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AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.1

Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the values of a, b and c in each case.
i) 8x + 5y – 3 = 0
Solution:
8x + 5y – 3 = 0
⇒ 8x + 5y + (- 3) = 0
Here a = 8, b = 5 and c = – 3

ii) 28x – 35y = – 7
Solution:
28x – 35y = – 7
⇒ 28x + (- 35) y + 7 = 0
Here a = 28, b = – 35 and c = 7

iii) 93x = 12- 15y
Solution:
93x = 12 – 15y
⇒ 93x + 15y -12 = 0
⇒ 93x + 15y + (- 12) = 0
Here a = 93, b = 15 and c = – 12

iv) 2x = – 5y
Solution:
2x = – 5y
⇒ 2x + 5y = 0
Here a = 2, b = 5 and c = 0

v) \(\frac{x}{3}+\frac{y}{4}=7\)
Solution:
\(\frac{x}{3}+\frac{y}{4}=7\)
⇒ \(\frac{x}{3}+\frac{y}{4}-7=0\)
⇒\(\frac{4 x+3 y-84}{12}=0\)
⇒ 4x + 3y – 84 = 0
Here a = 4, b = 3 and c = – 84

vi) y = \(-\frac{3}{2} x\)
Solution:
y = \(-\frac{3}{2} x\)
⇒ 2y = -3x
⇒ 3x + 2y = 0
Here a = 3, b = 2 and c = 0

vii) 3x + 5y = 12
Solution:
3x + 5y = 12
⇒ 3x + 5y + (- 12) = 0
Here a = 3, b = 5 and c = – 12

Question 2.
Write each of the following in the form of ax + by + c = 0 and find the values of a, b and c.
i) 2x = 5
Solution:
2x – 5 = 0
a = 2
b = 0
c = -5

ii) y – 2 = 0
Solution:
y – 2 = 0
a = 0
b = 1
c = – 2

iii) \(\frac{y}{7}\) = 3
Solution:
\(\frac{y}{7}\) = 3
y = 21
y – 21 = 0
a = 0
b = 1
c = -21

iv) x = \(-\frac{14}{13}\)
x = \(-\frac{14}{3}\)
⇒ 13x = – 14
⇒ 13x + 14 = 0
a = 13
b = 0
c = 14

Question 3.
Express the following statements as a linear equation in two variables,
i)The sum of two numbers is 34.
Solution:
x + y = 34; x, y are any two numbers ⇒ x + y – 34 = 0

ii) The cost of a ball pen is ?5 less than half the cost of a fountain pen.
Solution:
Let the cost of a fountain pen = x
Let the cost of ball pen = y
Then y = x – 5 or x – y – 5 = 0

iii) Bhargavi got 10 more marks than double of the marks of Sindhu. |l M)
Solution:
Let Sindhu’s marks = x
Bhargavi’s marks = y
Then by problem y = 2x + 10 or 2x – y + 10 = 0

iv) The cost of a pencil is ₹2 and one ball point pen costs ₹15. Sheela pays ₹100 for the pencils and pens she purchased.
Solution:
Giver: that cost of a pencil = ₹2
Cost of a ball point pen = ₹15
Let the number of pencils purchased = x
Let the number of pens purchased = y
Then the total cost of x – pencils = 2x
Then the total cost of y – pens = 15y
By problem 2x + 15y = 100

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.3

Question 1.
Plot the following points on the Cartesian plane whose x, y co-ordinates are given.

Solution:

Question 2.
Are the positions of (5, -8) and (-8, 5) Is same ? JustIfy your answer.
Solution:
The positions of (5, -8) and (-8, 5) are not same. They are two distinct points. (5, -8) lies at a distance of 5 units from Y – axis and 8 units from X – axis on down side of the origin. So it lies in Q₄. Where as (- 8, 5) lies in Q₂. The point is at a distance of 8 units from Y – axis on left side of the origin and 5 units from X – axis.

Question 3.
What can you say about the position of the points (1, 2), (1, 3), (1, – 4), (1, 0) and (1, 8), Locate on a graph sheet.
Solution:

All the given points lie on a line parallel to Y-axis at a distance of 1 cm.

Question 4.
What can you say about the position of the points (5, 4), (8, 4), (3, 4), (0, 4), (-4, 4), (-2,4)? Locate the points on a graph sheet and justify your answer.
Solution:
The points (5, 4), (8, 4), (3, 4), (0, 4), (- 4, 4) and (- 2, 4) all lie on a line parallel to X-axis at a distance of 4-units from it.

Question 5.
Plot the points (0, 3), (4, 3), (3, 4) (4, 0) in graph sheet. Join the points with straight lines to make a rectangle. Find the area of the rectangle.
Solution:

From the graph, area of the rectangle =12 square units (OR)
Length = 4 units ; breadth = 3 units
A = l/b = 4 × 3 = 12 sq. units.

Question 6.
Plot the points (2, 3), (6, 3) and (4, 7) in a graph sheet. Join them to make it a triangle. Find the area of the triangle.
Solution:

From the graph
Base of the triangle = 4 units
Height of the triangle = 4 units
∴ Area of the triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) × 4 × 4 = 8sq. units.

Question 7.
Plot at least six points in a graph sheet, each having the sum of its co-ordinates equal to 5. [Hint: (- 2, 7), (1, 4)………….]
Solution:
Given that (x co-ordinate) + (y co-ordinate) = 5
Let the points be

Question 8.
Look at the graph. Write the co-ordinates of the points
A, B, C, D, E, F. G. H, 1, J, K, L, M, N, O, P and Q.

Solution:
A(- 3, 4) ; B(0, 5) : C (3, 4) ; D (2, 4) ; E (2, 0) ;
F (3, 0) ; G (3, – 1) ; H (0, – 1) ; I (- 3, – 1) ;
J (- 3, 0) ; K (- 2, 0) ; L (- 2, 4) ; M (- 1, 0) ;
N (-1, 3); O (0, 0) ; P (1. 3) and Q (1, 0)

Question 9.
In a graph sheet plot each pair of points, join them by line segments.
i) (2, 5), (4, 7)
ii) (-3, 5) (-1,7)
iii) (-3, -4), (2, -4)
iv) (-3, -5) (2, -5)
v) (4, -2), (4, -3)
vi) (-2, 4), (-2, 3)
vii) (-2, 1), (-2, 0)
Now join the following pairs of points by straight line segments, in the same graph.
viii) (-3, 5), (-3, 4)
ix) (2, 5), (2, -4)
x) (2, -4), (4, -2)
xi) (2, -4), (4, -3)
xii) (4, -2), (4, 7)
xiii) (4, 7), (-1, 7)
xiv) (-3, 2), (2, 2)
Now you will get a surprise figure. What is it ?
Solution:

AP Board 9th Class Maths Solutions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.2

Question 1.
Write the quadrant in which the following points lie.
i) (- 2, 3)
ii) (5, – 3)
iii) (4, 2)
iv) (- 7, – 6)
v) (0, 8)
vi) (3, 0)
vii) (-4,0)
viii) (0, – 6)
Solution:
i) (- 2, 3) – Q₂ (second quadrant)
ii) (5, – 3) – Q₄ (fourth quadrant)
iii) (4, 2) – Q₁ (Iirst quadrant)
iv) (- 7, – 6) – Q₃ (third quadrant)
v) (0, 8) – on Y-axis
vi) (3, 0) – on X-axis
vii) (-4,0) – on X’ – axis
viii) (0, – 6) – on Y’: axis

Question 2.
Write the abscissae and ordinates of the following points.
i)(4,-8)
ii)(-5,3)
iii)(0,0)
iv)(5, 0)
v)(0, -8)
Note: Plural of abscissa is abscissae.
Solution:

Question 3.
Which of the following points lie on the axes ? Also name the axis.
i) (-5,-8)
ii) (0, 13)
iii) (4, – 2)
iv) (- 2,0)
v) (0, – 8)
vi) (7,0)
vii) (0,0)
Solution:
The points (ii) (0,13) ; (v) (0,- 8) lie on Y – axis.
The points (iv) (- 2, 0), (vi) (7, 0) lie on X – axis.
The point (vii) (0, 0) lie on both X – axis and Y – axis.
The points (i) (- 5, – 8); (iii) (4, – 2) do not lie on any axis.

