Category: Class 8

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 15 Playing with Numbers Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.2

Question 1.
If 345 A 7 is divisible by 3,supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
∴ 345A7 ⇒ 3 + 4 + 5 + A + 7 = 19 + A
19 + A = 3 x 7
⇒ A = 21 – 19 = 2 ⇒ A = 24 – 19 = 5

A + 19 = 3 x 8
⇒ A = 24 – 19 = 5

A + 19 = 3 x 9
⇒ A = 27 – 19 = 8

∴ A = {2,5,8}

Question 2.
If 2791 A,is divisible by 9, supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
∴ 2791A = 2 + 7 + 9 + 1 + A = 9 x 3
⇒ 19 + A = 9 x 3 = 27
⇒ A = 27 – 19 = 8
∴ A = 8

Question 3.
Write some numbers which are divisible by 2,3,5,9 and 10 also.
Solution:
90, 180, 270. are divisible by 2, 3, 5, 9 and 10.
[∵ The L.C.M. of 2, 3, 5, 9, 10 is 90]

Question 4.
2A8 is a number divisible by 2, what might be the value of A’?
Solution:
If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.
∴ 2A8 is divisible by 2 for any value of A.
∴ A = (0, 1, 2 ………………….9)

Question 5.
50B is a number divisible by 5, what might be the value of B?
Solution:
Given number is 50B.
The units digit of a number ¡s either ‘0’ or 5, then it is divisible by 5.
∴ 500 → \(\frac { 0 }{ 5 }\) (R = 0)
505 → \(\frac { 5 }{ 5 }\) (R = 0)
∴ B = {0, 5}

Question 6.
2P is a number which is divisible by 2 and 3, what is the value of P
Solution:
The given number is 2P.
If 2P is divisible by 2, 3 then 2P should be a multiple of 6. [ ∵ L.C.M. of 2, 3 is 6]
∴ 2P = 24, 30 ………….
24 → 2 + 4 → \(\frac { 6 }{ 3 }\) (R = 0)
∴ P = 4

Question 7.
54Z leaves remainder 2 when divided by 5 , and leaves remainder 1 when divided by 3, what is the value of Z’?
Solution:
If 54Z is divisible by 3 then the sum of the digits of the number is divisible by 3.
According to problem 54Z is divisible by 3 and leaves remainder 1’.
∴ 5 + 4 + Z = (3 x 4) + 1
= 9 + Z = 13
∴ Z = 4(or)
9 + Z = (3 x 5) + 1
9 + Z = 16
Z = 7
If 54Z is divisible by 5 then Z should be equal to either ‘0’ or ‘5’.
∴ 54(0 + 2) = 542 (Z = 2)
54(0 + 7) = 547 (Z = 7)
∴ From the above two cases
Z = 7
∵ 547 → \(\frac{7}{5}\)(R = 2)

Question 8.
27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?
Solution:
27Q is divided by 5 gives the remainder 3
Le.,27Q = 27 (0 + 3) = 273(Z = 3)(T)
= 27 (0 + 8) = 278 (Z = 8)
27Q is divided by 2 gives the remainder 1.
i.e., 27Q = 27(0 + 1) = 271 (Z = 1)
27Q = 27 (0 + 3) = 273 (Z = 3) (T)
∴ From above situations Z = 3
∴ 27Q = 273→ 2 + 7 + 3 → \(\frac{12}{3}\)(R = 0)
∴ 273 is divisible by 3 and gives the remainder 0’.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.1

Question 1.
Using divisibility rules, fmd which of the following numbers are divisible by 2,5,10 ( say
yes or no ) in the given table. What do you observe?

Solution:

Question 2.
Using divisibility tests, determine which of following numbers are divisible by 2
(a) 2144 (b) 1258 (c) 4336 (d) 633 (e) 1352
Solution:
If a number is divisible by 2 then the units digit of the number be 0, 2, 4, 6, 8.
∴ a) 2144, b) 1258, c) 4336 e) 1352 are divisible by ‘2’.

Question 3.
Using divisibility tests, determine which of the following numbers are divisible by 5
(a) 438750 (b) 179015 (c) 125 (d) 639210 (e) 17852
Solution:
If a number is divisible by 5 its units digit be either ‘0’ or 5.
∴ a) 438750, b) 179015, c) 125 d) 639210 are divisible by 5.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 10:
(a) 54450 (b) 10800 (c) 7138965 (d) 7016930 (e) 10101010
Solution:
If a number is divisible by 10 then its units digit must be 0’.
a) 54450, b) 10800, d) 7016930, e) 1010100 are divisible by 10.

Question 5.
Write the number of factors of the following’?
(a) 18 (b) 24 (e) 45 (d) 90 (e) 105
Solution:

Number	Factors                              ,	No.of factors
a) 18	1,2,3,6,9,18	6
b) 24	1, 2, 3, 4, 6, 8, 12, 24	8
c) 45	1,3,5,9,15,45	6
d) 90	1, 2, 3, 5, 6, 9,10, 15, 18, 30, 45, 90	12
e) 105	1,3, 5, 7,15,21,35,105	8

Question 6.
Write any 5 numbers which are divisible by 2,5 and 10.
Solution:
10, 20, 30, 40. are divisible by 2, 5 and 10
[∵ The L.C.M. of 2, 5, 10 is 10]

Question 7.
A number 34A is exactly divisible by 2 and leaves a remainder 1, when divided by 5, find A.
Solution:
If 34A Is divisible by 2 then the remainder should be equal to 0.
∴ A should be equal to 0, 2, 4, 6, 8.
∴ 340, 342, 344, 346, 348 are divisible by 2 and gives the remainder ‘0’.
Among these 346 is divisible by 5 and gives the remainder 1.
∴ 346 → \(\frac{6}{5}\) (R = 1)
∴ The value of A = 6.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volumes Exercise 14.2

Question 1.
Find the volume of the cuboid whose dimensions are given below.

Solution:

Question 2.
Find the capacity of the tanks with the following internal dimensions. Express the capacity in cubic meters and litres for each tank.

Solution:

Question 3.
What will happen to the volume of a cube if the length of its edge is reduced to half? Is the volume get reduced? If yes, how much?
Solution:
Volume of a cube of side (s) is V₁ = a³
If the length of the side is reduced by half then
s = \(\frac{\mathrm{a}}{2}\)
∴ Volume of cube (V₂ ) = s³

∴ V₂ = \(\frac { 1 }{ 8 }\) × V₁
∴ V₁ = 8V₂

Question 4.
Find the volume of each of the cube whose sides are.
(i) 6.4 cm
(ii) 1.3 m
(iii) 1.6 m.
Solution:
Volume of a cube(V) = a³

i) a = 6.4 cm
ii) a = 1.3 m
iii) a = 1.6 m

V = (6.4)³
Volume of a cube (V) = a³
= 6.4 × 6.4 × 6.4
= 262.144 cm³

V = (1.3)³
= 1.3 × 1.3 × 1.3
= 2.197 m³

V = (1.6)³
= 1.6 × 1.6 × 1.6
= 4.096 m³

Question 5.
How many bricks will be required to build a wall of 8 m long, 6m height and 22.5 cm thici if each brick measures 25 cm by 11.25 cm by 6 cm?
Solution:
The volume of a wall of measures
8 m × 22.5 cm × 6 m
(V₁) = l₁b₁h₁
= 8 m × 22.5 cm × 6 m
= 800 cm × 22.5 cm × 600 cm
The volume of a brick each measures
25 cm × 11.25 cm × 6 cm
(V₂) = l₂b₂h₂
= 25 × 11.25 × 6 cm³
∴ The no.of bricks will be required

