Maharashtra Board Class 8 Maths Chapter 7 Variation Set 7.1 Solutions

Question 1. Write the following statements using the symbol of variation.
(i) Circumference (c) of a circle is directly proportional to its radius (r).
(ii) Consumption of petrol (l) in a car and distance traveled by that car (d) are in direct variation.
Answer:
(i) \(c \propto r\)
(ii) \(l \propto d\)
In simple words: This question asks us to express relationships of direct proportionality using the symbol \( \propto \). When two quantities are directly proportional, it means they increase or decrease together at a constant rate.

🎯 Exam Tip: Understanding the symbol of variation (\( \propto \)) is fundamental for solving problems related to direct and inverse proportion. Make sure to clearly identify which quantities are being related.

Question 2. Complete the following table considering that the cost of apples and their number are in direct variation.

Answer:
The cost of apples (y) and their number (x) are in direct variation.
\(y \propto x\)
\( \implies y = kx \) ...(i)
where k is the constant of variation

(i) When, \(x = 1, y = 8\)
Substituting, \(x = 1\) and \(y = 8\) in (i), we get \(y = kx\)
\( \implies 8 = k \times 1\)
\( \implies k = 8\)
Substituting \(k = 8\) in (i), we get
\(y = kx\)
\( \implies y = 8x \) ...(ii)
This is the equation of variation

(ii) When, \(y = 56, x = ?\)
Substituting \(y = 56\) in (ii), we get
\(y = 8x\)
\( \implies 56 = 8x\)
\( \implies x = \frac{56}{8}\)
\( \implies x = 7\)

(iii) When, \(x = 12, y = ?\)
Substituting \(x = 12\) in (ii), we get
\(y = 8x\)
\( \implies y = 8 \times 12\)
\( \implies y = 96\)

(iv) When, \(y = 160, x = ?\)
Substituting \(y = 160\) in (ii), we get
\(y = 8x\)
\( \implies 160 = 8x\)
\( \implies x = \frac{160}{8}\)
\( \implies x = 20\)

Completed table:

In simple words: Since the cost of apples and their number are in direct variation, we found a constant of proportionality (k=8) and used it to complete the table. This means the cost is always 8 times the number of apples.

🎯 Exam Tip: Always calculate the constant of variation (k) first when dealing with proportionality problems. This constant is crucial for writing the equation of variation and then finding unknown values.

Question 3. If \(m \propto n\) and when \(m = 154, n = 7\). Find the value of m, when \(n = 14\).
Answer:
Given that,
\(m \propto n\)
\( \implies m = kn \) ...(i)
where k is constant of variation.

When \(m = 154, n = 7\)
Substituting \(m = 154\) and \(n = 7\) in (i), we get
\(m = kn\)
\( \implies 154 = k \times 7\)
\( \implies k = \frac{154}{7}\)
\( \implies k = 22\)

Substituting \(k = 22\) in (i), we get
\(m = kn\)
\( \implies m = 22n \) ...(ii)
This is the equation of variation.

When \(n = 14, m = ?\)
Substituting \(n = 14\) in (ii), we get
\(m = 22n\)
\( \implies m = 22 \times 14\)
\( \implies m = 308\)
In simple words: We first found the constant of variation (k=22) using the given values. Then, we used this constant to form the equation of variation, \(m = 22n\), and calculated m when n is 14.

🎯 Exam Tip: Double-check your calculations for the constant of variation as any error here will lead to incorrect final answers. Clearly state the constant of variation and the equation of variation.

Question 4. If n varies directly as m, complete the following table.

Answer:
Given, n varies directly as m
\( \implies n \propto m\)
\( \implies n = km \) ...(i)
where, k is the constant of variation

(i) When \(m = 3, n = 12\)
Substituting \(m = 3\) and \(n = 12\) in (i), we get
\(n = km\)
\( \implies 12 = k \times 3\)
\( \implies k = \frac{12}{3}\)
\( \implies k = 4\)
Substituting, \(k = 4\) in (i), we get
\(n = km\)
\( \implies n = 4m \) ...(ii)
This is the equation of variation.

(ii) When \(m = 6.5, n = ?\)
Substituting, \(m = 6.5\) in (ii), we get
\(n = 4m\)
\( \implies n = 4 \times 6.5\)
\( \implies n = 26\)

(iii) When \(n = 28, m = ?\)
Substituting, \(n = 28\) in (ii), we get
\(n = 4m\)
\( \implies 28 = 4m\)
\( \implies m = \frac{28}{4}\)
\( \implies m = 7\)

(iv) When \(m = 1.25, n = ?\)
Substituting \(m = 1.25\) in (ii), we get
\(n = 4m\)
\( \implies n = 4 \times 1.25\)
\( \implies n = 5\)

Completed table:

In simple words: We found the constant of variation (k=4) from the given values, forming the equation \(n = 4m\). Then we used this equation to fill in the missing values in the table for both n and m.

