Class 8

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 \(\left(\frac{\mathrm{n}}{2}\right)\) and \(\left(\frac{n}{2}+1\right)\) terms = Mean of \(\left(\frac{10}{2}\right)\) and \(\left(\frac{10}{2}+1\right)\) 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 = \(\frac{\Sigma x_{i}}{N}\) = \(\frac{80}{5}\) = 16
Mean of the deviations = \(\frac{-5}{5}\) = -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 = \(\frac{\text { Sum of the observations }}{\text { No. of the observations }}\)

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 = \(\frac{n+1}{2}\) = \(\frac{7+1}{2}\) = 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 = \(\frac{6+12+24}{3}\) = \(\frac{42}{3}\) = 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 = \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\) = 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 = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{6}{7}\) = 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 = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{23}{5}\) = 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 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.
Select a suitable identity and find the following products
(i) (3k + 4l)(3k + 4l)
(ii) (ax² + by²)(ax² + by²)
(iii) (7d – 9e)(7d – 9e)
(iv) (m² – n²)(m² + n²)
(v) (3t + 9s) (3t – 9s)
(vi) (kl – mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b – a)(2b +c)
Solution:
(3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².
=(3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²
= 3k × 3k + 24kl + 4l × 4l
= 9k² + 24kl + 16l²

ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².
= (ax²)² + 2 × ax² × by² + (by²)² [ ∵ (a + b)² = a² + 2ab + b²]
= ax² × ax² + 2abx²y² + by² × by²
= a²x⁴ + 2ab x²y² + b²y⁴

iii) (7d – 9e) (7d – 9e)
= (7d – 9e)² is in the form of (a – b)².
= (7d)² – 2 × 7d × 9e + (9e)² [ ∵ (a – b)² = a² – 2ab + b²]
= 7d × 7d – 126de + 9e × 9e
= 49d² – 126de + 81e²

iv) (m² – n²) (m² + n²) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a² – b²
∴ (m² + n²) (m² – n²) = (m²)² – (n²)² = m⁴ – n⁴

v) (3t + 9s) (3t – 9s) = (3t)² – (9s)² [ ∵ (a + b) (a – b) = a² – b² ]
= 3t × 3t – 9s × 9s
= 9t² – 81s²

vi) (kl – mn) (kl + mn) = (kl)² – (mn)² [ ∵(a + b) (a – b) = a² – b² ]
= kl × kl – mn × mn
= k²l² – m²n²

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a²x² + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6
= 36x² + 6x × 11 + 30
= 36x² + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a²x² + ax(c – b) – cb
(2b – a) (2b + c) = (2)²(b)² + 2b (c – a) – ca
= 4b² + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 304²
(ii) 509²
(iii) 992²
(iv) 799²
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 304² = (300 + 4)² is in the form of (a + b)².
∵ (a+b)² = a² + 2ab + b²
a = 300, b = 4
(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 509² = (500 + 9)²
a  = 500, b = 9
= (500)² + 2 × 500 × 9 + (9)²
[ ∵ (a + b)² = a² + 2ab + b²]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 992² = (1000 – 8)²
a = 1000, b = 8
= (1000)² – 2 × 1000 × 8 + (8)² [∵ (a-b)² = a² – 2ab + b²]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 799² = (800 – 1)²
a = 800, b = 1
= (800)² – 2 × 800 × 1 + (1)²
= 800 × 800 – 1600 + 1
= 6,40,000 – 1600 + 1
= 6,40,001 – 1600
= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
(a + b) (a – b) = a² – b²
∴ (300 + 4) (300 – 4) = (300)² – (4)²
= 300 × 300 – 4 × 4
= 90,000 – 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)
= (80)² – (3)² [ ∵ (a + b) (a – b) = a² – b²]
= 80 × 80 – 3 × 3
= 6400 – 9
= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)² + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)
= (205)² – (1)² [∵ (a + b)(a-b) = a² – b²]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 1 What is Science Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 1st Lesson What is Science

8th Class Biology 1st Lesson What is Science Textbook Questions and Answers

Improve Your Learning

Question 1.
What is Science?
Answer:
Science:

1. Science is the concerted human effort to understand the history of the natural world how the natural world works, with observable physical evidence.
2. Science is an organized study of knowledge which is based on experimentation.
3. Science is a tool for searching truths of nature.
4. Science is the way of exploring the world.

Question 2.
What are scientific methods ? Write the steps involved in a scientist’s work.
Answer:
Scientific Methods:

1. Scientists solve a problem or answer a question by using organized ways. They are called “Scientific Methods”.
2. The following are the steps involved in a scientific work.
   a) Step -1 : Ask questions.
   b) Step – 2 : Form hypothesis.
   c) Step – 3 : Plan experiments.
   d) Step – 4 : Conduct experiments.
   e) Step – 5 : Draw conclusions.

Question 3.
What is the way to find out solutions for the problems in scientific way?
Answer:
To find out solutions for the problems in scientific way we need to follow a sequential order. So we go through the following.

1. Identifying problem : Let us identify any problems from your surroundings.
   Ex: The bulb did not lit in the room.
2. Making hypothesis: List out different solutions which your think for the identifying problem.
   Ex: De filament, fuse failure, switch problem, wire problem.
3. Collecting information: To solve the identifying problem collect material, apparatus, Information, persons.
   Ex: Collect material like tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
4. Data analysis: Arrange the collected data or information to conduct experiment.
5. Experimentation: To prove selecting hypothesis conduct experiment.
   Ex: Observe filament of the bulb.
6. Result analysis: Analyzing the results to find out the solution for the problem based on the results you need to select another hypothesis to prove.
   Ex: Filament of the bulb is good in condition so we need to observe fuse.
7. Generalisation: Based on the experiment and its results explain the solution for the problem.
   Ex: Fuse is damaged so the bulb not glow, so we need to replace the fuse.

Question 4.
During investigation in science, name some rules to be followed for the safety.
Answer:

1. Think ahead: Study the steps of the investigation, so you know what to expect.
2. Be neat: Keep your work area clean.
3. Oops!: If you should spill or break something or get cut, tell your teacher right away.
4. Watch your eyes: Wear safety goggles anytime you are directed to do so.
5. Yuck!: Never eat or drink anything during a scientific activity.
6. Protect yourself from shocks: Be especially careful if an electric appliance is used. Be sure that electric cords are in a safe place where you can’t trip over them. Don’t ever pull a plug out an outlet by pulling on the cord.
7. Keep it clean: Always clean up when you have finished.

Question 5.
If you plan a test to find out how much water different brands of paper towels absorb, write down the steps of the experiment.
Answer:

1. Pour 1 liter of water into each of three beakers.
2. Put a towel from each of the three brands into a different beaker for 10 seconds.
3. Pull the towel out of the water, and let it drain back into the beaker for 5 seconds.
4. Measure the amount of water left in each beaker.
5. The towel put in the beaker in which less amount of water is left absorbs more water.

Question 6.
Collect information about scientists and their works and prepare a chart and paste it in your classroom.
Answer:

Question 7.
What are process skills ?
Answer:
Observe, compare, classify are the process skills.

1. Observe – Use the senses to learn about objects and events.
2. Compare – Identify characteristics of things or events to find out how they are alike and different.
3. Classify – Group or organize objects or events in categories based on specific characteristics.

Question 8.
Read the following.
Answer:
Endemic Species:

You may find that these animals are specifically found in certain regions of the world.
You are also aware of the fact that many plants and animals are widely distributed throughout the world. But some species of plants and animals are found restricted to some areas only. Plants or animal species found restricted to a particular area of a country are called Endemic Species.
Now answer the following questions.
a) Name an Endemic Species of our State.
Answer:
Tiger, Peacock and Crane.

b) You may notice that Kangaroo is endemic to Australia and Kiwi to New Zealand. Can you tell which among the above picures represent an endemic species of India ?
Answer:
Peacock and Tiger.

c) Name some other endemic species of India.
Answer:
Indian Lion, Indian Wolf, Great Indian bustard bill.

Question 9.
Reading the following poster.

Answer:
Biotic Components:
Producers – Mangrove, spirogyra, euglena, oscillatoria, blue green algae, ulothrix, etc.
Consumers – Shrimp, crab, hydra, protozoans, mussel, snails, turtle, daphnia, brittle word, tube worm, etc.
Decomposers – Detritus feeding bacteria, etc.
Abiotic components – Salt and fresh water, air, sunlight, soil, etc.

Question 10.
What is Generalisation? Give an example.
Answer:
Based on the experiment and its results explaining the solution for the problem is called Generalisation.
Example:

a) The bulb did not light in the room.
b) Identify the problem that may be defilament, fuse failure, switch problem, wire problem.
c) Then take tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
d) Observe the filament of the bulb.
e) If the filament of the bulb is good, then observe fuse.
f) As the fuse is damaged, we need to replace the fuse.

Based on this, the bulb did not lit in the room because the fuse is damaged. This is the generalisation in the above experiment.

Question 11.
What are the different types of writings used by scientists to describe what they are doing or learning?
Answer:

1. Informative writing, Narrative writing, Expressive writing, Persuasive writing are used by the Scientists.
2. In informative writing, scientists describe the observation, inferences and their conclusion.
3. In narrative writing the scientists describe about something, give examples or tell a story.
4. In expressive writing, they may write letters, poems, or songs.
5. In persuasive writing they write letters about important issues in science and also write about what they have learned about science which helps others to understand about their thinking.

Question 12.
How do scientists use numbers in their investigations?
Answer:

1. Measuring, interpreting data, using number sense are few methods used by scientists in their investigations.
2. Scientists make accurate measurements by using different measuring instruments like thermometer clocks, timers, rules, a spring scale, beakers, balance and other containers to measure liquids.
3. Scientists collect, organize, display and interpret data by using tables, charts and graphs.
4. Scientists compare and order numbers, compute with numbers shown on graphs and read the scales on thermometers, measuring cups, beakers and other tools.
5. Good scientists apply their maths skills to help them display and interpret the data they collect.

