Category: Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.5

Question 1.
Find the cube root of the following numbers by prime factorization method.
(i) 343
(ii) 729
(iii) 1331
(iv) 2744
Solution:

Question 2.
Find the cube root of the following numbers through estimation’?
(i) 512
(ii) 2197
(iii) 3375
(iv) 5832
Solution:
Step 1: Start making groups of three digits starting from the unit place.
i.e, \(\overline{512}\) First group is 512

Step 2: First group i.e 512 will give us the units digit of the cube root. As 512 ends with 2, then its cube root ends with 8 (2 x 2 x 2) So the units place of the cube root will be 8.

Step 3: Now take the second group i.e. 0. Which is 0³ < 1 < 2³.
So the least number is ‘0′.
∴ Tens digit of a cube root of a number be 0.
∴ \(\sqrt[3]{512}\) = 08 = 8

(ii) 2197
Step 1: Start making groups of three digits starting from the unit place.

Step 2: First group i.e., 197 will give us the units digit of the cube root.
As 197 ends with 7, its cube root ends with 3. ‘
[∵ 3 x 3 x 3 = 27]
∴ Its units digit is 7.

Step 3: Now take the second group i.e.,2
We know that i3 < 2 < 2
∴ The least number be 1.
∴ The required number is 13.
∴ \(\sqrt[3]{2197}=\sqrt[3]{13 \times 13 \times 13}=\sqrt[\not]]{(13)^{8}}\)
= 13

(iii) 3375
Step 1: Start making groups of three digits starting from the unit place.
i.e.;

Step 2: First group is 375. Its units digit is 5.
∴ The cube root is also ends with 5.
∴ The units place of the cube root will be 5.

Step 3: Now take the second group,
i.e., 3 we know that 1³ < 3³ <2³
∴ The least number is 1.
∴ The tens digit of a cube root will be 1.
∴ The required number = 15
\(\sqrt[3]{3375}=\sqrt[3]{15 \times 15 \times 15}=\sqrt[\not]{15^{\not 3}}=15\)

(iv) 5832
Step 1: Start making groups of three digits starting from the unit place.

Step 2: The units digit of 832 is 2.
∴ The cube root of the number ends with units digit 8.
[∵ 8 x 8 x 8 = 512]

Step 3: In the second group i.e., 5 lie between 1 and 6
i.e., 1³ < 5 < 2³
∴ The tens digit of a number will bel.
∴ The required number is 18.
∴ \(\sqrt[3]{5832}=\sqrt[3]{18 \times 18 \times 18}=\sqrt[3]{(18)^{3}}\)
= 18

Question 3.
State true or false?
(i) Cube of an even number is an odd number
(ii) A perfect cube may end with two zeros
(iii) If a number ends with 5, then its cube ends with 5
(iv) Cube of a number ending with zero has three zeros at its right
(v) The cube of a single digit number may be a single digit number.
(vi) There is no perfect cube which ends with 8
(vii) The cube of a two digit number may be a three digit number.
Solution:
(i) Cube of an even number is an odd number (F)
(ii) A perfect cube may end with two zeros (F)
(iii) If a number ends with 5, then its cube ends with 5. (T)
(iv) Cube of a number ending with zero has three zeros at its right. (T)
(v) The cube of a single digit number may be a single digit number. (F)
(vi) There is no perfect cube which ends with 8(F)
(vii) The cube of a two digit number may be a three digit number. (F)

Question 4.
Find the two digit number which is a square number and also a cubic number.
Solution:
The two digited square and cubic
number is 64
∴ 64 = 8 x 8 = 8² ⇒ \(\sqrt{64}\) = 8
64 = 4 x 4 x 4 = 4³ ⇒ \(\sqrt[3]{64}\) = 4

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.4

Question 1.
Find the cubes of the following numbers
(1) 8
(ii) 16
(iii) 21
(iv) 30
Solution:

Number	Cube Of a Number
i) 8	83 =  8 × 8 × 8 = 512
ii) 16	163 = 16 × 16 × 16 = 4096
iii) 21	213 = 21 × 21 × 21 = 9261
iv) 30	303 = 30 × 30 × 30 = 27000

Question 2.
Test whether the given numbers are perfect cubes or not.
(i) 243
(ii) 516
(iii) 729
(iv) 8000
(v)2700
Solution:

Number	Cube Of a Number	Yes / No
i) 243	3 × 3 × 3 × 3 × 3 = 35	No
ii) 516	2 × 2 × 3 × 43	No
iii) 729	9 × 9 × 9 = 93	Yes
iv) 8000	20 × 20 × 20 = (20)3	Yes
v) 2700	(30) × (30) × 3	No

Question 3.
Find the smallest number by which 8788 must be multiplied to obtain a perfect cube?
Solution:
The prime factorisation of 8788
= (2 × 2) × (13 × 13 × 13)

∴ From the above product 2 ¡s left in the triplet.
∴ 2 should be multiplied with 8788 we will get a perfect cube number.

Question 4.
What smallest number should 7803 be multiplied with so that the product becomes a perfect cube?
Solution:
The prime factorisation of 7803
= (3 × 3 × 3) × (17 × 17)

∴ From the above product 17 is left in
the triplet.
∴ 17 should be multiplied to 7803 then we will get a perfect cube number.

Question 5.
Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube’?
Solution:
The prime factorisation of 8640
= (2 × 2 × 2) × (2 × 2 × 2) × 5 × (3 × 3 × 3)
= (2)³ × (2)³ × 5 × (3)³

Question 6.
Ravi made a cuboid of plasticine ofdimensions 12cm, 8cm and 3cm. How many minimum number of such cuboids will be needed to form a cube’?
Solution:
The volume of a plasticine cuboid
= l × b × h
= 12 × 8 × 3
= 288 cm³
If the minimum no. of such cuboids will be needed to form a cube then its volume be less than 288 i.e., 216 cm³
∴ s³ = 216
s = \(\sqrt[3]{216}=\sqrt[3]{6 \times 6 \times 6}=\sqrt[3]{6^{3}}\) = 6
∴ The side of the cube 6 cm

Question 7.
Find the smallest prime number dividing the sum 3¹¹ +5¹³.
Solution:
The units digit in 3¹¹ is 7

∴ The units digit in 3¹¹ is 7
The units digit in 5¹³ is 5
7 + 5 = 12 is divided by a smallest prime number 2.
∴ The smallest prime number that divide the sum 311 + 513 = 2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.3

Question 1.
Find the square roots of the following numbers by division method.
(i) 1089
(ii) 2304
(iii) 7744
(iv) 6084
(v) 9025
Solution:
(i) 1089

(ii) 2304

(iii) 7744

(iv) 6084

(v) 9025

Question 2.
Find the square roots of the following decimal numbers.
(i) 2.56
(ii) 18.49
(iii) 68.89
(iv) 84.64
Solution:
(i) 2.56

(ii) 18.49

(iii) 68.89

(iv) 84.64

Question 3.
Find the least number that is to be subtracted from 4000 to make it perfect square
Solution:
Square root of 4000 by
Division Method:

∴ The least number 31 should be subtracted from 4000 we will get a perfect square number4
∴ 4000 – 31 = 3969
= \(\sqrt{3969}=\sqrt{63 \times 63}\) = 63

Question 4.
Find the length of the side of a square whose area is 4489 sq.cm.
Solution:
Area of a square (A) = 4489 sq.cms
A = s²
s² = 4489
s = \(\sqrt{4489}=\sqrt{67 \times 67}\) = 67cms.
∴ The side of a square (s) = 67cms.

