Let the speed of the person while going up be x km/h. Then, his speed during the downward return journey is (x + 10) km/h. Since Time = (Distance)/(Speed), the upward time is (150)/(x) h and the downward time is (150)/(x+10) h.

Let the speed of the person while going up be x km/h. Then, his speed during the downward return journey is (x + 10) km/h. Since Time = (Distance)/(Speed), the upward time is (150)/(x) h and the downward time is (150)/(x+10) h.

The upward journey took 2(1)/(2) hours = (5)/(2) hours more than the return journey, so (150)/(x) - (150)/(x+10) = (5)/(2). Factoring out 150 gives 150((x+10-x)/(x(x+10))) = (5)/(2), which simplifies to (1500)/(x(x+10)) = (5)/(2), leading to x(x+10) = 600, or x² + 10x - 1200 = 0.

The upward journey took 2(1)/(2) hours = (5)/(2) hours more than the return journey, so (150)/(x) - (150)/(x+10) = (5)/(2). Factoring out 150 gives 150((x+10-x)/(x(x+10))) = (5)/(2), which simplifies to (1500)/(x(x+10)) = (5)/(2), leading to x(x+10) = 600, or x² + 10x - 1200 = 0.

We split the middle term 10x as 40x - 30x [because 40x × (-30x) = -1200x² = (x²) × (-1200)]. Grouping in pairs gives x(x + 40) - 30(x + 40) = 0, so (x + 40)(x - 30) = 0.

The time taken by a person to travel an upward distance of $150\text{ km}$ was $2\frac{1}{2}\text{ hours}$ more than the time taken in the downward return journey. If he returned at a speed of $10\text{ km/h}$ more than the speed while going up, find the speeds in each direction.

Answer: The speed while going up is $30\text{ km/h}$ and the speed while returning downward is $40\text{ km/h}$ .

Step-by-step solution

Step 1: Define the unknown speed and expressions for time

Let the speed of the person while going up be $x\text{ km/h}$ . Then, his speed during the downward return journey is $(x + 10)\text{ km/h}$ . Since $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ , the upward time is $\frac{150}{x}\text{ h}$ and the downward time is $\frac{150}{x+10}\text{ h}$ .

Step 2: Formulate the quadratic equation

The upward journey took $2\frac{1}{2}\text{ hours} = \frac{5}{2}\text{ hours}$ more than the return journey, so $\frac{150}{x} - \frac{150}{x+10} = \frac{5}{2}$ . Factoring out $150$ gives $150\left(\frac{x+10-x}{x(x+10)}\right) = \frac{5}{2}$ , which simplifies to $\frac{1500}{x(x+10)} = \frac{5}{2}$ , leading to $x(x+10) = 600$ , or $x^2 + 10x - 1200 = 0$ .

Step 3: Solve by factorisation

We split the middle term $10x$ as $40x - 30x$ [because $40x \times (-30x) = -1200x^2 = (x^2) \times (-1200)$ ]. Grouping in pairs gives $x(x + 40) - 30(x + 40) = 0$ , so $(x + 40)(x - 30) = 0$ .

Step 4: Find the roots and conclude the speeds

Equating each factor to zero gives $x = 30$ or $x = -40$ . Since speed cannot be negative, we reject $x = -40$ . Therefore, the speed going up is $30\text{ km/h}$ , and the downward return speed is $30 + 10 = 40\text{ km/h}$ .

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