4 Marks Questions

Question 14 Marks

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

Answer

Let, $n$ be any positive integer. Since any positive is of the form $6q$ or $6q + 1$ or, $6q + 2$ or, $6q + 3$ or $6q + 4$ or $6q + 5.$
If $n - 6q,$ then

$n(n + 1)(n + 2) = [(6q)(6q + 1)(6q + 2)]$

$= 6[q(6q + 1)(6q + 2)]$

$= 6m$, which is divisible by $6$

If $n - 6q + 1,$ then

$n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)$

$= 6[(6q + 1)(3q + 1)(2q + 1)]$

$= 6m,$ which is divisible by $6$

If $n - 6q + 2,$ then

$n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)$

$= 6[(3q + 1)(2q + 1)(6q + 4)]$

$= 6m, $ which is divisible by $6$

If $n - 6q + 3, $ then

$n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)$

$= 6[(3q + 1)(3q + 2)(6q + 5)]$

$= 6m,$ which is divisible by $6$

If $n - 6q + 4,$ then

n$(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)$

$= 6[(6q + 4)(6q + 5)(q + 1)]$
$= 6m,$ which is divisible by $6$

If $n - 6q + 5,$ then

$n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)$

$= 6[(6q + 5)(q + 1)(6q + 7)]$

$= 6m,$ which is divisible by $6$

Hence, the product of three consecutive positive integer is divisible by $6$.

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Question 24 Marks

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

Answer

Let $a$ be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers $‘a’$ and $6,$ there exist non-negative integers $q$ and $r$ such that.
$ \Rightarrow a^3=(6 q+r)^3=216 q^3+r^3+3.6 q \cdot r(6 q+r) $
$ {\left[\because(a+b)^3=a^3+b^3+3 a b(a+b)\right]} $
$ \Rightarrow a^3=\left(216 q^3+108 q^2 r+18 q r^2\right)+r^3 \ldots(i)$
Where, $0 \leq \mathrm{r}<6$
Case I: When $r=0$, then putting $r=0$ in Eq. $(i),$ we get
$a^3=216 q^3=6\left(36 q^3\right)=6 m$
Where, $m=36 q^3$ is an integer.
Case II: Where $r=1$, then putting $r=1$ in
Eq. $(i),$ we get
$a^3=\left(216 q^3+108 q^3+18 q\right)+1=6\left(36 q^3+18 q^3+3 q\right)+1 $
$ a^3=6 m+1$
Where $m=\left(36 q^3+18 q^3+3 q\right)$ is an integer.
Case III: When $r=2$, then putting $r=2$ in
Eq. $(i),$ we get
$a^3=\left(216 q^3+216 q^2+72 q\right)+8 $
$ a^3=\left(216 q^3+216 q^2+72 q+6\right)+2 $
$ \Rightarrow a^3=6\left(36 q^3+36 q^2+12 q+1\right)+2=6 m+2$
Where, $m=\left(36 q^3+36 q^2+12 q+1\right)$ is an integer.
Case IV: when $\mathrm{r}=3$, then putting $\mathrm{r}=3$ in
Eq. $(i),$ we get
$a^3=\left(216 q^3+324 q^2+162 q\right)+27 $
$ =\left(216 q^3+324 q^2+162 q+24\right)+3 $
$ =6\left(36 q^3+54 q^2+27 q+4\right)+3=6 m+3$
Where, $m=\left(36 q^3+64 q^2+27 q+4\right)$ is an integer.
Case V : When $r=4$, then putting $r=4$ in
Eq. $(i),$ we get
$ a^3=\left(216 q^3+432 q^2+288 q\right)+64 $
$ a^3=6\left(36 q^3+72 q^2+48 q\right)+60+4 $
$ a^3=6\left(36 q^3+72 q^2+48 q+10\right) \text { is an integer. }$
Case VI: When $r=5$, then putting $r=5$ in
Eq. $(i),$ we get
$ a^3=\left(216 q^3+540 q^2+450 q\right)+125 $
$ \Rightarrow a^3=\left(216 q^3+540 q^2+450 q\right)+120+5 $
$ \Rightarrow a^3=6\left(36 q^3+90 q^2+75 q+20\right)+5 $
$ \Rightarrow a^3=6 m+5$
Where, $m=\left(36 q^3+90 q^2+75 q+20\right)$ is an integer.
Hence, the cube of a positive integer of the form $6q + r, q$ is an integer and $r = 0, 1, 2, 3, 4, 5$ is also of the forms $6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4$ and $6m + 5$ i.e., $6m + r.$