Question 5.
A gardener wishes to plant 8289 plants in the form of a square and found that there were 8 plants left. How many plants were planted in each row?
Solution:
No. of plants are planted = 8289 If 8289 plants are planted in a square shape, 8 plants are left.
Then remaining plants = 8289 – 8 = 8281

∴ No. of plants for each row = 91
∴ 8281 plants are planted in a square shape then no. of plants are planted for each row = 91

Question 6.
Find the least perfect square with four digits.
Solution:
The smallest number of 4 digits = 1000

∴ 24 should be added tö 1000 then 1000 + 24 = 1024
∴ The smallest 4 digited perfect square number is 1024.
[∵ \(\sqrt{1024}\) = 32]

Question 7.
Find the least number which must be added to 6412 to make it a perfect square?
Solution:

∴ The least number 149 should be added to 6412 then we will get a perfect square number.
∴ 6412 + 149 = 6561
∴ \(\sqrt{6561}=\sqrt{81 \times 81}\) = 81

Question 8.
Estimate the value of the following numbers to the nearest whole number
(i) \(\sqrt{97}\)
(ii) \(\sqrt{250}\)
(iii) \(\sqrt{780}\)
Solution:
ï) \(\sqrt{97}\) , 97 lie between the perfect
square numbers 81 and 100.
∴ 81 <97< 100
9² < 97 < 10²
=9 < \(\sqrt{97}\) < 10
∴ \(\sqrt{97}\) Is nearest to 10.
[∵ 97 is nearest to 100]

(ii) \(\sqrt{250}\), 250 lie between the perfect square numbers 225 and 256.
∴ 225 < 250 < 256
15² < 250 < 16²
= 15 < \(\sqrt{250}\) <16
∴ \(\sqrt{250}\) is nearest to 16.
[ ∵ 250 is nearest to 256]

ii) \(\sqrt{780}\), 780 lie between the perfect
square numbers 729 and 784.
∴ 729 < 780 < 784
27² < 780 < 28²
= 27< \(\sqrt{780}\) <28
∴ \(\sqrt{780}\) is nearest to 28.
[ ∵ 780 is nearest to 784]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.2

Question 1.
Find the square roots of the following numbers by Prime factorization method.
(1) 441
(ii) 784
(iii) 4096
(iv) 7056
Solution:

Question 2.
Find the smallest number by which 3645 must be multiplied to get a perfect square.
Solution:
The prime factorization of 3645
= (3 × 3) × (3 × 3) (3 × 3) × 5
∴ Deficiency of one ‘5’ is appeared in the above product.
∴ 3645 is multiplied with 5 then we will get a perfect square.

Question 3.
Find the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.
Solution:
The prime factorization of 2400
=(2 × 2) × (2 × 2) × 2 × (5 × 5) × 3
∴ 2,3 are needed to form a pair
∴ 2 × 3 = 6
∴ 6 should be multiplied with 2400 then we will get a perfect square number.
∴ 2400 × 6 = 14400
∴ \(\sqrt{14400}\) = 120

Question 4.
Find the smallest number by which 7776 is to be divided to get a perfect square.
Solution:

The prime factorization of 7776
=(2 × 2) × (2 × 2) × 2 × (3 × 3) × (3 × 3) × 3
∴ 2, 3 are needed to form a pair
∴ 2 × 3 = 6
∴ 7776 should be divided by 6 then we will get a perfect square number.

Question 5.
1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.
Solution:
Let the no. of trees planted in a garden for each row = x say.
No. of rows in the garden = x
∴ Total no. of trees in the garden = x × x = x²
According to the sum x² = 1521
x = \(\sqrt{1521}=\sqrt{39 \times 39}\) = 39
∴ No. of trees for each row = 39
No. of rows in the garden = 39

Question 6.
A school collected ₹ 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?
Solution:
Let the no. of students in a school = x say
The (amount) fee paid by each student = ₹ x
Amount collected by all the students
= x × x = x²
According to the sum
∴ x² = 2601
x = \(\sqrt{2601}=\sqrt{51 \times 51}\) = 51
∴ x = 51
∴ No. of students in the school = 51

Question 7.
The product of two numbers is 1296. If one number is 16 times the other, find the two numbers?
Solution:
Given that the product of two numbers = 1296.
Let the second number = x say
Then first number = 16 × x = 16x
∴ The product of two numbers
= x × 16x= 16x²
According to the sum
16x² = 1296
⇒ x² = \(\frac { 1296 }{ 16 }\) = 81
⇒ x² = 81
⇒ x = \(\sqrt{8} \overline{1}=\sqrt{9 \times 9}\) = 9
⇒ x = 9
∴ The first number = 16x
= 16 × 9
=144
The second number = x = 9

Question 8.
7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium’?
Solution:
Let the number of soldiers sat in an auditorium for each row = x say
∴ No. of rows in an auditorium = x
∴ Total no. of soldiers = x × x = x²
According to the sum,
x² = 7921
x = \(\sqrt{7921}=\sqrt{89 \times 89}\) = 89
∴ No. of rowS in an auditorium = 89

Question 9.
The area of a square field is 5184 m². Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.
Solution:
Area of a square field = 5184 m²
A = s² = 5184
:. s = \(\sqrt{5184}=\sqrt{72 \times 72}\) = 72
∴ s = 72
∴ Perimeter of the square field = 4 × s
= 4 × 72
= 288 m
According to the sum,
Perimeter of a rectangular field
= Perimeter of a square field = 288 m
Let the breadth of a rectangular field
= x m say
∴ Length = 2 × x = 2 × m
∴ Perimeter of the rectangular field
= 2 (1 + b)
= 2 (2x + x)
= 2 × 3x
= 6x
∴ 6x = 288 .
x = \(\frac { 288 }{ 6 }\)
x = 48
∴ Breadth of the rectangular field
= x = 48 m
Length of the rectangular field = 2x
= 2 × 48
=96m
∴ Area of the rectangular field
= l × b
= 96 × 48
= 4608 m²

