## Chapter 7 - Variation

**Practice set 7.1**

**1. Write the following statements using the symbol of variation.**

**1. Write the following statements using the symbol of variation.**

**(1) Circumference (c) of a circle is directly proportional to its radius (r). **

**Solution:**

**c ∝ r**

**(2) Consumption of petrol (l) in a car and distance travelled by that car (d) are in direct variation. **

**Solution:**

**l ∝ d**

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

**Solution:**

**x ∝ y … [k is constant of variation]**

**∴ x = ky**

** **

**When x = 1 then y = 8 … [Given]**

**∴ 1 = k × 8**

**∴ k = \(\large \frac {1}{8}\)**

** **

**The equation of variation is x = \(\large \frac {1}{8}\) y**

** **

**(i) When y = 56 then x = ?**

**Substituting y = 56 in equation of variation**

**x = \(\large \frac {1}{8}\) × 56**

**∴ x = 7**

** **

**(ii) When x = 12 then y = ?**

**Substituting x = 12 in equation of variation**

**∴ 12 = \(\large \frac {1}{8}\) × y**

**∴ 12 × 8 = y**

**∴ y = 96**

** **

**(iii) When y = 160 then x = ?**

**Substituting y = 160 in equation of variation.**

**∴ x = \(\large \frac {1}{8}\) × 160**

**∴ x = 20**

**3. If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14 **

**3. If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14**

**Solution:**

**m ∝ n … [Given]**

**∴ m = kn … [k is constant of variation]**

** **

**Substituting m = 154 and n = 7 we get,**

**154 = k × 7**

**∴ k = \(\large \frac {154}{7}\)**

**∴ k = 22**

** **

**The equation of variation is m = 22 × n**

** **

**Substituting n = 14 in equation of variation, we get,**

**m = 22 × 14**

**∴ m = 308**

** **

**Ans:** The value of m is 308

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

**Solution:**

**n ∝ m … [Given]**

**∴ n = km … [k is constant of variation] **

** **

**Substituting m = 3 and n = 12 we get,**

**12 = k × 3**

**\(\large \frac {12}{3}\) = k**

**∴ k = 4**

** **

**∴ The equation of variation is n = 4 × m**

** **

**(i) If m = 6.5 then n = ?**

**Substituting m = 6.5 in equation of variation**

**n = 4 × 6.5**

**∴ n = 26**

** **

**(ii) If n = 28 then m = ?**

**Substituting n = 28 in equation of variation.**

**28 = 4 m**

**∴ \(\large \frac {28}{4}\) = m**

**∴ m = 7**

** **

**(iii) If m = 1.25 then n = ?**

**Substituting m = 1.25 in equation of variation**

**n = 4 × 1.25**

**∴ n = 5**

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

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

**Solution:**

**y varies directly as square root of x … [Given]**

**y ∝ \(\sqrt{x}\)**

**y = k \(\sqrt{x}\) … [k is constant of variation]**

** **

**If x = 16 then y = 24 … [Given]**

**∴ 24 = k × \(\sqrt{16}\)**

**∴ 24 = k × 4**

**∴ \(\large \frac {24}{4}\) = k**

**∴ k = 6**

** **

**Constant of variation is 6**

**Equation of variation is y = 6 \(\sqrt{x}\)**

** **

**Ans:** The constant of variation is 6 and the equation of variation is y = 6 \(\sqrt{x}\)

**6. The total remuneration paid to labourers employed to harvest soyabeen is in direct variation with the number of labourers. If the remuneration of 4 labourers is ₹ 1000, find the remuneration of 17 labourers.**

**6. The total remuneration paid to labourers employed to harvest soyabeen is in direct variation with the number of labourers. If the remuneration of 4 labourers is ₹ 1000, find the remuneration of 17 labourers.**

