Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 5 Linear Equations in Two Variables Set 5.2 Solutions

Question 1. In an envelope there are some Rs.5 notes and some Rs.10 notes. Total amount of these notes together is Rs.350. Number of Rs.5 notes are less by 10 than twice the number of Rs.10 notes. Then find the number of Rs.5 and Rs.10 notes.
Answer: Solution: Let the number of Rs.5 notes be 'x' and the number of Rs.10 notes be 'y' Total amount of x notes of Rs. 5 = Rs.5x Total amount of y notes of Rs. 10 = Rs.10y \( \therefore \) Total amount = \( 5x + 10y \) According to the first condition, total amount of the notes together is Rs.350. \( \therefore 5x + 10y = 350 \) ...(i) According to the second condition, Number of 5 notes are less by 10 than twice the number of Rs. 10 notes. \( \therefore x = 2y - 10 \) \( \therefore x - 2y = -10 \) .....(ii) Multiplying equation (ii) by 5, \( 5x - 10y = -50 \) ...(iii) Adding equations (i) and (iii), \( 5x + 10y = 350 \) \( + 5x - 10y = -50 \) \( 10x = 300 \)
\( \therefore x = \frac{300}{10} \) \( \therefore x = 30 \) Substituting \( x = 30 \) in equation (ii), \( x - 2y = -10 \) \( 30 - 2y = -10 \) \( \therefore 30 + 10 = 2y \) \( \therefore 40 = 2y \)
\( \therefore y = \frac{40}{2} \) \( \therefore y = 20 \) There are 30 notes of Rs. 5 and 20 notes of Rs. 10 in the envelope. In simple words: We set up two linear equations based on the given information about the number of notes and their total value. By solving these simultaneous equations, we find that there are 30 Rs.5 notes and 20 Rs.10 notes.

🎯 Exam Tip: Always define your variables clearly at the start of the solution. Ensure you form both equations correctly based on the problem's conditions to avoid errors in calculation.

Question 2. The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3: 5. Find the fraction.
Answer: Solution: Let the numerator of the fraction be 'x' and its denominator be 'y'. Then, the required fraction is \( \frac{x}{y} \) According to the first condition, the denominator is 1 less than twice its numerator. \( \therefore y = 2x - 1 \) \( \therefore 2x - y = 1 \) ...(i) According to the second condition, if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.
\( \therefore \frac{x+1}{y+1} = \frac{3}{5} \) \( \therefore 5(x+1) = 3(y+1) \) \( \therefore 5x + 5 = 3y + 3 \) \( \therefore 5x - 3y = 3 - 5 \) \( \therefore 5x - 3y = -2 \) ......(ii) Multiplying equation (i) by 3, \( 6x - 3y = 3 \) ...(iii) Subtracting equation (ii) from (iii), \( 6x - 3y = 3 \) \( - (5x - 3y = -2) \) \( ----------- \) \( x = 5 \) Substituting \( x = 5 \) in equation (i), \( \therefore 2x - y = 1 \) \( \therefore 2(5) - y = 1 \) \( \therefore 10 - y = 1 \) \( \therefore y = 10 - 1 = 9 \)
\( \therefore \) The required fraction is \( \frac{5}{9} \). In simple words: We used two conditions to set up a system of linear equations with the numerator (x) and denominator (y). Solving these equations, we found the numerator to be 5 and the denominator to be 9, making the fraction 5/9.

🎯 Exam Tip: Be careful with algebraic manipulations, especially when cross-multiplying ratios and subtracting equations. Double-check the signs when combining terms.

Question 3. The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their present ages.
Answer: Solution: Let the present age of Priyanka be 'x' years and that of Deepika be 'y' years. According to the first condition, Priyanka's age + Deepika's age = 34 years \( \therefore x + y = 34 \) ...(i) According to the second condition, Priyanka is elder to Deepika by 6 years. \( \therefore x = y + 6 \) \( \therefore x - y = 6 \) .....(ii) Adding equations (i) and (ii), \( x + y = 34 \) \( + x - y = 6 \) \( ----------- \) \( 2x = 40 \) \( \therefore x = 20 \) Substituting \( x = 20 \) in equation (i), \( x + y = 34 \) \( \therefore 20 + y = 34 \) \( \therefore y = 34 - 20 = 14 \)
\( \therefore \) The present age of Priyanka is 20 years and that of Deepika is 14 years. In simple words: We defined Priyanka's age as 'x' and Deepika's as 'y', then formed two equations based on their age sum and age difference. Solving these equations, we found Priyanka is 20 years old and Deepika is 14 years old.

