Direct and Inverse Proportion Explained with Real-Life Examples
In mathematics, direct and inverse proportion are important concepts that help us solve real-life problems efficiently. In this blog, we’ll walk through two examples: the cost of onions and the number of workers completing a task. We will determine when to use division or multiplication based on the nature of the proportion.
Example 1: Direct Proportion
We’re given that 7 kg of onions cost ₹140. Our task is to find the cost of 12 kg of onions.
- Identify the relationship:
As the weight of onions increases, the cost also increases. Since both values increase, this is a direct proportion. - Set up the equation:
Let the cost of 12 kg of onions be X rupees. Using direct proportion, we divide the values:
- Solve the equation:
Cross multiply:
7×X=12×1407 \times X = 12 \times 1407×X=12×140
Simplify:
Reducing 140 ÷ 7 gives 20, so:
X=12×20=240X = 12 \times 20 = 240X=12×20=240
Answer: The cost of 12 kg of onions is ₹240.
Example 2: Inverse Proportion
Next, we have 5 workers who complete a job in 12 days, and we need to find how long it will take for 6 workers to do the same job.
- Identify the relationship:
When the number of workers increases, the time required to finish the work decreases. This is an inverse proportion. - Set up the equation:
Let the number of days for 6 workers be X. In inverse proportion, we multiply the values:
5×12=6×X5 \times 12 = 6 \times X5×12=6×X
Solve the equation:
Rearrange:
X=5×126X = \frac{5 \times 12}{6}
Simplify:
X=606=10X = \frac{60}{6} = 10
Answer: 6 workers will take 10 days to complete the task.
Conclusion
Direct and inverse proportions are powerful tools in solving various real-life problems. Watch the video below for a step-by-step explanation of these examples and more!
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Watch the full video below to get a better understanding of solving problems related to direct and inverse proportions. Subscribe to Wit Reach on YouTube for more free academic content, and sign up for our email newsletter for updates on new videos!
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