Class 8

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions InText Questions

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

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

Do this

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Try These

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

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

Think, Discuss and Write

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

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

AP State Syllabus AP Board 8th Class 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

Improve Your Learning

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:

Sexual Reproduction	Asexual Reproduction
1. Male and female gametes are formed.	1. No production of gametes.
2. Involves fusion of male and female gametes.	2. No fusion of gametes.
3. Involves two organisms.	3. Involves a single organism.
4. Offsprings have some characters from male parent and other from female parent. Some characters may not be present in either of the parents.	4. Produces offsprings that are identical to the parent.

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.

Hydra	Amoeba
1. Multicellular organism.	1. Unicellular organism.
2. Nucleus is absent.	2. Nucleus is present.
3. Buds are formed on the body surface.	3. A constriction is formed in the middle of the nucleus.
4. The bud increases in size and develops tentacles.	4. The constriction deepens divides the nucleus into two nuclei.
5. It grows in size.	5. A constriction is formed on the body wall of Amoeba in the middle.
6. This bud separates from the parent Hydra and lives independently.	6. The constriction deepens and divides the body of amoeba into two individuals (daughter amoebae).

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:

Animals that have external ears	Animals that do not have external ears
Deer	Fish
Leopard	Frog
Pig	Sparrow
Buffalo	Lizard
Giraffe	Crow
Elephant	Snake
Cat

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:

Name of Animals	Presence of external ears (Yes/No)	Presence of epidermal hairs on the skin/Feathers on their wings
Cow	Yes	Epidermal hair
Rat	Yes	Epidermal hair
Crow	No	Feathers on their wings
Pig	Yes	Epidermal hair
Fox	Yes	Epidermal hair
Hen	No	Feathers on their wings
Camel	Yes	Epidermal hair
Duck	No	Feathers on their wings
Frog	No	No hair, no feathers
Elephant	Yes	No hair
Buffalo	Yes	Epidermal hair
Pigeon	No	Feathers on their wings
Cat	Yes	Epidermal hair
Peacock	No	Feathers on their wings
Lizard	No	No feathers, no wings.

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)

Changes in the Nucleus/Body structure
1st diagram	Nucleus is round.
2nd diagram	Constriction in the centre of nucleus.
3rd diagram	Nucleus divides into two daughter nuclei. On the centre of the body surface a constriction is formed.
4th diagram	The constriction deepens.
5th diagram	The constriction still deepens and ready to cut into two parts.
6th diagram	Two daughter amoebae are formed.

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 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 9th Lesson Area of Plane Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures InText Questions and Answers.

8th Class Maths 9th Lesson Area of Plane Figures InText Questions and Answers

Do this

Question 1.
Find the area of the following figures:     [Page No. 200]
i)

Answer:
Area of a parallelogram = b × h = 7 × 4 = 28 sq.cm.

ii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 7 × 4
= 14 sq.cm.

iii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 5 × 4
= 10 sq.cm.

iv)

Answer:
Area of rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × (4+4) × (3+3)
[∴ d₁ = 4 + 4 = 8, d₂ = 3 + 3 = 6]
= \(\frac{1}{2}\) × 8 × 6
= 24 cm²

v)

Answer:
Area of a rectangle = l × b
= 20 × 14 = 280 sq.cm

vi)

Answer:
Area of a square = s²
= s × s
= 5 × 5 = 25 cm²

Question 2.
The measurements of some plane figures are given in the table below. However, they are incomplete. Find the missing information.     [Page No. 200]
Answer:

Question 3.
Find the area of the following trapezium.      [Page No. 204]
fig (i)

Answer:
Area of a trapezium

fig (ii)

Answer:
Area of a trapezium

Question 4.
Area of a trapezium is 16 cm². Length of one parallel side is 5 cm and distance between two parallel sides is 4 cm. Find the length of the other parallel side. Try to draw this trapezium on a graph paper and check the area.
[Page No. 204]
Answer:
Given that
Area of a trapezium = 16 sq.cm
Length of one of the parallel sides is a = 5 cm; h = 4 cm
Length of 2nd parallel side (b) = ?
A = \(\frac{1}{2}\)h(a + b)

Graph Sheet:

Area of parallelogram ABCD = 12 sq.cm + (S + P) + (Q + R) + (W + T) + (V + U)
= 12 + 1 + 1 + 1 + 1
= 12 + 4
= 16 sq.cm

Question 5.
ABCD is a parallelogram whose area is 100 sq.cm. P is any point insile the parallelogram (see fig.) find tie area of △APB + △CPD.       [Page No. 204]

Answer:
Area of parallelogram ABCD = 100 sq.cm
From the given figure,
ar (△APB) + ar (△CPD) = ar (△PD) + ar (△BPC)

Question 6.
The following details are noted in meters in the field book of a surveyor. Find the area of the fields.     [Page No. 213]
i)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of pentagonal field,
ii) AD is the diagonal.
iii) Now the area of the field = Areas of 4 triangles and a trapezium.
PQ = AQ – AP = 50 – 30 = 20
QD = AD – AQ = 140 – 50 = 90
RD = AD – AR = 140 – 80 = 60
Area of △APB:

Area of trapezium PBCQ:

Area of △QCD:

Area of △DER:

Area of △ERA:

∴ Area of the field = ar △APB + ar trapezium PBCQ + ar △QCD + ar △DER + ar △ERA
= 450 + 800 + 2250 + 1500 + 2000 = 7000 sq. units
ii)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of a pentagonal field.
ii) AC is the diagonal.
iii) The area of a field is equal to areas of 4 triangles and a trapezium.
QC = AC – AQ = 160 – 90 = 70
RC = AC – AR = 160 – 130 = 30
PR = AR – AP = 130 – 60 = 70
Area of △AQB:

Area of △QBC :

Area of △DRC :

Area of trapezium EPRD:

Area of △EPA :

∴ Area of the field = ar △AQB + ar △QBC + ar △DRC + ar trapezium EPRD + ar △EPA
= 2700 + 2100 + 450 + 2450 + 1200 = 8900 sq. units

Try these

Question 1.
We know that parallelogram is also a quadrilateral. Let us split such a quadrilateral into two triangles. Find their areas and subsequently that of the parallelogram. Does this process in turn with the formula that you already know?   [Page No. 209]

Answer:
Area of a parallelogram ABCD

Area of parallelogram ABCD
= base x height
= bh sq. units
(OR)

Area of parallelogram ABCD
= ar △ABC + ar △ACD
= \(\frac{1}{2}\) BC × h₁ + \(\frac{1}{2}\) AD × h₂
= \(\frac{1}{2}\) bh + \(\frac{1}{2}\) bh [∵ h₁ = h₂]
= bh sq. units.
∴ This process in turn with already known formula.

Question 2.
Find the area of following quadrilaterals.      [Page No. 213]
i)

Answer:
d = 6 cm, h₁ = 3 cm, h₂ = 5 cm
Area of a quadrilateral
= \(\frac{1}{2}\)d(h₁ + h₂)
= \(\frac{1}{2}\) × 6 (3 + 5) = 3(8) = 24 cm²

ii)

Answer:
d₁ = 7 cm; d₂ = 6 cm
Area of a rhombus A = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 7 × 6
= 7 × 3 = 21 cm²

iii)

Answer:
Area of a parallelogram (A) = bh
(∵ The given fig. is a parallelogram in which two opposite sides are parallel)
Area of a parallelogram = 2 ar AADC
= 2 × \(\frac{1}{2}\) × 8 × 2 = 16 Sq. cm.
[∵ Area of a parallelogram = ar △ADC + ar △ABC. But ar △ABC = ar △ADC]

Question 3.
i) Divide the following polygon into parts (triangles and trapezium) to find out its area.     [Page No. 214]

Answer:
FI is a diagonal of polygon EFGHI.
If perpendiculars GA, HB are drawn on the diagonal FI, then the given figure pentagon is divided into 4 parts.
∴ Area of a pentagon EFGHI = ar △AFG + ar AGHB + ar △BHI + ar △EFI.

NQ is a diagonal of polygon MNOPQR. Here the polygon is divided into two parts.
∴ Area of a hexagon MNOPQR = ar NOPQ + ar MNQR.

ii) Polygon ABCDE is divided into parts as shown in the figure. Find the area.     [Page No. 215].