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.1

Question 1.
What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:

Number	Square of the units digIt	Units digit of a squared number
i) 39	92 = 9 x 9 = 81	1
ii) 297	72 = 7 x 7 = 49	9
Iii) 5125	52 = 5 x 5 = 25	5
iv) 7286	62 = 6 x 6 = 36	6
v) 8742	22 = 2 x 2 = 4	4

Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256
(iv) 321
(v) 600
Solution:

Number	Prime factorizatlon	Perfect square numbers Yes/No
i) 121	121 = 11 x 11 = 112	yes
ii) 136	136 = 8 x 17 = 2 x 2 x 2 x 17	No
iii) 256	256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 28 = (24)2	yes
iv) 321	321 = 3 x 107	No
v) 600	600 = 120 x 5 = 12 x 10 x 5 = 2 x 2 x 2 x 3 x 5 x 5	No

Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number

Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:

Number	Units digit of square of a number	Even / Odd
(i) 431	12 = 1	1 , odd
(ii) 2826	62 = 36	6, odd
(iii) 8204	42 = 16	6 , even
(iv)17779	92 = 81	1 , odd
(v) 99998	82 = 64	4 , even

Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214

Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)² = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n²]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 9² = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 13² = 13 x 13 = 169

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.3

Question 1.
Sudhakar borrows ₹ 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments’?
Solution:
P = 15,000
R = 9%
T = 8 years

A = ₹ 25800
∴ His monthly payment = \(\frac{25800}{8 \times 12}\)
= ₹268.75
∴ Monthly he has to pay = ₹268.75

Question 2.
A TV was bought at a price of ₹ 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.
Solution:
The C.P. of T.V = ₹ 21,000.
After 1 year its value
= 21000 – 5% of 21000
=21000 – \(\frac { 5 }{ 100 }\) × 21000
= 21000 – 1050
= ₹19,950

Question 3.
Find the amount and the compound interest on ₹ 8000 at 5% per annum, for 2 years
compounded annually.
Solution:
P = ₹8000
R = 5%
The interest is compounded every year.
Then 2 time periods wII be occurred.
∴ n = 2

∴ Amount (A) = ₹8820
C.I = A – P
= 8820 – 8000 = ₹ 820

Question 4.
Find the amount and the compound interest on ₹ 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.
Solution:
P = ₹ 6500
R = 5%
T = 1 years
∴ \(\frac{\mathrm{PTR}}{100}=\frac{6500 \times 5 \times 1}{100}\) = 325
∴ A = P + I = 6500 + 325 = 6825
∴ P = 6825
(At the begining of 2,id year A=P)
R = 6%
T = 1 year
∴ \(\frac{\mathrm{PTR}}{100}=\frac{6825 \times 6 \times 1}{100}\) = 409.5
∴ A = P + I = 6825 + 409.5
∴ Amount = ₹ 7234.50
C.I. = A – P
= ₹ 7234.50 – 6500
= ₹734.50

Question 5.
Prathibha borrows ₹47000 from a fmance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period. Find: (a) How much
amount Prathibha should repay the finance company at the end of five years. (b) her equal
monthly repayments.
Solution:
P = ₹ 47000
R = 17%
T =5 years
∴ I = \(\frac{\mathrm{PTR}}{100}=\frac{47000 \times 5 \times 17}{100}\)
= ₹ 39,950

a) Amount to be paid
A = P + I
= 47000 + 39,950
= 86950
∴ Amount to be pay = ₹ 86950

b) In monthly equal instalments she has to pay

= 149.1
= ₹ 1450 (approx)

Question 6.
The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.
Solution:
The population of Hyderabad
= 68,09,000
If every year increase in 4.7%.
Then the population of the city in 2015
= 68,09,000 ( 1 + \(\frac{4.7}{100}\) )⁴
100 J
[ ∵ P = 6809000, R = 4.7 %, n = 4(2015 -2011)]
= 68,09,000 x \(\frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100}\)
= 81,82,199

Question 7.
Find Compound interest paid when a sum of ₹ 10000 is invested for 1 year and 3 months at 8\(\frac{1}{2}\) % per annum compounded annually.
Solution:
P = ₹10,000; R = 8\(\frac { 1 }{ 2 }\) % = \(\frac { 17 }{ 2 }\)%
T = 1 year

= 50 × 17 = 850
∴ I = ₹ 850
∴ A = P + I = 10,000 + 850
A = 10,850
∴ P = 10,850; R = \(\frac { 17 }{ 2 }\)% % ; T = 3 months

= ₹ 230.50
∴ Compound Interest
= 850 + 230.50
= ₹ 1080.50

Question 8.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the
difference in amounts he would be paying after 1\(\frac{1}{2}\) years, if the interest is (i) compounded annually (ii) compounded half yearly.
Solution:
P = ₹ 80,000; R = 10%;
T = 1 year
∴ \(\frac{\mathrm{PTR}}{100}\) = \(\frac{80000 \times 10 \times 1}{100}\)
= ₹8000
∴ A = P + I = 80000 + 8000
= ₹ 88,000

Interest on 6 months :
P = 88,000 ; R = 10% ; T = 6 Months
= \(\frac { 1 }{ 2 }\) year

i) The amount to be paid after 1 year 6 months = P + I
= 88000 + 4400
A₁ = ₹ 92,400

ii) He has to pay compounded on
every 6 months in 1 \(\frac { 1 }{ 2 }\) years
∴ 3 time periods will be occurred.
∴ n = 3
R = \(\frac { 10 }{ 2 }\) = 5% P = ₹ 80,000


A₂ = ₹ 92610
∴ Difference between the amounts = A₂ – A₁ = 92610 – 92400 = ₹ 210

Question 9.
I borrowed ₹ 12000 from Prasad at 6°/o per annum simple interest for 2 years. Had
I borrowed this sum at 6% per annum compounded annually, what extra amount would
I have to pay9
Solution:
Sum borrowed from Prasad
P = ₹ 12000
T = 2 years;
R = 6%

= ₹144O
A = P + I
A₁ = P + I = 12000 + 1440
= ₹13440
12000 + 1440 , = ₹ 13440
∴ He has to pay the amount after 2 years at the rate of 6% on C.I.
P = ₹12,000; R = 6%; n = 2 years

A₂ = ₹13483.2
∴ The difference between the C.I and S.I = 13483.2 – 13440
= ₹ 43.20

Question 10.
In a laboratory the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000
Solution:
No. of bacteria in a laboratory = 5,06,000
If they are increased at the rate of 2.5% per hour then their number after 2 hours

Question 11.
Kamala borrowed ₹ 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
Solution:
Kanala borrowed from bank = ₹ 26400
Rah of interest (R) =15%
n = 2 years

