Maharashtra Board Class 8 Maths Chapter 7 Variation Set 7.2 Solutions

Variation Class 8 Maths Chapter 7 Practice Set 7.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 7.2 Chapter 7 Solutions Answers

Question 1. The information about number of workers and number of days to complete a work is given in the following table. Complete the table.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका श्रमिकों की संख्या और किसी कार्य को पूरा करने के लिए आवश्यक दिनों की संख्या के बीच व्युत्क्रम संबंध को दर्शाती है। कुछ मान दिए गए हैं और छात्रों को यह समझने के लिए कि श्रमिकों की संख्या और दिनों की संख्या में क्या संबंध है, लापता मानों को ज्ञात करके तालिका को पूरा करना होगा।
Answer:Solution: Let, n represent the number of workers and d represent the number of days required to complete a work. Since, number of workers and number of days to complete a work are in inverse proportion. \( n \propto \frac{1}{d} \) \( \implies n = k \times \frac{1}{d} \) where k is the constant of variation. \( \implies n \times d = k \) ...(i) (i) When n = 30, d = 6 \( \implies \) Substituting n = 30 and d = 6 in (i), we get \( n \times d = k \) \( \implies 30 \times 6 = k \) \( \implies k = 180 \) Substituting k = 180 in (i), we get \( n \times d = k \) \( \implies n \times d = 180 \) ...(ii) This is the equation of variation (ii) When d = 12, n = ? \( \implies \) Substituting d = 12 in (ii), we get \( n \times d = 180 \) \( \implies n \times 12 = 180 \) \( \implies n = \frac{180}{12} \) \( \implies n = 15 \) (iii) When n = 10, d = ? \( \implies \) Substituting n = 10 in (ii), we get \( n \times d = 180 \) \( \implies 10 \times d = 180 \) \( \implies d = \frac{180}{10} \) \( \implies d = 18 \) (iv) When d = 36, n = ? \( \implies \) Substituting d = 36 in (ii), we get \( n \times d = 180 \) \( \implies n \times 36 = 180 \) \( \implies n = \frac{180}{36} \) \( \implies n = 5 \) The completed table is:

Number of workers	30	20	15	10	5
Days	6	9	12	18	36

In simple words: The number of workers and days to complete a work are in inverse proportion, meaning if one increases, the other decreases. We find the constant of variation by multiplying the given pairs (e.g., 30 workers * 6 days = 180), then use this constant to find the missing values.

🎯 Exam Tip: Always state the type of variation (direct or inverse) and derive the constant of variation (k) first. This constant is crucial for finding unknown values in similar problems and demonstrates a clear understanding of the concept. Make sure to present the final answer in a clear table format if the question originally provided one.

 

Question 2. Find constant of variation and write equation of variation for every example given below:
(i) \( p \propto \frac{1}{q} \); if p = 15 then q = 4.
(ii) \( z \propto \frac{1}{w} \); when z = 2.5 then w = 24.
(iii) \( s \propto \frac{1}{t^2} \); if s = 4 then t = 5.
(iv) \( x \propto \frac{1}{\sqrt{y}} \); if x = 15 then y = 9.

