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CBSE Questions for Class 10 Maths Real Numbers Quiz 3 - MCQExams.com
CBSE
Class 10 Maths
Real Numbers
Quiz 3
State whether true or false:
$$3+\sqrt{6}$$ is an irrational number.
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True
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False
Explanation
AS we know 3 is a rational number and $$\sqrt{6}$$ is also an irrational number
The addition of rational and irrational numbers is always an irrational number
So, 3 +
$$\sqrt{6}$$ is also a irrational number
Which of the following is an irrational number?
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$$0.14$$
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$$0.14 \overline{16}$$
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$$1.1 {416}$$
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$$0.4014001400014....$$
Explanation
The decimal expansion of a rational number is either terminating or non-terminating repeating.
(A) $$0.14$$ is terminating, so it is a rational number
(B) $$0.14\bar{16}=0.141616....$$
is also rational ( non-terminating repeating ), where digits $$16$$ are repeating.
(C) $$0.1416$$ is terminating, so it is a rational number
(D) $$0.4014001400014.....$$
is an irrational number because it is neither terminating nor repeating.
$$2-\sqrt {3}$$ is an irrational number.
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True
0%
False
Explanation
$$\sqrt{3}$$ is an irrational number.
Since, the difference of a rational and irrational number is irrational, hence, $$2-\sqrt{3}$$ is an irrational number.
Which of the following is always true
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$$irrational + irrational =irrational $$
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$$\dfrac{rational }{rational }=rational $$
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$$\dfrac{integer }{integer}=integer$$
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None of these
Explanation
Counter-example for A: $$(\sqrt{2}) + (4-\sqrt{2}) = 4$$
Counter-example for C: $$\dfrac{1}{2}=0.5$$
Proof for B:
Le
t $$q_1, q_2$$ be two rational numbers such that $$q_2\neq0$$.
As they are rational, they can be written as $$a/b, c/d$$ respectively for some integers $$a, b, c, d$$. $$(b,c,d\neq0)$$
$$\dfrac{q_1}{q_2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$$
Since, $$a,b,c,d$$ were integers, even $$ad$$ and $$bc(\neq0)$$ are integers and therefore, the above expression is rational.
$$\frac { 2 } { 2 + \sqrt { 3 } }$$ is an irrational number
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0%
True
0%
False
Explanation
Let if possible $$ ( 2 + \sqrt3 ) $$ is rational
then $$ 2 + \sqrt3 = \dfrac{a}{b} , \, b \neq = 0 $$
$$ \Rightarrow \sqrt3 = \dfrac{a}{b} - 2 \Rightarrow \sqrt3 = \dfrac{a-2b}{b} $$ . . . .eq.1
$$ \because a \, \& \, b $$ are integer.
$$ \because a - 2b $$ is also an integer.
$$ \Rightarrow \dfrac{a-2b}{b} $$ is rational.
LHS of the equation is the square root of a prime number.
So it is irrational and R.H.S. is rational.
Which is a contradiction. Because a rational number and an irrational number can be never equal.
So, our assumption i.e. $$2 + \sqrt 3 $$ is rational wrong.
Hence $$ ( 2 + \sqrt 3 ) $$ is an irrational number.
$$ \therefore $$ This is true.
$$ \therefore $$ option A is correct.
State whether the following statement is true or not:
$$\left( 3+\sqrt { 5 } \right) $$ is an irrational number.
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True
0%
False
Explanation
Let us suppose $$3+\sqrt 5$$ is rational.
Thus $$3+\sqrt 5$$ is in the form of $$\dfrac pq$$ where $$p$$ and $$q$$ are integers and $$q\neq0$$
$$\Rightarrow \sqrt5=\dfrac pq-3$$
$$\Rightarrow \sqrt5=\dfrac{p-3q}{q}$$
as $$p, q$$ and $$3$$ are integers $$\dfrac{p-3q}{q}$$ is a rational number.
$$\Rightarrow \sqrt 5$$ is a rational number.
But we know that $$\sqrt 5$$ is an irrational number.
So this is a contradiction.
This contradiction has arisen because of our wrong assumption that $$3+\sqrt 5$$ is a rational number.
Hence $$3+ \sqrt5$$ is an irrational number.