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Question 34 Marks

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

Answer

Let $a$ be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers $a$ and $6,$ there exist non-negative integers $q$ and $r$ such that
$a=6 q+r \text {, where } 0 \leq r<6 $
$ \Rightarrow a^2=(6 q+r)^2=36 q^2+r^2+12 q r $
$ {\left[\because(a+b)^2=a^2+2 a b+b^2\right]} $
$ \Rightarrow a^2=6\left(6 q^2+2 q r\right)+r^2 \ldots \text { (i) }$
Where, $0 \leq \mathrm{r}<6$
Case I: Where $r=0$, then putting $r=0$ in
Eq. $(i)$, we get
$a^2=6\left(6 q^2\right)=6 m$
Where, $m=6 q^2$ is an integer.
Case II: When $r=1$, then putting $r=1$ in
Eq. $(i),$ we get
$a^2=6\left(6 q^2+2 q\right)+1=6 m+$
Where, $m=\left(6 q^2+2 q\right)$ is an integer
Case III: When $r=2$, then putting $r=2$ in
Eq. $(i),$ we get
$a^2=6\left(6 q^2+4 q\right)+4=6 m+4$
Where, $m=\left(6 q^2+4 q\right)$ is an integer.
Case IV: When $\mathrm{r}=3$, then putting $\mathrm{r}=3$ in
Eq. $(i),$ we get
$a^2=6\left(6 q^2+6 q\right)+9 $
$ =6\left(6 q^2+6 q\right)+6+3 $
$ \Rightarrow a^2=6\left(6 q^2+6 q+1\right)+3=6 m+3$
Where, $m=\left(6 q^2+6 q+1\right)$ is an integer.
Case V: When $r=4$, then putting $r=4$ in
Eq. $(i),$ we get
$a^2=6\left(6 q^2+8 q\right)+16 $
$ =6\left(6 q^2+8 q\right)+12+4 $
$ \Rightarrow a^2=6\left(6 q^2+8 q+2\right)+4=6 m+4$
Where, $n=\left(6 q^2+8 q+2\right)+4=6 m+4$
Case VI: When $r=5$, then putting $r=5$ in
Eq. $(i),$ we get
$a^2=6\left(6 q^2+10 q\right)+25 $
$ =6\left(6 q^2+10 q\right)+24+1 $
$ \Rightarrow a^2=6\left(6 q^2+10 q+4\right)+1=6 m+1$
Where, $m=\left(6 q^2+10 q+1\right)$ is an integer.
Hence, the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer $m.$

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Question 44 Marks

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 $21.$
$8 = 2^3$
$15 = 3 × 5$
$21 = 3 × 7$
L.C.M of $8, 15$ and $21 = 23 × 3 × 5 × 7 = 840$
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 $21.$

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Question 54 Marks

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 that three persons step off together, their steps measure $80\ cm, 85\ cm$ and $90\ cm.$
The minimum distance each should walk is equal to L.C.M of $80, 85$ and $90$.
Prime factors of $80, 85$ and $90$ are
$ 80=2 \times 2 \times 2 \times 5=2^4 \times 5 $
$ 85=5 \times 17=5 \times 17 $
$ 90=2 \times 3 \times 3 \times 5=2 \times 3^2 \times 5 $
$ \text { L.C.M }(80,85,90)=2^4 \times 3^2 \times 5 \times 17 $
$ =16 \times 9 \times 85 $
$ =12240 \mathrm{~cm} $
$ \because 100 \mathrm{~cm}=1 \mathrm{~m} $
$ \therefore 12240 \mathrm{~cm}=122.40 \mathrm{~m} $
$ =122 \mathrm{~m} 40 \mathrm{~cm}$
So, the minimum distance each should walk is $122\ m\ 40\ cm.$

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Question 64 Marks

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

Answer

Since, $1,2$ and $3$ are the remainders of $1251, 9377$ and $15628,$ respectively.
Thus, after subtracting these remainders from the numbers.
We have the numbers, $1251 - 1 = 1250, 9377 - 2 = 9375$ and $15628 - 3 = 15625$ which is divisible by the required number.
Now, required number $= HCF$ of $1250, 9375$ and $15625 [$for the largest number$]$
By Euclid’s division algorithm,
$a = bq + r .....(i)$
$[\because$ dividend = divisor $×$ quotient $+$ remainder$]$
For largest number, put $a = 15625$ and $b = 9375$
$15625 = 9375 × 1 + 6250$
$⇒ 9375 = 6250 × 1 + 3125$
$⇒ 6250 = 3125 × 2 + 0$
$\therefore H.C.F (15625, 9375) = 3125$
Now, we take $c = 1250$ and $d = 3125,$ then again using Euclid's division algorithm,
$d = cq + r [$from Eq. $(i)]$
$⇒ 3125 = 1250 × 2 + 625$
$⇒ 1250 = 625 × 2 + 0$
$\therefore H.C.F (1250, 9375, 15625) = 625$
Hence, $625$ is the largest number which divides $1251, 9377$ and $15628$ leaving remainder $1, 2$ and $3$, respectively.

