On an airplane that was 2/3 full, 20% of the passengers were boys, one-fourth of the passengers were women, one-eighth of the passengers were girls, and there were 51 men. How many seats are on the plane

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- Nov 6th 2008, 12:02 PMnancymcwilliamspercents and fractions
On an airplane that was 2/3 full, 20% of the passengers were boys, one-fourth of the passengers were women, one-eighth of the passengers were girls, and there were 51 men. How many seats are on the plane

- Nov 6th 2008, 09:58 PMearboth
Let x denote the total number of seats in the plane.

Translate the sentence into numbers and operations. Keep in mind that $\displaystyle 20\% = \dfrac15$

$\displaystyle \left(\dfrac15 + \dfrac14 + \dfrac18\right)\cdot x+51=\dfrac23 x$

$\displaystyle \dfrac{23}{60} \cdot x +51=\dfrac23 \cdot x$

$\displaystyle 51=\dfrac{17}{60} x~\implies~x=\dfrac{51 \cdot 60}{17}=180\ seats$ - Nov 20th 2008, 12:00 PMnancymcwilliams
I really do not understand where some of the numbers came from in your answer

- Nov 20th 2008, 01:25 PMjahichuanna
the first thing you want to do is convert everything to either a fraction/ratio or a number

so it becomes

an airplane 2/3 full, 20% passengers were boys, 25% women, 12.5% girls, 51 men.

now, since you need 100%, you figure out what percent you have now: 20+25+12.5 = 57.5%

so 57.5% of the passengers WERE NOT men. That means that 42.5% WERE men. Since 42.5% were men, figure out how many passengers there are altogether. A percentage can become a decimal, e.g. 1% = .01; 42.5% = .425. Therefore, 42.5% of x = 51 becomes .425x=51. Solve for x and get 120. That means there were 120 total PASSENGERS on the plane. Since the plane is 2/3 full, do 2/3 y = 120. solve for y and get 180.

So there are 180 seats on the plane.

hope this helped you out. - Nov 20th 2008, 11:06 PMearboth
Not sure which numbers are not clear to you I'll show you where I've got the numbers from:

$\displaystyle

\overbrace{\left(\underbrace{\dfrac15}_{boys} + \underbrace{\dfrac14}_{women} + \underbrace{\dfrac18}_{girls}\right)}^{fractions\ of\ all\ seats}\cdot x+\underbrace{51}_{men}=\underbrace{\dfrac23 x}_{occupied\ seats}

$ - Nov 22nd 2008, 07:00 AMsyathish
- Nov 22nd 2008, 11:11 PMearboth
You are completely right. (In my first post I made a very advantageous error so I got the correct final result) In short:

Let t denote the total number of seats and x the number of occupied seats. Then you have a system of simultaneous equations:

$\displaystyle \left|\begin{array}{r}\dfrac23 t = x \\ \dfrac15 x+\dfrac14 x + \dfrac18 x + 51 = x\end{array}\right.$

You'll get x = 120 and t = 180 - Nov 24th 2008, 12:20 PMnancymcwilliams
It is the number of seats not passengers but i really appreaciate both answers. together I was able to figure the problem out.

- Dec 3rd 2008, 02:21 AMrepcvt