Category: Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volumes Ex 14.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volumes Exercise 14.1

Question 1.
There are two cuboidal boxes as shown in the given figure. Which box requires the less amount of material to make?

Solution:
Volume of a cuboid V₁ = lbh
= 60 × 40 × 50
V₁ = 1,20,000 cubic units.
Volume of a cube V₂ = (a)³
= (50)³ = 50 × 50 × 50
V₂ = 1,25,000 cubic units.
∴ The cuboidal box requires less amount of material.
∴ V₁ < V₂

Question 2.
Find the side of a cube whose surface area is 600 cm²
Solution:
Total surface area of a cube = 6a²
⇒ 6a² = 600
⇒ a² = [latex][latex]\frac { 600 }{ 6 }[/latex][/latex] = 100
⇒ a² = 100
⇒ a = √100 = 10
∴ The side of a cube (a) = 10 cm.

Question 3.
Prameela painted the outer surface of a cabinet of measures 1m × 2m × 1 .5m. Find the surface area she cover if she painted all except the bottom of the cabinet?
Solution:
The area of outer surface of a cabinet except the bottom of the cabinet will be equal to its lateral surface area.
I = lm,b = 2m, h = 1.5m.
A = 2h(l + b)
= 2 × 1.5(1 + 2)
= 3 × 3 = 9 m².

Question 4.
Find the cost of painting a cuboid of dimensions 20cm × 15 cm × 12 cm at the rate of 5 paisa per square centimeter.
Solution:
l = 20cm, b = 15cm, h = 12cm.
∴ Total surface area of a cuboid
A = 2 (lb + bh + lh)
=2(20 × 15 + 15 × 12 + 20 × 12)
= 2 (300 + 180 + 240)
= 2 × 720
= 1440 sq.cm.
The cost of painting a cuboid at the rate of 5 paisa per sq. cm for 1440 sq.cm.
= 1440 × 5 paisa
= 7200 paise
= ₹ [latex]\frac { 7200 }{ 100 }[/latex]
= ₹ 72

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions and Answers.

8th Class Maths 7th Lesson Frequency Distribution Tables and Graphs InText Questions and Answers

Do this

Question 1.
Here are the heights of some of Indian cricketers. Find the median height of the team.   [Page No. 154]

Answer:
The ascending order of heights is 5’3″, 5’5″, 57″, 5’8″, 5’9″, 571″, 571″, 6’0″, 6’0″, 6’7″
Number of players = 10 (is an even)
Median = Mean of [latex]\left(\frac{\mathrm{n}}{2}\right)[/latex] and [latex]\left(\frac{n}{2}+1\right)[/latex] terms = Mean of [latex]\left(\frac{10}{2}\right)[/latex] and [latex]\left(\frac{10}{2}+1\right)[/latex] terms = Mean of 5, 6 terms =

Question 2.
Ages of 90 people in an apartment are given in the adjacent grouped frequency distribution.

i) How many Class Intervals are there in the table?
ii) How many people are there in the Class Interval 21 – 30?
iii) Which age group people are more in that apartment?
iv) Can we say that both people the last age group (61-70) are of 61, 70 or any other age?    [Page No. 158]
Answer:
i) 7    ii) 17     iii) 31 – 40     iv) Yes, they are 62, 63, ……, 69

Question 3.
Long jump made by 30 students of a class are tabulated as

I. Are the given class intervals inclusive or exclusive?
II. How many students are in second class interval?
III. How many students jumped a distance of 3.01 m or more?
IV. To which class interval does the student who jumped a distance of 4.005 m belongs?    [Page No. 160]
Answer:
I. Inclusive
II. 7
III. 15 + 3 + 1 = 19
IV. 401 – 500

Question 4.
Calculate the boundaries of the class intervals in the above table.     [Page No. 160]
Answer:
Boundaries:
100.5 – 200.5
2005 – 300.5
300.5 – 400.5
400.5 – 500.5
500.5 – 600.5

Question 5.
What is the length of each class interval in the above table?     [Page No. 160]
Answer:
100

Question 6.
Construct the frequency polygons of the following frequency distributions.       [Page No. 174]
i) Runs scored by students of a class in a cricket friendly match.

ii) Sale of tickets for drama in an auditorium.

Answer:
i)

Steps of construction: Runs scored (Mid values of C.I.)
Step – 1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles, (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class. Also calculate their mid values (A and G) and mark on the axis. (Here, the first class is 10 – 20. So, to find the class preceding 10 – 20, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 60 – 70.
Step – 5: Join the first end point B to A and last end point F to G which completes the frequency polygon.
Frequency polygon can also be drawn independently without drawing histogram. For this, we require the midpoints of the class interval of the data.

ii)

Steps of construction:
Step -1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class.
Step – 5: To find the class preceding 2.5 – 7.5, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 32.5 – 37.5 like A, G.
Step – 6: Join A to B and G to F.
∴ The required ABCDEFG polygon is formed.

Try these

Question 1.
Give any three examples of data which are in situations or in numbers.      [Page No. 148]
Answer:
1) The data of 35 students who like different games:

2) The data of 35 students who like different colours:

3) The data of 35 students who like different fruits:

Question 2.
Prepare a table of estimated mean, deviations of the above cases. Observe the average of deviations with the difference of estimated mean and actual mean. What do you infer?
[Hint: Compare with average deviations]      [Page No. 151]
Answer:

Mean = [latex]\frac{\Sigma x_{i}}{N}[/latex] = [latex]\frac{80}{5}[/latex] = 16
Mean of the deviations = [latex]\frac{-5}{5}[/latex] = -1
Mean = Assumed mean + Mean of deviations = 17 + (-1) = 16
∴ Assumed mean, original arithmetic mean are equal.

Question 3.
Estimate the arithmetic mean of the following data.      [Page No. 153]
i) 17, 25, 28, 35, 40
ii) 5, 6, 7, 8, 8, 10 10, 10, 12, 12, 13, 19, 19, 19, 20
Verify your answers by actual calculations.
Answer:
i) 17, 25, 28, 35, 40
Assumed Mean = 35

A.M in general method = [latex]\frac{\text { Sum of the observations }}{\text { No. of the observations }}[/latex]

ii) 5, 6, 7, 8, 8, 10, 10, 10, 12, 12, 13, 19, 19, 19, 20
Assumed Mean = 10

Question 4.
Find the median of the data 24, 65, 85, 12, 45, 35, 15.       [Page No. 155]
Answer:
The ascending order of the data is 12, 15, 24, 35, 45, 65, 85
Number of observations (n) = 7 (odd)
∴ Median = [latex]\frac{n+1}{2}[/latex] = [latex]\frac{7+1}{2}[/latex] = 4th term
∴ Median = 35

Question 5.
If flie median of x, 2x, 4x is 12, then find mean of the data.       [Page No. 155]
Answer:
Given observations are x, 2x, 4x
∴ Median = 2x
According to the sum
2x = 12 ⇒ x = 6
2x = 2 × 6 = 12
4x = 4 × 6 = 24
∴ The mean of 6, 12, 24 = [latex]\frac{6+12+24}{3}[/latex] = [latex]\frac{42}{3}[/latex] = 14

Question 6.
If the median of the data 24, 29, 34, 38, x is 29 then the value of ‘x’ is
i) x > 38   ii) x < 29   iii) x lies in between 29 and 34   iv) none       [Page No. 155]
Answer:
Median of 24, 29, 34, 38, x is 29.
n = 5 is an odd.,
∴ Median = [latex]\frac{n+1}{2}[/latex] = [latex]\frac{5+1}{2}[/latex] = 3rd term
If x is less than 29, then only 29 should be a 3rd term.
∴ x < 29

Question 7.
Less than cumulative frequency is related to …….     [Page No. 165]
Answer:
Upper boundaries

Question 8.
Greater than cumulative frequency is related to ……..      [Page No. 165]
Answer:
Lower boundaries

Question 9.
Write the Less than and Greater than cumulative frequencies for the following data.       [Page No. 165]

Answer:

Question 10.
What is total frequency and less than cumulative frequency of the last class above problem? What do you infer?    [Page No. 165]
Answer:
The sum of the frequencies in the above distribution table = 30
Less than C.F of the last C.I = 30
∴ Sum of the observations = Less than C.F of last C.I.

Question 11.
Observe the adjacent histogram and answer the following questions.     [Page No. 169]

i) What information is being represented in the histogram?
ii) Which group contains maximum number of students?
iii) How many students watch TV for 5 hours or more?
iv) How many students are surveyed in total?
Answer:
i) The histogram represents students who watch the T.V.’s .(Duration of watching T.V).
ii) 4th class interval contains maximum number of students.
iii) 35 + 15 + 5 = 55
iv) Number of students are surveyed = 10 + 15 + 20 + 35 + 15 + 5 = 100

Think, discuss and write

Question 1.
Is there any change in mode, if one or two or more observations, equal to mode are included in the data?    [Page No. 155]
Answer:
If one or two or more observations equal to mode are included there will be no change in the mode.
Ex: The mode of 5, 6, 7, 8, 7, 9 is 7.
If 3, 7’s are added to above observations there will be no change in the mode.

