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At my fruit stand, I sold half my watermelons and half a watermelon more in the first hour. In the second hour, I sold half my remaining watermelons and half a watermelon more. Incredibly enough in the third hour, I AGAIN sold half my remaining watermelons and half a watermelon more. In the fourth hour, I sold my last 6 watermelons. How many did I start with?

 Nov 21, 2019
 #1
avatar+105370 
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Let  the total watermelons we start with  =  N

 

1st  hour sold  =  [   N/2 + 1/2]

 

Remaining  is  what we started with - number sold in 1st hour =   N - [ N/2 + 1/2 ]  =  (1/2)N -1/2

 

2nd hour sold =   (1/2)[ (1/2)N - 1/2] + 1/2 =  [ (1/4)N  + 1/4 ]

 

Remaining is   what we started with at the end of the first hour - what we sold in the second hour  =

[ (1/2)N - 1/2 ] -  [ (1/4)N + 1/4]  =  [ (1/4)N - 3/4 ]

 

In the third hour  we sold  

(1/2)  [ (1/4)N - 3/4 ]+ 1/2    =   [ (1/8)N + 1/8]

 

What we  have remaining is the what we started with at the end of the second hour - what we sold in the third hour  =[(1/4)N - 3/4]  - [ (1/8)N + (1/8) ] =   (1/8)N - 7/8

 

And the number remaining  = 6  ...so...

 

(1/8)N - 7/8  = 6      multiplty through by  8

 

N - 7  =  48

 

N  = 55   is what we started with

 

 

 

cool cool cool

 Nov 21, 2019
edited by CPhill  Nov 21, 2019
 #2
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+1

Let the starting number of watermelons=W
[1/2W - 1/2] - Melons left after the first hour.
[1/4W - 1/4 - 1/2] - After the second hour.
[1/8W - 1/8 - 1/4 - 1/2] - After the third hour.
[1/8W - 0.875] = 6, solve for W
W = 55

 Nov 22, 2019
 #3
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Thank you so much.

 Nov 22, 2019

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