State the following statement is True or False
$$\dfrac{987}{10500}$$ will have terminating decimal expansion.
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True
0%
False
Explanation
Terminating decimal expansion, because
$$ \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}$$
H.C.F. of $$26$$ and $$91$$ is:
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$$13$$
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$$2366$$
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$$91$$
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$$182$$
Explanation
Prime factors of $$26=2\times 13$$
Prime factors of $$91=7\times 13$$
As $$13$$ is the only common factor.
Therefore,
$$HCF(26,91)=13$$
The product of a non-zero rational and an irrational number is
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always irrational
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always rational
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rational or irrational
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one
Explanation
Let $$x$$ be a rational number and $$y$$ be an irrational number.
Let $$xy = a $$.
Let us assume that $$a$$ is rational.
Since, $$a$$ is rational it can be expressed as $$\dfrac{p}{q} $$, where $$p$$ and $$q$$ are integers.
Let $$ x =\dfrac{m}{n}$$ , where $$m$$ and $$n$$ are integers.
Now, $$xy = a $$
$$\dfrac{my}{n} = \dfrac{p}{q}$$.
On cross multipliying we get,
$$\Rightarrow y =\dfrac{pn}{qm}$$.
Now, $$pn$$ and $$qm$$ are integers.
Hence, $$\dfrac{pn}{qm}$$ is a rational number.
However, $$y$$ is irrational.
Hence, our assumption is incorrect.
Hence, the product of a non-zero rational and an irrational number is always an Irrational number.
So, option $$A$$.
According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers $$a$$ and $$b$$ with $$a > b$$ enables us to find the
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HCF
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LCM
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Decimal expansion
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Quotient
Explanation
According to Euclid's division lemma,
For each pair of positive integers $$a$$ and $$b$$, we can find unique integers $$q$$ and $$r$$ satisfying the relation
$$a = bq + r , 0 ≤ r < b$$
If $$r = 0$$ then q will be HCF of $$a$$ and $$b$$
The basis of the Euclidean division algorithm is Euclid’s division lemma.
To calculate the Highest Common Factor (HCF) of two positive integers $$a$$ and $$b$$ we use Euclid’s division algorithm.
HCF is the largest number which exactly divides two or more positive integers.
By exactly we mean that on dividing both the integers $$a$$ and $$b$$ by HCF, the remainder is zero.
So correct answer is option A
State True or False:
$$6+\sqrt{2}$$ is a rational number.
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True
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False
Explanation
Let's assume that $$6+\sqrt2$$ is rational.....
then
$$6+\sqrt2 = p/q $$
$$\sqrt2 =( p-6q)/(q) $$
now take $$p-6q$$ to be P and $$q$$ to be Q........where P and Q are integers
which means, $$\sqrt2= P/Q$$......
But this contradicts the fact that $$\sqrt2$$ is rational
So our assumption is wrong and $$6+\sqrt2$$ is irrational.
Euclids division lemma states that if a and b are any two $$+$$ ve integers, then there exists unique integers q and r such that :
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$$a=bq+r, 0 < r < b$$
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$$a=bq+r, 0 \leq r \leq b$$
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$$a=bq+r, 0 \leq r < b$$
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$$a=bq+r, 0 < b < r$$
Explanation
Division Algorithm: Euclids division lemme states that if a and b are any two positive integers then there exist unique integers q and r such that, $$ a = q b + r \, \, 0 \leq r < b $$
$$ \therefore $$ option C is correct.
Proof: We begin by proving that the set s = {a - xb | x is an integer: a-xb $$\geq$$ 0 }
is non-empty. To do this, if suffices to exhibit a value of x making a-xb nonnegative a value of x making a-xb nonnegative. Because the integer b $$\geq$$1 , we have |a| b $$\geq$$ |a| and so -
$$ a - (-|a|) b = a + |a| b \geq a + |a| \geq 0 $$
For the choice x = -|a|, then a - xb lies in s. This paves the way for an application of well ordering principle, from which we infer that the set S contains a smallest integer : call it r. By the definition S, $$\exists$$ an integer q satisfying , r = a - qb , $$0\leq r$$
we argue that r < b. If this were not this case, then r $$\geq$$ 0 and a - ( q + 1 ) b = ( a - q b ) - b = r - b $$\geq$$ 0
This implication is that the integer a - ( q + 1 ) b has the proper form to belong to the set S. But a - ( q + 1 ) b = r - b < r , leading to a contradiction of the choice of r as the smallest member of S. Hence r < b .
