A jet flew from tokyo to bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flight was 2 hours, what was the jets speed from bangkok to tokyo?
A jet flew from tokyo to bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flight was 2 hours, what was the jets speed from bangkok to tokyo?
Hello, Chr2010!
A jet flew from Tokyo to Bangkok, a distance of 4800km.
On the return trip, the speed was decreased by 200 km/hr.
If the difference in the times of the flight was 2 hours,
what was the jet's speed from Bangkok to Tokyo?
The jet flew 4800 km from Bangkok to Tokyo at $\displaystyle \,x$ km/hr.
. . This took $\displaystyle \dfrac{4800}{x}$ hours.
The jet flew 4800 km from Tokyo to Bangkok at $\displaystyle x + 200$ km/hr.
. . This took $\displaystyle \dfrac{4800}{x+200}$ hours.
The difference in time was 2 hours: .$\displaystyle \displaystyle \frac{4800}{x} - \frac{4800}{x+200} \;=\;2$
Multiply by $\displaystyle x(x+200)\!:\;\;4800(x+200) - 4800x \;=\;2x(x+200)$
This equation simplifies to: .$\displaystyle x^2 + 200x - 480,\!000 \:=\:0$
. . whoch factors: .$\displaystyle (x - 600)(x + 800) \:=\:0$
. . and has roots: .$\displaystyle x \:=\:600,\:\text{-}800$
The jet's speed from Bangkok to Tokyo was 600 km/hr.
Soroban started by choosing a letter to represent the quantity he wanted to find- in this case he let "x" represent the speed flying from Bangcock to Tokyo. You are told that the distance is 4800 km and should know that "speed= distance/time" so if we let $\displaystyle t_2$ be the time the flight took, $\displaystyle x= \frac{4800}{t_2}$ so, solving that for $\displaystyle t_2$, $\displaystyle t_2= \frac{4800}{x}$.
We don't know $\displaystyle t_2$ so we can use that equation alone to find x. But we also know that "On the return trip, the speed was decreased by 200 km/h". Since is the speed on the return trip, the speed on the first leg must be 200 more: x+ 200. The time to fly the same 4800 is $\displaystyle t_1= \frac{4800}{x+ 200}$. That flight was at a faster speed so of course, it takes less time- 2 hours less: $\displaystyle t_2- t_1$, the difference in times, is 2 hours:
$\displaystyle t_2- t_1= \frac{4800}{x}- \frac{4800}{x+ 200}= 2$
Get rid of the fractions by multiplying both sides of that equation by x(x+ 200) and you get the quadratic equation Soroban gave.