Class 8

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 4 Reproduction in Animals Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 4th Lesson Reproduction in Animals

8th Class Biology 4th Lesson Reproduction in Animals Textbook Questions and Answers

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Question 1.
Differentiate between:
a) Sexual Reproduction and Asexual Reproduction
b) Gametes and Zygote
c) External fertilization and Internal fertilization
d) Viviparous and Oviparous animals
Answer:
a) Sexual Reproduction and Asexual Reproduction:

b) Gametes and Zygote:

1. Millions of male gametes (sperms) are produced by the testes. These are microscopic and single celled. Sperm has a head, a middle piece and a tail.
   Ovary produces female gaffietes called ova. It is a single cell, (haploid)
2. The fusion of male and female garnet is called fertilization. The result of fertilization is the formation of a zygote. Zygote is a diploid cell. This develops mitorically and forms into an embryo, which further develops into a baby.

c) External fertilization and Internal fertilization:
The process of fertilization that occurs outside of an organism is called External fertilization.
E.g. Frog, Fish, Star fish, etc.
The process of fertilization that takes place inside the body of females is called Internal fertilization. E.g. Animals, Human beings.

d) Viviparous and Oviparous animals:
Animals which give birth to their offsprings are called Viviparous animals.
E.g. Animals, human beings.
Animals which lay eggs are called Oviparous animals.
E.g. Hen, duck, pigeon, etc.

Question 2.
Compare the reproduction in Hydra and Amoeba. Note down the differences in your notebook.
Answer:
Comparison:
Asexual Reproduction takes place in Hydra and Amoeba.

Question 3.
Why do fish and frog lay more number of eggs whereas cow and human beings usually give birth to only one at a time?
Answer:

1. Fish and frog lay many eggs to increase chance of survival of the offspring and the continuation or their generation.
2. They do not take care of their young ones making them prone to predators and may even be washed away by the water force.
3. Thus the more eggs produced, the greater the chances that some will grow to maturation.
4. Female frog and fish release their eggs in the water and male animals release their sperms in the water. As fertilization takes place in the water it is external ferlitization. There is no safety for the fertilized eggs in the water so these animals lay more number of eggs.
5. Whereas cow and human beings usually give birth to only one at a time and the internal fertilization takes place in these animals. There is safety for the embryo (the offspring) in the mother’s womb until it’s birth.

Question 4.
Can animals produce offsprings even without formation of zygotes, how? Explain with suitables example.
Answer:

1. Besides Asexual and Sexual Reprodution, there is other mode of reproduction called cloning.
2. Cloning is the production of an exact copy of a cell, any other living part, or a complete organism.
3. Cloning of an animal was successfully performed for the first time by Ian Wilmut and his colleagues at the Roslin Institute in Edinburgh, Scotland.
4. They successfully cloned a sheep named Dolly. Dolly was born on 5th July 1996 and was the first mammal to be cloned.
   
5. During the process of cloning dolly, a cell was collected from the mammary gland of a female Finn Dorset Sheep.
6. Simultaneously, an egg was obtained from Scottish black face ewe.
7. The nucleus was removed from the egg. Then the nucleus of the mammary gland cell from the Finn Dorset sheep was inserted into the egg of the Scottish black face ewe whose nucleus had been removed.
8. Thus the egg produced was implanted into the Scottish black face ewe. Development of this egg followed normally and finally Dolly was born.
9. Though Dolly was given birth by the Scottish black face ewe, it was found to be absolutely identified to the Finn Dorset sheep, from which the nucleus was taken.
   
10. Since the nucleus from the egg of the Scottish black face ewe was removed, Dolly did not show any character of the Scottish black face ewe.
    Dolly was a healthy clone of the Finn Dorset sheep and produced several off-springs of her own through normal sexual means.

Question 5.
How can you identify the animal is viviparous or oviparous?
Answer:

1. Animals giving birth to young ones have epidermal hair on their skin and external ears. These animals are called viviparous animals.
   E.g. Animals, human beings, etc.
2. The animals that lay eggs do not have epidermal hair or external ears. These animals are called oviparous animals.
   E.g. Hen, Duck, Pigeon, Parrot, etc.

Question 6.
Who am I?
a) I am formed by the fusion of male and female gametes.
Answer:
Zygote: Zygote is formed by the fusion of male and female gametes. This process is
called fertilization.

b) I am a gamete that has a tail and travel to fuse with female gametes.
Answer:
Male garnets or sperm or spermatozoa:
The structure of sperm has a head, a middle piece and a tail.

c) I am a fully developed embryo inside a mother’s body.
Answer:
Offspring or baby:
The zygote divides repeatedly to give rise a ball of cells. The cells then begin to form groups that develop into tissues and organs in the body. This developing structure is termed as an Embryo.
The embryo gets embedded in the wall of the uterus for further development. It develops in the uterus. It gradually develops body parts such as hands, legs, head, eyes, ears etc. From 3 months (12 weeks) of pregnancy the embryo is called FOETUS – After the completion of this period (about 270 – 280 days) a baby (offspring) is born. This is called gestation period.

Question 7.
State the reason why most of the terrestrial animals’, fertilization takes place internally.
Answer:

1. In animals like insects, reptiles, birds and mammals, the male animals deposit the sperms inside the body of the female animals, where fertilization occurs. This is called Internal Fertilization. This is most common in terrestrial animals.
2. In majority of the animals, sexes are separate and male and female animals are distinct. This is called sexual dimorphism and animals are said to be unisexual.
3. The external features by which the males and females can be distinguished are called Secondary sexual characters.
4. There are some animals in which male and female sex organs are present in the same animal. This is called Hermaphroditism and such animals are called Hermaphrodites or Bisexual.
5. Hermaphroditism is seen in some of the members of protozoa, Coelenterata, Platy- helminthes, Nematoda, Annelida and mollusca. In these animals sperm and ova are formed in the same animal. However self fertilization is prevented by several methods.

Question 8.
Observe the following figures and write the functions of them.

Answer:
a) Testes:
Testes are the male reproductive organs and produce male gametes known as sperms or spermatozoa. Testes are egg shaped. It is connected with a pair of seminal ducts through which sperms travel and ejaculate out with the help of penis.

b) Female Reproductive system. Oviduct or fallopian tube connected with ovary. Female reproductive system contains a pair ovaries, oviducts and also called fallopian tubes and uterus.
The ovary produces female gametes called ova or eggs. In human beings, a single matured egg is released into the oviduct by one of the ovaries every month.

The ovum which is a single cell released from ovary and enters into a tube called Fallopian Tube. The end of the tube is like a funnel with several finger like structures and is also ciliated. The movement of cilia help the movement of ovum through the fallopian tube into uterus.
c) Sperm:
Human sperms are minute, microscopic and motile they have a oval head, a neck, a middle piece and a long tail.
Head consists of a large haploid nucleus. Acrosome is present in the head, which helps in fertilization.

The neck is short middle piece has several mitochondria which produce energy required for the movements of sperms. Tail piece helps in the swimming of sperm to reach the ovum during fertilization.
d) Fusion of ovum and sperm:
(Fertilization) Fertilization is of internal type. Sperm reaches the ovum in the fallopian tube. Sperm nucleus enters the ovum which is haploid.
When sperm enters the ovum the membranes of ovum becomes thicken. So that another sperm can not penetrate the ovum. This prevents double fertilization of the ovum. During fertilization, the sperm and the ovum fuse to form a zygote.

This type of internal fertilization occurs in different organisms like insects, snakes, lizards, birds and mammals etc.

Question 9.
a) By taking help of given words label the following life cycle, (eggs, adult, pupa, larva)

Answer:

b) Explain the process of metamorphosis in housefly by taking help from in the given diagram.

Answer:
Metamorphosis in the house fly: A female housefly at a time lays about 120 to 160 eggs. The eggs are laid in garbage, on dung heaps, or on a decaying animal and vegetable matter. The life history consists of 1) egg 2) larva 3) pupa and 4) adult stages. Egg: The egg is white and cylindrical on one side. It has two ribbon like longitudinal thickenings. They hatch in about 24 hours into larva.
Larva: The larva is known as a Maggot. It is white in colour. The baby of the larva has 13 segments. It has a mouth and feeds on organic matter.
Pupa: The fully grown larva moves to a dry place in the dung and changes to a pupa. The pupa is dark brown and barrel shaped. In a week, the pupa changes into an adult or imago.
Houseflies spread, germs that cause diseases like typhoid, cholera, amoebic dysentery, tuberculosis. Our food should be kept covered from houseflies. The surroundings should be clean without garbage and dung heaps. Insecticides can be used to kill the house flies.

Question 10.
Match the following.
Group – A                                                Group – B
1. Oviparous                    (  )          A) Tadpole to adult
2. Metamorphosis           (  )          B) Birds
3. Embryo                        (  )          C) Fertilisation outside the body
4. External fertilization     (  )          D) Developed Zygote
Answer:
1) B
2) A
3) D
4) C

Question 11.
What would happen if all the organisms stop the process of reproduction?
Answer:

1. Without reproduction living organisms would not survive long.
2. Different species of living organisms die due to various reasons.
   E.g. Old age, diseases, accidents, etc.
3. Imagine the death of members of a species continues and new individuals of that species are not added.
4. A stage will come when that species will disappear.
5. To ensure the continuity of the species reproduction is important.
6. Depending on the available conditions in a community, different species will reproduce continuously and increase their numbers.
7. Reproduction in a species will therefore:
   a) Replace those species that die and
   b) Allow an increase in total numbers of the species under suitable conditions.

