Maharashtra Board Class 7 Chapter 8 Set 36 Algebraic Expressions and Operations on them Solutions

Question 1. Simplify (3x - 11y) - (17x + 13y) and choose the right answer.
(A) 7x - 12y
(B) -14x - 54y
(C) -3(5x + 4y)
(D) -2(7x + 12y)
Answer: (D) -2(7x + 12y) Hints: \( (3x - 11y) - (17x + 13y) = 3x - 11y - 17x - 13y \) \( = -14x - 24y \) \( = -2 \times 7x - 2 \times 12y \) \( = -2(7x + 12y) \)
In simple words: To simplify the expression, distribute the negative sign to the second parenthesis, then combine like terms (x with x, and y with y). Finally, factor out a common term if possible.

🎯 Exam Tip: Always pay close attention to distributing negative signs carefully, as this is a common source of error in algebraic simplification.

Question 2. The product of \( (23x^2y^3z) \) and \( (-15x^3yz^2) \) is __
(A) \( -34x^5y^4z^3 \)
(B) \( 34x^2y^3z^5 \)
(C) \( 145x^3y^2z \)
(D) \( 170x^3y^2z^3 \)
Answer: (A) \( -34x^5y^4z^3 \) Solution: To find the product, multiply the numerical coefficients and add the exponents of the same variables. \( (23x^2y^3z) \times (-15x^3yz^2) \) \( = (23 \times -15) \times (x^2 \times x^3) \times (y^3 \times y) \times (z \times z^2) \) \( = -345 \times x^{2+3} \times y^{3+1} \times z^{1+2} \) \( = -345x^5y^4z^3 \)
In simple words: When multiplying monomials, multiply the coefficients and add the powers of identical variables.

🎯 Exam Tip: Remember the rules of exponents for multiplication: \( a^m \times a^n = a^{m+n} \). Also, be careful with the sign of the product when multiplying positive and negative numbers.

Question 3. Solve the following equations:
(i) \( 4x + \frac{1}{2} = \frac{9}{2} \)
(ii) \( 10 = 2y + 5 \)
(iii) \( 5m - 4 = 1 \)
(iv) \( 6x - 1 = 3x + 8 \)
(v) \( 2(x - 4) = 4x + 2 \)
(vi) \( 5(x + 1) = 74 \)
Answer: Solution: (i) \( 4x + \frac{1}{2} = \frac{9}{2} \)
\( \implies 4x = \frac{9}{2} - \frac{1}{2} \)
\( \implies 4x = \frac{8}{2} \)
\( \implies 4x = 4 \)
\( \implies x = \frac{4}{4} \)
\( \implies x = 1 \) (ii) \( 10 = 2y + 5 \)
\( \implies 10 - 5 = 2y \)
\( \implies 5 = 2y \)
\( \implies \frac{5}{2} = y \)
\( \implies y = \frac{5}{2} \) (iii) \( 5m - 4 = 1 \)
\( \implies 5m = 1 + 4 \)
\( \implies 5m = 5 \)
\( \implies m = \frac{5}{5} \)
\( \implies m = 1 \) (iv) \( 6x - 1 = 3x + 8 \)
\( \implies 6x - 3x = 8 + 1 \)
\( \implies 3x = 9 \)
\( \implies x = \frac{9}{3} \)
\( \implies x = 3 \) (v) \( 2(x - 4) = 4x + 2 \)
\( \implies 2x - 8 = 4x + 2 \)
\( \implies -8 - 2 = 4x - 2x \)
\( \implies -10 = 2x \)
\( \implies x = \frac{-10}{2} \)
\( \implies x = -5 \) (vi) \( 5(x + 1) = 74 \)
\( \implies 5x + 5 = 74 \)
\( \implies 5x = 74 - 5 \)
\( \implies 5x = 69 \)
\( \implies x = \frac{69}{5} \)
In simple words: To solve linear equations, isolate the variable by performing inverse operations on both sides of the equation, maintaining balance.

🎯 Exam Tip: When solving equations, remember to distribute terms properly, combine like terms, and perform operations (addition, subtraction, multiplication, division) on both sides to keep the equation balanced.

