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CBSE Questions for Class 10 Maths Real Numbers Quiz 5 - MCQExams.com
CBSE
Class 10 Maths
Real Numbers
Quiz 5
The HCF of $$455$$ and $$42$$ using Euclid algorithm is
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$$7$$
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$$6$$
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$$5$$
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$$4$$
Explanation
According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
$$455=42\times 10+35$$
$$42=35\times 1+7$$
$$35=7\times 5+0$$
The HCF of $$455$$ and $$42$$ is $$7$$.
So, option A is correct.
Which of the following fractions will terminate when expressed as a decimal? (Choose all that apply.)
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$$\dfrac{1}{256}$$
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$$\dfrac{27}{100}$$
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$$\dfrac{100}{27}$$
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$$\dfrac{231}{660}$$
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$$\dfrac{7}{105}$$
Explanation
Recall that in order for the decimal expansion of a fraction to terminate, the fraction's denominator, in fully reduced form,
must have a prime factorization that consists of only $$2$$'s and/or $$5$$'s.
The denominator in $$A$$ is composed of only $$2$$'s, $$(256 = 2^8)$$.
The denominator in $$B$$ is composed of only $$2$$'s and $$5$$'s, $$(100=2^2\times 5^2)$$.
In fully reduced form, the fraction in $$D$$ is equal to $$\dfrac{7}{20}$$ and the denominator $$20$$ is composed of only $$2$$'s and $$5$$'s $$(20=2^2\times 5)$$.
By contrast, the denominator in $$C$$ has prime factors other than $$2$$'s and $$5$$'s $$(27 =3^3)$$, and in fully reduced form, the fraction in $$E$$ is equal to $$\dfrac{1}{15}$$, and $$15$$ has a prime factor other than $$2$$'s and $$5$$'s $$(15 = 3 \times 5)$$.
Therefore, options $$A$$, $$B$$ and $$D$$ are the correct answers.
$$\dfrac {1}{2} = 0.5$$
It is a terminating decimal because the denominator has a factor as ...........
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$$0$$
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$$1$$
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$$2$$
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$$6$$
Explanation
When the prime factorization of the denominator of a fraction has only factors of $$2$$
and factors of $$5$$
, we say the decimal as terminating decimal.
Here, t
he denominator has $$2$$ as a factor and hence it is terminating.
Therefore, $$C$$ is the correct answer.
Find the HCF of $$26$$ and $$455$$
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$$11$$
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$$12$$
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$$13$$
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none of the above
Explanation
$$26=13\times 2$$
$$455=5\times 13\times 7$$
HCF is $$13.$$
The HCF of the numbers $$0.48, 0.72$$ and $$0.108$$ is
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$$1$$
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$$12$$
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$$0.12$$
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$$0.012$$
Explanation
$$480=2^5\times 3\times 5$$
$$720=2^4\times 3^2\times 5$$
$$108=2^2\times 3^2$$
$$HCF=2^2\times 3=12$$
HCF of $$0.48, 0.72$$ and $$0.108 = 0.012$$
So, option $$D$$ is correct.
If the denominator of a fraction has factors other than $$2$$ and $$5$$, the decimal expression ..............
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repeats
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is that of a whole number
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has equal numerator and denominator
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terminates
Explanation
If there are prime factors in the denominator other than $$2$$ or $$5$$, then a block of digit repeats.
For example,$$\dfrac {1}{24} = \dfrac {1}{3\times 2\times 2\times 2}$$ (there is a factor of $$3$$, therefore decimal will repeat.)
$$=0.416666...$$
Therefore, $$A$$ is the correct answer.
