in its decimal expansion is 327.7081. What can you say about the prime factor of *q* , when this number is expressed in the form $\frac{p}{q}$? Give reasons.

#### Answer:

Since 327.7081 is a terminating decimal number, therefore, it can be represented as a rational number.

Also, its denominator must be in the form ${2}^{m}\times {5}^{n}$.

Thus,

$327.7081=\frac{3277081}{10000}=\frac{p}{q}\phantom{\rule{0ex}{0ex}}\therefore q=10000=2\times 2\times 2\times 2\times 5\times 5\times 5\times 5={2}^{4}\times {5}^{4}$

Hence, the prime factorization of *q* contains only factors of 2 and 5.

#### Page No 1.57:

#### Question 1:

The exponent of 2 in the prime factorisation of 144, is

(a) 4

(b) 5

(c) 6

(d) 3

#### Answer:

Using the factor tree for prime factorization, we have:

Therefore,

Thus, the exponent of 2 in 144 is 4.

Hence the correct choice is (*a*)**.**

#### Page No 1.57:

#### Question 2:

The LCM of two numbers is 1200. Which of the following cannot be their HCF?

(a) 600

(b) 500

(c) 400

(d) 200

#### Answer:

It is given that the LCM of two numbers is 1200.

We know that the HCF of two numbers is always the factor of LCM

Checking all the options:

(*a*) 600 is the factor of 1200.

So this can be the HCF.

(*b*) 500 is not the factor of 1200.

So this cannot be the HCF.

(*c*) 400 is the factor of 1200.

So this can be the HCF.

(*d*) 200 is the factor of 1200.

So this can be the HCF.

Hence the correct choice is (*b*).

#### Page No 1.57:

#### Question 3:

If *n* = 2^{3} ✕ 3^{4} ✕ 5^{4} ✕ 7, then the number of consecutive zeros in *n*, where *n* is a natural number, is

(a) 2

(b) 3

(c) 4

(d) 7

#### Answer:

Since, it is given that

*n* = ${2}^{3}\times {3}^{4}\times {5}^{4}\times 7$

$={2}^{3}\times {5}^{4}\times {3}^{4}\times 7\phantom{\rule{0ex}{0ex}}={2}^{3}\times {5}^{3}\times 5\times {3}^{4}\times 7\phantom{\rule{0ex}{0ex}}={\left(2\times 5\right)}^{3}\times 5\times {3}^{4}\times 7\phantom{\rule{0ex}{0ex}}=5\times {3}^{4}\times 7\times {\left(10\right)}^{3}$

So, this means the given number *n* will end with 3 consecutive zeroes.

#### Page No 1.57:

#### Question 4:

The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1

(b) 2

(c) 4

(d) 6

#### Answer:

Using the factor tree for prime factorization, we have:

Therefore,

The exponents of 2 and 7 are 2 and 2 respectively.

Thus the sum of the exponents is

Hence the correct choice is (*c*).

#### Page No 1.58:

#### Question 5:

The number of decimal place after which the decimal expansion of the rational number $\frac{23}{{2}^{2}\times 5}$will terminate, is

(a) 1

(b) 2

(c) 3

(d) 4

#### Answer:

We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which terminates after *k* places of decimals, where *k* is the larger of *m* and *n*.

This is given that the prime factorization of the denominator is of the form.

Hence, it has terminating decimal expansion which terminates after places of decimal.

Hence, the correct choice is (*b*).

#### Page No 1.58:

#### Question 7:

If two positive ingeters *a* and *b* are expressible in the form *a* = *pq*^{2} and *b* = *p*^{3}*q*; *p*, *q* being prime number, then LCM (*a*, *b*) is

(a) *pq*

(b) *p*^{3}*q*^{3}

(c) *p*^{3}*q*^{2}

(d) *p*^{2}*q*^{2}

#### Answer:

Two positive integers are expressed as follows:

*p* and *q* are prime numbers.

Then, taking the highest powers of *p *and *q* in the values for *a* and *b* we get:

LCM

Hence the correct choice is (*c*)*.*

#### Page No 1.58:

#### Question 8:

In Q.No. 7, HCF (*a*, *b*) is

(a) *pq*

(b) *p*^{3}*q*^{3}

(c) *p*^{3}*q*^{2}

(d) *p*^{2}*q*^{2}

#### Answer:

Two positive integers are expressed as follows:

*p* and *q* are prime numbers.

Then, taking the smallest powers of *p *and *q* in the values for *a* and *b* we get

HCF

Hence the correct choice is (*a*)*.*

#### Page No 1.58:

#### Question 9:

If two positive integers m and *n* are expressible in the form *m* = *pq*^{3} and *n* = *p*^{3}*q*^{2}, where *p*, *q* are prime numbers, then HCF (*m*, *n*) =

(a) *pq*

(b) *pq*2

(c) *p*^{3}*q*^{2}

(d) *p*^{2}*q*^{2}

#### Answer:

Two positive integers are expressed as follows:

*p* and *q* are prime numbers.

Then, taking the smallest powers of *p *and *q* in the values for *m* and *n* we get

HCF

Hence the correct choice is (*b*)*.*

#### Page No 1.58:

#### Question 10:

If the LCM of *a* and 18 is 36 and the HCF of *a* and 18 is 2, then *a* =

(a) 2

(b) 3

(c) 4

(d) 1

#### Answer:

LCM

HCF

We know that the product of numbers is equal to the product of their HCF and LCM.

Therefore,

Hence the correct choice is (*c*)*.*

#### Page No 1.58:

#### Question 11:

The HCF of 95 and 152, is

(a) 57

(b) 1

(c) 19

(d) 38

#### Answer:

Using the factor tree for 95, we have:

Using the factor tree for 152, we have:

Therefore,

HCF

Hence the correct choice is (*c*)*.*

#### Page No 1.58:

#### Question 12:

If HCF (26, 169) = 13, then LCM (26, 169) =

(a) 26

(b) 52

(c) 338

(d) 13

#### Answer:

HCF* *

We have to find the value for LCM

We know that the product of numbers is equal to the product of their HCF and LCM.