🎯 Exam Tip: When completing tables, it's efficient to first determine the constant of variation and the variation equation. This allows you to quickly calculate all missing values consistently.

Question 5. y varies directly as square root of x. When \(x = 16, y = 24\). Find the constant of variation and equation of variation.

Answer:
Given, y varies directly as square root of x.
\( \implies y \propto \sqrt{x}\)
\( \implies y = k\sqrt{x} \) ...(i)
where, k is the constant of variation.

When \(x = 16, y = 24\).
Substituting, \(x = 16\) and \(y = 24\) in (i), we get
\(y = k\sqrt{x}\)
\( \implies 24 = k\sqrt{16}\)
\( \implies 24 = 4k\)
\( \implies k = \frac{24}{4}\)
\( \implies k = 6\)
Substituting \(k = 6\) in (i), we get
\(y = k\sqrt{x}\)
\( \implies y = 6\sqrt{x}\)
This is the equation of variation.

The constant of variation is 6 and the equation of variation is \(y = 6\sqrt{x}\).
In simple words: Since y varies directly with the square root of x, we set up the equation \(y = k\sqrt{x}\). Using the given values, we solved for the constant k, which is 6. Then, we wrote the final equation of variation using this constant.

🎯 Exam Tip: Pay close attention to the exact relationship stated (e.g., "square root of x"). Correctly formulating the initial variation equation (\(y = k\sqrt{x}\)) is essential for accurate solutions.

Question 6. The total remuneration paid to laborers, employed to harvest soybean is in direct variation with the number of laborers. If remuneration of 4 laborers is Rs. 1000, find the remuneration of 17 laborers.
Answer:
Let, m represent total remuneration paid to laborers and n represent number of laborers employed to harvest soybean.
Since, the total remuneration paid to laborers, is in direct variation with the number of laborers.
\( \implies m \propto n\)
\( \implies m = kn \) ...(i)
where, k = constant of variation

Remuneration of 4 laborers is Rs. 1000.
i.e., when \(n = 4, m = \text{Rs. } 1000\)
Substituting, \(n = 4\) and \(m = 1000\) in (i), we get \(m = kn\)
\( \implies 1000 = k \times 4\)
\( \implies k = \frac{1000}{4}\)
\( \implies k = 250\)
Substituting, \(k = 250\) in (i), we get
\(m = kn\)
\( \implies m = 250n \) ...(ii)
This is the equation of variation.

Now, we have to find remuneration of 17 laborers.
i.e., when \(n = 17, m = ?\)
Substituting \(n = 17\) in (ii), we get
\(m = 250n\)
\( \implies m = 250 \times 17\)
\( \implies m = 4250\)
The remuneration of 17 laborers is Rs. 4250.
In simple words: We established a direct variation between remuneration and the number of laborers. We found the constant of variation (k=250) using the given data. With this, we formulated the equation of variation \(m=250n\) and calculated the remuneration for 17 laborers.

🎯 Exam Tip: Clearly define your variables (m for remuneration, n for laborers) at the start. This helps in setting up the correct variation equation and avoids confusion.

Maharashtra Board Class 8 Maths Chapter 7 Variation Practice Set 7.1 Intext Questions And Activities

Question 1. If the rate of notebooks is Rs. 240 per dozen, what is the cost of 3 notebooks? Also find the cost of 9 notebooks, 24 notebooks and 50 notebooks and complete the following table. (Textbook pg. no. 35)

Answer:
As the number of notebooks increases their cost also increases.
The number of notebooks and cost of notebooks are in direct proportion.

(i) \( \frac{12}{240} = \frac{3}{y} \)
\( \implies \frac{1}{20} = \frac{3}{y} \)
\( \implies y = 3 \times 20\)
\( \implies y = 60\)

(ii) \( \frac{12}{240} = \frac{9}{y} \)
\( \implies \frac{1}{20} = \frac{9}{y} \)
\( \implies y = 9 \times 20\)
\( \implies y = 180\)

(iii) \( \frac{12}{240} = \frac{24}{y} \)
\( \implies \frac{1}{20} = \frac{24}{y} \)
\( \implies y = 24 \times 20\)
\( \implies y = 480\)

(iv) \( \frac{12}{240} = \frac{50}{y} \)
\( \implies \frac{1}{20} = \frac{50}{y} \)
\( \implies y = 50 \times 20\)
\( \implies y = 1000\)

For 1 notebook: Since 12 notebooks cost Rs. 240, 1 notebook costs \( \frac{240}{12} = \text{Rs. } 20 \).

Completed table:

In simple words: We used the direct proportionality between the number of notebooks and their cost. Knowing that 12 notebooks cost Rs. 240, we established a ratio and then applied it to find the costs for 3, 9, 24, 50, and 1 notebook.

🎯 Exam Tip: For direct proportion problems involving tables, it's often easiest to find the unit cost or the constant of proportionality first. In this case, the cost per notebook is Rs. 20 (240/12). Then, multiply this unit cost by the number of notebooks to fill the table efficiently.