MCQ Questions for Class 10 Maths Real Numbers Quiz 4

$\frac {23}{2^35^2}$

0%
Terminating
0%
Non-terminating
0%
Ambiguous
0%
Data insufficient

Explanation

The given fraction,

$\displaystyle \frac {23}{2^{3}5^{2}}$ has its denominator composed of only

$2$ 's and

$5$ 's only.

Since the fraction is in the form of

${ 2 }^{ m }\times{ 5 }^{ n }$ where (

$n=3,m=2$ ),

the rational number

$\dfrac {23}{2^{3}5^{2}}$ will have a terminating decimal expansion.

State whether the following statement is True or False.

$\dfrac {15}{1600}$ has a terminating decimal expansion .

0%
True
0%
False

Explanation

Given,

$\displaystyle \dfrac {15}{1600}= \dfrac{15}{25\times8\times 8} =\dfrac {15}{5^{2}2^{6}}$

As it's denominator is in the form of

${ 2 }^{ m }\times { 5 }^{ n }$ where (

$n=6,m=2$ ).

So, the rational number

$\displaystyle \frac {15}{1600}$ has a terminating decimal expansion.

When the HCF of

$468$ and

$222$ is written in the form of

$468 x + 222y$ then the value of

$x$ and

$y$ is

0%
$x =-9 \ and \ y =19$
0%
$x =9 \ and \ y = -19$
0%
$x =9\ and \ y = 19$
0%
$x =-9 \ and \ y =- 19$

Explanation

According to the definition of Euclid's theorem,

$a = b \times q + r$ where

$0 \leq r < b .$

HCF of

$468$ and

$222$

$468 = \left(222 \times 2\right) + 24$

$222 = \left(24\times\ 9\right) + 6$

$24 = \left(6\times\ 4\right) + 0$

$\therefore HCF = 6$

$6 = 222 - \left(24\times\ 9\right)$

$= 222 - \left[\left(468 -222 \times 2\right) \times\ 9\right]$ [where

$468 = 222 \times 2 + 24$ ]

$= 222 - \left[468 \times 9 -222 \times 2 \times 9\right]$

$= 222 - \left(468 \times9\right) - \left(222\times 18\right)$

$= 222 + \left(222 \times 18\right) - \left(468 \times9\right)$

$= 222\left[1 + 18\right]- 468 \times 9$

$= 222 \times19- 468 \times 9$

$= 468 \times -9 + 222\times 19$

$\therefore x=-9$ and

$y=19$ .

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%
None of these

Explanation

Any rational number with its denominator is in the form of

$2^m\times 5^n$ , where

$m,n$ are positive integers and are terminating decimals. The examples are shown below:

$(i)\quad \dfrac { 15 }{ 8 } =\dfrac { 15 }{ 2\times 2\times 2 } =\dfrac { 15 }{ 2^{ 3 } } =\dfrac { 15\times 5^{ 3 } }{ 2^{ 3 }\times 5^{ 3 } } =\dfrac { 15\times 125 }{ 8\times 125 } =\dfrac { 1875 }{ 1000 } =1.875\\ (ii)\dfrac { 31 }{ 20 } =\dfrac { 31 }{ 2\times 2\times 5 } =\dfrac { 31\times 5 }{ 2\times 2\times 5\times 5 } =\dfrac { 155 }{ 100 } =1.55$

In both the examples, the result is a terminating decimal because terminating decimal is a decimal which can be expressed in a finite number of figures.

Hence, a rational number can be expressed as a terminating decimal if the denominator has factors

$2$ or

$5$ .

One and only one out of

$n, n + 4, n + 8, n + 12\ and \ n + 16$ is ......(where n is any positive integer)

0%
Divisible by 5
0%
Divisible by 4
0%
Divisible by 10
0%
Divisible by 12

Explanation

We know that any positive integer is of the form 5q, 5q + 1 or 5q + 2, 5q + 3 or 5q + 4 for some integer q and one and onlyone of these possibilities can occur. So, we have the following cases:
Case-I: When

$n=5q$
In this case, we have

$n=5q$ , which is divisible by 5
Now,

$n=5q$

$\Rightarrow n+4=5q+4$

$\Rightarrow n+4$ leaves remainder 4 when divided by 5

$\Rightarrow n+4$ is not divisible by 5.
Now

$n+8=5q+8=5(q+1)+3=5m+3$ , m is an integer.
Clearly, n+8 is not divisible by 5.
Again,

$n+12=5q+12=5(q+2)+2=5m+2$ , m in an integer.
Clearly n+12 is not divisible by 5.
Now

$n+16=5q+16=5(q+13)+1=5m+1$ , m is an integer

$\Rightarrow n+16$ is not divisible by 5
Thus, if n = 5q only one out of n, n + 4, n + 8, n +
12 and n + 16 is divlsible by 5,
Similarly, this result can be proved for the rest of .
the cases.

