algebra word problems

• May 3rd 2008, 03:31 PM
jscalalamoboy
algebra word problems
1. Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20, how many of each kind of coin did she have? 2. Yolanda had three times as many nickels as dimes. If the total value of her coins was$1, how many of each kind of coin did she have?
3. If 2/3 pound of coffee costs $4.84, what does one pound cost? 4. Peanuts worth$2.95/kg were mixed with cashews worth $6.25/kg to produce a mixture worth$5.00/kg. How many kilograms of each kind of nuts were used to make 33 kg of the mixture?
5. The smaller of two numbers is three more than -2 times the larger one. If four times the smaller number is subtracted from the larger number, the result is 51. Find the numbers.
6. The perimeter of a rectangle is 90 cm. The length is 15 cm more than 4 times the width. Find the dimensions of the rectangle.
7. Pure copper was mixed with a 12% alloy to produce an alloy that was 20% copper. How much of the pure copper and how much 12% alloy were used to produce 132 kg of the 20% alloy?
8. The cost of 5 boxes of small paper clups and 2 boxes of large paper clips is $49.30 seven boxes of small clips and 4 boxes of large ones cost$79.10. Find the cost of a box of each size of paper clips.
• May 3rd 2008, 06:09 PM
nppatel4428
1. 60 quarters,12 dimes
2.12 nickels, 4 dimes
3. Maybe 7.26?
4-----
5-------6.length 39 and width 6
• May 3rd 2008, 06:14 PM
jscalalamoboy
THANKS!
But would you mind showing work...?
• May 3rd 2008, 07:29 PM
Reckoner
Hello!

Quote:

Originally Posted by jscalalamoboy
1. Eugenia had five times as many quarters as dimes. If the total value of her coins was $16.20, how many of each kind of coin did she have? Let the number of quarters be $Q$ and the number of dimes $D$. Since Eugenia's total is$16.20, we have $0.25\,Q + 0.1\,D = 16.2$. But, we also know that there are five times as many quarters as dimes, so $Q=5\,D$. Now, we have two equations with two unknowns. Can you figure it out from here?

Quote:

Originally Posted by jscalalamoboy
2. Yolanda had three times as many nickels as dimes. If the total value of her coins was $1, how many of each kind of coin did she have? This is very similar to question 1. Try setting up the equations! Quote: Originally Posted by jscalalamoboy 3. If 2/3 pound of coffee costs$4.84, what does one pound cost?

Coffee costs $4.84 for every 2/3 of a pound. Therefore, $C=x\left(\frac{\4.84}{2/3\text{ lb}}\right)=x\left(\frac{3(\4.84)}{2\text{ lb}}\right)$, where $C$ is the cost and $x$ is amount. Quote: Originally Posted by jscalalamoboy 4. Peanuts worth$2.95/kg were mixed with cashews worth $6.25/kg to produce a mixture worth$5.00/kg. How many kilograms of each kind of nuts were used to make 33 kg of the mixture?

Suppose $p$ is the amount of peanuts, in kilograms, in the mixture, and $c$ is the amount of cashews. We know that the mixture is a combination of the peanuts and cashews, so $p+c=33\text{ kg}$. We can now set up a second equation using the costs of each substance:
$(\2.95)p +(\6.25)c = (\5.00)(33\text{ kg})$.
You should be able to take it from here.

Quote:

Originally Posted by jscalalamoboy
5. The smaller of two numbers is three more than -2 times the larger one. If four times the smaller number is subtracted from the larger number, the result is 51. Find the numbers.

Give a name to each number, and set up the equations described in the problem. This one should be straightforward.

Quote:

Originally Posted by jscalalamoboy
6. The perimeter of a rectangle is 90 cm. The length is 15 cm more than 4 times the width. Find the dimensions of the rectangle.

Again, use what you know about the relationship between a rectangle's length, width, and perimeter to set up a system of equations.

Quote:

Originally Posted by jscalalamoboy
7. Pure copper was mixed with a 12% alloy to produce an alloy that was 20% copper. How much of the pure copper and how much 12% alloy were used to produce 132 kg of the 20% alloy?

Call the mass of the pure copper $p$, and that of the alloy $a$. The mixture consists of the pure copper plus the alloy, so $p + a = 132\text{ kg}$. Also, 12% of the alloy and 20% of the mixture is copper, so we have:
$(100\%)p + (12\%)a = (20\%)132\text{ kg}$
$p + 0.12a=0.2(132)$.

Now solve the system.

Quote:

Originally Posted by jscalalamoboy
8. The cost of 5 boxes of small paper clups and 2 boxes of large paper clips is $49.30 seven boxes of small clips and 4 boxes of large ones cost$79.10. Find the cost of a box of each size of paper clips.

Try this on your own. Pick variables to represent the cost of each respective box size, and then use the problem description to set up a system of equations.