The HCF of two consecutive odd numbers is
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$$28$$
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$$30$$
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$$1$$
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$$5$$
Explanation
$$\textbf{Check for the HCF of consecutive odd numbers by writing down the first few odd numbers.}$$
$$\textbf{Step 1: Writing out the first few odd numbers}$$
$$\text{The first few odd numbers are 1, 3, 5, 7, 9, 11}$$
$$\textbf{Step 2: Finding out the HCF of consecutive odd numbers}$$
$$\text{It is evident that there is no common factors among any 2 consecutive odd numbers except 1}$$
$$\text{For example the HCF of 3 and 5 is 1, and that of 5 and 7 is also 1}$$
$$\textbf{Hence option C is correct}$$
Using the theory that any positive odd integers are of the form $$4 q + 1$$ or $$4 q + 3$$ where $$q$$ is a positive integer, If the quotient is $$4$$, the dividend is $$19$$ then what will be the remainder?
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$$17$$
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$$2$$
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$$1$$
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$$3$$
Explanation
Let $$a$$ be any positive integer.
We know by Euclid's algorithm, if $$a$$ and $$b$$ are two positive integers, there exist unique integers $$q$$ and $$r$$ satisfying, $$a= bq + r$$ where $$0 \leq r < b$$.
Take $$b= 4$$ we get,
$$ a = 4q + r, 0 \leq r < b$$.
Since $$0 \leq r < 4$$, therefore the possible remainders are $$0, 1, 2$$ and $$3$$.
That is, $$a$$ can be $$4q, 4q+1$$, or $$4q+2$$ or $$4q+3$$, where $$q$$ is the quotient.
Since $$a$$ is odd, $$a$$ cannot be $$4q$$ or $$4q + 2$$ as they are both divisible by $$2$$.
Therefore, any odd integer is of the form $$4q + 1$$ or $$4q + 3$$.
Given the quotient $$q=4$$ we get,
$$4q+1= 4 \times 4 +1 = 17$$, or $$4q + 3= 4 \times 4 + 3 = 19$$
Since dividend is $$19$$ the remainder is $$3.$$
Therefore$$,$$ $$D$$ is the correct answer.
There are five odd numbers $$1, 3, 5, 7, 9$$. What is the HCF of these odd numbers?
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$$2$$
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$$1$$
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$$6$$
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$$5$$
Explanation
Let us calculate and check what is the highest common factor for these numbers.
First finding the factors of individual numbers:
Factors of $$1=1$$
Factors of
$$3=1, 3$$
Factors of $$5=1, 5$$
Factors of $$7=1, 7$$
Factors of $$9=1, 3, 9$$
HCF is $$1$$ for the five odd numbers.
Therefore, $$B$$ is the correct answer.
The HCF of two consecutive even numbers is
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$$6$$
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$$3$$
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$$4$$
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$$2$$
Explanation
$$HCF$$ of two consecutive even numbers is always $$2$$. For example:
$$HCF$$ of $$2$$ and $$4$$ is $$2$$ and similarly,
$$HCF$$ of $$22$$ and $$24$$ is $$2$$ and we can do so on..
Find HCF of $$70$$ and $$245$$ using Fundamental Theorem of Arithmetic.
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$$35$$
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$$30$$
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$$15$$
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$$25$$
Explanation
Using Fundamental Theorem of Arithmetic we can write the given integers as a product of their prime factors,
$$70 =2\times 5\times 7$$
$$245=5\times 7 \times 7 $$
Hence, common factors of $$70$$ and $$245$$ are $$5$$ and $$7.$$
$$HCF = 5\times 7$$
$$=35$$
Hence, option $$A$$ is correct.
Three ropes are $$7\ m, 12\ m\ 95\ cm$$ and $$3\ m\ 85\ cm$$ long. What is the greatest possible length that can be used to measure these ropes?