Answer:Solution: (i) \( p \propto \frac{1}{q} \) ...[Given] \( \implies p = k \times \frac{1}{q} \) where, k is the constant of variation. \( \implies p \times q = k \) ...(i) When p = 15, q = 4 \( \implies \) Substituting p = 15 and q = 4 in (i), we get \( p \times q = k \) \( \implies 15 \times 4 = k \) \( \implies k = 60 \) Substituting k = 60 in (i), we get \( p \times q = k \) \( \implies p \times q = 60 \) This is the equation of variation. \( \implies \) The constant of variation is 60 and the equation of variation is \( pq = 60 \). (ii) \( z \propto \frac{1}{w} \) ...[Given] \( \implies z = k \times \frac{1}{w} \) where, k is the constant of variation, \( \implies z \times w = k \) ...(i) When z = 2.5, w = 24 \( \implies \) Substituting z = 2.5 and w = 24 in (i), we get \( z \times w = k \) \( \implies 2.5 \times 24 = k \) \( \implies k = 60 \) Substituting k = 60 in (i), we get \( z \times w = k \) \( \implies z \times w = 60 \) This is the equation of variation. \( \implies \) The constant of variation is 60 and the equation of variation is \( zw = 60 \). (iii) \( s \propto \frac{1}{t^2} \) ...[Given] \( \implies s = k \times \frac{1}{t^2} \) where, k is the constant of variation, \( \implies s \times t^2 = k \) ...(i) When s = 4, t = 5 \( \implies \) Substituting, s = 4 and t = 5 in (i), we get \( s \times t^2 = k \) \( \implies 4 \times (5)^2 = k \) \( \implies k = 4 \times 25 \) \( \implies k = 100 \) Substituting k = 100 in (i), we get \( s \times t^2 = k \) \( \implies s \times t^2 = 100 \) This is the equation of variation. \( \implies \) The constant of variation is 100 and the equation of variation is \( st^2 = 100 \). (iv) \( x \propto \frac{1}{\sqrt{y}} \) ...[Given] \( \implies x = k \times \frac{1}{\sqrt{y}} \) where, k is the constant of variation, \( \implies x \times \sqrt{y} = k \) ...(i) When x = 15, y = 9 \( \implies \) Substituting x = 15 and y = 9 in (i), we get \( x \times \sqrt{y} = k \) \( \implies 15 \times \sqrt{9} = k \) \( \implies k = 15 \times 3 \) \( \implies k = 45 \) Substituting k = 45 in (i), we get \( x \times \sqrt{y} = k \) \( \implies x \times \sqrt{y} = 45 \). This is the equation of variation. \( \implies \) The constant of variation is k = 45 and the equation of variation is \( x\sqrt{y} = 45 \).In simple words: For each given inverse variation relationship, we first express it as an equation using a constant 'k'. Then, we use the provided values of the variables to calculate the constant of variation 'k'. Finally, we rewrite the equation with the calculated 'k' to get the specific equation of variation.

🎯 Exam Tip: Always clearly identify the type of variation (direct, inverse, or joint) from the problem statement. The steps for finding the constant of variation (k) and the equation of variation are standard for all problems of this type. Ensure careful calculation, especially with squares or square roots.

 

Question 3. The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed?

Answer:Solution: Let x represent the number of apples in each box and y represent the total number of boxes required. The number of apples in each box are varying inversely with the total number of boxes. \( x \propto \frac{1}{y} \) \( \implies x = k \times \frac{1}{y} \) where, k is the constant of variation, \( \implies x \times y = k \) ...(i) If 24 apples are put in a box then 27 boxes are needed. i.e., when x = 24, y = 27 \( \implies \) Substituting x = 24 and y = 27 in (i), we get \( x \times y = k \) \( \implies 24 \times 27 = k \) \( \implies k = 648 \) Substituting k = 648 in (i), we get \( x \times y = k \) \( \implies x \times y = 648 \) ...(ii) This is the equation of variation. Now, we have to find number of boxes needed when, 36 apples are filled in each box. i.e., when x = 36, y = ? \( \implies \) Substituting x = 36 in (ii), we get \( x \times y = 648 \) \( \implies 36 \times y = 648 \) \( \implies y = \frac{648}{36} \) \( \implies y = 18 \) \( \implies \) If 36 apples are filled in a box then 18 boxes are required.In simple words: This is an inverse variation problem because more apples per box means fewer boxes are needed. We find the constant of variation (total apples) by multiplying the initial number of apples per box (24) by the number of boxes (27), which is 648. Then, we divide this constant by the new number of apples per box (36) to find the required number of boxes (18).