If AD = 8 cm, AH = 6 cm, AF = 3 cm and perpendiculars BF = 2 cm, GH = 3 cm and EG = 2.5 cm.
Answer:
Area of polygon ABCDE = ar △AFB + ar FBCH + ar △HCD + ar △AED

So, the area of polygon ABCDE = 3 + 7.5 + 3 + 10 = 23.5 sq.cm

iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm.   [Page No. 215].

NA, OD, QC and RB are perpendiculars to diagonal MP.
Answer:
Area of MNOPQR
= ar △MAN + ar ADON + ar △DOP + ar △CQP + ar BCQR + ar △MBR
Hence CP = MP – MC = 9 – 6 = 3 cm
BC = MC – MB = 6 – 4 = 2 cm
AB = MB – MA = 4 – 2 = 2 cm
DP = MP – MD = 9 – 7 = 2 cm
AD = MD – MA = 7 – 2 = 5 cm

= 2.5 + (2.5 × 5.5) + 3 + 3 + 4.5 + (2 × 2.5)
= 2.5 + 13.75 + 3 + 3 + 4.5 + 5
= 31.75 sq.cms

Think, discuss and write

Question 1.
A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?    [Page No. 213]
Answer:

No, we cannot divide a trapezium into two congruent triangles.
∵ From the adjacent figure,
△ABC ≆ △ADC

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.6 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.6

Question 1.
Find the sum of integers which are divisible by 5 from 1 to 100.
Solution:
Numbers which are divisible by 5 from 1 to 100 are 5, 10, 15, …………………95, 100.
∴ Sum of the above numbers = 5+10 + ……………..+ 95 + 100
= 5[1 + 2 + ………………….+ 20]
= 5 [ \(\frac{20 \times(20+1)}{2}\) ]
= \(\frac{5 \times 20 \times 21}{2}\) [∵ Sum of ‘n’ natural numbers = \(\frac{n(n+1)}{2}\) & n = 20 ]
= 1050

Question 2.
Find the sum of integers which are divisible by 2 from 11 to 50.
Solution:
. Numbers which are divisible by 2 from 11 to 50 are 12, 14,48, 50.
Sum of the numbers = 12 + 14 + ……….. + 48 + 50 ‘
= (2 + 4 + ……….. + 50) – (2 + 4 + ……….. + 10)
= 2(1 + 2 +……….. + 25) – 2 (1 + 2 + ……….. + 5)

= 25 × 26 – 5 × 6
= 650 – 30
= 620

Question 3.
Find the sum of integers which are divisible by 2 and 3 from 1 to 50.
Solution:
Numbers which are divisible by 2 and 3 i.-e., which are divisible by 6 from 1 to 50 are 6,12 …………….48.
Sum of the numbers = 6 + 12 + ……..+ 48
= 6(1 + 2 +……… + 8)
= 6 \(\left[\frac{8(8+1)}{2}\right]\)
= 3 × 8 × 9 = 216

Question 4.
(n³ – n) is divisible by 3. Explain the reason.
Solution:

∴ If n = 4, (n³ – n) is divisible by 3.
∴ (n³ – n) is divisible by all the values of n.
Method 2:
n³ – n = n(n² – 1)
= n(n + 1)(n – 1)
∴ (n³ – n) is divisible by ‘3’ for all the values of n.
[∵ (n – 1), n, (n + 1) are three consecutive odd numbers]

Question 5.
Sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’. Explain the reason.
Solution:
Sum of n’ consecutive odd numbers = n²
Since n is a factor of n², It Is divisible by ‘n’.

Question 6.
Is 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ divisible by 5? Explain.
Solution:
Sum of units digit of number 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹
= 1 + 8 + 7 + 4
= 20 → \(\frac{20}{5}\)(R = 0)
∴ 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ is divisible by 5.

Question 7.

