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AP Board Class 7 Maths Chapter 11 Exponents InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 11 Exponents InText Questions Book Answers

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AP Board Class 7 Maths Chapter 11 Exponents InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 11 Exponents InText Questions Book Answers

AP Board Class 7th Maths Chapter 11 Exponents InText Questions Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents InText Questions Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 7th Maths Chapter 11 Exponents InText Questions books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents InText Questions textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 7th Maths Chapter 11 Exponents InText Questions book solutions pdf online from this page.

Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents InText Questions Textbooks Solutions PDF

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Andhra Pradesh State Board Class 7th Maths Chapter 11 Exponents InText Questions Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	7th
Subject	Maths
Chapters	Maths Chapter 11 Exponents InText Questions
Provider	Hsslive

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AP Board Class 7th Maths Chapter 11 Exponents InText Questions Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 7th Maths Chapter 11 Exponents InText Questions Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Do This

Question 1.
Write the following in exponential form, (values are rounded off) (Page No. 212)
i) Total surface area of the Earth is 510,000,000 square kilometers.
Solution:
51 × 10⁷ = 3× 17 × 10⁷

ii) Population of Rajasthan is approximately 7,00,00,000.
Solution:
7 × 10⁷

iii) The approximate age of the Earth is 4550 million years.
Solution:
4550 millions = 4550 × 10,00,000 (v 1 million =10 lakhs)
= 455 × 10⁷ = 91 × 5 × 10⁷ = 5 × 7 × 13 × 10⁷

iv) 1000 km in meters.
Solution:
1 km = 1000 m
∴ 1000 km = 1000 × 1000 m = 10⁶

Question 2.
Express (i) 48951 (ii) 89325 in expanded form using exponents. (Page No. 212)
Solution:
i) 48951 = (4 × 10000) + (8 × 1000) + (9 × 100) + (5 × 10) + (1 × 1)
= (4 × 104) + (8× 103) + (9 × 102) + (5 × 1.0) + (1 × 1)

ii) 89325 = (8 × 10000) + (9 × 1000) + (3 × 100) + (2 × 10) + (5 × 1)
= (8 × 10⁴) + (9 × 10³) + (3 × 10²) + (2 × 10) + (5 × 1)

Question 3.
Is 3² equal to 2³ ? Justify. (Page No. 213)
Solution:
3² ≠ 2³
Since 3² = 3 × 3 = 9 and 2³ = 8
∴ 3² ≠ 2³

Question 4.
Write the following numbers in exponential form. Also state the
a) base b) exponent and c) how it is read.
i) 32 ii) 64 iii) 256 iv) 243 v) 49 (Page No. 213)
Solution:
i) 32 = 2 × 2 × 2 × 2 × 2 = 2⁵
Base = 2; exponent = 5; read as 2 raised to the power 5.
ii) 64 = 2 × 2× 2 × 2 × 2 × 2 = 2⁶
Base = 2; exponent = 6 and we read it as 2 raised to the power 6.
iii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁸
Base = 2, exponent = 8 and we read it as 2 raised to the power 8.
iv) 243 = 3 × 3 × 3 × 3 × 3 = 3⁵
Base = 3; exponent = 5 and we read it as 3 raised to the power 5.
v) 49 = 7 × 7 = 7²
= 7 is the base ; exponent = 2.

Question 5.
Write the expanded form of the following. (Page No. 213)
i) p⁷ ii) l⁴ iii) s⁹ iv) d⁶ v) z⁵
Solution:
i) p⁷ = p × p × p × p × p × p × p
ii) l⁴ = l × l × l × l
iii) s⁹ = s × s × s × s × s × s × s × s × s
iv) d⁶ = d × d × d × d × d × d
v) z⁵ = z × z × z × z × z

Question 6.
Write the following in exponential form. (Page No. 213)
i) a × a × a × ………………….l’ times
ii) 5 × 5 × 5 × 5 × ……………..’n’ times
iii) q × q × q × q × q ………………….15 times
iv) r × r × r × ………………….’b’ times
Solution:
i) a × a × a × ………………….’l’ times = a^l
ii) 5 × 5 × 5 × 5 × ……………..’n’ times = 5ⁿ
iii) q × q × q × q × q …………….15 times = q¹⁵
iv) r × r × r × ……………..’b’ times = r^b

Do This

Question 1.
Find the values of 2⁴, 2³ and 2⁷ and verify whether 2⁴ × 2³ = 2⁷. (Page No. 215)
2⁴ = 2 × 2 × 2 ×2 = 16;
2³ = 2 × 2 x 2 = 8
2⁷ = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
2⁴ × 2³ = 16 × 8 = 128 = 2⁷
2⁴ × 2³ = 2⁷

Question 2.
Find the values of 5², 5³ and 5⁵ and verify whether 5² × 5³ = 5⁵. (Page No. 215)
Solution:
5² = 5 × 5 = 25;
5³ = 5 × 5 × 5 = 125 and 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125
Now 52 × 53 = 25 × 125 = 3125 = 55
∴ 5² × 5³ = 5⁵