Therefore,

Hence the correct choice is (*c*)*.*

#### Page No 1.58:

#### Question 13:

If *a* = 2^{3} ✕ 3, *b *= 2 ✕ 3 ✕ 5, *c* = 3^{n} ✕ 5 and LCM (*a*, *b*, *c*) = 2^{3} ✕ 3^{2} ✕ 5, then *n* =

(a) 1

(b) 2

(c) 3

(d) 4

#### Answer:

LCM* *……(I)

We have to find the value for *n*

Also

We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.

Therefore, by applying this rule and taking we get the LCM as

LCM* *……(II)

On comparing (I) and (II) sides, we get:

Hence the correct choice is (*b*)*.*

#### Page No 1.58:

#### Question 14:

The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after

(a) one decimal place

(b) two decimal place

(c) three decimal place

(d) four decimal place

#### Answer:

We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which terminates after *k* places of decimals, where *k* is the larger of *m* and *n*.

This is given that the prime factorization of the denominator is of the form.

Hence, it has terminating decimal expansion which terminates after places of decimal.

Hence, the correct choice is (*d*)*.*

#### Page No 1.58:

#### Question 15:

If *p* and *q* are co-prime numbers, then *p*^{2} and *q*^{2} are

(a) coprime

(b) not coprime

(c) even

(d) odd

#### Answer:

We know that the co-prime numbers have no factor in common, or, their HCF is 1.

Thus,and have the same factors with twice of the exponents of p and *q* respectively, which again will not have any common factor.

Thus we can conclude that and are co-prime numbers.

Hence, the correct choice is (*a*)*.*

#### Page No 1.58:

#### Question 16:

Which of the following rational numbers have terminating decimal?

(i) $\frac{16}{225}$

(ii) $\frac{5}{18}$

(iii) $\frac{2}{21}$

(iv) $\frac{7}{250}$

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i) and (iv)

#### Answer:

(*i*) We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is not of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which does not have terminating decimal.

(*ii*) We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is not of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which does not have terminating decimal.

(*iii*) We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is not of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which does not have terminating decimal.

(*iv*) We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is of the form, where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which terminates after *k* places of decimals, where *k* is the larger of *m* and *n*.

Then, *x* has a decimal expression which will have terminating decimal after 3 places of decimal.

Hence the (*iv*) option will have terminating decimal expansion.

There is no correct option.

#### Page No 1.58:

#### Question 17:

If 3 is the least prime factor of number *a* and 7 is the least prime factor of number *b*, then the least prime factor of *a* +* b*, is

(a) 2

(b) 3

(c) 5

(d) 10

#### Answer:

Since

The least prime factor of has to be 2; unless is a prime number greater than 2.

Suppose is a prime number greater than 2. Then must be an odd number

o one of *a* or *b *must be an even number.

Suppose then that *a* is even. Then the least prime factor of *a* is 2; which is not *3* or *7*. So *a* can not be an even number nor can *b* be an even number. Hence can not be a prime number greater than 2 if the least prime factor of *a* is 3 and b is 7.

Thus the answer is 2.

Hence the correct choice is (*a*)*.*

#### Page No 1.58:

#### Question 18:

$3.\overline{)27}$ is

(a) an integer

(b) a rational number

(c) a natural number

(d) an irrational number

#### Answer:

We have,

Let

Then,

Subtract these to get

Thus, we can also conclude that all infinite repeating decimals are rational numbers.

Hence, the correct choice is (*b*)*.*

#### Page No 1.59:

#### Question 19:

The smallest number by which $\sqrt{27}$ should be multiplied so as to get a rational number is

(a) $\sqrt{27}$

(b) $3\sqrt{3}$

(c) $\sqrt{3}$

(d) 3

#### Answer:

Out of the given choices is the only smallest number by which if we multiply we get a rational number.

Hence, the correct choice is (*c*)*.*

#### Page No 1.59:

#### Question 20:

The smallest rational number by which $\frac{1}{3}$should be multiplied so that its decimal expansion terminates after one place of decimal, is

(a) $\frac{3}{10}$

(b) $\frac{1}{10}$

(c) 3

(d) $\frac{3}{100}$

#### Answer:

For terminating the decimal expansion after one place of decimal, the highest power of *m* and *n* in should be 1.

Let

We will get:

Thus, it is evident that we multiplied it by

Hence, the correct choice is (*a*)*.*

#### Page No 1.59:

#### Question 21:

If *n* is a natural number, then 9^{2n} − 4^{2n} is always divisible by

(a) 5

(b) 13

(c) both 5 and 13

(d) None of these

[Hint : 9^{2n} − 4^{2n} is of the form *a*^{2n} − *b*^{2n} which is divisible by both *a* − *b* and *a* + *b*. So, 9^{2n} − 4^{2n} is divisible by both 9 − 4 = 5 and 9 + 4 = 13.]

#### Answer:

We know that is always divisible by both and .

So, is always divisible by both and.

Hence, the correct choice is (*c*)*.*

#### Page No 1.59:

#### Question 22:

If *n* is any natural number, then 6^{n} − 5^{n} always ends with

(a) 1

(b) 3

(c) 5

(d) 7

[Hint: For any *n* ∈ N, 6^{n} and 5^{n} end with 6 and 5 respectively. Therefore, 6^{n} − 5^{n} always ends with 6 − 5 = 1.]

#### Answer:

We know that * *will end in 6

And* * will end in 5.

Now, always end with

Hence the correct choice in (*a*)*.*

#### Page No 1.59:

#### Question 23:

The LCM and HCF of two rational numbers are equal, then the numbers must be

(a) prime

(b) co-prime

(c) composite

(d) equal

#### Answer:

Let the two numbers be *a* and *b*.

(*a*) If we assume that the *a* and *b* are prime.

Then,

HCF

LCM

(*b*) If we assume that *a* and *b* are co-prime.

Then,

HCF

LCM

(*c*) If we assume that *a* and *b* are composite.

Then,

HCFor any other highest common integer

LCM

(*d*) If we assume that *a* and *b* are equal and consider *a*=*b*=*k*.