Question 13.
What is Hypothesis? What are variables?
Answer:

1. Making a statement about an expected outcome is called Hypothesis.
2. Variables are factors that can affect the outcome of the investigation.

Question 14.
What do the following persons do?
a) A geologist
b) A chemist
c) A biologist
d) An ecologist
e) A climatologist
Answer:
a) A geologist examines the distribution of fossils and makes observations to find patterns in natural phenomena.
b) A chemist observes the rate of a chemical reaction at a variety of temperatures.
c) A biologist observes the reaction of a particular tissue to various stimulants.
d) An ecologist observes the territorial behaviours of different animals and birds.
e) A climatologist collects data from weather balloons and makes observations basing on it.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions and Answers.

8th Class Maths 3rd Lesson Construction of Quadrilaterals InText Questions and Answers

Do this

Question 1.
Take a pair of sticks of equal length, say 8 cm. Take another pair of sticks of equal length, say 6 cm. Arrange them suitably to get a rectangle of length 8 cm and breadth 6 cm. This rectangle is created with the 4 available measurements. Now just push along the breadth of the rectangle. Does it still look alike? You will get a new shape of a rectangle Fig (ii), observe that the rectangle has now become a parallelogram. Have you altered the lengths of the sticks? No! The measurements of sides remain the same.

Give another push to the newly obtained shape in the opposite direction; what do you get? You again get a parallelogram again, which is altogether different Fig (iii). Yet the four measurements remain the same. This shows that 4 measurements of a quadrilateral cannot determine its uniqueness. So, how many measurements determine a unique quadrilateral? Let us go back to the activity!
You have constructed a rectangle with two sticks each of length 8 cm and other two sticks each of length 6 cm. Now introduce another stick of length equal to BD and put it along BD (Fig iv). If you push the breadth now, does the shape change? No!
It cannot, without making the figure open. The introduction of the fifth stick has fixed the rectangle uniquely, i.e., there is no other quadrilateral (with the given lengths of sides) possible now. Thus, we observe that five measurements can determine a quadrilateral uniquely. But will any five measurements (of sides and angles) be sufficient to draw a unique quadrilateral? (Page No. 60)
Answer:
Yes, any 5 individual measurements are needed to construct a quadrilateral.

Question 2.
Equipment (Page No. 61)
You need: a ruler, a set square, a protractor.
Remember: To check if the lines are parallel.
Slide set square from the first line to the second line as shown in adjacent figures.

Now let us investigate the following using proper instruments. For each quadrilateral,
a) Check to see if opposite sides are parallel.
b) Measure each angle.
c) Measure the length of each side.

Record your observations and complete the table below.
Answer:

Question 3.
Can you draw the angle of 60°?    (Page No. 63)

Answer:
Using a scale and compass,
we can construct 60°.

Question 4.
Construct the parallelogram above (Refer text book page no: 75) BELT by using other properties of parallelogram. (Page No. 75)
Answer:

We can construct a parallelogram using the measurements of a side, a diagonal and an angle.
BE = 5 cm ⇒ LT = 5 cm
∠B = 110° ⇒ ∠E = 180° – 110° = 70°
TE= 7.2 cm

Try these

Question 1.
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm?    (Page No. 70)
Answer:

In a parallelogram BATS, opposite sides are equal.
BA = ST = 5 cms
AT = BS 6 cms
AS = 6.5 cms

∴ So, we can construct BATS parallelogram. It needs only three measurements.

Question 2.
A student attempted to draw a quadrilateral PLAY given that PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm. But he was not able to draw it why ?
Try to draw the quadrilateral yourself and give reason. (Page No. 70)
Answer:
In a quadrilateral PLAY
PL = 3 cm LA = 4 cm AY = 4.5 cm
PY = 2 cm LY = 6 cm
Here YP + PL < YL [∵ 2 + 3 < 6 ⇒ 5 < 6]

But in a △YPL, the sum of two sides is less than the third side.
∴ We are unable to construct a quadrilateral PLAY [∵ YL > YP]
[∵ The arcs do not intersect which are drawn from L and P, also Y, P, L are collinear points]

Think, discuss and write

Question 1.
Is every rectangle a parallelogram? Is every parallelogram a rectangle?    (Page No. 63)
Answer:
Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Question 2.
Uma has made a sweet chikki. She wanted it to be rectangular. In how many different ways can she verify that it is rectangular?    (Page No. 63)
Answer:
If the sweet chikki is to made into a rectangular shape, she has to verify the following shapes:

1. Quadrilateral
2. Trapezium
3. Parallelogram

Question 3.
Can you draw the quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4.8 cm and diagonals AC = 5 cm, BD = 5.4 cm by constructing △ABD first and then fourth vertex ‘C’ ? Give reason.       (Page No. 72)
Answer:
We cannot construct △ABD. So, if we start first from △ABD, it is impossible to construct □ ABCD.
[∵ The length of \(\overline{\mathrm{AD}}\) is not given]

Question 4.
Construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm. Justify your result.      (Page No. 72)
Answer:

PQ = 3 cm
RS = 3 cm
PS = 7.5 cm
PR = 8 cm
SQ = 4 cm
With the given measurements △PQS is not possible to construct.
∵ PQ + QS < PS
The arcs which drawn from P and Q are not intersecting.
∴ We can’t obtain vertex ‘S’.
∴ Without vertex ‘S’ we can’t get a quadrilateral PQRS.

Question 5.
Can you construct the quadrilateral PQRS, if we have an angle of 100° at P instead of 75°? Give reason. (Page No. 74)
Answer:

PQ = 4 cm,
QR = 4.8 cm,
ZP = 100°,
ZQ = 100°,
ZR = 120°
∴ We can construct a quadrilateral with the given measurements.
Since the sum of 4 angles is equal to 360°.

Question 6.
Can you construct the quadrilateral PLAN if PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15° and ∠A = 140°?
(Draw a rough sketch in each case and analyse the figure). State the reasons for your conclusion. (Page No. 74)
Answer:
PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15°, ∠A = 140°

∴ With the given measurements it is not possible to construct a quadrilateral.

Question 7.
Do you construct the given quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75° by taking BC as base instead of AB? If so, draw a rough sketch and explain the various steps involved in the construction. (Page No. 77)
Answer:
AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75°

Construction Steps:

1. Construct a line segment with radius 4.5 cms as \(\overline{\mathrm{BC}}\)
2. With the centres B and C draw two rays with 100°, 75° respectively.
3. With the centres B and C, two arcs are drawn with radius 5 cm and 6 cm respectively. The arcs and the rays are intersected.
4. Let the intersecting points be keep as A, D.
5. Join A, D.
6. ∴ ABCD quadrilateral is formed.
   

Question 8.
Can you construct the given AC = 4.5 cm and BD = 6 cm quadrilateral (rhombus) taking BD as a base instead of AC? If not give reason. (Page No. 79)
Answer:

We can construct a rhombus taking BD as base instead of base AC.

Question 9.
Suppose the two diagonals of this rhombus are equal in length, what figure do you obtain? Draw a rough sketch for it. State reasons. (Page No. 79)
Answer:
In a rhombus if the two diagonals are equal then it becomes a square.
∴ ABCD is a square.
[∵ AB = BC = CD = DA Also AC = BD]

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 5th Lesson Attaining the Age of Adolescence

8th Class Biology 5th Lesson Attaining the Age of Adolescence Textbook Questions and Answers

Improve Your Learning

Question 1.
How is adolescence different from childhood?
Answer:

Question 2.
Write short notes on the following.
a) Secondary sexual characters
b) Adam’s Apple.
Answer:
a) Secondary Sexual characters:

1. In adolescence age some external changes have seen in boys and girls.
2. These are called secondary sexual characters.
3. Example: in boys facial hair, moustaches and beards begin to grow. Hair starts growing on the chest of boys.
4. In girls breast begin to develop.
5. In both boys and girls hair grows in the arm pits.

b) Adam’s Apple:

1. The Adam’s apple is actually a partial growth of our voice box or larynx.
2. The larynx is made up of nine cartilages, one of which is the largest, called thyroid cartilage.
   
3. Due to the elongation of the thyroid cartilage the Adam’s apple is formed. It protrudes out in front of the neck.
4. This is caused mainly by male hormone. Testosterone during adolescence.
5. As a result muscles or chords attached to the cartilage get loosened and thickened.
6. When air passes through these chords a hoarse sound is produced.
7. This is the reason for disturbance in voice in the stage of adolescence. At the end of this stage voice get perfect.

Question 3.
List out the changes in the body that take place at the age of adolescence.
(or)
What are the changes that could be observed in adolescence phase?
Answer:

1. During this adolescence changes occur in external, internal parts of the body.
2. They show interest to spend time with peers.
3. Girls voice becomes soft, pimples also may appear on the face by the activation of oil and sweat glands.
4. There is growth and maturity in reproductive system.
5. In boys voice becomes coarse. Pimples, acne may appear on the face. Facial hair like mustache begin to grow.

Question 4.
Match the following:
1. Testes                                  ( )           a. Estrogen
2. Endocrine gland                  ( )           b. Pituitary
3. Menarche                            ( )           c. Sperm
4.Female hormone                  ( )           d. First menstruation
Answer:
1. c
2. b
3. d
4. a

Question 5.
Why acne and pimples are common in adolescents?
Answer:

1. Naturally in adolescence boys and girls feel worried of their pimples and acne.
2. The reason is the secretions of sweat and sebaceous glands are very active in adolescence.
3. Because of increased activity of these glands in the skin, boys and girls get acne and pimples.