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$$35\ cm$$
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$$55\ cm$$
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$$1\ m$$
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$$65\ cm$$
Explanation
The given three ropes are $$7$$m, $$12$$ m $$95$$cm and $$3$$m$$85$$cm long. We know that $$1$$m=$$100$$cm, therefore,
The length of the respective ropes will be:
1st rope $$=7\times 100=700$$cm
2nd rope $$=(12\times 100)+95=1200+95=1295$$cm
3rd rope $$=(3\times 100)+85=300+85=385$$cm
Now, let us factorize the length of the ropes as follows:
$$700=2\times 2\times 5\times 5\times 7\\ 1295=5\times 7\times 37\\ 385=5\times 7\times 11$$
The highest common factor (HCF) is $$5\times 7=35$$
Hence, the greatest possible length that can be used to measure these ropes is $$35$$cm.
Identify a non-terminating repeating decimal.
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$$\dfrac{24}{1600}$$
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$$\dfrac{171}{800}$$
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$$\dfrac{123}{2^2 \times 5^3}$$
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$$\dfrac{145}{2^3 \times 5^2 \times 7^2}$$
Explanation
We know that, a fraction will have terminating decimal expansion only if its denominator is in the form of
$${2}^{m}\times{{5}^{n}}$$.
Let's check the denominator of each option:
$$(i)\ \dfrac{24}{1600}\\$$
$$1600=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5=2^6\times 5^2$$
Hence, the denominator is $$2^6\times 5^2$$
$$(ii)\ \dfrac{171}{800}\\$$
$$800=2\times 2\times 2\times 2\times 2\times 5\times 5=2^5\times 5^2$$
Hence, the denominator is $$2^5\times 5^2$$
$$(iii)\ \dfrac{123}{2^3\times 5^3}\\$$
Hence, the denominator is $$2^3\times 5^3$$
$$(iv)\ \dfrac{145}{2^3\times 5^2\times 7^2}\\$$
Hence, the denominator is $$2^3\times 5^2\times 7^2$$
Only option D does not have a denominator of the form $$2^m\times 5^n$$.
Hence, it will have a non-terminating decimal expansion. Further, the expansion will be repeating.
Hence, option D is correct.
A rational number can be expressed as a terminating decimal if the denominator has factors _________.
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$$2$$ or $$5$$
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$$2$$, $$3$$ or $$5$$
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$$3$$ or $$5$$
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Only $$2$$ and $$3$$
Explanation
We know, any rational number with its denominator is in the form of $${ 2 }^{ m }\times { 5 }^{ n }$$, where $$m,n$$ are positive integers are terminating decimals.
So, option $$A$$ is correct.
Which one of the following is not true?
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$$\sqrt{2}$$ is an irrational number
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If a is a rational number and $$\sqrt{b}$$ is an irrational number then $$a\sqrt{b}$$ is irrational number
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Every surd is an irrational number
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The square root of every positive integer is always irrational
Explanation
(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.
An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:
$$π = 3.141592…$$
$$\sqrt {2} = 1.414213…$$
Therefore,
$$\sqrt {2}$$ is an irrational number.
(b) Let us take a rational number $$a=\dfrac {2}{1}$$ and an irrational number
$$b=\sqrt {2}$$, then their product can be determined as:
$$a\times b=2\times \sqrt { 2 } =2\sqrt { 2 }$$ which is also an irrational number.
Therefore, if $$a$$ is a rational number and $$\sqrt {b}$$ is an irrational number than
$$a\sqrt {b}$$ is an irrational number.
(c)
A surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.
For example,
$$\sqrt {2}$$
is a surd since $$2$$ is rational and
$$\sqrt {2}$$
is irrational.
Surds are numbers left in root form
$$\sqrt {\ \ }$$
to express its exact value. It has an infinite number of non-recurring decimals.
Therefore, every surd is an irrational number.
(d) Let us take a positive integer $$4$$, now square root of $$4$$ will be:
$$\sqrt {4}=2$$ which is not an irrational number
Hence, the square root of every positive integer is not always irrational.
What is the HCF of $$7$$ and $$17$$?
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$$7$$
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$$1$$
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$$119$$
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$$17$$
A real number $$\displaystyle \frac{2^2 \times 3^2 \times 7^2}{2^5 \times 5^3 \times 3^2 \times 7}$$ will have _________.
