MCQ Questions for Class 10 Maths Real Numbers Quiz 5

The HCF of

$455$ and

$42$ using Euclid algorithm is

0%
$7$
0%
$6$
0%
$5$
0%
$4$

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.)

0%
$\dfrac{1}{256}$
0%
$\dfrac{27}{100}$
0%
$\dfrac{100}{27}$
0%
$\dfrac{231}{660}$
0%
$\dfrac{7}{105}$

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 ...........

0%
$0$
0%
$1$
0%
$2$
0%
$6$

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, the denominator has

$2$ as a factor and hence it is terminating.

Therefore,

$C$ is the correct answer.

Find the HCF of

$26$ and

$455$

0%
$11$
0%
$12$
0%
$13$
0%
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

0%
$1$
0%
$12$
0%
$0.12$
0%
$0.012$

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 ..............

0%
repeats
0%
is that of a whole number
0%
has equal numerator and denominator
0%
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

0%
$28$
0%
$30$
0%
$1$
0%
$5$

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?

0%
$17$
0%
$2$
0%
$1$
0%
$3$

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?

0%
$2$
0%
$1$
0%
$6$
0%
$5$

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

0%
$6$
0%
$3$
0%
$4$
0%
$2$

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.

0%
$35$
0%
$30$
0%
$15$
0%
$25$

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?

0%
$35\ cm$
0%
$55\ cm$
0%
$1\ m$
0%
$65\ cm$

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.

0%
$\dfrac{24}{1600}$
0%
$\dfrac{171}{800}$
0%
$\dfrac{123}{2^2 \times 5^3}$
0%
$\dfrac{145}{2^3 \times 5^2 \times 7^2}$

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 _________.

0%
$2$ or $5$
0%
$2$ , $3$ or $5$
0%
$3$ or $5$
0%
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?

0%
$\sqrt{2}$ is an irrational number
0%
If a is a rational number and $\sqrt{b}$ is an irrational number then $a\sqrt{b}$ is irrational number
0%
Every surd is an irrational number
0%
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

$17$ ?

0%
$7$
0%
$1$
0%
$119$
0%
$17$

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 _________.

0%
Terminating decimal
0%
Non-terminating decimal
0%
Non-terminating and non-repeating decimal
0%
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

0%
$6$
0%
$7$
0%
$8$
0%
$9$

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

0%
$5$
0%
$3$
0%
$2$
0%
$4$

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 ____________.

0%
$36$
0%
$18$
0%
$12$
0%
$6$

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

0%
Always irrational
0%
Always rational
0%
Can be both rational or irrational
0%
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?

0%
Also a rational number
0%
An irrational number
0%
Not a number
0%
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)

$5\sqrt { 2 }$

0%
True
0%
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.

0%
True
0%
False

Explanation

Here 2 is a rational number

and

$3\sqrt5$ is an irrational number and

$\text{difference of rational and irrational is always an irrational number}$

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.

0%
True
0%
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.

0%
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.

0%
True
0%
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.

0%
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 :

$p, q$ are prime positive integers, then

$\sqrt { p } +\sqrt { q }$ is an irrational number.

0%
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.

0%
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

CBSE Questions for Class 10 Maths Real Numbers Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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