Question 4.
Write the following based on the graph.
i) The ordinate of L
ii) The ordinate of Q
iii) The point denoted by (- 2,-2)
iv) The point denoted by (5, – 4)
v) The abscissa of N
vi) The abscissa of M

Solution:
i) – 7
ii) 7
iii) The point ’R’
iv) The point P
v) 4
vi) – 3

Question 5.
State true or false and write correct statement.
i) In the Cartesian plane the horizon-tal line is called Y – axis, if) in the Cartesian plane, the vertical line is called Y – axis.
iii) The point which lies on both the axes is called origin.
iv) The point (2, – 3) lies in the third quadrant.
v) (-5, -8) lies in the fourth quadrant.
vi) The point (- x, – y) lies in the first quadrant where x < 0; y < 0.
Solution:
i) False
Correct statement: In the Cartesian plane the horizontal line is called X – axis.
ii) True
iii) True
iv) False
Correct statement: The point (2, -3) lies in the fourth quadrant.
v) False
Correct statement: (- 5, – 8) lies in the third quadrant.
vi) True

Question 6.
Plot the following ordered pairs on a graph sheet. What do you observe ?
i) (1, 0), (3, 0), (- 2, 0), (- 5, 0), (0, 0), (5, 0), (- 6, 0)
ii) (0, 1), (0, 3), (0, – 2), (0, – 5), (0, 0), (0,5), (0,-6)
Solution:
i) All points lie on X – axis,
ii) All points lie on Y – axis.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.1

Question 1.
In a locality, there is a main road along North – South direction. The map is given below. With the help of the picture answer the following questions.

i) What is the 3rd object on the left side in street no. 3 while going in east direction ?
ii) Find the name of the 2nd house which is on right side of street 2 while going in east direction.
iii) Locate the position of Mr. K’s house.
iv) How do you describe the position of the post office ?
v) How do you describe the location of the hospital ?
Solution:
i) Water tank
ii) Mr. J’s house
iii) In street No. 2, 3rd house on right side.
iv) In street No. 4, the first house on right side.
v) In street No. 4, the last house on left side.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4

Question 1.
In the given triangles, find out ∠x, ∠y and ∠z

Solution:
In fig(i)
x° = 50° + 60°
(∵ exterior angle is equal to sum of the opposite interior angles)
∴ x= 110°

In fig (ii)
z° = 60° + 70°
(∵ exterior angle is equal to sum of the opposite interior angles)
∴ z = 130°

In fig (iii)
y° = 35° + 45° = 80°
(∵ exterior angle is equal to sum of the opposite interior angles)

Question 2.
In the given figure AS // BT; ∠4 = ∠5, \(\overline{\mathbf{S B}}\) bisects ∠AST. Find the measure of ∠1.

Solution:
Given AS // BT
∠4 = ∠5 and SB bisects ∠AST.
∴ By problem
∠2 = ∠3 …………..(1)
For the lines AS // BT
∠2 = ∠5 ( ∵alternate interior angles)
∴ In ΔBST
∠3 = ∠5 = ∠4
Hence ΔBST is equilateral triangle and each of its angle is equal to 60°.
∴∠3 = ∠2 = 60° [by eq. (1)]
Now ∠1 + ∠2 + ∠3 = 180°
∠1 + 60° + 60° = 180°
[ ∵ angles at a point on a line]
∴∠1 = 180° – 120° = 60°

Question 3.
In the given figure AB // CD; BC // DE then find the values of x and y.

Solution:
Given that AB // CD and BC // DE.
∴ 3x = 105° (∵ alternate interior angles for AB // CD)
x = \(\frac { 105° }{ 3 }\) = 35°
Also BC // DE
∴∠D = 105°
(∵ alternate interior angles)
Now in ΔCDE
24° + 105° + y = 180°
(∵ angle sum property)
∴ y = 180° – 129° = 51°

Question 4.
In the given figure BE ⊥ DA and CD ⊥ DA then prove that m∠1 = m∠3.

Solution:
Given that CD ⊥ DA and BE ⊥ DA.
⇒ Two lines CD and BE are perpendicular to the same line DA.
⇒ CD // BE (or)
∠D =∠E ⇒ CD // BE
(∵ corresponding angles for CD and BE and DA are transversal)
Now m∠1 = m∠3
(∵alternate interior angles for the lines CD // BE ; DB are transversal)
Hence proved.

Question 5.
Find the values of x, y for which lines AD and BC become parallel.

Solution:
For the lines AD and BC to be parallel x – y = 30° (corresponding angles) ……… (1)
2x = 5y ………….(2)
(∵ alternate interior angles)
Solving (1) & (2)

y = \(\frac{60}{3}\) = 20°
Substituting y = 20° in eq. (1)
x – 20° = 30°
⇒ x = 50°
∴ x = 50° and y = 20°

Question 6.
Find the values of x and y in the figure.

Solution:
From the figure y + 140° = 180°
(∵ linear pair of angles)
∴ y = 180° – 140° = 40°
And x° = 30° + y°
(∵ exterior angle = sum of the opposite interior angles)
x° = 30° + 40° = 70°

Question 7.
In the given figure segments shown by arrow heads are parallel. Find the values of x and y.

Solution:
From the figure
x° = 30° (∵ alternate interior angles)
y° = 45° + x° (∵ exterior angles of a triangle = sum of opp. interior angles)
y = 45° + 30° – 75°

Question 8.
In the given figure sides QP and RQ of ∠PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
Given that ∠SPR = 135° and ∠PQT =110°
From the figure
∠SPR + ∠RPQ = 180°
∠PQT + ∠PQR = 180°
[∵ linear pair of angles]
⇒ ∠RPQ = 180° – ∠SPR
= 180° – 135° = 45°
⇒ ∠PQR = 180° – ∠PQT
= 180°-110° = 70°
Now in APQR
∠RPQ + ∠PQR + ∠PRQ = 180°
[∵ angle sum property]
∴ 45° + ’70° + ∠PRQ = 180°
∴ ∠PRQ = 180°-115° = 65°

Question 9.
In the given figure ∠X = 62° ; ∠XYZ = 54°. In ΔXYZ. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respec-tively find ∠OZY and ∠YOZ.

Solution:
Given that ∠X = 62° and ∠Y = 54°
YO arid ZO are bisectors of ∠Y and ∠Z.
In ΔXYZ
∠X + ∠XYZ + ∠XZY = 180° .
62° + 54° + ∠XZY = 180°
=> ∠XZY = 180°- 116° = 64°
Also in Δ????OYZ
∠OYZ = 1/2 ∠XYZ = 1/2 x 54° = 27°
(∵ YO is bisector of ∠XYZ)
∠OZY = 1/2 ∠XZY = 1/2 x 64° = 32
(∵ OZ is bisector of ∠XYZ)
And ∠OYZ + ∠OZY + ∠YOZ = 180°
(∵ angle sum property, ΔOYZ)
⇒ 27 + 32° + ∠YOZ = 180°
⇒ ∠YOZ = 180° – 59° = 121°

Question 10.
In the given figure if AB // DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
Given that AB // DE, ∠CDE = 53°;
∠BAC = 35°
Now ∠E = 35°
( ∵ alternate interior angles)
Now in ∆CDE
∠C + ∠D + ∠E = 180°
(∵angle sum property, ACDE)
∴ ∠DCE + 53° + 35° = 180°
⇒ ∠DCE = 180° – 88° = 92°

Question 11.
In the given figure if line segments PQ and RS intersect at point T, such that∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
Given that ∠PRT = 40°; ∠RPT = 95°;
∠TSQ = 75°
In ∆PRT ∠P + ∠R + ∠PTR = 180°
(∵angle sum property)
95° + 40° + ∠PTR = 180°
⇒ ∠PTR = 180° – 135° = 45°
Now ∠PTR = ∠STQ
(∵ vertically opposite angles)
In ΔSTQ ∠S + ∠Q + ∠STQ = 180°
(∵ angle sum property)
75° + ∠SQT + 45° = 180°
∴ ∠SQT = 180° – 120° = 60°

Question 12.
In the given figure, ABC is a triangle in which ∠B = 50° and ∠C = 70°. Sides AB and AC are produced. If ∠ is the measure of angle between the bisec¬tors of the exterior angles so formed, then find ‘z’.