🎯 Exam Tip: When dealing with age problems, clearly state which variable represents whose age. The "elder by" condition often leads to a subtraction equation.

Question 4. The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Answer: Solution: Let the number of lions in the zoo be 'x' and the number of peacocks be 'y'. According to the first condition, the total number of lions and peacocks is 50. \( \therefore x + y = 50 \) ...(i) Lion has 4 legs and Peacock has 2 legs. According to the second condition, the total number of their legs is 140. \( \therefore 4x + 2y = 140 \) Dividing both sides by 2, \( 2x + y = 70 \) ...(ii) Subtracting equation (i) from (ii), \( 2x + y = 70 \) \( - (x + y = 50) \) \( ----------- \) \( x = 20 \) Substituting \( x = 20 \) in equation (i), \( x + y = 50 \) \( \therefore 20 + y = 50 \) \( \therefore y = 50 - 20 = 30 \)
\( \therefore \) The number of lions and peacocks in the zoo are 20 and 30 respectively. In simple words: We set up equations for the total number of animals and the total number of legs, assigning variables to lions and peacocks. By solving these equations, we determined there are 20 lions and 30 peacocks.

🎯 Exam Tip: Remember to account for the number of legs each animal has. This forms the basis of the second equation. Simplifying equations by dividing common factors can make calculations easier.

Question 5. Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs.4500 and after 10 years his monthly salary became Rs.5400, then find his original salary and yearly increment.
Answer: Solution: Let the original salary of Sanjay be 'x' and his yearly increment be 'y'. According to the first condition, after 4 years his monthly salary was Rs.4500 \( \therefore x + 4y = 4500 \) .....(i) According to the second condition, after 10 years his monthly salary became Rs.5400 \( \therefore x + 10y = 5400 \) ...(ii) Subtracting equation (i) from (ii), \( x + 10y = 5400 \) \( - (x + 4y = 4500) \) \( ----------- \) \( 6y = 900 \)
\( \therefore y = \frac{900}{6} \) \( \therefore y = 150 \) Substituting \( y = 150 \) in equation (i), \( x + 4y = 4500 \) \( \therefore x + 4(150) = 4500 \) \( \therefore x + 600 = 4500 \) \( \therefore x = 4500 - 600 = 3900 \)
\( \therefore \) The original salary of Sanjay is Rs.3900 and his yearly increment is Rs.150. In simple words: We modeled Sanjay's salary as a fixed amount plus a yearly increment, forming two linear equations from the given salary data at 4 and 10 years. Solving these equations, we found his original salary is Rs.3900 and his yearly increment is Rs.150.

🎯 Exam Tip: When dealing with incremental values, the "fixed" part is your constant, and the "increment" multiplied by the number of years is the variable component. Pay attention to the number of years for each condition.

Question 6. The price of 3 chairs and 2 tables is Rs.4500 and price of 5 chairs and 3 tables is Rs.7000, then find the price of 2 chairs and 2 tables.
Answer: Solution: Let the price of one chair be 'x' and that of one table be 'y'. According to the first condition, the price of 3 chairs and 2 tables is Rs.4500 \( \therefore 3x + 2y = 4500 \) ...(i) According to the second condition, the price of 5 chairs and 3 tables is Rs.7000 \( \therefore 5x + 3y = 7000 \) ...(ii) Multiplying equation (i) by 3, \( 9x + 6y = 13500 \) ....(iii) Multiplying equation (ii) by 2, \( 10x + 6y = 14000 \) ....(iv) Subtracting equation (iii) from (iv), \( 10x + 6y = 14000 \) \( - (9x + 6y = 13500) \) \( ----------- \) \( x = 500 \) Substituting \( x = 500 \) in equation (i), \( 3x + 2y = 4500 \) \( \therefore 3(500) + 2y = 4500 \) \( \therefore 1500 + 2y = 4500 \) \( \therefore 2y = 4500 - 1500 \) \( \therefore 2y = 3000 \)
\( \therefore y = \frac{3000}{2} \) \( \therefore y = 1500 \) \( \therefore \) Price of 2 chairs and 2 tables = \( 2x + 2y \) \( = 2(500) + 2(1500) \) \( = 1000 + 3000 = Rs.4000 \)
\( \therefore \) The price of 2 chairs and 2 tables is Rs.4000. In simple words: We established a system of two linear equations representing the cost of chairs (x) and tables (y) based on the given price combinations. After solving for 'x' and 'y' (Rs.500 per chair, Rs.1500 per table), we calculated the cost of 2 chairs and 2 tables to be Rs.4000.

