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Question 214792: Phillip has 10 coins in his pocket. Phillip has the same number of pennis as quarter and the same number of dimes as nickels. He also has one more nickel than he does pennies. If he has 97 cent how many coins of each type does he have?
Found 2 solutions by drj, ankor@dixie-net.com:
Answer by drj(1380)   (Show Source):
You can put this solution on YOUR website! Phillip has 10 coins in his pocket. Phillip has the same number of pennies as quarter and the same number of dimes as nickels. He also has one more nickel than he does pennies. If he has 97 cent how many coins of each type does he have?
Step 1. Let x be number of pennies, Let x be the number of quarter, let x+1 be number of nickels, and let x+1 also be number of dimes.
Step 2. x+x+x+1+x+1=10 or 3x+2=10 or 2x=8 or x=2 where we know the number of coins is 10.
Step 3. Check if equal to 97 cents.
 cents. ok!
Step 4. ANSWER: There are 2 pennies, 2 quarters, 3 nickels, and 3 dimes.
I hope the above steps were helpful.
For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Good luck in your studies!
Respectfully,
Dr J
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! Phillip has 10 coins in his pocket.
Phillip has the same number of pennies as quarters and the same number of dimes as nickels.
He also has one more nickel than he does pennies.
If he has 97 cent how many coins of each type does he have?
:
Let x = no. of pennies & quarters
Let y = no. of dimes & nickels
:
"He has one more nickel than pennies
y = x + 1
:
Total no. of coins
2x + 2y = 10
simplify, divide by 2
x + y = 5
:
we know: y = x+1
therefore
x = 2 ea pennies & quarters
y = 3 ea nickels & dimes
:
Check using $amt
.01(2) + .25(2) + .05(3) + .10(3) =
.02 + .50 + .15 + .30 = .97
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