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Terminating decimal
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Non-terminating decimal
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Non-terminating and non-repeating decimal
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Terminating repeating decimal
Explanation
Using the law of exponent,
$$\dfrac{2^{2}\times3^{2}\times7^{2}}{2^{5}\times5^{3}\times3^{2}\times7} = 2^{2-5}\times3^{2-2}\times5^{-3}\times7^{2-1}$$
$$=2^{-3}\times3^{0}\times5^{-3}\times7^{1}$$
$$= \dfrac{7}{2^{3}\times5^{3}}$$
$$= \dfrac{7}{1000}$$
$$= 0.007$$.
$$\therefore 0.007$$ is a terminating and non-repeating decimal.
That is, option $$A$$ is correct.
The greatest integer that divides $$358,\ 376$$ and $$ 232$$. The same remainder in each case is
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$$6$$
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$$7$$
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$$8$$
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$$9$$
Explanation
If the remainder is the same in each case and the remainder is not given, HCF of the differences of the
numbers is the required greatest number
Given: The greatest number that will divide 358,376 and 232 leaving the same remainder in each case.
To find: The number?
Solution :
First, we find the difference between these numbers.
The required numbers are
376 - 232 = 144
376-358= 18
358 - 232 =126
Now, We find the HCF of 18, 126, and 144
$$18 =2\times3^2$$
$$126= 2\times3^2\times7$$
$$144= 2^4\times3^2$$
$$HCF(18,126,144)= 2\times3\times3$$
Therefore, The required largest number is 18.
Find HCF of
25 and 55:
25 and
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$$5$$
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$$3$$
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$$2$$
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$$4$$
Explanation
Prime factors of $$25 = 5 \times 5$$
Prime factors of $$55 = 5 \times 11$$
Since the common factor is only $$5$$.
Therefore,
$$HCF(25,55)=5$$
Hence, the correct option $$A$$
The H. C. F. of $$252$$, $$324$$ and $$594$$ is ____________.
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$$36$$
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$$18$$
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$$12$$
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$$6$$
Explanation
H.C.F is the greatest number which divides each of them individually.
All the numbers are written in form of factors of prime numbers.
$$252$$ $$\rightarrow$$ $$2$$$$\times$$$$2$$$$\times$$$$3$$
$$\times$$
$$3$$
$$\times$$
$$7$$
$$324$$ $$\rightarrow$$ $$2$$
$$\times$$
$$2$$
$$\times$$
$$3$$
$$\times$$
$$3$$
$$\times$$
$$3$$
$$\times$$
$$3$$
$$594$$ $$\rightarrow$$ $$2$$
$$\times$$
$$3$$
$$\times$$
$$3$$
$$\times$$
$$3$$
$$\times$$
$$11$$
From this greatest factor is $$2$$
$$\times$$
$$3$$
$$\times$$
$$3$$ $$\rightarrow$$ $$18$$
Hence, option B is correct.
The product of two irrational numbers is
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Always irrational
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Always rational
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Can be both rational or irrational
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always an integer
Explanation
According to the properties of irrational numbers, the product of two irrational numbers can be both rational or irrational.
For example, let $$p=\sqrt 3$$ and $$q=\sqrt 3$$ be two irrational numbers
$$\therefore \ pq=\sqrt 3\times \sqrt 3=3$$
Here, $$3$$ is a rational number.
Now, let $$p=\sqrt 3$$ and $$q=\sqrt 2$$
$$\therefore \ pq=\sqrt 3\times \sqrt 2=\sqrt 6$$
Here, $$\sqrt6$$ is an irrational number.
Hence, the product can be both rational or irrational.
So, option C is correct.
$$\frac{2}{3}$$ is a rational number whereas $$\frac{\sqrt{2}}{\sqrt{3}}$$ is?