Question 6.
What can you suggest to your classmates to keep himself/herself clean and healthy?
Answer:

1. It would better to have bath atleast twice in a day.
2. All parts of the body and inner wears should be washed and cleaned every day.
3. If cleanliness is not maintained, there are chances of having fungal, bacterial and other unwanted infections.
4. Girls should take special care of cleanliness during the time of menstrual cycle.
5. Making use of disposable napkins.

Question 7.
If you have chance to talk with a doctor, what questions you would ask about adolescent emotions and changes in the body ?
Answer:
If I have chance to talk with a doctor, I would ask about

1. How to develop positive emotions like bravery, self confidence, happiness, satisfaction, appreciation gratitude, concern and forgiveness.
2. And also how to over come the negative emotions like anger, bitterness, dissatisfaction, sadness, anxiety, fear, shame and guilt; which are needed in adolescence.

Question 8.
Some mobile phones have auditory meter to measure frequency of produced sound. By using this phone measure your friend’s voice frequency one from each class VI to X. Report your findings.
Answer:

Question 9.
Write five suggestions to improve the performance of Red Ribbon club of your school.
Answer:

1. To instill life skills.
2. To ensure that every college going youth is equipped with conceptual knowledge about various basic health aspects.
3. To increase the capacity of educational system in teaching various basic health aspects.
4. To motivate youth and build their capacity as peer educators.
5. To promote voluntary blood donations.

Question 10.
Prepare a three minute speech on behavioural changes in adolescents.
Answer:

1. Adolescence is the growing age where we may observe some changes in behaviour.
2. They are very fast in taking decisions.
3. They do not want to be forced to do any work, behave peculiarly sometimes fast and sometimes frigid.
4. Adolescents prefers to spend more times before the mirror and like to use perfumes.
5. They do not want to listen to parents suggestions and feels friends are correct but not parents.
6. They search for identify from teachers and peer groups.
7. They want more independence in taking decisions.
8. Sometimes they feel shy and sometimes feel happy.
9. They try to get romantic relationships.
10. They are more inclined towards unhealthy habits.
11. The adolescents have attraction towards opposite sex.
12. The mind of an adolescent is full of zealous acts and urge to find reasons for several things around.
13. Emotionally they are in a turbulent state all the time they get new thoughts for their life activities.
14. An adolescent feel insecure while trying to adjust to the changes in the body and the mind.
15. They seek company of friends to share their feelings even if they are of opposite sex.

Question 11.
Nature prepares human body to reproduce her generations. What do you think of it?
Answer:

1. In females, the reproductive phase of life begins usually around 10 to 12 years of age.
2. And generally it lasts till the age of 40 – 50 years.
3. The ova begin to mature with the onset of adolescence.
4. One ovum matures and is released by one of the ovaries once in about 28 to 30 days.
5. During this period the wall of the uterus becomes thick so as to receive a fertilized egg.
6. This results in pregnancy and childs’ birth.
7. If fertilization does not occur, the released egg and thickened lining of the uterus will be released with some amount of blood in woman.
8. This is called menstrual cycle.
9. Thus nature prepares the female human body to reproduce generation after generation to continue human life on the earth.
10. This is the secret of nature and is nature’s wonderful phenomena.

Question 12.
You know that early marriage is a social taboo. Prepare some slogans to prevent this. (OR)
You know about that child-marriages are social evil. Your school students are conducting a rally to educate the society. Prepare some slogans on this.
Answer:

1. Avoid child marriage – Prevent childhood.
2. Let a child be a child – stop child marriage.
3. Child marriage – a loosing game.
4. Stop child marriage – stop child abuse
5. Childhood is not for motherhood.
6. Let girls be girls but not brides.

Question 13.
13 years old Swaroop always think of his height. Can he improve his height? What do you suggest him?
Answer:
The suggestion is to take nutritious food and to do body exercise regularly to improve his height.

Question 14.
Are you angry with your parents. How do you wish your parents to be?
Answer:
When insulted or threatened unfairly by parents we get angry on them. In our opinion a parent is

1. to be a good advisor to give advice how to control stress and strain, which is needed by the adolescent.
2. to be like a guide to give guidance how to behave with opposite sex.
3. to be like a friend to give good suggestions.
4. to be like a wellwisher and always stand behind us to lead us for bright future.

Question 15.
What are your expectations about your parents and teachers?
Answer:
Parents and teachers play an important role to develop the adolescents in to healthy, productive young ones to the nation. Parents feel to develop their children in to better ones than themselves. They must have good courage, confidence, boldness, and free to solve the problems. They should not be tense and worse. Parent is to be a friend and guide towards adolescents.
Teacher is not only a master is to be a captain or a leader. Every adolescent needs mental support. Teacher is the only people who give suitable suggestions to make them free from all mental stress.

8th Class Biology 5th Lesson Attaining the Age of Adolescence InText Questions and Answers

Question 1.
Some of you also may behave like this, Why?
Answer:
At the age of 13 – 19 years, some changes like voice becomes hoarse, not caring to follow the suggestions and advises of parents, shows restlessness and growing tall.
Because at this age children will be entering into a period which is called as Adolescence, where some changes in the behaviour is seen.

Question 2.
Have you noticed that you are growing?
Answer:
At the age of adolescence growth in height takes place about 18 years of age both boys and girls reach their maximum height.

Question 3.
Have you reached the age of“Adolescence”?
a) Is mustache growing on your upper lip?
b) Did your voice change?
c) Are hairs growing under arm pit?
d) Are there pimples or acne on your face?
e) Are you taking care of your face by applying powder and combing your hairs frequently?
f) Are you feeling shy when talking with opposite sex?
g) Are you not interested to play with opposite sex which you have done earlier?
h) Are you showing restlessness while your parents suggest to do something?
Answer:
For above all the questions the answer is ‘yes’ during adolescence changes occur in external, internal parts of the body.

8th Class Biology 5th Lesson Attaining the Age of Adolescence Activities

Activity – 1

Question 1.
Observing growth rate.
a) Observe the below table and given graph, answer the following questions.

i) What have you observed from the above table?
Answer:
We have observed the height attained by boys and girls in different ages.

ii) When does growth in height nearly stop?
Answer:
In boy growth in height nearly stops at the age of 18, in girls it stops at 17 years.

iii) Which period of age according to you is the fastest growing period for girls?
Answer:
The fastest growing period for girls is between 14-17 years.

iv) Which period of age is the fastest growing period for boys?
Answer:
The fastest growing period for boys is between 16 to 18 years.

v) Who do grow faster? How can you say?
Answer:
The girls grow faster than the boys. By seeing the above graph, about 17 years of age, girls reach their maximum height.

b) Sneha is 13 years old with 125 cm tall. At the end of the growth period likely to be use the following formulae and calculate the maximum height that Sneha will reach.

Answer:

Sneha’s present height = 125 cm
At the end of the growth period Sneha is likely to be 131.5 cm.

c) Table – 1 shows that girls grow faster than boys in their adolescent period. From a group with six students in your class. Measure the height and calculate the future heights in the following table:
Answer:

Activity – 2

Question 2.
Form five groups in your class. Select at least 15 students. Collect body measurement data of the selected 15 students.
Find an average body measurements for boys and girls separately.
Answer:

Activity – 3

Question 3.
Read the following check list. Put tick (✓) mark which points reflect your behaviour.
Answer:
Check List:

Think & Discuss

I. Read the following information and answer the following questions.
Some sections of people in our society believe that during the period of menstruation women are untouchable. So, they are asked to keep a distance from others. During this time girls may be ristricted from taking bath, cooking food or going to school. In that case they may lag behind in their studies. In some sections of the society even women are also forced to stay in the huts built at the outskirts of the village.

Question 1.
In what way this kind of discrimination is harmful for girls and women?
Answer:

1. During the period of menstruation women are treated as untouchable by some sections of people in our society.
2. Generally during this period they feel weak and uncomfortable physically.
3. By this kind of discrimination, mentally also they get hurt and feel that why they have born as woman.
4. By separation, it would be known to all by this the girl may feels shy, unable to move freely with others and she becomes dull in studies.

Question 2.
Several researches have been done to prove that all these are myths and there is no scientific reason behind these. The blood and egg that is discarded would give rise to a baby if fertilization took place. This is a biological phenomena.
So how can it be impure or unclean?
Answer:

1. If fertilization took place, the uterus receives a fertilized egg and this results in pregnancy and would give rise a baby which develops in the uterus.
2. If fertilization does not occur this cause bleeding in woman, which is the unwanted and waste to the uterus. So it can be treated as impure or unclean and may create some problems in uterus.
3. During menstrual period proper care regarding health and hygiene is needed rather than following myths.

