Maharashtra Board Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.5 Solutions

Question 1. Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Answer: Solution: Let the greater number be x and the smaller number be y. According to the first condition, x - y = 3 ...(i) According to the second condition, 3x + 2y = 19 ...(ii) Multiplying equation (i) by 2, we get 2x - 2y = 6 ...(iii) Adding equations (ii) and (iii), we get \[ \begin{array}{l} 3x + 2y = 19 \\ + 2x - 2y = 6 \\ \hline 5x \quad = 25 \end{array} \]
\( \implies x = \frac{25}{5} \)
\( \implies x = 5 \) Substituting x = 5 in equation (i), we get 5 - y = 3
\( \implies 5 - 3 = y \)
\( \implies y = 2 \) The required numbers are 5 and 2.
In simple words: We set up two linear equations based on the given conditions about the two numbers, solved them simultaneously to find the values of x and y, which represent the greater and smaller numbers, respectively.

🎯 Exam Tip: Clearly define your variables for the unknown numbers and ensure each condition translates accurately into a distinct linear equation for full marks.

Question 2. Complete the following.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत (rectangle) का चित्र है जिसमें उसकी भुजाओं की माप बीजीय व्यंजकों (algebraic expressions) के रूप में दी गई है। ऊपर की भुजा \(2x + y + 8\), नीचे की भुजा \(4x - y\), दाईं भुजा \(x + 4\), और बाईं भुजा \(2y\) है। चित्र में दो क्रियाशील बॉक्स भी हैं: एक 'Find the values of x and y.' और दूसरा 'Find my perimeter and area.'
Answer: Solution: Opposite sides of a rectangle are equal.
\( \implies 2x + y + 8 = 4x - y \)
\( \implies 8 = 4x - 2x - y - y \)
\( \implies 2x - 2y = 8 \)
\( \implies x - y = 4 \) ...(i) [Dividing both sides by 2] Also, \( x + 4 = 2y \)
\( \implies x - 2y = -4 \) ...(ii) Subtracting equation (ii) from (i), we get \[ \begin{array}{l} x - y = 4 \\ x - 2y = -4 \\ - \quad + \quad + \\ \hline \quad \quad y = 8 \end{array} \] Substituting y = 8 in equation (i), we get x - 8 = 4
\( \implies x = 4 + 8 \)
\( \implies x = 12 \) Now, length of rectangle = \( 4x - y \) = \( 4(12) - 8 \) = \( 48 - 8 \)
\( \implies \) Length of rectangle = 40 Breadth of rectangle = \( 2y = 2(8) = 16 \) Perimeter of rectangle = \( 2(\text{length} + \text{breadth}) \) = \( 2(40 + 16) \) = \( 2(56) \)
\( \implies \) Perimeter of rectangle = 112 units Area of rectangle = \( \text{length} \times \text{breadth} \) = \( 40 \times 16 \)
\( \implies \) Area of rectangle = 640 sq. units
\( \implies \) x = 12 and y = 8, Perimeter of rectangle is 112 units and area of rectangle is 640 sq. units.
In simple words: By equating the opposite sides of the rectangle, we formed a system of two linear equations with x and y. Solving these equations gave us the values of x and y, which were then used to calculate the actual length, breadth, perimeter, and area of the rectangle.

🎯 Exam Tip: Remember that opposite sides of a rectangle are equal. This property is key to forming the correct equations. Show all steps clearly, especially for the perimeter and area calculations, to avoid losing marks.

Question 3. The sum of father's age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Answer: Solution: Let the present ages of father and son be x years and y years respectively. According to the first condition, x + 2y = 70 ...(i) According to the second condition, 2x + y = 95 ...(ii) Multiplying equation (i) by 2, we get 2x + 4y = 140 ...(iii) Subtracting equation (ii) from (iii), we get \[ \begin{array}{l} 2x + 4y = 140 \\ 2x + y = 95 \\ - \quad - \quad - \\ \hline \quad \quad 3y = 45 \end{array} \]
\( \implies y = \frac{45}{3} \)
\( \implies y = 15 \) Substituting y = 15 in equation (i), we get x + 2(15) = 70
\( \implies x + 30 = 70 \)
\( \implies x = 70 - 30 \)
\( \implies x = 40 \) The present ages of father and son are 40 years and 15 years respectively.
In simple words: We assigned variables to the father's and son's ages, translated the two given conditions into linear equations, and solved them simultaneously to find their current ages.

🎯 Exam Tip: Pay close attention to wording like "twice the age" or "double the age" to correctly form the coefficients in your equations. Clearly label your variables for clarity.

Question 4. The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Answer: Solution: Let the numerator of the fraction be x and the denominator be y. Fraction = \( \frac{x}{y} \) According to the first condition, y = 2x + 4
\( \implies 2x - y = -4 \) ...(i) According to the second condition, \( (y - 6) = 12(x - 6) \)
\( \implies y - 6 = 12x - 72 \)
\( \implies 12x - y = 72 - 6 \)
\( \implies 12x - y = 66 \) ...(ii) Subtracting equation (i) from (ii), we get \[ \begin{array}{l} 12x - y = 66 \\ 2x - y = -4 \\ - \quad + \quad + \\ \hline 10x \quad = 70 \end{array} \]
\( \implies x = \frac{70}{10} \)
\( \implies x = 7 \) Substituting x = 7 in equation (i), we get 2(7) - y = -4 14 - y = -4
\( \implies 14 + 4 = y \)
\( \implies y = 18 \) Fraction = \( \frac{x}{y} = \frac{7}{18} \) The required fraction is \( \frac{7}{18} \).
In simple words: We defined the numerator as x and denominator as y, set up two equations based on the two given conditions about their relationship, solved for x and y, and then expressed the fraction.