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Also a rational number
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An irrational number
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Not a number
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A natural periodic number
State whether True or False :
All the following numbers are irrationals.
(i) $$\dfrac { 2 }{ \sqrt { 7 } } $$ (ii) $$\dfrac { 3 }{ 2\sqrt { 5 } }$$ (iii) $$4+\sqrt { 2 } $$ (iv) $$5\sqrt { 2 } $$
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True
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False
Explanation
In all of the above questions $$\sqrt7,\sqrt5,\sqrt2$$ are a $$irrational$$ numbers
And $$\text{the addition, subtraction, division and product between rational and irrational gives an irrational number}$$
So that all of the above are irrational numbers.
hence option A is correct.
State whether the given statement is True or False :
$$2-3\sqrt { 5 }$$ is an irrational number.
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True
0%
False
Explanation
Here 2 is a rational number
and
$$3\sqrt5$$ is an irrational number a
nd $$\text{difference of rational and irrational is always an irrational number}$$
So that $$2 - 3\sqrt5 $$ is an $$irrational$$ number
hence option A is correct.
State whether the given statement is True or False :
$$4-5\sqrt { 2 } $$ is an irrational number.
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True
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False
Explanation
$$\text{Here 4 is a rational number and }$$$$5\sqrt2$$ is a $$irrational$$ number
And $$\text{the difference of rational and irrational is always an irrational number}$$
So that $$(4-5\sqrt2)$$ is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
$$5-2\sqrt { 3 } $$ is an irrational number.
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True
0%
False
Explanation
$$\text{Here 5 is a rational number and }$$$$2\sqrt3$$ is a $$irrational$$ number
And $$\text{the difference of rational and irrational is always an irrational number}$$
So that $$(5 - 2\sqrt3)$$ is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
$$\sqrt { 3 } +\sqrt { 4 } $$ is an irrational number.
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True
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False
Explanation
$$\sqrt4= 2 \text{ is a rational number and }$$$$\sqrt3$$ is a $$irrational$$ number
And $$\text{the addition of rational and irrational is always an irrational number}$$
So that $$(\sqrt3+\sqrt4)$$ is an irrational number.
hence option A is correct.
$$2-\sqrt { 3 } $$ is an irrational number.
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True
0%
False
Explanation
We know that $$2$$ is a rational number and $$\sqrt3$$ is an irrational number.
Also difference between rational and irrational number is always irrational number.
So that $$(2-\sqrt3)$$ is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
If $$p, q $$ are prime positive integers, then $$\sqrt { p } +\sqrt { q } $$ is an irrational number.
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True
0%
False
Explanation
Let us asume $$\sqrt{p}+\sqrt{q}$$ are rational.
Given that $$p$$ and $$q$$ are prime positive integers.
$$\sqrt{p}+\sqrt{q}=\dfrac{a}{b}$$
Squaring on both sides, we get
$$p+q+2\sqrt{pq}=\left(\dfrac{a}{b}\right)^{2}$$
$$\sqrt{pq}=\dfrac{1}{2}\left[\left(\dfrac{a}{b}\right)^{2}-p-q\right] - (i)$$
We know that $$p$$ & $$q$$ are prime positive numbers
$$\Rightarrow \sqrt{p}$$, $$\sqrt{q}$$ and $$\sqrt{pq} $$ are irrational
So our assumption $$\sqrt{p}+\sqrt{q}$$ are rational is incorrect.
$$\Rightarrow\ \sqrt p+\sqrt q$$ is a irrational number if $$p,q$$ are prime positive numbers
State whether the given statement is True or False :
The number
$$6+\sqrt { 2 } $$ is
irrational.
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True
0%
False
Explanation
$$Here \ 6 \text{ is a rational number and }$$$$\sqrt2$$ is a $$irrational$$ number
And $$\text{the addition of rational and irrational is always an irrational number}$$
So that $$(6+\sqrt2)$$ is an irrational number.
hence option A is correct
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