Direct and Inverse Proportions Class 8 Important Questions: CBSE Maths Chapter 11

Very Short Answer Type Questions

1. If two quantities x and y are in direct proportion with each other, then:

(a) $\dfrac{{\text{x}}}{{\text{y}}}$ remains constant

(b) $x\times y$ remains constant

(c) ${\text{x}}\,{\text{ - }}\,{\text{y}}$ remains constant

(d) None of these

Ans: (a) $\dfrac{{\text{x}}}{{\text{y}}}$ remains constant. 

2. The cost of 5 metres of a particular quality of cloth is Rs.210. Find the cost of 2 metres of cloth of the same type.

(a) Rs. 84

(b) Rs. 60

(c) Rs. 90

(d) Rs. 100

Ans: Cost of 5 metre cloth = Rs. 210

Thus, cost of 2 metre cloth = $\dfrac{{2 \times 210}}{5}\, = \,{\text{Rs}}{\text{.84}}$

3. If X = 5Y, then X and Y vary ______ with each other.

Ans: They are directly proportional.

X = 5Y

${\text{X }} \propto {\text{ Y}}$

$\dfrac{{\text{X}}}{{\text{Y}}}\, = \,5$

4. If XY = 10 then X and Y vary _____ with each other.

Ans: Indirectly proportional.

5. Time taken to cover a distance by car and speed of the car are said to be in _______ variation.

Ans: Inversely

${\text{Speed  =  }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$

${\text{speed }} \propto \,\dfrac{1}{{{\text{time}}}}$

6. In the table state whether x and y vary directly or indirectly.

Ans: Since, as ‘x’ increases, ‘y’ also increases. 

Therefore, ‘x’ and ‘y’ vary directly.

7. If a car covers 80km in 5 litres of petrol, how much distance will it cover in 3 litres of petrol?

Ans: Given: In 5 litres of petrol, distance covered = 80km

Thus, in 1 litre of petrol, distance covered = $\dfrac{{80}}{5} = {\text{ 16km}}$

In, 13 litres of petrol, distance covered = $16\, \times \,13\, = \,{\text{208km}}$

Short  Answer Type Questions                                                        2 Mark

8. If 32 men can reap a field in 15 days. In how many days can 40 men reap the same field?

Ans: This situation is inverse variation (less men, more days)

Let $x=$ men,$y=$ no. of days $x_{1}=32, y_{1}=15$ $x_{2}=40, y_{2}=?$

Formula is $x_1 y_1=x_2 y_2$

$(32)(15)=(40) \mathrm{y} 2$

$\mathrm{y}_{2}=\dfrac{32 \times 15}{40}$

$\mathrm{y}_{2}=12$

Therefore, number of days $=12$

9. If 4 kg potatoes cost Rs. 60. What is the cost of 12kg of potatoes?

Ans: Given: cost of 4 kg potatoes = Rs. 60

Therefore, 1 kg potatoes cost Rs = $\dfrac{{60}}{4} = 15$

Thus, 12kg potatoes cost = $15\, \times \,12\, = \,{\text{Rs}}{\text{.180}}$

10. Find the value of x and y if x : y = 2 : 3 and 2 : x = 1 : 2.

Ans: Given: 2 : x = 1 : 2

$\Rightarrow {\text{ }}\dfrac{2}{{\text{x}}} = \dfrac{1}{2}$

 $\Rightarrow \,{\text{x  =  2}} \times {\text{2}} $

 $\Rightarrow {\text{x = 4}} $

  ${\text{x : y  =  2 : 3}} $

 $ {\text{4 : y  =  2 : 3}} $

  $\dfrac{4}{{\text{y}}}\, = \,\dfrac{2}{3} $

  ${\text{y  =  }}\dfrac{{4 \times 3}}{2} $

  ${\text{y  =  6}} $ 

11. If 2 : 3 = x : 51. Find ‘x’.

Ans:$\dfrac{2}{3}\, = \,\dfrac{{\text{x}}}{{51}} $

   $\Rightarrow \,{\text{x  =  }}\dfrac{{2 \times 51}}{3} $

   $\Rightarrow {\text{x  =  34}} $ 

Short  Answer Type Questions                                                        3 Mark

12. If x and y are in inverse proportion. Find the value of a, b and c in the table.

Ans:

$\mathrm{x} \propto \dfrac{1}{\mathrm{y}} \Rightarrow \mathrm{x_1} \mathrm{y_1}=\mathrm{x_2} \mathrm{y_2}$

1. $25 \times 3=15 \times \mathrm{a} \Rightarrow \mathrm{a}=\dfrac{75}{15}=5$

2. $25 \times 3=\mathrm{b} \times 4 \Rightarrow \mathrm{b}=\dfrac{75}{4}=18.75$

3. $25 \times 3=10 \times \mathrm{c} \Rightarrow \mathrm{c}=\dfrac{75}{10}=7.5$

13. The scale of a map is given as 1 : 80000000. Two places A and B on the map are 3 cm apart. What is the actual distance between A and B? If C and D are at a distance of 3200 km, then find the distance between them on map?