Next we turn to the task of showing the uniqueness of q and r. Suppose that a has two representation of the desired form say,
$$ a = q b + r = q^{'}b + r^{'} $$
Where $$ 0 \leq r < b, \, 0 \leq r^{'} < b . $$ Then $$r^{'}- r = b (q - q^{'}) $$ and, owing to the fact that
$$ | r^{'} - r | = b | q - q^{'} | $$ upon adding two inequalities $$-b < -r \leq 0 $$ and $$ 0 \leq r^{'} < b $$ we obtain $$ -b < r^{'}-r < b \Rightarrow |r^{'} - r | < b $$
Thus, $$b|q-q^{'}| < b $$ which yields $$ 0 \leq |q-q^{'} | < 1 $$
As,
$$ |q-q^{'} | $$ is positive, the only possibility is that
$$ |q-q^{'} | = 0 $$ where $$q = q^{'} : $$ this is turns give $$ r = r^{'} $$ and ends the proof.
Write whether every positive integer can be of the form $$4q + 2$$, where $$q$$ is an integer.
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Yes
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No
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Ambiguous
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Data Insufficient
Explanation
No, all positive integers cannot be written in the form of $$ 4q+2$$
Because, $$4q+2=2\left( 2q+1 \right) $$
Therefore, $$4q+2$$ is an even number.
So, we can't write odd numbers in the form $$4q+2$$ (where $$q$$ is an integer)
Also $$2q+1$$ is an odd number, hence the maximum power of $$2$$ that divides $$4q+1$$ is $$1$$. Therefore, we can't represent the numbers divisible by $$4$$ this form.
If $$d$$ is the $$HCF$$ of 45 and 27, then $$x, y$$ satisfying $$d=27x+45y$$ are :
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$$x=2, y=1$$
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$$x=2, y=-1$$
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$$x=-1, y=2$$
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$$x=-1, y=-2$$
Explanation
Applying Euclid's division lemma to $$27$$ and $$45$$, we get
$$45 = 27 \times 1 + 18$$ ...(1)
$$27 = 18 \times 1 + 9$$ ...(2)
$$18 = 9 \times 2 + 0$$ ...(3)
Since the remainder is zero, therefore, last divisor $$9$$ is the $$HCF$$ of $$27$$ and $$45$$.
From (2), we get
$$9 = 27 - 18 \times 1$$
$$ = 27 - \left(45 - 27 \times 1\right) \times 1 \ \ \ \ \ \ \left[using (1)\right]$$
$$ = 27 - 45 \times 1 + 27 \times 1 \times 1$$
$$ = 27 - 45 + 27$$
$$ = 54 - 45$$
$$\Rightarrow 9 = 27 \times 2 - 45 \times 1$$ ...(4)
Comparing (4) with $$d = 27x + 45y$$, we get
$$d = 9, x = 2 \ and\ y = -1$$
Say true or false:
A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.
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0%
True
0%
False
Explanation
Let the positive integer $$n$$ is of the form $$3q, 3q+1,$$ and $$ 3q+2$$
If $$n=3q$$
Squaring both sides, we get,
$$=>{ n }^{ 2 }=9{ q }^{ 2 }$$
$$=>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right) $$
$$=>{ n }^{ 2 }=3m$$, where $$m=3{ q }^{ 2 }$$
Now, if $$n=3q+1$$
$$=>{ n }^{ 2 }={ \left( 3q+1 \right) }^{ 2 }$$
$$=>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1$$
$$=>{ n }^{ 2 }={ 3q\left( 3q+2 \right) }+1$$
$$=>{ n }^{ 2 }=3m+1 ,$$ where $$ m=q\left( 3q+2 \right) $$
Now, if $$n=3q+2$$
$$=>{ n }^{ 2 }={ \left( 3q+2 \right) }^{ 2 }$$
$$=>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4$$
$$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+4$$
$$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+3+1$$
$$=>{ n }^{ 2 }=3m+1$$ where $$m=\left( 3{ q }^{ 2 }+4q+1 \right) $$
Hence, $${ n }^{ 2 }$$ integer is of the form $$3m$$ and $$3m+1$$ not $$3m+2$$
H.C.F. of $$6, 72$$ and $$120$$ is:
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$$6$$
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$$2$$
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$$3$$
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$$1$$
Explanation
Factors of 6 are $$1,2,3,6$$
Factors of 72 are $$1,2,3,4,6,8,9,12,18,24,36,72$$
Factors of 120 are $$1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$$
The highest factor in all the three is $$6$$ .
