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Kenneth has $5. He spends g cents everyday. How much money does he have left after one week?

a. Express your answer in cents.
b. Express your answer in dollars.
 Mar 4, 2014
 #1
avatar+2353 
0
mathcalc:

Kenneth has $5. He spends g cents everyday. How much money does he have left after one week?

a. Express your answer in cents.
b. Express your answer in dollars.



Note that 1 cent is $ 0.01

Then $5 = 500 cents,

Given that a week has 7 days the answer is

500 cents - 7*g cents (where * stands for times or multiplied by) after one week

or similarly $5 - 0.01 * g * 7 after one week
 Mar 4, 2014
 #2
avatar+201 
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I did try using same way you explained & i was confused at last part & I still didn't get it (last portion).
 Mar 5, 2014
 #3
avatar+118609 
0
mathcalc:

Kenneth has $5. He spends g cents everyday. How much money does he have left after one week?

a. Express your answer in cents.
b. Express your answer in dollars.



Hi mathcalc (I liked Jyoti better - never mind, i can't always have everything my own way.)

So he spends g cents every day. That is 7 x g cents every week. and this can be written as 7*g or just 7g (there is an invisable multiply sign)

Since he spends 7g cents and he had 500 cents to start with. He must be left with (500-7g )cents. This is probably the bit you already understood.

Now there are different ways to do this. I am going to do it a little differently from reinout but I can explain his way as well if you want me to.

There are 100 cents in a dollar so
If you have 100cents, that is the same as $1 (100 divided by 100=1)
If you have 200cents, that is the same as $2 (200 divided by 100=2)
If you have 150 cents, that is the same as $1.50 (150 divided by 100=1.5)
if you have m cents, that is the same as m divided by 100cents [and this can be written as (m/100 ) dollars.]
If you have (500-7g) cents that must be the same as (500-7g) divided by 100 (and this can be written as (500-7g)/100 dollars [I got this just by following the pattern]
This looks different from reinout's answer, but it is really the same.

Whenever i get hard problems with horrible numbers or algebra, I try to make it easier by swapping the hard numbers with easy numbers and then see what I would do. When I know what to do with the easy numbers I can then use the same method with the hard numbers.
It makes it easier this way.

Now, would you like me to explain reinout's method as well. Just take a little time to think about mine first. okay?
 Mar 5, 2014
 #4
avatar+201 
0
God morning. Thank you much for reply. I got it.
 Mar 5, 2014

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