Question 3.
If young generation is trapped into such unhealthy habits, what will be the
future of our country ? What are its effects?
Answer:

1. If young generation is trapped into unhealthy habits like consuming tobacco (gutkha, cigarettes, cigar, beedi) they addicted to such social evil.
2. Todays children are tomorrows citizens so it should be avoided.
3. A famous psychiatrist Stanly Hall stated that adolescence is the age of stress and strain. By getting proper guidance from teachers, parents and elders, the adolescents be able to lead a happy meaning full life and they will save the future of the country.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation InText Questions

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

8th Class Maths 12th Lesson Factorisation InText Questions and Answers

Do this

Question 1.
Express the given numbers in the form of product of primes. [Page No. 267]
(i) 48      (ii) 72      (iii) 96
(i) 48
Answer:
48 = 2 × 2 × 2 × 2 × 3

ii) 72
Answer:
72 = 2 × 2 × 2 × 3 × 3

iii) 96
Answer:
96 = 2 × 2 × 2 × 2 × 2 × 3

Question 2.
Find the factors of following:      [Page No. 268]
(i) 8x²yz     (ii) 2xy (x + y)       (iii) 3x + y³z
Answer:
i) 8x²yz = 2 × 2 × 2 × x × x × y × z
ii) 2xy (x + y) = 2 × x × y × (x + y)
iii) 3x + y³z = (3 × x) + (y × y × y × z)

Question 3.
Factorise:      [Page No. 270]
(i) 9a² – 6a
(ii) 15a³b – 35ab³
(iii) 7lm – 21lmn
Answer:
(i) 9a² – 6a = 3 × 3 × a × a – 2 × 3 × a
= 3 × a (3a – 2)
∴ 9a² – 6a = 3a (3a – 2)

ii) 15a³b – 35ab³
= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b
= 5 × a × b [3 × a × a – 7 × b × b]
= 5ab [3a² – 7b²]

iii) 7lm – 21lmn
= 7 × l × m – 7 × 3 × m × n × l
= 7 × l × m [1 – 3n]
= 7lm [1 – 3n]

Question 4.
Factorise:
i) 5xy + 5x + 4y + 4
ii) 3ab + 3a + 2b + 2 [Pg. No. 271]
Answer:
i) 5xy + 5x + 4y + 4
= (5xy + 5x) + (4y + 4)
= 5x(y + 1) + 4(y + 1)
= (y + 1) (5x + 4)

ii) 3ab + 3a + 2b + 2
= [3 × a × b + 3 × a] + [2 × b + 2]
= 3 × a [b + 1] + 2 [b + 1]
= (b + 1) (3a + 2)

Think, Discuss and Write

While solving some problems containing algebraic expressions in different operations, some students solved as given below. Gan you identify the errors made by them? Write correct answers.     [Page No. 279]
Question 1.
Srilekha solved the given equation as shown below.
3x + 4x + x + 2x = 90
9x = 90 Therefore x = 10 What could say about the correctness of the solution?
Can you identify where Srilekha has gone wrong?
Answer:
Srilekha’s solution is wrong,
∵ 3x + 4x + x + 2x = 90
10x = 90
x = \(\frac{90}{10}\)
∴ x = 9

Question 2.
Abraham did the following.      [Page No. 280]
For x = -4, 7x = 7 – 4 = -3.
Answer:
Abraham’s solution is wrong.
∴ If x = -4
⇒ 7x = 7(-4) = -28

Question 3.
John and Reshma have done the multiplication of an algebraic expression by the following methods: verify whose multiplication is correct.      [Page No. 280]

Answer:

∴ John’s solutions are wrong and Reshma’s solutions are correct.

Question 4.
Harmeet does the division as (a + 5) ÷ 5 = a + 1 His friend Srikar done the same (a + 5) ÷ 5 = a/5 + 1 and his friend Rosy did it this way (a + 5) ÷ 5 = a Can you guess who has done it correctly? Justify!     [Page No. 280]
Answer:
The solutions of Harmeet, Rosy are wrong.
(a + 5) ÷ 5 = \(\frac{a+5}{5}\)
= \(\frac{a}{5}\) + \(\frac{5}{5}\)
= \(\left(\frac{a}{5}+1\right)\)
∴ Srikar had done it correctly.

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.

Solution:

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 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m² – n² and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a² + 8a – 27a – 36
= 6a² – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x² – xy – 4xy + 2y²
= 2x² – 5xy + 2y²

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k²l – l²k + klm – l²m

iv) m² – n² and m + n
(m² – n²) (m + n) = m²(m + n) – n²(m + n)
= m³ + m²n – n²m – n³

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab² – a²b)
(iv) (p³ + q³)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x² – 5xy + 3x²y + 2xy – 5y² + 3xy²
= 2x² – 5y² – 3xy + 3x²y + 3xy²

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k²l² + k²lm – lm²n + kl²m – klm²

iii) (a – 2b + 3c) (ab² – a²b) = a(ab² – a²b) – 2b(ab² – a²b) + 3c(ab²– a²b)
= a²b² – a³b – 2ab³ + 2a²b² + 3ab²c – 3a²bc
= 3a²b² – a³b – 2ab³ + 3ab²c – 3a²bc

iv) (p³ + q³) (p – 5q + 6r) = p³(p – 5q + 6r) + q³(p – 5q + 6r)
= p⁴ – 5p³q + 6p³r + pq³ – 5q⁴ + 6rq³
= p⁴ – 5q⁴ – 5p³q + 6p³r + pq³ + 6rq³

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m² – mn + n²)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y² – 3xy – 9xy + 9x² + x² – 3xy + xy – 3y²
= 10x² – 14xy

ii) (m + n) (m²– mn + n²)
= m(m² – mn + n²) + n(m² – mn + n²)
= m³ – m²n + n²m + nm² – mn² + n³
= m³ + n³

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a² – 2ab + 5ac – ab + 2b² – 5bc – 2a² + 2ab + 2ac – 3ac + 3bc + 3c² + 12ac – 18a² – 30ab + 2bc – 3ab – 5b²
= – 19a² – 3b² – 34ab + 16ac + 3c²

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p²q² – pq²r + p²qr + pq²r – q²r² + pqr² – p²r – pqr + pr² – p²q – pq² + pqr
= p²q² – q²r² + p²qr + pqr² – p²r + pr² – p²q – pq ²

Question 4.
If a, b, care positive real numbers such that \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) ,find the value of \(\frac{(a+b)(b+c)(c+a)}{a b c}\)
Solution:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) = k then
\(\frac{a+b-c}{c}\) = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 9 Production and Management of Food From Animals Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 9th Lesson Production and Management of Food From Animals

8th Class Biology 9th Lesson Production and Management of Food From Animals Textbook Questions and Answers

Improve Your Learning

Question 1.
One honey bee hive consists of different types of bees. What are they? How they differ from each other?
Answer:

1. Honey bee species are social insects, which lives in colonies.
2. A honey bee colony consists of three types of bees.
3. One queen, several thousands of workers and few hundreds of drones.
4. There is only one queen bee in a colony. The primary function of a queen is to lay eggs (800 – 1200 eggs per day). The life span of queen is two to three years.
5. A worker has 5-6 weeks and the drone has 57 days of life span.
6. There are sterile female which are called workers in the hive. These bees attend to indoor duties during first three weeks during first three weeks of their lives such as secretion of royal jelly, feeding of the brood.
7. After three weeks they attend outdoor duties like collecting nector, pollen and water.
8. Drones are the male members of the colony. They are very lazy and unable to gather food. Their main duty is participating in mating, after this they die.

Question 2.
Make a list of characters of local variety buffaloes, which gives good quantity of milk in your village.
Answer:

1. Buffaloes are most common in India and other Asian countries.
2. In general, buffaloes give more milk than cows and buffaloes milk has more fat than the cow’s milk.
3. Buffaloes are more resistant to diseases than cows.
4. There are about seven different breeds of buffaloes, Murrah, Bhadwari, Jaffrabadi, Surthi, Mehsana, Nagpri and Nili Ravi in our country.
5. Each breed of buffalo is found in specific regions of the country and they differ in their body colour, shape and length of horns and in the shape and size of their forehead.
6. Of these different breeds of buffaloes, Murrah breed of buffaloes have been recognised by Government of India, as the best native milk yielding breed.
7. It yields about 8 litres of milk/day/animal and about 1800 – 2200 litres per year.

Question 3.
Explain the process of hatching eggs under broody hen in rural areas.
Answer:

1. Hatching of eggs is an interesting process.
2. The common village hen has less care and attention.
3. So it’s productivity is low.
4. It has the capacity to lay 15 to 20 eggs an egg per day respectively.
5. After that the hen becomes a broody state (wish to hatch the egg).
6. In this process the people of the house arrange the eggs in such a way that the broody hen can sit comfortably on the eggs to hatch.
7. The hen sits on the eggs stretching its wings to provide heat to the eggs, to develop chicks inside the eggs.
8. It takes 21 days to the chicks emerges out by breaking the egg shell. This process is called incubation.
9. Almost all the eggs hatch into chicken but some eggs which do not get sufficient heat from the broody hen become rotten.

Question 4.
Write about the accessory products produced in animal husbandry.
Answer:
Milk, meat, eggs, wool are the accessory products of animal husbandry.

Question 5.
What is estuaries, how they are suitable for both marine and river fish to live?
Answer:

1. Estuary is the place where river joins the sea usually called mouth of the river. In this area fresh water (with low salt content from the river) is mixed up with sea water (with high salt content).
2. When the wave action is strong or during high tide, large amount of sea water mixes with river water. In fact sea flows deep into the river.
3. During low tide and when the wave action is weak, less amount of sea water mixes with river water.
4. Because of this, salt content of the water in the estuary changes very rapidly.
5. Animals living in the estuary should tolerate and adopt to rapid changes in the salt content of water.

Question 6.
If you have a chance to visit milk chilling centre, what doubts would you like to clarify? Please list out them.
Answer:

1. Chilling of milk means rapid cooling of raw milk to sufficiently to low temperature. So that the growth of microorganisms present in the milk is checked.
   Doubts to clarify
2. What would happen if the temperature of the milk should be reduced to less than 10°C preferably 3-4°C, in the chilling process?
3. Whether the growth of the microorganisms will be controlled as the required nutrients and the growth conditions will be favourable to them?
4. If the growth of the organisms will not be checked what would happen to the milk. Any changes would happen? Is it fit for consumption?
5. If milk has to be transported to longer distances, considerable time is involved between production and heating process.
6. During this period milk should be protected from spoilage by the action of micro-organisms the chilling process therefore considered is necessary.

Question 7.
Collect news from news papers about milk production and impurities in milk.
Prepare a note and display it on wall magazine.
Answer:
Milk Production:

1. Our government treats producing milk as an industry.
2. Generally farmers rear 1 to 5 cattle in small scale at their homes to produce milk.
3. Among cows, traditional species give 2-5 liters of milk in a day. Murrah species are reared in most of the districts in our state.
4. They give up to 8 liters of milk per day. Haryana, Jaferabad, Nagapuri are the traditional variety of cows which give good quantity of milk.
5. Jersy (England) and Holstein (Denmark) are the foreign varieties. They give 25 liters of milk per day.
6. These foreign varieties are cross breed with our traditional local varieties. They give 8-20 liters milk per day.
7. Cows play vital role in total milk production of our country.
8. Out of milk produced in our country 60% is used to prepare cheese, cova, ghee, curd, milk powder and other milk products.
9. The milk produced in dairy forms is collected from house holds and pasteurized.

Impurities in Milk

1. Addition of pure or impure water.
2. Addition of colouring matter.
3. Addition of preservatives like Sodium bicarbonate, Borax or Boric Acid, Salicylic Acid and formaldehyde.
4. Addition of substances used for thickening after dilution with water, e.g. Flour, arrowroot, chalk, carbonate of magnesium.
5. Sugar is added to raise the specific gravity of diluted skimmed milk.
6. Milk is contaminated by
   a) improper or poisonous food eaten by the animal.
   b) Poor condition of animal due to nursing.
   c) Contamination of disease germs from the cub.
   d) Absorption of bad odours.
   e) If milk has been diluted it becomes pale and blueish, so milk and cream are artificially coloured with anilines or other pigments. Annotto is the common dye used to milk cream and butter.

Question 8.
Collect information about sea weeds, sea kelp from your school library and write a note with examples.
Answer:
Sea Weeds:
Sea weeds constitute an important marine resource and are found along the Rocky intertidal and subtidal regions of the coasts of India. The Sundarbans, the Chilka lake, the deltas of Godavari and Krishna. Gulf of Mannar. Palk bay Gujarath coast and around Lakshadweep, Andaman and Nikobar Islands are areas rich in sea weeds. They are used for human consumption, as cattle and poultry feed, as manure and for industrial purposes as the source phyco colloids like Agar-agar.

Sea Kelps:
Kelps are the largest sea weeds belonging to the brown algae (phaeophyceae) in the order laminariales.
Kelps grow in under water in shallow oceans, it has a high rate of growth and it’s decay is quite efficient in yielding methane as well as sugars that can be converted into ethanol.
It has been proposed that large open ocean, kelp farms could serve as a source as renewable source of energy.
Unlike some biofuels, such as corn, ethanol, Kelp energy avoids “food verses fuel” issues and does not require irrigation.

Question 9.
Observe nearby poultry farm and findout how do they export eggs to market? What material is used for transportation?
Answer:

1. The most important component in the marketing of eggs is to handle them with care at the time of collection, transportation and in the sale counter.
2. To avoid breakage of eggs special designs are planned in the poultry forms. They are packed in tailor made egg trays which can be piled up one over the other after inserting eggs in each tray.
3. These trays protect the eggs from both vertical and horizontal friction and avoid breakage during transportation.
4. The design of the transport vehicles for the chicken are also specially designed cages to allow the requisite freedom to the birds.

Question 10.
Observe a dry honey bee hive and how the bees built it. Draw a picture. How does it look like?
Answer:

Honey Bee hive is a construction of six sided (hexagon) wax compartments made by bees to store honey and eggs.
Wax is produced by the honey bees is known as bee’s wax or honey wax. Wax is used in the production of cosmetics, shoe polish, candles and leather industry.

Question 11.
Agriculture and animal husbandry are both sides of the same coin. How can you justify this?
Answer:

1. Farmers adopt different methods of management for getting better yields in agriculture. In the same way, care is required in the management of reading animals also.
2. Since long ago, man used animals not only for obtaining food but also for agriculture, transportation.
3. People living in rural areas used to domesticate animals like cows, buffaloes, bullocks, goats, sheep, pigs, hens etc., supplying of nutritious food accommodating clear and hygienic, shelters for animals is very important issue in animal husbandry.
4. Generally villages send their cattle to rear in their fields. Where grass is easily avail-able. These animals provide him with food (milk, meat, eggs etc) and clothing. Man in turn protected, fed and took care of these domestic animals.
5. The selective relationship between man and the domestic animals has become in-dispensable – man cannot live without them and these animals cannot survive in wild.
6. Thus, the relationship is mutually dependent and beneficial.

Question 12.
How do you appreciate the uses of cattle?
Answer:

1. Cattle are not only for our food, even its excretions like dung and urine also we make them use. Cow dung is used as a cooking fuel, sanitizing cleanser, construction material, insulation and water proofing for walls and floors in rural houses, a cultural symbol in religious worship, the raw material for producing organic compost and generating electricity.
2. The urine of cows is considered an elixir of life and is used as a natural remedy for liver and heart conditions as well as for enhancing mental and physical strength and increasing longevity.
3. The utilization of cow dung and urine is a perfect example of sustainable living. An understanding of the use of cow dung and urine by the rural Indian population can illustrate the indigenous knowledge associated with these materials and alternative sources of materials for electricity generation as well as cost-effective and environmentally friendly fuel and medicines.
4. Even the ash formed from burning of dung can be used as a cleaning agent for household utensils or used as a fertilizer.
5. Cow manure contains several plant nutrients including nitrogen, potassium, calcium and magnesium. The composition of cow manure makes it ideal for several uses, including fuel, fertilizer and medicine.

Question 13.
What makes you amazing in division of work in Honey bee colony? Support your answer.
Answer:

1. Honey bees lives in colonies. It consists of three types of bees.
2. One queen bee, several thousands of worker bees and few hundreds of drones or male bees.
3. The primary function of a queen bee is to lay 800 – 1200 eggs per day.
4. The worker bees are the sterile female bees. They attend to indoor duties like secretion of royal jelly, feeding of the brood during first three weeks.
   After three weeks they attend outdoor duties like collecting nector, pollen and water.
5. Drones are the male members of the colony. They are lazy. Their main duty is participating in mating. After this they die.
6. Thus the division of work is amazing in honey bee colony.

Question 14.
Conversion of agricultural lands into fish ponds leads to food crisis and environmental pollution. Write your opinion to conduct in debate on this issue.
Answer:

1. Fish culture is not new to our country. Even today in several villages, fishes are grown in small ponds in the backyard of the houses. Besides this, growing of fishes in rice fields is also more common in villages. In fact fish production is considered as second crop for farmers, the primary crop being rice.
2. Fish culture is sometimes practised in combination with a rice crop, so that fish are grown in the water in the paddy field.
3. Growing fish in paddy field is also multi-utilitarian practice. The reason for this are the increasing use of inorganic fertilizers and insecticides in paddy fields which cause deleterious effects on fish and predation for birds, snakes etc.
4. Cultivating fish in paddy fields lower diseases like stem borers on paddy.
5. If the agricultural fields are converted to fish pond, food grains would not be sufficient for the growing population. This leads to food crisis.
6. But ours is a agricultural country. So agriculture will be done in the crop seasons, and in the unseason they can convert to fish pond.
7. The places where it is near to the water sources farmers can utilize these places for growing fish.

8th Class Biology 9th Lesson Production and Management of Food From Animals InText Questions and Answers

Question 1.
What are the food items that are obtained from animals?
Answer:
Milk, meat, eggs, honey.

Question 2.
Do we get our food only from domesticated animals? List out the food that is obtained from animals.
Answer:
We domesticated only such animals which were helpful to us. Buffalo, Cow etc. are reared for milk. Hens, goats, sheep for meat, ducks for eggs.

Question 3.
Do you know the period from which wild animals were being tamed?
Answer:

Question 4.
Why did early man domesticate only some of the animals?
Answer:
Domestic animals provide him food, (milk, meat, eggs), clothing (skin of the animals). The early man realised the capabilities of these animals, tamed and domesticated them to help him in his daily activities.

Question 5.
Why he did not domesticated animals like elephant, tiger, lion etc. or birds like eagle and owl?
Answer:
They are the wild animals and it is difficult to tame them. There is no use of these animals for his daily activities.

Question 6.
Do all the persons who won agriculture fields also rear cattle?
Answer:
All most all the persons who won agriculture fields, rear cattle also.

Question 7.
Is there any relation between agriculture and cattle rearing or animal husbandry?
Answer:
Cattle provide him food, help in agriculture and transport, man provide food for the animals from the agricultural fields.
Collect the following information from your calls.

Question 8.
Number of families in agriculture
Answer:
Less families

Question 9.
Number of families in agriculture along with animal husbandry.
Answer:
Almost all families.

Question 10.
Number of families in Animal husbandry alone.
Answer:
Less families.

LET US DO Cattle Rearing:

Question 11.
Form a group with four or five students in your class. Discuss about the reasons. Why does a farmers rear cattle?
Answer:
Farmers believe that animal husbandry is part and parcel of agriculture.

Question 12.
Where do people rear their cattle in your village?
Answer:
Generally people send their cattle to rear at the places where grass is easily available.

Question 13.
What are the cattle here?
Answer:
Cows, buffaloes, goats, sheep.

Question 14.
At which places fodder is available?
Answer:
Fodder is available in fields and open places.

Question 15.
What are places where water is available?
Answer:
Canals, ponds, wells etc.

Question 16.
Are there any differences between rearing of cows, buffaloes, goats and sheep?
Answer:
All the animals are herbivorous animals. Farmers use bullocks in agricultural practices like ploughing.

Question 17.
What are the major problems that cattle rearers generally face?
Answer:
Cattle sheds become unclean because of the remains of fodder, dung and urine. Care should be taken to prevent the growth of lice and mytes on cattle’s body.

Question 18.
Make a list of agricultural practices by using bullocks and the buffaloes.
Answer:
Ploughing, to draw the leveller, transporting agricultural goods etc.

Question 19.
Think in which way this practice is helpful to the farmer as well as field crops?
Answer:
Sheep and goats provide him meat and wool, the dung and urine becomes good manure to the field crops.