Question 4. Rakesh's age is less than Sania's age by 5 years. The sum of their ages is 27 years. How old are they?
Answer: Solution: Let the age of Rakesh be x years.
Sania's age = (x + 5) years. According to the given condition, \( x + (x + 5) = 27 \)
\( \implies 2x + 5 = 27 \)
\( \implies 2x = 27 - 5 \)
\( \implies 2x = 22 \)
\( \implies x = \frac{22}{2} \)
\( \implies x = 11 \) Sania's age = \( x + 5 = 11 + 5 = 16 \) years
The ages of Rakesh and Sania are 11 years and 16 years respectively.
In simple words: Represent the ages using a variable and set up an equation based on the given relationships and sum. Solve the equation to find their individual ages.

🎯 Exam Tip: Clearly define your variables before setting up the equations in word problems. Double-check your calculations, especially when transposing terms across the equals sign.

Question 5. When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted?
Answer: Solution: Let the number of jambhul trees planted be x.
Number of ashoka trees = x - 60 According to the given condition, \( x + x - 60 = 200 \)
\( \implies 2x = 200 + 60 \)
\( \implies 2x = 260 \)
\( \implies x = \frac{260}{2} \)
\( \implies x = 130 \)
130 jambhul trees were planted.
In simple words: Assign a variable to the unknown quantity, form an equation from the given conditions about the relationship between the two types of trees and their total, and solve for the variable.

🎯 Exam Tip: Always re-read the question after finding the value of the variable to ensure you are answering what was asked (e.g., number of jambhul trees vs. ashoka trees).

Question 6. Shubhangi has twice as many 20-rupee notes as she has 50-rupee notes. Altogether, she has 2700 rupees. How many 50-rupee notes does she have?
Answer: Solution: Let the number of 50-rupee notes with Shubhangi be x.
Number of 20-rupee notes = 2x Total amount with Shubhangi = Number of 50-rupee notes \( \times \) 50 + Number of 20-rupee notes \( \times \) 20 \( = x \times 50 + 2x \times 20 \) \( = 50x + 40x \) \( = 90x \) According to the given condition, \( 90x = 2700 \)
\( \implies x = \frac{2700}{90} \)
\( \implies x = 30 \)
Shubhangi has 30 notes of 50 rupees.
In simple words: Define variables for the count of each type of note, set up an equation representing the total value based on the number of notes and their denominations, then solve for the variable representing the 50-rupee notes.

🎯 Exam Tip: Be careful to distinguish between the *number* of notes and their *value*. The total amount is the sum of (number of notes \( \times \) denomination) for each type of note.

Question 7. Virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make?
Answer: Solution: Let the runs made by Rohit be x.
Runs made by Virat = 2x According to the given condition, \( x + 2x = 200 - 2 \)
\( \implies 3x = 198 \)
\( \implies x = \frac{198}{3} \)
\( \implies x = 66 \)
Runs made by Virat = \( 2x = 2 \times 66 = 132 \)
The runs made by Virat and Rohit are 132 and 66 respectively.
In simple words: Represent Rohit's runs as 'x', then Virat's as '2x'. Form an equation where their combined runs equal 2 less than 200, and solve to find each player's score.

🎯 Exam Tip: Carefully translate the word problem into mathematical expressions, especially phrases like "twice as many" and "2 less than a double century."

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions And Operations On Them Practice Set 36 Intext Questions And Activities

Question 1. Solve the following equations. (Textbook pg. no. 59)
(i) \( x + 7 = 4 \)
(ii) \( 4p = 12 \)
(iii) \( m - 5 = 4 \)
(iv) \( \frac{t}{3} = 6 \)
Answer: Solution: (i) \( x + 7 = 4 \)
\( x + 7 - 7 = 4 - 7 \) ....(Subtracting 7 from both sides)
\( \implies x + 0 = -3 \)
\( \implies x = -3 \) (ii) \( 4p = 12 \)
\( \frac{4p}{4} = \frac{12}{4} \) ....(Dividing both sides by 4)
\( \implies p = 3 \) (iii) \( m - 5 = 4 \)
\( m - 5 + 5 = 4 + 5 \) .... (Adding 5 to both sides)
\( \implies m + 0 = 9 \)
\( \implies m = 9 \) (iv) \( \frac{t}{3} = 6 \)
\( \frac{t}{3} \times 3 = 6 \times 3 \) .... (Multiplying both sides by 3)
\( \implies t = 18 \)
In simple words: To solve basic linear equations, apply the inverse operation (addition, subtraction, multiplication, or division) to both sides of the equation to isolate the variable.

🎯 Exam Tip: Remember to always perform the same operation on both sides of the equation to maintain equality. This is the fundamental principle for solving any equation.