= 32 × 2 × 100 = 6400

Question 6.
A cuboid is 25 cm long, 15 cm broad, and 8 cm high . How much of its volume will differ from that of a cube with the edge of 16 cm’?
Solution:
Volume of a cuboid (V₁) of measures
= 25 cm, b = 15 cm, h = 8 cm.
V₁ = 25 × 15 × 8 = 3000 cm³
Volume of a cube of measure side (s) = 16 cm is
V₂ = (s)³ = (16)³ = 16 × 16 × 16
= 4096 cm³
The difference between their volumes
= V₂ – V₁
= 4096 – 3000
= 1096 cm³

Question 7.
A closed box is made up of wood which is 1cm thick .The outer dimensions of the box is 5 cm × 4 cm × 7 cm. Find the volume of the wood used.
Solution:
The volume of a box formed with outer measures 5 cm × 4 cm × 7 cm
V₁ = l × b × h
= 5 × 4 × 7
∴ V₁ = 140 cm³
Inner measures
= l – 2w, b – 2w, h – 2w
= (5 – 2 × 1), (4 – 2 × 1), (7 – 2 × 1)
= (5 – 2), (4 – 2), (7 – 2)
= 3 cm, 2 cm, 5 cm
∴ Volume of a box formed with inner measures
V₂ = (l – 2w)(b – 2w)(h – 2w)
= 3 × 2 × 5 = 30 cm³
∴ The volume of wood used = V₁ – V₂
= 140 – 30 = 110 cm³

Question 8.
How many cubes of edge 4cm, each can be cut out from cuboid whose length, breadth and height are 20 cm, 18 cm and 16 cm respectively
Solution:
The volume of a cuboid formed with the measures 20 cm × 18 cm × 16 cm
(V₁) = l₁b₁h₁ = 20 × 18 × 16
Volume of a cube (V₂) = s³
s = 4 cm (given)
∴ V₂ = (s)³ = (4)³= 4 × 4 × 4 cm³
∴ No.of cubes are required
= \(\frac{V_{1}}{V_{2}}=\frac{20 \times 18 \times 16}{4 \times 4 \times 4}\)
= 90

Question 9.
How many cuboids of size 4 cm × 3 cm × 2 cm can be made from a cuboid of size 12 cm x 9cm x 6cm?
Solution:
Volume of a cuboid of measures 12 cm × 9 cm × 6 cm
V₁ = l × b × h = 12 × 9 × 6
Volume of the smaller cuboid of measures 4 cm × 3 cm × 2 cm
V₂ = l₂b₂h₂ = 4 × 3 × 2
∴ No.of cuboids are made
= \(\frac{V_{1}}{V_{i}}=\frac{12 \times 9 \times 6}{4 \times 3 \times 2}\) = 27

Question 10.
A vessel in the shape of a cuboid is 30 cm long and 25 cm wide. What should be its height to hold 4.5 litres of water ?
Solution:
Length of a cuboidal vessel (l) = 30 cm
breadth (b) = 25 cm
height (h) = ?
The volume of water m a cuboidal vessel = 4.5 Lts.
= 4.5 × 1000 cm³
= 4500 cm³
∴ l × b ×h = 4500
⇒ 30 × 25 × h = 4500
⇒ h = \(\frac{4500}{30 \times 25}\)
∴ h = 6 cm
∴ Height of the vessel (h) = 6 cm

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

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

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

Do this

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

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

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

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

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

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

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

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

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

ii) Sale of tickets for drama in an auditorium.

Answer:
i)

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

ii)

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

Try these

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

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

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

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

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

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

A.M in general method = \(\frac{\text { Sum of the observations }}{\text { No. of the observations }}\)

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

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

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

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

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

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

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

Answer:

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

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

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

Think, discuss and write

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution:

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

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

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

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

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

Solution:

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

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

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

Question 7.
Complete the following table

Solution:

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

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

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

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

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

(iii) Write the names of the objects given below

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

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

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

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