**Solution:**

**Let the total remuneration paid be m and the number of labourers be n.**

**m ∝ n … [Given]**

**∴ m = kn … [k is constant of variation]**

** **

**If n = 4 then m = 1000 … [Given]**

**∴ 1000 = k × 4**

**∴ \(\large \frac {1000}{4}\) = k**

**∴ k = 250**

** **

**∴ The equation of variation is m = 250 × n**

** **

**If n = 17 then m = ?**

**Substituting the value of n in equation of variation.**

**m = 250 × 17**

**∴ m = 4250**

** **

**Ans:** The remuneration paid to 17 labourers is ₹4250.

**Practice set 7.2**

**Practice set 7.2**

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

**Solution:**

**More the number of workers then less days are required to complete the work.**

** **

**So, the number of workers and days vary in inverse proportion.**

** **

**Let the number of workers be m and the number of days be n.**

**m ∝ \(\large \frac {1}{n}\)**

**∴ m × n = k … [k is constant of variation]**

** **

**When m = 30 then n = 6 … [Given]**

**30 × 6 = k**

**∴ k = 180**

** **

**The equation of variation is m × n = 180**

** **

**(i) If n = 12 then m = ?**

**Substituting n = 12 in equation of variation.**

**m × 12 = 180**

**∴ m = \(\large \frac {180}{12}\)**

**∴ m = 15**

** **

**(ii) If m = 10 then n = ?**

**Substituting m = 10 in equation of variation.**

**10n = 180 **

**∴ n = \(\large \frac {180}{10}\)**

**∴ n = 18**

** **

**(iii) If n = 36 then m = ?**

**Substituting n = 36 in equation of variation**

**m × 36 = 180**

**∴ m = \(\large \frac {180}{36}\)**

**∴ m = 5**

**2. Find constant of variation and write equation of variation for every example given below.**

**2. Find constant of variation and write equation of variation for every example given below.****(1) p ∝ \(\large \frac {1}{q}\) ; if p = 15 then q = 4 **

**Solution:**

**p ∝ \(\large \frac {1}{q}\) **

**∴ p × q = k …[k is constant of variation]**

** **

**Substituting p = 15 and q = 4**

**15 × 4 = k**

**k = 60**

** **

**Constant of variation = 60**

**Equation of variation = p × q = 60**

** **

**Ans:** The constant of variation is 60 and the equation of variation is p × q = 60.

** **

**(2) z ∝ \(\large \frac {1}{w}\) ; when z = 2.5 then w = 24**

**Solution:**

**z ∝ \(\large \frac {1}{w}\) **

**∴ z × w = k … [k is constant of variation]**

** **

**Substituting z = 2.5 and w = 24, we get,**

**2.5 × 24 = k**

**∴ k = 60**

** **

**Constant of variation = 60**

**The equation of variation is z × w = 60**

** **

**Ans:** The constant of variation is 60 and the equation of variation is z × w = 60.

** **

**(3) s ∝ \(\large \frac {1}{t²}\) ; if s = 4 then t = 5 **

**Solution:**

**s ∝ \(\large \frac {1}{t²}\)**

**s × t² = k …[k is constant of variation]**

** **

**Substituting s = 4 and t = 5**

**4 × 5² = k**

**∴ k = 4 × 25**

**∴ k = 100**

** **

**Constant of variation = 100**

**Equation of variation is s × t² = 100**

** **

**Ans:** The constant of variation is 100 and the equation of variation is s × t² = 100.

** **

**(4) x ∝ \(\large \frac {1}{\sqrt{2}}\) ; if x = 15 then y = 9**

**Solution:**

**x ∝ \(\large \frac {1}{\sqrt{2}}\)**

**∴ x × \(\sqrt{y}\) = k … [k is constant of variation]**

** **

**Substituting x = 15 and y = 9**

**15 × \(\sqrt{9}\) = k**

**∴ k = 15 × 3**

**∴ k = 45**

** **

**Constant of variation = 45**

**Equation of variation is x × \(\sqrt{y}\) = 45**

** **

**Ans:** The constant of variation is 45 and the equation of variation is x × \(\sqrt{y}\) = 45.

**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 ?**

**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 ?****Solution:**

**Let the number of apples filled in each box be x and number of boxes be y.**

**More the apples filled in each box, less number of boxes will be required. **

**So, this is an example of inverse variation.**

**x ∝ \(\large \frac {1}{y}\) **

**∴ xy = k … [k is constant of variation]**

**Substituting x = 24 and y = 27**

**24 × 27 = k**

**k = 648**

**The equation of variation is x × y = 648**

**Substituting x = 36 in the equation of variation.**

**36 × y = 648**

**∴ y = \(\large \frac {648}{36}\)**

**∴ y = 18**

**Ans:** 18 boxes will be needed.

**4. Write the following statements using symbol of variation .**

**4. Write the following statements using symbol of variation .****(1) The wavelength of sound (l) and its frequency (f) are in inverse variation.**