Then,

HCF

LCM

Hence the correct choice is (*d*)*.*

#### Page No 1.59:

#### Question 24:

If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is

(a) 203400

(b) 194400

(c) 198400

(d) 205400

#### Answer:

Let the HCF be *x* and the LCM of the two numbers be *y*.

It is given that the sum of the HCF and LCM is 1260

* *…… (*i*)

And, LCM is 900 more than HCF.

* *…… (*ii*)

Substituting (*ii*) in (*i*), we get:

Substituting * * in (*ii*), we get:

We also know that the product the two numbers is equal to the product of their LCM and HCF

Thus the product of the numbers

Hence, the correct choice is (*b*)*.*

#### Page No 1.59:

#### Question 25:

The remainder when the square of any prime number greater than 3 is divided by 6, is

(a) 1

(b) 3

(c) 2

(d) 4

[Hint: Any prime number greater than 3 is of the from 6*k* ± 1, where *k* is a natural number and (6*k* ± 1)^{2} = 36*k*^{2} ± 12*k* + 1 = 6*k*(6*k* ± 2) + 1]

#### Answer:

Any prime number greater than 3 is of the form*, *where *k *is a natural number.

Thus,

When*, * is divided by 6, we get, and remainder as 1.

Hence, the correct choice is (*a*)*.*

#### Page No 1.59:

#### Question 26:

For some integer *m*, every integer is of the form

(a) *m* (b) *m* + 1 (c) 2*m* (d) 2*m*+ 1

#### Answer:

It is known that, even integers are ...,−4, −2, 0, 2, 4,...

Observe that, all of the even numbers are the multiple of 2.

So, even numbers can be written as 2*m*, where, *m* is an integer.

*m* can be ...,−2, −1, 0, 1, 2,...

∴ 2*m *can be ...,−4, −2, 0, 2, 4,...

Hence, the correct answer is option C.

#### Page No 1.59:

#### Question 27:

For some integer *q *, every odd integer is of the form

(a) *q * (b) *q *+ 1 (c) 2*q * (d) 2*q *+ 1

#### Answer:

We know that, all numbers that are not the multiple of 2 are odd numbers.

Odd integers are ...,−3, −1, 1, 3, 5,...

So, odd numbers can be written as 2*m* + 1, where *m* is an integer.

*m* can be ..., −2, −1, 0, 1, 2,...

∴ 2*m* + 1 can be ..., −3, −1, 1, 3,...

Hence, the correct answer is option D.

#### Page No 1.59:

#### Question 28:

${n}^{2}-1$ is divisible by 8 , if n is

(a) an integer (b) a natural number

(c) an odd integer (d) an even integer

#### Answer:

*m*+ 1 or 4

*m*+ 3 for some integer

*m*.

*n*= 4

*m*+ 1,

${n}^{2}-1={\left(4m+1\right)}^{2}-1=16{m}^{2}+8m+1-1=16{m}^{2}+8m=8m\left(2m+1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}-1\mathrm{is}\mathrm{divisible}\mathrm{b}y8.\phantom{\rule{0ex}{0ex}}\mathrm{When}n=4m+3,\phantom{\rule{0ex}{0ex}}{n}^{2}-1={\left(4m+3\right)}^{2}-1=16{m}^{2}+24m+9-1=16{m}^{2}+24m+8=8\left(2{m}^{2}+3m+1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {n}^{2}-1\mathrm{is}\mathrm{divisible}\mathrm{by}8.$

*n*

^{2}– 1 is divisible by 8 if

*n*is an odd positive integer.

#### Page No 1.59:

#### Question 29:

The decimal expansion of the rational number $\frac{33}{{2}^{2}\times 5}$ will terminate after

(a) one decimal place (b) Two decimal places

(c) three decimal places (d) more than 3 decimal places

#### Answer:

Consider the rational number$\frac{33}{{2}^{2}\times 5}$.

Denominator of the above rational number is of the form ${2}^{m}\times {2}^{n}$, so the rational number has terminating decimal expansion.

$\frac{33}{{2}^{2}\times 5}=\frac{33\times 5}{{2}^{2}\times {5}^{2}}=\frac{165}{100}=1.65$

So, the decimal expansion of the given rational number terminates after two decimal places.

Hence, the correct answer is option B.

#### Page No 1.60:

#### Question 30:

If two positive integers* a *and* b* are written as $a={x}^{3}{y}^{2}$ and $b=x{y}^{3};x,y$ are prime numbers , then HCF(*a*,*b*) is

(a)* xy * (b) *xy*^{2 } (c)* x*^{3}*y*^{3 } (d) *x*^{2}*y*^{2 }

#### Answer:

It is given that,

$a={x}^{3}{y}^{2}=x\times x\times x\times y\times y\phantom{\rule{0ex}{0ex}}b=x{y}^{3}=x\times y\times y\times y\phantom{\rule{0ex}{0ex}}\mathrm{HCF}\left(a,b\right)=\mathrm{HCF}\left({x}^{3}{y}^{2},x{y}^{3}\right)=x\times y\times y=x{y}^{2}$

Hence, the correct answer is option B.

#### Page No 1.60:

#### Question 31:

The largest number which divides 70 and 125, leaving remainders 5 and 8 , respectively,is

(a) 13 (b) 65 (c) 875 (d) 1750

#### Answer:

It is given that on dividing 70 by the required number, there is a remainder of 5. This means that 70 − 5 = 65 is exactly divisible by the required number.

Similarly, 125 − 8 = 117 is exactly divisible by the required number.

The required number is HCF of 65 and 117.

By Euclid's division algorithm,

$117=65\times 1+52\phantom{\rule{0ex}{0ex}}\Rightarrow 65=52\times 1+13\phantom{\rule{0ex}{0ex}}\Rightarrow 52=13\times 4+0$

Here, the remainder is zero. Therefore, the HCF of 65 and 117 is 13.

So, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.

Hence, the correct answer is option A.

#### Page No 1.60:

#### Question 32:

The product of a non-zero rational number and an irrational number is

(a) always irrational

(b) always rational

(c) rational or irrational

(d) one

#### Answer:

The product of a non-zero rational number and an irrational number is always an irrational number.

Hence, the correct option is (a).

