AP 9th Class Maths Chapter 1 Number System Important Questions

These AP 9th Class Maths Important Questions 1st Lesson Number System will help students prepare well for the exams.

AP Board Class 9 Maths 1st Lesson Number System Important Questions

9th Class Maths Number System 2 Marks Important Questions

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has:
i) \(\frac{1}{11}\)
ii) 4\(\frac{1}{8}\)
Solution:
i) \(\frac{1}{11}\)

The decimal expansion is non-terminating repeating.

ii) 4\(\frac{1}{8}\) = \(\frac{4 \times 8+1}{8}\) = \(\frac{32 + 1}{8}\) = \(\frac{33}{8}\)

∴ 4\(\frac{1}{8}\) = 4.125
The decimal expansion is terminating.

Question 2.
Simplify the following by rationalising the denominator \(\frac{6-4 \sqrt{3}}{6+4 \sqrt{3}}\).
Solution:

Question 3.
Simplify :
(125)^2/3 – \(\sqrt{25}\) 5⁰ × \(\left(\frac{1}{225}\right)^{\frac{-1}{2}}\)
Solution:

= 5² – 5 × 15 = 25 – 75 = -50

Question 4.
Find two rational numbers between 3 and 4.
Solution:
\(\frac{3}{1}\) = \(\frac{6}{2}\) = \(\frac{9}{3}\)
\(\frac{4}{1}\) = \(\frac{4}{2}\) = \(\frac{12}{3}\)
(\(\frac{3}{1}\)) = \(\frac{9}{3}\) < \(\frac{10}{3}\) < \(\frac{11}{3}\) < \(\frac{12}{3}\) = (\(\frac{4}{1}\))
Rational numbers between 3 and 4 are \(\frac{10}{3}\), \(\frac{11}{3}\).

Question 5.
Multiply 6√2 with 2√3.
Solution:
6√2 × 2√3
= \(12 \sqrt{2 \times 3}\)
= 12√6

Question 6.
Divide 11√8 by √2.
Solution:
\(\frac{11 \sqrt{8}}{\sqrt{2}}\) = \(\frac{11 \sqrt{4 \times 2}}{\sqrt{2}}\) = \(\frac{11 \sqrt{4} \cdot \sqrt{2}}{x^2}\)
= 11 × 2 = 22

9th Class Maths Number System 4 Marks Important Questions

Question 1.
Express \(3.65 \overline{1}\) in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0.
Solution:
Let x = \(3.65 \overline{1}\)
then x = 3.65111 ….. ________ (1)
Multiplying both sides of (1) by 100 we get
100 x = 365.111 ….. _________ (2)
Multiplying both sides of by 10, we get
1000 x = 3651.1111 ….. ________ (3)
Subtracting (2) from (3), we get
900 x = 3286
x = \(\frac{3286}{900}\)
x = \(\frac{1643}{450}\)
⇒ \(3.65 \overline{1}\) = \(\frac{1643}{450}\)
which is in the form of \(\frac{p}{q}\),
where p = 1643 and q = 450 (≠ 0) are integers.

Question 2.
If x = 3 + 2√2 , then find the value of x + \(\frac{1}{x}\).
Solution:
x = 3 + 2√2 _______ (1)

Adding (1) and (2) we get
x + \(\frac{1}{x}\) = (3 + 2√2) + (3 – 2√2) = 6

Question 3.
If \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) = a + b√6 , find ‘a’ and ‘b’.
Solution:

⇒ 5 – 2√6 = a + b√6
∴ a = 5, b = – 2

Question 4.
Simplify :
i) (3 + √3)²
ii) √3(\(\sqrt{27}\) + √3)
Solution:
i) (a + b)² = a² + 2ab + b²
(3 + √3)² = 3² + 2 × 3 × √3 + (√3)²
= 9 + 6√3 + 3 = 12 + 6√3

ii) √3 (\(\sqrt{27}\) + √3) = √3 × \(\sqrt{27}\) + √3 × √3
= \(\sqrt{81}\) + √9
= 9 + 3
= 12