Solution:
Given that ∠B = 50°; ∠C = 70°
Angle between bisectors of exterior angles B and C is ∠.
From the figure
50° + 2x = 180°
70° + 2y = 180°
(∵ linear pair of angles)
∴ 2x= 180°-50°
2x= 130°
x = \(\frac{130}{2}\)
= 65°

2y= 180°-70°
2y= 110°
x = \(\frac{110°}{2}\)
= 55°

Now in ΔBOC
x + y + ∠ = 180° (∵ angle sum property)
65° + 55° + ∠ = 180°
z = 180° -120° = 60°

Question 13.
In the given figure if PQ ⊥ PS; PQ // SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:
Given that PQ ⊥ PS ; PQ // SR
∠SQR = 28°, ∠QRT = 65°
From the figure
∠QSR = x° (∵ alt. int. angles for the lines PQ // SR)
Also 65° = x + 28° (∵ ext. angles = sum of the opp. interior angles)
∴ x° = 65° – 28° = 37°
And x° + y° = 90°
[ ∵ PQ ⊥ PS and PQ // SR. ⇒ ∠P = ∠S]
37° + y = 90°
∴ y = 90° – 37° = 53°

Question 14.
In the given figure ΔABC side AC has been produced to D. ∠BCD = 125° and ∠A: ∠B = 2:3, find the measure of ∠A and ∠B

Solution:
Given that ∠BCD = 125°
∠A : ∠B = 2 : 3
Sum of the terms of the ratio
∠A : ∠B = 2 + 3 = 5
We know that ∠A + ∠B = ∠BCD
(∵ exterior angles of triangle is equal to sum of its opp. interior angles)
∴ ∠A = \(\frac{2}{5}\) x 125° = 50°
∠B = \(\frac{3}{5}\) x 125° = 75°

Question 15.
In the given figure, it is given that, BC // DE, ∠BAC = 35° and ∠BCE = 102°. Find the measure of 0 ∠BCA i0 ∠ADE and iii) ∠CED.

Solution:
Given that BC // DE ; ∠BAC = 35°;
∠BCE = 102°

i) From the figure
102° + ∠BCA = 180°
(∵ linear pair of angles)
∴ ∠BCA = 180° – 102° = 78°

ii) ∠ADE + ∠CBD = 180°
(∵ interior angles on the same side of the transversal)
∠ADE + (78° + 35°) = 180°
(∵ ∠CBD = ∠BAC + ∠BCA)
∴ ∠ADE = 180° – 113° = 67°

iii) From the figure .
∠CED = ∠BCA = 78°
(∵ corresponding angles)

Question 16.
In the given figure, it is given that AB = AC; ∠BAC = 36°; ∠ADB = 45° and ∠AEC = 40°. Find i) ∠ABC
i) ∠ACB iii) ∠DAB iv) ∠EAC.

Solution:
Given that AB = AC; ∠BAC = 36°,
∠ADB = 45°, ∠AEC = 40°
(i) & (ii)
In ∆ABC ; AB = AC
⇒ ∠ABC = ∠ ACB
And 36° + ∠ABC + ∠ACB = 180°
(∵ angle sum property)
∴ ∠ABC = \(\frac{180^{\circ}-36^{\circ}}{2}=\frac{144^{\circ}}{2}=72^{\circ}\)
∠ACB = 72°

iii) From the figure
∠ABD + ∠ABC = 180°
∠ABD = 180° – 72° = 1086
In ΔABD
∠DAB + ∠ABD + ∠D = 180°
∠DAB + 108° + 45° = 180°
∠DAB = 180° – 153° = 27°

iv) In ΔADE
∠D + ∠A + ∠E = 180°
45° + ∠A + 40° = 180°
⇒ ∠A = 180° -85° = 95°
But ∠A = ∠DAB + 36° + ∠EAC
95° = 27°, + 36° + ∠EAC
∴ ∠EAC = 95° – 63° = 32°

Question 17.
Using information given in the figure, calculate the values of x and y.

Solution:
From the figure In ∆ACB
34° + 62° + ∠ACB = 180°
(∵ angle sum property)
∴ ∠ACB = 180° – 96° = 84° .
And x + ∠ACB = 180°
(∵ linear pair of angles) .
∴ x + 84° = 180°
x = 180°-84° = 96°
(OR)
x = 34° + 62° = 96°
( ∵ x is exterior angle, ∆ABC)
y = 24° + x°
= 24° + 96° = 120°
(∵ y is exterior angle, ∆DCE)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.4

Question 1.
Give the graphical representation of the following equation
a) on the number line and b) on the Cartesian plane.
l) x = 3
ii) y + 3 = 0
iii) y = 4
iv) 2x – 9 = 0
v) 3x + 5 = 0
Solution:
i) x = 3 is a line parallel to Y-axis at a distance of 3 units on the right side of the origin.
ii) y + 3 = 0 y = – 3 is a line parallel to X-axis, below the origin.
iii) y = 4 is a line parallel to X-axis at a distance of 4 units above the origin.
iv) 2x – 9 = 0
⇒ x = \(\frac{9}{2}\) = 4.5 is a line parallel to Y-axis at a distance of 4.5 units, right side of the zero.
v) 3x + 5 = 0
⇒ 3x = -5 x = \(\frac{-5}{3}\) is a line parallel to Y – axis at a distance of \(\frac{5}{3}\)units on the left side of the origin.

x = 3

y + 3 = 0

y = 4

2x – 9 = 0

3x + 5 = 0

Question 2.
Give the graphical representation of 2x – 11 = 0 as an equation in i) one variable ii) two variables
Solution:
2x – 11 = 0

Question 3.
Solve the equation 3x + 2 = 8x – 8 and represent the solution on
i) the number line ii) the Cartesian plane.
Solution:
Given that 3x + 2 = 8x – 8
3x – 8x = – 8 – 2
– 5x = -10
x = \(\frac{-10}{-5}\) = 2

Question 4.
Write the equation of the line parallel to X-axis and passing through the point i) (0, – 3) ii) (0,4) iii) (2, – 5) iv) (3,4)
Solution:
i) The given point is (0, – 3)
Equation of a line parallel to X-axis is y = k
∴ Required equation is y = – 3 or y + 3 = 0

ii) The given point is (0, 4)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = 4ory-4 = 0

iii) The given point is (2, – 5)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = -5 or y + 5 = 0

iv) The given point is (3, 4)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = 4 or y – 4 = 0

Question 5.
Write the equation of the line parallel to Y-axis passing through the point
i) (- 4, 0)
ii) (2,0)
iii) (3, 5)
(iv) (- 4, – 3)
Solution:
Equation of a line parallel to Y-axis is x = k
∴ The required equations are
i) Through the point (- 4, 0) ⇒ the equation is x = – 4 or x + 4 = 0
ii) Through the point (2, 0) ⇒ the equation isx = 2orx-2 = 0
iii) Through the point (3, 5) ⇒ the equation isx = 3orx-3 = 0
iv) Through the point (- 4,-3) ⇒ the equation is x = – 4 or x + 4 = 0

Question 6.
Write the equation of three lines that are
1) Parallel to the X-axis
Solution:
y = 3
y = -4
y = 6

ii) Parallel to the Y-axis
Solution:
x = – 2
x = 3
x = 4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.1

Question 1.
In ΔABC, ∠ABC = 90°; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of ΔADB

Solution:
ΔADB = \(\frac { 1 }{ 2 }\) ΔABC [ ∵ AD is a median of ΔABC]
\(\frac { 1 }{ 2 }\) = [ \(\frac { 1 }{ 2 }\) AB x BC]
= \(\frac { 1 }{ 4 }\) x 12 x 6.5
= 19.5 cm²

Question 2.
Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR =17 cm.
[Hint: PQRS has two parts]