After 4 rpnths the amount will be ₹ 34914
∴ P = 34914; R = 15%; T = 4 months
T = \(\frac { 4 }{ 12 }\) year
= \(\frac { 1 }{ 3 }\) year
∴ \(I=\frac{P T R}{100}=\frac{34914 \times 15 \times \frac{1}{3}}{100}\)
= ₹1745.7
∴ Kamala has to pay the amount after 2 years and 4 months to the bank = 34914 + 1745.7
= ₹36659.7

Question 12.
Bharathi borrows an amount of ₹ 12500 at 12% per annum for3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
Bharathi borrowed the sum
P = ₹12500
R = 12%
T = 3 years
S. I (I) = \(\frac { PTR }{ 100 }\)
= \(\frac{12500 \times 12 \times 3}{100}\)
= 125 × 36
= 4500
After 3 years she has to pay
(A₁)= P + I
= 12500 + 4500 .
A₁ = ₹17,000
Madhuri has to pay the amount on


A₂ = 16637.5
∴ A₁ > A₂
A₁ – A₂ = 17000 – 16637.5
= ₹ 362.5
∴ Bharathi has to pay ₹ 362.5 more than Madhuri.

Question 13.
Machinery worth ₹ 10000 depreciated by 5%. Find its value after 1 year.
Solution:
The value of machinery after 1 year on 5% depreciation

= 95 × 100
= ₹ 9500

Question 14.
Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.
Solution:
Present population of a city = 12,00,000 If its population increases at the rate of 4%, then the population after 2 years

= 120 × 104 × 104
= 12,97,920

Question 15.
Calculate compound interest on ₹ 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?
Solution:
compounded quarterly then 4 time periods will be there in 1 year.
∴ n = 4
C.I. on ₹ 1000 over a period of 1 year at
10% per annum A = P (1 + \(\frac{\mathrm{R}}{100}\) )ⁿ
P = 1000; n = 4; R = \(\frac{10}{4}=\frac{5}{2}\) %

= ₹ 1103.81
A = ₹ 1103.81
C.I. for 1 year
= 1103.81 – 1000
= ₹ 10.81

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.2

Question 1.
In the year 2012, it was estimated that there were 36.4 crore Internet users worldwide. In
the next ten years, that number will be increased by 125%. Estimate the number of Internet
users worldwide in 2022.
Solution:
Internet users in the year 2012
= 36.4 crores.
The number will be increased by next
10 years = 125%
∴ The no. of internet users in the year 2022
= 36.4 + 125% of 36.4
= 36.4 + \(\frac { 125 }{ 100 }\) × 36.4
= 36.4 + 45.5
= 81.9 crores.

Question 2.
A owner increases the rent of his house by 5% at the end of each year. If currently its rent is ₹ 2500 per month, how much will be the rent after 2 years’?
Solution:
Present house rent = ₹ 2500 If the owner increases the rent by 5% on every year then the rent of the house after 2 years

= 2500 × \(\frac { 21 }{ 20 }\) x \(\frac { 21 }{ 20 }\)
= ₹ 2756.25

Question 3.
On Monday, the value of a company’s shares was ₹ 7.50. The price increased by 6% on Tuesday, decreased by 1.5% on Wednesday, and decreased by 2% on Thursday. Find the value of each share when trade opened on Friday.
Solution:
The value of the share when trade opened on Friday
= ₹ 7.674

Question 4.
With most of the Xerox machines. you can reduce or enlarge your original by entering a percentage for the copy. Reshma wanted to enlarge a 2 cm by 4 cm drawing. She set the Xerox machine for 150% and copied her drawing. What will be the dimensions of the
copy of the drawing be’?
Solution:
Length of the copy = 2 cm
breadth = 4 cm
If the length is increase in 150% then its
measure = 150 % of 2 cm
= \(\frac { 150 }{ 100 }\) × 2 = 1.5 × 2 = 3 cm

If the breadth is increase in 150% then
its measure = 150 % of 4 cm 150
= \(\frac { 150 }{ 100 }\) × 4 = 1.5 × 4 = 6 cm
100
∴ New length = 3 cm
breadth = 6 cm

Question 5.
The printed price of a book is ₹ 150. And discount is 15%. Find the actual amount to be paid.
Solution:
The printed price of a book = ₹ 150
Discount % = 15%
∴ Discount = 15% of 150
= \(\frac { 15 }{ 100 }\) × 150 = ₹22.5
∴ The C.P. of a book = 150 – 22.5
= ₹127.50/-

Question 6.
The marked price of an gift item is ₹ 176 and sold it for ₹ 165. Find the discount percent.
Solution:
Marked price of a gift = ₹176
S.P. = 165
Discount = 176 – 165 = ₹ 11

Question 7.
A shop keeper purchased 200 bulbs for ₹10 each. However 5 bulbs were fused and put
them into scrap. The remaining were sold at ₹12 each. Find the gain or loss percent.
Solution:
The C.P. of 200 bulbs at the rate of ₹10 for each = 200 × 10 = ₹ 2000
If 5 bulbs are fused then remaining are
= 200 – 5 = 195
∴ TheS.P. of 195 bulbs at the rate of ₹12 for each = 195 × 12 = ₹ 2340
∴ S.P. > C.P.
∴ Profit = S.P.-C.P.
= 2340 – 2000 = 340
Profit = \(\frac { Profit }{ C.P }\) × 100 = \(\frac { 340 }{ 2000 }\) × 100
Profit = 17%

Question 8.
Complete the following table with appropriate entries (Wherever possible)

Solution:

Question 9.
A table was sold for ₹2,142 at a gain of 5%. At what price should it be sold to gain 10%.
Solution:
S.P. of a table = ₹ 2142
Profit = 5%

∴ The C.P. of buyyer = ₹ 2040
Profit % = 10%

∴ S.P. = ₹2244

Question 10.
Gopi sold a watch to Ibrahim at 12% gain and Ibrahim sold it to John at a loss of 5%. If John paid ₹1,330, then find how much did Gopi sold it?
Solution:
. Let Gopi’s cost price = ₹100
Gain = 12%
∴ Gopi’s selling price to Ibrahim or Ibrahim’s cost price = ₹100 + ₹12 = ₹112
∴ Ibrahim’s loss = 5%
∴ Ibrahim’s selling price =
\(112\left(\frac{100-5}{100}\right)=\frac{112 \times 95}{100}\) = ₹106.40
For ₹100 we get = ₹106.40
For ₹1330 how much we get ?