Find the number of rectangles of the given figure?
Solution:

∴ No.of rectangles in the given figure = 1 + 2 + 3 + 4 + 5 + 6 = 21

Question 8.
Rahul’s father wants to deposit sorne amount of money every year on the day of Rahul’s birthday. On his 1st birth day Rs.100, on his 2nd birth day Rs.300, on his 3 birth day Rs.600, on his 4th birthday Rs. 1000 and so on. What is the amount deposited by his father on Rahul’s 15th birthday.
Solution:

Rahul’s father deposits on every year 200, 300, 400 more than before year.
Then he deposits ₹ 10,500 on 14th birthday.
∴ The amount deposits on 15th birthday
= 10,500 + 1,500
= ₹ 12,000/-

Question 9.
Find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Solution:
Sum of the numbers which are divisible by 2 from 1 to 100
= 2 + 4 + ……….. + 100
= 2(1 + 2 + ………… +50)
= 2 × \(\frac{50 \times(50+1)}{2} \)
= 50 × 51 = 2550
Sum of the numbers which are dMsible by 5froin I to 100
= 5 + 10 + ……….. + 100
= 5(1 + 2 +……….. +20)
= 5 × \(\frac{20 \times(20+1)}{2}\)
=5 × 10 × 21
=1050

Sum of the numbers which are.divisible by both 2 and 5 = 2550 + 1050 =3600
∴ Sum ol the numbers which are divisible by 2 or 5 from 1 to 100
= 10 + 20 + ………..+ 100 ( L.C.M of 2, 5 is 10)
=10(1 + 2 + ………..+ 10)
= 10 × \(\frac{10 \times(10+1)}{2}\)
= 5 × 10 × 11 .
= 550
∴ The sum of required numbers 3600—550 3050

Question 10.
Find the sum of integers from 11 to 1000 which are divisible by 3.
Solution:
Sum ol the numbers which are divisible by 3 from lito 1000
= 12 + 15+ ……….. +099
= 3(4 + 5 + ……….. +333)
= 3(1 + 2 + ……….. + 333) – 3(1 + 2+3)

= 999 × 167 – 9 × 2
= 166833 – 18
= 166815

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.4

Question 1.
Check whether 25110 is divisible by 45.
Solution:
The given number = 25110
If 25110 is divisible by 45 then it should be divisible by 5 and 9.

∴ The number 25110 is divisible by 45

Question 2.
Check whether 61479 is divisible by 81.
Solution:
If 61479 is divisible by 81 then it is divisible by 9.
If the sum of the digits of a number is dívisible by 9 then the entire number is divisible by 9.
∴ 61479 → 6 + 1 + 4 + 7 + 9 → \(\frac { 27 }{ 9 }\) (R = 0)
∴ 61479 is divisible by 81. [∵ 9 is factor of 81]

Question 3.
Check whether 864 is divisible by 36? Verif,’ whether 864 is divisible by all the factors of 36 ?
Solution:
864 is divisible by 2 and 3.
∴ 864 is divisible by 6.
∴ 864 is divisible by 36 [ ∵ 6 is the factor of 36]
∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18. 36.


∴ 864 is divisible by all the factor of 36.

Question 4.
Check whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?
Solution:
756 is divisible by 2 and 3.
∴ 756 is divisible by 6.
2a + 3b + c = 2 x 7 + 3 x 5 + 6 = 14 + 15 + 6 → \(\frac { 35 }{ 7 }\) (R = 0)
∴ 756 is divisible by 7.
∴ 756 is divisible by 42. [ ∵ 6, 7 are the factors of 42]
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42.


∴ 756 is divisible by all the factor of 42.

Question 5.
Check whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by product of 11 and 7?
Solution:

Question 6.
Check whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?
Solution:

Question 7.
Check whether 456 and 618 are divisible by 6’? Also check whether 6 divides the sum of 456 and 618 ‘?
Solution:

Question 8.
Check whether 876 and 345 are divisible by 3. Also check whether 3 divides the difference of 876 and 345?
Solution:

Number	Divisible by 3	Y/N	Difference is divisible by 3 Y/N
876	8 + 7 + 6 → \(\frac { 21 }{ 3 }\) (R = 0)	Yes	876 – 345 = 531
345	3 + 4 + 5 → \(\frac { 12 }{ 3 }\) (R = 0)	Yes	The difference of 876, 345 is divisible by 3.
531	5 + 3 + 1 → \(\frac { 9 }{ 3 }\) (R = 0)	Yes

Question 9.
Check whether 2² + 2³+2⁴ is divisible by 2 or 4 or by both 2 and 4’?
Solution:

∴ 2² + 2³+2⁴ is divisible by both 2 and 4.

