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Understanding Compound Proportion: Detailed Notes And Solved MCQs

This educational post offers a clear and concise explanation of compound proportion, complete with detailed notes and solved multiple-choice questions (MCQs). Designed for students, it provides step-by-step solutions to enhance understanding of both direct and inverse proportions, making it a valuable resource for exam preparation and academic success.    

Prakash Joshi
updated: 16 Aug 2024

Introduction:

Compound Proportion (also called Joint Proportion) deals with comparing two or more quantities that are linked by ratios. When multiple ratios are involved, we use the concept of compound proportion to find the unknown value, ensuring that the relationships between the different quantities are maintained. 


Types of Proportions:

  1. Direct Proportion: Two quantities are in direct proportion if they increase or decrease together in such a way that their ratio remains constant.

    • Example: More hours worked → More pay earned.
  2. Inverse Proportion: Two quantities are in inverse proportion if one quantity increases while the other decreases, such that their product remains constant.

    • Example: More workers → Less time to complete a job.
  3. Compound Proportion: It is the combined effect of two or more direct or inverse proportions working simultaneously. The unknown quantity is calculated by using a combination of these proportions.


Steps to Solve Compound Proportion Problems:

  1. Identify the type of proportion (direct or inverse) involved between the known quantities and the unknown quantity.
  2. Set up the ratios for each proportion.
  3. Formulate an equation involving all proportions and solve for the unknown quantity.

Formula for Compound Proportion:

Suppose two or more quantities A1,A2,...,AnA_1, A_2, ..., A_nA1​,A2​,...,An​ are in proportion to other quantities B1,B2,...,BnB_1, B_2, ..., B_nB1​,B2​,...,Bn​, respectively, and we need to find an unknown quantity xxx. We use the relation:

x=(A1B1)×(A2B2)×...×(AnBn)×(constant)x = \left(\frac{A_1}{B_1}\right) \times \left(\frac{A_2}{B_2}\right) \times ... \times \left(\frac{A_n}{B_n}\right) \times \text{(constant)}x=(B1​A1​​)×(B2​A2​​)×...×(Bn​An​​)×(constant)

This equation allows us to solve for the unknown by considering all the factors involved.




Solved MCQs / Problems:


Problem 1:

If 8 workers can finish a task in 12 days, how many days will it take for 6 workers to complete the same task, assuming the rate of work remains constant?

  • a) 8 days
  • b) 10 days
  • c) 14 days
  • d) 16 days

Solution:

Here, we have an inverse proportion because fewer workers will take more time to complete the task. We set up the inverse proportion as:

8 workers6 workers=New time12 days\frac{8 \text{ workers}}{6 \text{ workers}} = \frac{\text{New time}}{12 \text{ days}}6 workers8 workers​=12 daysNew time​

Cross-multiplying gives:

New time=86×12=16 days\text{New time} = \frac{8}{6} \times 12 = 16 \text{ days}New time=68​×12=16 days

Answer: d) 16 days


Problem 2:

A car covers 120 km in 3 hours. How long will it take to cover 200 km at the same speed?

  • a) 4 hours
  • b) 5 hours
  • c) 6 hours
  • d) 7 hours

Solution:

This is a direct proportion because more distance requires more time. The ratio of distance to time is constant.

120 km3 hours=200 kmx hours\frac{120 \text{ km}}{3 \text{ hours}} = \frac{200 \text{ km}}{x \text{ hours}}3 hours120 km​=x hours200 km​

Cross-multiplying gives:

x=200×3120=5 hoursx = \frac{200 \times 3}{120} = 5 \text{ hours}x=120200×3​=5 hours

Answer: b) 5 hours


Problem 3:

If 12 machines can produce 240 units in 8 hours, how many units can 15 machines produce in 10 hours, working at the same rate?

  • a) 300 units
  • b) 450 units
  • c) 400 units
  • d) 500 units

Solution:

This is a compound proportion because the number of machines and the number of hours both change. We use both direct proportions (machines and hours) together.

Let the number of units be xxx.

12 machines15 machines×8 hours10 hours=240 unitsx units\frac{12 \text{ machines}}{15 \text{ machines}} \times \frac{8 \text{ hours}}{10 \text{ hours}} = \frac{240 \text{ units}}{x \text{ units}}15 machines12 machines​×10 hours8 hours​=x units240 units​

Simplifying,

1215×810=240x\frac{12}{15} \times \frac{8}{10} = \frac{240}{x}1512​×108​=x240​

Cross-multiply and solve for xxx:

x=240×15×1012×8=450 unitsx = \frac{240 \times 15 \times 10}{12 \times 8} = 450 \text{ units}x=12×8240×15×10​=450 units

Answer: b) 450 units


Problem 4:

10 pumps can empty a tank in 6 hours. How many hours would it take 8 pumps to empty the tank?

  • a) 5 hours
  • b) 7.5 hours
  • c) 8 hours
  • d) 9 hours

Solution:

Since the number of pumps is inversely proportional to the time taken:

10 pumps8 pumps=6 hoursx hours\frac{10 \text{ pumps}}{8 \text{ pumps}} = \frac{6 \text{ hours}}{x \text{ hours}}8 pumps10 pumps​=x hours6 hours​

Cross-multiplying gives:

x=108×6=7.5 hoursx = \frac{10}{8} \times 6 = 7.5 \text{ hours}x=810​×6=7.5 hours

Answer: b) 7.5 hours


Problem 5:

A recipe requires 5 cups of flour to make 10 cakes. How many cakes can be made with 8 cups of flour?

  • a) 12 cakes
  • b) 14 cakes
  • c) 16 cakes
  • d) 18 cakes

Solution:

This is a direct proportion since more flour will make more cakes.

5 cups8 cups=10 cakesx cakes\frac{5 \text{ cups}}{8 \text{ cups}} = \frac{10 \text{ cakes}}{x \text{ cakes}}8 cups5 cups​=x cakes10 cakes​

Cross-multiplying gives:

x=10×85=16 cakesx = \frac{10 \times 8}{5} = 16 \text{ cakes}x=510×8​=16 cakes

Answer: c) 16 cakes


Conclusion:

Compound proportion is an essential mathematical concept that applies in real-world situations involving multiple related variables. By analyzing the proportional relationships and solving the equations systematically, we can solve complex problems involving direct and inverse proportions.




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