Ans: Given: scale of map = 1 : 80000000

Thus, 1 unit on map shows 80000000 units in the real world.

If A and B are 3 cm apart on map,

Actual distance =

 ${\text{3cm }} \times {\text{ 80000000}} $

  ${\text{ =  240000000}} $

  ${\text{ =  2400 km}} $ 

If C and D are at a distance of 32km apart, then

${\text{3200km  =  3200 }} \times {\text{ 1000 }} \times {\text{ 100cm}} $

  ${\text{ =  320000000 cm}} $ 

Therefore, on the map it should be $ = \,\dfrac{{320000000}}{{80000000}}\, = \,{\text{4cm}}$

14. There are 50 students in a hostel. The food provision for them is for 15 days. How long will their provision last if 5 students leave the group?

Ans: It is inverse variation since no. of students increases as no. of days food provision provided increases.

Let ' $x$ ' be the no. of students And ' $\mathrm{y}$ ' be the number of days 

Given: $\mathrm{x}_{1}=50, \mathrm{y}_{1}=15$

$\mathrm{x}_{2}=50-5=45, \mathrm{y}_{2}=? $

$\mathrm{x_1} \mathrm{y_1}=\mathrm{x_2} \mathrm{y_2}$

$50 \times 15=45 \times \mathrm{y_2}$

$\mathrm{y}_{2}=\dfrac{50 \times 15}{45}$

$\mathrm{y}_{2}=16.66=17$ days

15. A workforce of 210 men with a supervisor can finish a certain piece of work in 5 months. How many extra men must he employ if he want to complete job in just 2 months?

Ans:  Let the extra men employed be ‘x’

Number of men(x):       210           x

Months(y):                         5             2

Since, men hired and time required are inversely proportional, we have

$\mathrm{x_1} \mathrm{y_1}=\mathrm{x_2}\mathrm{y}_{2}$

$210 \times 5=\mathrm{x} \times 2 $

$\mathrm{x}=\dfrac{210 \times 5}{2}=525$

Thus, extra men needed $=525-210=315$.

16. Ranjith has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs.50 on each machines?

Ans: Let the no. of machines he can buy if a discount of Rs. 50 is offered on each machine be ‘x’.

Number of Machines(x):                         75                             x

Price of Each Machine(y):                       200                           150

Since the discount is Rs.50, the cost of each machine will be 200 – 50  = 150.

It is the inverse proportion as if the price of a machine is less, the more machines he can buy.

$75\, \times \,200\, = \,{\text{x}}\, \times {\text{ 150}} $

$\Rightarrow \,{\text{x  =  }}\dfrac{{75\, \times \,200}}{{150}}\, = \,\dfrac{{15000}}{{150}} $

 $\Rightarrow \,{\text{x  =  100}} $ 

17. A worker is paid Rs. 420 for 2 days work. If his total income of the month is Rs. 1750, For how many days did he work?

Ans: It is direct variation. More wages, more days of work.

Let ‘x’ be the amount paid and ‘y’ be the number of days.

Amount paid(x)     :   Rs.420       Rs.1750     

Number of days(y):   12                ?

$\dfrac{\mathrm{x}_1}{\mathrm{y}_1}=\dfrac{\mathrm{x}_2}{\mathrm{y}_2}$

$\dfrac{420}{12}=\dfrac{1750}{\mathrm{y}_2} $

$\mathrm{y}_{2}=\dfrac{1750 \times 12}{420} $

$\mathrm{y}_{2}=50 \text { days }$

18. Abdul takes 75 steps to cover a distance of 50m. How much distance will it cover in 375 steps?

Ans: It is direct variation as the number of steps increases, the distance covered will be more.

Let ‘x’ be the number of steps and ‘y’ be the distance covered.

