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AP Board Class 7 Maths Chapter 9 Algebraic Expressions Unit Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Book Answers

AP Board Class 7 Maths Chapter 9 Algebraic Expressions Unit Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Book Answers
AP Board Class 7 Maths Chapter 9 Algebraic Expressions Unit Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Book Answers


AP Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise book solutions pdf online from this page.

Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks Solutions PDF

Andhra Pradesh State Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks. These Andhra Pradesh State Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.

Andhra Pradesh State Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Books Solutions

Board AP Board
Materials Textbook Solutions/Guide
Format DOC/PDF
Class 7th
Subject Maths
Chapters Maths Chapter 9 Algebraic Expressions Unit Exercise
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AP Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Question 1.
Fill in the blanks:
(i) The constant term in the expression a + b + 1 is ………………..
Answer:
1.

(ii) The variable in the expression 3x – 8 is ………………..
Answer:
x

(iii) The algebraic term in the expression 2d – 5 is ………………..
Answer:
2d

(iv) The number of terms in the expression ……………….. p2 – 3pq + q is
Answer:
3

(v) The numerical coefficient of the term – ab is …………………..
Answer:
– 1

Question 2.
Write below statements are True or False:
(i) 3𝑥9𝑦 is a binomial.
Answer:
False

(ii) The coefficient of b in – 6abc is – 6a.
Answer:
False

(iii) 5pq and – 9qp are like terms.
Answer:
True

(iv) The sum of a + b and 2a + 7 is 3a + 7b.
Answer:
False

(v) When x = – 2, then the value of x + 2 is 0.
Answer:
True.

Question 3.
Identify like terms among the following:
3a, 6b, 5c, – 8a, 7c, 9c, – a, 23b, 7𝑐9𝑎2.
Answer:
Given terms are
3a, 6b, 5c, – 8a, 7c, 9c, – a, 23b, 7𝑐9𝑎2
Like terms: 3a, – 8a, – a, 𝑎2
6b, 23b
5c, 7c, 9c, 7𝑐9

Question 4.
Arjun and his friend George went to a stationary shop. Arjun bought 3 pens and 2 pencils whereas George bought one pen and 4 pencils. If the price of each pen and pencil is ₹ x and ₹ y respectively, then find the total bill amount in x and y.
Answer:
Given cost of each pen is ₹ x and cost of each pencil is ₹ y .
Arjun bought 3 pens and 2 pencils. Cost of 3 pens = 3 × ₹ x = ₹ 3x
Cost of 2 pencils = 2 × ₹y = ₹ 2y
Amount paid by Arjun = 3x + 2y = ₹ (3x + 2v)
George bougth one pen and 4 pencils.
Cost of 1 pen = 1 × ₹ x = ₹ x
Cost of 4 pencils = 4 × ₹ y = ₹ 4y
Amount paid by George = x + 4y = ₹(x + 4y)
Total bill amount
= Arjun amout + George amount
= (3x + 2y) + (x + 4y)
= 3x + 2y + x + 4y
= 3x + x + 2y + 4y
= (3 + 1)x + (2 +4)y
∴ Total bill amount = ₹ (4x + 6y)

Question 5.
Find the errors and correct the following :
(i) 7x + 4y = 11xy
Answer:
7x and 4y are not like terms and different variables x, y.
So, we should not add the coefficients.

(ii) 8a2 + 6ac = 14a3c
Answer:
8a2 and 6ac are not like terms. So, we should not add the coefficients.

(iii) 6pq2 – 9pq2 = 3pq2
Answer:
6pq2 – 9pq2 = (6 – 9)pq2
= – 3pq2

(iv) 15mn – mn = 15
Answer:
15mn – mn = (15 – 1) mn
= 14 mn

(v) 7 – 3a = 4a
Answer:
7 and 3a are not like terms.
So, we should not subtract the coefficient 7 and 3.

Question 6.
Add the expressions
(i) 9a + 4, 2 – 3a
Answer:
Given expressions are 9a + 4; 2 – 3a
Write the given expressions in standard form.
9a + 4, – 3a + 2.
The sum = (9a + 4) + (- 3a + 2)
= 9a + 4 – 3a + 2
= (9a – 3a) + (4 + 2)
= (9 – 3)a + 6
= 6a + 6

(ii) 2m – 7n, 3n + 8m, m + n
Answer:
Given expressions are
2m – 7n, 3n + 8m, m + n
Write the given expressions in standard form.
2m – 7n, 8m + 3n, m + n
The sum
= (2m 7n) + (8m + 3n) + (m + n)
= 2m – 7n + 8m + 3n + m + n
= (2m + 8m + m) + (- 7n + 3n + n)
= (2 + 8 + 1)m + (- 7 + 3 + 1)n
= 11 m + (- 3)n
= 11 m – 3n

