+0  
 
0
65
2
avatar

A man goes into a store and says to the owner: "Give me as much money as I have with me and I will spend $10."

It is done and the man repeats the operation at a second and a third store, after which he has no money left.

How much did he start with?

 Aug 23, 2020
 #1
avatar+93 
+1

1. Let x be how much he started with.

How Much He Had Left in Store 1: x+x-10

How Much He Had Left in Store 2: (2x-10)+(2x-10)-10=2x+2x-10-10-10=4x-30

How Much He Had Left in Store 3: (4x-30)+(4x-30)-10=4x+4x-30-30-10=8x-70

 

2. It's implied that 8x-70=0, so 8x must equal 70 after adding 70 on both sides of the equation.

 

3. Divide 8 on both sides of the equation and we get x=70/8, or x=$8.75.

 

Answer: The man started with $8.75. smiley

 Aug 23, 2020
edited by joliel3  Aug 23, 2020
 #2
avatar
0

2 a - 10 = b
2 b - 10 = c
2 c - 10 = 0, solve for a, b, c

 

a=35/4 =$8.75 - what he started with

b=15/2 =$7.50 - what he had at 2nd store

c = $5.00 - what he had at the 3rd store.

 Aug 23, 2020

46 Online Users