Number of steps(x):     75      375

Distance covered(y):   50m     ?

$\dfrac{\mathrm{x}_1}{\mathrm{y}_1}=\dfrac{\mathrm{x}_2}{\mathrm{y}_2}$

$\dfrac{75}{50}=\dfrac{375}{\mathrm{y}_2} $

$\mathrm{y}_{2}=\dfrac{375 \times 50}{75} $

$\mathrm{y}_{2}=250 \mathrm{~m}$

19. If the weights of 8 sheets of paper be 45 grams. How many sheets would weigh $1\dfrac{1}{2}$kg?

Ans: It is direct variation as more number of sheets implies more weight.

Number of sheets(x):   6       ?

Number of hours(y):   45     $1\dfrac{1}{2}$kg = 1500g [1 Kg = 1000g, 1.5Kg = 1500g]

$\dfrac{\mathrm{x}_1}{\mathrm{y}_1}=\dfrac{\mathrm{x}_2}{\mathrm{y}_{2}} $

$\dfrac{6}{45}=\dfrac{\mathrm{x}_2}{1500}$

$\mathrm{x}_{2}=\dfrac{6 \times 1500}{45}$

$\mathrm{x}_{2}=200$

20. 20 pumps can empty a reservoir is 12 hours. In how many hours can 45 such pumps do the same work?

Ans: It is inverse variation as it takes less hours if the number of pumps are more.

Number of pumps(x):   20      45

Number of hours(y):     12       ?

$x_1 y_{1}=x_2 y_2 $

$20 \times 12=45 \times y_2 $

$y_{2}=\dfrac{20 \times 12}{45} $

$y_{2}=5.33=5 \dfrac{1}{3} \text { hours }$

Long  Answer Type Questions                                                                         5 Mark  

21. A water tanker can finish a certain journey in 10 hours at the speed of 38 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?

Ans: Given: speed = 38 km/hr, time = 10 hours.

Distance covered = speed $ \times $ time = $38\, \times \,10\, = \,380\,{\text{km}}$

Speed(x)            :     38km/hr          ?

Time taken(y)   :     10 hours        8 hours

It is an inverse variation.

$\mathrm{x}_1 \mathrm{y}_1=\mathrm{x}_2 \mathrm{y}_2 $

$38 \times 10=\mathrm{x}_2 \times 8 $

$\mathrm{x}_{2}=\dfrac{38 \times 10}{8} $

$\mathrm{x}_{2}=47.5 \mathrm{~km} / \mathrm{hr}$

Thus, the speed is increased by $47.5-38=9.5 \mathrm{~km} / \mathrm{hr}$.

22. 1000 children in a hostel had enough food for 28 days. After 4 days, some children were shifted to other hostel. As a result, the food now lasted for 32 days. How many students were shifted?

Ans: Given: 1000 students in a hostel had enough food for 28 days.

Let ‘x’ be the number of students shifted.

Number of students(x):   1000      1000-x

Number of days:                 28           32

It is an inverse variation: as the number of students increases, food remains for less number of days.

${x_1 y_1  =  x_2 y_2} $

$1000\times 28=(1000-x)\times 32 $

$1000-x = \dfrac{1000\times 28}{32}=875 $

$x = 1000-875=125 $ 

Therefore, the number of students shifted = 125.

23. The amount of extension in the length of the elastic string directly varies as the weight hung on it. If a weight of 500 gm produces an extension of 3 cm, then what weight would produce an extension of 36.2 cm. Write the solution in Kg.

Ans: Weight(x)                      :   200gm      7

Extension in length(y) :    3 cm        36.2 cm

It is a direct variation.

$\dfrac{x_1}{y_1}= \dfrac{x_2}{y_2} $

$\dfrac{{200}}{3}\, = \,\dfrac{{{\text{x2}}}}{{36.2}} $

$x_2 = \dfrac{{200\, \times \,36.2}}{3} $

$x_2 = \dfrac{{7240}}{3} $

$x_2 = 2.41kg  $

24. Find ‘a’ in the following table when

(a) x, y vary directly

(b) x, y vary inversely.

Ans: 

a. when x and y vary directly.

