Question

If Deepesh had walked \(20 \mathrm{~km} / \mathrm{h}\) faster he would have saved 1 hour in the distance of \(600 \mathrm{~km}\). What is the usual speed of Deepesh? (a) 100 (b) 120 (c) 150 (d) none of these

Short answer

Answer: (d) none of these

Step-by-step solution

Define variables

Let \(s\) be Deepesh's usual speed (in \(\mathrm{km}/\mathrm{h}\)), and let \(t\) represent the usual time he takes to cover \(600\mathrm{~km}\) (in hours).

Write the equation for distance

As we know, the distance is equal to speed times time, so we can write the equation for Deepesh's usual distance, which is 600 km: \(s \cdot t = 600 \mathrm{~km}\)

Write the equation for the faster speed

If Deepesh increases his speed by \(20 \mathrm{~km}/\mathrm{h}\), then his speed would become \((s + 20)\). As a result, he would complete the journey in \((t-1)\) hours. The distance is still the same, so we can write the equation for the faster speed: \((s + 20) \cdot (t - 1) = 600 \mathrm{~km}\)

Solve the system of equations

Now we have two equations: 1. \(s \cdot t = 600\) 2. \((s + 20)(t - 1) = 600\) To solve this system of equations, we can express either \(s\) or \(t\) from the first equation and substitute it into the second equation. Let's express \(t\): \(t = \frac{600}{s}\) Substituting into the second equation: \((s + 20) \left( \frac{600}{s} - 1 \right) = 600\)

Simplify and solve for s

Expanding the equation, we get: \((s+20)(\frac{600}{s}-1) = \frac{600(s+20)-s(s+20)}{s} = \frac{600s+12000-1000-20s}{s}= \frac{580s+12000}{s}=600\) Now, we can cross multiply: \(580s + 12000 = 600s\) Now, solving for s: \(20s = 12000\) \(s = 600\) However, the options are a) 100, b) 120, c) 150, and d) none of these. Since our answer does not match any of the options, the correct answer is: (d) none of these

Key concepts

Quantitative Aptitude

At its core, quantitative aptitude involves the ability to understand and solve problems involving numerical and logical reasoning. It's a pillar of competitive exams and aptitude tests, where understanding concepts like speed, time, and distance is essential. In our example, we applied quantitative aptitude to determine a person's typical speed by setting up an equation using given information.

To excel in quantitative aptitude, especially in speed-time-distance problems, students must be adept at creating relationships between these entities and forming equations that represent real-world scenarios. Using algebra as a tool, solving for unknown variables becomes a straightforward process of manipulating these equations.

Algebraic Equations

Algebraic equations are fundamental tools used to represent problems where you seek to determine unknown quantities. In the context of the given exercise, we used an algebraic equation, specifically a linear equation in two variables, to represent the relationship between Deepesh's speed (\(s\text{ km/h}\)) and the time (\(t\text{ hours}\)) taken to cover a distance of 600 km.

An equation like \(s \times t = 600\) offers a clear, concise way to show that Deepesh’s speed multiplied by the time he travels equals the distance. Mastering the use of these equations is crucial to solving many real-world problems efficiently. Algebra trains the mind to think logically and to approach complex problems in a structured manner, breaking them down into smaller, more manageable parts.

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. In our speed-time-distance problem, we had two linear equations constituting a system. Solving a system of equations like this usually involves finding a common variable to eliminate or substitute, allowing you to isolate the remaining variable.

The exercise demonstrates the substitution method, where one equation is solved for a particular variable, which is then inserted into the other equation. For instance, \(s \times t = 600\) and \( (s + 20) \times (t - 1) = 600\) are two equations that form our system. Solving for one variable (\(t\text{, in this case}\) and plugging it into the second equation simplifies the problem to having one unknown (\(s\text{, in this case}\). Systems of equations are not limited to two variables and can involve multiple variables and equations, requiring different methods like elimination, substitution, or matrix operations to solve.