SOLUTION: A child has 40 coins worth more than $6.00. If the cois are dimes and quarters only, what can you conclude about the number of quarters?
I can set up a problem with .10x + .2
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-> SOLUTION: A child has 40 coins worth more than $6.00. If the cois are dimes and quarters only, what can you conclude about the number of quarters?
I can set up a problem with .10x + .2
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Question 52766This question is from textbook
: A child has 40 coins worth more than $6.00. If the cois are dimes and quarters only, what can you conclude about the number of quarters?
I can set up a problem with .10x + .25y > 6.00 but this does not address the 40 coins, so I know something is wrong. It doesn't seem like 40-.10x+.25y > 6.00 works either.
Thanks This question is from textbook
You can put this solution on YOUR website! A child has 40 coins worth more than $6.00. If the coins are dimes and quarters only, what can you conclude about the number of quarters?
:
We can make two equations:
:
"child has 40 coins (dimes & quarters)
d + q = 40
and they are worth over $6
.10d + .25q > 6
:
Using the 1st equation, arrange so we are solving for q (quarters)
d + q = 40
d = [40 - q]
:
Substitute [40-q] for d in the 2nd equation:
.10[40-q] + .25q > 6
4 - .10q + .25q > 6
.15q > 6 - 4
q > 2/.15
q > 13.33
q => 14 has to be an integer number
Check:
14 quarters + 26 dimes
3.50 + 2.60 = $6.10 [fulfills the requirement of > $6]
:
Try 13 quarters + 27 dimes
3.25 + 2.70 = $5.95 less than the required $6
