Brain Teasers
I Will Spend $10
A man goes into a store one day, and says to the cashier, "If you can match and give me what I have in my pocket, I will spend $10 in your store."
The cashier matched and gave the man the money, and the man spent $10. The man then went to another store, said the same thing, and did the same. He then went to a third store, said the same thing, and did the same. After leaving the third store, he had no money left. How much money did he start out with?
The cashier matched and gave the man the money, and the man spent $10. The man then went to another store, said the same thing, and did the same. He then went to a third store, said the same thing, and did the same. After leaving the third store, he had no money left. How much money did he start out with?
Answer
He started out with $8.75.8.75+8.75=17.50-10.00=$7.50
7.50+7.50=15.00-10.00=$5.00
5.00+5.00=10.00-10.00=$0
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Comments
...interesting idea.
That was confusing.
I like this one - rather simple though. It comes down to a simple linear equation to solve, 8x - 70 = 0, where x is the initial amount.
[[x*2 + (-10)]*2 + (-10)]*2 + (-10)]]] = 0
8x - 70 = 0
x = 70/8
Or you could just simply think "the only way to have your money doubled before spending exactly $10 is if you had $5", and keep working backwards from there.
8x - 70 = 0
x = 70/8
Or you could just simply think "the only way to have your money doubled before spending exactly $10 is if you had $5", and keep working backwards from there.
Oops, brackets the wrong way:
[[[x*2 + (-10)]*2 + (-10)]*2 + (-10)] = 0
[[[x*2 + (-10)]*2 + (-10)]*2 + (-10)] = 0
it took me a while but i finally got it!
I liked it. I didn't think to make an equation though. Next time I'll try that first.
Good one Jessica! It was too hard for me though. I never thought of doing it that way!
nice one used a different method thought. I worked on the outcome going back:
x=original amount
shop 1 money=x (initial amt)
shop 2 money=y (initial amt)
shop 3 money=z (initial amt)
#3 shop
(z+z)-10 = 0
2z=0+10
(2z)/2 = 10/2
z=5
#2 shop
(y+y)-10 = 5
2y=5+10
(2y)/2 = 15/2
y=7.5
#1 shop
(x+x)-10 = 7.5
2x=7.55+10
(2x)/2 = 17.5/2
x=8.75
-although I would like to do this if a shopkeeper would really double my money
x=original amount
shop 1 money=x (initial amt)
shop 2 money=y (initial amt)
shop 3 money=z (initial amt)
#3 shop
(z+z)-10 = 0
2z=0+10
(2z)/2 = 10/2
z=5
#2 shop
(y+y)-10 = 5
2y=5+10
(2y)/2 = 15/2
y=7.5
#1 shop
(x+x)-10 = 7.5
2x=7.55+10
(2x)/2 = 17.5/2
x=8.75
-although I would like to do this if a shopkeeper would really double my money
Hey Jessie! It is me, Bri. I loved this teaser, but I am not exactly in Enriched Math at my school. Okay, I am, but this is still a bit on the hardish size. But Jess, keep up the good work, and keep these confusing ones coming to witness my head explode! Just kidding. Keep upi the good work, and don't forget to e-mail me!
The man needs to have $10 - .5*10 = $5 when entering the last store and 10 - .5^2*10 when entering the next to last store and so on. So 10 - .5^3*10 = 10 - .125*10 = 8.75.
Working backwards from the last $10 is definitely the best strategy.
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