Question 2.
Make a frequency distribution of the following series.
1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7.    [Page No. 161]
Answer:
The range of the observations = Highest value – Least value
∴ Range = 7 – 1 = 6
If number of classes = 7 then
Class Interval = [latex]\frac{\text { Range }}{\text { No. of classes }}[/latex] = [latex]\frac{6}{7}[/latex] = 0.8 (approx.)

Question 3.
Construct a frequency distribution for the following series of numbers.
2, 3, 4, 6, 7, 8, 9, 9, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 16, 17, 18, 18, 19, 20, 20, 21, 22, 24, 24, 25. (Hint: Use inclusive classes)      [Page No. 161]
Answer:
Range = Maximum value – Minimum value = 25 – 2 = 23
Class Interval = [latex]\frac{\text { Range }}{\text { No. of classes }}[/latex] = [latex]\frac{23}{5}[/latex] = 4.6 = 5 (approx.) [∵ No. of classes = 5]

Question 4.
What are the differences between the above two frequency distribution tables?      [Page No. 161]
Answer:
The class intervals of first frequency distribution table are exclusive class intervals. The C.I’s of 2nd frequency distribution table are inclusive class intervals.

Question 5.
From which of the frequency distributions we can write the raw data again?      [Page No. 161]
Answer:
Classes

Question 6.
All the bars (or rectangles) in a bar graph have     [Page No. 168]
a) same length b) same width c) same area d) equal value
Answer:
b) same width

Question 7.
Does the length of each bar depend on the lengths of other bars in the graphs?     [Page No. 168]
Answer:
No

Question 8.
Does the variation in the value of a bar affect the values of other bars in the same graph?      [Page No. 168]
Answer:
No

Question 9.
Where do we use vertical bar graphs and horizontal bar graphs?     [Page No. 168]
Answer:
Vertical and horizontal bar graphs are used to present the equal widths corresponding to the given frequencies.

Question 10.
Class boundaries are taken on the X-axis. Why not class limits?      [Page No. 172]
Answer:
The difference between upper and lower boundaries gives the class interval i.e., we take class boundaries on X-axis.

Question 11.
Which value decides the width of each rectangle in the histogram?      [Page No. 172]
Answer:
Class Interval

Question 12.
What does the sum of heights of all rectangles represent?     [Page No. 172]
Answer:
Sum of the frequencies.

Question 13.
How do we complete the polygon when there is no class preceding the first class?       [Page No. 173]
Answer:
The frequency of preceding class should be taken as ‘0’ (zero) then it should be joined.

Question 14.
The area of histogram of a data and its frequency polygon are same. Reason how?       [Page No. 173]
Answer:
Because both the figures are constructed on the basis of mid values of class intervals.

Question 15.
Is it necessary to draw histogram for drawing a frequency polygon?       [Page No. 173]
Answer:
No need.

Question 16.
Shall we draw a frequency polygon for frequency distribution of discrete series?       [Page No. 173]
Answer:
No, we can’t.

Question 17.
Histogram represents frequency over a class interval. Can it represent the frequency at a particular point value?            [Page No. 175]
Answer:
Yes, the histogram represents the frequency at a particular point value. Since the length of a histogram represents the value of its corresponding frequency (length of the frequency).

Question 18.
Can a frequency polygon give an idea of frequency of observations at a particular point?       [Page No. 175]
Answer:
Yes, we can identify the frequency of observation with a frequency polygon at a particular point. Since the height of the polygon is equal to frequency of polygon.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D Ex 13.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D Exercise 13.2

Question 1.
Count the number of faces , vertices , and edges of given polyhedra and verify Euler’s formula.

Solution:

Question 2.
Is a square prism and cube are same? explain.
Solution:
All cubes are square prisms, but converse is not true. (i.e.,) All square prisms are either cubes or, not.

Question 3.
Can a polyhedra have 3 triangular faces only? explain.
Solution:
Any polyhedra can’t have 3 triangular faces because the triangular pyramids are formed starts with 4 faces. So it does not exist.

Question 4.
Can a polyhedra have 4 triangular faces only? explain.
Solution:
Yes, a triangular pyramid have 4 triangular faces.

Question 5.
Complete the table by using Euler’s formula.

F	8	5	?
V	6	?	12
E	?	9	30

Solution:

F	8	5	20
V	6	6	12
E	12	9	30

i) E = V + F- 2 = 8 + 6- 2 = 12
ii) V = E + 2- F = 9 + 2- 5 = 6
iii) F = E + 2- V = 30 + 2-12 = 20

Question 6.
Can a polyhedra have 10 faces ,20 edges and 15 vertices?
Solution:
No. of faces = 10
No. of edges = 20
No. of vertices = 15
According to Euler’s formula E = V + F – 2
⇒ 20 = 15 + 10 – 2
20 = 25 – 2
20 = 23 (False)
∴ A polyhedra doesn’t exist with 10 faces, 20 edges, 15 vertices.

Question 7.
Complete the following table

Solution:

Question 8.
Name the 3-D objects or shapes that can be formed from the following nets.

(i) Hexagonal pyramid
(ii) Cuboid
(iii) Pentagonal pyramid
(iv) Cylinder
(v) Cube
(vi) Hexagonal pyramid
(vii) Trapezoid

Question 9.
Draw the following diagram on the check ruled book and fmd out which of the following diagrams makes cube?
(i)

Solution:
The diagrams which makes cubes are a, b, c, e.

(ii) Answer the following questions.
(a) Name the polyhedron which has four vertices, four faces’?
(b) Name the solid object which has no vertex?
(c) Name the polyhedron which has 12 edges’?
(d) Name the solid object which has one surface’?
(e) How a cube is different from cuboid?
(f) Which two shapes have same number of edges, vertices and faces?
(g) Name the polyhedron which has 5 vertices and 5 faces’?
Solution:
(a) Tetrahedron
(b) Sphere
(c) Cube/Cuboid
(d) Sphere
(e) Cube is a regular polyhedron where cuboid is not.
(f) Cube, Cuboid
(g) Square pyramid

(iii) Write the names of the objects given below

Solution:
(a) Octagonal prism
(b) Hexagonal prism
(c) Triangular prism
(d) Pentagonal prism

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D Ex 13.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D Exercise 13.1

Question 1.
Draw the following 3-D figures on isometric dot sheet.

Solution:

Question 2.
Draw a cuboid on the isometric dot sheet with the measurements 5 units × 3 units × 2 units.
Solution:

Question 3.
Find the number of unit cubes in the following 3-D figures.

Solution:

Figure	No. of cubes
i)	2 + 3 = 5
ii)	2 × 4 + 1 = 9
iii)	4 + 16 = 20
iv	1 + 4 + 9 = 14

Question 4.
Find the areas of the shaded regions of the 3-D figures given in question number 3.

Solution:

Figure	Area of the shaped regions
i)	3 × 1 × 1 =3 Sq. Units.
ii)	4(2 × 1) + 1 = 9 Sq. Units.
iii)	4 + (16 – 8) = 4 + 8= 12 Sq. Units.
iv)	1 + (4 – 1) ÷ (9 – 4) = 1 + 3 + 5 = 9 Sq.Units.

Question 5.
Consider the distance between two consecutive dots to be 1 cm and draw the front view, side view and top view of the following 3-D figures.

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.4

Question 1.
Find the errors and correct the following mathematical sentences
(i) 3(x – 9) = 3x – 9
(ii) x(3x+2) = 3x² + 2
(iii) 2x+3x = 5x²
(iv) 2x + x + 3x = sx
(v) 4p + 3p + 2p + p – 9p = 0
(vi) 3x + 2y = 6xy
(vii) (3x)² + 4x +7 = 3x² + 4x +7
(viii) (2x)² + 5x = 4x + 5x = 9x
(ix) (2a + 3)² = 2a² + 6a +9
(x) Substitute x -3 in
(a) x² + 7x + 12 (- 3)² + 7(-3) + 12 = 9 + 4 + 12 = 25
(b) x² – 5x + 6(-3)² – 5(-3) + 69 – 15 + 6 = 0
(c) x² +5x = (-3)² + 5(3) + 6 = -9 – 15 = -24
(xi) (x – 4)² = x² – 16
(xii) (x + 7)² = x² +49
(xiii) (3a + 4b)(a – b)= 3a² – 4a²
(xiv) (x + 4) (x + 2) = x² + 8
(xv) (x – 4) (x – 2) = x²– 8
(xvi) 5x³ ÷ 5 x³ = 0
(xvii) 2x³ + 1 ÷ 2x³ = 1
(xviii) 3x + 2 ÷ 3x = [latex]\frac{2}{3 x}[/latex]
(xix) 3x + 5 ÷ 3 = 5
(xx) [latex]\frac{4 x+3}{3}[/latex] = x + 1
Solution:
(i) 3(x – 9) = 3x – 9
3(x – 9) = 3x – 9
⇒ 3x – 3 x 9 = 3x – 9
⇒ 3x – 27 = 3x – 9
⇒ – 27 ≠ – 9
∴ The given sentence is wrong. Correct sentence is 3(x – 9) = 3x – 27.