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Question 74 Marks

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

Answer

To Prove: that if a positive integer is of the form $6q + 5$ then it is of the form $3q + 2$ for some integer $q,$ but not conversely.
Proof: Let $n = 6q + 5$
Since any positive integer n is of the form of $3k$ or $3k + 1, 3k + 2$
If $q = 3k$
Then, $n = 6q + 5$

$⇒ n = 18k + 5(q = 3k)$

$⇒ n = 3(6k + 1) + 2$

$⇒ n = 3m + 2($where $m = (6k + 1))$

If $q = 3k + 1$
Then, $n = (6q + 5)$

$⇒ n = (6(3k + 1) + 5)(q = 3k + 1)$

$⇒ n = 18k + 6 + 5$

$⇒ n = 18k + 11$

$⇒ n = 3(6k + 3) + 2$

$⇒ n = 3m + 2($where $m = (6k + 3))$

If $q = 3k + 2$
Then, $n = (6q + 5)$

$⇒ n = (6(3k + 2) + 5)(q = 3k + 2)$

$⇒ n = 18k + 12 + 5$

$⇒ n = 18k + 17$

$⇒ n = 3(6k + 5) + 2$

$⇒ n = 3m + 2($where $m = (6k + 5))$

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

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Question 84 Marks

$105$ 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
$140 = 105 × 1 + 35$
$105 = 35 × 3 + 0.$
Therefore, $H.C.F. $ of $105$ and $140 = 35$
Now, we consider $35$ and $175.$
By applying Euclid’s division lemma
$175 = 35 × 5 + 0.$
Therefore, $H.C.F.$ of $105, 140$ and $175 = 35$
Hence, the number of animals went in each trip is $35.$

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Question 94 Marks

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

Answer

Given that two positive integers $408$ and $1032$ and $1032 > 408.$
So, appliying Euclid’s division algorithem
$1032 = 408 × 2 + 216 …..(i)$
$408 = 216 × 1 + 192 …..(ii)$
$216 = 192 × 1 + 24 …..(iii)$
$192 = 24 × 6 + 0 …..(iv)$
So, $HCF$ of 408 and $1032$ is divison of eq. $(iv)$ and remainder of eq. $(iii)$ i.e. $24.$
$HCF (408, 1032) = 1032m - 408 × 5$
$⇒ 24 = 1032m - 2040$
$⇒ 1032m = 2040 + 24$
$⇒ 1032m = 2064$
$\Rightarrow\text{m} = \frac{2064}{1032}$
$⇒ m = 2$
Thus, $m = 2.$

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Question 104 Marks

Show that one and only one out of $n, n + 4, n + 8, n + 12$ and $n + 16$ is divisible by $5,$ where $n$ is any positive integer.

Answer

Consider the numbers $n, (n + 4), (n + 8), (n + 12)$ and $(n + 16),$ where $n$ is any positive integer.
Suppose $n = 5q + r,$ where $0 \leq r < 5$
$n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4$
$($By Euclid's division algorithm$)$
Case I:
When $n = 5q.$
$n = 5q$ is divisible by $5.$
$n + 4 = 5q + 4$ is not divisible by $5.$
$n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3$ is not divisible by $5$.
$n + 12 = 5q + 10 + 2 = 5(q + 2) + 2$ is not divisible by $5.$
$n + 16 = 5q + 15 + 1 = 5(q + 3) + 1$ is not divisible by $ 5.$
Case II:
When $n = 5q + 1.$
$n = 5q + 1$ is not divisible by $5.$
$n + 4 = 5q + 1 + 4 = 5(q + 1)$ is not divisible by $5.$
$n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4$ is not divisible by $5.$
$n + 12 = 5q + 1 + 12 = 5(q + 2) + 3$ is not divisible by $5.$
$n + 16 = 5q + 1 + 16 = 5(q + 3) + 2$ is not divisible by $5.$
Case III:
When $n = 5q + 2.$
$n = 5q + 2$ is not divisible by $5.$
$n + 4 = 5q + 2 + 4 = 5(q + 1) + 1$ is not divisible by $5.$
$n + 8 = 5q + 2 + 8 = 5(q + 2)$ is not divisible by $5.$
$n + 12 = 5q + 2 + 12 = 5(q + 2) + 4$ is not divisible by $5.$
$n + 16 = 5q + 2 + 16 = 5(q + 3) + 3$ is not divisible by $5.$
Case IV:
When $n = 5q + 3.$
$n = 5q + 3$ is not divisible by $5.$
$n + 4 = 5q + 3 + 4 = 5(q + 1) + 2$ is not divisible by $5.$
$n + 8 = 5q + 3 + 8 = 5(q + 2) + 1$ is not divisible by $5.$
$n + 12 = 5q + 3 + 12 = 5(q + 3)$ is not divisible by $5.$
$n + 16 = 5q + 3 + 16 = 5(q + 3) + 4$ is not divisible by $5.$
Case V:
When $n = 5q + 4.$
$n = 5q + 4$ is not divisible by $5.$
$n + 4 = 5q + 4 + 4 = 5(q + 1) + 3$ is not divisible by $5.$
$n + 8 = 5q + 4 + 8 = 5(q + 2) + 2$ is not divisible by $5.$
$n + 12 = 5q + 4 + 12 = 5(q + 3) + 1$ is not divisible by $5.$
$n + 16 = 5q + 4 + 16 = 5(q + 4)$ is not divisible by $5.$
Hence, in each case, one and only one out of $n, n + 4, n + 8, n + 12$ and $n + 16$ is divisible by $5.$

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Question 114 Marks

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

Answer

We have, $3 × 5 × 7 + 7 = 105 + 7 = 112$
Now, $112 = 2 × 2 × 2 × 2 × 7 = 2^4× 7$
So, it is the product of prime factors $2$ and $7.$ i.e., it has more than two factors.
Hence, it is a composite number.

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Question 124 Marks

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

Answer

Let $a = 2q + 3$ and $b = 2q + 1$ be two positive odd integers such that $a > b$
Now, $\frac{\text{a}+\text{b}}{2}=\frac{2\text{q}+3+2\text{q}+1}{2}=\frac{4\text{q}+4}{2}=2\text{q}+2=$ an odd number
and $\frac{\text{a}-\text{b}}{2}=\frac{(2\text{q}+3)-(2\text{q}+1)}{2}=\frac{2\text{q}+3-2\text{q}-1}{2}=\frac{2}{2}=1=$ an odd number
Hence one of the two numbers $\frac{\text{a}+\text{b}}{2}$ and $\frac{\text{a}-\text{b}}{2}$ is odd and the other is even for any two positive odd integer.

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Question 134 Marks

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

Answer

Let, $n$ be any positive integer
We know that positive integer $n$ is a in the form of $3q, (3q + 1)$ and $(3q + 2)$, where q is some integer
Case 1: When $n=3 q$
Squaring of both sides
$ \Rightarrow n^2=(3 q)^2 $
$ \Rightarrow n^2=9 q^2 $
$ \Rightarrow n^2=3 q(3 q) $
$ \Rightarrow n^2=3 m[\text { where } m=q(3 q)]$
Case 2: When $\mathrm{n}=(3 q+1)$
Squaring of both sides
$ \Rightarrow n^2=(3 q+1)^2 $
$ \Rightarrow n^2=9 q^2+6 q+1 $
$ \Rightarrow n^2=3 q(3 q+2)+1 $
$ \Rightarrow n^2=3 m+1[\text { Where } m=q(3 q+2)]$
Case 3: When $n=(3 q+2)$
Squaring of both sides
$ \Rightarrow n^2=(3 q+2)^2 $
$ \Rightarrow n 2=9 q^2+12 q+4 $
$ \Rightarrow n^2=9 q^2+12 q+3+1 $
$ \Rightarrow n^2=3\left(3 q^2+4 q+1\right)+1 $
$ \Rightarrow n^2=3 m+1\left[\text { Where } m=\left(3 q^2+4 q+1\right)\right]$
Thus, them case $1,$ case $2$ and case $3,$ it is proved that square of any positive odd integer is of the form $3m, (3m + 1)$ but not of the form $(3m + 2).$

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Question 144 Marks

The following table gives the number of boys of a particular age in a class of $40$ students. Calculate the mean age of the students

Age (in yeare)	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$3$	$8$	$10$	$10$	$5$	$4$

Answer

Given;

Age (in years): $x_i$	$15$	$16$	$17$	$18$	$19$	$20$
No.of students: $f_i$	$3$	$8$	$10$	$10$	$5$	$4$

First of all prepare the frequency table in such a way that its first column consist of the values of the variate$(x_i)$ and the second column the corresponding frequencies$(f_i).$
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing$(f_ix_i).$
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$.

