Rational or Irrational Numbers MCQ Quiz - Objective Question with Answer for Rational or Irrational Numbers - Download Free PDF

Given:

If p and q are perfect squares, then √(p/q) is always a rational number.

Formula used:

If p = a² and q = b², where a and b are integers, then:

√(p/q) = √(a²/b²) = a/b

Calculation:

Let p = 16 (4²) and q = 9 (3²):

√(p/q) = √(16/9)

⇒ √(16/9) = √(4²/3²)

⇒ √(4²/3²) = 4/3

Since 4/3 is rational, the statement holds true.

∴ The correct answer is option 3).

Given:

Find Greatest among \(\sqrt[3]{19}, \sqrt[6]{13}, \sqrt[4]{17}\)\(\sqrt[3]{15}\)

Calculation:

\(\sqrt[3]{19}, \sqrt[6]{13}, \sqrt[4]{17}\) \(\sqrt[3]{15}\)

Take LCM of  powers i.e. 3, 6, 4, and 3 

∴ LCM = 12 

So, 19⁴, 13^2, 17³ and 15⁴ 

⇒ 194 >154 > 173 > 132

The Greatest among is \(\sqrt[3]{19}\)

Calculation:

Converting all the fraction to the decimal form,

5/8 = 0.625

9/10 = 0.9 

Checking the option which is not in this range

7/9 = 0.78

11/20 = 0.55

16/21 = 0.76

9/11 = 0.81

From the options 0.55 is not in the range of 0.625 and 0.9

∴ 11/20 does not lie between 5/8 and 9/11

Given :

Rationalise the expression \(\dfrac{4a^2}{\sqrt{4a^2+b^2}+b}\)

Calculation :

\(\dfrac{4a^2}{\sqrt{4a^2+b^2}+b}\)

⇒  \(\dfrac{4a^2\times (\sqrt{4a^2+b^2}-b)}{(\sqrt{4a^2+b^2}+b)\times (\sqrt{4a^2+b^2}-b }\)

⇒  \(\dfrac{4a^2\times (\sqrt{4a^2+b^2}-b)}{4a^2+b^2-b^2 }\)

⇒ \(\dfrac{4a^2\times (\sqrt{4a^2+b^2}-b)}{4a^2}\)

⇒ \(\sqrt{4a^2+b^2}-b\)

∴ The answer is \(\sqrt{4a^2+b^2}-b\) .

Given:

\(\frac{1 + \sqrt{3}}{1 - \sqrt{3}} + \frac{1 - \sqrt{3}}{1 + \sqrt{3}}\)

Formula Used:

To simplify the expression, rationalize both terms.

Calculation:

First, simplify the first fraction:

\(\frac{1 + \sqrt{3}}{1 - \sqrt{3}}\)  by multiplying numerator and denominator by \((1 + \sqrt{3})\)

⇒ \(\frac{(1 + \sqrt{3})^2}{(1 - \sqrt{3})(1 + \sqrt{3})}\)

⇒ \(\frac{1 + 2\sqrt{3} + 3}{1^2 - (\sqrt{3})^2}\)

⇒ \(\frac{4 + 2\sqrt{3}}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}\)

Next, simplify the second fraction:

\(\frac{1 - \sqrt{3}}{1 + \sqrt{3}}\) by multiplying numerator and denominator by \((1 - \sqrt{3})\)

⇒ \(\frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})}\)

⇒ \(\frac{1 - 2\sqrt{3} + 3}{1^2 - (\sqrt{3})^2}\)

⇒ \(\frac{4 - 2\sqrt{3}}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3}\)

Now, add the two simplified results:

⇒ \((-2 - \sqrt{3}) + (-2 + \sqrt{3})\)

⇒\( -2 - \sqrt{3} - 2 + \sqrt{3}\)

⇒ -4

The value of the expression is -4.

Given:

\(0.4\overline6-0.5\overline{89} +0.3\overline{33}\)

Concept used:

0.ab̅  = (ab - a)/90 

0.ab̅c̅ = (abc - a)/990

Calculation:

 \(0.4\overline6-0.5\overline{89} +0.3\overline{33}\)

⇒ (46 - 4)/90 - (589 - 5)/990 + (333 - 3)/990

⇒ 42/90 - 584/990 + 330/990

⇒ 42/90 - 254/990 

⇒ (462 - 254)/990

⇒ 208/990 

According to this formula

0.ab̅c̅ = (abc - a)/990

\(0.2\overline{10}\) = (210 - 2)/990

∴ The value of \(0.4\overline6-0.5\overline{89} +0.3\overline{33}\) is equal to \(0.2\overline{10}\).

