Solving the Cycling Speed Problem

This question asks us to find the original speed at which Krishna cycled a certain distance, given information about how a change in speed affects the time taken.

Let's define the variables:

• Let $D$ be the distance Krishna cycled. We are given $D = 90$ km.
• Let $v$ be the original speed in km/hr. This is what we need to find.
• Let $t$ be the original time taken in hours.

The relationship between distance, speed, and time is: Distance = Speed $\times$ Time, or $D = v \times t$.

Setting Up Equations Based on the Problem

From the given information, we can form two equations:

1. The actual trip: Krishna cycled 90 km at speed $v$ in time $t$.
   So, $90 = v \times t$ (Equation 1)
2. The hypothetical trip: If Krishna cycled 3 km/h slower, his new speed would be $(v-3)$ km/hr. He would have taken 5 more hours, so the new time would be $(t+5)$ hours. The distance is still 90 km.
   So, $90 = (v-3) \times (t+5)$ (Equation 2)

Solving the Equations to Find the Original Speed

We have a system of two equations with two variables ($v$ and $t$). We can solve this system.

From Equation 1, we can express $t$ in terms of $v$: $t = \frac{90}{v}$

Now, substitute this expression for $t$ into Equation 2:

$90 = (v-3) \left(\frac{90}{v} + 5\right)$

Let's expand the right side of the equation:

$90 = v \times \frac{90}{v} + v \times 5 - 3 \times \frac{90}{v} - 3 \times 5$

$90 = 90 + 5v - \frac{270}{v} - 15$

Simplify the equation by subtracting 90 from both sides:

$0 = 5v - \frac{270}{v} - 15$

To eliminate the fraction, multiply the entire equation by $v$ (assuming $v \neq 0$, which must be true for cycling speed):

$0 \times v = (5v) \times v - \left(\frac{270}{v}\right) \times v - (15) \times v$

$0 = 5v^2 - 270 - 15v$

Rearrange the terms to form a standard quadratic equation $av^2 + bv + c = 0$:

$5v^2 - 15v - 270 = 0$

We can simplify this quadratic equation by dividing all terms by 5:

$\frac{5v^2}{5} - \frac{15v}{5} - \frac{270}{5} = 0$

$v^2 - 3v - 54 = 0$

Now, we need to solve this quadratic equation for $v$. We can try factoring. We look for two numbers that multiply to -54 and add up to -3. The numbers are 6 and -9.

So, we can factor the equation as:

$(v + 6)(v - 9) = 0$

This gives two possible solutions for $v$:

• $v + 6 = 0 \implies v = -6$
• $v - 9 = 0 \implies v = 9$

Since speed cannot be a negative value in this context, the actual speed at which Krishna cycled must be $v = 9$ km/hr.

Verification of the Solution

Let's check if this speed satisfies the conditions in the problem.

• Actual speed = 9 km/hr. Time taken = Distance / Speed = 90 km / 9 km/hr = 10 hours.
• If speed was 3 km/hr slower, new speed = 9 - 3 = 6 km/hr. Time taken = Distance / Speed = 90 km / 6 km/hr = 15 hours.

The difference in time is $15 - 10 = 5$ hours, which matches the problem statement (he would have taken 5 more hours).

Therefore, the speed at which Krishna actually cycled is 9 km/hr.

Revision Table: Speed, Distance, and Time

These fundamental formulas are crucial for solving problems involving motion.

Additional Information: Solving Quadratic Equations

In this problem, we solved the quadratic equation $v^2 - 3v - 54 = 0$ by factoring. A quadratic equation in the standard form $ax^2 + bx + c = 0$ can also be solved using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation $v^2 - 3v - 54 = 0$, we have $a=1$, $b=-3$, and $c=-54$. Plugging these values into the formula:

$v = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-54)}}{2(1)}$

$v = \frac{3 \pm \sqrt{9 + 216}}{2}$

$v = \frac{3 \pm \sqrt{225}}{2}$

$v = \frac{3 \pm 15}{2}$

This gives two possible values for $v$:

• $v = \frac{3 + 15}{2} = \frac{18}{2} = 9$
• $v = \frac{3 - 15}{2} = \frac{-12}{2} = -6$

Again, we discard the negative solution because speed cannot be negative. So, $v = 9$ km/hr.

Factoring is often quicker if the numbers are simple, but the quadratic formula always works for any quadratic equation.