Question 11.
Madhu and Kavitha purchased a new house for ₹3,20,000. Due to some economic
problems they sold the house for ₹2,80,000. Find (a) The loss incurred (b) the loss percentage.
Solution:
C.P. of a house = ₹ 3,20,000
S.P. of a house = ₹ 2,80,000
∴ C.P. > S.P.
a) Loss
= C.P. – S.P.
= 3,20,000 – 2,80,000 = 40,000

Question 12.
A pre-owned car show-room owner bought a second hand car for ₹ 1,50,000. He spent ₹20,000 on repairs and painting, then sold it for ₹ 2,00,000. Find whether he gets profit or loss. If so, what percent?
Solution:
After repair, the C.P of a car
= 1,50,000 + 20,000 = 1,70,000
S.P. of a ear = ₹ 2,00,000
∴ S.P. > C.P.
∴ Profit = S.P.-C.P.
= 2,00,000-1,70,000= 30,000
Profit = \(\frac { Profit }{ C.P }\) × 100
\(\frac { 30,000 }{ 1,70,000 }\) × 100 = Profit = \(\frac { 300 }{ 17 }\)
Profit% = 17.64%

Question 13.
Lalitha took a parcel from a hotel to celebrate her birthday with her friends. It was billed with ₹ 1,450 including 5% VAT. Lalitha asked for some discount, the hotel owner gave 8% discount on the bill amount. Now find the actual amount that lalitha has to pay to the hotel owner
Solution:
After allowing 5% VAT, the total bill = ₹ 1450
If 8% discount is allowed on bill, then
Discount = 8% of 1450
\(\frac { 8 }{ 100}\) × 1450 = ₹116
Discount = ₹116
∴ Lalitha has to pay the bill = 1450 -116
= ₹ 1334

Question 14.
If VAT is included in the price, find the original price of each of the following.

Solution:

Question 15.
Find the buying price of each of the following items when a sales tax of 5% is added on them.
(1) a towel of ₹50 (ii) Two bars of soap at ₹35 each.
Solution:
Given that Sales tax = 5%
(i) Cost of a towel = ₹ 50
Sales Tax = 5% of 50
= \(\frac { 5 }{ 100 }\) x 50 = \(\frac { 5 }{ 2 }\) = ₹ 2.50
∴ C.P. = Net Price + Sales
Tax = 50 + 2.50 = ₹ 52.50

(ii) The cost of two soaps at the rate of
₹ 35 each = 2 × 35 = ₹ 70
Sales Tax = 5% of 70
= \(\frac { 5 }{ 100 }\) × 70 = \(\frac { 7 }{ 2 }\) = ₹ 3.50
∴ C.P. = Net Price + Sales
Tax = 70 + 3.50 = ₹ 73.50

Question 16.
A Super-Bazar prices an item in rupees and paise so that when 4% sales tax is added, no rounding is necessary because the result is exactly in ‘n’ rupees, where ‘n’ is a positive integer. Find the smallest value of ’n’.
Solution:
Let the cost price = x say
∴ If x is increased 4% sales tax is added then
x + 4% of x = n
x + \(\frac { 4 }{ 100 }\) × x = n
\(\frac { 140x }{ 100 }\) = n
x = x = n × \(\frac { 100 }{ 104 }\) = \(\frac{25 \times \mathrm{n}}{26}\)
∴ n should be a least multiple of 26, then only the value of the article should be represented in only rupees.

[∵n = 13, 26, 39. from them 13 should betaken]
:. Required value of the article
=12.50 + \(\frac { 4 }{ 100 }\) × 12.5
12.50 + 0.5 = ₹13

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.1

Question 1.
Find the ratio of the following
(i) Smita works in office for 6 hours and Kajal works for 8 hours in her office. Find the
ratio of their working hours.
Solution:
The ratio of working hours of smita and kajal = 6:8
= (2 × 3 ) : (2 × 4) = 3 : 4

(ii) One pot contains 8 litre of milk while other contains 750 milliliter.
Solution:
8lit : 750ml
8 × 1000 : 750
= \([latex]\frac { 8000 }{ 750 }\)[/latex] = 32 : 3

(iii) speed of a cycle is 15km/h and speed of the scooter is 30km/h.
Solution:
The ratio of speeds of a cycle and a sector
= 15 : 30 = (15 × 1) : 15 × 2 = 1 : 2

Question 2.
If the compound ratio of 5:8 and 3:7 is 45:x. Find the value of x.
Solution:
The compound ratio of 5:8 and 3:7
= \(\frac{5}{8} \times \frac{3}{7}=\frac{15}{56}\)
According to the sum
15 : 56 = 45 : x
∴ x = 168

Question 3.
If the compound ratio of 7:5 and 8:x is 84:60. Find x.
Solution:
The compound ratio of 7:5 and 8:x
= \(\frac{7}{5} \times \frac{8}{x}=\frac{56}{5 x}\)
According to the sum
56 : 5x = 84 : 60

∴ x = 8

Question 4.
The compound ratio of 3:4 and the inverse ratio of 4:5 is 45:x. Find x.
Solution:
The inverse ratio of 4:5 is 45 : x
The compound ratio of 3:4 and 5 : 4
= 45 : x

⇒ x = 16 × 3 = 48
∴ x = 48

Question 5.
In a primary school there shall be 3 teachers to 60 students. If there are 400 students
enrolled in the school, how many teachers should be there in the school in the same ratio?
Solution:
No. of teachers are required for 400 students at the rate of 3 teachers to 60
students are ⇒ 60 : 3 400 : x

∴ x = 20

Question 6.
In the given figure, ABC is a triangle. Write all possible ratios by A
taking measures of sides pair wise.
8cm 10cm
(Hint: Ratio of AB : BC =8 : 6)

Solution:

In ΔABC
AB : BC = 8 : 6 = 4:3
⇒ BC : AB = 6 : 8 = 3: 4
BC : CA = 6 : 10 = 3 : 5
⇒ CA : BC = 10 : 6 = 5 : 3
CA : AB = 10:8=5:4
⇒ AB : CA = 8: 10 = 4: 5

Question 7.
If 9 out of 24 students scored below 75% marks in a test. Find the ratio of student scored below 75% marks to the student scored 75% and above marks.
Solution:
Out of 24 students who got below 75% of marks = 9
Who got 75% and above marks =24 – 9 = 15
∴ The ratio between no. of students
who got less than 75% of marks and
who got 75% and above marks
= 9 : 15 =(3 × 3):(3 × 5) = 3 : 5

Question 8.
Find the ratio of number of vowels in the word’ MISSISSIPPI’ to the number of consonants in the simplest form.
Solution:
No. of vowels in the word MI S S SS! PPI = 4 (IIII)
No. of consonants in that word = 7 (MSSSSPP)
∴ The ratio between vowels and consonants = 4: 7

Question 9.
Rajendra and Rehana own a business. Rehana receives 25% of the profit in each month. If
Rehana received ₹ 2080 in particular month, what is the total profit in that month?
Solution:
Total Profit = x say
25% of x = 2080
⇒ \(\frac{25}{100}\) × x = 2080
⇒ \(\frac{x}{4}\) = 2080
⇒ x = 2080 × 4
∴ x = ₹ 8320

Question 10.
In triangle ABC, AB = 2.2 cm, BC = 1.5 cm and AC = 2.3 cm. In triangle XYZ, XY = 4.4cm, YZ = 3cm and XZ = 4.6cm. Find the ratio AB:XY, BC:YZ, AC:XZ. Are the lengths of corresponding sides of ΔABC and ΔXYZ are in proportion?
[Hint : Any two triangles are said to be in proportion, if their corresponding sides are in the
same ratio]
Solution:

∴ The corresponding sides of both the triangles are in proportion.
∴ ΔABC ~ ΔXYZ

Question 11.
Madhuri went to a super market. The price changes are as follows. The price of rice reduced by 5% jam and fruits reduced by 8% and oil and dal increased by 10%. Help Madhuri to find the changed prices in the given table.

