Question

An aeroplane first flew with a speed of $440$ km/h and covered some distance. It still had to cover $770$ km less than what it had already covered, and it flew this remainder at $660$ km/h. The average speed for the entire flight was $500$ km/h. Find the total distance.

• A. $3250$ km
• B. $2750$ km
• C. $4400$ km
• D. $1375$ km

Hint:

For mixed–speed journeys, write both legs with one variable and apply $V_{\text{avg}}=\dfrac{D_{\text{total}}}{T_{\text{total}}}$. Multiply by the LCM of denominators to avoid fractions.

Answer 1

The Correct Option is B

Step 1: Express the two legs with one variable.
Let the first–leg distance be $x$ km. Then the second–leg distance is $x-770$ km.
Total distance $D= x+(x-770)=2x-770$. Step 2: Use the average–speed relation.
Total time $T=\dfrac{x}{440}+\dfrac{x-770}{660}$. Given $\dfrac{D}{T}=500$: \[ \frac{2x-770}{\frac{x}{440}+\frac{x-770}{660}}=500. \] Step 3: Solve cleanly by clearing denominators.
$\text{LCM}(440,660)=1320$: \[ \frac{2x-770}{\frac{5x-1540}{1320}}=500 \Rightarrow 1320(2x-770)=500(5x-1540). \] \[ 2640x-1{,}016{,}400=2500x-770{,}000 \Rightarrow 140x=246{,}400 \Rightarrow x=1760. \] Thus $D=2x-770=3520-770=2750$ km.
\[ \boxed{2750\ \text{km}} \]