Question 12.
Kavita found a tadpole in a pond. She collected it carefully and put it in an aquarium supposing it as a fish. After some days what did she find and why?
Answer:

1. The larva that emerges from the eggs, known as tadpole, have oval bodies and long, vertically flattened tail and are fully aquatic.
2. Tadpole lack eyelids and have cartilaginous skeleton. They take respiration through external gills, later it develops internal gills. It looks like a fish at this stage completely. The vertically flattened tails use for swimming.
3. Tadpole lack true teeth, but the jaws two elongated parallel rows of small structures called keradonts in their upper jaw and three rows of keradonts in the lower jaw as same as the fish has.
4. Tadpole are typically herbivorous, feeding mostly on algae, including diatoms, filtered the water through the gills.
5. The tadpoles may be as short as a week during metamorphosis.

Question 13.
Collect information from your library or from other sources like internet and discuss the life cycle of Honeybees in the symposium at your school.
Answer:
In the life cycle of butterfly there are four stages.

1. Egg
2. Larva
3. Pupa
4. Adult.

The cycle of changes that takes place from egg to adult is called metamorphosis.

1) Egg: The egg is the first stage in the butterfly. They are very small and round. The female butterfly lays eggs on or near the plants.
2) Larva: The larva hatches from the egg. Butterfly larva are usually called Caterpillar. Caterpillars spend most of their time eating. Butterfly do all their growing when they are caterpillars, and food gives them the energy and body building materials they need. A caterpillar’s exoskeleton can’t stretch or grow, so the caterpillar sheds its skin or molts, several times as it grows.
3) Pupa: When the caterpillar has finished growing, it forms from the outside, the pupa looks as if it’s resting. But inside, every part of the caterpillar is changing. Most of it’s organs and other body parts like head, thorax and abdomen, 3 pairs of legs, 2 pairs of wings, a pair of compound eyes, the antennae, a proboscis etc. are formed. Butterfly pupa are called chrysalises.
4) Adult: When the pupa has finished changing, it molts one last time and emerges as an adult butterfly. The adult emerges with its wings folded up against its body. The adult is the stage when butterfly mate the reproduce. Females lay their eggs on plants or other surfaces and the cycle starts all over again.

Question 14.
Sketch the diagrams of male and female reproductory systems.
Answer:

Question 15.
Draw labelled diagram of life history of frog and identify forms are herbivores.
Answer:

Parts:

1. Egg
2. Embryo before hatching
3. Hatched tadpole
4. Tadpole attached to water plant
5. Tadpole with external gills
6. Developing tadpole
7. Tadpole with fore and hind limbs
8. Tadpole changing into frog
9. Frog

Question 16.
How would you appreciate Ritwik’s work when he kept back the pigeon squab in the ventilator? If you were in Ritwik’s place what would you do?
Answer:

1. If i were in Ritwik’s place, I would like to show kindness towards the pigeon squab by keeping back the pigeon squab in the ventilator. He took great care towards it.
2. Research their needs and do what makes them happiest.
3. Check that we are not inadvertently supporting animal cruelty, which are in our surroundings.
4. Leave room for wild life habitates in the own yard by providing birds with feeders and bird bath.
5. Create a clean environment for the birds and animals.
6. Cut the usage of plastic so they can not be danger to wild life.
7. Appreciate wild life and learn more about it but do not approach them or attempt to resque them.
8. Never tolerate birds or animal cruelty. Report suspected cruelty to the authorities.
9. In still compasion in the children by demonstrating kindness and using positive training methods for the pets.
10. Keep them vaccinations’ current and visit veterinarian regularly.

Project work

Note: This project work needs patience and carefulness. Teachers should be cautious while doing this project. Care should be taken at the time of collection of eggs of frogs. From a nearby pond or slow flowing streams. If eggs are not available, you need not to worry. You can start your project after collecting Tadpoles.
To conduct this project you require:

1. Wide mouth transparent bottle / tub
2. Transparent glass
3. Dropper
4. Petridish
5. Some pebbles
6. Magnifying lens

Answer:
Step -1: Go to a nearby pond or a slow flowing stream where usually sewage stagnates during rainy season. Collect few eggs of a frog with the help of wide mouthed bottle as shown in the figure. While collecting eggs, take care that the clusters of eggs are not disturbed and isolated.

Step – 2: After collecting eggs, take a tub of 15 cm depth and a radius of 8-10 cms. Transfer the eggs along with the weeds and algae that you have collected from the pond into the tub. Carefully observe the eggs. You will find a blackish part in the middle of the eggs. That is the embryo of the frog.

Step – 3: Observe the tub daily and note down the changes in your observation book. Draw diagrams after observing for atleast once in three days.

CHANGES TAKES PLACE FROM EGG TO ADULT IN FROG

Step – 4:

Step – 5:
Try to answer these questions after your observation.
1. How many days did it take for the eggs to hatch?
Answer:
It takes 10 to 15 days for the eggs to hatch.

2. How does the tadpole look like?
Answer:
The tadpole looks like a fish.

3. When did you find gill slits in a tadpole?
Answer:
19 to 22 days.

4. On which dates did you observe?
Answer:
Heart: 28th to 30th dates.
Intestine: 31st to 3rd (31 to 33 days)
Bones: 4th to 6th (34 to 36 days)
Rectum: 31st to 3rd (31 to 33 days)
Hindlimbs: 4th to 6th (34 to 36 days)
Forelimbs: 7th to 9th (37 to 39 days)

Step – 6:

1. When did gill slits disappear?
Answer:
37 to 39 days.

2. When did the tail completely disappear?
Answer:
42 to 44 days.

3. How many days did it take for a tadpole to transform into an adult frog?
Answer:
It takes 45 to 46 days to take for a tadpole to transform into an adult frog.

8th Class Biology 4th Lesson Reproduction in Animals InText Questions and Answers

Question 1.
Do all eggs hatch into nestlings?
Answer:
Yes, all eggs hatch into nestlings.

Question 2.
Can there be pigeons if there were no eggs?
Answer:
If there were no eggs there can not be no pigeons.

Question 3.
Can there be eggs if there were no pigeons?
Answer:
If there were no pigeons, there cannot be no eggs.

Question 4.
Do all animals lay eggs?
Answer:
All animals do not lay eggs.

Question 5.
Are there any animals that give birth to young ones?
Answer:
Animals like cat, dog etc., give birth to their young ones.

Question 6.
How can we identify which animals lay eggs and which give birth to young ones?
Answer:
Animals that lay eggs do not have epidermal hair or external ears.
E.g. Crow, Pigeon, Parrot etc.
Animals giving birth to young ones have epidermal hair on their skin and external ears. E.g. Cow, Buffalo, Dog, Cat etc.

Question 7.
Are there any patterns in nature that give clues to modes of reproduction?
Answer:
There are two types of reproduction.

1. Asexual reproduction and
2. Sexual reproduction.

Question 8.
Names of some animals are listed below. Observe carefully and fill the table. Deer, Leopard, Pig, Fish, Buffalo, Giraffe, Frog, Sparrow, Lizard, Crow, Snake, Elephant, Cat.
Answer:

You can also add some more names of animals you know to this table.

Question 9.
Think how animals could hear without external ears?
Answer:
Though the animals do not have ears to hear, they can sense the surrounding by its body.

Question 10.
Read the names of animals given below and try to fill the table given below.
Cow, Rat, Crow, Pig, Fox, Hen, Camel, Duck, Frog, Elephant, Buffalo, Pigeon, Cat, Peacock, Lizard. You can also add a few more animals to this list.
Answer:

8th Class Biology 4th Lesson Reproduction in Animals Activities

Activity – 1

Question 1.
Draw the diagram of Hydra. Compare it with the figure below recall what you have observed in the first slide ?

Compare slide 1 & 2 to observe which part of it’s body develops a swelling?
Answer:
The body wall develops swelling. Observe all the remaining slides.
a) What have you observed in slide/picture 1, 2 and 3?
Answer:
Picture 1, 2 the body surface of hydra has smooth surface.
Picture 3, a swelling is formed on it’s body surface.

b) What is the main difference between slide 1 and 2 as well as 3 and 4?
Answer:
Slide 1 and 2 Hydra body is smooth.
Slide. 3, a swelling is formed. Swelling increases in size, tentacles are formed which is called bud.
Slide 4, the bud is cut off and separated from parent Hydra and can live individually.

c) What does swelling (bulge) develop into?
Answer:
The swelling develops into a bud.

Activity – 2

Question 2.
Observe the given diagram carefully and fill the following table:

i)

ii) How many amoebae are formed at the end ?
Answer:
Two amoebae are formed at the end.
Male flower – its parts:

1. Calyx (sepals)
2. Corolla (Petals)
3. Androecium (stamen)(male part)
4. Pollen grain (male gametes)

Female Flower – Its parts.