Question 20.
Where is veterinary hospital located in your area?
Answer:
It is located at the place where it is convenience to the people to bring their animals for check.

Question 21.
Who are working there and what do they do?
Answer:
The employees working in veterinary hospital are a veterinary doctor or animal husbandary assistant, a compounder and attenders. They provide treatment and health care for the cattle.

Question 22.
Meet a nearby veterinary doctor or animal husbandry assistant. Collect infor¬mation about common diseases in cattle and prepare a note on them.
Answer:

1. Gali Kuntu is a common and dangerous disease in cows and buffaloes.
2. Sheep and goats suffer from worm infections (Nattala Vyadhi)
3. Growth of lice and mytes on cattle’s body.
4. Some parasitic diseases cause damage to liver intestine.
5. Viral and bacterial diseases also effect on milk production.

Milk Production:

Question 23.
From which animals we get maximum milk production?
Answer:
Cows, buffalows

Question 24.
In which areas people use camel milk?
Answer:
Desert areas.

Question 25.
Have you ever see people taking donkeys milk? Why was it preffered?
Answer:
Yes. Donkey milk which is Rich in immunoglobulin helps human body from many viral and bacterial infections. These are mostly found in Telangana regions especially in Adilabad. The secret behind the glowing skin of Egyptian princess Cleopatra’s is donkey milk.

Question 26.
What are the types of fodder generally farmers feed the cattle with?
Answer:

1. They supply fodder from their agricultural fields.
2. They also feed the cattle with hay, green and dry grass, oil seed cakes of ground nut.

Question 27.
How farmers preserve fodder for cattle after harvesting?
Answer:
After harvesting farmers preserve fodder by arranging into heaps. This heap will be used for the cattle throughout the year.

Pasteurization :

Question 28.
Is there a milk collecting centre in your village?
Answer:
Krishna Milk Union, Vijaya Dairy Milk centres are located in our village.

Question 29.
How do they collect milk and export?
Answer:
The milk produced in dairy farm is collected from households and cattle rearers and export the milk in chilling form (cooling condition).

Question 30.
Do you know how they decide cost of milk?
Answer:
The cost of milk is decided due to fat content.

Question 31.
Do you know in which month the rate of milk production is high? Why?
Answer:
September to November the milk production is high.

Question 32.
Why the milk production is higher during those months than remaining year ? Discuss with your friends and find out the reasons.
Answer:
Milk production is slightly higher in the November and December months. Because

1. In these months food availability is rich and intaking of good food increase the milk production.
2. the climate is cool and enough hot. This sunny weather increase the milk production.
3. water resources are well in these months. Availability of food and water leads to milk production.
4. mostly animals delivered in July and August months and give milk before the summer. So the availability of milk is higher in November and December months.

Selection Procedure:

Question 33.
What care should be taken while buying cattle for milk production?
Answer:
The following points should be kept in mind.

1. Select high milk producing varieties, either traditional or hybrid.
2. Observe 2 to 3 days for average milk production.
3. Number yielding size, health, eating fodder.
4. Consult veterinary doctor, official of Director of Animal husbandry.

Some of rural people are experts in identifying high producing varieties.

Question 34.
What occasions, They decorate their cattle in your village?
Answer:
Pongal, Onam etc.

Question 35.
Do they respond when called by names ? Do you have any such experience with your pets ?
Answer:
We see the pet dogs climbing our body, sitting beside us, moves it’s tail and licks our feet when we call.

Poultry:

Question 36.
Collect information about Biogas Productions from your school library or internet and write notes on Biogas.
Answer:

1. In recent years, an alternate and better method is used to obtain energy from cattle dung and from the excreta of other animals including man and some types of organic waste material from agriculture, homes and industries.
2. This is by anaerobic fermentation of waste, to produce a gas which can be used as fuel. As this gas is produced from biological wastes, this is called BIO GAS.
3. Biogas is a mixture of gases – methane, carbondioxide and small amounts of hydro-gen, nitrogen and hydrogen sulphide. About 200 cc of biogas gives about 900 K.Cal of energy.
4. Production of biogas occurs in three stages. In the first stage, aerobic bacteria are allowed to degrade the complex molecules into simple molecules.
5. In the second stage, the simple molecules are fermented anaerobically to produce organic acids mostly to acetic acid.
6. In the final stage methane producing bacteria act on the acetic acid under anaerobic condition produce methane. The gases produced are collected into specially constructed chambers and supplied to users.
7. Use of bio gas for domestic purposes (cooking) lighting of street lights etc.
8. When bio gas is burnt, it does not pollute the environment.
9. The left over material after the production of bio gas can be used as manure in agriculture.

Question 37.
Are the hens reared in the poultry is same as our traditional varieties reared by farmers in the village?
Answer:
Farmers rear cocks and hens in villages. Most of these are local varieties (Natukollu). Poultry farms are of two types. One is for production of eggs (layers) and other for meat (Broilers).

Question 38.
Think and discuss Is genetically modified food useful or not?
Answer:
Natural wild varieties grow fully in 5 to 6 years. But broilers grow fully in just 6 to 8 weeks. This happens due to genetic modifications in the hen. So genetically modified food is useful.

Question 39.
Do you know Chicken 65? Why is this called so?
Answer:
This preparation is prepared by A.M. Buhari, in South Indian food industry in Chennai, in the year 1965. So this is called Chicken-65.

Question 40.
Have you heard about cock fight during some festival seasons. Think and discuss in your class about this type of practices which show human cruelty towards animals.
Answer:

At the time of Pongal festival, in some places like India and Tamilnadu, the birds are equipped with either metal spurs or knives tightly tied to the legs in the area where the cock’s natural spur has been partially removed. In this cock fight both the cocks would be wounded, severely bleed, this leads to the death of the birds. By watching this cock fight show, people get enjoyment. This shows human cruelty towards animals.

Question 41.
Do you know how many days a hen spends to hatch it’s eggs?
Answer:
A hen spends 21 days to hatch its eggs.

Question 42.
Prepare a detailed note on hatching eggs by observing at your village. If you need, please draw pictures also.
Answer:

Hatching of eggs into chick is called as incubation period. The full incubation period for an egg is 21 days. During this time the hen sits on her eggs and maintains a temperature of 100°F, which is needed to ensure proper embryonic development.

Question 43.
Egg is a nutritious food. Collect information about various nutrients in egg and write a note on them in your notebook.
Answer:
The nutrients present in egg are is as follows:
Choline: Healthy cell membranes
Vitamin – B: Folate and Riboflavin – converts food we eat into energy. Folate also reduces homocysteine levels and prevent birth defects.
Vitamin – A: Vision and healthy skin.
Vitamin – E and Vitamin – C also present
Luteine and Zeaxanthin: Found in yellow pigment of yolk, prevent macular degeneration.
NECC:
If you want to be healthy person eat egg everyday. This is the slogan of National Egg Co-ordination Committee. Egg is chief nutritious food which is easily available for all.

APICULTURE:

Question 44.
In what way honey bees are helpful in pollination?
Answer:
Honey bees are helpful in pollination, there after growth of food grains.

Question 45.
Generally where do you find honey bee hives in your surroundings?
Answer:
Those plants which contain nectar and pollen liked by bees are called bee flora. We find honey bee hives on the fruit trees like citrus, apple, guava, tamarind, cultivated field crops like mustered, gingelly, wheat, cotton, sunflower, vegetable plants like beans, bendi, brinjal, Trees like acacia, neem, sal and bushes, shrubs.

Question 46.
In which seasons we find honey bee hives?
Answer:
In winter and summer seasons we find honey bee hives.

Question 47.
Collection of honey from hive is a careful activity. Write a note on how people collect honey from hives. What did they do for this ?
Answer:

1. People take risk to collect honey from hives.
2. They burn some plants or wood to make smoke. They cover their body with a thick blanket.
3. They put the smoking material into a container. They take it over to the hive.
4. Working slowly and carefully, they wait for the honey bees to leave the bee hive.
5. Carefully they cut the hive with a sharp knife.
   By pressing the hive they can collect honey.

Question 48.
Ask your parents/teacher how a bear hunts bee hives for honey?
Answer:

1. Before going for hibernation (winter sleep) the bear builts fat reserves in its body and prepares itself for the winter season.
2. The bear attracted towards the bee hive and honey.
3. When it finds bee hive it drives away the bees with its forelimbs.
4. It has the advantage that its body has thick fur and avoids bee sting.
5. It chews the bee hive along with some bees in the hive swallow honeys spits out the waste.
6. This is how a bear hunts bee hives for honey.

FISHERIES:

Question 49.
Write a list of fishes that are available in your surroundings. Just write local names only.
Answer:
Marrel (Korramenu), Katla (Jella), Katrana (Bochu), Rahu (Mosu), Seer (Vanjiram) are the local varieties.

Question 50.
Do you know how to catch fish in a pond?
Answer:
Fishing rod will be helpful to catch fish.

Question 51.
How to catch fish in a large scale?
Answer:
Nylan nets will be useful to catch fish in large scale.

Question 52.
Think what will happen if mechanized fishing continuous for a long run.
Answer:
If mechanized fishing continuous at last we find no fish (extinction)

Question 53.
Ask your teacher what are the uses of Oysters?
Answer:
Oysters are the marine molluscs which produce pears. Economically pearls are important.

Question 54.
Tuna is an important fish which is available in our marine area. Collect information about Tuna and in what way it is important?
(OR)
Poultry Emu culture /Fish forms / Apiculture. Visit any one of the above industries. Get the information from farmers and prepare a note on this.
Answer:

1. Poultry are domesticated birds kept by human for the eggs and meat.
2. Generally chickens, ducks, geese, turkeys, quail are cultivated in poultry.
3. Chicken poultry’s are very common in our areas.
4. Here two types of hen’s are cultivated. Layer’s for eggs, and broiler’s for meat.
5. Poultry’s are arranged in stair manner, which save the lot of place.
6. Modern feeds for poultry consists such as soyabean oil, mineral supplements and vitamins.
7. Daily they collect the eggs and parcel in egg cases to supply.
8. Broiler chickens are domesticated and specially meant for meat production.
9. Poultry become a major economic source in rural areas.
10. It provides work and earning for rural people.