**Ans: **

**l ∝ \(\large \frac {1}{f}\)**

** **

**(2) The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp. **

**Ans: **

**I ∝ \(\large \frac {1}{d²}\)**

**5. x ∝ \(\large \frac {1}{y}\) and when x = 40 then y = 16. If x = 10, find y.**

**5. x ∝ \(\large \frac {1}{y}\) and when x = 40 then y = 16. If x = 10, find y.****Solution:**

**x ∝ \(\large \frac {1}{y}\)**

**∴ x × \(\sqrt{y}\) = k … [k is constant of variation] **

** **

**Substituting x = 40 and y = 16, we get,**

**40 \(\sqrt{16}\) = k**

**∴ k = 40 * 4**

**∴ k = 160**

** **

**∴ The equation of variation is x × \(\sqrt{y}\) = 160 **

** **

**Substituting x = 10 in equation of variation, we get,**

**10 × \(\sqrt{y}\) = 160 **

**∴ \(\sqrt{y}\) = \(\large \frac {160}{10}\)**

**∴ \(\sqrt{y}\) = 16**

**∴ \(\sqrt{y²}\) = 16²**

**∴ y = 256**

** **

**Ans:** The value of y is 256.

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

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

**x varies inversely as y … [Given]**

**x ∝ \(\large \frac {1}{y}\)**

**∴ x × \(\sqrt{y}\) = k …[k is constant of variation] **

** **

**Substituting x = 15 and y = 10, we get,**

**15 × 10 = k**

**∴ k = 150**

** **

**The equation of variation is x × y = 150 **

** **

**Substituting x = 20 in equation of variation, we get,**

**20 × y = 150**

**y = \(\large \frac {150}{20}\)**

**∴ y = 7.5**

** **

**Ans:** The value of y is 7.5

**Practice set 7.3**

**Practice set 7.3****1. Which of the following statements are of inverse variation?**

**1. Which of the following statements are of inverse variation?****(1) Number of workers on a job and time taken by them to complete the job.**

**Solution:**

**This is an example of inverse variation.**

** **

**(2) Number of pipes of same size to fill a tank and the time taken by them to fill the tank.**

**Solution:**

**This is an example of inverse variation.**

** **

**(3) Petrol filled in the tank of a vehicle and its cost**

**Solution:**

**This is an example of direct variation.**

** **

**(4) Area of circle and its radius.**

**Solution:**

**This is an example of direct variation.**

**2. If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?**

**2. If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?****Solution:**

**Let the number of hours be t and number of workers be n.**

**Number of workers = n = 15**

**Time = t = 48 hours**

** **

**The number of hours to build a wall is inversely proportional to the number of workers.**

** **

**∴ t ∝ \(\large \frac {1}{n}\)**

**∴ tn = k … [k is constant of variation]**

** **

**Substituting t = 48 and n = 15, we get**

**48 × 15 = k**

**∴ k = 720**

** **

**∴ The equation of variation is t × n = 720**

** **

**Substituting t = 30 in equation of variation, we get,**

**30n = 720**

**∴ n = \(\large \frac {720}{30}\)**

**∴ n = 24**

** **

**Ans:** 24 workers can build the wall in 30 hours.

**3. 120 bags of half litre milk can be filled by a machine within 3 minutes. Find the time to fill such 1800 bags ?**

**3. 120 bags of half litre milk can be filled by a machine within 3 minutes. Find the time to fill such 1800 bags ?****Solution:**

**Let the number of bags be n and time required to fill them be t.**

**Number of bags = n = 120**

**Time = t = 3 hours**

** **

**More the number of bags to be filled by machine, more time is required. **

**So, this is an example of direct variation.**

** **

**∴ n ∝ t**

**∴ n = kt … [k is constant of variation]**

** **

**Substituting n = 120 and t = 3, we get,**

**120 = k × 3**

**∴ k = \(\large \frac {120}{3}\)**

**∴ k = 40**

** **

**The equation of variation is n = 40 × t**

** **

**Substituting n = 1800 in equation of variation, we get,**

**1800 = 40t**

**∴ t = \(\large \frac {1800}{40}\)**

**∴ t = 45**

** **

**Ans:** 1800 bags of milk can be filled by the machine in 45 minutes.

**4. A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in 7 \(\large \frac {1}{2}\) hours?**

**4. A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in 7 \(\large \frac {1}{2}\) hours?****Solution:**

**Let the speed of the car be s and time required to cover some distance be t.**

** **

**There is an inverse variation between speed and time.**

** **

**∴ s ∝ \(\large \frac {1}{t}\)**

**∴ st = k … [k is constant of variation]**

** **

**Substituting s = 60 and t = 480 … [1 hour = 60 minutes]**

** **

**∴ 60 × 480 = k**

**∴ k = 28800**

** **

**t = 7 \(\large \frac {1}{2}\) hours**

**∴ t = 7 × 60 + 30**

**∴ t = 450 minutes**

** **

**The equation of variation is s × t = 28800**

** **

**Substituting t = 450 minutes in equation of variation, we get,**

**s × 450 = 28800**

**∴ s = \(\large \frac {28800}{450}\)**

**∴ s = 64**

** **

**Hence, if speed of car is 64 km/hr then the same distance can be covered in 7 \(\large \frac {1}{2}\) hours**

** **

**∴ Increase in speed = 64 – 60 **

**∴ Increase in speed = 4 km/hrs. **

** **

**Ans:** Speed of the car should be increased by 4 km/hr.