#### Page No 1.60:

#### Question 33:

The HCF and LCM of 12, 21, 15 respectively are

(a) 3, 140

(b) 12, 420

(c) 3, 420

(d) 420, 3

#### Answer:

Here,

$12={2}^{2}\times 3$

$21=3\times 7$

$15=3\times 5$

Therefore, HCF(12, 21, 15) = 3 and

LCM(12, 21, 15) = ${2}^{2}\times 3\times 5\times 7=420$

Hence, the correct answer is option C.

#### Page No 1.60:

#### Question 34:

Euclid's division lemma states that for two positive integers *a *and* b *, there exist unique integers *q *and *r *such that* a* = *bq* + *r* , where * r *must satisfy

(a) 1 < *r* < * b* (b) 0 < *r* $\le $* b* (c) 0 $\le $ *r* < * b* (d) 0 < *r* < * b*

#### Answer:

Euclid's division lemma states that for two positive integers *a* and *b*, there exists unique integers *q *and* r* such that *a* = *bq* +* r*, where *r *must satisfy $0\le r<b$.

Hence, the correct answer is option C.

#### Page No 1.60:

#### Question 1:

If *a* and *b* are two positive co-prime integers such that *a* = 12*b*, then HCF (*a*, 12) = _________.

#### Answer:

If *a* and *b* are two positive co-prime integers

⇒ There is no common factor in *a* and *b* except 1.

⇒ HCF (*a*, 12) = HCF (12*b*, 12) (∵ *a* = 12*b*)

= 12

Hence, HCF (*a*, 12) = 12 .

#### Page No 1.60:

#### Question 2:

If two positive integers *m* and *n* are expressible in the form *m* = *a*^{2} *b*^{3} and *n* =* a*^{3}*b*^{2}, where *a, b* are prime numbers, then HCF (*m, n*) = _______ and LCM (*a, b*) = _______.

#### Answer:

It is given that, *m* = *a*^{2}*b*^{3} and *n* =* a*^{3}*b*^{2}, where *a, b* are prime numbers.

HCF (*m, n*) = HCF (*a*^{2}*b*^{3}*, a*^{3}*b*^{2})

= The lowest of indices of *a* and *b*

= * a*^{2}*b*^{2}

Hence, HCF (*m, n*) is *a*^{2}*b*^{2}.

LCM (*m, n*) = LCM (*a*^{2} *b*^{3}*, a*^{3}*b*^{2})

= The lowest of indices of *a* and *b*

= * a*^{3}*b*^{3}

Hence, LCM (*m, n*) is *a*^{3}*b*^{3}.

Hence, HCF (*m, n*) = *a*^{2}*b*^{2} and LCM (*m, n*) = *a*^{3}*b*^{3}.

__Disclaimer__: In the question, we need to find LCM (*m*, *n*) instead of LCM (*a*, *b*).

#### Page No 1.60:

#### Question 3:

If the HCF and LCM of two positive integers *a* and* b* are *x *and *y* respectively, then $\frac{{x}^{2}{y}^{2}}{{a}^{2}{b}^{2}}$ = _______.

#### Answer:

Product of two numbers = LCM × HCF

⇒* a* × *b* = *y* × *x*

⇒ *ab* = *xy* ...(1)

Now,

$\frac{{x}^{2}{y}^{2}}{{a}^{2}{b}^{2}}=\frac{{\left(xy\right)}^{2}}{{\left(ab\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(ab\right)}^{2}}{{\left(ab\right)}^{2}}\left(\mathrm{From}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=1$

Hence, $\frac{{x}^{2}{y}^{2}}{{a}^{2}{b}^{2}}$ = __ 1 __.

#### Page No 1.60:

#### Question 4:

A positive integer *m* when divided by 11 gives remainder 6. If 4*m* + 5 is divided by 11, the remainder is _________.

#### Answer:

We know that,

Dividend = Divisor × Quotient + Remainder

⇒* m* = 11 × *q* + 6 , where* m* is the dividend and *q* is the quotient

⇒* *4*m* = 11 × 4*q* + 24

⇒* *4*m* = 11 × 4*q* + 22 + 2

⇒* *4*m* = 11 × (4*q* + 2) + 2

Thus, if* *4*m* is divided by 11, we get 2 as the remainder.

Therefore, If 4*m* + 5 is divided by 11, the remainder is 2 + 5 = 7.

Hence, If 4*m* + 5 is divided by 11, the remainder is __ 7 __.

#### Page No 1.60:

#### Question 5:

If HCF (306, 657) = 9, then LCM (306, 657) __________.

#### Answer:

we know that,

Product of two numbers = LCM × HCF

⇒ 306 × 657 = LCM × 9

$\Rightarrow \mathrm{LCM}\times 9=306\times 657\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=\frac{306\times 657}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=34\times 657\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=22338$

Hence, LCM (306, 657) is __22,338__.

#### Page No 1.60:

#### Question 6:

The LCM of the smallest prime number and the smallest odd composite number is ___________.

#### Answer:

Smallest prime number is 2.

Smallest odd composite number is 9.

Therefore, LCM (2, 9) is 18.

Hence, The LCM of the smallest prime number and the smallest odd composite number is __18__.

#### Page No 1.61:

#### Question 7:

The sum of the exponents of prime factors in the prime factorisation of 250 is _________.

#### Answer:

The prime factorisation of 250 are:

250 = 2 × 5 × 5 × 5

= 2^{1} × 5^{3}

The sum of exponents is 1 + 3 = 4.

Hence, The sum of the exponents of prime factors in the prime factorisation of 250 is __ 4 __.

#### Page No 1.61:

#### Question 8:

The LCM of the smallest prime number and the smallest composite number is _________.

#### Answer:

Smallest prime number is 2.

Smallest composite number is 4.

Therefore, LCM (2, 4) = 4

Hence, The LCM of the smallest prime number and the smallest composite number is __ 4 __.

#### Page No 1.61:

#### Question 9:

6^{n} can not end with digit 0 for _________ value of *n.*

#### Answer:

The prime factors of 6 are 2 and 3.