State TRUE or FALSE

$\displaystyle\frac{(2+\sqrt{2})(3-\sqrt{5})}{(3+\sqrt{5})(2-\sqrt{2})}$ is Rational.

0%
True
0%
False

Explanation

$\displaystyle \frac { (2+\sqrt { 2 } )(3-\sqrt { 5 } ) }{ (3+\sqrt { 5 } )(2-\sqrt { 2 } ) } =\frac { { (2+\sqrt { 2 } ) }^{ 2 }{ (3-\sqrt { 5 } ) }^{ 2 } }{ (9-5)(4-2) } =\frac { (4+2+4\sqrt { 2 } )(9+5-6\sqrt { 5 } ) }{ 8 } \\$

$\displaystyle =\frac { (6+4\sqrt { 2 } )(14-6\sqrt { 5 } ) }{ 8 } =\frac { (3+2\sqrt { 2 } )(7-3\sqrt { 5 } ) }{ 2 } =\frac { (21-9\sqrt { 5 } +14\sqrt { 2 } -6\sqrt { 10 } ) }{ 2 } \\ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebraic\quad equation\quad in\quad numerator\quad \\ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\\$
The given statement is false.

The square of any positive odd integer for some integer

$m$ is of the form

0%
$7m+1$
0%
$8m+1$
0%
$8m+3$
0%
$7m+2$

Explanation

We know that any positive odd integer

$n$ is of the form

$4q + 1$ or

$4q + 3$ where

$q$ is some integer.

When,

$n=4q+1$

$n^2=(4q+1)^2$

$=16q^2+8q+1$

$=8q(2q+1)+1$

$=8m+1$ Where

$m=q(2q+1)$

$=>n^2$ is in the form

$8m+1$

When

$n=4q+3$

$n^2=(4q+3)^2$

$=16q^2+24q+9$

$=16q^2+24q+8+1$

$=8(2q^2+3q+1)+1$

$=8m+1$ Where

$m=(2q^2+3q+1)$

$=>n^2$ is in the form

$8m+1$

Hence,

$n^2$ is in the form

$8m+1$ if

$n$ is an odd positive integer.

If any positive even integer is of the form

$4q$ or

$4q + 2$ , then

$q$ belongs to:

0%
whole number
0%
irrational number
0%
Both A & B
0%
none of these

Explanation

Let

$a$ be any positive even integer and

$b=4$ .Then by division algorithm,

$a=4q+r$ for some integer

$q \ge 0$ and

$r=0,1,2,3$
So,

$a=4q$ or,

$4q+1$ ,

$4q+2$

$4q+3$
Because

$0 \ge r \ge 4$
Now,

$4q$ i.e

$2(2q)$ is an even number

$\therefore$

$4q+1$ is an odd number

$4q+2$ i.e.

$2(2q+1)$ is an even number

$\therefore (4q+2)+1=4q+3$ is an odd number
Thus, We can say that any positive even integer can be written as in the form of

$4q, 4q+2$ where

$q$ is the whole number

The number

$\displaystyle\frac{3-\sqrt{3}}{3+\sqrt{3}}$ is

0%
Rational
0%
Irrational
0%
Both
0%
Can't say

Explanation

$Here,\quad we\quad will\quad carry\quad out\quad rationalization.\quad \\ \frac { 3-\sqrt { 3 } }{ 3+\sqrt { 3 } } =\frac { 3-\sqrt { 3 } }{ 3+\sqrt { 3 } } x\frac { 3-\sqrt { 3 } }{ 3-\sqrt { 3 } } =\frac { { (3-\sqrt { 3 } ) }^{ 2 } }{ (3+\sqrt { 3) } (3-\sqrt { 3 } ) } =\frac { 9+3-6\sqrt { 3 } }{ 9-3 } =\frac { 12-6\sqrt { 3 } }{ 6 } =\frac { 2-\sqrt { 3 } }{ 1 } \\ Since\quad \sqrt { 3 } is\quad irrational\quad number\quad and\quad subtraction\quad of\quad rational\quad and\quad irrational\quad is\quad irrational.\\ The\quad given\quad expression\quad is\quad irrational.\\ \quad$