Question 55.
What is blue revolution? What are its effects? Discuss in your classroom.
Answer:
In the recent times, the Department of Agriculture and Cooperation (Blue Revolution) in the country. These schemes include the development of fresh water aquaculture through Fish Farmers Development Agencies (FFDAs) with development of brackish water aquaculture through Brackish Water Fish Farmer’s Development Agencies (BFDAs’ A shrimp and fish culture project is being implemented with World Bank assistance for the development of Shrimp culture in the states of Andhra Pradesh, Odisa and West Bengal.

8th Class Biology 9th Lesson Production and Management of Food From Animals Activities

Activity – 1

Question 1.
Form a group of 5 or 6 students. Collect different types hens and find their characters. If you want to know more details, about you need to ask hen rearers or poultry farmers in your village. Do not forget to collect information about the feed and diseases, treatment by using local technology.
Answer:

1. Production and rearing of hens on a large scale is generally called poultry. So billion hens are reared world wide for eggs and chicken.
2. Farmers rear cocks and hens in villages. Most of these are local varieties (Natukollu). Natural country varieties are good for hatching purpose, Aseel, Kadaknath, Chittagang, Longshan, Bursa are the pure local varieties.
3. Aseel (Berisa Kodi) the Indian traditional variety is meant for fighting because of its pugnacity, high stamina and majestic gait.
4. Feeding: The balanced feed must have nutrients, proteins, carbohydrates, fats, minerals, vitamins and water in sufficient quantities. Minerals like calcium, phosphorus, iodized salt, manganese and zinc are useful for better yield of eggs and meat.
5. Diseases: Poultry chicken are known to suffer from bacterial, fungal and viral diseases. Fowl cholera, salmonellasis and coryza are the common bacterial diseases. Fowl pox and Rani kher are the dreaded viral diseases.
   Fowl mite, chicken mite, fleas, licks, lice etc., are known to be present on the external surface of the poultry chicken and cause diseases and are therefore called as external parasites.
   Round worm, tape worm and coccidiosis are categorized as internal parasites.
6. Prevention and control: Prevention is always better than cure. For most of the diseases vaccines are now available. Newly born chicken should be vaccinated. A wide range of antibiotics, particularly broad spectrum sulfa drugs are widely used for treatment for poultry diseases. When a bird is identified to be suffering from a disease, it should be immediately to be isolated.

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:

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

Solution:

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 8th Class Maths Solutions 11th Lesson Algebraic Expressions InText Questions

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

8th Class Maths 11th Lesson Algebraic Expressions InText Questions and Answers

Do this

Question 1.
Find the number of terms in following algebraic expressions.
5xy², 5xy³ – 9x, 3xy + 4y – 8, 9x² + 2x + pq + q.    [Page No. 248]
Answer:

Question 2.
Take different values for x and find values of 3x + 5.     [Page No. 248]
Answer:
If x = 1 then 3x + 5 = 3(1) + 5 = 3 + 5 = 8
If x = 2 then 3x + 5 = 3(2) + 5 = 6 + 5 = 11
If x = 3 then 3x + 5 = 3(3) + 5 = 9 + 5 = 14

Question 3.
Find the like terms in the following: ax²y, 2x, 5y², -9x², -6x, 7xy, 18y².    [Pg. No. 249]
Answer:
Like terms are (2x, – 6x) (5y², 18y²).

Question 4.
Write 3 like terms for 5pq².     [Pg. No. 249]
Answer:
Like terms of 5pq² are – 3pq², pq², \(\frac{\mathrm{pq}^{2}}{2}\)etc.,

Question 5.
If A = 2y² + 3x – x², B = 3x² – y² and C = 5x² – 3xy then find          [Pg. No. 250]
(i) A + B (ii) A – B (iii) B + C (iv) B – C (v) A + B + C (vi) A + B – C
Answer:
A = 2y² + 3x – x², B = 3x² – y², C = 5x² – 3xy
i) A + B = (2y² + 3x – x²) + (3x² – y²)
= (2y² – y²) + 3x + (3x² – x²)
∴ A + B = y² + 3x + 2x² = 2x² + 3x + y²

ii) A – B = (2y² + 3x – x²) – (3x² – y²)
= 2y² + 3x – x² – 3x² + y²
∴ A – B = 3y² + 3x – 4x²

iii) B + C = (3x² – y²) + (5x² – 3xy)
= 3x² + 5x² – y² – 3xy
∴ B + C = 8x² – y² – 3xy

iv) B – C = (3x² – y²) – (5x² – 3xy)
= 3x² – y² – 5x² + 3xy
∴ B – C = – 2x² – y² + 3xy

v) A + B + C = A + (B + C)
= (2y² + 3x – x²) + (8x² – y² – 3xy)
= (8x² – x²) + (2y² – y²) + 3x – 3xy
∴ A + B + C = 7x² + y² + 3x – 3xy

vi) A + B – C = A + (B – C)
= (2y² + 3x – x²) + (-2x² – y² + 3xy)
= (2y² – y²) + (-x² – 2x²) + 3x + 3xy :
∴ A + B – C = y² – 3x² + 3x + 3xy

Question 6.
Complete the table:       [Page No. 253]

Answer:

Question 7.
Check whether you always get a monomial when two monomials are multiplied.        [Page No. 253]
Answer:
Yes, the product of two monomials is always a monomial.
Ex: 2xy × 5y = 10xy is a monomial.

Question 8.
Product of two monomials is a monomial? Check.     [Pg. No. 253]
Answer:
Yes, the product of two monomials is a monomial.
∵ 2x × y = 2xy

Question 9.
Find the product: (i) 3x(4ax + 8by) (ii) 4a²b(a – 3b) (iii) (p + Sq²) pq (iv) (m³ + n³) 5mn²       [Pg. No. 255]
Answer:
i) 3x (4ax + 8by) = 3x × 4ax + 3x × 8by
= 12ax² + 24bxy

ii) 4a²b (a – 3b) = 4a²b × a – 4a²b × 3b
= 4a³b – 12a²b²

iii) (p + 3q²) pq = p × pq + 3q² × pq
= p²q + 3pq³

iv) (m³ + n³) 5mn² = m³ × 5mn² + n³ × 5mn²
= 5m⁴n² + 5mn⁵

Question 10.
Find the number of maximum terms in the product of a monomial and a binomial?       [Pg. No. 255]
Answer:
The no.of terms in the product of a monomial and a binomial are two (2).

Question 11.
Find the product:       [Pg. No. 257]
(i) (a – b) (2a + 4b)
(ii) (3x + 2y) (3y – 4x)
(iii) (2m – l)(2l – m)
(iv) (k + 3m)(3m – k)
Answer:
i) (a – b) (2a + 4b) = a(2a + 4b) – b(2a + 4b)
= (a × 2a + a × 4b) – (b × 2a + b × 4b)
= 2a² + 4ab – (2ab + 4b²)
= 2a² + 4ab – 2ab – 4b²
= 2a² + 2ab – 4b²

ii) (3x + 2y) (3y – 4x) = 3x(3y – 4x) + 2y(3y – 4x)
= 9xy – 12x² + 6y² – 8xy
= xy – 12x² + 6y²

iii) (2m – l) (2l – m) = 2m(2l – m) – l(2l – m)
= 2m × 2l – 2m × m – l × 2l + l × m
= 4lm – 2m² – 2l² + lm
= 5lm – 2m² – 2l²

iv) (k + 3m) (3m – k) = k(3m – k) + 3m(3m – k)
= k × 3m – k × k + 3m × 3m – 3m × k
= 3km – k² + 9m² – 3km
= 9m² – k²

Question 12.
How many number of terms will be there in the product df two binomials?        [Page No. 257]
Answer:
No. of terms in the product of two binomials are 4.
Ex: (a + b) (c + d) = ac + ad + be + bd

Question 13.
Verify the following are identities by taking a, b, c as positive integers.    [Pg. No. 260]
(i) (a – b)² = a² – 2ab + b²
(ii) (a + b) (a – b) = a² – b²
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Answer:
i) (a – b)² = a² – 2ab + b²
a = 3, b = 1
⇒ (3 – 1)² = (3)² – 2 × 3 × 1 + 1²
⇒ (2)² = 9 – 6 + 1
⇒ 4 = 4
∴ (i) is an identity,

ii) (a + b) (a – b) = a² – b²
a = 2, b = 1
⇒ (2 + 1) (2 – 1) = (2)² – (1)²
⇒ 3 × 1 = 4 – 1
⇒ 3 = 3
∴ (ii) is an identity.

iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
a = 1, b = 2, c = 0
⇒ (1 + 2 + 0)² = 1² + 2² + 0² + 2 × 1 × 2 + 2 × 2 × 0 + 2 × 0 × 1
⇒ (3)² = 1 + 4 + 0 + 4 + 0 + 0
⇒ 9 = 1 + 4 + 4
⇒ 9 = 9
∴ (iii) is an identity.

Question 14.
Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab.        [Pg. No. 260]
i) What do you observe? Is LHS = RHS?
ii) Take different values for x, a and b for verification of the above identity.
iii) Is it always LHS = RHS for all values of a and b?
Answer:
i) (x + a) (x + b) = x² + (a + b)x + ab
x = 2, a = 1, b = 3 then
⇒ (2 + 1) (2 + 3) = 2² + (1 + 3)² + 1 × 3
⇒ 3 × 5 = 4 + 4x² + 3
⇒ 15 = 4 + 8 + 3 ⇒ 15 = 15
∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then
⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2
⇒ 1 × 2 = 0 + 0 + 2
⇒ 2 = 2
∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.

