MHB Finding the Speed of an Aeroplane - Math Help Forum

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The discussion revolves around solving a math problem related to finding the speed of an airplane, with the answer given as 553.6 km/h. Participants are attempting to form equations based on the relationship between speed, distance, and time, specifically using the formula for time as distance divided by speed. One user struggles with setting up the correct equation, while another clarifies that the time taken at different speeds must be equated properly. The conversation highlights the importance of correctly manipulating the equations to arrive at the solution, emphasizing the need for accurate algebraic operations. The thread illustrates the collaborative effort in solving mathematical problems and clarifying misunderstandings.
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question is attached.
the answer at the back of the book is 553.6 kmph

this is for my maths homework and I am stuck on this question.:(
 

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wrightarya said:
question is attached.
the answer at the back of the book is 553.6 kmph

this is for my maths homework and I am stuck on this question.:(

What have you tried so far?
 
If the speed is x km/h then the time taken is 1000/x hours.

If the speed is x-120 then the time taken is 1000/x hours and this is equal to the other time plus 0.5 hours.

i don't know how to form an equation =/
 
wrightarya said:
If the speed is x km/h then the time taken is 1000/x hours.

If the speed is x-120 then the time taken is 1000/x hours and this is equal to the other time plus 0.5 hours.

i don't know how to form an equation =/

Actually if the speed is (x - 120) then the speed is 1000/(x - 120).

Now as you have stated, this is equal to the other time plus half an hour, so

$\displaystyle \begin{align*} \frac{1000}{x - 120} &= \frac{1000}{x} + \frac{1}{2} \end{align*}$

Can you now solve for x?
 
Prove It said:
Actually if the speed is (x - 120) then the speed is 1000/(x - 120).

Now as you have stated, this is equal to the other time plus half an hour, so

$\displaystyle \begin{align*} \frac{1000}{x - 120} &= \frac{1000}{x} + \frac{1}{2} \end{align*}$

Can you now solve for x?

i tried solving it like this:

1000/x-120 = 2000 + x
1000= (2000+x) (x-120)
1000= x^2 + 1880x - 240000
= x^2 + 1880x - 241000
substituted into quadratic formula
x=120 or -2000

dont know if i did that right, but its not what the answer says in the back of the book...
 
wrightarya said:
i tried solving it like this:

1000/x-120 = 2000 + x

No, $\displaystyle \begin{align*} \frac{1000}{x} + \frac{1}{2} \end{align*}$ is not $\displaystyle \begin{align*} 2000 + x \end{align*}$, it's $\displaystyle \begin{align*} \frac{2000 + x}{2x} \end{align*}$, so when you cross multiply you should get

$\displaystyle \begin{align*} \frac{1000}{x - 120} &= \frac{1000}{x} + \frac{1}{2} \\ \frac{1000}{x - 120} &= \frac{2000 + x}{2x} \\ 1000 \left( 2x \right) &= \left( 2000 + x \right) \left( x - 120 \right) \end{align*}$

Go from here...
 
wrightarya said:
If the speed is x km/h then the time taken is 1000/x hours.

If the speed is x-120 then the time taken is 1000/x hours and this is equal to the other time plus 0.5 hours.

i don't know how to form an equation =/
Finding speed of an aeroplane - The Student Room

You almost quoted me correctly. ;)
 

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