Question 510402: Natalie had some nickels, Dirk had some dimes, and Quincy had some quarters. Dirk has 5 more dimes than Quincy has quarters. If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, they will all have the same amount of money. How many coins did each originally have?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Natalie had some nickels, Dirk had some dimes, and Quincy had some quarters.
Dirk has 5 more dimes than Quincy has quarters.
If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, they will all have the same amount of money.
How many coins did each originally have?
:
Let n = no. nickels
let d = no. of dimes
let q = no. 0f quarters
:
"Dirk has 5 more dimes than Quincy has quarters."
d = q + 5
:
"If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, "
Results:
Natalie amt:
.05n - .05 + .25
.05n + .20
Dirk amt:
.10d + .05 - .10
.10d - .05
Quin amt:
.25q + .10 - .25
.25q - .15
:
"They all have the same amt of money"
.25q - .15 = .10d -.05
.25q = .10d - .05 + .15
.25q = .10d + .10
We know,"Dirk has 5 more dimes than Quincy has quarters." d = q+5, replace d
.25q = .10(q+5) + .10
.25q = .10q + .50 + .10
.25q - .10q = .60
.15q = .60
q = 
q = 4 quarters
then
d = 4 + 5
d = 9 dimes
Find n
.05n + .20 = .10(9) - .05
.05n = .90 - .05 - .20
.05n = .65
n = 
n = 13 nickels
:
Summarize, originally, 13 nickels, 9 dimes, 4 quarters
:
You can check this finding the amt each had after all these coin transfers, they each had 85 cents
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