Item	Original price/kg	Changed price
Rice	₹ 30
Jam	₹ 100
Apples	₹ 280
Oil	₹ 120
Dal	₹ 80

Solution:

Item	Original price/kg	Changed price
Rice	₹ 30	₹28.50
Jam	₹ 100	₹ 92
Apples	₹ 280	₹ 257.6
Oil	₹ 120	₹ 132
Dal	₹ 80	₹ 88

Question 12.
There were 2075 members enrolled in the club during last year. This year enrolment is
decreased by 4%.
(a) Find the decrease in enrolment.
(b) How many members are enrolled during this year?
Solution:
No. of persons are enrolled in the last year = 2075
Present year no. of persons are enrolled
= 4% less than the previous year.
a) Decrease in enrolment = 4% of 2075

b) No.of members are enrolled this
year = 2075 – 4% of 2075
=2075 – 83 = 1992

Question 13.
A farmer obtained a yielding of 1720 bags of cotton last year. This year she expects her crop to be 20% more. How many bags of cotton does she expect this year?
Solution:
During the last year yielding the bags of
cotton = 1720
If she expects 20% crop to be more then
=20% of 1720 .
= \(\frac{20}{100}\) × 1720
= 2 × 172
= 344 bags
Her expectation of total bags
= 1720 + 344
= 2064

Question 14.
Points P and Q are both in the line segment AB and on the same side of its midpoint. P divides AB in the ratio 2: 3, and Q divides AB in the ratio 3 :4. If PQ =2, then find the length of the line segment AB.
Solution:
Given that ‘C’ is the midpoint of line segment AB.
Here ‘P’ divides AB inthe ratio 2 : 3
‘Q’ divides AB in the ratio 3: 4

PQ =2 cm [Given]
PQ = QB – PB
= 4 – 3 = 1 part = 2cm
∴ AB = AQ + QB [with respect to Ql
AB = AP+ PB [with respect to P]
L.C.M. of 5, 7 parts = 35 parts
∴ Length of AB 35 parts
= 35 × 2[ ∵ part = 2cm]
= 70cm

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers Exercise 4.2

Question 1.
Express the following numbers in the standard form.
(i) 0.000000000947
Solution:
= \(\frac{947}{1000000000000}\) = 947 × 10^-12

(ii) 543000000000
Solution:
= 543 × 1000000000 = 543 × 10⁹

(iii) 48300000
Solution:
= 483 × 100000 = 483 × 10⁵

(iv) 0.00009298
Solution:
= \(\frac{9298}{100000000}\)
= 9298 × 10^-8

(v) 0.0000529
Solution:
= \(\frac{529}{10000000}\)
= 529 × 10^-7

Question 2.
Express the following numbers in the usual form.
(i) 4.37 × 10⁵
Solution:
= 4.37 × 100000
= 437000

(ii) 5.8 × 10⁷
Solution:
= 5.8 × 10000000

(iii) 32.5 × 10^-4
Solution:
= \(\frac{32.5}{10^{4}}=\frac{32.5}{10000}\)
= 0.00325

(iv) 3.71529 × 10⁷
Solution:
= 3.71529 × 10000000
= 37152900

(v) 3789 × 10^-5
Solution:
= \(\frac{3789}{10^{5}}=\frac{3789}{100000}\)
= 0.03789

(vi) 24.36 × 10^-3
Solution:
= \(\frac{24.36}{10^{3}}=\frac{24.36}{1000}\)
= 0.02436

Question 3.
Express the following information in the standard form
(i) Size of the bacteria is 0.0000004 m
Solution:
= [klatex]\frac{4}{10000000}[/latex] m = 4 x 10^-7 m

(ii) The size of red blood cells is 0.000007mm
Solution:
= \(\frac{7}{1000000}\) = 7 × 10^-6

(iii) The speed of light is 300000000 m/sec
Solution:
= 3 × 10,00,00,000 = 3 × 10⁸ m/sec

(iv) The distance between the moon and the earth is 384467000 m(app)
Solution:
= 384467 × 1000 m
= 384467 × 10³

(v) The charge of an electron is 0.0000000000000000016 coulombs
Solution:
= 0.0000000000000000016
= \(\frac{16}{10000000000000000000}\)
= \(\frac{16}{10^{19}}\)
= 16 × 10^-19 coulombs

(vi) Thickness of a piece of paper is 0.0016 cm
Solution:
= 0.0016 cm = \(\frac{16}{10000}\)
= \(\frac{16}{10^{4}}\)
= 16 × 10^-4 cm

(vii) The diameter of a wire on a computer chip is 0.000005 cm
Solution:
= 0.000005 cm = \(\frac{5}{1000000}\) cm
= \(\frac{5}{10^{6}}\) cm = 5 × 10^-6 cm

Question 4.
In a stack, there are 5 books, each of thickness 20 mm and 5 paper sheets each of thickness 0.016mm. What is the total thickness of the stack.
Solution:
The thickness of 5 books of a pack
= (5 books × their thickness)
= 5(papers × their thickness)
= (20 mm × 5) + (0.016 mm × 5)
= 100 mm + 0.080 mm
= (100 + 0.080) mm
= 100.08 mm
= 1.0008 × 10² mm

Question 5.
Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why? Justify your argument.
(i) x^-3 × x^-2 = x^-6
Solution:
x^-3 × x^-2 = x^-6
= x^-3+(-2) = x^-6 [∵ a^m × aⁿ = a^{m + n} ]
= x^-5 = x^-6
= -5 = -6(False)
∴ In this case do not agree with Rakesh solution.

(ii) \(\frac{X^{3}}{X^{2}}\) = x⁴
Solution:
⇒ x^3-2 = x⁴ [∵ \(\frac{a^{m}}{a^{n}}\) = a^{m – n} ]
⇒ x¹ = x⁴
[∵Here bases are equal, so exponents are also equal]
⇒ 1 = 4 (It is false)
∴ I do not agree with Rakesh solution.