Given:

0.135135....

Concept used:

The numbers of the form (p/q), where q ≠ 0 and p and q is integer is known as rational number.

Calculation:

Let x = 0.135135....      ----(1)

Multiply equation (1) by 1000, we have

1000x = 135.135....      ----(2)

Subtract equation (1) from equation (2), we have

1000x - x = (135.135...) - (0.135135....)

⇒ 999x = 135

⇒ x = 135/999

⇒ x = 45/333

⇒ x = 5/37

∴ The 0.135135.... can be written as 5/37 in the form of p/q.

Concept Used:-

Irrational numbers are those numbers which can not be written in the form of p/q. Where p and q are integer and q is not equal to zero.

Key Points

•  Sum or difference of two irrational numbers can be rational or irrational.
• Product or division of two irrational numbers can be rational or irrational.

Explanation:-

Suppose there are two irrational numbers \(\sqrt{3}\) and \(-\sqrt{3}\). The sum of these two numbers is 0.

\(\sqrt{3}+(-\sqrt{3})=0\)

Here, 0 is the rational number. So, the sum of two irrational number is a rational number.

Now let there are two irrational numbers \(\sqrt{3}\) and \(\sqrt{3}\). The sum of these two numbers is,

\(\sqrt{3}+\sqrt{3}=2\sqrt{3}\)

Here, \(2\sqrt{3}\) is an irrational number. So, the sum of two irrational number is an irrational number.

So, the sum of two irrational numbers may be a rational or an irrational number.

Now, we know that real number is the number which can be both rational or irrational number. So we can say that the sum of two irrational number is always a real number.

Thus, the correct option is 3.  

105/112 = (15 × 7) / (16 × 7) = 15/16

Concept used:

\(.BC\overline{DEF}\) = \(\frac{BCDEF - BC}{99900}\)

Calculation:

\(.45\overline{235}\)

⇒ \(\frac{45235-45}{99900}\)

⇒ \(\frac{45190}{99900}\)

⇒ \(\frac{4519}{9990}\)

∴ The correct answer is \(\frac{4519}{9990}\).

Formula used:

If we have number in this form say \(0.a \bar b\) 

Then, \(0.a \bar b\) = \(ab - a\over 90\) 

In this  the digit without bar is subtracted from the number 

Now, this fraction is in the form of p/q 

Calculation: 

Here, we have \(0.2\bar7\) 

As, here we have bar on only one digit 

Also, 2 is without bar so it will get subtracted from 27 in numerator 

So, \(0.2\bar7\) = \(27 -2\over90\) = 25/90 = 5/18 

Now, 5/18 is is in the form of p/q

Hence, it can be expressed in the form of p/q is 5/18 .

Calculation:

Rational number - A number which is in the form of p/q 

According to the given option

⇒ \(\sqrt {{3^2} + {4^2}} \) = √25 = 5 is a rational number

⇒ \( √ {12.96} \) = 3.6 is a rational number

⇒ √125 = 5√5 is not a rational number

⇒ √900 = 30 is a rational number

∴ √125 is not a rational number

Given:

\(0.3\overline {35} \)

Calculation:

Let x = \(0.3\overline {35} \)          → (1)

As two numbers are repeated, we'll multiply both sides by 100.

⇒ 100x = 33.535

Subtracting (1) from this, we get

⇒ 100x – x = 33.535 – 0.335

⇒ 99x = 33.200

⇒ x = \(\frac{33.2}{99}\) = \(\frac{332}{990}\)

Therefore, the fractional representation of \(0.3\overline {35} \) is \(\frac{332}{990}\).

⇒ 11025 = 5² × 21²

⇒ 6025 = 5² × 241

⇒ 9025 = 5² × 19²

⇒ 3025 = 5² × 11²

Given

√ 5 and  √7

Concept

Rational numbers are those numbers which are either terminating, non terminating or recurring.

Calculation

√5 = 2.236 and √7 = 2.64

rational number lies between the 2.33... and 2.64...

so, only \(2{2\over5}\) is the number which lies between 2.236 and 2.64

∴ \(2{2\over5}\) is a rational number between \(\sqrt{5}\) and \(\sqrt{7}\)