Question 7.
Subtract:
(i) – y from y
Answer:
– y from y = y – (-y) = y + y = 2y

(ii) 18 pq from 25pq
Answer:
18pq from 2 5pq
= 25 pq – 18 pq
= (25 – 18) pq = 7pq

(iii) 6t + 5 from 1 – 9t
Answer:
Given expressions are (6t + 5), (1 – 9t)
Write the given expressions in the standard form (6t + 5); (- 9t + 1)
(6t + 5) from (- 9t + 1)
= (- 9t + 1) – (6t + 5)
= – 9t + 1 – 6t – 5
= (- 9t – 6t) + (1 – 5)
= (- 9 – 6) t + (- 4) = – 15t – 4

Question 8.
Simplify the following :
(i) t + 2 + t + 3 + t + 6- t- 6 + t
Answer:
Given t + 2 + t + 3+ t + 6 – t – 6 + t

= 3t + 5

(ii) (a + b + c) + (2a + 3b – c) – (4a + b – 2c)
Answer:
Given (a + b + c) + (2a + 3b – c) – (4a + b – 2c)
= a + b + c + 2a + 3b – c – 4a – b + 2c
= (a + 2a – 4a) + (b + 3b – b) + (c – c + 2c)
= (1 + 2 – 4)a + (1 + 3 – 1)b + (1 – 1 + 2)c
= (- 1) a + 3b + 2c
= – a + 3b + 2c

(iii) x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)
Answer:
Given x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)
= x + y + 1 + x + 2 + y + 3 + x + 4 + y + 5
= (x + x + x) + (y + y + y) + (1 + 2 + 3 + 4 + 5)
= 3x + 3y + 15

Question 9.
The perimeter of a triangle is 8x2 + 7x – 9 and two of its sides are x2 – 3x + 4, 2x2 + x – 9 respectively, then find third side.
Answer:
Let the sides of triangle are A, B, C.
A = x2 – 3x + 4; B = 2x2 + x – 9 ; C = ?
Perimeter = 8x2 + 7x – 9
Perimeter of the triangle = A + B + C
To get the third side (C). subtract sum of A and B from the perimeter.
∴ C = Perimeter – (A + B)
So,
A + B = (x2 – 3x + 4) + (2x2 + x – 9)
= x2 – 3x + 4 + 2x2 + x – 9
= x2 + 2x2 – 3x + x +4 – 9
= (1 + 2) x2 + (- 3 + 1) x – 5
A + B = 3x2 – 2x – 5
Additive inverse of A + B is – (A + B)
– (A + B) = – (3x2 – 2x – 5) .
– (A + B) = – 3x2 + 2x + 5
C = perimeter + [- (A + B)]
= (8x2 + 7x – 9) + (- 3x2 + 2x + 5)
= 8x2 + 7x – 9 – 3x2 + 2x + 5
= 8x2 – 3x2 + 7x + 2x – 9 + 5
= (8 – 3) x2 + (7 + 2) x – 4
C = 5x2 + 9x – 4
∴ Third side is 5x2 + 9x – 4.

Question 10.
The perimeter of a rectangle is 2a3 – 4a2 – 12a + 10, if length is 3a2 – 4, find its breadth.
Answer:
Given length of rectangle l = 3a2 – 4 and breadth b = ?
Perimeter of a rectangle = 2a3 – 4a2 – 12a + 10
Perimeter of a rectangle
= 2(l + b)
= 2a3 – 4a2 – 12a + 10

12 × 2(a3 – 2a2 – 6a + 5)
⇒ l + b = (a3 – 2a2 – 6a + 5)
⇒ l + b – l = (a3 – 2a2 – 6a + 5) – l
∴ b = (a3 – 2a2 – 6a + 5) – l
Additive inverse of 1 is – 1 = – (3a2 – 4)
∴ – l = – 3a2 + 4 .
b = (a3 – 2a2 – 6a + 5) + (- 1)
= (a3 – 2a2 – 6a + 5) + (- 3a2 + 4)
= a3 – 2a2 – 6a + 5 – 3a2 + 4
= a3 – 2a2 – 3a2 – 6a + 5 + 4
= a3 + (- 2 – 3) a2 – 6a + 9
= a3 + (- 5) a2 – 6a + 9
∴ Breadth of rectangle
= a3 – 5a2 – 6a + 9


AP Board Textbook Solutions PDF for Class 7th Maths


Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks for Exam Preparations

Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbook Solutions can be of great help in your Andhra Pradesh Board Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise exam preparation. The AP Board STD 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 7th Maths Chapter 9 Algebraic Expressions Unit Exercise Books State Board syllabus with maximum efficiency.

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Students can download the Andhra Pradesh Board Class 7 Maths Chapter 9 Algebraic Expressions Unit Exercise Answers PDF from the links provided above.

Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

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