Solution:
Area of ΔQPS = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 9 x 12
= 54cm²
In ΔQPS
QS² = PQ² + PS²
QS = \(\begin{aligned}
\sqrt{12^{2}+9^{2}} &=\sqrt{144+81} \\
&=\sqrt{225}=15
\end{aligned}\)
Area of ΔQSR =\(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 15 x 8 = 60 cm²
∴ □PQRS = ΔQPS + ΔQSR
= 54 + 60= 114 cm²

Question 3.
Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle.
[Hint: ABCD has two parts]

Solution:
Area of trapezium 1
= \(\frac { 1 }{ 2 }\) (sum of parallel sides) x (distance between the parallel sides)
= \(\frac { 1 }{ 2 }\) (a + b) h
From the figure, a = 3 + 3 = 6 cm
b = 3 cm
(∵ Opp. sides of rectangle)
h = 8 cm
∴ A = \(\frac { 1 }{ 2 }\)(6 + 3)x8 = 36cm²

Question 4.
ABCD is a parallelogram. The diago-nals AC and BD intersect each other at O. Prove that ar (ΔAOD) = ar (ΔBOQ. [Hint: Congruent figures have equal area]

Solution:
Given that □ABCD is a parallelogram.
Diagonals AC and BD meet at ‘O’.
In ΔAOD and ΔBOC
AD = BC [ ∵ Opp. sides of a ||gm]
AO = OC [ ∵ diagonals bisect each
OD = OB other]
ΔAOD = ΔBOC [S.S.S. congruence]
∴ ΔAOD = ΔBOC (i.e., have equal area)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

Question 1.
Simple the following expressions.
i) (5 + √7) (2 + √5)
Solution:
(5 + √7) (2 + √5)
= 10 + 5√5 + 2√7 + √35

ii) (5 + √5) (5 – √5)
Solution:
(5 + √5) (5 – √5)
= 5² + (√5)²
= 25 – 5 = 20

(iii) (√3 + √7)²
Solution:
(√3 + √7)²
= (√3)² + (√7)² + 2(√3)(√7)
= 3 + 7 + 2√21
= 10 + 2√21

iv) (√11 – √7) (√11 + √7)
= (√11)² – (√7)²
= 11 – 7 = 4

Question 2.
Classify the following numbers as rational or irrational.
i) 5 – √3
ii) √3 + √2
iii) (√2 – 2)²
iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
v) 2π
vii) (2 +√2) (2 – √2)
Solution:
i) 5 – √3 – irrational
ii) √3 + √2 – irrational
iii) (√2 – 2)² – irrational
iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) – rational
v) 2π – Transcendental number. (not irrational)
vi) \(\frac{1}{\sqrt{3}}\)– irrational
vii) (2 +√2) (2 – √2) – rational

Question 3.
In the following equations, find whether variables x, y, z etc., represents rational or irrational numbers.
i) x² = 7
ii) y² = 16
iii) z² = 0.02
iv) u² = \(\frac{17}{4}\)
v) w² = 27
vi) t⁴ = 256
Solution:
i) x² = 7
⇒ x = √7 is an irrational number.
ii) y² = 16 ⇒ y = 4 is a rational number.
iii) z² = 0.02 ⇒ z = \(\sqrt{0.02}\) is an irrational number.
iv) u² = \(\frac{17}{4}\) ⇒ x = \(\frac{\sqrt{17}}{2}\) is an irrational number.
v) w² = 27 ⇒ w = \(3 \sqrt{3}\) an irrational number.
vi) t⁴ = 256 ⇒ t² = \(\sqrt{256}\) = 16
⇒ t = \(\sqrt{16}\) = 4 is a rational number

Qeustion 4.
The ratio of circumference to the diameter of a circle c/d is represented by π. But we say that π is an irrational number. Why?

Question 5.
Rationalise the denominators of the following.
i) \(\frac{1}{3+\sqrt{2}}\)
Solution:

ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Solution:

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

iv) \(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}\)
Solution:

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

ii) \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
Solution:

iii) \(\frac{1}{3 \sqrt{2}-2 \sqrt{3}}\)
Solution:

iv) \(\frac{3 \sqrt{5}-\sqrt{7}}{3 \sqrt{3}+\sqrt{2}}\)
Solution:

Question 7.
Find the value of \(\frac{\sqrt{10}-\sqrt{5}}{2 \sqrt{2}}\) upto three decimal places. (take \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732 and \(\sqrt{5}\) = 2.236).
Solution:

Question 8.
Find
i) 64^1/6
Solution:
= (2⁶)^1/6
= 6

ii) 32^1/5
Solution:
32^1/5
= (2⁵)^1/5
= 2

iii) 625^1/4
625^1/5
= (5⁴)^1/4
= 5

iv) 16^3/2
Solution:
16^3/2
= (4²)^3/2

v) 243^2/5
Solution:
243^2/5
= (3⁵)^2/5

vi) (46656)^-1/6
Solution:

Question 9.
Simplify \(\sqrt[4]{81}-8 \sqrt[3]{343}+15 \sqrt[5]{32}+\sqrt{225}\)
Solution:

Question 10.
If ‘a’ and ‘b’ are rational numbers, find the values of a and b in each of the following equations.
i) \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)
Solution:
Given that \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)
Rationalising the denominator we get

Comparing 5 + 2√6 with a + b√6
We have a = 5 and b = 2

ii) \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)
Solution:
Given that \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)
Rationalising the denominator we get

AP Board 9th Class Maths Solutions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.3

Question 1.
Visualise \(2.8 \overline{74}\) on the number line, using successive magnification.
Solution:

Question 2.
Visualise \(5 . \overline{28}\) on the number line, upto 3 decimal places.
Solution:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

Question 1.
Classify the following numbers as rational or irrational.
i) \(\sqrt{27}\)
ii) \(\sqrt{441}\)
iii) 30.232342345
iv) 7.484848
v) 11.2132435465
vi) 0.3030030003
Solution:
i) \(\sqrt{27}\) – irrational number
ii) \(\sqrt{441}\) = 21 – rational
iii) 30.232342345 – irrational number
iv) 7.484848 – rational number
v) 11.2132435465 – irrational number
vi) 0.3030030003 – irrational number

Question 2.
Explain with an example how irrational numbers differ from rational numbers ?
Solution:
Irrational numbers can’t be expressed in \(\frac { p }{ q }\) form where p and q are integers and q ≠ 0.
E.g.\(\sqrt{2}, \sqrt{3} ; \sqrt{5}, \sqrt{7}\) etc.
Where as a rational can be expressed in \(\frac { p }{ q }\) form
E.g. :- -3 = \(\frac { -3 }{ 1 }\) and \(\frac { 5 }{ 4 }\) etc.

Question 3.
Find an irrational number between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\). How many more there may be ?
Solution :
The decimal forms of \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) are
\(\frac{5}{7}=0 . \overline{714285} \ldots ., \frac{7}{9}=0.7777 \ldots \ldots=0 . \overline{7}\)
∴ An irrational between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) is 0.727543…………
There are infinitely many irrational numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\).

Question 4.
Find two irrational numbers between 0.7 and 0.77.
Solution:
Two irrational numbers between 0.7 and 0.77 can take the form
0.70101100111000111…………. and 0.70200200022……………

Question 5.
Find the value of √5 uPto 3 decimal places.
Solution:

[√5 is not exactly equal to 2.2350679………….. as shown ¡n calculators]

Question 6.
Find the value of √7 upto six decimal places by long division method.
Solution:

Question 7.
Locate \(\sqrt{\mathrm{10}}\) on number line.

Step – 1 : Draw a number line.
Step – 2 : Draw a rectangle OABC at zero with measures 3 x 1. i.e., length 3 units and breadth 1 unit.
Step – 3 : Draw the diagonal OB.
Step – 4 : Draw an arc with centre ‘O’ and radius OB which cuts the number line at D.
Step – 5 : ‘D’ represents \(\sqrt{\mathrm{10}}[latex] on the number line.

Question 8.
Find atleast two irrational numbers between 2 and 3.
Solution:
An irrational number between a and b is Tab [latex]\sqrt{\mathrm{ab}}\) unless ab is a perfect square.
∴ Irrational number between 2 and 3 is √6

∴ Required irrational numbers are 6^1/2, 24^1/4

Method – II:
Irrational numbers between 2 and 3 are of the form 2.12111231234………….. and 3.13113111311113…….