(ii) x(3x+2) = 3x² + 2
x(3x + 2) = 3x² + 2
⇒ x × 3x + x × 2 = 3x² + 2
⇒ 3x² + 2x ≠ 3x² + 2
∴ The given sentence is wrong.
Correct sentence is x(3x + 2) = 3x² + 2x.

(iii) 2x+3x = 5x²
2x + 3x = 5x²
⇒ 5x = 5x²
⇒ x ≠ x²
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(iv) 2x + x + 3x = 5x
2x + x + 3x = 5x
⇒ 6x = 5x
⇒ 6 ≠ 5
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(v) 4p + 3p + 2p + p – 9p = 0
4p + 3p + 2p + p – 9p = 0
⇒ 10p – 9p = 0
⇒ p = 0
It is not possible
∴ The given sentence is wrong. Correct sentence is
4p + 3p + 2p + p – 9p – p = 0

(vi) 3x + 2y = 6xy
3x + 2y = 6xy
a + b ≠ ab
∴ The given sentence is wrong.
Correct sentence is 3x x 2y = 6xy.

(vii) (3x)² + 4x +7 = 3x² + 4x +7
(3x)² + 4x +7 = 3x² + 4x +7
⇒ (3x)² = 3x²
⇒ 9x² = 3x²
⇒ 9 = 3
It is not possible
∴ The given sentence is wrong. Correct sentence is
(3x)²+ 4x + 7 = 9x² + 4x + 7.

(viii) (2x)² + 5x = 4x + 5x = 9x
(2x)² + 5x = 4x + 5x = 9x
⇒ 4x² + 5x = 4x + 5x
⇒ 4x² = 4x
⇒ x² = x
⇒ x ≠ √x
∴ The given sentence is wrong. Correct sentence is (2x)² + 5x = 4x² + 5x.

(ix) (2a + 3)² = 2a² + 6a +9
(2a + 3)² = 2a² + 6a +9
⇒ (2a)² + 2 × 2a × 3 + 3² = 2a² + 6a + 9
⇒ 4a² + 12a + 9 = 2a²+ 6a + 9
⇒ 4a² – 2a² = 6a – 12a
⇒ 2a² = – 6a
⇒ 2a ≠ 6
∴ The given sentence is wrong.
Correct sentence is
(2a + 3)² = 4a² + 12a + 9.

(x) Substitute x -3 in
(a) x² + 7x + 12 (- 3)² + 7(-3) + 12 = 9 + 4 + 12 = 25
x² + 7x + 12 = (- 3)² + 7 (- 3) + 12
= 9 – 21 + 12
= 21 – 21
= 0 25 (False)

(b) x² – 5x + 6(-3)² – 5(-3) + 69 – 15 + 6 = 0
x² – 5x + 6 = (-3)² – 5 (- 3) + 6
= 9 + 15 + 6
= 30 ≠ 0 (False)

(c) x² +5x = (-3)² + 5(3) + 6 = -9 – 15 = -24
x² + 5x = (- 3)² + 5 (- 3)
= 9 – 15 = – 6 ≠ 24 (False)

(xi) (x – 4)² = x² – 16
(x – 4)² = x² – 16 = (x)² – (4)²
(a – b)² ≠ a² – b²
∴ (x-4)² ≠ (x)² – (4)²
∴ The given sentence is wrong.
Correct sentence is (x – 4)² = x² – 8x + 16.

(xii) (x + 7)² = x² +49
(x + 7)² = x² + 49 = (x)² + (7)²
(a + b)² ≠ a² + b²
∴ (x+7)² ≠ (x)² – (7)²
∴ The given sentence is wrong.
Correct sentence is (x + 7)² = x² + 14x + 49.

(xiii) (3a + 4b)(a – b)= 3a² – 4a²
3a(a – b) + 4b(a – b) = 3a² – 4²
3a² – 3ab + 4ab – 4b² = – a²
3a² + ab – 4b² ≠ a²
∴ The given sentence is wrong. Correct sentence is
(3a + 4b) (a – b) = 3a² + ab – 4b²

(xiv) (x + 4) (x + 2) = x² + 8
(x + 4) (x + 2) = x² + 8
⇒ x² + 6x + 8 = x² + 8
⇒ 6x ≠ 0
Here ’6x’ term is missing in R.H.S.
∴ The given sentence is wrong. Correct sentence is
(x + 4)(x + 2) = x² + 6x + 8.

(xv) (x – 4) (x – 2) = x²– 8
(x – 4) (x – 2) = x² – 8
⇒ x² – 6x + 8 ≠ x² – 8
∴ The given sentence is wrong. Correct sentence is
(x – 4) (x – 2) = x² – 6x + 8

(xvi) 5x³ ÷ 5 x³ = 0
5x³ ÷ 5 x³ = 0
⇒ x^3-3 = 0
⇒ x⁰ = 0
∴ 1 ≠ 0 (∵ but x° = 1)
∴ The given sentence is wrong. Correct sentence is 5x³ ÷ 5x³ = 1.
In the denominator the term T is missing. .•. The given sentence is wrong. Correct sentence is

(xvii) 2x³ + 1 ÷ 2x³ = 1
2x³ + 1 ÷ 2x³ = 1
⇒ [latex]\frac{2 x^{3}+1}{2 x^{3}}[/latex] = 1
In the denominator the term T is missing.
∴ The given sentence is wrong. Correct sentence is
2x³ + 1 ÷ 2x³ = 1 + [latex]\frac{1}{2 \mathrm{x}^{3}}[/latex]

(xviii) 3x + 2 ÷ 3x = [latex]\frac{2}{3 x}[/latex]
3x + 2 ÷ 3x = [latex]\frac{2}{3 x}[/latex]
⇒ [latex]\frac{3 x+2}{3 x}=\frac{2}{3 x}[/latex]
⇒ 1 + [latex]\frac{2}{3 x}=\frac{2}{3 x}[/latex] ⇒ 1 ≠ 0
∴ The given sentence is wrong. Correct sentence is 3x + 2 ÷ 3x = 1 + [latex]\frac{2}{3 x}[/latex]

(xix) 3x + 5 ÷ 3 = 5
⇒ [latex]\frac{3 x+5}{3}[/latex] = 5
⇒ [latex]\frac{3 x}{3}+\frac{5}{3}[/latex] = 5 ⇒ x + [latex]\frac{5}{3}[/latex] ≠ 5
∴ It is a wrong sentence.
Correct sentence is 3x + 5 ÷ 3 = x + [latex]\frac{5}{3}[/latex]

(xx) [latex]\frac{4 x+3}{3}[/latex] = x + 1
[latex]\frac{4 x+3}{3}[/latex] = x + 1
⇒ [latex]\frac{4 \mathrm{x}}{3}+\frac{3}{3}[/latex] = x + 1
⇒ [latex]\frac{4 \mathrm{x}}{3}[/latex] + 1 ≠ x + 1
∴ It is a wrong sentence.
Correct sentence is [latex]\frac{4 x+3}{3}=\frac{4 x}{3}+1[/latex]

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots InText Questions and Answers.

8th Class Maths 6th Lesson Square Roots and Cube Roots InText Questions and Answers

Do this

Question 1.
Find the perfect squares between (i) 100 and 150 (ii) 150 and 200      [Page No. 124]
Answer:
i) The perfect squares between 100 and 150 are = 121, 144
ii) Perfect squares between 150 and 200 = 169, 196

Question 2.
Is 56 a perfect square? Give reasons.      [Page No. 124]
Answer:
Product of primes of 56 = 8 × 7 = (2 × 2) × 2 × 7
56 is not a perfect square. Since it can’t be written as product of two same numbers.

Question 3.
How many non perfect square numbers are there between 9² and 10²?      [Page No. 128]
Answer:
No. of non perfect square numbers between 9² and 10² are
= 2 × base of first number = 2 × 9 = 18
They are 82, 83, ……. 99.

Question 4.
How many non perfect square numbers are there between 15² and 16²?     [Page No. 128]
Answer:
No. of non perfect square numbers between 15 and 16 are = 2 × base of first number = 2 × 15 = 30
They are 226, 227, ……. 255,

Question 5.
Check whether the following numbers form pythagorean triplet.     [Page No. 129]
(i) 2, 3, 4
(ii) 6, 8, 10
(iii) 9, 10, 11
(iv) 8,15, 17
Answer:

Question 6.
Take a pythagorean triplet. Write their multiples. Check whether these multiples form a pythagorean triplet.      [Page No. 129]
Answer:
3, 4, 5 are pythagorean triplets.