Question 3.
Simplify the following using the formula a^m × aⁿ = a^{m + n}(Page No. 216)
i) 3¹¹ × 3⁹ ii) p⁵ × p⁸
Solution:
i) 3¹¹ × 3⁹ = 3¹¹⁺⁹ = 3²⁰
ii) p⁵ × p⁸ = p⁵⁺⁸ = p¹³

Question 4.
Find the appropriate number in place of the symbol’?’in the following. (Page No. 216)
Let ‘k’ be any non-zero integer.
i) k³ × k⁴ = k^?
Solution:
i) k³ × k⁴ = k^?
as k³ × k⁴ = k³⁺⁴ = k⁷ the value of ‘?’ = 7

ii) k¹⁵ × k^? = k³¹
as k¹⁵ × k^? = k^15+?
but k^{15 + ?} = k³¹
Since bases are equal we equate the exponents
∴ 15 + ? = 31
(i.e„) ? = 31 – 15 = 16

Question 5.
Compute 3⁶, cube of 3² and verify whether (3²)³ = 3⁶. (Page No. 216)
Solution:
3⁶ = 3 × 3 × 3 × 3 × 3 × 3 = 729.
cube of 3² = (3²)³ = 9³ = 9 × 9 × 9 = 729
Now (3²)³ = 3² × 3² × 3² = 9 × 9 × 9 = 729
3⁶ = 3 × 3 × 3 × 3 × 3 × 3 = 9 × 9 × 9
(3²)³ = 3⁶

Question 6.
Simplify the following using the law a^m × b^m = (ab)^m(Page No. 218)
i) (2 × 3)⁴
ii) x^p × y^p
iii) a⁸ × b⁸
iv) (5 × 4)¹¹
Solultion:
i) (2 × 3)⁴ = 2⁴ × 3 ⁴ = (2 × 2 × 2 × 2) × (3 × 3 × 3 ×3) = 16 × 81 = 1296
ii) x^p × y^p = (x . y)^p
iii) a⁸ × b⁸ = (a.b)⁸
iv) (5 × 4)¹¹ = 5¹¹ × 4¹¹ = 5¹¹ × (2 × 2)¹¹
= 5¹¹ × 2¹¹ × 2¹¹ = (5 × 2)¹¹ × 2¹¹ = 10¹¹ × 2¹¹

Question 7.
Write the following, by using 𝐚−𝑛=1𝐚𝑛 with positive exponents. (Page No. 219)
Solution:

Question 8.
Simplify and write in the form of a^m-n or 1𝐚𝐧−𝐦
i) 138135
ii) 34314
Solution:

Question 9.
Fill the appropriate number in the box. (Page No. 222)

Solution:

Question 10.
Complete the following (Page No. 223)

Solution:

Question 11.
Write in expanded form. (Page No. 224)
i) a^-5
ii) (-a)⁴
iii) (-7)^-5
iv) (-a)^m
Solution:

Question 12.
Write in exponential form. (Page No. 224)
i) (-3) × (-3) × (-3)
ii) (-b) × (-b) × (-b) × (-b)
iii) (1−2)×(1−2)×(1−2) ………………….’m’ times
Solution:
i) (-3) × (-3) × (-3) = (-3)³
ii) (-b) × (-b) × (-b) × (-b) = (-b)⁴
iii) (1−2)×(1−2)×(1−2) ………………….’m’ times = (−12)𝑚 or (-2)^-m

Do This

Question 1.
Write the following in exponential form using prime factorization. (Page – 214)
i) 2500 ii) 1296 iii) 8000 iv)6300
Solution:
i) 2500 = 2 × 1250 = 2 × 2 × 625
= (2 × 2) × 5 × 125
= (2 × 2) × 5 × 5 × 25
= (2 × 2) × (5 × 5 × 5 × 5)
= 2² × 5⁴

ii) 1296 = 2 × 648 = 2 × 2 × 324 = 2 × 2 × 2 × 162
= (2 × 2 × 2 × 2) × 81
= (2 × 2 × 2 × 2 ) × 3 × 27
= (2 × 2 × 2 × 2 ) × 3 × 3 × 9
= (2 × 2 × 2 × 2 ) × ( 3 × 3 × 3 × 3 )
= 2⁴ × 3⁴

iii) 8000 = 2 × 4000 = 2 × 2 × 2000 = 2 × 2 × 2 × 1000
= 2 × 2 × 2 × 2 × 500
= 2 × 2 × 2 × 2 × 2 × 250
= (2 × 2 × 2 × 2 × 2 × 2) × 125
= (2 × 2 × 2 × 2 × 2 × 2) × 5 × 25
= (2 × 2 × 2 × 2 × 2 × 2) × ( 5 × 5 × 5)
= (2⁶ × 5³)

iv) 6300 = 2 × 3150 = 2 × 2 × 1575
= (2 × 2) × 3 × 525
= 2 × 2 × 3 × 3 × 175
= (2 × 2) × (3 × 3) × 5 × 35
= (2 × 2) × (3 × 3) × (5 × 5) × 7
= 2² × 3² × 5² × 7

AP Board Textbook Solutions PDF for Class 7th Maths

Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents InText Questions Textbooks for Exam Preparations

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Wednesday, October 19, 2022