State true or false:
$$\sqrt{2}$$ is not a rational number.
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True
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False
Explanation
Let us assume that $$\sqrt{2}$$ is a rational number
$$\Rightarrow \sqrt{2}=\dfrac{p}{q}\left [ \dfrac{p}{q}\ is\ in\ simplest\ form\ \right ]$$
$$\Rightarrow 2q^{2}=p^{2}$$
$$p^{2}$$ is even $$\Rightarrow p$$ is even $$\Rightarrow p=2k$$
$$2q^{2}=4k^{2}\Rightarrow q^{2}=2k^{2}\Rightarrow q^{2}$$ is even
$$\Rightarrow q$$ is even.
$$\Rightarrow p,q$$ have 2 as a common factor which is contradiction to assumption.
$$\therefore \sqrt{2}=\dfrac{p}{q}$$ is a false
$$\Rightarrow \sqrt{2}$$ is not a rational.
Use Euclid's division lemma to find the HCF of the following
16 and 176
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$$16$$
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$$14$$
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$$10$$
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$$12$$
Explanation
by using Euclid's division lemma,
$$a=bq+r$$ ; $$r$$ greater than or equal to 0, and $$r$$ less than $$b$$
$$176=(16)(11)+0, H.C.F=16$$
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
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1 < r < b
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$$0 < r \leq b$$
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$$0\leq r < b$$
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0 < r < b
Explanation
If $$r$$ must satisfy$$ 0 \le r < b$$
Proof,
$$..,a-3b,a-2b,a-b,a,a+b,a+2b,a+3b,..$$
clearly it is an arithmetic progression with common difference $$b$$ and it extends infinitely in both directions.
Let $$r$$ be the smallest non-negative term of this arithmetic progression.Then,there exists a non-negative integer $$q$$ such that,
$$a-bq=r$$
$$=>a=bq+r$$
As,$$r$$ is the smallest non-negative integer satisfying the result.Therefore, $$0\quad \le \quad r\quad \le \quad b$$
Thus, we have
$$a=bq_1+r_1$$, $$0 \le r_1 \le b$$
If HCF of $$210$$ and $$55$$ is of the form $$(210) (5) + 55 y$$, then the value of $$y$$ is :
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$$-19$$
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$$-18$$
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$$5$$
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$$55$$
Explanation
HCF of 210 and 55 is 5
Given HCF = $$(210)(5)+55y$$
$$5 =(210)(5)+55y$$
$$55y$$=$$-1045$$
On solving this equation we get $$y=-19$$
So, correct answer will be option A
Use Euclid's division lemma to find the HCF of $$40$$
and $$248$$.
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$$4$$
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$$10$$
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$$6$$
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$$8$$
Explanation
Here, $$a=248$$ and $$b=40$$
by using Euclid's division lemma,
$$a=bq+r$$ ; $$0\le r<b$$
$$248=40\times 6 +8$$
$$40=8\times 5+0$$
$$\therefore H.C.F (40,248)=8$$
hence option $$D$$ is correct.
The decimal expansion of the rational number $$\displaystyle\frac{23}{2^{3}5^{2}}$$, will terminate after how many places of decimal?
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1
0%
3
0%
4
0%
5
Explanation
$$\displaystyle\frac{23}{2^{3}5^{2}}$$
= $$\displaystyle\frac{23}{2(2)^{2}(5)^{2}}$$
= $$\displaystyle\frac{23}{2(2\times5)^{2}}$$
= $$\displaystyle\frac{23}{2(10)^{2}}$$
= $$\displaystyle\frac{11.5}{100} = 0.115$$
Therefore, the decimal expansion of the rational number $$\displaystyle\frac{23}{2^{3}5^{2}}$$
will terminate after three places of decimal.