$\Rightarrow {6}^{n}={2}^{n}\times {3}^{n}$

Since, 5 is not the factor of 6^{n}

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

Hence, 6

^{n}can not end with digit 0 for

__any__value of

*n.*

#### Page No 1.61:

#### Question 10:

The ratio between the HCF and LCM of 5, 15 and 20 is __________.

#### Answer:

Prime factorisation of 5, 15 and 20 is

5 = 5

15 = 3 × 5

20 = 2 × 2 × 5

Thus, LCM of 5, 15 and 20 is 2 × 2 × 3 × 5 = 60

and HCF of 5, 15 and 20 is 5.

$\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{HCF}}{\mathrm{LCM}}=\frac{5}{60}=\frac{1}{12}$

Hence, The ratio between the HCF and LCM of 5, 15 and 20 is __1 : 12__.

#### Page No 1.61:

#### Question 11:

The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is _______.

#### Answer:

It is given that the first number is completely divided by 2 the quotient is 33,

∴ First number = 2 × 33 + 0 = 66

Let the other number be *x*.

Now,

Product of two numbers = LCM × HCF

⇒ 66 × *x* = 264 × 33

$\Rightarrow x=\frac{264\times 33}{66}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{264}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=132$

Hence, the other number is __132__.

#### Page No 1.61:

#### Question 12:

If the prime factorisation of a natural number *n *is 2^{4 }× 3^{4} × 5^{3} × 7, then the number of consecutive zeros in *n, *is _________.

#### Answer:

To calculate the number of consecutive zeros, we need to find the number of pairs of 2 and 5 to form a product of 10.

Here, *n* = 2^{4 }× 3^{4} × 5^{3} × 7

= 2 × 2 × 2 × 2^{ }× 3 × 3 × 3 × 3 × 5 × 5 × 5 × 7

= 2 × 5 × 2 × 5^{ }× 2 × 5 × 2 × 3 × 3 × 3 × 3 × 7

= 10 × 10^{ }× 10 × 2 × 3 × 3 × 3 × 3 × 7

= 2^{ }× 3^{4} × 7 × 10^{3}

Hence, the number of consecutive zeros in *n, *is __ 3 __.

#### Page No 1.61:

#### Question 13:

If 2520 = 2^{a} × 3^{b} × 5^{c }× 7^{d}, then^{ }*a + b *– 2*c* – 3*d* = _________.

#### Answer:

Prime factorisation of 2520 is

2520 = 2^{3} × 3^{2} × 5^{1 }× 7^{1}

Therefore, *a* = 3, *b* = 2, *c* = 1 and *d* = 1.

^{ }*a + b *– 2*c* – 3*d* = 3 + 2 – 2(1) – 3(1)

= 5 – 2 – 3

= 5 – 5

= 0

Hence, *a + b *– 2*c* – 3*d* = __0__.

#### Page No 1.61:

#### Question 14:

If *n* is a natural number, then the number of consecutive zeros in 7* ^{n}*, is _______.

#### Answer:

To calculate the number of consecutive zeros, we need to find the number of pairs of 2 and 5 to form a product of 10.

But, 2 and 5 both are not the factors of 7* ^{n}*.

Hence, the number of consecutive zeros in 7

*, is*

^{n}__0__.

#### Page No 1.61:

#### Question 15:

Two numbers are in the ratio 21 : 17. If their HCF is 5, the numbers are ________ and ________.

#### Answer:

Let the two numbers be 21*x* and 17*x*.

LCM of (21*x*, 17*x*) = 17 × 21 × *x*

We know that,

Product of two numbers = LCM × HCF

$\Rightarrow 21x\times 17x=17\times 21\times x\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow 21\times 17{x}^{2}=17\times 21\times 5x\phantom{\rule{0ex}{0ex}}\Rightarrow 21\times 17x=17\times 21\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{17\times 21\times 5}{21\times 17}\phantom{\rule{0ex}{0ex}}\Rightarrow x=5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{first}\mathrm{number}\mathrm{is}21\times 5=105\mathrm{and}\mathrm{second}\mathrm{number}\mathrm{is}17\times 5=85.$

Hence, the numbers are __105__ and __85__.

#### Page No 1.61:

#### Question 16:

If the least prime factors of two positive integers *a *and *b* are 5 and 13 respectively, then the least prime factor of *a + b*, is _________.

#### Answer:

It is given that the least prime factors of two positive integers *a *and *b* are 5 and 13 respectively.

Thus, *a* = 5*p* and *b* = 13*q* , where p and q are some natural numbers.

Now, *a + b *= 5*p +* 13*q*

Since, 5*p* and 13*q* both are odd numbers

and sum of two odd numbers is even.

Therefore, *a + b *is an even number.

Hence, the least prime factor of *a + b*, is __ 2 __.

#### Page No 1.61:

#### Question 17:

If 2^{3} × 3^{a} ×* b* × 7 is the prime factorisation of 2520, then 5*a* + 2*b* = _________ .

#### Answer:

Prime factorisation of 2520 is

2520 = 2^{3} × 3^{2} × 5^{1 }× 7^{1}

Therefore, *a* = 2 and *b* = 5

^{ }5*a* + 2*b *= 5(2) + 2(5)

= 10 + 10

= 20

Hence, 5*a* + 2*b* = __20__.

#### Page No 1.61:

#### Question 18:

Given that LCM (91, 26) = 182, then HCF (91, 26) = _________.

#### Answer:

We know that,

Product of two numbers = LCM × HCF

⇒ 91 × 26 = 182 × HCF

$\Rightarrow \mathrm{HCF}\times 182=91\times 26\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{HCF}=\frac{91\times 26}{182}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{HCF}=\frac{26}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{HCF}=13$

Hence, HCF (91, 26) = __13__.

#### Page No 1.61:

#### Question 19:

The decimal expansion of $\frac{17}{8}$ will terminate after _______ places of decimal.

#### Answer:

$\frac{17}{8}=\frac{17}{{2}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{17\times 5\times 5\times 5}{{2}^{3}\times {5}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{85\times 5\times 5}{{10}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{425\times 5}{1000}\phantom{\rule{0ex}{0ex}}=\frac{2125}{1000}\phantom{\rule{0ex}{0ex}}=2.125$

Hence, the decimal expansion of $\frac{17}{8}$ will terminate after __three__ places of decimal.