Question 5.
w = \(\sqrt{\frac{4}{36}}\), x = \(\sqrt[3]{\frac{8}{125}}\), y = \(\sqrt{6+\sqrt{4}}\) and z = \(\sqrt{\sqrt{4}+\sqrt{9}}\)
Based on the above information, answer the following questions:
i) Which of the following are rational numbers?
A) w only
B) w and x only
C) y and z only
D) z only
Answer:
B) w and x only

ii) Which of the following are irrational numbers?
A) w only
B) w and x only
C) y and z only
D) z only
Answer:
C) y and z only

iii) Which of the following can be represented as a terminating decimal ?
A) w
B) x
C) y
D) z
Answer:
B) x

iv) Which of the following can be represented as on a terminating non-recurring decimal ?
A) w
B) x
C) y
D) z
Answer:
A) w

v) Which of the following can be represented as a non-terminating non-recur-ring decimal ?
A) w
B) x
C) y
D) None of these
Answer:
C) y

Question 6.
If √2 = 1.414, √3 = 1.732, then find the value of \(\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}\)
Solution:
Given that √2 = 1.414 and √3 = 1.732

= \(\frac{39.2}{19}\) ≈ 2.06

Question 7.
If x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) = and y = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\), then find the value of x² + y².
Solution:
Given that x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
R.F. of √3 – √2 is √3 + √2

R.F. of √3 + √2 is √3 – √2

∴ x² + y² = (5 + 2√6 )² + (5 – 2√6 )²
= 25 + 20√6 + 24 + 25 – 20√6 + 24
= 98

9th Class Maths Number System 8 Marks Important Questions

Question 1.
Visualise 3.765 on number line.
Solution:
Step 1 : Draw a number line and zoom in the gap between 3 and 4 into 10 equal divisions and name them 3.1, 3.2, 3.3, ….. 3.9, 4

Step 2 : Again repeat the same process by zooming in the gap between 3.7 and 3.8 into 10 equal parts and lable them 3.71, 3.72, etc.

Step 3 : Again repeat the Same process by zooming in the gap between 3.76 and 3.77 into 10 equal divisions and lable them as 3.761, 3.762, 3.763 and 3.770.

Thus we can locate 3.765 as shown above on the number line.
Thus we need to repeat the process n’ times if decimal expansion has n-digits in its decimal part.

Question 2.
Visualise \(4 . \overline{26}\) on number line up to 4 decimal places.
Solution:
We are asked to visualise \(4 . \overline{26}\) up to 4 decimal plaes, it means 4.2626.
Step 1 : Given 4.2626 is more than 4 and less than 5.
So, we zoom in the gap between 4 and 5 into 10 equal divisions which will be 4.1, 4.2, 4.3,……. as shown below.

Step 2 : 4.26 is greater than 4.2 and less than 4.3. So, we zoom in the gap between 4.2 and 4.3 into 10 equal parts and name them 4.21 and 4.22, 4.23, 4.24,

Step 3 : And 4.262 is greater than 4.26 and less than 4.27. So, we zoom in the gap between 4.26 and 4.27 into 10 equal divisions and lable them 4.261, 4.262,………
4.269, 4.27 as shown below.

Step 4 : It is clear 4.2626 is greater than 4.262 and less than 4.263. So, we zoom in the gap between 4.262 and 4.263 into 10 equal divisions and lable them 4.2621, 4.2622, 4.2623,……….
, 4.2621 4.2623 4.2625 4.2627 .

Thus we can locate 4.2626 as shown above on number line.

Question 3.
Write operations on real numbers :
Answer:

1. Rational number + Rational number = Rational number
   It means sum of two rational numbers is always a rational number.
2. Difference of two rational numbers is also a rational number.
3. Product of two rational numbers is always a rational number.
4. Division of two rational numbers is always a rational number.
5. Sum of two irrational numbers is rational or irrational.
6. Difference of two irrational numbers is not always an irrational number.
7. Product of two irrational numbers is not always an irrational may be rational also.
8. Division of two irrational numbers is not always an irrational It may be either rational or irrational.