🎯 Exam Tip: Be careful when translating word problems involving fractions. Clearly distinguish between numerator and denominator and how changes affect their relationship in the equations.

Question 5. Two types of boxes A, B, are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weights 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Answer: Solution: Let the weights of box of type A be x kg and that of box of type B be y kg. 1 ton = 1000 kg
\( \implies \) 10 tons = 10000 kg According to the first condition, 150x + 100y = 10000
\( \implies 3x + 2y = 200 \) ...(i) [Dividing both sides by 50] According to the second condition, 260x + 40y = 10000
\( \implies 13x + 2y = 500 \) ...(ii) [Dividing both sides by 20] Subtracting equation (i) from (ii), we get \[ \begin{array}{l} 13x + 2y = 500 \\ 3x + 2y = 200 \\ - \quad - \quad - \\ \hline 10x \quad = 300 \end{array} \]
\( \implies x = \frac{300}{10} \)
\( \implies x = 30 \) Substituting x = 30 in equation (i), we get 3(30) + 2y = 200 90 + 2y = 200 2y = 200 - 90 2y = 110
\( \implies y = \frac{110}{2} \)
\( \implies y = 55 \) The weights of box of type A is 30 kg and that of box of type B is 55 kg.
In simple words: We converted the truck capacity to kilograms, then formulated two linear equations based on the two loading scenarios described. Solving these equations simultaneously allowed us to determine the individual weights of type A and type B boxes.

🎯 Exam Tip: Pay attention to unit conversions (tons to kg) at the beginning of the problem. Simplify your equations by dividing by common factors to make calculations easier and reduce errors.

Question 6. Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance Vishal travelled by bus.
Answer: Solution: Let the distance Vishal travelled by bus be x km and by aeroplane be y km. According to the first condition, x + y = 1900 ...(i) Time = \( \frac{\text{Distance}}{\text{Speed}} \)
\( \implies \) Time required to cover x km by bus = \( \frac{x}{60} \) hr Time required to cover y km by aeroplane = \( \frac{y}{700} \) hr According to the second condition, \( \frac{x}{60} + \frac{y}{700} = 5 \)
\( \implies \frac{700x + 60y}{60 \times 700} = 5 \)
\( \implies 700x + 60y = 5 \times 60 \times 700 \)
\( \implies 70x + 6y = 21000 \) ...(ii) [Dividing both sides by 10] Multiplying equation (i) by 6, we get 6x + 6y = 11400 ...(iii) Subtracting equation (iii) from (ii), we get \[ \begin{array}{l} 70x + 6y = 21000 \\ 6x + 6y = 11400 \\ - \quad - \quad - \\ \hline 64x \quad = 9600 \end{array} \]
\( \implies x = \frac{9600}{64} \)
\( \implies x = 150 \) The distance Vishal travelled by bus is 150 km.
In simple words: We assigned variables to the distances covered by bus and aeroplane. We formed one equation from the total distance and another from the total time, using the formula time = distance/speed. Solving these two equations simultaneously gave us the distance travelled by bus.

🎯 Exam Tip: When dealing with distance, speed, and time problems, remember the formula `Time = Distance/Speed`. Ensure your equations correctly represent both the total distance and total time conditions.

Question 1. There are some instructions given below. Frame the equations from the information and write them in the blank boxes shown by arrows. (Textbook pg. no. 20)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रवाह आरेख (flow diagram) है जो सार्थिक (Sarthak) और साक्षी (Sakshi) की वर्तमान आयु x और y के संदर्भ में तीन अलग-अलग शर्तों को दर्शाता है। प्रत्येक शर्त से एक रैखिक समीकरण बनता है: सार्थिक की आयु साक्षी की आयु के दोगुने से 8 कम है (x = 2y - 8); उनकी वर्तमान आयु का योग 25 है (x + y = 25); और 4 साल पहले साक्षी की आयु सार्थिक की आयु से 3 साल कम थी ((y - 4) = (x - 4) - 3)।
Answer: The framed equations are:
(i) Sarthak's age is less by 8 than double the age of Sakshi: \( x = 2y - 8 \) \( x - 2y = -8 \)
(ii) The sum of present ages of Sarthak and Sakshi is 25: \( x + y = 25 \)
(iii) 4 years ago Sakshi's age was 3 years less than Sarthak's age at that time: \( (y - 4) = (x - 4) - 3 \) \( x - y = 3 \)
In simple words: By assigning 'x' to Sarthak's age and 'y' to Sakshi's age, we translated the three descriptive conditions provided in the diagram into three distinct linear equations.

🎯 Exam Tip: Accurately translating word problems into algebraic equations is crucial. Pay close attention to keywords like "less by", "double", "sum", and phrases indicating past or future ages to form correct expressions.