1. Calyx (Sepals)
2. Corolla (Petals)
3. Gynoecium (Ovary) female part
4. Ovules (Future seeds)

1) What would happen if fusion of sperm and ova doesn’t takes place?
Answer:
If fusion of sperm and ova doesn’t takes place fertilization would not happen.

2) Why animals give birth to their babies?
Answer:
To continue their species on the earth.

3) What happens if each couple give birth to more than two babies?
Answer:
The population increases.

4) Is it necessary to control population?
Answer:
The rapidly increasing population is posing a number of problems as our resources in nature do not increase proportionately. On the other hand they diminish. So it is necessary to control population.

Activity – 3

Question 3.
Observation of resemblance in Parents & Children.
Table given below will help you to note the similar and dissimilar. Fill in the table.

You can ask your teacher and know why sometimes no characters match with your father or mother.
Answer:

1. The ability of an organism to produce a new generation of individuals of the same species is called reproduction.
2. That means the characteristics of parental organisms are being transferred to their next generation in the process of reproduction.
3. It involves the transmission of genetic material (chromosomes) from the parental generation to the next generation.
4. In some methods of reproduction the genetic material of the parent and the offspring next generation will be exactly same.
5. Whereas in some methods the characters from two parents (male and female) recombine to form a new individual.
6. In this process some characters of one parent and remaining characters from the other parent are seen in the offsprings. Some characters will be new which are not seen in either of the parents.
7. This happens because of chromosome recombination. The process of reproduction ensures continuation of race and the perpetuation of characteristics of the species and particularly the parental organisms.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.1

Question 1.
Find the product of the following pairs:
(i) 6, 7k
(ii) – 31, – 2m
(iii) -5t² – 3t²
(iv) 6n, 3m
(v) – 5p², – 2p
Solution:
The product of 6, 7k = 6 × 7k = 42k
ii) The product of – 3l, – 2m = (- 3l) × (- 2m) = 6/m
iii) The product of – 5t², – 3t² = (- 5t²) × (- 3t²) = 15t⁴
iv) The product of 6n, 3m = 6n × 3m = 18mn
v) The product of – 5p², – 2p = (- 5p²) × (- 2p) = 10p³

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials
(i) xy, x²y , xy, x
(ii) a, b, ab, a³ b, ab³
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) – 3a, 4ab, – 6c, d
Solution:
i) The product of xy, x²y, xy, x = xy × x²y × xy × x
= x⁵ × y³= x⁵y³

ii) The product of a, b, ab, a³b, ab³ = a × b × ab × a³b × ab³
= a⁶ × b⁶ = a⁶ b⁶

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k³ × l³ × m³ =k³l³m³

iv) The product of pq, pqr, r = pq × pqr × r
= p² × q² × r² – p²q²r²

v) The product of – 3a, 4ab, – 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a² × b × c × d
= 72a²bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=
Solution:
ABC = xy × yz × zx = x²y²z²

Question 6.
If P = 4x², T = 5x and R = 5y, then \(\frac{\mathrm{PTR}}{100}\) =
Solution:
\(\frac{P^{\prime} \Gamma R}{100}=\frac{4 x^{2} \times 5 x \times 5 y}{100}=\frac{100 x^{3} y}{100}\) = x³ y

Question 7.
Write some monomials of your own and find their products.
Solution:
The product of,some monomials is given below :
i) abc × a²bc = a³b²c²
ii) xy × x²z × yz² = x³y²z³
iii) p × q × r = p³q³r³

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers InText Questions and Answers.

8th Class Maths 15th Lesson Playing with Numbers InText Questions and Answers

Do this

Question 1.
Write the place value of numbers underlined.     (Pg. No: 312)
i) 29879   ii) 10344   iii) 98725
Answer:
i) 29879
Place value of 8 = 8 × 100 = 800
Place value of 2 = 2 × 10,000 = 20,000
ii) 10344
Place value of 4 = 4 × 1 = 4
Place value of 3 = 3 × 100 = 300
iii) 98725
Place value of 5 = 5 × 1 = 5
Place value of 8 = 8 × 1000 = 8,000

Question 2.
Write the following numbers in expanded form,        (Pg. No: 313)
i) 65    ii)    74    iii) 153    iv) 612
Answer:
Number Expanded form
i) 65 = 60 + 5 = (6 × 10¹) + (5 × 10⁰)
ii) 74 = 70 + 4 = (7 × 10¹) + (4 × 10⁰)
iii) 153 = 100 + 50 + 3 = (1 × 10²) + (5 × 10¹) + (3 × 10⁰)
iv) 612 = 600 + 10 + 2 = (6 × 10²) + (1 × 10¹) + (2 × 10⁰)

Question 3.
Write the following in standard notation.       (Pg. No: 313)
i) 10 × 9 + 4     ii) 100 × 7 + 10 × 4 + 3
Answer:
Expanded form General form
i) 10 × 9 + 4 = 90 + 4 = 94
ii) 100 × 7 + 10 × 4 + 3 = 700 + 40 + 3 = 743

Question 4.
Fill in the blanks.       (Pg. No: 313)
Answer:
i) 100 × 3 + 10 × ——— + 7 = 357 (5)
ii) 100 × 4 + 10 × 5 + 1 = ——— (451)
iii) 100 × ——— + 10 × 3 + 7 = 737 (7)
iv) 100 × ——— + 10 × q + r = \(\overline{\mathrm{pqr}}\) (p)
v) 100 × x + 10 × y + z = ——— (\(\overline{\mathrm{xyz}}\))
Do you know?

Question 5.
The number 8281807978777675747372717069686766656463626160595857565554535251504948474645444342414039383736353433323130292827262524232221201918 1716151413121110987654321 is written by starting at 82 and writing backwards to 1 and see that it is a prime number.        (Pg. No: 313)
Answer:
No.of digits in the given number are 155.

Question 6.
Write all the factors of the following numbers.       (Pg. No: 314)
a) 24    b) 15   c) 21   d) 27   e) 12   f) 20   g) 18   h) 23   i) 36
Answer:
a) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
b) Factors of 15 = 1, 3, 5, 15
c) Factors of 21 = 1, 3, 7, 21
d) Factors of 27 = 1, 3, 9, 27
e) Factors of 12 = 1, 2, 3, 4, 6, 12
f) Factors of 20 = 1, 2, 4, 5, 10, 20
g) Factors of 18 = 1, 2, 3, 6, 9, 18
h) Factors of 23 = 1, 23
i) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

Question 7.
Write first five multiples of given numbers     (Pg. No: 314)
a) 5   b) 8   c) 9
Answer:
a) Multiples of 5 = 5, 10, 15, 20, 25
b) Multiples of 8 = 8, 16, 24, 32, 40
c) Multiples of 9 = 9, 18, 27, 36, 45

Question 8.
Factorize the following numbers into prime factors.    (Pg. No: 314)
a) 72    b) 158   c) 243
Answer:
a) 72 = 2 × 2 × 2 × 3 × 3
b) 158 = 2 × 7 × 9
c) 243 = 7 × 7 × 7

Question 9.
Check whether the following given numbers are divisible by 10 or not.   (Pg. No: 315)
a) 3860   b) 234   c) 1200   d) 10³   e) 10 + 280 + 20
Answer:
a) 3860, c) 1200, d) 103, e) 10 + 280 + 20 are divisible by ’10’.
[∵ the units digit of above numbers is ‘0’]
b) 234, is not divisible by 10.
[∵ its unit digit is 4]

Question 10.
Check whether the given numbers are divisible by 10 or not.    (Pg. No: 315)
a) 10¹⁰   b) 2¹⁰   c) 10³ + 10¹
Answer:
a) 10¹⁰ = 10000000000
b) 2¹⁰ = 1024
c) 10³ + 10¹ = 1000 + 10 = 1010
∴ a) 10¹⁰, c) 10³ + 10¹ are divisible by ’10’.
[∵ Their units digits are ‘0’.]
b) 1024 is not divisible by 10.
[∵ Its units digit is 4.]

Question 11.
Check whether the given numbers are divisible by 5 or not      (Pg. No: 315)
a) 205   b) 4560    c) 402    d) 105    e) 235785
Answer:
a) 205, b) 4560, d) 105, e) 235785 are divisible by 5.
[∵ The units digit of the above numbers are either 0 (or) 5.]
c) 402 is not divisible by 5.
[∵ Its units digit is 2.]

Question 12.
Check whether the given numbers which are divisible by 3 or 9 or by both,      (Pg. No: 318)
a) 3663    b) 186    c) 342    d) 18871    e) 120    f) 3789    g) 4542    h) 5779782
Answer:

Question 13.
Check whether the given numbers are divisible by 6 or not.      (Pg. No: 320)
a) 1632    b) 456     c) 1008     d) 789     e) 369    f) 258
Answer:

Question 14.
Check whether the given numbers are divisible by 6 or not.     (Pg. No: 320)
a) 458 + 676    b) 6³    c) 6² + 6³    d) 2² × 3²
Answer:

Question 15.
Can you arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in an order so that the number formed by first two digits is divisible by 2, the number formed by first three digits is divisible by 3, the number formed by first four digits is divisible by 4 and so on upto nine digits?
Solution: The order 123654987 looks promising check and verify.     (Pg. No : 320)
Answer:

∴ This number can’t continue upto ‘9’.
→ 123654987

∴ The given number 123654987 is not divisible by all the numbers like 2, 3, 4, 5,……… 9.

Question 16.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.
a) 464    b) 782     c) 3688    d) 100     e) 1000    f) 387856    g)4⁴     h) 8³ (Pg. No: 321)
Answer:
If a number is divisible by 4 then the last two digits of the number must be divisible by 4.
If the last 3 digits of a number is divisible by 8 then it is divisible by 8.

Question 17.
Check whether the given numbers are divisible by 7. (Pg. No: 322)
a) 322     b) 588     c) 952     d) 553    e) 448
Answer:

All the given numbers are divisible by ‘7’.

Question 18.
Check whether the given numbers are divisible by 11.    (Pg. No: 323)
i) 4867216      ii) 12221     iii) 100001
Answer:
If the difference between the sum of digits of odd places and even places is divisible by 11, then entire number is divisible by 11.

Question 19.
Take different pairs of numbers and check the above four rules.      (Pg. No: 325)
Answer:
a) Consider a factor of 36, say 9.
Factors of 9 are 1,3,9.
∴ 36 is divisible by 1, 3, 9.
∴ 36 is also divisible by all the factors of 9.
b) Let us consider a number 60. It is divisible by 5 and 6. It is also divisible by 5 x 6 = 30 Where 5, 6 are co-primes.
c) Take two numbers 25, 30. These numbers are both divisible by 5.
The number 25 + 30 = 55 is also divisible by 5.
d) Take two numbers 36, 54. These numbers are both divisible by 9.
Their difference i.e., 54 – 36 = 18 is also divisible by 9.

Question 20.
144 is divisible by 12. Is it divisible by the factors of 12? Verify.     (Pg. No : 325)
Answer:
Factors of 12 = 1, 2, 3, 4, 6, 12.
If 12 is a factor of 144 then 144 is divisible by all the factors of 12.

Question 21.
Check whether 2³ + 2⁴ + 2⁵ is divisible by 2. Explain.       (Pg. No : 325)
Answer:
2³ + 2⁴ + 2⁵ = 8 + 16 + 32 = 56 is an even.
∴ 56 is divisible by 2.

Question 22.
Check whether 33 – 32 is divisible by 3. Explain    (Pg. No : 325)
Answer:
33 – 32 = 27 – 9 = 18 → 1 + 8 = 9 ⇒ \(\frac{9}{3}\) (R = 0)
∴ It is divisible by ‘3’.

Question 23.
Check the result if the numbers chosen were       (Pg. No : 328)
i) 37    ii) 60    iii) 18   iv) 89
Answer:
i) If the digits are interchanged in 37 then it becomes as 73.
∴ 37 + 73 = 110 → \(\frac{110}{11}\) (R = 0)
It is divisible by ’11’.

Question 24.
In a cricket team there are 11 players. The selection board purchased 10x + y T-shirts to players. They again purchased ‘10y + x’ T-shirts and total T-shirts were distributed to players equally. How many T-shirts will be left over after they distributed equally to 11 players ? How many each one will get?     (Pg. No : 328)
Answer:
No.of players in the team = 11
No.of T- shirts are purchased at first = 10x + y
No. of T – shirts are purchased for the 2nd time = 10y + x
Sum of the T – shirts = (10x + y) + (10y + x)
= 11x + 11y = ll(x + y)
∴ 11(x + y) T – shirts are distributed among 11 players then each will get ll(x + y)
\(\frac{11(x + y)}{11}\) = x + y
Remaining T – shirts = Purchased T – shirts – 11 (No.of T-shirts got by each)
= 11(x + y) – 11(x + y)
= 0

Question 25.
In a basket there are ‘10a + b’ fruits (a ≠ 0 and a > b). Among them ‘10b + a’ fruits are rotten. The remaining fruits distributed to 9 persons equally. How many fruits are left over after equal distribution? How many fruits would each child get?      (Pg. No: 328)
Answer:
No. of fruits in a basket = 10a + b
No. of fruits are rotten = 10b + a
Remaining fruits to be distributed = (10a + b) – (10b + a)
= 10a + b – 10b – a
= 9a – 9b = 9(a – b)
∴ 9(a – b) fruits are distributed among ‘9’ Children
then each will get = 9(a – b) ÷ 9 = \(\frac{9(a – b)}{9}\) = (a – b)
No. of fruits left over after distribution
= Total no. of fruits distributed – No.of fruits got by each
= 9(a – b) – 9(a – b) = 0

Question 26.
Check in the above activity with the following numbers.      (Pg. No: 329)
i) 657    ii) 473     iii) 167    iv) 135
Answer:

Question 27.
If 21358AB is divisible by 99, find die values of A and B.     (Pg. No: 331)
Answer:
If 21358AB is divisible by 99, then it is divisible by 9 and 11.
If 21358AB is divisible by 9 then the sum of the digits is divisible by 9.
2 + 1 + 3 + 5 + 8 + A + B = 9 × 3 say
⇒ 19 + A + B = 27
⇒ A + B = 27-19 = 8
A + B = 8 …… (1)
If 21358AB is divisible by ‘ll’ then the difference of sum of even and odd digits will be divisible by’ll’.
2 1 3 5 8 A B
∴ (2 + 3 + 8 + B) – (1 + 5 + A) = 11 × 1 say
⇒ 13 + B – 6 – A = 11
⇒ B – A = 11 – 7 = 4 ……. (2)
From (1) & (2) A = 2, B = 6
∴ The required number is 21358AB = 2135826 which is divisible by 99.

Question 28.
Find the values of A and B of file number 4AB8 (A, B are digits) which is divisible by 2, 3, 4, 6, 8 and 9.       (Pg. No: 331)
Answer:
Given number is 4AB8.
4AB8 → \(\frac{8}{2}\) (R = 0) so, it is divisible by ‘2’.
4AB8 → If it is divisible by ‘3’, sum of all the digits should be a multiple of ‘3’. .
∴ 4 + A + B + 8 = 3 or 6 or 9 or 12 or 15 …….

4AB8 → If it is divisible by 9, sum of all the digits should be a multiple of ‘9’.
∴ 4 + A + B + 8 = 9 or 18 or 27 or 36
⇒ A + B + 12 = 9 ∣18∣ 27∣ 36 ……. (3)
From (1) & (3)
A + B + 12 = 9 or 18 say
If A + B + 12 = 9
A + B = 9 – 12 = -3
It is impossible
If A + B + 12 = 18
A + B = 18 – 12 = 6
∴ A + B = 6
If A = 4 & B = 2
4AB8 = 4428
4AB8 → 4428 → \(\frac{428}{8}\) (R ≠ 0)
∴ A = 4 & B = 2 are not possible.
If A = 2& B = 4
4AB8 → 4248 → \(\frac{248}{8}\) (R = 0)
∴ A = 2 and B = 4

Question 29.
By using the above method check whether 7810364 is divisible by 4 or not.        (Pg. No: 333)
Answer:
Given number = 7810364

Sum of product of place values and remainders of place values = 0 + 0 + 0 + 0 + 0 + 12 + 4
→ \(\frac{16}{4}\) (R = 0)
∴ 7810364 is divisible by ‘4’.

Question 30.
By using the above method check whether 963451 is divisible by 6 or not.     (Pg. No: 333)
Answer:
The given number = 963451

Sum of product of place values and remainders of place values
= 36 + 24 + 12 + 16 + 20 + 1 → \(\frac{109}{6}\) (R ≠ 0)
∴ 963451 is not divisible by ‘6’.

Try these

Question 1.
In the division 56 Z ÷ 10 leaves remainder 6, what might be the value of Z.     (Pg. No: 315)
Answer:
Let 56Z, Z = 0, 1,2, 3, 4, ….. , 9 say.
To obtain remainder ‘6’ when divided by 10, Z = 6
\(\frac{566}{10}\) = \(\frac{560+6}{10}\)
Remainder is 6.

∴ Z = 6

Question 2.
If 4B ÷ 5 leaves remainder 1, what might be the value of B?    (Pg. No : 316)
Answer:
If 4B is divided by 5 then remainder should be ‘1’,
∴ B = {0, 1, 2, ….. , 9}
i.e., 40, 41, 42, …… , 49
From the above numbers we have to take 41 and 46.
41 and 46 are divided by 5 and leaves the remainder 1.
∴ B = {1, 6}

Question 3.
If 76C ÷ 5 leaves remainder 2, what might be the value of C?     (Pg. No: 316)
Answer:
To get remainder 2, when 76C is divided by 5 take C = {0, 1,……, 9}.
If C = 2, 7 then
76C = 762 or 767 are divided by 5 leaves the remainder 2.

Question 4.
“If a number is divisible by 10, it is also divisible by 5.” Is the statement true? Give reasons.     (Pg. No : 316)
Answer:
The given statement is true.
∵ When a number is divisible by ’10’, then its units digit should be ‘0’.
Similarly the units digit of a number is 5 or 0, then it is divisible by 5.
∴ The number which is divisible by 10 is also divisible by 5.