Question 15.
Consider (x + p) (x + q) = x + (p + q)x + pq.
(i) Put q instead of ‘p’ what do you observe?
(ii) Put p instead of ‘q’ what do you observe?
(iii) What identities you observed in your results?            [Pg. No. 261]
Answer:
i) (x + p) (x + q) = x² + (p + q)x + pq …… (1)
Substitute q instead of p in (1).
⇒ (x + q) (x + q) = x² + (q + q)x + q × q
⇒ (x + q)² = x² + 2qx + q²

ii) Substitute ‘p’ instead of q in (1).
⇒ (x + p) (x + p) = x² + (p + p)x + p × p
⇒ (x + p) = x² + 2px + p²

iii) ∴ I observe the following identities.
(x + q)² = x² + 2qx + q²
(x + p)² = x² + 2px + p²

Question 16.
Find: (i) (5m + 7n)²
(ii) (6kl + 7mn)²
(iii) (5a² + 6b²)²
(iv) 302²
(v) 807²
(vi) 704²
(vii) Verify the identity: (a – b)² = a² – 2ab + b², where a = 3m and b = 5n.         [Pg. No. 261]
Answer:
i) (5m + 7n)² is in the form of (a + b)².
(a + b)² = a² + 2ab + b² [a = 5m, b = 7n]
(5m + 7n)² = (5m)² + 2 × 5m × 7n + (7n)²
= (5m × 5m) + 70 mn + 7n × 7n
= 25m² + 70mn + 49n²

ii) (6kl + 7mn)²
We know that (a + b)² = a² + 2ab + b²
∴ (6kl + 7mn)² = (6kl)² + 2 × 6kl × 7mn + (7mn)²
= 36 k²l² + 84 klmn + 49 m²n²

iii) (5a² + 6b²)²
a = 5a², b = 6b²
(5a² + 6b²)² = (5a²)² + 2 × 5a² × 6b² + (6b²)²
= (5a² × 5a²) + 60a²b² + (6b² × 6b²)
= 25a⁴ + 60a²b² + 36b⁴

iv) (302)² = (300 + 2)²
a = 300, b = 2
∴ (300 + 2)² = (300)² + 2 × 300 × 2 + (2)²
= (300 × 300) + 1200 + (2 × 2)
= 90,000 + 1200 + 4
= 91,204

v) (807)² = (800 + 7)²
a = 800, b = 7
∴ (800 + 7)² = (800)² + 2 × 800 × 7 + (7)²
= (800 × 800) + 11,200 + (7 × 7)
= 6,40,000 + 11,200 + 49
= 6,51,249

vi) (704)² = (700 + 4)²
a = 700, b = 4
∴ (700 + 4)² = (700)² + 2 × 700 × 4 + 4²
= (700 × 700) + 5600 +(4 × 4)
= 4,90,000 + 5600 + 16
= 4,95,616

vii) (a – b)² = a² – 2ab + b² …… (1)
Substitute a = 3m, b = 5n in (1).
LHS = (3m – 5n)² = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
RHS = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
∴ LHS = RHS

Question 17.
Find:
(i)(9m – 2n)²
(ii) (6pq – 7rs)²
(iii) (5x² – 6y²)²
(iv) 292²
(v) 897²
(vi) 794²        [Pg. No. 262]
Answer:
i) (9m – 2n)² is in the form of (a – b)².
(a – b)² = a² – 2ab + b²
(9m – 2n)² = (9m)² – 2 × 9m × 2n + (2n)²
= (9m × 9m) – 36mn + (2n × 2n)
= 81m² – 36mn + 4n²

ii) (6pq – 7rs)²
a = 6pq, b = 7rs
(6pq – 7rs)² = (6pq)² – 2 × 6pq × 7rs + (7rs)²
= (6pq × 6pq) – 84pqrs + (7rs × 7rs)
= 36p²q² – 84pqrs + 49r²s²

iii) (5x² – 6y²)² = (5x²)² – 2 × 5x² × 6y² + (6y²)²
= (5x² × 5x²) – 60x²y² + (6y² × 6y²)
= 25x⁴ – 60x²y² + 36y⁴

iv) (292)² = (300 – 8)²
a = 300, b = 8
∴ (300 – 8)² = (300)² – 2 × 300 × 8 + (8)² = (300 × 300) – 4800 + (8 × 8)
= 90,000 – 4800 + 64
= 90,064 – 4800
= 85,264

v) (897)² = (900 – 3)²
= (900)² – 2 × 900 × 3 + (3)²
= 8,10,000 – 5400 + 9
= 8,10,009 – 5400
= 8,04,609

vi) (794)² = (800 – 6)²
= (800)² – 2 × 800 × 6 + (6)²
= 6,40,000 – 9600 + 36
= 6,40,036 – 9600
= 6,30,436

Question 18.
Find:
(i) (6m + 7n) (6m – 7n)
(ii) (5a + 10b) (5a – 10b)
(iii) (3x² + 4y²) (3x² – 4y²)
(iv) 106 × 94
(v) 592 × 608
(vi) 92² – 8²
(vi) 984² – 16²      [Pg. No. 262]
Answer:
i) (6m + 7n) (6m -,7n) is in the form of (a + b) (a – b). (a + b) (a – b) = a² – b²,
here a = 6m, b = 7n
(6m + 7n) (6m – 7n) = (6m)² – (7n)²
= 6m × 6m – 7n × 7n
= 36m² – 49n²

ii) (5a + 10b) (5a – 10b) = (5a)² – (10b)² [∵ (a + b) (a – b) = a² – b²]
= 5a × 5a – 10b × 10b
= 25a² – 100b²

iii) (3x² + 4y²) (3x² – 4y²)
= (3x²)² – (4y²)²
= 3x² × 3x² – 4y² × 4y²
= 9x⁴-16y⁴ [∵ (a + b) (a – b) = a² – b²]

iv) 106 × 94 = (100 + 6) (100 – 6)
= 100² – 6² = 100 × 100 – 6 × 6 [∵ (a + b) (a- b) = a²– b²]
= 10,000 – 36
= 9,964

v) 592 × 608 = (600 – 8) (600 + 8)
= (600)² – (8)²
= 600 × 600 – 8 × 8
= 3,60,000 – 64
= 3,59,936

vi) 92² – 8² is in the form of a² – b² = (a + b) (a – b).
92² – 8² = (92 + 8)(92 – 8)
= 100 × 84
= 8400

vii) 9842 – 162 = (984 + 16) (984 – 16)
= (1000) (968) [∵ (a + b)(a – b) = a² – b²]
= 9,68,000

Try These

Question 1.
Write an algebraic expression using speed and time; simple interest to be paid, using principal and the rate of simple interest.    [Pg. No. 251]
Answer:
Distance = speed × time
d = s × t

Question 2.
Can you think of two more such situations, where we can express in algebraic expressions?     [Pg. No. 251]
Answer:
Algebraic expressions are used in the following situations:
i) Area of a triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) bh
ii) Perimeter of a rectangle = 2(length + breadth) = 2(l + b)

Think, Discuss and Write

Question 1.
Sheela says the sum of 2pq and 4pq is 8p²q² is she right? Give your explanation.      [Pg. No. 249]
Answer:
The sum of 2pq and 4pq = 2pq + 4pq = 6pq
According to Sheela’s solution it is 8p²q².
6pq ≠ 8p²q²
Sheela’s solution is wrong.

Question 2.
Rehman added 4x and 7y and got 1 lxy. Do you agree with Rehman?     [Pg. No. 249]
Answer:
The sum of 4x and 7y
= (4x) + (7y)
= 4x + 7y ≠ 11xy
I do not agree with Rehman’s solution.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.2

Question 1.
Complete the table:

Solution:

Question 2.
Simplify: 4y(3y + 4)
Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y² + 6y

Question 3.
Simplify x(2x² – 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0
Solution:
x(2x² – 7x + 3)
= x × 2x² – x × 7x + x × 3
= 2x³ – 7x² + 3x
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
=9 – 3 – 3 + 1 = 4
∴(a – b)² = 4sq.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 4.
Add the product: a(a – b), b(b – c), c(c – a)
Solution:
a(a – b) + b(b – c) + c(c – a)
=a × a – a × b + b × b – b × c + c × c – c × a
=a² – ab + b² – bc + c² – ca
=a² + b² + c² – ab – bc – ca

Question 5.
Add the product: x(x + y – r), y(x – y+r), z(x – y – z)
Solution:
x(x + y – r) +y(x – y + r) + z(x – y – z)
= x² + xy – xr + xy – y² + yr + zx – yz – z²
= x² – y² – z² + 2xy – xr + yr + zx – yz

Question 6.
Subtract the product of 2x(5x – y) from product of 3x(x+2y)
Solution:
3x(x + 2y) – 2x(5x – y)
=(3x × x + 3x × 2y)-(2x × 5x – 2x × y)
= 3x² + 6xy – (10x² – 2xy)
= 3x² + 6xy- 10x² + 2xy
= 8xy – 7x²

Question 7.
Subtract 3k(5k – l + 3rn) from 6k(2k + 3l – 2rn)
Solution:
6k(2k + 3l – 2m) – 3k(5k – l + 3m)
= 12k²+ 18kl – 12km – 15k² + 3kl – 9km
= -3k² + 21kl – 21km

Question 8.
Simplify: a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
Solution:
a²(a – b + c) + b²(a + b – c) – c²(a – b – c)
= a³ – a²b + a²c + ab² + b³ – b²c – ac² + bc² + c³
= a³ + b³ + c³ – a²b + a²c + ab² – b²c – ac² – bc²