Use Euclid's division lemma to find the HCF of the following
27727 and 53124
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$$233$$
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$$200$$
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$$154$$
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$$212$$
Explanation
Here, $$a=53124$$ and $$b=27727$$
by using Euclid's division lemma,
$$a=bq+r$$ ; $$0<r<b$$
$$53124=(27727)(1)+25397$$
$$27727=(25397)(1)+2330$$
$$25397=(2330)(10)+2097$$
$$2330=(2097)(1)+233$$
$$2097=(233)(9)+0$$
H.C.F is $$233$$
Use Euclid's division lemma to find the HCF :
65 and 495.
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$$5$$
0%
$$10$$
0%
$$15$$
0%
$$0$$
Explanation
According to Euclid’s Division Lemma if we have two positive integers $$a$$ and $$b$$, then there exists unique integers $$q$$ and $$r$$ which satisfies the condition $$a = bq + r$$ where $$0 ≤ r < b$$.
$$HCF$$ is the largest number which exactly divides two or more positive integers.
By Euclid's division lemma, we mean that on dividing both the integers $$a$$ and $$b$$ , the remainder is zero.
The given integers are $$a=65$$ and $$b=495$$.
Clearly $$495 > 65$$.
So, we will apply Euclid’s division lemma to
$$65$$ and $$495$$
, we get,
$$495 = (65\times 7) + 40$$
Since the remainder $$40≠ 0$$. So we again apply the division lemma to the divisor $$65$$ and remainder $$40$$. We get,
$$65 = (40\times 1) + 25$$
Again
the remainder $$25≠ 0$$, so
applying the division lemma to the new divisor $$40$$ and remainder $$25$$. We get,
$$40 = (25\times 1) + 15$$
Now, again
the remainder $$15≠ 0$$, so
applying the division lemma to the new divisor $$25$$ and remainder $$15$$. We get,
$$25 = (15\times 1) + 10$$
Again
the remainder $$10≠ 0$$, so
applying the division lemma to the new divisor $$15$$ and remainder $$10$$. We get,
$$15 = (10\times 1) + 5$$
Again
the remainder $$5≠ 0$$, so
applying the division lemma to the new divisor $$10$$ and remainder $$5$$. We get,
$$10 = (5\times 1) + 0$$
Finally we get the remainder $$0$$ and the divisor is $$5$$.
Hence, the $$HCF$$ of
$$65$$ and $$495$$
is
$$5$$.
Use Euclid's division lemma to find the $$HCF$$ of the following numbers:
$$8068$$ and $$12464$$
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$$4$$
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$$2$$
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$$0$$
0%
$$1$$
Explanation
Here, $$a=12464$$ and $$b=8068$$
So, by using Euclid's division lemma,
$$a=bq+r$$ ; $$0<r<b$$
$$\begin{aligned}{}12464 &= (8068)(1) + 4396\\8068 &= (4396)(1) + 3672\\4396& = (3672)(1) + 724\\3672 &= (724)(5) + 52\\724& = (52)(13) + 48\\52 &= (48)(1) + 4\\48 &= (4)(12) + 0\end{aligned}$$
Hence, $$HCF$$ of $$8068$$ and $$12464$$ is equal to $$4.$$
Without doing any actual division, find which of the following rational numbers have terminating decimal representation :
(i) $$\displaystyle \dfrac{7}{16}$$ (ii) $$\displaystyle \dfrac{23}{125}$$
(iii) $$\displaystyle \dfrac{9}{14}$$ (iv) $$\displaystyle \dfrac{32}{45}$$
(v) $$\displaystyle \dfrac{43}{50}$$ (vi) $$\displaystyle \dfrac{17}{40}$$
(vii) $$\displaystyle \dfrac{61}{75}$$ (viii) $$\displaystyle \dfrac{123}{250}$$
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(i), (iii), (v), (vi) and (vii)
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(i), (ii), (v), (vi) and (viii)
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(i), (iii), (v), (vi) and (viii)
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(i), (ii), (v), (vi) and (vii)
Explanation
The rational not having denominator as multiple of $$2^{m}\times 5^{n}$$
will be non terminating.