Question 4.
Write rational, irrational combine operations :
Answer:

1. Sum of a rational number and an irrational number is always irrational.
2. Difference of a rational and an irrational number is always an irrational.
3. The product of a (non-zero) rational and an irrational number is always an irrational number.
4. The quotient of a non-zero rational number with an irrational number is irrational.

Question 5.
Show how √5 can be represented on the number line.
(or)
Locate √5 on a number line.
Solution:
Representation of √5 on the number line : Consider a unit square OABC and transfer it onto the number line making sure that the vertex ‘O’ coincides with zero.
Then, OB = \(\sqrt{1^2+1^2}\) = √2

Representation of √5
Construct BD of unit length perpendicular to OB.
Then, OD = \(\sqrt{(\sqrt{2})^2+1^2}\) = √3
Construct DE of unit length perpendicular to OD.
Then OE = \(\sqrt{(\sqrt{3})^2+1^2}\) = √4 = 2
Construct EF of unit length perpendicular to OE.
Then OF = \(\sqrt{2^2+1^2}\) = √5
Using a compass, with centre ‘O’ and radius OF. Draw an arc which intersects the number line in the point R. Then R corresponds to √5 and OR = √5
Construct the “Square root spiral” :

Constructing Square root spiral.

Question 6.
Locate √7 on the number line.
Solution:
Mark the distance 7 from a fixed point A on a given line l to obtain a point B such that AB = 7 units.

From B mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O.

Draw a semi-circle with centre ‘O’ and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.
Then, BD = √7

Now, let us treat BC as the number line with B as zero, C as 1 and so on. Draw an arc with centre B and radius BD which intersects the number line in E.
Then, E represents √7 on the number line.

Question 7.
Prove that \(\frac{1}{2+\sqrt{3}}\) + \(\frac{2}{\sqrt{5}-\sqrt{3}}\) + \(\frac{1}{2-\sqrt{5}}\) = 0.
Solution:

Question 8.
If x = \(\frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}\), then find the value of qx² – 2px + q.
Solution:


Squaring on both sides, we get
(qx – p)² = p² – q²
q²x² – 2pqx + p² = q²
q²x² – 2pqx + q² = 0
qx² – 2px + q = 0

Question 9.
Evaluate \(\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}\) it being given that √5 = 2.236 and \(\sqrt{10}\) = 3.162.
Solution:

= \(\frac{5(\sqrt{10}+\sqrt{5})}{10-5}\)
= \(\frac{5(\sqrt{10}+\sqrt{5})}{5}\)
= \(\sqrt{10}\) + √5
= 3.162 + 2.236
= 5.398

Question 10.
If \(\frac{9^n \times 3^2 \times\left(3^{\frac{-n}{2}}\right)^{-2}-27^n}{3^{3 \mathrm{~m}} \times 2^3}\) = \(\frac{1}{27}\), prove that m – n = 1.
Solution:

⇒ 3^{3n – 3m} = 3^-3
⇒ 3n – 3m = -3
⇒ n – m = -1
∴ m – n = 1
Hence proved.

Question 11.
Represent \(\sqrt{4.2}\) on the number line.
Solution:
Make a line segment AB = 4.2 cm on number line. Further, take BC of 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre. Draw a perpendicular to line OC passing through the point B.

Let if intersect the semi-circle at E. Taking B as centre and BE as radius, draw an arc intersecting number line at F. BF is \(\sqrt{4.2}\).

