Speed, Time and Distance Problems with Quadratic Equations

This lesson works through two Class 10 word problems on speed, time and distance. In each one we use the relationship between distance, speed and time to build a quadratic equation, then solve it by factorisation and again with the quadratic formula.

The basic formulas

The three forms of the speed relationship are:

$\text{Distance} = \text{Speed} \times \text{Time}$

$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Problem 1: speed of the train

A train travels $360$ km at a uniform speed. If the speed had been $5$ km/h more, it would have taken $1$ hour less for the same journey. Find the speed of the train.

Let the speed be $x$ km/h. Then the time taken is:

$\frac{360}{x}$

With the speed $5$ km/h more, the new speed is $x + 5$ and the new time is:

$\frac{360}{x + 5}$

The new time is $1$ hour less, so:

$\frac{360}{x} - \frac{360}{x + 5} = 1$

Taking $360$ out and combining the fractions:

$360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1$

$360 \cdot \frac{(x + 5) - x}{x(x + 5)} = 1$

$\frac{360 \times 5}{x(x + 5)} = 1$

So $1800 = x(x + 5)$, which gives the quadratic:

$x^2 + 5x - 1800 = 0$

By factorisation

We need two numbers with product $-1800$ and sum $+5$. Since the product is negative, look for numbers whose difference is $5$: these are $45$ and $40$. The equation factors as:

$(x + 45)(x - 40) = 0$

So $x = -45$ or $x = 40$. Speed cannot be negative, so $x = 40$.

By the quadratic formula

With $a = 1$, $b = 5$, $c = -1800$:

$b^2 - 4ac = 5^2 - 4(1)(-1800) = 25 + 7200 = 7225$

$\sqrt{7225} = 85$

$x = \frac{-5 \pm 85}{2}$

This gives $x = \tfrac{80}{2} = 40$ or $x = \tfrac{-90}{2} = -45$. Rejecting the negative value, $x = 40$.

The speed of the train is $40$ km/h.

Finding a square root by long division

To check $\sqrt{7225}$, group the digits in pairs from the right: $72$ and $25$. The largest square not exceeding $72$ is $64 = 8^2$, so the first digit is $8$ and the remainder is $72 - 64 = 8$. Bring down $25$ to get $825$. Double the $8$ to get $16$, then find a digit $d$ so that $(160 + d) \times d \le 825$. Here $165 \times 5 = 825$ exactly, so the next digit is $5$ and:

$\sqrt{7225} = 85$

Problem 2: express and passenger trains

An express train takes $1$ hour less than a passenger train to travel $132$ km between Mysore and Bangalore. The express train's average speed is $11$ km/h more than the passenger train's. Find the average speed of each train.

Let the passenger train's speed be $x$ km/h, so the express train's speed is $x + 11$ km/h. Their times are:

$\text{Passenger: } \frac{132}{x}, \qquad \text{Express: } \frac{132}{x + 11}$

The express train takes $1$ hour less:

$\frac{132}{x} - \frac{132}{x + 11} = 1$

$132 \cdot \frac{(x + 11) - x}{x(x + 11)} = 1$

$\frac{132 \times 11}{x(x + 11)} = 1$

So $1452 = x(x + 11)$, giving:

$x^2 + 11x - 1452 = 0$

By factorisation

We need two numbers with product $-1452$ and sum $+11$: these are $44$ and $-33$. So:

$(x + 44)(x - 33) = 0$

Then $x = -44$ or $x = 33$. Rejecting the negative value, $x = 33$.

By the quadratic formula

With $a = 1$, $b = 11$, $c = -1452$:

$b^2 - 4ac = 11^2 - 4(1)(-1452) = 121 + 5808 = 5929$

$\sqrt{5929} = 77$

$x = \frac{-11 \pm 77}{2}$

This gives $x = \tfrac{66}{2} = 33$ or $x = \tfrac{-88}{2} = -44$. Rejecting the negative value, $x = 33$.

So the passenger train's average speed is $33$ km/h and the express train's is $33 + 11 = 44$ km/h.

Key takeaways

• Use $\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$ and the given time difference to form a quadratic equation.
• Solve by factorisation when the numbers are easy to spot; switch to the quadratic formula quickly if they are not.
• A speed must be positive, so always reject the negative root.