$ \dfrac{x_1}{y_1}= \dfrac{x_2}{y_2} $

$ \dfrac{2}{{10}}\, = \,\dfrac{5}{{\text{a}}} $

 $ {\text{a  =  }}\dfrac{{50}}{2}\, = \,25 $ 

b. When x and y vary inversely

$ {x_1 y_1  =  x_2 y_2} $

$ 2\, \times \,10\, = \,5\, \times \,{\text{a}} $

${\text{a  =  }}\dfrac{{20}}{5} $

$ {\text{a  =  4}} $

25. Which of the following quantities vary directly or indirectly with each other

a. Number of pens and their cost

b. Distance travelled (at constant speed) and petrol used.

c. Number of men available and time taken to do a job.

d. Area of land and its price.

e. wages y and hours of work x.

Ans:

a. As pens increase, cost increases – direct variation.

b. As distance travelled increases, the amount of petrol increases – direct variation.

c. Number of men decreases, time taken increases – Inverse variation.

d. Direct variation.

e. Direct variation.

5 Important Formulas of CBSE Class 8 Maths Chapter 11 Direct and Inverse Proportions

Common Mistakes Students Make While Solving Class 8 Maths Chapter 11 - Direct and Inverse Proportions Problems:

Here are common mistakes students make while solving problems in Class 8 Maths Chapter 11 - Direct and Inverse Proportions:

1. Confusing Direct and Inverse Proportions: Students often mix up the concepts of direct and inverse proportions. It's crucial to grasp when one variable increases with the other (direct) and when it decreases (inverse).

2. Misinterpreting Proportionality Constants: Students sometimes miscalculate or misinterpret the constant of proportionality. Pay attention to its role in maintaining the relationship between variables.

3. Skipping Units in Ratios: Ignoring units while setting up ratios can lead to errors. Always include units to ensure accurate proportionality.

4. Overlooking Cross Multiplication: When solving proportions, students may forget to cross-multiply. This step is essential for finding the value of unknowns correctly.

5. Neglecting Unitary Method: Forgetting to apply the unitary method to solve problems related to proportions can hinder accurate solutions. Ensure consistency in units throughout.

6. Not Verifying Answers: Students might forget to check their solutions back into the original problem. Always verify answers to confirm their correctness in the given context.

Being mindful of these common pitfalls can help students navigate Direct and Inverse Proportions with greater accuracy and understanding.

Tips and Tricks to Solve Chapter 11 - Direct and Inverse Proportions with Ease!

Here are 8 tips and tricks to solve "Chapter 11 - Direct and Inverse Proportions" with ease:

1. Understand the Concept: Grasp the difference between direct and inverse proportions. In direct, one variable increases as the other increases, and in inverse, one decreases as the other increases.

2. Identify the Proportionality Constant: Pay attention to the constant that relates the two variables. It's crucial for setting up accurate proportions.

3. Include Units in Ratios: Always include units when setting up ratios. This helps in maintaining consistency and ensuring correct proportionality.

4. Cross-Multiply Carefully: When dealing with proportions, cross-multiplication is key. Double-check your calculations to avoid errors.

5. Apply Unitary Method: Use the unitary method to solve problems related to proportions. It simplifies calculations and ensures accuracy.

6. Verify Answers: After finding solutions, don't forget to check them back into the original problem. This step ensures the correctness of your answers in the given context.

7. Practice Regularly: Practice solving various problems regularly to reinforce your understanding of direct and inverse proportions. It builds confidence and familiarity.

8. Use Real-Life Examples: Relate problems to real-life situations to enhance your understanding. This makes the concepts more tangible and relatable.

By incorporating these tips, you can approach "Direct and Inverse Proportions" with confidence and tackle problems with ease!

What are the Benefits of Important Questions from Vedantu for Class 8 Maths Chapter 11 - Direct and Inverse Proportions 

• Focus on key topics for efficient studying.

• Prepares students for exams and reduces anxiety.

• Reinforces understanding of fundamental concepts.

• Teaches effective time management.

• Enables self-assessment and progress tracking.

• Strategic approach for higher scores.

• Covers a wide range of topics for comprehensive understanding.

• Supports exam preparation and boosts confidence.

Conclusion

Reviewing all the crucial questions for Class 8 Maths Chapter 11 - Direct and Inverse Proportions provides students with a solid understanding of the chapter's topics. By learning these concepts, students can solve practical problems involving proportionality in real-life scenarios. The extra and important questions for Class 8 Maths Chapter 11 - Direct and Inverse Proportions engage in a concept-focused discussion, encompassing all chapter themes. This question-and-answer method proves time-saving during exam prep, offering an efficient way to revise the chapter and enhance understanding. Practising these important questions streamlines preparation and boosts confidence for the upcoming exams.

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