Question 9.
State whether the following statements are true or false. Justify your answers.
Solution:

1. Every irrational number is a real number – True (since real numbers consist of rational numbers and irrational numbers)
2. Every rational number is a real number – True (same as above)
3. Every rational number need not be a rational number – False (since all rational numbers are real numbers).
4. \(\sqrt{n}\) is not irrational if n is a perfect square – True. (since by definition of an irrational number).
5. \(\sqrt{n}\) is irrational if n is not a perfect square – True. (same as above)
6. All real numbers are irrational – False (since real numbers consist of rational

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

Let a ΔABC be right angled at ∠B.
Then ∠A + ∠C = 90°
(i.e.,) ∠A and ∠C are both acute.
Now, ∠A < ∠B ⇒ BC < AC
Also ∠C < ∠B ⇒ AB < AC
∴ AC, the hypotenuse is the longest side.

Question 2.
In the given figure, sides AB and AC of ΔABC are extended “to points P and Q respectively. Also ∠PBC < ∠QCB. Show that AC > AB.

Solution:
From the figure,
∠PBC = ∠A + ∠ACB
∠QCB = ∠A + ∠ABC
Given that ∠PBC < ∠QCB
⇒∠A + ∠ACB < ∠A + ∠ABC
⇒ ∠ACB < ∠ABC
⇒ AB < AC
⇒ AC > AB
Hence proved.

Question 3.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Given that ∠B < ∠A; ∠C < ∠D
∠B < ∠A ⇒ AO < OB [in ΔAOB] ……………… (1)
∠C < ∠D ⇒ OD < OC [in ΔCOD]…… (2)
Adding (1) & (2)
AO + OD < OB + OC
AD < BC
Hence proved.

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A >∠C and ∠B > ∠D.

Solution:

Given that AB and CD are the smallest and longest sides of quadrilateral ABCD.
From the figure,
In ΔBCD
∠1 > ∠2 [∵ DC > BC] ………………(1)
In ΔBDA
∠4 > ∠3 [∵ AD > AB] ………….(2)
Adding (1) & (2)
∠1 + ∠4 > ∠2 + ∠3
∠B > ∠D
Similarly,
In ΔABC, ∠6 < ∠7 [ ∵AB < BC] ……………….(3)
In ΔACD
∠5 < ∠8 …………. (4)
Adding (3) & (4)
∠6 + ∠5 < ∠7 + ∠8
∠C < ∠A ⇒ ∠A > ∠C
Hence proved.

Question 5.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that
∠PSR > ∠PSQ.

Solution:
Given that PR > PQ;
∠QPS =∠RPS
PR> PQ
∠Q > ∠R
Now ∠Q +∠QPS > ∠R + ∠RPS
⇒ 180° – (∠Q + ∠QPS) < 180° – (∠R + ∠RPS)
⇒ ∠PSQ < ∠PSR ⇒ ∠PSR > ∠PSQ
Hence proved.

Question 6.
If two sides of a triangle measure 4 cm and 6 cm find all possible measurements (positive integers) of the third side. How many distinct triangles can be obtained ?
Solution:
Given that two sides of a triangle are 4 cm and 6 cm.
∴ The measure of third side > Differ-ence between other two sides.
third side > 6 – 4
third side > 2
Also the measure of third side < sum of other two sides
third side <6 + 4 < 10
∴ 2 < third side <10
∴ The measure of third side may be 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm
∴ Seven distinct triangles can be obtained.

Question 7.
Try to construct a triangle with 5 cm, 8 cm and 1 cm. Is it possible or not ? Why ? Give your justification.
Solution:
As the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.

Andhra Pradesh AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors World Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class Maths Solutions Chapter 5 Multiples and Factors

Question 1.
Observe which numbers are not divisible?

Answer: Which of the above numbers are exactly divisible by 2?
Answer:
2410, 1282, 3784, 6728, 5466

b. Observe the units place of the numbers which are divisible by 2.
Answer:
0, 2, 4, 6, 8.

c. Are all these numbers even numbers ? Yes / no. (Yes)
Answer:
So a number is divisible by 2, if the digit at its one place is either 0/2/4/6/8.

Do these: (TextBook Page No.62)

Question 1.
Circle the following numbers which are divisible by 2.

Answer:

Question 2.
Write any 5, four-digit numbers which arc divisible by 2 ?
Answer:
25680, 45,622, 78,964, 87,766 and 97.678

Exercise 1:

Question 1.
Find the numbers which are divisible by 2. Write the reason for the numbers which are not divisible?
a) 3458
b)56745
c) 3850
d) 6736
e) 6733
f) 3394
Answer:
3458, 3850, 6736 and 3394 are divisible by 2. Remaining numbers are not divisible.
Reason : The digits of the numbers of ones places is 3 and 5.

Question 2.
Find the numbers which are divisible by 5 and 19. Write the reason for the numbers which are not divisible ?
a) 3568
b) 3540
c) 6585
d) 7550
e) 4235
f) 7200
g) 7865
h) 5880
i) 4440
j) 8198
k) 8645
Answer:
b, d, f, h, j are divisible by both 5 and 10. Remaining numbers are not divisible by 5 and 10.
Reason : The numbers which have n’t zero at their ones place.

Question 3.
The numbers below are divisible by 5. Fill in the blanks with suitable digit.
a) 786_
Answer:
786_ (0/5)

b)560_
Answer:
560_(0/5)

c)785_
Answer:
785_(0/5)

d) 555_
Answer:
555_ (0/5)

e) 586_
Answer:
586(0/5)

f) 786_
Answer:
786_(0/5)

g) 584_
Answer:
584_(0/5)

h) 100_
Answer:
100_(0/5)

Question 4.
Write any 5 numbers which are exactly divisible by 2 and 5.
Answer:
2540, 62570, 250, 367280 and 764520.

Question 5.
Write any 5 numbers which are exactly divisible by 2,5 and 10 ?
Answer:
86540, 79980, 89960, 45570 and 76540.

Do these: (TextBook Page No.65)

Question 1.
Circle the number which is exactly divisible by 3 and 9 and write correct reason.

Answer:

Reason : If the digital root of the given number is 9, then the number is exactly divisible by 3 and 9.

Question 2.
Write any 5 numbers which are exactly divisible by 3 and 9?
Answer:
1350, 1476, 0342, 1539 and 1629.

Do these: (TextBook Page No.66)

Question 1.
Circle the numbers which are divisible by 4. Give the reason, if is not divisible by 4.

Answer:

a, b, d, f and g are divisible by 4. Remaining numbers last two digits is not divisible by 4. So they are not divisible by 4.

Question 2.
Write the missing number in the blank to make the number exactly divisible by 4.
a) 323_
Answer:
323_ (2 / 6)

b) 304_
Answer:
304_ (0)

c) 58_6
Answer:
58_ 6 (1/3)

d) 53_ _
Answer:
53_ _ (04 / 0812 / 16 / 20 / 24 / 28 / 36/ 40)

e) 65_ _
Answer:
65_ _
(04 / 08 / 12 / 16 / 20 / 24 / 28/ 36 / 40).

Do these (Textbook Page No. 67)

Question 1.
Check whether the following numbers are divisible by 6 or not.
1) 210
2) 162
3) 625
4) 120
5) 156
Answer:
210, 162, 120 and 156 are divisible by ’6’.

Question 2.
Change the digits of the following numbers to make them divisible by 6.
1) 543
2) 231
3) 5463
4) 1002
5) 4815
Answer:

Do these: (TextBook Page No.68)

Question 1.
Find the following numbers which are divisible by 8.
a) 2456
b) 3971
c) 824
d) 923
e) 2780
f) 93624
g) 76104
Answer:
a) In 2456, 456 is divisible by 8. So, 2456 is divisible by 8.
b) In 3971, 971 is not divisible by 8. So, 3971 is not divisible by 8.
c) 824 is divisible by 8.
d) 923 is not divisible by 8.
e) In 2780, 780 is not divisible by 8. So, 2780 is not divisible by 8.
f) In 93624, 624 is divisible by 8. So, 93624 is divisible by 8.
g) In 76104, 104 is divisible by 8. So, 76104 is divisible by 8.