Question 5.
“If a number is divisible by 5, it is also divisible by 10.” Is the statement is true or false? Give reasons.     (Pg. No : 316)
Answer:
The given statement is false.
∵ If a number is divisible by 5, then its units digit must be ‘5’ or ‘0’. But in case of 10, it . must be ‘0’ only.
∴ The number which is divisible by ‘5’ is need not be divisible by ’10’.

Question 6.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.     (Pg. No : 321)
a) 4² × 8² b) 10³ c) 10⁵ + 10⁴ + 10³ d) 4³ + 4² + 4¹ – 2²
Answer:

Question 7.
Take a four digit general number, make the divisibility rule for ‘7’.     (Pg. No : 322)
Answer:
Let the 4 – digited number be ‘abcd’ say.
The remainders when divided by ‘7’,

∴ If (6a + 2b + 3c + d) is divisible by 7 then the 4 – digited number be divisible by ‘7’.

Question 8.
Check your rule with the number 3192 which is a multiple of 7.      (Pg. No : 322)
Answer:
The given number is 3192
⇒ a = 3, b = 1, c = 9, d = 2
6a + 2b + 3c + d = 6 × 3 + 2 × l + 3 × 9 + 2.
= 18 + 2 + 27 + 2
= 49 → \(\frac{49}{7}\) (R = 0)
∴ 3192 is divisible by 7’according to my law.

Question 9.
1) Verify whether 789789 is divisible by 11 or not.     (Pg. No: 323)
2) Verify whether 348348348348 is divisible by 11 or not.
3) Take an even palindrome i.e. 135531 check whether this number is divisible by 11 or not.
4) Verify whether 1234321 is divisible by 11 or not.
Answer:

Question 10.
Check whether 1576 × 1577 × 1578 is divisible by 3 or not.     (Pg. No : 325)
Answer:
The given number is 1576 × 1577 × 1578.
The product of any 3 consecutive numbers is divisible by ‘3’.
Ex : 4 × 5 × 6 = 120 → \(\frac{120}{3}\) (R = 0)
∴ 1576 × 1577 × 1578 is divisible by ’3’.

Question 11.
Check the above method applicable for the divisibility of 11 by taking 10-digit number.     (Pg. No : 326)
Answer:
The largest 10 – digited number = 9,99,99,99,999
D C B A
∴ 9/999/999/999
⇒ B + D = 9 + 999 = 1008
A + C = 999 + 999 = 1998
∴ (A + C) – (B + D) = 990 → \(\frac{990}{11}\) (R = 0)
∴ The largest 10 – digited number should be divisible by 11 according to this method.

Question 12.
Take a three digit number and make the new numbers by replacing its digits as (ABC, BCA, CAB). Now add these three numbers. For what numbers the sum of these three numbers is divisible?      (Pg. No : 329)
Answer:

Question 13.
If YE × ME = TTT find the numerical value of Y + E + M + T.
[Hint: TTT = 100T + 10T + T = T(111) = T(37 × 3)]      (Pg. No: 332)
Answer:
TTT = 100T + 10T + T
= T(111)
= T(37 × 3)
∴ YE × ME = T(37 × 3)
∴ T ={1, 2, 3, ….., 9}
But T = {3, 6, 9} are multiples of 3.
T(37 × 3) = 3(111), 6(111), 9(111) are divisible by 3.
∴ YE × ME = 333∣666∣999
YE × ME = 999 = 27 × 37
∴ Y = 2, M = 3, E = 7, T = 3
∴ Y + E + M + T = 2 + 3 + 7 + 3 = 15

Question 14.
If cost of 88 articles is A733B, find the values of A and B.       (Pg. No: 334)
Answer:
If A733B is divisible by 88 then it is divisible by 8 × 11.
Divisibility of 11:
⇒ A733B → (A + 3 + B) – (7 + 3) = 0
⇒ A + B = 7 ……. (1)
Divisibility of 8:
⇒ A733B ⇒ \(\frac{33B}{8}\)
∴ \(\frac{336}{8}\) (R = 0) (If B = 6 then it is divisible by 8)
∴ B = 6 ……. (2)
From (1), (2)
∴ A = 1, B = 6

Question 15.
Check whether 456456456456 is divisible by 7, 11 and 13.     (Pg. No: 334)
Answer:
∴ The given number = 456456456456
456456456456 = 456 (1001001001)
= 456 × (7 × 11 × 13) × (1000001)
∴ 456456456456 is divisible by 7, 11 and 13.

Think, Discuss and Write

Question 1.
Find the digit in the units place of a number if it is divided by 5 and 2 leaves the remainders 3 and 1 respectively.   (Pg. No: 316)
Answer:
If a number is divided by 5 and 2 leaves the remainders 3 and 1 respectively, then its units digit be 3.
Ex: \(\frac{13}{5}\) ⇒ (R = 3), \(\frac{13}{2}\) ⇒ (R = 1)
\(\frac{23}{5}\) ⇒ (R = 3), \(\frac{23}{2}\) ⇒ (R = 1)
∴ The unit’s digit of a required number be 3.

Question 2.
Take a two digit number reverse the digits and get another number. Subtract smaller number from bigger number. Is the difference of those two numbers is always divisible by 9?      (Pg. No : 328)
Answer:

Question 3.
1) Can we conclude 10²ⁿ – 1 is divisible by both 9 and 11? Explain.     (Pg. No: 333)
2) Is 10²ⁿ⁺¹ – 1 is divisible by 11 or not? Explain.
Answer:

Question 4.
Verify a⁵ + b⁵ is divisible by (a + b) by taking different natural numbers for ‘a’ and ‘b’.    (Pg. No : 334)
Answer:

∴ a⁵ + b⁵ is divisible by (a + b).
∴ (a⁵ + b⁵) is divisible by (a + b) for all the values of a, b.

Question 5.
Can we conclude (a²ⁿ⁺¹ + b²ⁿ⁺¹) is divisible by (a + b)?      (Pg. No : 334)
Answer:

a²ⁿ⁺¹ + b²ⁿ⁺¹ is divisible by (a + b) for all the values of ‘n’.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable InText Questions and Answers.

8th Class Maths 2nd Lesson Linear Equations in One Variable InText Questions and Answers

Do this

Question 1.
Which of the following are linear equations:        [Page No. 35]
i) 4x + 6 = 8
ii) 4x – 5y = 9
iii) 5x² + 6xy – 4y² = 16
iv) xy + yz + zx = 11
v) 3x + 2y – 6 = 0
vi) 3 = 2x + y
vii) 7p + 6q + 13s = 11
Answer:
(i), (ii), (v), (vi), (vii) are the linear equations.

Question 2.
Which of the following are simple equations?        [Page No. 36]
i) 3x + 5 = 14
ii) 3x – 6 = x + 2
iii) 3 = 2x + y
iv) \(\frac{x}{3}\) + 5 = 0
v) x² + 5x + 3 = 0
vi) 5m – 6n = 0
vii) 7p + 6q + 13s = 11
viii) 13t – 26 = 39
Answer:
(i), (ii), (iv), (viii) are the simple equations.
Since these are all in the form of ax + b = 0.

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

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

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

Improve Your Learning

Question 1.
State reasons why wheat is cultivated in Kharif.
Answer:

1. The crops grown in the rainy season are termed as Kharif in the months of June to October.
2. If we cultivate wheat crop in the month of July it takes 8 – 10 weeks for growing.
3. After that flowering will take place. By that time, it would be October.
4. Then the night duration extends more than 12\(\frac{1}{2}\) hours. Wheat plants flowering takes place only in long night durations.
5. Crop production is based on the flowering of plant. If flowering of plant is more, the crop production also is more.
6. Wheat is important cereal crop gained a lot through Green Revolution by developing high yielding hybrid strains.

Question 2.
Ramaiah’s field is flattened. Somaiah’s field has many up and downs. Who will get more crop ?
Answer:

1. Generally the fields have a lot of ups and downs even after ploughing. So a leveller is used for levelling the soil.
2. By levelling the soil, it becomes flattened, water and nutrients can be reached to every part of the land. It also helps in sowing seeds and planting.
3. Because Ramaiah’s field is flattened, he will get more produce than Somaiah.

Question 3.
What are the advantages of ploughing?
Answer:
Before growing crops ploughing the soil properly is necessary.
Advantages:

1. Ploughing loosens the soil and it helps in easier transportation of air and water.
2. Water is stored deeply for a long time as the soil is soft.
3. Roots penetrate in the deep and can respire well as the air enters easily into the soil.
4. Friendly microorganisms and earthworms can grow well when the soil is soft.
5. Some foe microorganisms die due to the sun rays.

Question 4.
Treating with fungicides before sowing the seed is necessary. Why?
Answer:

1. Sometimes farmers wash seeds with chemicals to protect from pests.
2. Medication is done to keep away the seeds from the harmful microorganisms like bacteria, fungi etc.
3. So, generally farmers treat the seeds with fungicides before sowing the seeds before making them to germinate.

Question 5.
Why do farmers dry the paddy crop after cutting them from fields?
Answer:

1. Naturally food produce can be damaged by fungi, pests, rats and bacteria.
2. If moisture is also there in the grains, it helps to develop moulds (fungi).
3. Such grains neither germinates nor suitable to eat.
4. To overcome this problem farmers dry the grains for 2-3 days in sun.
5. After drying they keep the grains in jute bags and preserve them in godowns.