From 6,8,10
⇒ 10² = 8² + 6²
⇒ 100 = 64 + 36
⇒ 100 = 100 (T)
From 9, 12, 5
⇒ 15² = 9² + 12²
⇒ 225 = 81 + 144
⇒ 225 = 225 (T)
∴ The multiples of pythagorean triplets are also pythagorean triplets.

Question 7.
By subtraction of successive odd numbers And whether the following numbers are perfect squares or not.        [Page No. 131]
(i) 55 (ii) 90 (iii) 121
Answer:
(i) √55
Step 1 → 55 – 1 = 54 (1st odd number be subtracted)
Step 2 → 54 – 3 = 51 (2nd odd number be subtracted)
Step 3 → 51 – 5 = 46 (3rd odd number be subtracted)
Step 4 → 46 – 7 = 39 (4th odd number be subtracted)
Step 5 → 39 – 9 = 30 (5th odd number be subtracted)
Step 6 → 30 – 11 = 19 (6th odd number be subtracted)
Step 7 → 19 – 13 = 6 (7th odd number be subtracted)
∴ 55 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

ii) √90
Step 1 → 90 – 1 =89 (1st odd number be subtracted)
Step 2 → 89 – 3 = 86 (2nd odd number be subtracted)
Step 3 → 86 – 5 = 81 (3rd odd number be subtracted)
Step 4 → 81 – 7 = 74 (4th odd number be subtracted)
Step 5 → 74 – 9 = 65 (5th odd number be subtracted)
Step 6 → 65 – 11 = 54 (6th odd number be subtracted)
Step 7 → 54 – 13 = 41 (7th odd number be subtracted)
Step 8 → 41 – 15 = 26 (8th odd number be subtracted)
Step 9 → 26 – 17 = 9 (9th odd number be subtracted)
∴ 90 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

iii) √121
Step 1 → 121 – 1 = 120 (1st odd number is subtracted)
Step 2 → 120 – 3 = 117 (2nd odd number is subtracted)
Step 3 → 117 – 5 = 112 (3rd odd number is subtracted)
Step 4 → 112 – 7 = 105 (4th odd number is subtracted)
Step 5 → 105 – 9 = 96 (5th odd number is subtracted)
Step 6 → 96 – 11 = 85 (6th odd number is subtracted)
Step 7 → 85 – 13 = 72 (7th odd number is subtracted)
Step 8 → 72 – 15 = 57 (8th odd number is subtracted)
Step 9 → 57 – 17 = 40 (9th odd number is subtracted)
Step 10 → 40 – 19 = 21 (10th odd number is subtracted)
Step 11 → 21 – 21 = 0 (11th odd number is subtracted)
∴ At the 11th step, the difference of consecutive odd numbers is ‘0’
121 is a perfect square number.
∴ √121 = [latex]\sqrt{11 \times 11}[/latex] = 11 (∵ It ends at 11th step)

Question 8.
Which of the following are perfect cubes?     [Page No. 143]
(i) 243    (ii) 400    (iii) 500   (iv) 512     (v) 729
Answer:

∴ 512 and 729 are perfect cubes.

Try These

Question 1.
Guess and give reason which of the following numbers are perfect squares. Verify from the above table. (Refer table in Text Page no: 124)         [Page No. 124]
(i) 84   (ii) 108   (iii) 271   (iv) 240    (v) 529
Answer:
(i), (ii), (iii), (iv) are not perfect squares.
(v) 529 = 23 × 23
∴ 529 is a perfect square number.

Question 2.
Which of the following have one in its units place?     [Page No. 125]
(i) 126²    (ii) 179²    (iii) 281²     (iv) 363²
Answer:

Number	Square of units digit	Units digit of a number
i) 1262	(6)2 = 36	6
ii) 1792	(9)2 = 81	1
iii) 2812	(1)2 = 1	1
iv) 3632	(3)2 = 9	9

Question 3.
Which of the following have 6 in the units place?
(i) 116²    (ii) 228²    (iii) 324²    (iv) 363²        [Page No. 125]
Answer:
i) 1162 ⇒ (6)² = 36 units digit = 6
ii) 2282 ⇒ (8)² = 64 units digit = 4
iii) 3242 ⇒ (4)² = 16 units digit = 6
iv) 3632 ⇒ (3)² = 9 units digit = 9
∴ Numbers which are having ‘6’ in its unit’s digit are: (i) 116² (iii) 324²

Question 4.
Guess, how many digits are there in the squares of i) 72   ii) 103    iii) 1000        [Page No. 125]
Answer:

Question 5.

27 lies between 20 and 30
27² lies between 20² and 30²
Now find what would be 27² from the following perfect squares.      [Page No. 125]
(i)329      (ii) 525     (iii) 529    (iv) 729
Answer:
The value of (27)² = 27 × 27 = 729

Question 6.
Rehan says there are 37 non square numbers between 9² and 11². Is he right? Give your reason.       [Page No. 128 ]
Answer:
No. of (integers) non perfect square numbers between 9² and 11²
= 82, 83, ……. 100 …… 120 = 39
But 100 is a perfect square number.
∴ Required non perfect square numbers are = 39 – 1 = 38
∴ No, his assumption is wrong.

Question 7.
Is 81 a perfect cube?      [Page No. 140]
Answer:
81 = 3 × 3 × 3 × 3 = 3⁴
No, 81 is not a perfect cube.
[∵ 81 can’t be written as product of 3 same numbers.]

Question 8.
Is 125 a perfect cube?       [Page No.140]
Answer:
125 = 5 × 5 × 5 = (5)³
Yes, 125 is a perfect cube.
[∵ It can be written as product of 3 same numbers]

Question 9.
Find the digit in units place of each of the following numbers.      [Page No. 141]
(i) 75³   (ii) 123³    (iii) 157³    (iv) 198³    (v) 206³
Answer:

Number	Cube of a units digit	Units digit
i) 753	53= 125	5
ii) 1233	33 = 27	7
iii) 1573	73 = 343	3
iv) 1983	83 = 512	2
v) 2063	63 = 216	6

Think, Discuss and Write

Question 1.
Vaishnavi claims that the square of even numbers are even and that of odd are odd. Do you agree with her? Justify.  [Page No. 125]
Answer:
The square of an even number is an even
∵ The product of two even numbers is always an even.
Ex: (4)² = 4 × 4 = 16 is ah even.
The square of an odd number is an odd.
∵ The product of two odd numbers is an odd number.
Ex: 11² = 11 × 11 = 121 is an odd.

Question 2.
Observe and complete the table:      [Page No. 125]

Answer:

Question 3.
How many perfect cube numbers are present between 1 and 100,1 and 500,1 and 1000?     [Page No. 140]
Answer:
Perfect cube numbers between 1 and 100 = 8, 27, 64
Perfect cube numbers between 1 and 500 = 8, 27, 64, 125, 216, 343
Perfect cube numbers between 1 and 1000 = 8, 27, 64, 125, 216, 343, 512, 729

Question 4.
How many perfect cubes are there between 500 and 1000?      [Page No. 140]
Answer:
Perfect cubes between 500 and 1000 = 512 and 729

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.3

Question 1.
Carry out the following divisions
(i) 48a³ by 6a
(ii) 14x³ by 42x³
(iii) 72a³b⁴c⁵ by 8ab²c³
(iv) 11xy²z³ by 55xyz
(v) -54l⁴m³n² by 9l²m²n²
Solution:
(i) 48a³ by 6a
48a³ ÷ 6a
= [latex]\frac{6 \times 8 \times a \times a^{2}}{6 \times a}[/latex]
= 8a²

(ii) 14x³ by 42x³
= 14x³ ÷ 42x³

(iii) 72a³b⁴c⁵ by 8ab²c³

(iv) 11xy²z³ by 55xyz
11xy²z³ ÷ 55xyz

(v) -54l⁴m³n² by 9l²m²n²
-54l⁴m³n² ÷ 9l²m²n²

= -6l²m

Question 2.
Divide the given polynomial by the given monomial
(i) (3x² – 2x) ÷ x
(ii) (5a³b – 7ab³) ÷ ab
(iii) (25x⁵ – 15x⁴) ÷ 5x³
(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²
(v) 15 (a³b²c² – a²b³c² + a²b²c³ ) ÷ 3abc
(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)
(vii) ([latex]\frac{2}{3}[/latex] a² b² c²+ [latex]\frac{4}{3}[/latex] a b² c³) ÷ [latex]\frac{1}{2}[/latex]abc
Solution:
(i) (3x² – 2x) ÷ x

(ii) (5a³b – 7ab³) ÷ ab

(iii) (25x⁵ – 15x⁴) ÷ 5x³

= 5x² – 3x (or) x(5x – 3)

(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²

= 2l² – 3l² + 4l = l(2l² – 3l + 4)

