College Algebra Word Problems

[Chino]

Anyone wanna give these a shot while I work on them. I think I can figure them out but if you post your equation to it and how you get the answer It may help me out alot I literally suck at math word problems.

1.The distance from first to second base in indoor baseball is 33 feet less than the corresponding distance in softball. The square bounded by the bases of an indoor baseball diamond is 2871 feet less in area than that bounded by the bases of a softball diamond. Find the distance from one base to the other in indoor baseball



2. An inlaid checkerboard has a mahogany border 2 inches wide. The area of the border is 176 square inches. Find the length of the board


3. last one is.... The sum of the digits of two-digit numeral is 8. The number with the digits interchanged is 7 times the tens digit of the original number. Find the original number. (Hint. let u represent the units digit and t represent the tens digit, you need to solve two simultaneous equations..


I know the first two are quadratic based word problems... Anyone who can help out I greatly appreciate it!

[Ralph]

Originally posted by Chino
Anyone wanna give these a shot while I work on them. I think I can figure them out but if you post your equation to it and how you get the answer It may help me out alot I literally suck at math word problems.

1.The distance from first to second base in indoor baseball is 33 feet less than the corresponding distance in softball. The square bounded by the bases of an indoor baseball diamond is 2871 feet less in area than that bounded by the bases of a softball diamond. Find the distance from one base to the other in indoor baseball



I havent done those type of probleems in a long time.

Let L= Distance bettween indoor bases.
Let X= Distance bettween softball bases

We know:

L= x-33 and L^2=x^2-2871

Set them equal to eachother. (x-33)^2=x^2-2871

Solve for x. x=60 Then knowing that The distance between indoor bases(L) is L=x-33. Plug in the x and you get L.

23 Feet

[Ralph]

Originally posted by Chino
2. An inlaid checkerboard has a mahogany border 2 inches wide. The area of the border is 176 square inches. Find the length of the board

Ill let you do this one. Ok. If the area of the border is 176 and its 2 inches thick. Think of it as 2 squares. If you subtract the area of the inner square from area of the bigger one you will get the area of the border.

So set the lenght of the board = X. So the area of the whole chess board is "x^2". If there is a 2 inch border what is the lenght of the side of the smaller square?[May help to draw a diagram. ]

Once you find what your two areas are in terms of X.

[Big Area] - [Inner area] = 176

Solve for X.

[coolguy8769]

bravo ralph!! when i looked at i was like:confused: but i just didnt look at it in math terms (im not in school im on a snow day(week) dont make me think)

chino i plan on goin to WVU how is it? do you live on or off campus

[Chino]

Originally posted by Ralph
I havent done those type of probleems in a long time.

Let L= Distance bettween indoor bases.
Let X= Distance bettween softball bases

We know:

L= x-33 and L^2=x^2-2871

Set them equal to eachother. (x-33)^2=x^2-2871

Solve for x. x=60 Then knowing that The distance between indoor bases(L) is L=x-33. Plug in the x and you get L.

23 Feet

Ralph 60-33 = 27 feet not 23 feet... any help on the next two would be greatly appreciated. I know how to solve equations the hard part for me is getting the equation...

[Ralph]

Originally posted by Chino
Anyone wanna give these a shot while I work on them. I think I can figure them out but if you post your equation to it and how you get the answer It may help me out alot I literally suck at math word problems.

I hate word problems too. But make sure you get alot of practice with them cause they will keep coming back.

[Ralph]

Originally posted by Chino
Ralph 60-33 = 27 feet not 23 feet... any help on the next two would be greatly appreciated. I know how to solve equations the hard part for me is getting the equation...

TYPO. Did you read the response for #2. Let me know if u need me to elaborate.

[Chino]

Originally posted by Ralph
TYPO. Did you read the response for #2. Let me know if u need me to elaborate.

working on it now, so the border will be x-2 right? so I have x, x^2 and x-2?

[Chino]

Ralph if you have aim, it would help me out tremendously, my name is ex416xcracer thanks bro im stuck on the equation... blahhhhhhhh

[Ralph]

Originally posted by Chino
working on it now, so the border will be x-2 right? so I have x, x^2 and x-2?