Question 12.
Simplify the following expressions.
i) (3 + √3)(2 + √2)
ii) (5 + √5)(5 – √5)
iii) (√3 + √7)²
iv) (\(\sqrt{11}\) + √7)(\(\sqrt{11}\) – √7)
Solution:
i) (3 + √3)(2 + √2)
= 6 + 3√2 +2√3 + √6

ii) (5 + √5)(5 – √5)
= 25 – 5 = 20

iii) (√3 + √7)²
= 3 + 7 + 2\(\sqrt{21}\)
= 10 + 2\(\sqrt{21}\)

iv) (\(\sqrt{11}\) + √7)(\(\sqrt{11}\) – √7)
= 11 – 7 = 4

Question 13.
Represent 2.019 on number line using the method of successive magnification.
Solution:

AP 9th Class Maths 1st Lesson Important Questions and Answers Real Numbers

Question 1.
Represent \(\frac{-13}{5}\) on number line.
Solution:
Draw number line, having, -3, -2, -1,0, 1 integers on it.

→ Divide each unit as 5 equal parts on either side of zero.
→ From these count 13 parts on left side of the zero, which indicates \(\frac{-13}{5}\)

Question 2.
Find any two irrational numbers between 0.5 and 0.55.
Answer:
Two irrational numbers between 0.5 and 0.55 are (0.52515345 ……………….. and
0.541656475 …………….. ).

Question 3.
Simplify (25)^3/2 × (625)^-1/4
Solution:
Given that (25)^3/2 × (625)^-1/4
= (5²)^3/2 × (5⁴)^-1/4
= \(5^{2 \times \frac{3}{2}} \times 5^{4 \times\frac{-1}{4}}\) = 5³ × 5^-1 = 5^3-1 = 5² – 25.

Question 4.
Find two rational numbers between 0.2 and 0.3.
Solution:
The rational numbers between 0.2 and 0.3 is 0.25 and 0.28.
It means \(\frac{25}{100}, \frac{28}{100}=\frac{1}{4}, \frac{7}{25}\)

Question 5.
Convert \(\sqrt[3]{5}\) from the radical form to the exponential form.
Solution:
As \(\sqrt[n]{a^{m}}=(a)^{\frac{m}{n}}\) = \(\sqrt[n]{a}=(a)^{\frac{1}{n}}\)
\(\sqrt[3]{5}=(5)^{\frac{1}{3}}\)

Question 6.
Find the value of \(\text { (625) } \frac{-1}{4} \text { . }\) .
Solution:

Question 7.
Simplify: \(\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{3}-\sqrt{2}}\)
Solution:

Question 8.
Prove that if x is odd, then x² is also odd.
Solution:
x is an odd number, x = 2k + 1
{definition of odd}
x = 2k + 1
squaring on both sides
(x)² = (2k + 1)²
x² = 4k² 4k + 1 = 2(2k² + 2k) + 1 = 2l + 1
x² = 2l + 1 {l = 2k² + 2k}
∴ x² is an odd number.

Question 9.
Express \(5 . \overline{25}\) in from where q ≠ 0, p and q are co primes.
Solution:
Let x = \(5 . \overline{25}\)
x – 5.25 25 25 ……………… (1)
Multiply by 100 to (1)
100 x = 525.25 25 25 ………….. (2)
Subtract (1) from (2)

Question 10.
Express as a decimal number.
Solution:
For the correct process of division.
For expressing = 0.5222222 …………
= \(0.5 \overline{2}\)

Question 11.
If = a + b√3, then find the value of a³ + b³
The Rationalising Factor of 7 – 4√3 and √3 – 2 is 7 + 4√3 and √3 + 2

= 7 + 4√3 – √3 – 2
= 5 + 3√3 = a + b√3
Here a = 5,b = 3
∴ a³ + b³ = (5)³ + (3)³ = 125 + 27 = 152

Question 12.
Visualize the value of √5 upto 3 decimals on a number line, using successive magnification.
Solution:

Question 13.
If x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) and y = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\), then find the value of x² + y².
Solution:
Given that x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
R.F. of √3 – √2 is √3 + √2

x² + y² = (5 + 2√6 )² + (5 – 2√6 )²
= 25 + 20√6 + 24
+ 25 – 20√6 + 24
= 98

Question 14.
If √2 = 1.414, √3= 1.732, then find then value of \(\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}\).
Solution:
Given that √2 = 1.414 and √3 = 1.732