(1) $$\dfrac{7}{16}=\dfrac{7}{2\times2\times2\times2}-$$ denominator is multiple of $$2^{m}\times 5^{n}$$
hence terminating
.
(2) $$\dfrac{23}{125}=\dfrac{23}{5\times5\times5}$$ -- denominator is multiple of $$2^{m}\times 5^{n}$$
hence terminating.
(3)
$$\dfrac{9}{14}=\dfrac{9}{7\times2}$$
-- denominator is not multiple of $$2^{m}\times 5^{n}$$
hence non-terminating
(4)
$$\dfrac{32}{45}=\dfrac{32}{3\times3\times5}$$
-- denominator is not multiple of $$2^{m}\times 5^{n}$$
hence non-terminating.
(5)
$$\dfrac{43}{50}=\dfrac{43}{5\times5\times2}$$
-- denominator is multiple of $$2^{m}\times 5^{n}$$
hence terminating.
(6)
$$\dfrac{17}{40}=\dfrac{17}{2\times2\times2\times5}$$
-- denominator is multiple of $$2^{m}\times 5^{n}$$
hence terminating.
(7)
$$\dfrac{61}{75}$$
-- denominator is not multiple of $$2^{m}\times 5^{n}$$
hence non-terminating
.
(8)
$$\dfrac{123}{250}=\dfrac{123}{5\times5\times5\times2}$$
-- denominator is multiple of $$2^{m}\times 5^{n}$$
hence terminating.
(i), (ii), (v), (vi) and (viii) will have terminating decimal.
The given pair of numbers $$ 231, 396$$ are __________ .
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multiple of each other
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factors of each other
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co-prime to each other
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having a difference of $$120$$ among each other
Explanation
Euclid's division lemma,
$$a=bq+r$$ ; $$0\le r<b$$
Applying Euclid's Algorithim to
$$231 , 395$$
$$395 = 231\times 1+164$$
$$231 = 164 \times 1+67$$
$$164 = 67 \times 2+30$$
$$67 = 30 \times 2 +7 $$
$$30 = 7 \times 4 +2$$
$$7 = 2 \times 3+1$$
$$2 = 1 \times 2+0$$
The given numbers have only $$1$$ as common factor.
Hence, the given numbers are co-prime.
State: True or False
$$3+2\sqrt 5$$ is an irrational number.
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0%
True
0%
False
Explanation
Let us assume, to the contrary, that $$3+2\sqrt{5}$$ is rational.
That is, we can find coprime integers $$a$$ and $$b$$ (b does not equal to 0) such that $$3+2\sqrt{5}=\dfrac{a}{b}$$.
Therefore, $$\dfrac{a}{b} - 3=2\sqrt{5}$$
$$\dfrac{a-3b}{b}=2\sqrt{5}$$
$$\dfrac{a-3b}{2b}=\sqrt{5}$$
$$\dfrac{a}{2b}-\dfrac{3}{2}=\sqrt{5}$$
Since $$a$$ and $$b$$ are integers and rational, we get $$\dfrac{a}{2b}-\dfrac{3}{2}$$ is rational, and so $$\dfrac{a-3b}{2b}=\sqrt{5}$$ is rational.
But this contradicts the fact that $$\sqrt{5}$$ is irrational.
This contradiction has arisen because of our incorrect assumption that $$3+2\sqrt{5}$$ is rational.
So, we conclude that $$3+2\sqrt{5}$$ is irrational.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
$$\dfrac {29}{343}$$
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Terminating
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Non-terminating
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Ambiguous
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Data insufficient
Explanation
Given, $$\displaystyle \frac {29}{343}= \frac {29}{7^{3}}$$
As it is not in the form of $${ 2 }^{ m }\times { 5 }^{ n }$$.
So, the rational number $$\displaystyle \frac {29}{343}$$ has a non terminating decimal expansion
If these numbers form
positive odd integer
6q+1, or 6q+3 or 6q+5 for some q then q belongs to:
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integers
0%
irrational
0%
real
0%
none of these
Explanation
We know that product and sum of integers will give integer.
Given
$$6q+1, 6q+3 , 6q+5$$ are integers so $$6q$$ will be integer and then q will be integer.
So correct answer will be option A.
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