(v) 15 (a³ b² c² – a² b³ c² + a² b² c³ ) ÷ 3abc

= 5[a x abc – b x abc + c x abc ]
= 5abc [a – b + c]

(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)

= -[p² – 3pq – 2q²]
= 2² + 3pq – p²

(vii) ([latex]\frac{2}{3}[/latex] a²b²c²+ [latex]\frac{4}{3}[/latex] ab²c³) ÷ [latex]\frac{1}{2}[/latex]abc

Question 3.
Workout the following divisions:
(i) (49x -63) ÷ 7
(ii) 12x (8x – 20,) ÷ 4(2x – 5)
(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)
(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)
(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)
(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)
Solution:
(i) (49x -63) ÷ 7

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

(iii) 11a³ b³ (7c – 35) ÷ 3a² b² (c – 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x² + 7x + 10) ÷ 9(x + 4)

4 ( x² + 7x + 10)
= 4 ( x² + 5x + 2x + 10)
= 4 [x( x + 5) +2(x + 5)]
= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4.
Factorize the expressions and divide them as directed:
(i) (x² + 7x + 12) ÷ (x + 3)
(ii) (x² – 8x + 12) ÷ (x – 6)
(iii) (p² + 5p + 4,) (p + l)
(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
(v) 151m (2p² – 2q²) ÷ 3l(p + q)
(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
Solution:
(i) (x² + 7x + 12) ÷ (x + 3)
(x² + 7x + 12) ÷ (x + 3)
x² + 7x + 12 = x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)

(ii) (x² – 8x + 12) ÷ (x – 6)
(x² – 8x + 12) ÷ (x – 6)
x² – 8x + 12 = x² – 6x – 2x + 12
= x(x – 6) – 2(x – 6)
= (x – 6) (x – 2)
∴ (x² – 8x + 12) 4 (x – 6)
= [latex]\frac{(x-6)(x-2)}{(x-6)}[/latex] = x – 2

(iii) (p² + 5p + 4,) (p + 1)
p² + 5p + 4 = p² + p + 4p + 4
= p(p + 1) + 4(p + 1)
= (p + 1) (p + 4)
(p² + 5p + 4) ÷ (p + 1)
= [latex]\frac{(p+1)(p+4)}{(p+1)}[/latex] = p + 4

(iv) 15ab(a² – 7a + 10) ÷ 3b(a – 2)
15ab (a² – 7a + 10) ÷ 3b (a – 2)
15ab (a² – 7a + 10) = 15ab (a² – 5a – 2a + 10)
= 15ab [(a² – 2a) – (5a -10)]
= 15ab [a(a – 2) – 5(a – 2)]
= 15ab(a – 2)(a – 5)
∴ 15ab (a² – 7a + 10) ÷ 3b (a – 2)

(v) 151m (2p² – 2q²) ÷ 3l(p + q)
15lm (2p² – 2q²) ÷ 3l (p + q)
15lm (2p² – 2q²) = 15lm x 2(p² – q²)
= 30lm (p + q) (p – q)
∴ 15lm(2p² – 2q²) ÷ 3l(p + q)

(vi) 26z³(32z² – 18,) ÷ 13z² (4z – 3)
26z³(32z² – 18) ÷ 13z² (4z – 3)
26z³(32z² – 18) = 26z³ (2 x 16z² – 2 x 9)
= 26z³ x 2 [16z³ – 9]
= 52z³ [(4z)³ – (3)³]
= 52z³ (4z + 3) (4z – 3)
∴ 26z³ (32z² – 18) ÷ 13z² (4z – 3)

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion InText Questions and Answers.

8th Class Maths 5th Lesson Comparing Quantities Using Proportion InText Questions and Answers

Do this

Question 1.
How much compound interest is earned by investing Rs. 20000 for 6 years at 5% per annum compounded annually? (Page No. 114)
Answer:
P = Rs. 20,000; R = 5%; n = 6 years

∴ Compound Interest = Amount – Principal = 26802 – 20,000
∴ C.I. = Rs. 6802 /-

Question 2.
Find compound interest on Rs. 12600 for 2 years at 10% per annum compounded annually.    (Page No. 114)
Answer:
P = Rs. 12,600; R = 10%; n = 2 years

∴ Compound Interest = Amount – Principal = 15,246 – 12,600
∴ C.I. = Rs. 2646 /-

Question 3.
Find the number of conversion times the interest is compounded and rate for each.
i) A sum taken for 1[latex]\frac{1}{2}[/latex] years at 8% per annum is compounded half yearly.
ii) A sum taken for 2 years at 4% per annum is compounded half yearly.     (Page No. 115)
Answer:
Compound interest will be calculated for every 6 months.
There will be 3 periods in 1[latex]\frac{1}{2}[/latex] year.
∴ n = 3
∴ Rate of interest for half yearly = [latex]\frac{1}{2}[/latex] × 8% = 4%
∴ R = 4%; n = 3
ii) C.I. should be calculated for every 6 months.
There will be 4 time periods in 2 years.
∴ n = 4
∴ Rate of interest for half yearly = [latex]\frac{1}{2}[/latex] × 4% = 2%
∴ n = 4 ; R = 2%

Try These

Question 1.
Find the ratio of gear of your bicycle.       (Page No. 96)

Count the number of teeth on the chain wheel and the number of teeth for the sprocket wheel.
{number of teeth on the chain wheel} : {number of teeth of sprocket wheel}
This is called gear ratio. Write how many times sprocket wheel turns for every time the chain wheel rotates once.
Answer:
The ratio between the rotations of chain wheel and sprocket wheel is 4 : 1.

Question 2.
Collect newspaper cuttings related to percentages of any five different situations.  (Page No. 96)
Answer:
1) Bharti to sell 5% stake for $ 1.2b:
New Delhi, May 3: The country’s largest telecom operator Bharti Airtel said on Friday that it will sell 5 per cent stake to Doha – based Qatar Foundation Endowment (QFE) for $1.26 billion (Rs. 6,796 crores) to fund its future growth plans.
The deal will bring cash for the company at a time when its balance sheet is stretched and there is threat of Bharti Airtel having to pay hefty fees to regulatory authorities as government is re-looking at past policies.

2) Indian Firms Mop – Up Down By 36% In FY13:
New Delhi: Indian companies raised nearly Rs. 31,000 crore from the public issuance of equity and debt in 2012 – 13, a slump of 36 per cent from the preceding year.
According to latest data available with market regulator Sebi (Securities and Exchange Board of India), a total of Rs. 30,859 crore worth of fresh capital were mopped – up from equity and debt market during 2012 – 13, which was way below than Rs. 48,468 crore garnered in 2011 -12. Going by the statistics, it was mostly debt market that was leveraged to meet the funding requirements of businesses in the past fiscal as compared to capital raised through sale of shares through instruments like initial public offering (IPO) and rights issue. A total of Rs. 15,386 crore were raised from the debt market via 11 issues in 2012 – 13, much lower than Rs. 35,611 crore garnered through 20 issues in the preceding fiscal.

3) IT – ITeS sector employs 2.97m people in FY13:
New Delhi: The total number of professionals working in India’s $100 billion IT – information technology enabled services (IT – ITeS) sector grew by 7 per cent to 2.97 million in the last fiscal, Parliament was informed on Friday.
The IT – ITeS sector, which contributes about 8 per cent to the country’s economy, provided employment to 2.77 million professionals in 2011 -12 fiscal, minister of state for communications and IT Milind Deora said. “The Indian IT – ITeS industry has been progressively growing and is able to secure new projects from various foreign coun¬tries,” Mr Deora said. During the 2012 -13 fiscal, 6,40,000 professionals were employed in the domestic market.

4) For RBI, it’s not all is well yet:
Slashes repo rate by 0.25%; rules out any more cuts; raises red flag on CAD DC Correspondent Mumbai, May 3:
The RBI cut the repo rate (rate at which it lends to banks) by a quarter per cent on Friday to 7.25 per cent from 7.75 per cent, but this will not be passed on to the consumers by way of lower personal loans for housing etc., immediately according to bankers.
It also raised the growth rate from 5.2 per cent projected in January to 5.7 per cent for 2013 -14 and lowered the inflation rate to 5.5 per cent for the year.
RBI governor Dr D. Subbarao said based on the current and prospective assessment of various economic factors and the dismal 4.5 per cent lowest growth rate in the last quarter, it was decided to cut the policy rate by 25 basis points.

5) Markets sink on RBI’s Bearish outlook on rate:
DC Correspondent Mumbai, May 3:
In a highly volatile trading session, the markets retreated from their three month high led by interest rate sensitive banking, auto and real estate sector stocks after the Reserve Bank of India (RBI) cautioned that the room for further monetary policy easing is limited.
The Sensex closed 19,575.64, sliding 160.13 points or 0.81 per cent while the Nifty dropped 55.35 points or 0.92 per cent to end the week at 5,944.