Exercise 2:

Question 1.
Circle the following numbers which are divisible by 2 (by using divisibility rule).
3624 3549 7864 8420 8500 8646 5007 7788
Answer:

Question 2.
Find out which of the following numbers are divisible by 6.
1276 43218 71218 71826 4734 3743
Answer:
(i) Given number = 1276
Ones place is 6 that is even.
1276 is divisible by 2.
Digital root of 1276 is 1 + 2 + 7 + 6 = 16
16 is not divisible by 3.
So, 1276 is not divisible by 6.

(ii) Given number = 43218
Ones place is 8 that is even.
43218 is divisible by 2.
Digital root of 43218 is 4 + 3 + 2 + 1 + 8 = 18
18 is divisible by 3.
So, 43218 is divisible by ‘6’.

(iii) Given number = 71218
Ones place is 8 that is even.
71218 is divisible by ‘2’
Digital root of 71218 is
7 + 1 + 2 + 1 + 8 = 19
19 is not divisible by ‘3’
So, 71218 is divisible by ‘6’.

(iv) Given number = 71826
Ones place is ‘6’ that is even.
71826 is divisible by ‘2’
Digital root of 71826 is 7 + 1 + 8 + 2 + 6 = 24
24 is divisible by ‘3’
So, 71826 is divisible by’6′.

(v) Given number = 4734
Ones place is ‘4’ that is even.
4734 is divisible by ‘2’
Digital root of 4734 is 4 + 7 + 3 + 4 = 18
18 is divisible by’3′
So, 4734 is divisible by ‘6’.

(vi) Given number = 3743
Ones place is ‘3’ that is odd.
3743 is not divisible by ‘2’
So, 3743 is not divisible by ‘6’.

Question 3.
The number 50 9 is exactly divisible by 9. Fill the with the correct number.
Answer:
Given number 50 19 is exactly divisible by 9.
So, digital root is 5 + 0 + ? + 1 + 9 = 15 + ?
= 15 + 3 = 18 = 1 + 8 = 9
∴ 9 is divisible by ‘9’.

Question 4.
The number 4 468 is exactly divisible by 6. Fill the with the correct number.
Answer:
Given number 4 468 is exactly divisible by 6. So, digital roots is 4 + + 4 + 6 + 8 = 22 + 2 = 24
24 is divisible by 3.
So, 4 2 468 is divisible by 6.

Question 5.
Fill the blanks with suitable digits. So that it can be divisible by 2 and 10.
678_ 588_ 388_ 222_
364_ 786_ 666_ 788_
Answer:
Given numbers are divisible by ’10’ they have ‘0’ in its units place and that number also divisible by ‘2’.
6780 5880 3880 2220
3640 7860 6660 7880

Question 6.
Find the numbers which are divisible by 4 and 8.
2104 726352 1800 32256 52248 25608
Answer:
Observe the last two digits of the given numbers 2104, 726352, 1800, 32256, 52248 and 25608.
There are 04, 52, 00, 56, 48 and 08 at the end of the numbers.
Hence all these numbers are multiples of 4.
So, 2104, 726352, 1800, 32256, 52248 and 25608 are divisible by 4.

Divisible by 8 :
i) Given number 2104
Last three digits of given 104 is divisible by 8.
So, 2104 is divisible by 8.

ii) Given number 726352
Last three digits of given 352 is di-visible by 8.
So, 726352 is divisible by 8.

iii) Given number 1800
Last three digits of given 800 is divisible by 8.
So, 1800 is divisible by 8.

iv) Given number 32256
Last three digits of given 256 is divisible by 8.
So, 32256 is divisible by 8.

v) Given number 52248
Last three digits of given 248 is divisible by 8.
So, 52248 is divisible by 8.

Question 7.
Try whether the numbers are divisible by 2, 3, 4, 5, 6, 8, 9 and 10.
a) 333
b) 128
c) 225
d) 7535
e) 8289
f) 99483
g) 67704
h) 67713
i) 9410
j) 67722
k) 20704
l) 35932
m) 85446
n) 90990
o) 18540
Answer:
a) Given number is 333
Digital root of 333 is 3 + 3 + 3 = 9
Hence it is multiple of 3 and 9.
So, 333 is divisible by 3 and 9.

b) Given number is 128.
Ones place is 8 that is even.
128 is divisible by 2.
Last two digits of the given number is 28 is multiple of 4.
So, 128 is divisible by 4.
128 ÷ 8 = 16.
So, 128 is divisible by 8
Hence 128 is divisible by 2, 4 and 8.

c) Given number is 225
Ones place is ‘5’, so it is divisible by 5.

d) Given number is 7535
Ones place is 5, so, it is divisible by 5.

e) Given number is 8289.
Digital root of 8289 is 8 + 2 + 8 + 9 = 27 = 2 + 7 = 9
9 is multiple of 3 and 9.
So, 8289 is divisible by 3 and 9.

f) Given number is 99483.
Digital root of 99483 is 9 + 9 + 4 + 8 + 3 = 33 = 3 + 3 = 6
6 is multiple of 3, so, 99483 is divisible by 3.

g) Given number is 67704
Ones place is 4 that is even.
67704 is divisible by 2.
Digital root of 67704 is 6 + 7 + 7 + 0 + 4 = 24 = 2 + 4 = 6.
6 is multiple of 3, so, 67704 is divisible by 3.
The number is divisible by 2 and 3, so, 67704 is divisible by 6.
Last two digits of the number is 04 is divisible by 4. So, 67704 is divisible by 4.
Last three digits of the number is 704 is divisible by 8.
So, 67704 is divisible by 8.
∴ 67704 is divisible 2, 3,4, 6 and 8.

h) Given number is 67713
Digital root of 67713 is 6 + 7 + 7 + 1 + 3 = 24 = 2 + 4 = 6
6 is multiple of 3.
So, 67713 is divisible of 3.

i) Given number is 9410
Last digit of given number is ‘0’ that is even.
So, 9410 is divisible by 2, 5 and 10.

j) Given number is 67722
Ones place is 2 that is even.
67722 is divisible by 2.
Digital root of 67722 is 6 + 7 + 7 + 2 + 2 = 24 = 2 + 4 = 6
‘6’ is divisible by 3.
So, 67722 is divisible by 6.

k) Given number is 20704.
Ones place is 4 that is even.
20704 is divisible by 2.
Last two digits of the number is 04 is divisible by 4.
So, 20704 is divisible by 4.
Last three digits of the number is 704 is divisible by 8. So, 20704 is divisible by 8.
∴ 20704 is divisible 2, 4 and 8

l) Given number is 35932
Ones place is 2 that is even.
35932 is divisible by 2.
Last two digits of the number is 32
32 is multiple of 4.
So, 35932 is divisible by 4.
∴ 35932 is divisible 2 and 4

m) Given number is 85446
Ones place is 6 that is even.
85446 is divisible by 2.
Digital root of 85446 is 8 + 5 + 4 + 4 + 6 = 27 = 2 + 7 = 9
∴ ‘9’ is multiple of 3 and 9.
So, 85446 is divisible by 3 and 9.
∴ 85446 is divisible by 2,3 and 9.

n) Given number is 90990
Last digit is ‘0’.
So, it is divisible by 2,5 and 10.

o) Given number is 18540 Last digit is ‘0’.
So it is divisible by 2, 5 and 10.

Question 8.
Find the missing digit that would make each number divisible by the given number.
a) 395 _ by 10
Answer:
395 0

b)24305 _ by 9
Answer:
24305 4

c) 69839 _ by 3 and 9
Answer:
698391

d) 271 _ 8 by 6
Answer:
2710 8

e) 20710 _
Answer:
20710_

f) 5027 _ 5 by 3 and 5
Answer:
50273 5

g) 145 _ 2 by 8
Answer:
14512

h) 92048 _ by 2
Answer:
92048 _ (0/2/4/6/8)

i) 23405 _ by 5
Answer:
(0/5)

Answer: For eiven ‘e’ problem divisible by is not given. So, the problem is not solved.

Question 9.
Find the smallest number that is to be added to 289279, so that it can be divisible by 8.
Answer:
Given number is 289279.
Last three digits of given number is 279.
279 is not divisible by 8, but 280 is divisible by 8.
289279 is added by 1 it is divisible by 8.

