The World of Numbers — CBSE Class 9 Maths Important Questions

Answer: (c) $\sqrt{7}$√7.

$\sqrt{16}=4$√16=4 is a whole number, $0.\overline{3}=\tfrac13$0.3=13 is rational, and $\tfrac{22}{7}$227 is a ratio of integers. Only $\sqrt7$7 is not a perfect square, giving a non-terminating, non-recurring decimal — it is irrational.

Answer: (b) terminating or recurring.

Every rational number $\tfrac{p}{q}$pq has a decimal expansion that either terminates or repeats a block of digits. Non-terminating, non-recurring decimals belong to irrational numbers.

Answer: (b) $\sqrt5 + 2$5 + 2.

Multiply top and bottom by the conjugate $\sqrt5+2$5+2: $\dfrac{1}{\sqrt5-2}\times\dfrac{\sqrt5+2}{\sqrt5+2} = \dfrac{\sqrt5+2}{(\sqrt5)^2-2^2} = \dfrac{\sqrt5+2}{5-4} = \sqrt5+2.$1/5-2×5+2/5+2 = 5+2/(5)^2-2^2 = 5+2/5-4 = 5+2.

Answer: (b) $4$4.

Write $32 = 2^5$32 = 2^5. Then $(2^5)^{2/5} = 2^{5\times \frac{2}{5}} = 2^2 = 4.$(2^5)^2/5 = 2^5× 2/5 = 2^2 = 4.

Answer: (c) A is true but R is false. $\sqrt2 + (-\sqrt2) = 0$2 + (-2) = 0, which is rational, so A is true. But the sum of two irrationals is not always irrational (this example itself gives a rational $0$0), so R is false.

Convert to a common denominator: $\dfrac13 = \dfrac{4}{12}$13 = 4/12 and $\dfrac12 = \dfrac{6}{12}$12 = 6/12. A number between them is $\dfrac{5}{12}$5/12.

For a second one, use larger equivalents: $\dfrac13 = \dfrac{40}{120}$13 = 40/120, $\dfrac12 = \dfrac{60}{120}$12 = 60/120, so $\dfrac{50}{120} = \dfrac{5}{12}$50/120 = 5/12... choose instead $\dfrac{9}{24}$9/24 where $\dfrac13=\dfrac{8}{24}$13=8/24, $\dfrac12=\dfrac{12}{24}$12=12/24. Thus $\dfrac{5}{12}$5/12 and $\dfrac{9}{24}=\dfrac{3}{8}$9/24=3/8 both lie between them.

Express each surd in terms of $\sqrt5$5:

$\sqrt{45} = \sqrt{9\times5} = 3\sqrt5$√45 = √9×5 = 35, and $\sqrt{20} = \sqrt{4\times5} = 2\sqrt5$√20 = √4×5 = 25.

So the expression $= 3\sqrt5 - 3(2\sqrt5) + 4\sqrt5 = 3\sqrt5 - 6\sqrt5 + 4\sqrt5 = \sqrt5.$= 35 - 3(25) + 45 = 35 - 65 + 45 = 5.

Let $x = 0.\overline{36} = 0.363636\ldots$x = 0.36 = 0.363636

Since two digits repeat, multiply by $100$100: $100x = 36.3636\ldots$100x = 36.3636

Subtract: $100x - x = 36.3636\ldots - 0.3636\ldots$100x - x = 36.3636 - 0.3636, so $99x = 36$99x = 36.

Thus $x = \dfrac{36}{99} = \dfrac{4}{11}$x = 36/99 = 4/11 (dividing top and bottom by $9$9).

Hence $0.\overline{36} = \dfrac{4}{11}$0.36 = 4/11.

Multiply numerator and denominator by the conjugate $\sqrt7 + \sqrt3$7 + 3:

$\dfrac{3}{\sqrt7-\sqrt3}\times\dfrac{\sqrt7+\sqrt3}{\sqrt7+\sqrt3} = \dfrac{3(\sqrt7+\sqrt3)}{(\sqrt7)^2-(\sqrt3)^2}.$3/7-3×7+3/7+3 = 3(7+3)/(7)^2-(3)^2.