Question 15.
If \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}\) = a + b√3, then find the values of a and b.
Solution:
Given that \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}\)
Now rationalise the denominator, (Rational Factor of 7+4√3 is 7 – 4√3 )
= \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}\)
= \(\frac{35-20 \sqrt{3}+14 \sqrt{3}-8 \times 3}{(7)^{2}-(4 \sqrt{3})^{2}}\)
= \(\frac{11-6 \sqrt{3}}{49-48}\) = 11 – 6√3
a + b√3 – 11 – 6√3 =11 + (-6)√3
.’. a = 11, b = -6.

Question 16.
Visualise \(4.6 \overline{7}\) on number line upto 3 decimal places, using successive magnification.
Solution:

Question 17.
Represent -√3 and √3 on number line.
Solution:

Question 18.
Find the value of ‘a’ and ‘b’, if \(\frac{(\sqrt{5}-\sqrt{3})}{2 \sqrt{5}+3 \sqrt{3}}\) = a + b\(\sqrt{15}\)
Solution:
The Rationalising Factor of \(\frac{(\sqrt{5}-\sqrt{3})}{2 \sqrt{5}+3 \sqrt{3}}\)
is 2√5 – 3√3 , multiplying the numerator and denominator with 2√5 – 3√3.

Question 19.
Represent 2.019 on number line using the method of successive magnification.
Solution:

Question 20.
Simplify:
\(15 \sqrt[5]{32}+\sqrt{225}-8 \sqrt[3]{343}+\sqrt[4]{81}\)
Solution:
= \(15 \sqrt[5]{32}+\sqrt{225}-8 \sqrt[3]{343}+\sqrt[4]{81}\)
= \(15 \sqrt[5]{2^{5}}+\sqrt{15^{2}}-8 \sqrt[3]{7^{3}}+\sqrt[4]{3^{4}}\)
= (15 × 2) + 15 – (8 × 7) + 3
= 30 + 15 – 56 + 3 = – 8.

Question 21.
Express \(5.12 \overline{3}\) in \(\frac{\mathbf{p}}{\mathbf{q}}\) form where p and q are co-primes.
Solution:
Let x = \(5.12 \overline{3}\)
=> x = 5.12333 ……………. (1)
Here periodicity is 1.
So, we have to multiply (1) with 10.on both sides.
=> 10 × x = 10 × 5.12333 ……………….
10 x = 51.2333. …………… (2)
(2) – (1)

x = \(\frac{46.11}{9}=\frac{46.11}{9}\) × \(\frac{100}{100}=\frac{4611}{900}\)
∴ x = \(\frac{4611}{900}\)

Question 22.
The length of a rectangular field is \(\sqrt[3]{5}+\sqrt[3]{2}\) Find the measure of its breadth such that the area of the rectangle is a rational number.
Solution:
Given that the length of a rectangular
field (l) = \(\sqrt[3]{5}+\sqrt[3]{2}\).
If we want to get area of rectangle as a rational number, then we should take the breadth as Rationalising Factor of length.

Question 23.
Simplify:
\(\sqrt[4]{81}-8 \cdot \sqrt[5]{243}+15 \cdot \sqrt[5]{32}+\sqrt{225}\)
Solution:
\(\sqrt[4]{81}-8 \cdot \sqrt[5]{243}+15 \cdot \sqrt[5]{32}+\sqrt{225}\)
= \(\sqrt[4]{3^{4}}-8 \cdot \sqrt[5]{3^{5}}+15 \cdot \sqrt[5]{2^{5}}+\sqrt{15^{2}}\)
= \(\left(3^{4}\right)^{\frac{1}{4}}-8 \cdot\left(3^{5}\right)^{\frac{1}{5}}+15\left(2^{5}\right)^{\frac{1}{5}}+\left(15^{2}\right)^{\frac{1}{2}}\)
= 3 – 8.3 + 15.2 + 15
= 3 – 24 + 30+ 15
= 48 – 24 = 24