Do these: (TextBook Page No. 69 & 70)

Question 1.
Write first ten multiples of the following.
a) 3 b) 5 c) 8 d) 9 e) 10
Answer:
a) Ten multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
b) Ten multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
c) Ten multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
d) Ten multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
e) Ten multiples of 10 = 10, 20, 30, 40, 50, 60, 70, 80, 90, 100

Question 2.
Find out the multiples of 2, 3, 5 from 1 to 20. Write separately.
Answer:
First 1 to 20 multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
First 1 to 20 multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
First 1 to 20 multiples of 5 = 5, 10, 15,20

Question 3.
Write down the first 10 multiples of 7.
Answer:
First 10 multiples of 7 is = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70.

Question 4.
Find out the multiples of 7, 8, 10 from the following numbers and write separately.
20, 14, 45, 24, 32, 35, 90, 8, 7, 10, 441, 385
Answer:
Multiples of 7 from given = 7, 14, 35, 385, 441
Multiples of 8 from given = 8, 24, 32
Multiples of 10 from given = 10, 20, 90

Question 5.
Find out the numbers which are not the multiples of 3.

Answer:
8 26 32 28 88 48

Question 6.
Write the odd multiples of 9 less than hundred.
Answer:
9, 27, 45, 63, 81, 99.

Do these: (TextBook Page No.71)

Question 1.
Write first 10 multiples of the following numbers and list the com-mon multiples.
a) 2 and 4
b) 4 and 12
c) 6 and 8
d) 5 and 10
Answer:
a) Multiples of 2 : 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Multiples of 4 : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Common multiples of 2 and 4 : 4, 8, 12, 16, 20

b) Multiples of 4 : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Common multiples of 4 and 12 : 12, 24, 36

c) Multiples of 6 : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
Multiples of 8 : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
Common multiples of 6 and 8 : 24, 48

d) Multiples of 5 : 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 10 : 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
Common multiples of 5 and 10 : 10, 20, 30, 40, 50.

Do this: (TextBook Page No. 72)

Find the LCM for the following sets of numbers.
Answer:
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …………..
Multiples of 15: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …………
Common multiples of 12 and 15 = 60, 120
Least common multiple of 12 and 15 = 60.

2) Multiples of 16: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160
Multiples of 20 : 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, ………..
Common multiples of 16 and 20 = 80, 160.
Least common multiple of 16 and 20 = 80

3) Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ……….., 120, ………..
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …………
Multiples of 20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, …………
Least common multiple of 16 and 20 = 80
Common multiples of 8, 12 and 20 = 120
Least common multiple of 8, 12 and 20 = 120

4) Multiples of 15 : 15, 30, 45, 60, 75, 90, 105, 120, 135, 150 ……….
Multiples of 20 : 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
Common multiples of 15 and 20 = 60, 120
Least common multiple of 15 and 20 = 60

5) Multiples of 6 : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …………
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ………..
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Common multiples of 6, 9 and 12 = 36
Least common multiple of 6, 9 and 12 = 36.

Try this: (TextBook Page No. 72)

Find the LCM for the following pairs of numbers. What do you ob-serve.
1) 15, 30
2) 4, 16
3) 5, 15
4) 6, 18
Answer:
1) Multiples of 15 : 15, 30, 45.60. 75.90, 105, 120, 135,150
Multiples of 30 : 30,60, 90. 120, 150, 180, 210, 240, 270, 300, …………
Common multiples of 15 and 30
30, 60, 90, 120, 150, ……………..
Least common multiples of 15 and 30 = 30

2) Multiples of 4 :
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …………
Multiples of 16 : 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, …………..
Common multiples of 4 and 16 = 16, 32, 48 …………
LCM of 4 and 16 = 16

3) Multiples of 5 : 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, …………..
Multiples of 15 : 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, ………….
Common multiples of 5 and 15 = 15, 30, 45, 60
LCM of 5 and 15 = 15

4) Multiples of 6 :
6, 12, 18,24, 30, 36,42,48, 54, 60, ……………
Multiples of 18 :
18, 36, 54, 72,90, 108, 126, 144, 162, 180, ……….
Common multiples of 6 and 18 = 18, 36, 54
LCM of 6 and 18 = 18
Observation :
In a given pair of numbers, if one of them is multiple of other number than the biggest number is LCM of the number.

Do these: (TextBook Page No.75)

Question 1.
Find all the factors of the following numbers.
a) 21
b) 38
c) 72
d) 96
Answer:
a) 21 = 1 × 21 = 3 × 7 = 7 × 3 = 21 × 1
Thus, all the factors of 21 are : 1, 3, 7 and 21.

b) 38 = 1 × 38
= 2 × 19
Thus, all the factors of 38 are : 1, 2, 19 and 38.

c) 72 = 1 × 72
= 2 × 36
= 3 × 24
= 4 × 18
= 6 × 12
= 8 × 9
Thus, all the factors of 72 are : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

d) 96 = 1 × 96
= 2 × 48
= 3 × 32
= 4 × 24
= 6 × 16
= 8 × 12
Thus, all the factors of 96 are : 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96.

Question 2.
Find out whether the first number is a factor of the second number.
a) 14; 322
b) 26; 832
c) 35; 425
d) 56; 3500
e) 8; 48
f) 14; 37
g) 15; 75
h) 12; 72
Answer:
a) Yes, 14 is a factor of 322.
b) Yes, 26 is a factor of 832.
c) No, 35 is not a factor of 425.
d) No, 56 is not a factor of 3500.
e) Yes, 8 is a factor of 48.
0 No, 14 is not a factor of 37.
g) Yes, 15 is a factor of 75.
h) Yes, 12 is a factor of 75.

Question 3.
What are the factors of 66 ?
Answer:
Factors of 66 = 1 × 66
= 2 × 33
= 3 × 22
= 6 × 11
Factors are 1, 2, 3, 6, 11, 22, 33 and 66.

Question 4.
Write all the even factors of 64.
Answer:
Factors of 64 = 1 × 64 = 2 × 32
= 4 × 16 = 8 × 8
Even Factors of 64 are : 2, 4, 8, 16, 32 and 64.

Question 5.
List out the numbers, which are prime / composite below 20.
Answer:

Fun activity:

Answer the following questions:

Question 1.
What are the prime numbers between 1 to 10 ?
Answer:
2, 3, 5, 7.

Question 2.
What are the prime numbers between 10 to 20 ?
Answer:
11, 13, 17, 19.

Question 3.
What are the prime numbers between 20 to 50 ?
Answer:
23, 29, 31, 37, 41, 43, 47.

Question 4.
How many prime numbers are there between 1 to 50 ?
Answer:
15 prime numbers are there. (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 49).

Question 5.
What are the prime numbers between 50 to 100 ?
Answer:
53, 59, 61, 67, 71, 73, 79, 83, 87, 89, 97.

Question 6.
How many prime numbers are there between 50 to 100 ?
Answer:
10 prime numbers are there between 50 to 100.

Question 7.
Do you observe any speciality in these prime numbers ? What is it ?
Answer:
All numbers are odd, ‘2’ is only one even prime.

Question 8.
Are the all prime numbers even or odd ?
Answer:
All are odd except 2.

Do these: (TextBook Page No.78)

a) 52
b) 100
c) 88
d) 96
e) 90
Answer:
a)

∴ Prime factorisation of 52 = 2 × 2 × 13

b)

∴ Prime factorisation of 100 = 2 × 2 × 5 × 5

c)

∴ Prime factorisation of 88 = 2 × 2 × 2 × 11

d)

∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

e)

∴ Prime factorisation of 90 = 2 × 3 × 3 × 5

Question 2.
The prime factorisation of 12 × 15 is
Answer:

∴ 12 × 15 = 2 × 2 × 3 × 3 × 5.