Question 3.
Find the compound ratios of the following. (Page No. 99)
a) 3 : 4 and 2 : 3
b) 4 : 5 and 4 : 5
c) 5 : 7 and 2 : 9
Answer:
Compound ratio of a : b and c : d is ac : bd.

Question 4.
Give examples for compound ratio from daily life.     (Page No. 99)
Answer:
Examples for compound ratio from daily life:
i) To compare the ratio of tickets of 8th class students (Boys & Girls) is 3:4 and the ratio of tickets of 7th class students is 4 : 5.
ii) The comparision between two situations is 4 men can do a piece of work in 12 days, the same work 6 men can do in 8 days.
iii) Time – distance – speed.
iv) Men – days – their capacities etc.

Question 5.
Fill the selling price for each.     (Page No. 104)

Answer:

Question 6.
i) Estimate 20% of Rs. 357.30 ii) Estimate 15% of Rs. 375.50      (Page No. 105)
Answer:

ii) 15% of 375.50 = [latex]\frac{15}{100}[/latex] × 375.50 = 15 × 3.7550 = Rs. 56.325

Question 7.
Complete the table.     (Page No. 105)

Answer:

Think, discuss and write

Question 1.
Two times a number is 100% increase in the number. If we take half the number what would be the decrease in percent?    (Page No. 101)
Answer:
Increase percent of 2 times of a number = [latex]\frac{(2-1)}{1}[/latex] × 100 = 1 × 100 = 100%
Half of the number = 1 – [latex]\frac{1}{2}[/latex] = [latex]\frac{1}{2}[/latex]
Decrease in percent = [latex]\frac{\frac{1}{2}}{1}[/latex] × 100 = [latex]\frac{1}{2}[/latex] × 100 = 50%

Question 2.
By what percent is Rs. 2000 less than Rs. 2400? Is it the same as the percent by which Rs. 2400 is more than Rs. 2000? (Page No. 101)
Answer:
Decrease in percent of Rs. 2000 less than Rs. 2400

Increase in percent of Rs. 2400 more than Rs. 2000

Question 3.
Preethi went to a shop to buy a dress. Its marked price is Rs. 2500. Shop owner gave 5% discount on it. On further insistence, he gave 3% more discount. What will be the final discount she obtained? Will it be equal to a single discount of 8%? Think, discuss with your friends and write it in your notebook. (Page No. 105)
Answer:
Marked price of a dress selected by Preethi = Rs. 2500
After allowing 5% of discount then S.P = M.P. – Discount%
= 2500 – [latex]\frac{5}{100}[/latex] × 2500 = 2500 – 125 = Rs. 2375
Again 3% discount is allowed on Rs. 2375 then
S.P = 2375 – 3% of 2375
= 2375 – [latex]\frac{3}{100}[/latex] × 2375 = 2375 – 71.25 = Rs. 2303.75
If 8% discount is allowed then S.P =

The S.P’s of both cases are not equal.
Discount on 5% + Discount on 3% = 125 + 71.25 = Rs. 196.25
Discount on 8% = Rs. 200
∴ Discounts are not equal which are obtained by Preethi.

Question 4.
What happens if cost price = selling price. Do we get any such situations in our daily life?
It is easy to find profit % or loss% in the above situations. But it will be more meaningful if we express them in percentages. Profit % is an example of increase percent of cost price and loss % is an example of decrease percent of cost price. (Page No. 106)
Answer:
If selling price is equal to cost price then either profit or loss will not be occurred.
In our daily life S.P. will not be equal to C.P. Then profit or loss will be occurred.
∴ Profit % = [latex]\frac{\text { Profit }}{\text { C.P. }}[/latex] × 100;
Loss % = [latex]\frac{\text { Loss }}{\text { C.P. }}[/latex] × 100.

Question 5.
A shop keeper sold two TV sets at Rs. 9,900 each. He sold one at a profit of 10% and the other at a loss of 10%. Oh the whole whether he gets profit or loss? If so what is its percentage? (Page No. 108)
Answer:
S.P of each T.V = Rs. 9,900
S.P of both T.Vs = 2 × 9,900 = Rs. 19,800
10% profit is allowed on first then C.P. =

10% loss is allowed on second then C.P.

C.P. of both T.V.’s = 9000 + 11000 = Rs. 20,000
Here C.P > S.P then loss will be occurred.
∴ Loss = C.P – S.P = 20000 – 19,800 = 200

Question 6.
What will happen if interest is compounded quarterly? How many conversion periods will be there? What about the quarter year rate – how much will it be of the annual rate? Discuss with your friends. (Page No. 115)
Answer:
Here C.I will be calculated for every 3 months. So, 4 time periods will be occurred in 1 year.
Rate of Interest (R) = [latex]\frac{R}{4}[/latex] [∵ [latex]\frac{12}{3}[/latex] = 4]

A = P[latex]\left[1+\frac{R}{400}\right]^{4}[/latex]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.2

Question 1.
Factorise the following expression
i) a² + 10a +25
ii) l² – 16l + 64
iii) 36x² + 96xy + 64y²
iv) 25x² + 9y² – 30xy
v) 25m²– 40mn + 1 6n²
vi) 81x² – 198 xy + 12ly²
vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
viii) l⁴ + 4l²m² + 4m⁴
Solution:
i) a² + 10a +25
= (a)² + 2 × a × 5 + (5)²
It is in the form of a² + 2ab + b²
a² + 2ab + b²= (a + b)²
∴ a² + 10a + 25 = (a + 5)² = (a + 5) (a + 5)

ii) l² – 16l + 64
l² – 16l + 64
= (l)² – 2 × l × 8 + (8)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ l² – 16l + 64 = (l – 8)² = (l – 8) (l – 8)

iii) 36x² + 96xy + 64y²
36x² + 96xy + 64y²
= (6x)² + 2 × 6x × 8y + (8y)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a + b)²
∴ 36x² + 96xy + 64y²
= (6x + 8y)² = (6x + 8y) (6x + 8y)

iv) 25x² + 9y² – 30xy
25x² + 9y² – 30xy
= (5x)² + (3y)² – 2 × 5x × 3y
It is in the form of a² + b² – 2ab
a² + b² – 2ab = (a – b)²
∴ 25x² + 9y² – 30xy
= (5x – 3y)² = (5x – 3y) (5x – 3y)

v) 25m²– 40mn + 1 6n²
25m² – 40mn + 16n²
= (5m)² – 2 × 5m × 4n + (4n)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 25m² – 40mn + 16n²
= (5m – 4n)²
= (5m – 4n) (5m – 4n)

vi) 81x² – 198 xy + 12ly²
81x² – 198xy + 121y²
= (9x)² – 2 × 9x × 11y + (11y)²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ 81x² – 198xy + 121y²
= (9x – 11y)² – (9x – 11y) (9x – 11y)

vii) (x+y)² – 4xy
(Hint : first expand ( x + y)² )
= (x + y)² – 4xy
= x² + y² + 2xy – 4xy
= x² + y² – 2xy = (x – y)² = (x – y)(x – y)

viii) l⁴ + 4l²m² + 4m⁴
l⁴ + 4l²m² + 4m⁴
= (l²)² + 2 × l² × 2m² + (2m²)²
It is in the form of a² + 2ab + b²
a² + 2ab + b² = (a – b)²
∴ l⁴ + 4l²m² + 4m⁴
= (l² + 2m²)² = (l² + 2m²) (l² + 2m²)

Question 2.
Factorise the following
i) x² – 36
ii) 49x² – 25y²
iii) m² – 121
iv) 81 – 64x²
v) x²y² – 64
vi) 6x² – 54
vii) x² – 81
viii) 2x -32 x⁵
ix) 81x⁴ – 121x²
x) (p² – 2pq + q²)-r²
xi) (x+y)² – (x-y)²
Solution:
i) x² – 36
x² – 36
⇒ (x)² – (6)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
∴ x² – 36 = (x + 6) (x – 6)

ii) 49x² – 25y²
= (7x)² – (5y)²
= (7x + 5y) (7x – 5y)

iii) m² – 121
m² -121
= (m)² – (11)²
= (m + 11) (m – 11)

iv) 81 – 64x²
81 – 64x²
= (9)² – (8x)²
= (9 + 8x) (9 – 8x)

v) x²y² – 64
= (xy)² – (8)²
= (xy + 8)(xy – 8)

vi) 6x² – 54
6x² – 54
= 6x² – 6 x 9 ‘
= 6(x² – 9)
= 6[(x)² – (3)²]
= 6(x + 3) (x – 3)

vii) x² – 81
x² – 81
= x² – 9²
= (x + 9 )(x – 9)

viii) 2x – 32 x⁵
2x – 32 x⁵
= 2x – 2x x 16x⁴
= 2 x (1 – 16x⁴)
= 2x [1²) – (4x²)²]
= 2x (1 + 4x²) (1 – 4x²)
= 2x (1 + 4x²) [(1⁵ – (2x)²]
= 2x (1 + 4x²) (1 + 2x) (1 – 2x)

ix) 81x⁴ – 121x²
81x⁴ – 121x²
– x² (81² – 121)
= x²[(9x)² – (11)²]
= x² (9x + 11) (9x -11)

x) (p² – 2pq + q²)-r²
(p² – 2pq + q²) – r²
= (p – q)² – (r)² [∵ p² – 2pq + q² = (p – q)²]
= (p – q + r) (p – q – r)

xi) (x + y)² – (x – y)²
(x + y)² – (x – y)²
It is in the form of a² – b²
a = x + y, b = x- y
∴ a² – b² =(a + b)(a-b)
= (x + y + x – y) [x + y- (x – y)]
= 2x [x + y-x + y]
= 2x x 2y = 4xy