Question 6.
Give some examples of plants that grow after replanting.
Answer:
Seeds that are broadcast in a plot, grow into seedlings. When these plants grow to certain height, farmers pick out the plants (seedlings) from the plots, make bundles and are sown in proper distances. This is called transplantation (replanted)
E.g.: Rice (Sri Vari), Tobacco, Onions, Chillies etc.

Question 7.
Rahim removed weeds in his crop field, but David did not. Guess who get more yield. Why?
Answer:

1. Rahim get more yield than David. Because
2. Weeds are the unwanted or undesirable plants which grow in the fields and compete with crop plants for water, nutrients, light and space.
3. Because of these plants the crop plants may not grow properly. So they should be removed.
4. Otherwise the yield of the crop will be reduced.
5. Weeds give shelter for insects, pests and microorganisms and serve as a host for them.
6. Weeds are capable of germinating and growing fastern than crop plants. They flower and form seeds much earlier than the crop plants.
7. Some weeds disperse pollen grain to air which in turn causes respiratory diseases.

Question 8.
What is natural manure? How to prepare it? Give two examples.
Answer:
Natural manure: A manure made by the decomposition of plants and animal (organic) wastes is called natural manure or natural fertilizer or Bio Fertilizer.
Preparation:

1. These fertilizers are formed by decomposing plant and animal wastes.
2. In rural areas farmers keep these plant and animal wastes outside the village in open space.
3. Some bacteria like Azatobacter, Nitrobacter decompose and it becomes manure which contain nutrients.
4. Wherever the manure is added to the soil, there it provides nutrients to the plants. Examples : Vermi compost, Dung fertilizer.

Question 9.
Why do farmers plough their field during summer?
Answer:

1. Ploughing loosens the soil and it helps in easier transportation of air and water.
2. In summer temperature is very high. So the soil becomes dry. Then the soil becomes very loosly.
3. Some foe microorganisms die due to the sunrays.

Question 10.
Rajendar cultivated cotton crop in his field. He did not get sufficient yield. Can you guess the reasons? (OR)
A farmer cultivated cotton crop in his field. He did not get sufficient yield. Guess any four causes for it.
Answer:

1. Farmers in our state generally purchase seeds in the nearby market. The grains that are available in the packets play vital role in agriculture.
2. Sometimes the rate of germination of the seeds is not up to the mark, which was labelled on the packet.
3. Sometimes never germinate too.
4. At times, seeds grow into plants but they be sterile.
5. And sometimes the crop may be attacked by the larva of spotted brown boll-worm and pinkish boll-worms.
6. The larva of brown catter pillar sucks the juice from the leaves. The buds and the fruits of cotton plants drop off from the infected plants.
7. The larva of pinkish boll worm of cotton make hole in the stem, flower buds, flowers and fruits. As a result flower buds, cotton bolls drop off.
8. May be these reasons Rajendar did not get sufficient yield.

Question 11.
I am a plant. I grow in crop fields. Farmers pluck me. I do not know the reason. Can you tell who am I ?
Answer:

1. We observe some other plants growing along with the crop plants These are undesirable plants called weeds. They should be removed immediately.
2. The weeds, compete with the crop plants for nutrients, water and light, so he crop plants may not grow properly. This is the reason why they (weeds) should be removed.
3. Besides competition for food, light and water they also work as a carrier for different diseases. They also serve as host for different pests.
4. Some weeds disperse pollen grains to air which in turn causes respiratory diseases.
5. So the farmers pluck the weeds from crop plants by using different methods.

Question 12.
What do you observe in the experiment of dropping a fist of Bengal gram seeds in water?
a) What are the differences you observed in both the seeds?
Answer:
We can observe some seeds floated on water, the remaining sank under the water. The seeds which are floated are wrinkled and rough shaped but the sank seeds are round and smooth.
The floated seeds are light in weight but the sank seeds are more in weight.

b) Do you know why the floated seeds are lighter in weight?
Answer:
The floated seeds are not healthy, so they are lighter in weight.

c) Which seeds germinate well? Why?
Answer:
The seeds which sank in water germinate well because they are healthy.

d) Which seeds would not germinate properly? Why?
Answer:
The seeds which are wrinkled and rough would not germinate properly. Because the cotyledon inside the seed would not develop healthy.

Question 13.
Go to your nearest fertilizer shop and collect the information about chemical fertilizers and fill the table. Copy the following table in your notebook.
Answer:

Question 14.
Prepare a flow chart from ploughing to yielding in paddy.
Answer:

Question 15.
How do you appreciate the irrigation systems used in the drought prone areas?
Answer:

1. This method is employed when the availability of water is poor.
2. As the water reaches the plants drop by drop this is called Drip irrigation.
3. A long tube followed by small tubes attached to a motor. The tubes are made holes. So the water comes out from the tube.
4. The holes are arranged in such a way that it provide water exactly at the place where plant roots could receive water.
5. The man’s best technical method of utilizing the water in farming where the conditions are pravailing and in the areas where the availability of water is scanty.

Question 16.
Narendra sprayed over dose of pesticides on his cotton crop. Ramesh says it is a hazard to biodiversity and crop yield. Can you support Ramesh? How?
Answer:

1. In agriculture pests damage the crops. Almost all crops are generally effected by pests.
2. Wheat, Paddy and suagarcane are generally affected by fungal diseases. Groundnut is affected by Tikka disease. The catterpillars of spotted brown boll worm and Pinkish boll worm affect the crop.
3. A wide variety of agricultural and garden pesticides are available. A few derived from neem tobacco and chrysanthemum (Chamanthi) are less dangerous to other living organisms.
4. A wide variety of inorganic and organic pesticides are commonly used D.D.T. (Dichloro diphenoxy Trichloro ethane) BHC (Benzene Hexa Chloride), Chlordane, Endrin, Aldrin, Endosulfan and Diazinon pesticides are usually dusted or sprayed on crops or put in the soil.
5. But pesticides should not use unwisely. They get into the bodies of plants and animals in the soil and water. When these plants are eaten by animals like fish the pesticide get into their bodies.
6. A bird that eat the fish might get a concentrated lethal dose.
7. D.D.T. also accumulate in the egg shells, weakening them and making the shells break before hatching. It is observed D.D.T. is present in the milk of buffaloes and cows.
8. In this way pesticides are passed down the food chain and accumulate in the bodies of higher animals including human beings causing sickness and sometimes death.

Question 17.
Venkatesh observed the irrigation method for paddy field. He wanted to follow the same practice for his Maize crop. What suggestions do you give him?
Answer:

1. Paddy is grown as a Kharif or a Rabi crop. It requires high temperature of 22°C to 32°C and heavy rain fall. It is cultivated heavily in Kharif season.
2. Maize is cultivated in both Kharif and Rabi seasons but heavily in Kharif season. This requires high temeprature (35°C) with moderate rainfall.
3. So Venkatesh can follow the ame practice for his maize crop as the requirements of both the crops are almost same.

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

Question 1.
Look at the picture given below and write the constituents in it.

Answer:
Nitrogen (20%), Phosphorus (5%), Potash (10%)

Question 2.
Which manure is beneficial?
Answer:
Natural manure is beneficial.

Question 3.
Let us compare both, which manure is beneficial.

Observe the table carefully, discuss with your teacher and conclude which fertilizer is best to the farmers and why?
Answer:
Natural fertilizer is the best because this is made by the decomposition of plants and animal (organic) waste. Deposits of humus layer is found in the soil with less amount of Nitrogen, Phosphorus and potash deposits in the soil.
a) When do farmers irrigate the land?
Answer:
After applying manure farmers irrigate the land.

b) List out the water resources of your village.
Answer:
Wells, ponds, canals, tanks are the water resources.

c) Are they useful to your farmers?
Answer:
Yes. They are useful.

d) In what way the farmers of your village get water to the fields ?
Answer:
Farmers irrigate their fields either manually using bullocks or by using pumps.

Question 4.
What are the reasons for high production in Japan?
Answer:
Japan has cooler temperature.

Question 5.
What are the reasons for low production in India?
Answer:
High temperature and uncertainty of rainfall.

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

Activity – 1

Question 1.
CROPS IN INDIA:
Observe the following India map.

a) Are there many crops that are grow in most of the parts of our country? What are they?
Answer:
India is a unique position to grow almost every possible crop. It is the land of producing a variety of cereals like paddy, wheat, jowar, bajra, maize and ragi, pulses, spices, fruits, vegetables, oil seeds, fibre crops etc.

b) Why such crops are grown all over the country?
Answer:
India is an agricultural country and people derive their livelihood from agriculture. Agriculture is the back bone of Indian economy.

c) From the above map, which of them are grown in your village?
Answer:
Rice, pulses, banana, vegetables and leafy vegetables
a) Country: India
b) State: Andhra Pradesh
c) Your village: Khajipalem

d) How many days are required for getting the crop?
Answer:
Nearly 120 days.

e) Is time period for all crops are same?
Answer:
The time peirod for all the crops are not same. They are different from one crop to another.

f) Which crop needs more duration?
Answer:
Rice and Wheat.