Question 3.
Match the following:

Answer:

Question 4.
5 × 2 × 3 × 3 is the prime factorisation of _________
Answer:
90

Do this: (TextBook Page No.79)

Question 1.
Find the common factors of the following numbers and represent in the diagram.
a) 6 and 12
b) 12 and 20
c) 9 and 18
d) 11 and 22
Answer:
a) Factors of 6 = 1, 2, 3, 6
Factors of 12 = 1, 2, 3, 4, 6, 12

b) Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 20 = 1, 2, 4, 5, 10, 20

c) Factors of 9 = 1, 3, 9
Factors of 18 = 1, 2, 3, 6, 9, 18

d) Factors of 11 = 1, 11
Factors of 22 = 1,2, 11, 22

Do this: (TextBook Page No.80)

Find the HCF of the following pairs of numbers by writing common factors.
1) 21 and 28
2) 34 and 20
3) 33 and 39
4) 16 and 36
5) 12 and 18
6) 80 and loo
Answer:
1) Factors of 21 = 1, 3, 7, 21
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factors of 21 and 28 = 1, 7
HCF of 21 and 28 = 7.

2) Factors of 34 = 1, 2, 17, 34
Factors of 20 = 1, 2, 4, 5, 10, 20
Common factors of 34 and 20 = 1, 2
HCF of 34 and 20 = 1, 2.

3) Factors of 33 = 1, 3, 11, 33
Factors of 39 = 1, 3, 13, 39
Common factors of 33 and 39 = 1, 3
HCF of 33 and 39 = 3.

4) Factors of 16 = 1, 2, 4, 8, 16
Factors of 36 = 1, 2, 3, 4, 9, 12, 18, 36
Common factors of 16 and 36 = 1, 2, 4

5) Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 18 = 1, 2, 3, 6, 9, 18
Common factors of 12 and 18 = 1,2, 3, 6
HCF of 12 and 18 = 6

6) Factors of 80 = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
Common factors of 80 and 100 = 1, 2, 4, 5, 10, 20
HCF of 80 and 100 = 20.

Try this: (TextBook Page No.80)

Find the HCF for the following pair of numbers. What do you observe ?
1) 4, 16
2) 4, 12
3) 5, 15
4) 14, 42
Answer:
1) Factors of 4 = 1, 2, 4
Factors of 16 = 1,2, 4, 8, 16
Common factors of 4 and 16 = 1, 2, 4
HCF of 4 and 16 = 4.

2) Factors of 4 = 1, 2, 4
Factors of 12 = 1,2, 3, 4, 6, 12
Common factors of 4 and 12 = 1, 2, 4
HCF of 4 and 12 = 4.

3) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Common factors of 5 and 15 = 1, 5
HCF of 5 and 15 = 5.

4) Factors of 14 = 1, 2, 7
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 22
Common factors of 14 and 42 = 1, 2, 7
HCF of 14 and 42 = 7

Observation :
In pair of numbers, if one of them is multiple of the other. The smallest number is the HCF of the pair of number.

Do these: (TextBook Page 82)

Question 1.
Find LCM and HCF by prime factorisation method for the following.
a) 15, 48
b) 18, 42, 48
c) 15, 25, 30
d) 10, 15, 25
e) 15, 18, 36, 20
Answer:
a) Given numbers 15 and 48.

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Common factors = 1 × 3
Other factors = 5 × 2 × 2 × 2 × 2
LCM = 1 × 3 × 5 × 2 × 2 × 2 × 2 = 240
HCF = 1 × 3 = 3.

b) Given numbers 18, 42 and 48.

Prime factorisation of 18 = 1 × 2 × 3 × 3
Prime factorisation of 42 = 1 × 2 × 3 × 7
Common factors = 1 × 2 × 3
Other factors = 3 × 7 × 2 × 2 × 2
LCM = 1 × 2 × 3 × 3 × 7 × 2 × 2 × 2 = 1008
HCF = 1 × 2 × 3 = 6

c) Given numbers 15, 25 and 30

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 25 = 1 × 5 × 5
Prime factorisation of 30 = 1 × 2 × 3 × 5
Common factors = 1 × 5
Other factors = 3 × 3 × 5 × 2
LCM = 1 × 5 × 3 × 3 × 5 × 2 = 450
HCF = 1 × 5 = 5.

d) Given numbers 10, 15 and 25

Prime factorisation of 10 = 1 × 2 × 5
Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 25 = 1 × 5 × 5
Common factors = 1 × 5 = 5
Other factors = 2 × 3 × 5 = 30
LCM = 5 × 30 = 150
HCF = 5

e) Given numbers 15, 18, 36 and 20

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 18 = 1 × 2 × 3 × 3
Prime factorisation of 36 = 1 × 3 × 3 × 2 × 2
Prime factorisation of 20 = 1 × 2 × 2 × 5
Common factors = 1
Other factors = 3 × 5 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 5 = 194,400
LCM = 194, 400
HCF = 1.

Question 2.
Find LCM and HCF by division method.
a) 16, 28, 36
b) 12, 18, 42
c) 30, 75, 90
d) 24, 32, 48
e) 12, 15, 18

Answer:
a) Given numbers 16, 28 and 36

LCM = 2 × 2 × 4 × 7 × 9 = 1008
HCF = 2 × 2 = 4

b) Given numbers 12, 18 and 42

LCM = 2 × 3 × 2 × 3 × 7 = 252
HCF = 2 × 3 = 6

c) Given numbers 30, 75, 90

LCM = 5 × 3 × 2 × 5 x 6
HCF = 5 × 3 = 15

d) Given numbers 24, 32, 48

LCM = 4 × 2 × 3 × 4 × 6 = 576
HCF = 4 × 2 = 8

e) Given numbers 12, 15 and 18

LCM = 3 × 2 × 2 × 5 × 6 = 360
HCF = 3

Exercise 3:

Solve the following word-problems:

Question 1.
There are some fruits in a basket. If we arrange 4 or 6 or 8 or 10 fruits in a pile, no fruits are left in the bas¬ket. What is the minimum number of fruits in the basket ?

Answer:
Arrangement of fruits in a pile are 4 or 6 or 8 or 10
We need to find the LCM of 4, 6, 8, 10

LCM = 2 × 2 × 1 × 3 × 2 × 5 = 120
The minimum number of fruits in the basket = 120.

Question 2.
Ramu has 16 blue marbles and 12 white ones. If he wants to arrange them in identical groups without leaving any marbles, what is the maximum number in each group Ramu can make?

Answer:
Number of marbles Ramu has = 16 and 12
We need to find the HCF of 12 and 16

HCF = 4
Maximum number of marbles Ramu can make = 4.

Question 3.
Two Neon lights are turned on at the same time. One blinks for every 4 seconds and other blinks for every 6 seconds. In 60 seconds how many times will they blink at a time ?

Answer:
Blink time of lights = 4 sec and 6 sec
LCM of 4 and 6 =
LCM = 2 × 2 × 3 = 12 sec
The lights blinks at a time for every 12 sec.
Given time = 60 sec
Number of times will the lights blink = 60 ÷ 12 = 5 times.

Question 4.
There are 40 girls and 32 boys, who want to participate in state-level games competition. If each team must have the same number of girls and the same number of boys.
i) What is the maximum number in each team that can participate in state-level games ?
ii) How many boys and girls will be on each team?

Answer:
Number of boys 32 and girls participate in the game is 32 and 40
i) To find maximum number in each team can participate in state-level game, we need to find
HCF of 32 and 40

∴ HCF = 2 × 2 × 2 = 8
The maximum number in each team = 8 numbers.

ii) To find out boys and girls will be an each team, we find LCM of 32 and 40.

∴ LCM of 32 and 40 = 2 × 2 × 2 × 4 × 5 = 160.

Question 5.
Find the least number of sheets of paper required to make notebooks containing 32 sheets or 40 sheets or 48 sheets without a single sheet leaving behind.

Answer:
To find the least number of sheets of paper required to make note books, we need to find LCM of 32, 40, 48.

LCM = 2 × 2 × 2 × 4 × 5 × 6 = 960
∴ 960 of sheets of paper required to make notebooks containing given sheets.

Question 6.
What is the least number of chairs needed for an auditorium so that they can be arranged either 27 in a row or 33 in a row?

Answer:
To find least number of chairs needed for an auditorium are need to find LCM of 27 and 33.

LCM = 3 × 9 × 11 = 297.
∴ 297 chairs needed for an auditorium.