Question 3.
Factorise the expressions
(i) lx² + mx
(ii) 7y² + 35Z²
(iii) 3x⁴ + 6x³y + 9x²Z
(iv) x² – ax – bx + ab
(v) 3ax – 6ay – 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab – b² + 12ac – 2bc
(viii) p²q – pr² – pq + r²
(ix) x (y + z) -5 (y + z)

(i) lx² + mx
lx² + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y² + 35z²
7y²+ 35z²
= 7 × y² + 7 × 5 × z²
= 7(y² + 5z²)

(iii) 3x⁴ + 6x³y + 9x²Z
3x⁴ + 6x³y + 9x²Z
= 3 × x² × x² + 3 × 2 × x × x² × y + 3 × 3 × x² × z
= 3x² (x² + 2xy + 3z)

(iv) x² – ax – bx + ab
x² – ax – bx + ab
= (x² – ax) – (bx – ab)
= x(x – a) – b(x – a)
= (x – a) (x – b)

(v) 3ax – 6ay – 8by + 4bx
3ax – 6ay – 8by + 4bx
= (3ax – 6ay) + (4bx – 8by)
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

(vi) mn + m + n + 1
mn + m + n + 1
= (mn + m) + (n + 1)
= m (n + 1) + (n + 1)
= (n + 1) (m + 1)

(vii) 6ab – b² + 12ac – 2bc
6ab – b² + 12ac – 2bc
= (6ab – b²) + (12ac – 2bc)
= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)
= b [6a – b] + 2c [6a – b]
= (6a – b) (b + 2c)

(viii) p²q – pr² – pq + r²
p²q – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= P(pq – r²) – (pq – r²) × 1
= (pq – r²)(p – 1)

(ix) x (y + z) -5 (y + z)
= x(y + z) – 5(y + z)
= (y + z) (x – 5)

Question 4.
Factorise the following
(i) x⁴ – y⁴
(ii) a⁴ – (b + c)⁴
(iii) l² – (m – n)²
(iv) 49x² – [latex]\frac{16}{25}[/latex]
(v) x⁴ – 2x²y² + y⁴
(vi) 4 (a + b)² – 9 (a – b)²
Solution:
= (x²)² – (y²)² is in the form of a² – b²
a² – b² = (a + b) (a – b)
x⁴ – y⁴ = (x² + y²)(x² – y²)
= (x² + y²)(x + y)(x – y)

(ii) a⁴ – (b + c)⁴
a⁴ – (b + c)⁴
= (a²)² – [(b + c)²]²
= [a² + (b + c)²] [a² – (b + c)²] ,
= [a² + (b + c)²] (a + b + c) [a – (b + c)]
= [a² + (b + c)²] (a + b + c) (a – b – c)

(iii) l² – (m – n)²
l² – (m – n)²
= (l)² – (m – n)²
= [l + m – n] [l – (m – n)]
= [l + m -n] [l – m + n]

(iv) 49x² – [latex]\frac{16}{25}[/latex]
= (7x)² – ([latex]\frac{4}{5}[/latex])²
= (7x+ ([latex]\frac{4}{5}[/latex]) (7x – ([latex]\frac{4}{5}[/latex])

(v) x⁴ – 2x² y² + y⁴
= (x² )² – 2x² y² + (y² )²
It is in the form of a² – 2ab + b²
a² – 2ab + b² = (a – b)²
∴ x⁴ – 2x² y² + y⁴ = (x² – y² )²
= [(x)² – (y)² ]²
= [(x + y) (x – y)]²
= (x + y)² (x – y)²
[∵ (ab)^m = a ^m . bⁿ ]

(vi) 4 (a + b)² – 9 (a – b)²
4 (a + b)² – 9 (a – b)²
= [2(a + b)]² – [3(a – b)]²
= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]
= (2a + 2b + 3a – 3b) (2a + 2b – 3a + 3b)
= (5a – b) (5b – a)

Question 5.
Factorise the following expressions
(i) a²+ 10a + 24
(ii) x² +9x + 18
(iii) p² – 10q + 21
(iv) x² – 4x – 32
Solution:
(i) a²+ 10a + 24
a² + 10a + 24 .
= a² + 6a + 4a + 24
= a x a + 6a + 4a + 6 × 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a² + 10a + 24

∴ a² + 10a + 24 = (a + 6) (a + 4)

(ii) x² + 9x + 18
x² + 9x + 18
= (x + 3) (x + 6)

∴ x² + 9x + 18 = (x + 3) (x + 6)

(iii) p² – 10q + 21
p² – 10p + 21
= (P – 7) (p – 3)

∴ p² – 10p + 21 = (p – 7)(p – 3)

(iv) x² – 4x – 32
x² – 4x – 32
= (x – 8) (x + 4)

∴ x² – 4x – 32 = (x – 8) (x + 4)

Question 6.
The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Solution:
Perimeter of a triangle
= AB + BC + CA = 48
⇒ c + a + b = 48

The solutions of Harmeet, Rosy are wrong.

∴ Srikar had done it correctly.
⇒ 21 + a + b = 48
⇒ a + b = 48 – 21 = 27
∴ The lengths of a, b should be 10, 17
∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]
∴ 10 + 17 > 2
27 > 21 (T).
∴ The length of the shortest side is 10 cm.

Question 7.
Find the values of ‘m’ for which x² + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.
Solution:
Given equation is x² + 3xy + x + my – m ……….(1)
Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).
Then (x + 3y + a) (x + 0y + b)
= x² + 0xy + bx + 3xy + 0y² + 3by + ax + 0y + ab
= x² + bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x² + 3xy + x + my – m
= x² + (a + b)x + 3xy + 3by + ab
Equating the like terms on both sides,
ab = – m ………….. (3)
(a + b)x = x ⇒ a + b = 1 ……………. (4)
3by = my ⇒ 3b = m ⇒ b = [latex]\frac{\mathrm{m}}{3}[/latex]
Substitute ‘b’ value in equation (4),
a = [latex]1-\frac{m}{3}=\frac{3-m}{3}[/latex]
ab = -m
[ ∵ from (3)]
put a & b value then ,
[latex]\left(\frac{3-m}{3}\right)\left(\frac{m}{3}\right)[/latex] = -m
[latex]\frac{3 \mathrm{~m}-\mathrm{m}^{2}}{9}[/latex]= -m
⇒ 3m – m² = – 9m
⇒ m² – 12m = 0
⇒ m(m – 12) = 0
⇒ m = 0 (or) m = 12
lf m = 12

∴ b = [latex]\frac{12}{3}[/latex] = 4&a = [latex]\frac{3-\mathrm{m}}{3}=\frac{3-12}{3}[/latex]
= [latex]\frac{-9}{3}[/latex] = -3
∴ Linear factors are (x + 3y – 3), (x + 4) If m = 0
b = [latex]\frac{0}{3}[/latex] = 0 & a = [latex]\frac{3-0}{3}=\frac{3}{3}[/latex] = 1
∴ Linear factors are (x + 3y + 1), x.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers InText Questions and Answers.

