Proportional Reasoning - 1 | IT

Question 13

Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle at a speed of 50 km/h. If he drives at 75 km/h, how long will it take him to reach Kanpur? Can we form this problem as a proportion—

Would it take Puneeth’s father more time or less time to reach Kanpur? Think about it.

Solution

Understand the Question

When the distance between two places is fixed, speed and time are inversely proportional: $\text{Speed} \times \text{Time} = \text{Distance} = \text{Constant}$
As speed increases, the time taken to cover the same distance decreases (takes less time).
Therefore, this relationship cannot be written as a direct proportion ( $S_1 : T_1 :: S_2 : T_2$ ), but rather as an inverse relation ( $S_1 \times T_1 = S_2 \times T_2$ ).

Step 1 · Calculate the Total Distance

Given:

Initial speed, $S_1 = 50\text{ km/h}$
Initial time taken, $T_1 = 2\text{ hours}$

$\text{Distance } (D) = S_1 \times T_1$

\begin{aligned} D &= 50\text{ km/h} \times 2\text{ hours} \\[0.6em] &= 100\text{ km} \end{aligned}

Step 2 · Calculate the New Time Taken

Let the new speed be $S_2 = 75\text{ km/h}$ and the new time taken be $T_2$ .

Since distance remains constant at $100\text{ km}$ : $S_1 \times T_1 = S_2 \times T_2$

$100 = 75 \times T_2$

\begin{aligned} T_2 &= \dfrac{100}{75}\text{ hours} \\[0.6em] &= \dfrac{4}{3}\text{ hours} \\[0.6em] &= 1\text{ hour} + \left(\dfrac{1}{3} \times 60\right)\text{ minutes} \\[0.6em] &= 1\text{ hour } 20\text{ minutes} \end{aligned}

Since the speed increased from $50\text{ km/h}$ to $75\text{ km/h}$ , it will take less time ( $1\text{ hour } 20\text{ minutes}$ compared to $2\text{ hours}$ ).

This cannot be written as a direct proportion like $50 : 2 :: 75 : x$ , because speed and time are inversely proportional.

Answer

It will take Puneeth’s father less time (1 hour 20 minutes).

We cannot form this problem as a direct proportion because speed and time are inversely proportional for a constant distance.

Common Mistakes

Direct Proportion Trap: Setting up a direct proportion $\dfrac{50}{2} = \dfrac{75}{T_2}$ gives $T_2 = 3\text{ hours}$ , which incorrectly implies that going faster takes more time. For a constant distance, always use inverse proportion ( $S_1 T_1 = S_2 T_2$ ).
Fraction of an Hour Error: Misinterpreting $\dfrac{1}{3}\text{ hour}$ as $30\text{ minutes}$ or $0.3\text{ hours}$ instead of $\dfrac{1}{3} \times 60 = 20\text{ minutes}$ .

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Q13

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Q14

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Q15

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Q16

Context: You have 12 countable objects or counters to share between the two of you.

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