Using Euclid's division algorithm, find H.C.F.of

$56, 96$ and

$404.$

0%
$8$
0%
$16$
0%
$12$
0%
$4$

Explanation

Let us begin by choosing any two number out of any three number.
Say

$56$ and

$96$
As

$96>56$ , by applying Euclid's division lemma to

$56$ and

$96$ we have,

$96=56\times 1+40$
Since remainder

$40 \neq 0.$

So,applying Euclid's division lemma to

$56$ and

$40$ we have,

$56=40\times 1+16$
Since remainder

$16 \neq 0$

So,applying Euclid's division lemma to

$40$ and

$16$ we have,

$40=16\times 2+8$
Since remainder

$8 \neq 0$

So, applying Euclid's division lemma to

$16$ and

$8$ we have,

$16=8\times 2+0$
Since remainder is zero. Hence,divisor

$8$ is the H.C.F of

$56$ and

$96.$
Now,
Again, applying Euclid's division lemma on the H.C.F of the two number and remaining number.
Since, the H.C.F of

$56$ and

$96$ is

$8$ and the remaining number is

$404.$

So, by applying Euclid's division lemma on

$8$ and

$404$ we have,

$404=8\times 50+4$
Since remainder

$4 \neq 0$ So,applying Euclid's division lemma to

$8$ and

$4$ we have,

$8=4\times 2+0$
Hence, remainder is zero.
Hence, remainder

$4$ is the H.C.F of

$8$ and

$404$
Hence, H.C.F. of

$404, 96$ and

$56$ is

$4.$

If the square of an odd positive integer can be of the form

$6q + 1$ or

$6q + 3$ for some

$q$ then q belongs to:

0%
integers
0%
irrational
0%
real
0%
none of these

Consider the following statements :
1.

$\displaystyle \frac{1}{22}$ can not be written as terminating decimal

$\displaystyle \frac{2}{15}$ can be written as a terminating decimal

$\displaystyle \frac{1}{16}$ can be written as a terminating decimal

Which of the statements given above is/are correct ?

0%
$1$ only
0%
$3$ only
0%
$1$ and $3$
0%
$2$ and $3$

Explanation

$\displaystyle \frac{1}{22} = 0.04545454545$ is not a terminating decimal.

$\displaystyle \frac{2}{15} = 0.133333333$ is not a terminating decimal.

$\displaystyle \frac{1}{16} = 0.0625$ is a terminating decimal.

Hence, statement

$1$ and

$3$ are correct.

What is the H.C.F. of two co-prime numbers ?

0%
$1$
0%
$0$
0%
$2$
0%
none of these

Explanation

${\textbf{Step - 1: Find the required answer using definition of co-prime numbers.}}$

${\text{By definition of co-prime numbers,}}$

${\text{Two numbers having only 1 as their common factor are called co-prime numbers.}}$

${\text{So, HCF of two co-prime numbers is 1.}}$

${\textbf{Hence, the correct option is A.}}$

If the H.C.F. of

$A$ and

$B$ is

$24$ and that of

$C$ and

$D$ is

$56,$ then the H.C.F. of

$A, B, C$ and

$D$ is

0%
$4$
0%
$12$
0%
$8$
0%
$3$

Explanation

Given the H.C.F. of

$A$ and

$B$ is

$24$ and that of

$C$ and

$D$ is

$56.$
Then the H.C.F. of

$A, B, C$ and

$D$ is the HCF of

$24$ and

$56$ which is

$8.$

The HCF of 136, 170 and 255 is:

0%
$13$
0%
$15$
0%
$17$
0%
$1$

Explanation

Given numbers are

$136,\ 170\text{ and }255$ .

Representing the given numbers as the product of their prime factors:

$136=2\times 2\times 2\times 17=2^3\times 17$

$170=2\times 5\times 17$

$255=3\times 5\times 17$

We can see that

$17$ is the only factor common between the three numbers.

Hence, HCF of

$136,\ 170\text { and }255$ is

$17$

Apply Euclid"s theorem for

$17, 5$ .

0%
$17=5\times 3+2$
0%
$17=5\times 2+7$
0%
$17=5\times 4-3$
0%
None of the above

Explanation

According to the definition of Euclid's theorem,

$a = b \times q + r$ where

$0 \leq r < b .$

so,

$17=5\times 3+2$

The HCF of

$135$ and

$225$ using Euclid algorithm is

0%
$41$
0%
$42$
0%
$43$
0%
$45$

Explanation

According to the definition of Euclid's theorem,

$a = b \times q + r$ where

$0 \leq r < b .$

$225=135\times 1+90$ [by Euclid"s algorithm]

$135=90\times 1+45$

$90=45\times 2+0$

$\therefore$ HCF of

$135$ and

$225$ is

$45$ .

