Question 967840: Nancy has 14.80 in change. She has three times as many nickels as quarters. she has two more quarters then dimes.How many coins of each type she has
Found 3 solutions by CubeyThePenguin, greenestamps, josgarithmetic:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! 5n + 10d + 25q = 1480
n = 3q
q = d + 2
5(3(d+2)) + 10d + 25(d+2) = 1480
15d + 30 + 10d + 25d + 50 = 1480
50d = 1400
d = 28
---> 28 dimes, 30 quarters, 90 nickels
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
Presumably a solution using formal algebra is needed. If so, take the time to set up the problem using a single variable (requiring a single equation) instead of separate variables for the three kinds of coins, requiring three equations.
x = # of quarters
3x = # of nickels
x-2 = # of dimes
The total value is $14.80 = 1480 cents:
Solve using basic algebra; I leave that to you.
You can get some very good mental exercise by solving the problem informally, using logical reasoning and simple mental arithmetic.
(1) Loan Nancy 2 more dimes so that the number of dimes is equal to the number of quarters. Now the total value of the coins is 1500 cents; and for each quarter she has 1 dime and 3 nickels.
(2) Group the coins in to groups of 1 quarter, 1 dime, and 3 nickels. The value of each of those groups is 25+10+15=50 cents.
(3) The number of those groups needed to make the total of 1500 cents is 1500/50 = 30. So there are 30 quarters, 30 dimes, and 90 nickels.
(4) Now have Nancy give you back the 2 dimes you loaned her, to find the coins she has are...
ANSWER: 30 quarters, 90 nickels, 28 dimes
CHECK: 30(25)+90(5)+28(10)= 750+450+280 = 1480
Answer by josgarithmetic(39630)  (Show Source):
|
|
|