🎯 Exam Tip: Identify the relationship type (direct or inverse) correctly first. Clearly define your variables (e.g., x for apples per box, y for number of boxes) and follow through with a clear calculation of the constant of variation. Precision in arithmetic is key to arriving at the correct final answer.

 

Question 4. Write the following statements using symbol of variation.
1. The wavelength of sound (l) and its frequency (f) are in inverse variation.
2. The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.

Answer:Solution: 1. \( l \propto \frac{1}{f} \) 2. \( I \propto \frac{1}{d^2} \)In simple words: Inverse variation means that as one quantity increases, the other decreases proportionally. If it's inversely proportional to a power (like square), then we write 1 divided by that power of the variable.

🎯 Exam Tip: Focus on accurately translating verbal descriptions into mathematical symbols. Pay close attention to keywords like "inversely proportional," "square," or "square root," as these dictate the form of the variation symbol and the variable's power in the denominator.

 

Question 5. \( x \propto \frac{1}{\sqrt{y}} \) and when x = 40 then y = 16. If x = 10, find y.

Answer:Solution: Given that, \( x \propto \frac{1}{\sqrt{y}} \) \( \implies x = k \times \frac{1}{\sqrt{y}} \) where, k is the constant of variation. \( \implies x \times \sqrt{y} = k \) ...(i) When x = 40, y = 16 \( \implies \) Substituting x = 40 and y = 16 in (i), we get \( x \times \sqrt{y} = k \) \( \implies 40 \times \sqrt{16} = k \) \( \implies k = 40 \times 4 \) \( \implies k = 160 \) Substituting k = 160 in (i), we get \( x \times \sqrt{y} = k \) \( \implies x \times \sqrt{y} = 160 \) ...(ii) This is the equation of variation. When x = 10, y = ? \( \implies \) Substituting, x = 10 in (ii), we get \( x \times \sqrt{y} = 160 \) \( \implies 10 \times \sqrt{y} = 160 \) \( \implies \sqrt{y} = \frac{160}{10} \) \( \implies \sqrt{y} = 16 \) \( \implies y = 256 \) ... [Squaring both sides]In simple words: We are given an inverse variation where x is proportional to 1 over the square root of y. First, we find the constant of variation 'k' using the given values (x=40, y=16), which turns out to be 160. Then, we use this constant and the new value of x (x=10) in the variation equation to solve for y, finding y = 256.

🎯 Exam Tip: When dealing with square roots in variation problems, remember to square both sides of the equation at the final step to solve for the variable itself, not just its square root. Clearly state each step, especially the calculation of the constant of variation.

 

Question 6. x varies inversely as y, when x = 15 then y = 10, if x = 20, then y = ?

Answer:Solution: Given that, x varies inversely as y. \( x \propto \frac{1}{y} \) \( \implies x = k \times \frac{1}{y} \) where, k is the constant of variation. \( \implies x \times y = k \) ...(i) When x = 15, y = 10 \( \implies \) Substituting, x = 15 and y = 10 in (i), we get \( x \times y = k \) \( \implies 15 \times 10 = k \) \( \implies k = 150 \) Substituting, k = 150 in (i), we get \( x \times y = k \) \( \implies x \times y = 150 \) ...(ii) This is the equation of variation. When x = 20, y = ? \( \implies \) substituting x = 20 in (ii), we get \( x \times y = 150 \) \( \implies 20 \times y = 150 \) \( \implies y = \frac{150}{20} \) \( \implies y = 7.5 \)In simple words: Since x and y vary inversely, their product (x * y) is a constant. We calculate this constant 'k' using the initial values (15 * 10 = 150). Then, using this constant and the new value of x (20), we find y by dividing the constant by x (150 / 20 = 7.5).

🎯 Exam Tip: For inverse variation, the product of the two varying quantities is always constant. Clearly establish the constant of variation (k) first, as it forms the basis for finding any unknown values. Double-check your calculations for fractions and decimals.