(iii) (x²)³ = (x²)³ = x⁸
Solution:
(x²)³ = \(x^{2^{3}}\) = x⁸
⇒ (x²)³ = \(x^{2^{3}}\)
⇒ x^{2 × 3} = \(x^{2^{3}}\)
⇒ x⁶ = x^{2 × 2 × 2}
⇒ x⁶ = x⁸
[∵ Bases are equal, so exponents are also equal]
⇒ 6 = 8
It is false
∴ Rakesh solution is wrong.

(iv) x^-2 = √x
Solution:
⇒ x^-2 = x^1/2
⇒ -2 = 1/2 (it is false)
∴ I don’t agree with Rakesh solution.

(v) 3x^-1 = \(\frac{1}{3 x}\)
Solution:
⇒ 3x^-1 = \(\frac{1}{3 x}\)
⇒ 3 × 3 = \(\frac{x}{x}\)
⇒ x⁰ = 9
⇒ 1 = 9
It is false
∴ I don’t agree with Rakesh solution.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers Exercise 4.1

Question 1.
Simplify and give reasons
(i) 4^-3
(ii) (-2) ⁷
(iii) \(\left(\frac{3}{4}\right)^{-3}\)
(iv) (-3)^-4
Solution:
(i) 4^-3 \(\frac{1}{4^{3}}=\frac{1}{64}\) [∵ a^-n = \(\frac{1}{\mathrm{a}^{\prime \prime}}\)

(ii) (-2) ⁷ = -(2) ⁷ = -128
[∵ 7 is an odd number]
[∵ (-a)ⁿ = -(aⁿ) if ‘n’ is odd]

(iii) \(\left(\frac{3}{4}\right)^{-3}\) = \(\frac{3^{-3}}{4^{-3}}=\frac{4^{3}}{3^{3}}=\left(\frac{4}{3}\right)^{3}\)

(iv) (-3)^-4 = \(\frac{1}{(-3)^{4}}\) [∵a^-n = \(\frac{1}{a^{n}}\)
= \(\frac{1}{(3)^{4}}\) [∵ 4 is even ]
= \(\frac{1}{81}\)

Question 2.
Simplify the following:
(i) \(\left(\frac{1}{2}\right)^{4} \times\left(\frac{1}{2}\right)^{5} \times\left(\frac{1}{2}\right)^{6}\)
(ii) (-2)⁷ x (-2)³ x (-2)⁴
(iii) 4⁴ x \(\left(\frac{5}{4}\right)^{4}\)
(iv) \(\left(\frac{5^{-4}}{5^{-6}}\right)\) x 5³
(v) (-3) ⁴ x 7⁴
Solution:
(i) \(\left(\frac{1}{2}\right)^{4} \times\left(\frac{1}{2}\right)^{5} \times\left(\frac{1}{2}\right)^{6}\)
\(\left(\frac{1}{2}\right)^{4+5+6}=\left(\frac{1}{2}\right)^{15}=\frac{1}{2^{15}}\)
[∵ a^m x aⁿ = a^{m + n}]

(ii) (-2)⁷ x (-2)³ x (-2)⁴
(-2)^{7 + 3 + 4} = (-2) ¹⁴ = 2 ¹⁴
[∵ (-a)ⁿ = aⁿ is even]

(iii) 4⁴ x \(\left(\frac{5}{4}\right)^{4}\)
4⁴ x \(\left(\frac{5}{4}\right)^{4}\) = 5⁴
[ ∵ \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}\) ]

(iv) \(\left(\frac{5^{-4}}{5^{-6}}\right)\) x 5³
5^-4 x (5⁶ x 5³) [ ∵ \(\frac{1}{a^{-n}}=a^{n}\)
= 5^-4 x 5⁶⁺³ [ ∵ a^m x aⁿ = (a)^m+n
= 5^-4 x 5⁹
= 5^(-4)+9 = 5⁵

(v) (-3) ⁴ x 7⁴
= 3⁴ x 7⁴[.4isevennumber]
=(3 x 7)⁴ [:amxbm=(ab)m]
= (21)⁴

Question 3.
Simplify
(i) \(2^{2} \times \frac{3^{2}}{2^{-2}} \times 3^{-1}\)
(ii) (4^-1 x 3^-1) ÷ 6^-1
Solution:
(i) \(2^{2} \times \frac{3^{2}}{2^{-2}} \times 3^{-1}\)
= 2² x 2² x 3² x 3^-1
= 2²⁺² x 3^{2 + ( – 1)}
=2⁴ x 3¹ = 16 x 3 = 48

(ii) (4^-1 x 3^-1) ÷ 6^-1

Question 4.
Simplify and give reasons
(i) (4⁰ + 5^-1) x 5² x \(\frac{1}{3}\)
Solution:

(ii) \(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{4}\right)^{-3} \times\left(\frac{1}{5}\right)^{-3}\)
Solution:
= \(\left(\frac{1}{2} \times \frac{1}{4} \times \frac{1}{5}\right)^{-3}\)
= \(\left(\frac{1}{40}\right)^{-3}\) [∵ a^m x b^m x c^m = (abc)^m
= (40)³ [ ∵]\(\frac{1}{a^{-n}}\) = aⁿ ]

(iii) (2^-1 + 3^-1 + 4^-1) x \(\frac{3}{4}\)
Solution:

(iv) \(\frac{3^{-2}}{3}\) x (3⁰ – 3^-1
Solution:

(v) 1 + 2^-1 + 3^-1 + 4⁰
Solution:

(vi) \(\left[\left(\frac{3}{2}\right)^{-2}\right]^{2}\)
Solution:

Question 5.
Simplify and give reasons
(i) \(\left[\left(3^{2}-2^{2}\right) \div \frac{1}{5}\right]^{2}\)
Solution:
\(\left.\left[(9-4) \div \frac{1}{5}\right)\right]^{2}\)
= \(\left[5 \times \frac{5}{1}\right]^{2}\) = (5²)² 5⁴ = 625 [∵ (a^m)ⁿ = a^mn]

(ii) ((5²)³ x 5⁴) ÷ 5⁶
Solution:

Question 6.
Find the value of ’n’ in each of the following:
(i) \(\left(\frac{2}{3}\right)^{3} \times\left(\frac{2}{3}\right)^{5}=\left(\frac{2}{3}\right)^{\mathrm{n}-2}\)
Solution:

Here bases are equal, so exponents are
also equal.
⇒ n – 2 = 8
⇒ n = 8 + 2 = 10
∴ n = 10

