Challenging algebra problem (for some of us)

[S3rJ]

Anna and Mark pooled their money and went to a garage sale. In 30 minuites they spent half their money. They now have as many pennies (cents) as they had dollars before. They now have half as many dollars as they had pennies (cents) before.

How much did they spend at the sale?



call me a posting noob if you want but this is hard stuff for some of us

[xDdRmAnIaCx]

I think this is technically impossible.

There must be some number in the equation for it to be solved.

Like how much they had before. Or how many coins they have all together.

I *think* its impossible.

Now someone come and school me.

Maniac.

[MiniNeo]

Er... lets see

Amount of pennies = a
Amount of dollars = b

Before, they have a+100b pennies(cents)
After, since the number interchanged, they have b+100a pennies (cents)
and since they spent half their money:

(a+100b)/2 = b+100a
a+100b = 2b+200a
98b = 199a

Work it out from there I guess..

[S3rJ]

it is indeed possible

its algebra there for u wont get **** and will have to solve everything *Ugh* symbolically


i need this extra credit so someone figure it out!


wow dude lmao u posted at the exact same time as me

thx ur smart

[MiniNeo]

Hi ^_^

[Z3ratul]

I think you explained the problem wrong... that, or I'm thinking it's impossible.

EDIT: Or I'm just interpreting your words wrong. Regardless... =/

[S3rJ]

zomg ur right

ok i fixed it srry guys

[Z3ratul]

Originally Posted by S3rJ

Anna and Mark pooled their money and went to a garage sale. In 30 minuites they spent half their money. They now have as many pennies (cents) as they had dollars before. They now have half as many dollars as they had pennies (cents) before.

How much did they spend at the sale?



call me a posting noob if you want but this is hard stuff for some of us

K...

A = Amount of dollars before
B = Amount of dollars after

A = B/2 (half as much money as before)
A = 100B (cents now as dollars before)
100A = B/2 (half as many dollars now than cents before)

I see contradictions... impossible? Or am I just a math noob? Again, I think I'm interpreting the problem wrong... they pooled their money together, awesome. What then?

[T0rajir0u]

From the beginning we must assume only pennies and dollars are involved.

Let p_0, d_0 be the amount of pennies and dollars before. Let p_1, d_1 be the amount of pennies and dollars after. The problem gives

p_1 = d_0
2d_1 = p_0
(p_0 + 100d_0) = 2(p_1 + 100d_1)

And from this we do a little substitution and find

(d_1 / 2 + 100p_1) = 2(p_1 + 100d_1)
196p_1 = 399 d_1
28p_1 = 57d_1

gcd(28, 57) = 1. Here we must assume the number of pennies is less than 100. (Edit: In case anyone wasn't clear on this, not assuming so means there are infinite solutions, each of which is an integer multiple of this base case.) Then the solution is given by

p_1 = 57, d_1 = 28

Making the amount spent $28.57.

[purplepopcorn]

thx thx, I'm in serj's class too, so i kinda need it. ;;>.>

[S3rJ]

i have no clue who u are kid quit stalking me or ill call the cops

[Twilight_Overlord]

This is soooooo not a CT thread.

[Tasselfoot]

locked. DO NOT POST YOUR HOMEWORK FOR HELP IN CT.

next person to post their retardedly easy math homework in CT for someone else to do it for them will get a 7 day ban.