So, option

$D$ is correct.

Find the HCF of

$867$ and

$255$ using Euclid theorem.

0%
$50$
0%
$51$
0%
$52$
0%
$53$

Explanation

The given numbers are

$867$ and

$255$

According to the definition of Euclid's theorem,

$a = b \times q + r$ where

$0 \leq r < b .$

$867=255\times 3+102$

$255=102\times 2+51$

$102=51\times 2+0$

The HCF of

$867$ and

$255$ is

$51$ .

So, option

$B$ is correct.

The HCF of 75 and 180 is equal to

0%
$11$
0%
$12$
0%
$13$
0%
$15$

Explanation

The given numbers are

$75$ and

$180$ .

To find out: The HCF of the given numbers.

We can represent the given numbers as the product of their prime factors as:

$75=3\times 5\times 5$

$180=2\times 2\times 3\times 3\times 5$

We know that, the HCF of two numbers is the product of the factors common between them.

Here,

$3$ and

$5$ are the factors common between

$75$ and

$180$ .

So, HCF

$=3\times 5$

$=15$

Hence, the HCF of

$75$ and

$180$ is

$15$ .

If we apply Euclid"s division algorithm for

$20$ using

$8$ then the correct answer will be:

0%
$20=8\times 2+4$
0%
$20=8\times 4+0$
0%
$20=8\times 3+3$
0%
none of the above

Explanation

Here, Euclid"s division algorithm is used to represent

$20$ using

$8.$

The only option in which L.H.S

$=$ R.H.S is option A.

i.e.

$20 = 2 \times 8 + 4$

$=16+4=20$

So, option A is correct.

The HCF of

$196$ and

$38416$ using Euclid algorithm is

0%
$191$
0%
$192$
0%
$193$
0%
$196$

Explanation

Let us first state Euclid’s division algorithm:

To obtain the HCF of two positive integers, say

$c$ and

$d$ , with

$c > d$ , follow the steps below:

Step 1 : Apply Euclid’s division lemma, to

$c$ and

$d$ . So, we find whole numbers,

$q$ and

$r$ such that

$c = dq + r, 0 ≤ r < d$ .

Step 2 : If

$r = 0$ ,

$d$ is the HCF of

$c$ and

$d$ . If

$r ≠ 0$ , apply the division lemma to

$d$ and

$r$ .

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Now, let us apply the Euclid’s division algorithm to the given numbers

$196$ and

$38416$ as follows:

$38416=196\times 196+0$

Since the remainder is

$0$ .

Hence, the HCF of

$196$ and

$38416$ is

$196$ .

$\displaystyle d=\frac { 1 }{ { 2 }^{ 3 }\times { 5 }^{ 7 } }$ is expressed as a terminating decimal, how many non zero digits will

$d$ have?

0%
One
0%
Two
0%
Three
0%
Seven
0%
Ten

Explanation

Given,

$d=\cfrac{1}{2^3\times5^7 }$ .

Multiply and divide by

${2^4}$ ,

$\Rightarrow d=\cfrac{2^4}{2^3 \times 5^7 \times 2^4}$

$\cfrac{16}{2^7 \times 5^7 }$

$\cfrac{16}{10^7 }$

$0.0000016$

Hence,

$d$ will have two non-zero digits,

$1$ and

$6$ , when expressed as a decimal.

Therefore, option

$B$ is the correct answer.

The LCM of 54 90 and a third number is 1890 and their HCF is 18 The third number is

0%
36
0%
180
0%
126
0%
108

Explanation

Given the LCM two numbers 54, 90 and third number is 1890 and HCF is 18

Let the number is 18x because one factor is also 18 the common factor HCF

Then factor 54,90 ,18 =

$18\times 3,18\times 5,18\times 18\times x$

$\therefore 18\times 3\times 5\times x=1890\Rightarrow 270x=1890\Rightarrow x=7$

Then third number is

$18\times 7=126$

The the HCF of

$248$ and

$492$ is equal to

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

Explanation

$248=2^3 \times 31$

$492=2^2 \times 3 \times 41$

HCF is

$4.$

For finding the greatest common divisor of two given integers. A method based on the division algorithm is used called ............

