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

Calculation:

\(\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} - \sqrt {1 - x} }}\) × \(\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}\)

⇒ \((1 + x) + (1 – x) + 2√(1 – x^2) \over(1 + x) – (1 – x)\)

⇒ \(2 + 2√(1 – (√3/2)^2)\over2x\)

⇒ \(2 + 2 × (1/2)\over 2 × (√3/2) \)

⇒ (3/√3) × (√3/√3)  

⇒ √3  

Given:

The repeating decimal is 8.27272...

Formula used:

Let x = repeating decimal, then manipulate the equation to eliminate the repeating part.

Calculation:

Let x = 8.27272...

⇒ 100x = 827.27272... (Multiply by 100 to shift the repeating part)

⇒ 100x - x = 827.27272... - 8.27272... (Subtract to remove the repeating part)

⇒ 99x = 819

⇒ x = 819 / 99

⇒ x = 91 / 11

∴ The correct answer is option (4).

Given:

2x + y = 256 

4x - y = 16

Formula used:

(Xm)n = Xmn

Calculation:

2x + y = (16)² 

2x + y = 2⁸

^{x + y = 8  ---- (1)}

4x - y = 16

2^{2x - 2y} = 2⁴

2x - 2y = 4

x - y = 2  ----(2)

By adding  (1) & (2)

we get, x = 5 

Answer is 5.

Additional Information

Xm × Xn  = X m+n

Xm ÷ Xn  = X m-n

(Xm)n = Xmn

Given :

x = \(\frac{ 5 ^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2}{2(2 +√3) - (2√3+ 3)}\)

Formula used :

1² + 2² + 3² + ........ + n² = \(\frac{n (n + 1) (2n + 1)}{6}\)  ----- (1)

Calculations :

To find the value of 52 + 62 + 72 + 82 + 92 + 102 

We can apply 

Using equation (1), we get [12 + 22 + 32 + 42 + ........ + 102] - [12 + 22 + 32 + 42]

So, 12 + 22 + 32 + 42 + ........ + 10² =\(\frac{10 (10 + 1) (20 + 1)}{6}\)

⇒ 385       ---- (2)

⇒ 12 + 22 + 32 + 4² = \(\frac{4 (4 + 1) (8 + 1)}{6}\)

⇒ 30       ----- (3)

Equation (1) can be written as

⇒ x =  \(\frac{ 5 ^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2}{2(2 +√3) - (2√3+ 3)}\)

⇒ x =  \(\frac{(1^2+ 2^2+ 3^2+...... + 10^2)- (1^2+ 2^2+ 3^2+ 4^2)}{2(2 +√3) - (2+ √3)}\)

Using equation (2) and (3) we get

\(x = 385- 30/[2(2 +√3) - (2+ √3)]\)

⇒ 355

Now, we have to find \(√{(x/5 + 10)}\)

Putting the value of x, we get

⇒ \(√{(355/5 + 10)}\)

⇒ \(√({71 + 10})\)

⇒ √81 = 9

∴ The value of square root of (x/5 + 10) is 9.

Given:

\(2.\overline{32} + 3.\overline{57} - 4.\overline{63} \)

Concept used:

In the numerator, take the difference between the number formed by all the digits after decimal points (repeated digits will be taken only once) and the number formed by non repeating digits. In the denominator, place as many nines as there are repeating digits and after nine put as one zeros as the number of non repeating digits.

Calculations:

\(2.\overline{32} + 3.\overline{57} - 4.\overline{63} \)

= (2 + \(\dfrac{32 - 0}{99}\)) + (3 + \(\dfrac{57 - 0}{99}\)) - (4 + \(\dfrac{63 - 0}{99}\))

= (\(2\dfrac{32}{99}\)) + (\(3\dfrac{57}{99}\)) - (\(4\dfrac{63}{99}\))

Now, 2 + 3 - 4 + (\(\dfrac{32}{99}\) + \(\dfrac{57}{99}\) - \(\dfrac{63}{99}\))

= 1 + \(\dfrac{32 + 57 - 63}{99}\)

= 1 + \(\dfrac{26}{99}\)

= \(\dfrac{1 × 99 + 26}{99}\)

= \(\dfrac{125}{99}\)

∴ The answer is \(\dfrac{125}{99}\).

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.  

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 .

105/112 = (15 × 7) / (16 × 7) = 15/16

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 \(\frac{332}{990}\).

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 \(\sqrt{7}\)

⇒ 11025 = 5² × 21²

⇒ 6025 = 5² × 241

⇒ 9025 = 5² × 19²

⇒ 3025 = 5² × 11²