🎯 Exam Tip: When using the elimination method, choose multipliers that make the coefficients of one variable equal so they can cancel out. Remember to answer the specific question asked, which in this case was the price of 2 chairs and 2 tables, not just the price of one of each.

Question 7. The sum of the digits in a two-digit number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Answer: Solution: Let the digit in unit's place be 'x' and the digit in ten's place be 'y'.
 

According to the first condition. the sum of the digits in a two-digit number is 9 \( x + y = 9 \) ...(i) According to the second condition, the number obtained by interchanging the digits exceeds the original number by 27 \( \therefore 10x + y = 10y + x + 27 \) \( \therefore 10x - x + y - 10y = 27 \) \( \therefore 9x - 9y = 27 \) Dividing both sides by 9, \( x - y = 3 \) .......(ii) Adding equations (i) and (ii), \( x + y = 9 \) \( + x - y = 3 \) \( ----------- \) \( 2x = 12 \)
\( \therefore x = \frac{12}{2} \) \( \therefore x = 6 \) Substituting \( x = 6 \) in equation (i), \( x + y = 9 \) \( \therefore 6 + y = 9 \) \( \therefore y = 9 - 6 = 3 \) \( \therefore \) Original number = \( 10y + x = 10(3) + 6 \) \( = 30 + 6 = 36 \)
\( \therefore \) The two digit number is 36. In simple words: We used two conditions about the digits of a two-digit number to form a system of equations. The first condition was the sum of digits, and the second related to the number formed by interchanging the digits. Solving these equations revealed the number to be 36.

🎯 Exam Tip: Remember that for a two-digit number with tens digit 'y' and units digit 'x', the number is \( 10y + x \). When digits are interchanged, the new number is \( 10x + y \). This is a common setup for such problems.

Question 8. In \( \triangle ABC \), the measure of \( \angle A \) is equal to the sum of the measures of \( \angle B \) and \( \angle C \). Also the ratio of measures of \( \angle B \) and \( \angle C \) is 4 : 5. Then find the measures of angles of the triangle.
Answer: Solution: Let the measure of \( \angle B \) be 'x°' and that of \( \angle C \) be 'y°'. According to the first condition, \( m\angle A = m\angle B + m\angle C \) \( \therefore m\angle A = x° + y° \) In \( \triangle ABC \), \( m\angle A + m\angle B + m\angle C = 180° \) ...[Sum of the measures of the angles of a triangle is 180°] \( \therefore (x + y) + x + y = 180 \) \( \therefore 2x + 2y = 180 \) Dividing both sides by 2, \( x + y = 90 \) ...(i) According to the second condition, the ratio of the measures of \( \angle B \) and \( \angle C \) is 4 : 5.
\( \therefore \frac{x}{y} = \frac{4}{5} \) \( \therefore 5x = 4y \) \( \therefore 5x - 4y = 0 \) .......(ii) Multiplying equation (i) by 4, \( 4x + 4y = 360 \) ...(iii) Adding equations (ii) and (iii), \( 5x - 4y = 0 \) \( + 4x + 4y = 360 \) \( ----------- \) \( 9x = 360 \)
\( \therefore x = \frac{360}{9} \) \( \therefore x = 40 \) Substituting \( x = 40 \) in equation (i), \( x + y = 90 \) \( \therefore 40 + y = 90 \) \( \therefore y = 90 - 40 \) \( \therefore y = 50 \) \( \therefore m\angle A = x° + y° = 40° + 50° = 90° \)
\( \therefore \) The measures of \( \angle A \), \( \angle B \) and \( \angle C \) are 90°, 40°, and 50° respectively. In simple words: We used the properties of angles in a triangle and the given ratio to form two linear equations. By solving these equations, we found the measures of angles B and C to be 40° and 50° respectively, which then allowed us to calculate angle A as 90°.

🎯 Exam Tip: Always remember the fundamental property that the sum of angles in a triangle is 180°. Clearly defining angle measures with variables 'x' and 'y' helps in setting up the equations correctly.

