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math

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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
+111321
+1

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

Nov 21, 2019
edited by CPhill  Nov 21, 2019
#2
+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
0

Thank you so much.

Nov 22, 2019