#### Page No 1.61:

#### Question 20:

One of the three consecutive positive integers is always divisible by ________.

#### Answer:

Let the three consecutive numbers be *n*, *n* + 1 and *n* + 2.

Whenever a number is divided by 3, the remainder obtained is 0, 1, or 2.

Therefore, *n* = 3*p *or *n* = 3*p* + 1 or *n* = 3*p* + 2, where *p* is any natural number.

Now,

If *n* = 3*p*,* *then n is divisible by 3

If *n* = 3*p* + 1

⇒ *n* + 2 = 3*p* + 1 + 2 = 3*p* + 3 = 3(*p* + 1), then (*n* + 2) is divisible by 3

If *n* = 3*p* + 2

⇒ *n* + 1 = 3*p* + 2 + 1 = 3*p* + 3 = 3(*p* + 1), then (*n* + 1) is divisible by 3

Thus, we can say that one of the three numbers *n*, *n* + 1, *n* + 2 is always divisible by 3.

Hence, One of the three consecutive positive integers is always divisible by __3__.

#### Page No 1.61:

#### Question 21:

6^{n} cannot end with digit 5 for _______ value of *n*.

#### Answer:

The prime factors of 6 are 2 and 3.

$\Rightarrow {6}^{n}={2}^{n}\times {3}^{n}$

Since, 5 is not the factor of 6^{n}

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

Hence, 6

^{n}cannot end with digit 5 for

__any__value of

*n*.

#### Page No 1.61:

#### Question 22:

The value of the remainder, *r*, when a positive integer '*a*' is divided by 3, are ________.

#### Answer:

Let a positive integer '*a*' is divided by 3,

*a* = 3*q* + *r*, where *q* and *r* both are natural numbers and $0\le r<3$.

Since, *r* is a natural number and $0\le r<3$

Therefore, *r* can be 0, 1 or 2

Hence, the value of the remainder, *r*, when a positive integer '*a*' is divided by 3, are __0, 1, 2__.

#### Page No 1.61:

#### Question 23:

12^{n} cannot end with the digit 0 to 5 for _______ value of *n*.

#### Answer:

The prime factors of 12 are 2 and 3.

$12={2}^{2}\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow {12}^{n}={2}^{2n}\times {3}^{n}$

Since, 5 is not the factor of 12^{n}

Therefore, for any value of *n*, 12* ^{n }*will not be divisible by 5.

Hence, 12

^{n}cannot end with the digit 0 or 5 for

__any__value of

*n*.

__Disclaimer__: The question is: 12

^{n}cannot end with the digit 0 or 5 for _______ value of

*n*.

#### Page No 1.61:

#### Question 24:

Two numbers are in the ratio 3 : 4 and their LCM is 120. The sum of the numbers is _________.

#### Answer:

Let the two numbers be 3*x* and 4*x*.

HCF of (3*x*, 4*x*) = *x*

We know that,

Product of two numbers = LCM × HCF

$\Rightarrow 3x\times 4x=120\times x\phantom{\rule{0ex}{0ex}}\Rightarrow 12{x}^{2}=120x\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=120\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{120}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{first}\mathrm{number}\mathrm{is}3\times 10=30\mathrm{and}\mathrm{second}\mathrm{number}\mathrm{is}4\times 10=40.$

Sum of the numbers = 30 + 40 = 70.

Hence, the sum of the numbers is __70__.

#### Page No 1.61:

#### Question 25:

The LCM of 2*x*, 5*x* and 7*x* is _________, where* x* is a positive integer.

#### Answer:

LCM of 2*x*, 5*x* and 7*x *= 2 × 5 × 7 × *x*

= 70*x*

Hence, the LCM of 2*x*, 5*x* and 7*x* is __70 x__.

#### Page No 1.61:

#### Question 1:

State Euclid's division lemma.

#### Answer:

Euclid’s Division Lemma:

Let *a * and *b* be any two positive integers.

Then, there exist unique integers *q* and *r* such that

,

If then .

Otherwise, *r *satisfies the stronger inequality.

#### Page No 1.61:

#### Question 2:

State Fundamental Theorem of Arithmetic.

#### Answer:

FUNDAMENTAL THEOREM OF ARITHMETIC:

Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique except for the order in which the prime factors occur.

While writing a positive integer as the product of primes, if we decide to write the prime factors in ascending order and we combine the same primes, then the integer is expressed as the product of powers of primes and the representation is unique.

So,we can say that every composite number can be expressed as the products of powers distinct primes in ascending or descending order in a unique way.

#### Page No 1.61:

#### Question 3:

Write 98 as product of its prime factors.

#### Answer:

Using the factor tree for prime factorization, we have:

Therefore,

#### Page No 1.61:

#### Question 4:

Write the exponent of 2 in the price factorization of 144.

#### Answer:

Using the factor tree for prime factorization, we have:

Therefore,

Hence the exponent of 2 in 144 is .

#### Page No 1.61:

#### Question 5:

Write the sum of the exponents of prime factors in the prime factorization of 98.

#### Answer:

Using the factor tree for prime factorization, we have:

Therefore,

The exponents of 2 and 7 are 1 and 2 respectively.

Hence the sum of the exponents is .

#### Page No 1.62:

#### Question 6:

If the prime factorization of a natural number *n* is 2^{3} ✕ 3^{2} ✕ 5^{2} ✕ 7, write the number of consecutive zeros in *n*.

#### Answer:

Since, it is given that

Hence the number of consecutive zeroes are.

#### Page No 1.62:

#### Question 7:

If the product of two numbers is 1080 and their HCF is 30, find their LCM.

#### Answer:

It is given that the product of two numbers is 1080.

Let the two numbers be *a* and *b*.

Therefore,

HCF is 30.

We need to find the LCM

We know that the product of two numbers is equal to the product of the HCF and LCM.

Thus,

Hence the LCM is .

#### Page No 1.62:

#### Question 8:

Write the condition to be satisfied by *q* so that a rational number $\frac{p}{q}$ has a terminating decimal expansions.