0%
Euclid's division algorithm
0%
Euclid's addition algorithm
0%
Euclid's subtraction algorithm
0%
Euclid's multiplication algorithm

Explanation

Euclid's division algorithm is a way to find the HCF of two numbers by using Euclid's division lemma. It states that if there are any two integers

$a$ and

$b$ , there exists

$q$ and

$r$ such that it satisfies the given condition

$a = bq + r$ where

$0 ≤ r < b$ ..
Therefore,

$A$ is the correct answer.

Which of the following is terminating decimal?

$\cfrac{23}{90}, \cfrac{111}{148}, \cfrac{29}{145}, \cfrac{1}{6}$

0%
$\cfrac{111}{148}$
0%
$\cfrac{23}{90}$
0%
$\cfrac{29}{145}$
0%
$\cfrac{1}{6}$

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.

In fully reduced form, the denominator in

$A$ is

$4$ ,composed of only

$2$ 's,

$(4 = 2^2)$ .

In fully reduced form, the denominator in

$C$ is

$5$ , composed of only

$5$ 's,

$(5 = 5^1)$ .

Hence, they have terminating decimals

$0.75$ and

$0.2$ , respectively.

By contrast, the denominator in

$B$ has prime factors other than

$2$ 's and

$5$ 's

$(90 =5\times2\times\times 3^2)$ ,

and the fraction in

$D$ has the denominator

$6$ , composed of prime factors other than

$2$ 's and

$5$ 's

$(6 =2\times3)$ .

Therefore, options

$A$ and

$C$ are the correct answers.

If the denominator of a fraction has only factors of

$2$ and factors of

$5$ , the decimal expression .............

0%
has equal numerator and denominator
0%
becomes a whole number
0%
does not terminate
0%
terminates

Explanation

When the prime factorization of the denominator of a fraction has only factors of

$2$ and factors of

$5$ , we can always express the decimal as terminating decimal. Reason is we can always multiply the denominator with required factors and make it to the power of

$10$ which will always result in a terminating decimal.
For example,

$\dfrac {1}{25} = \dfrac {4}{25\times 4}=\dfrac {4}{100}=0.04$
Therefore,

$D$ is the correct answer.

$\sqrt {5}$ is an ......... number.

0%
rational
0%
whole
0%
integer
0%
irrational

Explanation

$\sqrt {5} = \dfrac {a}{b}$

$b\sqrt {5} = a$

$(a$ and

$b$ are co-prime i.e. they have no common factors

$) ...(1)$

$5b^{2} = a^{2}$ (squaring both sides)
Therefore

$5$ divides

$a^{2}$
As per Fundamental Theorem of Arithmetic,

$5$ divides

$a.$
Let's take it as

$a = 5c,$

$5b^{2} = 25 c^{2}$

$b^{2} = 5c^{2}$
As per Fundamental Theorem of Arithmetic,

$5$ divides

$a.$
So

$a$ and

$b$ have

$5$ as a common factor but

$a$ and

$b$ have only

$1$ common factor

$1$ from equation

$(1),$ so it is not rational.
So, we conclude that

$\sqrt {5}$ is irrational.
Therefore,

$D$ is the correct answer.

............. states that for any two positive integers

$a$ and

$b$ we can find two whole numbers

$q$ and

$r$ such that

$a = b \times q + r$ where

$0 \leq r < b .$

0%
Euclid's addition lemma
0%
Euclid's subtraction lemma
0%
Euclid's multiplication lemma
0%
Euclid's division lemma

Explanation

Let the two positive integers be

$a$ , a dividend and

$b$ , a divisor.

We can find two whole numbers,

$q$ and

$r$ such that

$q$ is the quotient and

$r$ is the remainder.

So,

$a = b \times q + r$ where

$0 \leq r < b.$

This is Euclid's Division Lemma.

Hence, option D is correct.

The H.C.F. of the numbers

$16.5, 0.90$ and

$15$ is

0%
$16.5$
0%
$0.90$
0%
$15$
0%
$0.3$

Explanation

The given numbers are

$16.5, 0.90$ and

$15$

Multiplying each number by

$100$ we get

$1650,90$ and

$1500$

Let us consider the numbers

$1650,90$ and

$1500$

By prime factorization, we get,

$1650 = 2\times 3\times 5^{2}\times 11$

$90 = 2\times 3^{2} \times 5$

$1500 = 2^{2} \times 3\times 5^{3}$

H.C.F. of

$1650, 90$ and

$1500$ is

$2\times 3\times 5 = 30$

Therefore, H.C.F. of

$16.5, 0.90$ and

$15$ will be

$\dfrac{30}{100}$

$=0.30$

So, option

$D$ is correct.

CBSE Questions for Class 10 Maths Real Numbers Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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