Question 9. Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to \( \frac{1}{3} \) of the larger part. Then find the length of the larger part.
Answer: Solution: Let the length of the smaller part of the rope be 'x' cm and that of the larger part be 'y' cm. According to the first condition, total length of the rope is 560 cm. \( \therefore x + y = 560 \) ...(i) Twice the length of the smaller part = \( 2x \) \( \frac{1}{3} \) rd length of the larger part = \( \frac{1}{3}y \) According to the second condition, \( 2x = \frac{1}{3}y \) \( \therefore 6x = y \) \( \therefore 6x - y = 0 \) ......(ii) Adding equations (i) and (ii), \( x + y = 560 \) \( + 6x - y = 0 \) \( ----------- \) \( 7x = 560 \)
\( \therefore x = \frac{560}{7} \) \( \therefore x = 80 \) Substituting \( x = 80 \) in equation (ii), \( 6x - y = 0 \) \( \therefore 6(80) - y = 0 \) \( \therefore 480 - y = 0 \) \( \therefore y = 480 \)
\( \therefore \) The length of the larger part of the rope is 480 cm. In simple words: We used two equations: one for the total length of the rope (x + y = 560) and another for the relationship between the two parts (\( 2x = \frac{1}{3}y \)). Solving these equations allowed us to find the lengths of the smaller (80 cm) and larger (480 cm) parts.

🎯 Exam Tip: Ensure that the variables are correctly assigned to "smaller part" and "larger part". Converting the word problem "twice the length... is equal to 1/3 of..." into a precise mathematical equation is crucial.

Question 10. In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Answer: Solution: Let us suppose that Yashwant got 'x' questions right and 'y' questions wrong. According to the first condition, total number of questions in the examination are 60. \( \therefore x + y = 60 \) ...(i) Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer. \( \therefore \) He got \( 2x - y \) marks. According to the second condition, he got 90 marks. \( 2x - y = 90 \) ... (ii) Adding equations (i) and (ii), \( x + y = 60 \) \( + 2x - y = 90 \) \( ----------- \) \( 3x = 150 \)
\( \therefore x = \frac{150}{3} \) \( \therefore x = 50 \) Substituting \( x = 50 \) in equation (i), \( 50 + y = 60 \) \( \therefore y = 60 - 50 = 10 \)
\( \therefore \) Yashwant got 10 questions wrong. In simple words: We set up a system of equations: one for the total number of questions (correct + wrong = 60) and another for the total marks (2 marks for correct - 1 mark for wrong = 90). Solving these equations, we found Yashwant answered 10 questions incorrectly.

🎯 Exam Tip: Be mindful of the signs for marks-per-question. Correct answers add marks (+2x), while incorrect answers subtract marks (-1y).

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations In Two Variables Practice Set 5.2 Intext Questions And Activities

Question 1. The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year? (Textbook pg. no. 89)
Answer: Solution: Step 1: Read the given word problem carefully and try to understand it. Step 2: Make assumptions using two variables x and y. Let the number of males in previous year be 'x' and the number of females be 'y'. Step 3: From the given information, form mathematical statements using the above variables. According to the first condition, the total population of town was 50,000. \( \therefore x + y = 50000 \) ...(i) Male population increased by 5%. \( \therefore \) Number of males = \( x + 5\% \) of \( x \), \( = x + x \times \frac{5}{100} \) \( = \frac{100x+5x}{100} \) \( = \frac{105x}{100} \) Female population increased by 3%. \( \therefore \) Number of females = \( y + 3\% \) of \( y \) \( = y + y \times \frac{3}{100} \) \( = \frac{100y+3y}{100} \) \( = \frac{103y}{100} \) According to the second condition, in a year population became 52020
\( \therefore \frac{105}{100} x + \frac{103}{100} y = 52020 \) \( \therefore 105x + 103y = 5202000 \) ...(ii) Multiplying equation (i) by 103, \( 103x + 103y = 5150000 \) ...(iii) Step 4: Here, we use elimination method. Subtracting equation (iii) from (ii), \( 105x + 103y = 5202000 \) \( - (103x + 103y = 5150000) \) \( ----------- \) \( 2x = 52000 \)
\( \therefore x = \frac{52000}{2} \) \( \therefore x = 26000 \) Substituting \( x = 26000 \) in equation (i), \( \therefore 26000 + y = 50000 \) \( \therefore y = 50000 - 26000 \) \( \therefore y = 24000 \) \( \therefore \) Number of males = \( x = 26000 \) \( \therefore \) Number of females = \( y = 24000 \) Step 5: Write the answer. The number of males and females in the previous year were 26,000 and 24,000 respectively. Step 6: Verify your result using smart check. In simple words: We used the initial total population and the percentage increase for males and females to set up a system of linear equations. By solving these equations using the elimination method, we found that there were 26,000 males and 24,000 females in the previous year.

🎯 Exam Tip: When dealing with percentage increases, remember to add the increase to the original amount (e.g., \( x + 5\% \) of \( x \)). Large numbers in calculations require careful attention to avoid arithmetic errors. The elimination method is effective for solving such systems.