#### Answer:

We need to find the condition to be satisfied by *q* so that a rational number has a terminating decimal expression.

For the terminating decimal expression, we should have a multiple of 10 in the denominator.

Hence, the prime factorization of *q* must be of the form, where *m* and *n* are non-negative integers.

#### Page No 1.62:

#### Question 9:

Write the condition to be satisfied by *q* so that a rational number $\frac{p}{q}$has a terminating decimal expansion.

#### Answer:

We need to find the condition to be satisfied by *q* so that a rational number has a non-terminating decimal expression.

For the terminating decimal expression, we should not have a multiple of 10 in the denominator.

Hence, the prime factorization of *q* must not be of the form, where *m* and *n* are non-negative integers.

#### Page No 1.62:

#### Question 10:

Complete the missing entries in the following factor tree.

#### Answer:

We need to fill the values for *a* and *b* in the following factor tree:

It is clear from the factor tree above that

Also,

Thus, the missing entries are * *and.

#### Page No 1.62:

#### Question 11:

The decimal expansion of the rational number $\frac{43}{{2}^{4}\times {5}^{3}}$ will terminate after how many places of decimals?

#### Answer:

We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is of the form , where *m* and *n* are non-negative integers.

*x* has a decimal expression which terminates after *k* places of decimals, where *k* is the larger of *m* and *n*.

This is clear that the prime factorization of the denominator is of the form .

Hence, it has terminating decimal expansion which terminates after places of decimal.

#### Page No 1.62:

#### Question 12:

Has the rational number $\frac{441}{{2}^{2}\times {5}^{7}\times {7}^{2}}$a terminating or a nonterminating decimal representation?

#### Answer:

We have,

Theorem states:

Let be a rational number, such that the prime factorization of *q* is not of the form , where *m* and *n* are non-negative integers.

Then, *x* has a decimal expression which is non-terminating repeating.

This is clear that the prime factorization of the denominator is not of the form.

Hence, it has non-terminating decimal expansion.

#### Page No 1.62:

#### Question 13:

Write whether $\frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt{5}}$on simplification gives a rational or an irrational number.

#### Answer:

Let us simplify

is rational number

#### Page No 1.62:

#### Question 14:

What is an algorithm?

#### Answer:

Algorithm is a step-by-step procedure for calculations.

For example:

__Euclid’s Division Algorithm__: In order to compute the HCF of two positive integers say *a* and *b*, with by using Euclid’s algorithm, we follow the following steps:

__STEP I__ : Apply Euclid’s Division Lemma to *a* and *b *and obtain whole numbers and ,such that ,

__STEP II__: If* *, *b* is the HCF of *a* and *b.*

__STEP III__: If , apply Euclid’s division lemma to *b* and * *and obtain whole numbers and ,such that

__STEP IV__: If , then is the HCF of *a* and *b.*

__STEP V__: If , then apply Euclid’s division lemma to and * *and continue the above process till the remainder is zero. The divisor at this stage i.e; , or the non-zero remainder at the previous stage is the HCF of *a* and *b.*

#### Page No 1.62:

#### Question 15:

What is a lemma?

#### Answer:

A proven statement used as a stepping-stone toward the proof of another statement is called lemma.

For example:

Euclid’s Division Lemma: Let *a * and *b* be any two positive integers.

Then, there exist unique integers *q* and *r* such that

,

If then .

Otherwise, *r *satisfies the stronger inequality.

#### Page No 1.62:

#### Question 16:

If *p* and *q* are two prime number, then what is their HCF?

#### Answer:

It is given that *p* and *q* are two prime numbers; we have to find their HCF.

We know that the factors of any prime number are 1 and the prime number itself.

For example, let * *and* *

Thus, the factors are as follows

And

Now, the HCF of 2 and 3 is 1.

Thus the HCF of *p* and *q *is* ** *.

#### Page No 1.62:

#### Question 17:

If *p* and *q* are two prime number, then what is their LCM?

#### Answer:

It is given that *p* and *q* are two prime numbers; we have to find their LCM.

We know that the factors of any prime number are 1 and the prime number itself.

For example, let * *and* *

Thus, the factors are as follows

And

Now, the LCM of 2 and 3 is * * .

Thus the HCF of *p* and *q *is* ** .*

#### Page No 1.62:

#### Question 18:

What is the total number of factors of a prime number?

#### Answer:

We know that the factors of any prime number are 1 and the prime number itself.

For example, let * *

Thus, the factors are as follows

Hence, the total number of factors of a prime number is* *

#### Page No 1.62:

#### Question 19:

What is a composite number?

#### Answer:

A composite number is a positive integer which has a divisor other than one or itself.

In other words a composite number is any positive integer greater than one that is not a prime number.

#### Page No 1.62:

#### Question 20:

What is the HCF of the smallest composite number and the smallest prime number?

#### Answer:

The smallest composite number is 4

The smallest prime number is 2

Thus, the HCF of and is .

#### Page No 1.62:

#### Question 21:

HCF of two numbers is always a factor of their LCM (True/False).

#### Answer:

HCF of two numbers is always a factor of their LCM

__True__

__Reason__:

The HCF is a factor of both the numbers which are factors of their LCM.

Thus the HCF is also a factor of the LCM of the two numbers.

#### Page No 1.62:

#### Question 22:

π is an irrational number (True/False).

#### Answer:

Hereis an irrational number

__True__

Reason:

Rational number is one that can be expressed as the fraction of two integers.

Rational numbers converted into decimal notation always repeat themselves somewhere in their digits.

For example, 3 is a rational number as it can be written as 3/1 and in decimal notation it is expressed with an infinite amount of zeros to the right of the decimal point. 1/7 is also a rational number. Its decimal notation is 0.142857142857…, a repetition of six digits.

However cannot be written as the fraction of two integers and is therefore irrational.

Now,

Thus, it is irrational.

#### Page No 1.62:

#### Question 23:

The sum of two prime number is always a prime number (True/ False).

#### Answer:

The sum of two prime numbers is always a prime number.

__False__

__Reason__:

Let us prove the above by taking an example.