Question 10.
Check whether 32² is divisible by 4 or 8 or by both 4 and 8’?
Solution:

32² is divisible by 4 and 8

Question 11.
If A679B is a 5-dit number is divisible by 72 find ‘A’ and ‘B”?
Solution:
If A679B is divisible by 72 then it should be divisible by 8 and 9.
[ ∵ 8, 9 are the factors of 72]
A679B is divisible by 9 then
A + 6 + 7 + 9 + B = A + B + 22 = 27 (= 9 x 3)
=A + B = 5 ……………. (1)
A679B → \(\frac{79 \mathrm{~B}}{8}\) [From B (2,4,6,8) we take B = 2]
= \(\frac{792}{8}\) (R = 0)
∴ B = 2
From (1) ⇒ A + 2 = 5
∴ A = 3, B = 2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.3

Question 1.
Check whether the given numbers are divisible by ‘6’ or not?
(a) 273432
(b) 100533
(c) 784076
(d) 24684
Solution:
if a number is divisible by ‘6’, it has to be divisible by 2 and 3.

Question 2.
Check whether the given numbers are divisible by ‘4’ or not?
(a) 3024
(b) 1000
(c) 412
(d) 56240
Solution:

Number	Divisible by 4	Yes/No
a) 3024	3024 → \(\frac { 24 }{ 4 }\) (R = 0)	Yes
b) 1000	1000 → \(\frac { 0 }{ 4 }\) (R = 0)	Yes
c) 412	412 →  \(\frac { 12 }{ 4 }\) (R = 0)	Yes
d) 56240	56240 →  \(\frac { 40 }{ 4 }\) (R = 0)	Yes

Question 3.
Check whether the given numbers are divisible by ‘8’ or not?
(a) 4808
(b) 1324
(c) 1000
(d) 76728
Solution:

Number	Divisible by 4	Yes/No
a) 4808	4808 → \(\frac { 808 }{ 8 }\) (R = 0)	Yes
b) 1324	1324 → \(\frac { 324 }{ 8 }\) (R ≠ 0)	No
c) 1000	1000 →  \(\frac { 0 }{ 8 }\) (R = 0)	Yes
d) 76728	76728 →  \(\frac { 728 }{ 8 }\) (R = 0)	Yes

Question 4.
Check whether the given numbers are divisible by ‘7’ or not?
(a) 427
(b) 3514
(e) 861
(d) 4676
Solution:

Question 5.
Check whether the given numbers are divisible by ‘11’ or not?
(a) 786764
(b) 536393
(c) 110011
(d) 1210121
(e) 758043
(f) 8338472
(g) 54678
(h) 13431
(i) 423423
(j) 168861
Solution:

Question 6.
If a number is divisible by ‘8’, then it also divisible by ‘4’. also Explain?
Solution:
If a number is divisible by 8 it ¡s also divisible by 4.
∴ If a number is divisible by 8, then it ¡s also divisible by the factors of 8.
Factors of 8 = 1, 2, 4, 8.
∴ The number which is divisible 8, is also divisible by 4.

Question 7.
A 3-digit number 4A3 is added to another 3-digit number 984 to give four digit number 13B7, which is divisible by 11. Find (A + B).
Solution:
The given 3 – digited numbers are = 4A3, 984
∴ 4A3 + 984 = 13B7. If It is divisible by 11 then,
⇒ 1 3 B 7
(1 + B) – (3 + 7)
⇒ (B+1) – 10 = 0 ⇒ B – 9 = 0
∴ B = 9

⇒ A + 8 = 9 ⇒ A = 9 – 8 = 1
∴ A = 1
A + B= 1+9
∴ A + B = 10

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions and Answers.

8th Class Maths 8th Lesson Exploring Geometrical Figures InText Questions and Answers

Do this

Question 1.
Identify which of the following pairs of figures are congruent.     [Page No. 184]

Answer:
The congruent figures are (1, 10), (2, 6, 8), (3, 7), (12, 14), (9, 11), (4, 13).