(ii) (-3)ⁿ⁺¹ x (-3)⁵ = (-3)³
Solution:
⇒(-3)ⁿ⁺¹⁺⁵ = (-3)^-4 [∵ a^m x aⁿ = a^m+n ]
⇒ (-3)ⁿ⁺⁶ = (-3)^-4
⇒ n + 6 = -4
⇒ n = -4 – 6 = -10
⇒ n = -10

(iii) 7²ⁿ⁺¹ ÷ 49 = 7³
Solution:

⇒ 7^2n+1-2 = 7³ [ ∵ \(\frac{a^{m}}{a^{n}}=a^{m-n}\) ]
⇒ 7^{2n – 1}= 7³
⇒ 2n – 1 = 3
⇒ 2n = 3 + 1 = 4
⇒ n = \(\frac{4}{2}\)
∴ n = 2

Question 7.
Find ’x’ if 2^-3 = \(\frac{1}{2^{x}}\)
Solution:
2^-3 = \(\frac{1}{2^{x}}\) = 2^-x
⇒ 2^-3 = 2^-x [ \(\frac{1}{a^{n}}\) = a^-n ]
⇒ -x = -3
∴ x = 3

Question 8.
Simplify \(\left[\left(\frac{3}{4}\right)^{-2} \div\left(\frac{4}{5}\right)^{-3}\right] \times\left(\frac{3}{5}\right)^{-2}\)
Solution:

Question 9.
If m = 3 and n = 2 find the value of
(i) 9m² – 10n³
(ii) 2m² n²
(iii) 2m³ + 3n² – 5m²n
(iv) mⁿ – n^m
Solution:
1) 9m² – 10n³
= 9(3)² – 10(2)³
= 9 x 9 – 10 x8
= 81 – 80 = 1

(ii) 2m² n²
= 2(3)² (2)²
= 2 x 9 x 4 = 72

(iii) 2m³ + 3n² – 5m²n
= 2(3)³ + 3(2)² – 5(3)²(2)
= (2 x 27) + (3 x 4) – (5 x 9 x 2)
= 54 + 12 – 90
= 66 – 90 = – 24

(iv) mⁿ – n^m
= 3² – 2³
= 3 x 3 – 2 x 2 x 2
= 9 – 8 = 1

Question 10.
Simplify and give reasons \(\left(\frac{4}{7}\right)^{-5} \times\left(\frac{7}{4}\right)^{-7}\)
Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.3

Construct the quadrilateral with the measurements given below :

Question a).
Quadrilateral GOLD: OL = 7.5 cm, GL = 6 cm, LD = 5 cm, DG = 5.5 cm and OD = 10 cm.
(Ex 3.3, Page No. 72)
Solution:
In a quadrilateral GOLD, Rough Diagram
OL = 7.5 cm, GL = 6 cm
LD = 5 cm, DG = 5.5 cm, OD = 10 cm

Construction Steps:

1. Draw a line segment \(\overline{OL}\) equal to radius 7.5 cm.
2. With the centres O, L draw arcs with radius 10 cm and 5 cm respectively. These two arcs meet at point ‘D’.
3. With the centres L, D draw arcs equal to 6 cm and 5.5 cm respectively. These two arcs meet at point ‘G’.
4. Join O, G and L, G. Also join O, D and L, D and G, D.
5. ∴ The required quadrilateral GOLD is formed.

Question b).
Quadrilateral PQRS: PQ = 4.2 cm, QR = 3 cm, PS = 2.8 cm, PR = 4.5 cm and QS = 5 cm.
Solution:


In a quadrilateral PQRS
PQ = 4.2 cm PS = 2.8 cm
QR = 3 cm PR = 4.5 cm
QS = 5 cm

Construction Steps:

1. Draw a line segment \(\overline{P Q}\) with radius 4.2 cm.
2. With the centres P, Q draw arcs equal to the radius 4.5 cm, 3 cm respectively. These two arcs meet at point R’. Join P, R and Q, R.
3. With the centres Q, P draw arcs equal to the radii 5 cm and 2.8 cm respectively. These two arcs meet at point ‘S’.
4. Join P, S and Q, S and S, R.
5. ∴ The required PQRS quadrilateral is formed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Ex 3.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals Exercise 3.2

Construct quadrilateral with the measurements given below:

Question (a).
Quadrilateral ABCD with AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm andAC= 7cm
Solution:
In Quadrilateral ABCD with AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and AC = 7 cm.

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{AB}}\) with radius 4.5 cms.
2. With the centres A, B draw arcs equal to 7 cm and 5.5 cm respectively. The intersection of these two arcs keep as ‘C’.
3. Join A, C and B, C.
4. With centres C, A draw arcs equal to 4 cm, 6 cm respectively. These intersecting point is keep as ’D’.
   Join D, C and A, D.
5. ∴ The required quadrilateral ABCD is formed.

Question (b).
Quadrilateral PQRS with PQ = 3.5 cm, QR = 4 cm, RS = 5 cm, PS = 4.5 cm and QS= 6.5 cm
Solution:
In a quadrilateral PQRS,
PQ = 3.5 cm, QR = 4 cm, RS = 5 cm,
PS – 4.5 cm, QS = 6.5 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{PQ}}\) with radius 3.5 cm.
2. With the centres P, Q draw arcs equal to 4.5 cm and 6.5 cm respectively.
3. These two arcs meet at point ‘S’.
4. With the centres S, Q draw arcs with radius 5 cm, 4 cm respectively. These two arcs intersected at point ‘R’.
5. Join P, S; Q, S; S, R and Q, R.
6. ∴ The required quadrilateral PQRS is formed.

Question (c).
Parallelogram ABCD with AB = 6cm, CD = 4.5 cm and BD = 7.5 cm
Solution:
In a parallelogram ABCD; AB = 6 cm, BC = 4.5 cm, BD = 7.5 cm
AB = CD (;cm
BC = AD = 4.5 cm
BD = 7.5 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{AB}}\) with radius 6 cms.
2. With the centres A, B draw arcs with radius 4.5 cm, 7.5 cm respectively. These two arcs meet at point ‘D’.
3. With the centres D, B draw arcs with radius 6 cm, 4.5 cm respectively. These two arcs meet at point ‘C’.
4. Join A, D and B, C and D, C and B, D.
5. ∴ The required parallelogram ABCD is formed.

Question (d).
Rhombus NICE with NI = 4 cm and IE = 5.6 cm
Solution:
In a rhombus NI = IC = CE = NE = 4 cm, IE = 5.6 cm

Construction Steps:

1. Draw a line segment \(\overline{\mathrm{NI}}\) with radius 4 cm.
2. With the centres N, I draw two arcs with radius 4 cm, 5.6 cm respectively. These two arcs meet at point E’.
3. With the centres E, I draw arcs with radius 4 cm. These two arcs meet at point ‘C’.
4. Join N, E and I, E. Also join E, C and I, C.
5. .’. The required rhombus NICE is formed.