Let the two given prime numbers be 2 and 7.

Thus, their sum, i.e; 9 is not a prime number.

Hence the above statement is false

#### Page No 1.62:

#### Question 24:

The product of any three consecutive natural number is divisible by 6 (True/False).

#### Answer:

The product of any three natural numbers is divisible by 6.

__True__

__Reason__:

Let the three consecutive natural numbers be 1,2 and 3.

Their product is 6, which is divisible by 6

Let the other set of three consecutive natural numbers be 3, 4 and 5.

Their product is 60, which is divisible by 6

#### Page No 1.62:

#### Question 25:

Every even integer is of the form 2*m*, where *m* is an integer (True/False).

#### Answer:

Every even integer is of the form 2*m*, where *m* is an integer (True/False)

__True__

__Reason__:

Let the various values of *m* as -1, 0 and 9.

Thus, the values for 2m become -2, 0 and 18 respectively.

#### Page No 1.62:

#### Question 26:

Every odd integer is of the form 2*m* − 1, where *m* is an integer (True/False).

#### Answer:

Every odd integer is of the form, where *m* is an integer (True/False)

__True__

__Reason__:

Let the various values of *m* as -1, 0 and 9.

Thus, the values for become -3, -1 and 17 respectively.

These are odd integers.

#### Page No 1.62:

#### Question 27:

The product of two irrational numbers is an irrational number (True/False).

#### Answer:

The product of two irrational numbers is an irrational number (True/False)

__False__

Reason:

Let us assume the two irrational numbers be and

Sometimes, it is and sometimes it isn't.

And are both irrational as their product is

Nowand are both irrational but their product, is rational (in fact, it equals 4)

#### Page No 1.62:

#### Question 28:

The sum of two irrational number is an irrational number (True/False).

#### Answer:

The sum of two irrational numbers is an irrational number (True/False)

False

__Reason__:

However, is not rational because there is no fraction, no ratio of integers that will equal. It calculates to be a decimal that never repeats and never ends. The same can be said for . Also, there is no way to write as a fraction. In fact, the representation is already in its simplest form.

To get two irrational numbers to add up to a rational number, you need to add irrational numbers such as and . In this case, the irrational portions just happen to cancel out leaving: which is a rational number (i.e. 2/1).

#### Page No 1.62:

#### Question 29:

For what value of *n*, 2^{n} ✕ 5^{n} ends in 5.

#### Answer:

We need to find the value of *n*, for which ends in 5.

Clearly,

Also, all the values of *n* will make end in 0.

Thus, there is no value of *n* for which ends in 5.

#### Page No 1.62:

#### Question 30:

If *a* and *b* are relatively prime numbers, then what is their HCF?

#### Answer:

It is given that *a* and *b* are two relatively prime numbers; we have to find their HCF.

We know that two numbers are relatively prime if they don’t have any common divisor.

Also, the factors of any prime number are 1 and the prime number itself.

For example, let*a = 7*and

*b*= 20

Thus, the factors are as follows

*a* = 7 × 1

And

*b* = 2^{2} × 5 × 1

Now, the HCF of 7 and 20 is 1.

Thus the HCF of *a* and *b** *is* ** .*

#### Page No 1.62:

#### Question 31:

If *a* and *b* are relatively prime numbers, then what is their LCM?

#### Answer:

It is given that *a* and *b* are two relatively prime numbers; we have to find their LCM.

We know that two numbers are relatively prime if they don’t have any common divisor.

Also, the factors of any prime number are 1 and the prime number itself.

For example, let *a* = 7 and *b* = 20

Thus, the factors are as follows

*a* = 7 × 1

And

b = 2^{2} × 5 × 1

Now, the LCM of 7 and 20 is 140.

Thus the HCF of *a* and *b** *is *ab**.*

#### Page No 1.62:

#### Question 32:

Two numbers have 12 as their HCF and 350 as their LCM (True/False).

#### Answer:

Two numbers have 12 as their HCF and 350 as their LCM (True/False).

False.

__Reason__:

We know that HCF should divide LCM.

But, the HCF 12 does not divide the LCM 350.

#### Page No 1.63:

#### Question 33:

Find after how many places of decimal the decimal form of the number $\frac{27}{{2}^{3}.{5}^{4}.{3}^{2}}$ will terminate.

#### Answer:

$\frac{27}{{2}^{3}.{5}^{4}.{3}^{2}}=\frac{{3}^{3}}{{2}^{3}.{5}^{4}.{3}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{{2}^{3}.{5}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{3}{8\times 625}\phantom{\rule{0ex}{0ex}}=\frac{3}{5000}\phantom{\rule{0ex}{0ex}}=\frac{0.003}{5}\phantom{\rule{0ex}{0ex}}=0.0006$

Hence, after four places of decimal the decimal form of the number $\frac{27}{{2}^{3}.{5}^{4}.{3}^{2}}$ will terminate.

#### Page No 1.63:

#### Question 34:

Express 429 as the product of its prime factors.

#### Answer:

The prime factorisation of 429 is

429 = 3 × 11 × 13

#### Page No 1.63:

#### Question 35:

Two positive integers *a* and *b* can be written as *a* = *x*^{3}*y*^{2} and *b* =* xy*^{3}, where* x, y* are prime numbers. Find LCM (*a, b*).

#### Answer:

It is given that, *a* = *x*^{3}*y*^{2} and *b* =* xy*^{3}, where* x, y* are prime numbers.

LCM (*a, b*) = LCM (*x*^{3}*y*^{2}*, xy*^{3})

= The highest of indices of *x* and *y*

= * x*^{3}*y*^{3}

Hence, LCM (*a, b*) is * x*^{3}*y*^{3}.

#### Page No 1.63:

#### Question 36:

If HCF (336, 54) = 6, find LCM (336, 54).

#### Answer:

we know that,

Product of two numbers = LCM × HCF

⇒ 336 × 54 = LCM × 6

$\Rightarrow \mathrm{LCM}\times 6=336\times 54\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=\frac{336\times 54}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=336\times 9\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LCM}=3024$

Hence, LCM (336, 54) is 3024.

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