Question 204021This question is from textbook Introductory Algebra
: I have tried...for several hours. :( I absolutely can not translate this word problem and solve it correctly. I truly appreciate any help with this.
PROBLEM:
A storekeeper goes to the bank to get $10 worth of change. She requests twice as many quarters as half dollars, twice as many dimes as quarters, three times as many nickles as dimes and no pennies or dollars. How many of each coin did the storekeeper get?
This question is from textbook Introductory Algebra
Found 3 solutions by jsmallt9, RAY100, Earlsdon:
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! In solving a word problem you often use one or more variables and then one or more equations to solve it. Generally the number of variables dictates the number of equations. So it is preferable to use as few variables as possible.
You can reduce the number of variables you use if you decide what one variable will represent and then try to express as many of the other unknowns in the problem using expressions involving that variable. And creating these expressions is usually easier if your variable represents the lowest (or near to the lowest) unknown in the problem.
In your problem, I hope it is clear that the number of half dollars would be the lowest number in the problem. (Please read and think about the problem until this becomes clear.) So we will use our variable, x, to represent the number of half dollars. Now let's see how many of the other unknowns we can express in terms of x:
Let x = the number of half dollars.
Since she asks for "twice as many quarters as half dollars":
The number of quarters = 2*x = 2x
Since she asks for "twice as many dimes as quarters":
The number of dimes = 2*(2x) = 4x
Since she asks for "three times as many nickels as dimes":
The number of nickels = 3*(4x) = 12x
We have succesfully expressed the numbers of all the different coins in terms of a single variable! We only need one equation, then, to solve the problem.
The equation will state that the sum of the values of these coins adds up to $10. Therefore we will need expressions for the values of the coins.
After some thought it we should be able to come up with the correct expressions.
Since 1 half dollar is worth $0.50 and 2 half dollars is worth $1.00 and 3 half dollars is worth $1.50, we should be able to figure out that the value of the half dollars is $0.50 times the number of half dollars. Since we have "x" half dollars then the value of these half dollars is x*0.50 = 0.5x
Similarly we can determine that...
the value of 2x quarters is (2x)*0.25 = 0.5x
the value of 4x dimes is (4x)*0.10 = 0.4x
the value of 8x nickels is (12x)*0.05 = 0.6x
We're now ready to write our equation and solve it. Again, the equation will express the fact the the sum of the values of all these coins is $10.00:
0.5x + 0.5x + 0.4x + 0.6x = 10
Simplifying by adding like terms on the left:
2x = 10
Divide both sides by 2:
x = 5
Now we can use our expressions to answer the question.
Since x represents the number of half dollars, she got 5 half dollars.
Since 2x represents the number of quarters, she got 2(5) or 10 quarters.
Since 4x represents the number of dimes, she got 4(5) or 20 dimes.
Since 12x represents the number of nickels, she got 12(5) or 60 nickels.
Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! Let n= number of half dollars
n*50=50n,,,,for half dollars
2n*25 =50n,,,,for quarters
4n *10=40n,,,,for dimes
12n*05=60n,,,for nickles
,
Total = 200n,,,,this equals $10=1000 cents
.
200n=1000,,n=5
.
Now half dollars = 5,,,,,,$2.50
,,,,,quarters =2n=10,,,,,,,,$2.50
,,,,,dimes =4n=20,,,,,,,,,,,$2.00
,,,,,nickels =12n=60,,,,,,,,$3.00
TOTAL,,,,,,,,,,,,,,,,,,,,,,$10.00
Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Well, the first step is to assign variables to each of the unknowns in the problem:
Let:
H = # of half-dollars.
Q = # of quarters.
D = # of dimes.
N = # of nickels.
Now to translate the English prose to algebraic statements:
Q = 2H "...twice as many quarters as half-dollars,..."
D = 2Q "...twice as many dimes as quarters,..."
N = 3D "...three times as many nickels as dimes..."
Now we can write an equation relating all four varables (H, Q, D, and N) to the total amount tendered at the bank ($10.00)
For example, the dollar-value of H half dollars can be expressed as: (0.5)H, so...
(0.5)H+(0.25)Q+(0.1)D+(0.05)N = $10.00
Now the trick is to reduce the number of variables from four down to one! It really doesn't matter which variable you choose but try to pick the easiest path to getting it.
I picked the variable H, the number of half-dollars.
Here we go!
N = 3D substitute D = 2Q to get:
N = 3(2Q) = 6Q substitute Q = 2H
N = 3(2(2H)) or (3(4H)) so...
N = 12H
D = 2Q substitute Q = 2H, so...
D = 2(2H) = 4H
So now we have:
Q = 2H
D = 4H
N = 12H so we can substitute into the first equation to get:
(0.5)H+(0.25)(2H)+(0.1)(4H)+(0.05)(12H) = 10 Add the H's together:
(0.5)H+(0.5)H+0.4H+0.6H = 10
2H = 10 Divide both sides by 2.
H = 5 and...
Q = 2H = 10 and...
D = 4H = 20 and...
N = 12H = 60, so...
The store-keeper got:
5 Half-dollars.
10 Quarters.
20 Dimes.
60 Nickels.
Now let's check to see if this all adds up tp $10.00
(0.5)5+(0.25)(10)+(0.1)(20)+(0.05)(60) = 2.5+2.5+2+3 = 10 OK!
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