8th Class Maths 4th Lesson Exponents and Powers InText Questions and Answers

Do this

Question 1.
Simplify the following.   (Page No. 81)
i) 3⁷ × 3³
ii) 4 × 4 × 4 × 4 × 4
iii) 3⁴ × 4³
Answer:
(i) 3⁷ × 3³ = 3⁷ + 3³ = 3¹⁰       [∵ a^m × aⁿ = a^m+n]
(ii) 4 × 4 × 4 × 4 × 4 = 4⁵      [∵ a × a × a × ……. m times = a^m]
(iii) 3⁴ × 4³ = 3⁴⁺³ = 3⁷      [∵ a^m × aⁿ = a^m+n]

Question 2.
The distance between Hyderabad and Delhi is 1674.9 km by rail. How would you express this in centimetres? Also express this in the scientific form.     (Page No. 81)
Answer:
Distance from Hyderabad to Delhi is
= 1674.9 km = 1674.9 × 1000 m = 1674900 mts
= 1674900 × 100 cm
= 167490000 cm
= 16749 × 10⁴ cm

Question 3.
What is 10^-10 equal to?     (Page No. 83)
Answer:
10^-10 = [latex]\frac{1}{10^{10}}[/latex]      [∵ a^-n = [latex]\frac{1}{a^{n}}[/latex]]

Question 4.
Find the multiplicative inverse of the following. (Page No. 83)
Answer:

Question 5.
Expand the following numbers using exponents. (Page No. 84)
Answer:
i) 543.67
= (5 × 100) + (4 × 10) + (3 × 10⁰) + [latex]\left(\frac{6}{10}\right)[/latex] + [latex]\left(\frac{7}{10^{2}}\right)[/latex]
= (5 × 10²) + (4 × 10) + (3 × 10⁰) + (6 × 10^-1) + (7 × 10^-2)   [∵ aⁿ = a^-n]

ii) 7054.243
= (7 × 1000) + (0 × 100) + (5 × 10) + (4 × 10⁰) + [latex]\left(\frac{2}{10}\right)[/latex] + [latex]\left(\frac{4}{100}\right)[/latex] + [latex]\left(\frac{3}{1000}\right)[/latex]
= (7 × 10³) + (0 × 10²) + (5 × 10¹) + (4 × 10⁰) + (2 × 10^-1) + (4 × 10^-2) + (3 × 10^-3)

iii) 6540.305
= (6 × 1000) + (5 × 100) + (4 × 10) + (0 × 10⁰) + [latex]\left(\frac{3}{10}\right)[/latex] + [latex]\left(\frac{0}{100}\right)[/latex] + [latex]\left(\frac{5}{1000}\right)[/latex]
= (6 × 10³) + (5 × 10²) + (4 × 10¹) + (0 × 10⁰) + (3 × 10^-1) + (0 × 10^-2) + (5 × 10^-3)

iv) 6523.450
= (6 × 1000) + (5 × 100) + (2 × 10) + (3 × 10⁰) + [latex]\left(\frac{4}{10}\right)[/latex] + [latex]\left(\frac{5}{100}\right)[/latex] + [latex]\left(\frac{0}{1000}\right)[/latex]
= (6 × 10³) + (5 × 10²) + (2 × 10¹) + (3 × 10⁰) + (4 × 10^-1) + (5 × 10^-2) + (0 × 10^-3)

Question 6.
Simplify and express the following as single exponent.    (Page No. 85)
(i) 2^-3 × 2^-2
(ii) 7^-2 × 7⁵
(iii) 3⁴ × 3^-5
(iv) 7⁵ × 7^-4 × 7^-6
(v) m⁵ × m^-10
(vi) (-5)^-3 × (-5)^-4
Answer:
(i) 2^-3 × 2^-2 = 2^(-3)+(-2) = 2^-5 = [latex]=\frac{1}{2^{5}}[/latex] = [latex]\frac{1}{2 \times 2 \times 2 \times 2 \times 2}[/latex] = [latex]\frac{1}{32}[/latex] [∵ a^m × aⁿ = a^m+n]
(ii) 7^-2 × 7⁵ = 7^-2+5 = 7³ = 343
(iii) 3⁴ × 3^-5 = 3^4+(-5) = 3^-1 = [latex]\frac{1}{3}[/latex] [∵ a^-n = [latex]\frac{1}{a^{n}}[/latex]]
(iv) 7⁵ × 7^-4 × 7^-6 = 7^5+(-4)+(-6) = 7^5-10 = 7^-5 = [latex]=\frac{1}{7^{5}}[/latex]
(v) m⁵ × m^-10 = m^5+(-10) = m^-5 = [latex]=\frac{1}{m^{5}}[/latex]
(vi) (-5)^-3 × (-5)^-4 = (-5)^(-3)+(-4) = (-5)^-7 = [latex]\frac{1}{(-5)^{7}}[/latex] = -[latex]\frac{1}{5^{7}}[/latex]

Question 7.
Change the numbers into standard form and rewrite the statements.      (Page No. 93)
i) The distance from the Sun to Earth is 149,600,000,000 m
Answer:
149,600,000,000 m = 1496 × 10⁸ m

ii) The average radius of the Sun is 695000 km
Answer:
695000 km = 695 × 10³ km

iii) The thickness of human hair is in the range of 0.005 to 0.001 cm.
Answer:
0.005 to 0.001 cm
= [latex]\frac{5}{1000}[/latex] to [latex]\frac{1}{1000}[/latex] cm = 5 × 10^-3 to 1 × 10^-3 cm

iv) The height of Mount Everest is 8848 m
Answer:
8848 m, itself is a standard form.

Question 8.
Write the following numbers in the standard form.      (Page No. 93)
The standard form of the following numbers are
Answer:
(i) 0.0000456 = [latex]\frac{456}{10000000}[/latex] = 456 × 10^-7
(ii) 0.000000529 = [latex]\frac{529}{1000000000}[/latex] = 529 × 10⁹
(iii) 0.0000000085 = [latex]\frac{85}{10000000000}[/latex] = 85 × 10¹⁰
(iv) 6020000000 = 602 × 10000000 = 602 × 10⁷
(v) 35400000000 = 354 × 100000000 = 354 × 10⁸
(vi) 0.000437 × 10⁴ = [latex]\frac{437}{1000000}[/latex] × 10⁴
= 437 × 10^-6 × 10⁴
= 437 × 10^(-6)+4
= 437 × 10^-2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.1

Question 1.
Find the common factors of the given terms in each.

(i) 8x, 24
(ii) 3a, 2lab
(iii) 7xy, 35x²y³
(iv) 4m², 6m², 8m³
(v) 15p, 20qr, 25rp
(vi) 4x², 6xy, 8y²x
(vii) 12 x²y, 18xy²
Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x²y³
7xy = 7 × x × y
35x²y³ = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x²y³
= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m², 6m², 8m³
4m² = 2 × 2 × m × m
6m² = 2 × 3 × m × m
8m³ = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m² , 6m² , 8m³
= 2, m, m², 2m, 2m².

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x², 6xy, 8y²x
4x² = 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y²x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x², 6xy, 8xy² = 2, x, 2x.

vii) 12x²y, 18xy²
12x²2y = 2 × 2 × 3 × x × x × y
18xy² = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x²y, 18xy²
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2.
Factorise the following expressions
(i) 5x² – 25xy
(ii) 9a² – 6ax
(iii) 7p² + 49pq
(iv) 36 a²b – 60 a²bc
(v) 3a²bc + 6ab²c + 9abc²
(vi) 4p² + 5pq – 6pq²
(vii) ut + at²
Solution:
(i) 5x² – 25xy
= 5 x × x × – 5 × 5 × x × y
= 5 × x [x – 5 × y]
= 5x [x – 5y]

ii) 9a² – 6ax
= 3 × 3 × a × a – 2 × 3 × a × x
= 3a [3a – 2x]

iii) 7p² + 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a²b – 60a²bc
= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 – 5c]
= 12a²b [3 – 5c]

v) 3a²bc + 6ab²c + 9abc²
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p² + 5pq – 6pq²
= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q
= p [4p + 5q – 6q²]

vii) ut + at²
= u × t + a × t × t = t [u + at]

Question 3.
Factorise the following:
(i) 3ax – 6xy + 8by – 4bx
(ii) x³ + 2x² + 5x + 10
(iii) m² – mn + 4m – 4n
(iv) a³ – a²b² – ab + b³
(v) p²q – pr² – pq + r²
Solution:
i) 3ax – 6xy + 8by – 4ab
= (3ax – 6xy) – (4ab – 8by)
= (3 × a × x – 2 × 3 × x × y)
– (4 ×a × b – 4 × 2 × b × y)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y)(3x – 4b)

ii) x³ + 2x² + 5x + 10
= (x³ + 2x²) + (5x +10)
= (x² × x + 2 × x²) + (5 × x + 5 × 2)
= x²(x + 2) + 5(x + 2)
= (x + 2) (x² + 5)

iii) m² – mn + 4m – 4n
= (m² – mn) + (4m – 4n)
= (m × m – m × n) + (4 × m – 4 × n)
= m(m – n) + 4(m – n)
= (m – n) (m + 4)

iv) a³ – a²b² – ab + b³
= (a³ – a²b²) – (ab – b³)
= (a² × a – a² × b²) – (a × b – b × b²)
= a²(a – b²) – b(a – b²)
= (a – b²) (a² – b)

v) p²1 – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= p(pq – r²) – (pq – r²) × 1
= (p – 1) (pq – r²)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a – b)² ≡ a² – 2ab+ b² geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)² = a² – 2ab + b² [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)² = 4 sq. units
[∵ (3 – 1 )² = 2² = 4]

ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS = a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a² – b² geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a² – b² = 9 – 4= 5sq . units
[ ∵ 3² – 2² = 9 – 4 = 5]

ii)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a² – b² = 3 sq. units
[∵ (2² – 1²) = 4 – 1 = 3]