Check Again.

The border runs all around;)

[Chino]

Originally posted by Ralph
Check Again.

The border runs all around;)
x-8? so 176 = (x-8)x

[Chino]

Originally posted by Ralph
Check Again.

The border runs all around;)

(x-2)^2

[Ralph]

Originally posted by Chino
x-8? so 176 = (x-8)x

Draw a picture

You got the First part.

So it was [x^2 - (x-?)^2=176]

[Ralph]

Im not the clearest of writers. How long is the Red line in terms of the Blue (x)

[Chino]

Originally posted by Ralph
Im not the clearest of writers. How long is the Red line in terms of the Blue (x)

x-4

[Chino]

24?

[Ralph]

Originally posted by Chino
x-4

Yup. Now Solve for x

[Ralph]

Originally posted by Chino
24?

Got it. unit is 24 sq. in

[Chino]

Originally posted by Ralph
Got it. unit is 24 sq. in

now on to the last one??

[Ralph]

Originally posted by Chino
3. last one is.... The sum of the digits of two-digit numeral is 8. The number with the digits interchanged is 7 times the tens digit of the original number. Find the original number. (Hint. let u represent the units digit and t represent the tens digit, you need to solve two simultaneous equations..

Well just use the advice the problem gave us. The Problem is looking for a double digit number. The "tens digit of the number is = t and let u = the single digit. (For example, in 93: T=9 and U=3)

Ok so the problem tells us that the sum of the two digits is 8. Hence we get u+t=8.

The second part says. If we interchange the digits. (93 becomes 39), THAT number is equal to 7 times t.

So. If we have T and U. We would write the number in terms of t and u. [10u+t] . So we now know that 10u+t=7t.

Now solve the two equations you got for t and then set them equal to eachother which would then eliminate the t and lets you solve for U. once you have U plug that value in U + T = 8 and solve for T.

Now that you have T and U. The answer is 10T + U


Let me know if u ant me to clarify anything

[Ralph]

This is a really dumb problem. I know it might be confusing.

[Chino]

Originally posted by Ralph
Well just use the advice the problem gave us. The Problem is looking for a double digit number. The "tens digit of the number is = t and let u = the single digit. (For example, in 93: T=9 and U=3)

Ok so the problem tells us that the sum of the two digits is 8. Hence we get u+t=8.

The second part says. If we interchange the digits. (93 becomes 39), THAT number is equal to 7 times t.

So. If we have T and U. We would write the number in terms of t and u. [10u+t] . So we now know that 10u+t=7t.

Now solve the two equations you got for t and then set them equal to eachother which would then eliminate the t and lets you solve for U. once you have U plug that value in U + T = 8 and solve for T.

Now that you have T and U. The answer is 10T + U


Let me know if u ant me to clarify anything


so my two equations are as follows u+t=7 and 10u+t=7t
when i solve the first equation i get T=7-u
the second equation I get t=10/6u

that doesn't make sense to me the fraction part???

[Ralph]

Originally posted by Chino
so my two equations are as follows u+t=7 and 10u+t=7t
when i solve the first equation i get T=7-u
the second equation I get t=10/6u

that doesn't make sense to me the fraction part???

You get t=8-u and t=(10u)/6

both equations are equal to t.
This means that (8-U)=(10u)/6

Now solve for u

48 - 6u = 10u
16u= 48
U=3

Now that you know U=3 plug it in [t=8-u]

T=8-3
T=5

So T=5 and U=3

So the number is 53. [10t+u]

[Chino]

Thank you much Ralph, that problem was dumb I concur lol! I can do finance and accounting no problem but when it comes to word problems i'm stumped! Thanks again I greatly appreciate it!

[Ralph]

Originally posted by Chino
Thank you much Ralph, that problem was dumb I concur lol! I can do finance and accounting no problem but when it comes to word problems i'm stumped! Thanks again I greatly appreciate it!

No problem. It was making the commercials in the Reno911 marathon pass pretty quickly