Can you help me with this? I asked everyone at school, mom, and dad. They all dont know.?

Question

This math homework assignment is so confusing. NO one gets it. i asked everyone i know that has this assignment and they dont get it either.
Heres the problem

The Crayon Problem

Hannah loves to color. She is also very skilled and can color with both hands at the same time. When she uses one of her big fat crayons, it wears down to nothing after 15 hours. Her skinny crayon wears down after coloring for 6 hours. Hannah enters a coloring marathon. SHe was given two crayons of the same height (one fat and one skinny). She began coloring with both crayons, one in each hand, at 10:30 AM. At what time will the skinny crayon be half the height of the fat crayon?



I am not looking for an answer. I just do not understand how to do the problem. If it helps i am in Algebra 1 and our unit is Linear Functions.

Thank you so much!!!

Sharon: I copied it exactly from the handout. That website is not from my school district and is not very close to my problem. thanks anyways

Answer 1

Hi Savvy.
Working on it.
Driving me mad.
But will get you an answer.
Will edit my answer in a few.
Keep looking and will give you the
equation to figure it. Hang in there....

SAVVY,, THESE equations may help

S = skinny crayon
F = fat crayon

If S lasts 6 hours, then
the usage of the crayon in one hour
is S - S/6 after one hour

After T hours of continous usage
the length of the S crayon would
be
S - (S/6)*T

Same applies to the Fat Crayon of F.

The only time S and F are of equal length is
at time 0 = the beginning.
Thereafter use the following:
S(1-T/6)=F(1-T/15)

the question you need ask yourself is

When is S the 1/2 of F?
in other words for what T?

I think the folowing will provide that answer
(F/2)(1-T/6) = F(1-T/15)
I´m not quite sure, but think it will do it.
Let us know!
Good luck!

Answer 2

Linear functions are functions that have x as the input variable, and x is raised only to the first power.

i found this ..... it looks like there was a piece of info left out.

http://staff.pausd.org/~mkitch/papers/crayoneasiest.htm

Linear Functions Name:
Easiest Due: August 27, 2004

The Crayon Problem
Gavin learned to color this summer. He is has become very skilled and can color with both hands at the same time. When he uses one of his big fat crayons, it wears down by 2 inches every hour. His skinny crayons wear down 3 inches every hour. Gavin entered the baby coloring marathon. He was given two crayons each one foot (12 inches) long (one fat and one skinny). He began coloring with both crayons, one in each hand at 10:30 a.m. At what time will the skinny crayon be half the height of the fat crayon?

Bonus
Write an equation to show how much of each crayon has been used if the fat crayon wears down 2 inches every hour and the skinny crayon wears down 3 inches every hour. Each Crayon should begin as a foot long. The time and the length of the crayons should be variables.

hope this helps!

Answer 3

Ok, you know the skinny crayon takes six hours to wear down and the fat crayon takes 15 hours. Since the fat crayon wears down so slowly, but it does wear down, the answer is NOT three hours.

I suggest you start by dividing the height v. the time it takes to wear down on both crayons. The fat crayon loses 1/15th in height for each hour. The skinny crayon loses 1/6th.

You'll have to figure out the rest since its been 36 years since I took algebra.

Answer 4

maybe a different way? one uses 1/15 per hour, the other 1/6:

look at percentage used.
one uses x% while the other uses 2.5x% (15/6=2.5)

so, if crayon 1 used x%, then the remaining section would be
[1-x%]
you want to know when that equals twice the other crayon, which reduced at 2.5x%:
(1-x%)=2(1-2.5x%)
this means: remaining longer crayon = twice remaining shorter crayon. solving:
1-x% = 2-5x%
1+4x=2
4x=1
x=1/4=.25%
crayon 1 used .25% x 15hrs = 3.75hrs
in 3.75 hrs, crayon 2 used 3.75/6=62.5%
SO:
crayon 1 has 75% remaining (used 25%)
crayon 2 has 37.5% remaining (used 62.5%)
so crayon 2 is now 1/2 the length of crayon 1
3.75 hours past 10:30am equals 2:15pm

BUT: i'm no math whiz. it's worth checking a more conventional method. this answer presupposes that the crayons were equal in length to start.

Answer 5

OMG - This is driving me crazy. Someone tell us how to figure this!!!!

Trying to make a linear chart - I don't think the skinny crayon will ever be half the height of the fat crayon. When the skinny crayon is gone, the fat crayon will be half the size it started. The skinny will never be half the height of the fat.

Answer 6

this is a typical word problem giveing you too much information to be more confuseing.

the inportent info is this

the crayons are of equal hight.
the fat crayon will last 15 hours
the skinny crayon will last 6 hours.
she began at 10:30.

what you need to do if to figure out at what point the skinny crayon will be reduced to half the size of the fat crayon.
you have to factor in that they will each decrease in size over time but the skinny one will decrease much faster

Answer 7

My husband is a whiz at math, and says this is a easy problem to figure out. I think it has something to do with the pressure of your own hand. I believe its a formula on how you figure this out. My husband figured it out this way:
"There isnt enough information. You have to know the height of the crayon to get the correct information. You cant figure it out cause of lack of information. So it depends on the height of the crayon!"
Yep, ya need more numbers! lol :)

Answer 8

I think it's a trick question. If she uses them at the same time, it will take the longest time of 15hrs (cause she is using both hands). So 10:30am - 1:30am the next day

Answer 9

well i'm stumped on writing an equation to this problem so i just did it with the other method....drawing diagrams with trial and error!

and i get three hours..but that is on the other problem of where it is 12in and the rate is 2in/hr and 3in/hr

_1 2 3_________
|x|x|x|x|x|x| | | | | | | <-------fat crayon
__1 2 3______
|x|x|x|x|x|x|x|x|x| | | | <-------skinny crayon

well...there is more than one way to solve a problem...i draw if i'm stumped on word problems.

**i'm giving this credit to my little darling sister who should have sufficent math history (finished calculus bc her junior year of high school). yet...she used the physics technique to solve this problem -_-()

Answer 10

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