Age (in years):$ x_i$	No. of students: $f_i$	$f_ix_i$
$15$	$3$	$45$
$16$	$8$	$128$
$17$	$10$	$170$
$18$	$10$	$180$
$19$	$5$	$95$
$20$	$4$	$80$
	$\sum\text{f}_\text{i}=40$	$\sum\text{f}_\text{i}\text{x}_\text{i}=698$

We know that mean, $\overline{X}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\overline{X}=\frac{698}{40}$
$=17.45$
Hence, the mean age of the students $=17.45 \ \text{years}$

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Question 154 Marks

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

Answer

Denominator of the rational number $\frac{257}{5000}$ is $5000.$
Now, factors of $5000=2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5=(2)^3 \times(5)^4$, which is of the type $2^m \times 5^n$, where $m = 3$ and $n = 4$ are non-negative integers.
$\therefore$ Rational number $=\frac{257}{5000}=\frac{257}{2^3\times5^4}\times\frac{2}{2}$
$[$Since, multiplying numerator and denominater by $2]$
$=\frac{514}{2^4\times5^4}=\frac{514}{(10)^4}$
$=\frac{514}{10000}=0.0514$
Hence, which is the required decimal expansion of the rational $\frac{257}{5000}$ and it is also a terminating decimal number.

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Question 164 Marks

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

Answer

Let us assume that $\sqrt{5}+\sqrt{3}$ is a rational number.
$\therefore\ \sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$ where, $a$ and $b$ are positive co-prime numbers.
$\sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \sqrt{5}=\frac{\text{a}}{\text{b}}-\sqrt{3}$
Squaring both sides,
$\Rightarrow(\sqrt{5})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{3}\Big)^2$
$\Rightarrow5=\Big(\frac{\text{a}}{\text{b}}\Big)^2+3-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow5-3=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow2=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow\frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-2$
$\Rightarrow\ \frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)$
$\Rightarrow\ \sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)\frac{\text{b}}{2\text{a}}$
$\Rightarrow\sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{2\text{ab}}\Big)$
We know that $\sqrt{3}$ is an irrational number.
This contradicts our assumption that $\sqrt{5}+\sqrt{3}$ is a rational number.

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Question 174 Marks

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

Answer

By Eudid's division Algorithm
$a = bm + r,$ where $0 ≤ r < b$
Put $b = 4$
$a = 4m + r,$ where $0 ≤ r < 4$
If $r = 0,$ then $a = 4m$
If $r = 1,$ then $a = 4m + 1$
If $r = 2, $ then $a = 4m + 2$
If $r = 3,$ then $a = 4m + 3$
Now, $(4 \mathrm{~m})^2=16 \mathrm{~m}^2$
$=4 \times 4 \mathrm{~m}^2$
$=4 q$ where $q$ is some integer
$ (4 m+1)^2=(4 m)^2+2(4 m)(1)+(1)^2 $
$ =16 m^2+8 m+1 $
$ =4\left(4 m^2+2 m\right)+1 \text { where } 4 m^2+2 m=q$
$=4 q+1$ where $q$ is some integer
$(4 m+3)^2=(4 m)^2+2(4 m)(3)+(3)^2 $
$ =16 m^2+24 m+9 $
$ =16 m^2+24 m+8+1 $
$ =4\left(4 m^2+6 m+2\right)+1 $
$ =4 q+1, \text { where } q \text { is some integre }$
Hence, the square of any positive in teger is of the form $4q$ or $4q + 1$ for some integer $m.$

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Question 184 Marks

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

Answer

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

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Question 194 Marks

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

Answer

Let us asssume that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is rational.
Then there exist positive co-primes a and b such that
$\sqrt{\text{p}}+\sqrt{\text{q}}=\frac{\text{a}}{\text{b}}$
$\sqrt{\text{p}}=\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}$
$(\sqrt{\text{p}})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}\Big)^2$
$\text{p}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}+\text{q}$
$\text{p}-\text{q}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\Big(\frac{\text{a}}{\text{b}}\Big)^2-(\text{p}-\text{q})=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}\bigg)\Big(\frac{\text{b}}{2\text{a}}\Big)=\sqrt{\text{q}}$
$\sqrt{\text{q}}=\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{2\text{ab}}\bigg)$
Here we see that $\sqrt{\text{q}}$ is rational number which is a contradiction as we know that $\sqrt{\text{q}}$ is an irrational number
Hence $\sqrt{\text{q}}+\sqrt{\text{q}}$ is irrational

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