Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Algebraic Expressions (Exercise 7.4)
Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in
the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Page 2
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in
the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= -2x
2
+ 2y
2
- 2xy - 3x
2
- 3y
2
+ 3xy
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
Page 3
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in
the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= -2x
2
+ 2y
2
- 2xy - 3x
2
- 3y
2
+ 3xy
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= a – [b – {a – (b – 1) + 3a}]
= a – [b – {a – b + 1 + 3a}]
= a – [b – {4a – b + 1}]
= a – [b – 4a + b – 1]
= a – [2b – 4a – 1]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
Page 4
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in
the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= -2x
2
+ 2y
2
- 2xy - 3x
2
- 3y
2
+ 3xy
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= a – [b – {a – (b – 1) + 3a}]
= a – [b – {a – b + 1 + 3a}]
= a – [b – {4a – b + 1}]
= a – [b – 4a + b – 1]
= a – [2b – 4a – 1]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
10. 2a – [4b – {4a – 3(2a – b)}]
Solution:
Given 2a – [4b – {4a – 3(2a – b)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
11. -a – [a + {a + b – 2a – (a – 2b)} - b]
Solution:
Given -a – [a + {a + b – 2a – (a – 2b)} - b]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + {- 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – [- a + 2b]
On simplifying, we get
= – a + a – 2b
= – 2b
Page 5
Exercise 7.4 Page No: 7.20
Simplify each of the following algebraic expressions by removing grouping symbols.
1. 2x + (5x – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in
the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 3x – (y – 2x)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Solution:
Given - 2(x
2
- y
2
+ xy) - 3(x
2
+y
2
- xy)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in
the parentheses when we remove them. Therefore, we have
= -2x
2
+ 2y
2
- 2xy - 3x
2
- 3y
2
+ 3xy
On rearranging,
= -2x
2
- 3x
2
+ 2y
2
- 3y
2
- 2xy + 3xy
On simplifying, we get
= -5x
2
- y
2
+ xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square
brackets.
Then we get,
= a – [b – {a – (b – 1) + 3a}]
= a – [b – {a – b + 1 + 3a}]
= a – [b – {4a – b + 1}]
= a – [b – 4a + b – 1]
= a – [2b – 4a – 1]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square
brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
10. 2a – [4b – {4a – 3(2a – b)}]
Solution:
Given 2a – [4b – {4a – 3(2a – b)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
11. -a – [a + {a + b – 2a – (a – 2b)} - b]
Solution:
Given -a – [a + {a + b – 2a – (a – 2b)} - b]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= – a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + {- 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – [- a + 2b]
On simplifying, we get
= – a + a – 2b
= – 2b
12. 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Solution:
Given 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 2x – 3y – [3x – 2y – {x – z – (x – 2y)})
= 2x – 3y – [3x – 2y – {x – z – x + 2y}]
= 2x – 3y – [3x – 2y – {- z + 2y}]
= 2x – 3y – [3x – 2y + z – 2y]
= 2x – 3y – [3x – 4y + z]
On simplifying, we get
= 2x – 3y – 3x + 4y – z
= - x + y – z
13. 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Solution:
Given 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Then we get,
= 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]
= 5 + [x – {y – 4x + 4} – {x – y + 2}]
= 5 + [x – y + 4x – 4 – x + y – 2]
= 5 + [4x – 6]
= 5 + 4x – 6
= 4x – 1
14. x
2
- [3x + [2x - (x
2
- 1)] + 2]
Solution:
Given x
2
- [3x + [2x - (x
2
- 1)] + 2]
First we have to remove the parentheses, then remove braces, and then the square
brackets.
Read More
|
76 videos|344 docs|39 tests
|
FAQs on Algebraic Expressions (Exercise 7.4) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What are algebraic expressions? |  |
Ans. Algebraic expressions are mathematical expressions that consist of variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. These expressions are used to represent relationships between quantities and can be simplified or evaluated using specific values for the variables.
| 2. How do you simplify algebraic expressions? | |
Ans. To simplify algebraic expressions, you need to combine like terms by adding or subtracting them. Like terms have the same variables raised to the same powers. You can also use the distributive property to simplify expressions by multiplying and simplifying terms within parentheses. The goal is to simplify the expression as much as possible by removing any unnecessary parentheses or combining like terms.
| 3. What is the difference between an equation and an expression? | |
Ans. An equation is a statement that shows the equality between two expressions. It contains an equal sign (=) and represents a balance or relationship between the two sides of the equation. On the other hand, an expression is a mathematical phrase that may contain variables, constants, and operations but does not have an equal sign. Expressions can be simplified or evaluated, while equations are solved to find the values of the variables that satisfy the equality.
| 4. How do you evaluate algebraic expressions? | |
Ans. To evaluate an algebraic expression, you substitute the given values for the variables and simplify the expression using the order of operations (PEMDAS/BODMAS). Start by replacing each variable with its corresponding value, perform any necessary calculations (such as addition, subtraction, multiplication, and division), and simplify the expression until you obtain a numerical value.
| 5. Can algebraic expressions be solved? | |
Ans. Algebraic expressions cannot be "solved" in the same way as equations because they do not contain an equal sign. However, you can simplify or evaluate algebraic expressions by substituting values for the variables. If you have an equation that contains an algebraic expression, you can solve the equation to find the values of the variables that make the equation true.
Related Exams
About this Document
|
4.71/5 Rating |
|
Dec 27, 2024 Last updated |
Document Description: RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) covers topics
like and RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Algebraic Expressions (Exercise 7.4).
Introduction of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Algebraic Expressions (Exercise 7.4)
In this doc you can find the meaning of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,Viva Questions
,Summary
,Sample Paper
,Extra Questions
,Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,MCQs
,video lectures
,Objective type Questions
,ppt
,Important questions
,study material
,Exam
,past year papers
,Free
,Algebraic Expressions (Exercise 7.4) RD Sharma Solutions | Mathematics (Maths) Class 7
,mock tests for examination
,Semester Notes
,shortcuts and tricks
,practice quizzes
,Previous Year Questions with Solutions
;
Additional Information about RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) for Class 7 Preparation
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Free PDF Download
The RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4)
The importance of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Notes
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Algebraic Expressions (Exercise 7.4).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Class 7 Questions
The "RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) on the App
Students of Class 7 can study RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Algebraic Expressions (Exercise 7.4),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Algebraic Expressions (Exercise 7.4) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup