SOLUTION: I need help with this impossible problem Jake has 6 coins. Ellie has 12 coins. There is a total of 5 quarters. Jake has more than $1 Jake's coins are worth three times as mu

Question 302052: I need help with this impossible problem
Jake has 6 coins.
Ellie has 12 coins.
There is a total of 5 quarters.
Jake has more than $1
Jake's coins are worth three times as much as Ellie's coins.
How much money does Jake have?
How much money does Ellie have?
My niece's teacher apparently wants us to show our work (my niece is in the 2nd grade and she somehow has this problem for a test. she's failing most of her tests and I want to find out why.) There is supposed to be a non-algebraic way to solve this problem since she is only in the 2nd grade. I hope you can reply soon.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
2nd grade? I was taking remedial sandbox and advanced crayons.
This is a problem where she has to come to some logical
conclusions and put them together to get what they're
looking for
They don't say anything about it, but any number of pennies, nickels,
and dimes may be used
I focused on "Jake has more than $1"
(case #1) Jake could have 3 quarters and 3 dimes, giving him $1.05
(case #2) or, he could have 4 quarters, and it doesn't matter what the other 2 coins are
(case #3) 0r, he could have 5 quarters, and it doesn't matter what the other coin is
Now I look at "Jake's coins are worth 3 times Ellie's coins"
So, Jakes coins must be evenly divisible by 3
for case #1: 1.05/3 = .35
So, if Ellie has 12 coins, and I know 2 of them are quarters, she's already got
$.50, so case #1 doesn't work
for case #2: Jake could have:
$1.20 (4 quarters and 2 dimes)
$1.10 (4 quarters and 2 nickels)
$1.15 (4 quarters , a nickel and a dime)
$1.11 (4 quarters, a dime and a penny)
$1.06 (4 quarters, a nickel, and a penny)
$1.02 (4 quarters and 2 pennies)
The right one has to be evenly divisible by 3
$1.20/3 = .40
$1.11/3 = .37
$1.02/3 = .34
Now Ellie has 1 quarter, so her 11 remaining coins must add up to
.40 - .25 = .15
or
.37 - .25 = .12
or
.34 - .25 = .09
There's no way to get the last 2 with 11 coins, but
.15 = 10 pennies and a nickel
So, she ends up with 1 quarter, 10 pennies, and 1 nickel
when Jake has 4 quarters and 2 dimes
Now I just have to deal with case #3
Jake has either:
5 quarters and a dime = $1.35
5 quarters and a nickel = $1.30
5 quarters and a penny = $1.26
$1.35/3 = .45
$1.26/3 = .42
Nellie has 12 coins to make $.45
That would be 5 pennies, 6 nickels, and 1 dime
I can't seem to make $.42 with 12 coin, but no quarters
So, what I finally come up with is:
--------------------------------
When Jake has 4 quarters and 2 dimes,
Nellie has 1 quarter 1 nickel, 10 pennies
---------------------------------
When Jake has 5 quarters and 1 dime,
Nellie has 1 dime, 6 nickels, and 5 pennies
---------------------------------
Whew! What grade is she in?
This was hard for me. I'm still not sure I've
got all the answers and I've got 2 years of college
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