MP Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.5

MP Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational –

1. 2 – √5
2. (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\)
3. \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \)
4. \(\frac{1}{sqrt { 2 }}\)
5. 2π

Solution:

1. 2 – \(\sqrt{5}\) is an irrational number
2. (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\) = 3 is a rational number
3. \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \) = \(\frac{2}{7}\) is a rational number
4. \(\frac{1}{\sqrt { 2 }}\) = is an irrational number
5. 2π is an irrational number.

Question 2.
Simplify each of the following expressions –

1. (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))
2. (3 + \(\sqrt{3}\) ) (3 – \(\sqrt{3}\))
3. (\(\sqrt{5}\) + \(\sqrt{2}\))²
4. (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))

Solution:
1. (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))
= 6 + 3\(\sqrt{2}\) + 2\(\sqrt{3}\) + \(\sqrt{6}\)

2. (3 + \(\sqrt{3}\)) (3 – \(\sqrt{3}\))
= (3)² – (\(\sqrt{3}\) )² = 9 – 3 = 6

3. (\(\sqrt{5}\) + \(\sqrt{2}\) )²
= (\(\sqrt{5}\))² + 2\(\sqrt{5}\) x \(\sqrt{2}\) + (\(\sqrt{2}\))²
= 5 + 2\(\sqrt{10}\) + 2 = 7 + 2\(\sqrt{10}\)

4. (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))
= (\(\sqrt{5}\))² – (\(\sqrt{2}\))² = 5 – 2 = 3

Question 3.
Recall, 7t is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = \(\frac{c}{d}\). This seems to contradiet the fact n is irrational. How will you resolve this contradiction?
Solution:
No contradiction will be there. Whenever we measure any length with any device, we only get an approx, rational value and so cannot realise that either c or d is irrational.

Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Solution:
Steps:

1. Draw a line segment AC = 9.3 cm and extend it to B such that CB – 1 cm.
2. Draw the perpendicular bisector of AS and mark the mid point O of AB.
3. With O as centre and OB as radius, draw a semicircle.
4. At point C, draw perpendicular CD which intersect the semicircle at D such that CD = \(\sqrt{9.3}\).
5. With C as centre and CD as radius, draw an arc which intersect AB produced at E.

Proof:
OA = OB = \(\frac{10.3}{2}\) = 5.15cm
OD = 5.15 cm
OC = OB – BC = 5.15 – 1 = 4.15 cm
CD = \(\sqrt { OD^{ 2 }-OC^{ 2 } } \)
= \(\sqrt{9.3}\) cm

Question 5.
Rationalise the denominators of the following:

1. \(\frac{1}{\sqrt { 7 }}\)
2. \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
3. \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
4. \(\frac{1}{\sqrt{7}-2}\)

Solution:
1. \(\frac{1}{\sqrt { 7 }}\)

2. \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

3. \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

4. \(\frac{1}{\sqrt{7}-2}\)

Laws of Exponents for real numbers:
If a, n and m are natural numbers, then

1. a^m x aⁿ = a^m+n
2. (a^m)ⁿ = a^mn
3. \(\frac { a^{ m } }{ b^{ n } } \) where m > n
4. a^m x b^m – (ab)^m

Example 1.

1. 2³ x 2⁴ = 2³⁺⁴ = 2⁷
2. (3²)³ = 3^2×3 = 3⁶
3. \(\frac { 5^{ 5 } }{ 5^{ 2 } } \) = 5^5-2 = 5³
4. 2³ x 3³ = (2 x 3)³ = 6³

Example 2.
1. 6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)}
Solution:
6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)} = 6^{\(\frac{5}{5}\)} = 6

2. 6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)}
Solution:
6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)} = (6 x 7)^{\(\frac{1}{2}\)} = 42^{\(\frac{1}{2}\)}

3. \(\frac{5^{6 / 7}}{5^{2 / 3}}\)
Solution:

Example 3.
Simplify

Solution:

MP Board Class 9th Maths Solutions