The denominator $= 7 - 3 = 4$= 7 - 3 = 4.

So the value is $\dfrac{3(\sqrt7+\sqrt3)}{4} = \dfrac{3\sqrt7 + 3\sqrt3}{4}.$3(7+3)/4 = 37 + 33/4.

First find $\dfrac{1}{a}$1/a by rationalising:

$\dfrac{1}{3+2\sqrt2}\times\dfrac{3-2\sqrt2}{3-2\sqrt2} = \dfrac{3-2\sqrt2}{3^2-(2\sqrt2)^2} = \dfrac{3-2\sqrt2}{9-8} = 3 - 2\sqrt2.$1/3+22×3-22/3-22 = 3-22/3^2-(22)^2 = 3-22/9-8 = 3 - 22.

Therefore
$a + \dfrac{1}{a} = (3+2\sqrt2) + (3-2\sqrt2) = 6.$a + 1/a = (3+22) + (3-22) = 6.

The irrational parts cancel, leaving the rational value $6$6.

Use the mean-proportional (semicircle) method for $\sqrt{4.5}$√4.5.

Steps:

1. Draw a line and mark $A$A. From $A$A, measure $AB = 4.5$AB = 4.5 units.
2. From $B$B, measure $BC = 1$BC = 1 unit along the same line, so $AC = 4.5 + 1 = 5.5$AC = 4.5 + 1 = 5.5 units.
3. Find the midpoint $O$O of $AC$AC and draw a semicircle with $AC$AC as diameter.
4. At $B$B, draw a perpendicular to $AC$AC meeting the semicircle at $D$D. Then $BD = \sqrt{4.5}$BD = √4.5.
5. With $B$B as centre and radius $BD$BD, draw an arc cutting the number line at $P$P; then $P$P represents $\sqrt{4.5}$√4.5.

Why it works: In a semicircle, the perpendicular from a point on the diameter is the geometric mean of the two segments: $BD = \sqrt{AB\times BC} = \sqrt{4.5\times 1} = \sqrt{4.5}.$BD = √AB× BC = √4.5× 1 = √4.5.

Factor powers of $2$2 in the numerator. Note $2^{n+4} = 2^n\cdot 2^4 = 16\cdot 2^n$2^n+4 = 2^n· 2^4 = 16· 2^n and $2\cdot 2^n = 2\cdot 2^n$2· 2^n = 2· 2^n.

Numerator $= 16\cdot 2^n - 2\cdot 2^n = (16 - 2)\,2^n = 14\cdot 2^n.$= 16· 2^n - 2· 2^n = (16 - 2)\,2^n = 14· 2^n.

Denominator $= 2\cdot 2^{n+3} = 2\cdot 2^n\cdot 2^3 = 2\cdot 8\cdot 2^n = 16\cdot 2^n.$= 2· 2^n+3 = 2· 2^n· 2^3 = 2· 8· 2^n = 16· 2^n.

So the expression $= \dfrac{14\cdot 2^n}{16\cdot 2^n} = \dfrac{14}{16} = \dfrac{7}{8}.$= 14· 2^n/16· 2^n = 14/16 = 7/8.

The factor $2^n$2^n cancels, leaving the constant $\dfrac{7}{8}$7/8.

(i) Side $= \sqrt{50}$= √50 cm. Since $50$50 is not a perfect square, $\sqrt{50}$√50 is irrational.

(ii) $\sqrt{50} = \sqrt{25\times 2} = 5\sqrt2$√50 = √25× 2 = 52 cm.

(iii) $7.07$7.07 is a terminating decimal, equal to $\dfrac{707}{100}$707/100, so it is rational. It is only an approximation of the irrational exact value $5\sqrt2$52.

(iv) Two tiles side by side give length $2\times 5\sqrt2 = 10\sqrt2$2× 52 = 102 cm.