Activity – 2

Question 2.
DURATION OF CROP:
a) Write the information in the table.

1. Example for long term crops:
Answer:
Jowar, red gram.

2. Example for short term crops:
Answer:
Pulses like green gram, black gram etc., and onions.

Activity – 3

Question 3.
WHERE DO CROPS ARE GROWN.
Discuss in groups and make a list of these things for the following table.

1. In which season do you find more varieties of vegetables in the market?
Answer:
Rainy season.

2. Generally farmers grow varieties of vegetables during rainy season. Can you guess the reason?
Answer:
In rainy season ponds, wells, rivers, ditches are pooled with water.

3. If we cultivate wheat in the month of the November what will happen?
Answer:
We get hot climate from February onwards. It is suitable for maturing the grains. That is the reason wheat is cultivated in the Rabi season only.

Activity – 4

Question 4.
PRODUCTION OF PADDY:
a) Go and collect the information through your nearest farmer and fill the following table.

a) In which season farmers get more benefits?
Answer:
Kharif season

b) Are there any other crops which are growing both Kharif and Rabi seasons?
Answer:
Paddy, wheat and maize

c) In which seasons farmers generally get good quality of seeds.
Answer:
Kharif and Rabi

d) The quantity of grains is higher in Kharif season than Rabi season. Do you agree this? Give your reasons.
Answer:
The climate, (the temperature, humidity with abundance of water supply) will be suitable in Kharif season

e) Do you know about third crop?
Answer:
The third crop season known as Zayad, grown in the months of April, May and June.

Activity – 5

Question 5.
SELECTION OF SEEDS:
Take some water in glass. Drop a fist of seeds in it. You cam observe some seeds will float on water. Collect those seeds and observe with hand lens and comparing with seeds sink under the water. Write your observations in the table.

a) What are the differences you observed in both seeds?
Answer:
Some seeds sank in water, some seeds floated.

b) Do you know why the floated seeds are light in weight?
Answer:
They are unhealthy seeds.

Activity – 6

Question 6.
GERMINATION AND SELECTION:
Show both the seeds in different pots and provide water uniformly, observe the growth of the plants in two pots and make a report.
1. Which seeds germinate well? Why?
Answer:
The seeds which are smooth and round germinate well because they are healthy seeds. Biology

2. Which seeds do not germinate properly? Why?
Answer:
The seeds which are wrinkled and rough do not germinate properly because they are unhealthy.

3. Were all the seeds were tested like this?
Answer:
The crop plant like Rice, wheat etc.

4. Do you know how the paddy seeds germinate?
Answer:
There are different stages in sprouting of the soaked rice seeds before it is planted.

5. Observe a sprout of paddy. Cam you say which part become root? Which part become shoot im the picture?
Answer:
Coleoptile become shoot and the part beneath the ground is root.

Activity – 7

Question 7.
SOWING METHODS
Collect information from the nearby farmers and fill in the table.

1. Why the seedlings are replanted at proper distance?
Answer:
They get water, mineral and sunlight equally when they are replanted at proper distance.

2. Do all the crops grow when replanted? Why not?
Answer:
Mostly all plants will grow.

Activity – 8

Question 8.
CROPS AND DISEASES:
Form a group with 4 to 5 of your classmates, visit nearby field, discuss with farmers about diseases effected by, and how to control them. If you do not know the name of the disease, write its local name or its characters.

1. Do all the farmers use the same pesticides for the same crop?
Answer:
For different crops different pesticides are used.

2. Is there any disease that you find in all fields?
Answer:
No

3. Where do they buy pesticides?
Answer:
From government and private agencies.

4. What are the appliances that they use to spray pesticides?
Answer:
Sprayer or dusters.

5. Did you find any other living organisms dying along with pests due to pesticides ? What are they ?
Answer:
Yes. Caterpillars, sparrows etc.

Activity – 9

Question 9.
IDENTIFICATION OF PESTS:
Observe the plants in a nearby field or in your school garden. Closely observe the leaves and stems to collect the following information. If the character is present put a ‘S’ mark and if there is no character put ‘X’ mark.
Name of the plant/crop : Rice (Blast of Rice)
Place: Prakasam

a. Do all the leaves of plant have spots?
Answer:
Yes, all the leaves of plant have spots.

b. Draw the leaf with these spots.
Answer:

c. What is your reason for the leaves which have cutting edges?
Answer:
Eaten by grasshoppers.

d. Do you find any twilted leaves with insects? How are they?
Answer:
Infection occurs on leaf sheaths.

e. Are the scars on the stems is same as spots on leaves?
Answer:
Yes, the scars on the stems is same as spots on leaves.

f. Collect powdery substance of the spots on leaves and observe under microscope. Write down your observations.
Answer:
Fungus produce small spores known as conidia.

Activity – 10

Question 10.
PEST CONTROLING PRACTICES:
In your village farmers control pests by using different pesticides and insecticides for different crops. For this they use different practices. Ask your elders the names of pesticides that they use in the following pest controlling practices.
Answer:

1. Spraying: Endrin, Diethane, M-45, Eldrine.
2. Dusting: Aldrin, D.D.T.
3. Put in the soil: Zinc, Sulphur, Phosphorous, fluorine
4. Burning and picking are also the practices where they use these: Sugarcane, citrus
5. Bio pesticides: Neem water

Observe the following pest controlling practices
a. Which of the above practices is good?
Answer:
A farmer remove the affected leaves from the plant and burnt them.

b. Why do you think so?
Answer:
By burning the affected leaves the pests will be controlled.

c. Why did the farmer use two pesticides at a time?
Answer:
At the first time the pests will be controlled by spraying pesticide, but if we use unwisely, pests become resistant to the pesticides.

d. What will we do to solve the problem?
Answer:
Pesticides will be used as per the requirement for the disease.

e. Farmers add manure to the soil.
What they used to add?
Answer:
They used to add nutrients to the soil.

f. Have you a compost pit in your school / house?
Answer:
Yes, there is a compost pit in our school. All the waste materials like dry leaves, fruits peel etc. will dump into the pit.

Activity – 11

Question 11.
WHEN SHOULD FARMERS IRRIGATE THE HELD?
Consult to the farmers and fill the table with the information to how and when they provide water to various crops.

a. Are all the crops provided with equal amount of water?
Answer:
No. Irrigation should be done according to nature of the soil and the type of crop to be grown.

b. Why do farmers provide more water to the summer crops?
Answer:
Summer season is the hottest climate so the crops which grow in summer requires high quantity of water.

Weedling:
a. Why should they (weeds) be removed?
Answer:
The weeds compet with the primary crops for nutrients, water and light because of these plants crop plants may not grow properly. So they should be removed.

Question 12.
How sprinklers and drip system are used and write down their merits and demerits.
Answer:
When the availability of water is poor, drip irrigation system is used. In this the water reaches the plant drop by drop through sprinklers. So that water comes out from the sprinklers wetting exactly the place of the roots of the plant.
Advantages:

1. Maximum use of available water.
2. No water being available to weeds.
3. Maximum crop yield.
4. Efficiency use of fertilizers.
5. Less weed growth.
6. Low labour and low operation cost.
7. No soil erosion.
8. Improved infiltration in soil with low in take.

Disadvantages:

1. Sensitivity to clogging.
2. Moisture distribution problem.
3. Salinity hazards
4. High cost compared to furrow.
5. High skills is required for design, install and operation.

Activity – 12

Question 13.
Ask your nearby nursery and know the weeds that grow in different crops. Make a table in your notebook.

Harvesting of paddy:
a. If the paddy is not dried well enough. What will happen?
Answer:
If moisture is there in paddy grains it helps to develop moulds (fungi). Such grains neither germinate nor suitable to eat.

b. Where do farmers harvest the crops in your village ?
Answer:
Farmers generally used to harvest by using traditional methods.

c. Is harvesting same for all crops?
Answer:
Yes.

Activity – 13

Question 14.
Find out the methods of harvesting in and around our village and fill the table.
Answer:

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
Answer:
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)
Answer:

Question 3.
Complete the following table.     (Page No. 9)
Answer:

Question 4.
Complete the following table.      (Page No. 13)
Answer:

Question 5.
Complete the following table.      (Page No. 16)
Answer:

Question 6.
Complete the following table.      (Page No. 17)
Answer:

Question 7.
Represent – \(\frac{13}{5}\) on the number line.     (Page No. 22)
Answer:
Representing – \(\frac{13}{5}\) on the number line.