Question 2.
Look at the following pairs of figures and find whether they are congruent. Give reasons. Name them.    [Page No. 185]

Answer:
i) △ABC, △PQR
∠A = ∠Q Angle
There is no information about other angles (or) sides.
But if we overlap each other, they coincide.
∴ △ABC ≅ △PQR

ii) From △PLM, △QNM
PL = QN (S)
LM = MN (S)
PM = QM (S)
By S.S.S congruency, these two triangles are congruent.
∴ △PLM ≅ △QNM

iii) From △LMN, △PQR
NL ≠ PQ,LM ≠ QR, NM ≠ RP [∵ The corresponding angles are not given]
∴ △LMN ≆ △PQR

iv) From fig. ABCD is a parallelogram and LMNO is a rectangle.
In any case a rectangle and a parallelogram are not congruent.
∴ ▱ ABCD ≆ □ DLMNO

v) Both the circles are having same radii,
i.e., r₁ = r₂ = 2 units
∴ The given circles are congruent to each other.

Question 3.
Identify the out line figures which are similar to those given first.    [Page No. 186]

Answer:
The similar figures are (a) (ii), (b) (ii).

Question 4.
Draw a triangle on a graph sheet and draw its dilation with scale factor 3. Are those two figures are similar?      [Page No. 191]
Answer:
Step – 1: Draw a △ PQR and choose the center of dilation C which is not on the triangle. Join every vertex of the triangle from C and produce.

Step – 2: By using compasses, mark three points P’, Q’ and R’ on the projections
so that
CP’ = k(CP) = 3CP
CQ’ = 3 CQ
CR’ = 3 CR

Step- 3: Join P’Q’,Q’R’and R’P’.
Notice that △P’Q’R’ ~ △PQR

Question 5.
Try to extend the projection for any other diagram and draw squares with scale factor 4, 5. What do you observe? [Page No. 191]
Answer:
Sometimes we need to enlarge 10 the figures say for example while making cutouts, and sometimes we reduce the figures during designing. Here in every case the figures must be similar to the original. This means we need to draw enlarged or reduced similar figures in daily life. This method of drawing enlarged or reduced similar figure is called ‘Dilation’.

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
Every vertex A, B, C, D are joined from the sign ‘O’ and produced to 4 times the length upto A, B, C and D respectively. Then A, B, C, Dare joined to form a square which 4 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and
\(\frac{OA’}{OA}\) = \(\frac{4}{1}\) = 4 is called scale factor.

Question 6.
Draw all possible lines of symmetry for the following figures.     [Page No. 193]

Answer:

Try these

Question 1.
Stretch your hand, holding a scale in your hand vertically and try to cover your school building by the scale (Adjust your distance from the building). Draw the figure and estimate height of the school building.      [Page No. 189]
Answer:
Illustration: A girl stretched her arm towards a school building, holding a scale vertically in her arm by standing at a certain distance from the school building. She found that the scale exactly covers the school building as in figure. If we compare this illustration with the previous example, we can say that

By measuring the length of the scale, length of her arm and distance of the school building, we can estimate the height of the school building.

Question 2.
Identify which of the following have point symmetry.     [Page No. 196]
1.

2. Which of the above figures are having symmetry ?
3. What can you say about the relation between line symmetry and point symmetry?
Answer:
1. The figures which have point symmetry are (i), (ii), (iii), (v).
2. (i), (iii), (v).
3. Number of lines of symmetry = Order of point symmetry.

Think, discuss and write

Question 1.
What is the relation between order of rotation and number of axes of symmetry of a geometrical figure?     [Page No. 195]
Answer:
The line which cuts symmetric figures exactly into two halves is called line of symmetry. The figure is rotated around its central point so that it appears two or more times as original. The number of times for which it appears the same is called the order of rotation.

From the above table number of lines of symmetry = Number of order of rotation.

Question 2.
How many axes of symmetry does a regular polygon has? Is there any relation between number of sides and order of rotation of a regular polygon?      [Page No. 195]
Answer:
Number of sides of a regular polygon are n. Then its lines of symmetry are also n.

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

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

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

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

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

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

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

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

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

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

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

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

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

Question 5.

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

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

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

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

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

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

Think, Discuss and Write

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

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

Answer:

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

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