Try These

Question 1.
Hamid says \(\frac{5}{3}\) is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
Answer:
I would not agree with Hamid’s argument. Since \(\frac{5}{3}\) is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion \(\frac{5}{3}\), 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
Answer:
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) \(\frac{2}{3}\), \(\frac{7}{4}\) are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
Answer:
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = \(\frac{3}{5}\) is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = \(\frac{1}{2}\) is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) \(\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}\)
B) \(\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}\)
Answer:
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

Answer:
i) A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{5}{5}\) = 1, D = \(\frac{7}{5}\), E = \(\frac{8}{5}\), F = \(\frac{10}{5}\) = 2.
ii) S = \(\frac{-6}{4}\), R = \(\frac{-6}{4}\), Q = \(\frac{-3}{4}\), P = \(\frac{-1}{4}\)

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Answer:
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
Answer:
The number T is multiplicative inverse of itself.
∵ 1 × \(\frac{1}{1}\) = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
          (Page No. 15)
Answer:
The reciprocal of ‘0’ is \(\frac{1}{0}\).
But the value of \(\frac{1}{0}\) is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\), \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\)
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
Answer:
i) \(\frac{7}{5}\) = 0.4,
\(\frac{3}{4}\) = 0.75,
\(\frac{23}{10}\) = 2.3,
\(\frac{5}{3}\) = 1.66… = \(1 . \overline{6}\),
\(\frac{17}{6}\) = 2.833… = \(2.8 \overline{3}\),
\(\frac{22}{7}\) = 3.142
ii) From the above decimals \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\) are terminating decimals.
While \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\) are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals \(0 . \overline{9}\), \(14 . \overline{5}\) and \(1.2 \overline{4}\) to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Answer:
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or \(0 . \overline{9}\) = 1
Second Method:
\(0 . \overline{9}\) = 0 + \(0 . \overline{9}\)
= 0 + \(\frac{9}{9}\)
= 0 + 1
= 1

Let x = \(14 . \overline{5}\)
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = \(1.2 \overline{4}\)
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

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

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

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

Do this

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

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

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

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

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

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

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

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

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

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

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

∴ 512 and 729 are perfect cubes.

Try These

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

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

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

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

Question 5.

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

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

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

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

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

Think, Discuss and Write

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

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

Answer:

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

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

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.4

Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ \(\frac{1}{\text { No. of days }}\)

⇒ Compound ratio of 8:12 and 20: 15
= \(\frac{8}{12}=\frac{20}{15}\) = \(\frac{8}{9}\) …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ \(\frac{480}{x}=\frac{8}{9}\)
⇒ x = \(\frac{480 \times 9}{8}\) = ₹540
∴ The cost of required rice is ₹ 540

Method – II :
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = No. of men
D₁ = No .of days
W₁ = Cost of rice
∴ M₁ = 8
D₁ = 20
W₁ = ₹ 480
M₂ = 12
D₂ = 15
W₂ = ? (x)

⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-

Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
∴ M₁ = 10
D₁ = 5
W₁ = 75
M₂ = 15
D₂ = ?
W₂ = 45

∴ x = 2
∴ No. of days are required = 2

Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M₁D₁H₁ = M₂D₂H₂
∴ M₁ = 24
D₁ = 15 days
H₁ = 8 hrs
M₂ = 20
D₂ = ?
H₂ = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9

∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]

Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = 175
D₁ = 36
W₁ = 3150
M₂ = ?
D₂ = 24
W₂ = 3900

∴ No. of workers are required = 325

Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M₁D₁H₁ = M₂D₂H₂
M₁ = 14
D₁ = 12 days
H₁ = 6
M₂ = 4
D₂ = ?
H₂ = 7
⇒ 14 × 12 × 6 = 4 × x × 7

⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]

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

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

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

(ii) 14x³ by 42x³
= 14x³ ÷ 42x³

(iii) 72a³b⁴c⁵ by 8ab²c³

(iv) 11xy²z³ by 55xyz
11xy²z³ ÷ 55xyz

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

= -6l²m

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

(ii) (5a³b – 7ab³) ÷ ab

(iii) (25x⁵ – 15x⁴) ÷ 5x³

= 5x² – 3x (or) x(5x – 3)

(iv) (4l⁵ – 6l⁴ + 8l³) ÷ 2l²

= 2l² – 3l² + 4l = l(2l² – 3l + 4)

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

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

(vi) 3p³– 9p²q – 6pq²) ÷ (-3p)

= -[p² – 3pq – 2q²]
= 2² + 3pq – p²

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

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

(ii) 12x (8x – 20,) ÷ 4(2x – 5)

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

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

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

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

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

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

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

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

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

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

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

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x₁y₁ = x₂ y₂
⇒ 8 × 5 = 10 × x
⇒ x = \(\frac{8 \times 5}{10}\) = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x₁y₁ = x₂ y₂ (Let y₂ = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = \(\frac{500 \times 70}{700}\) = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x₁y₁ = x₂ y₂ let y₂ = x (say)
= 36 × 12 = 9 × x
x = \(\frac{36 \times 12}{9}\) = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , Let y₂ = x (say)
⇒ \(\frac{28}{2}\) = \(\frac{56}{x}\)
⇒ x = \(\frac{56}{14}\)
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x₁y₁ = x₂ y_{2 ,} Let x₂ = x (say)
⇒ 16 × 10 = x × 8
⇒ x = \(\frac{16 \times 10}{8}\)= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1\(\frac { 1 }{ 2 }\) hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 5 × 1\(\frac { 1 }{ 2 }\) = x x 1\(\frac { 1 }{ 2 }\)
⇒ 5 × \(\frac { 3 }{ 2 }\) = x x \(\frac { 1 }{ 2 }\)
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 15 × 48 = x × 30
⇒ x = \(\frac{15 \times 48}{30}\) = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 45 × 8 = x × 6
⇒ \(\frac{45 \times 8}{6}\)
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ \(\frac{\mathrm{x}}{\mathrm{y}}\)

Let x₁ = 100x, x₂ = 112x
(∵ It increases 12%)
y₁ = 100y, y₂ = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = \(\frac{(x+1)(x+1)}{(x+2)}\)
∴ k = \(\frac{(x+1)^{2}}{(x+2)}\)

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.
What do you observe? Write your observations in your note books

Solution:

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions InText Questions and Answers.

8th Class Maths 10th Lesson Direct and Inverse Proportions InText Questions and Answers

Do this

Question 1.
Write five more such situations where change in one quantity leads to change in another quantity.     [Page No. 231]
Answer:
The change in one quantity leads to change in another quantity will see in the following situations.

1. If speed increases then time decreases.
2. In a family, the number of persons are increased then their consumption will also increases.
3. If water consumption increases then water levels decreases.
4. If the capacity of worker’s increases then time decreases.
5. If thickness of a wire increases then its resistance decreases.

Question 2.
Write three situations where you see direct proportion.      [Page No. 233]
Answer:

1. The relation between number of students to number of teachers.
2. Number of buffaloes to their consumption of grass.
3. Number of workers to length of wall.

Question 3.
Let us consider different squares of sides 2, 3, 4 and 5 cm. Find the areas of the squares and fill the table.

What do you observe? Do you find any change in the area of the square with a change in its side? Further, find the ratio between the area of a square to the length of its side. Is the ratio same? Obviously not.
∴ This variation is not a direct proportion.     [Page No. 233]
Answer:

From the above table the ratios are not equal.
∴ So the change is not in direct proportion.
If the measure of side of a square will be change then its area also be changed.

Question 4.
The following are rectangles of equal breadth on a graph paper. Find the area for each rectangle and fill in the table.


Is the area directly proportional to length?      [Page No. 233]

Answer:
Yes, the area is directly proportional to its length.

Question 5.
Take a graph paper make same rectangles of same length and different width. Find the area for each. What can you conclude about the breadth and area?        [Page No. 233]
Answer:

Area of first rectangle (A₁) = 3 × 1 = 3 sq. cm.
Area of second rectangle (A₂) = 3 × 2 = 6 sq. cm.
∴ The relation between the areas of rectangle and breadths is in direct proportion.
[∵ \(\frac{1}{3}\) = \(\frac{2}{6}\)]

Question 6.
Measure the distance in the given map and using that calculate actual distance between (i) Vijayawada and Visakhapatnam, (ii) Tirupati and Warangal. (Scale is given)        [Page No. 235]

Answer:
i) The distance between Vijayawada and Visakhapatnam = 2 cm
According to the sum
1 cm = 300 km then 2 cm = ?
1 …… 300
2 …… ? (x)
⇒ x = 2 × 300 = 600 km
The distance between the above two cities is 600 km.
ii) The distance between Tirupathi and Warangal = 3 cm
But given that 1 cm = 300 km
3 cm = ? (x)
x = 3 × 300 = 900 km
∴ The distance between Tirupathi and Warangal = 900 km.

Question 7.
Write three situations where you see inverse proportion.      [Page No. 238]
Answer:
i) Time – work capacity
ii) Speed – distance
iii) Time – speed

Question 8.
To make rectangles of different dimensions on a squared paper using 12 adjacent squares. Calculate length and breadth of each of the rectangles so formed. Note down the values in the following table.

What do you observe? As length increases, breadth decreases and vice-versa (for constant area).
Are length and breadth inversely proportional to each other?      [Page No. 238]
Answer:
In a rectangle if length is increases then breadth is decreases and vise-versa.
∴ Length and breadth of a rectangle are in inverse proportion.

Think, discuss and write

Question 1.
Can we say that every variation is a proportion.
A book consists of 100 pages. How do the number of pages read and the number of pages left over in the book vary?

What happened to the number of left over pages, when completed pages are gradually increasing? Are they vary inversely? Explain.         [Page No. 239]
Answer:
In every situation number of pages read and number of pages left over in the book are in inverse proportion.
If number of pages read are increases then number of pages left are decreases.