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vonsmiley
06-22-2006, 04:15 PM
Hi :)
I was wondering if some one can help me with this problem I'm in trouble from the set up :(

The problem says: The barn on Sue's small farm was built 10 years before the garden shed and twenty years after the house. Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed. What is the present age of each building?

Barn Shed house

if the shed is 'x' and the barn is x-10 what is the relationship with the shed and the house, if the barn and the house are (x-10) + 20? or is all that wrong? I don't know what I'm doing!

tkhunny
06-22-2006, 04:46 PM
The problem says: The barn on Sue's small farm was built 10 years before the garden shed and twenty years after the house. Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed. What is the present age of each building?You may think you are letting the notation help you, but you are not. Do not be afraid of defining more than one variable.

B = Age of Barn
S = Age of Shed
H = Age of House

"The barn on Sue's small farm was built 10 years before the garden shed and twenty years after the house."

B = S + 10
B = H - 20

"Twenty years ago"

That's

B-20 = Age of Barn
S-20 = Age of Shed
H-20 = Age of House

"the age of the house was the same as the combined ages of the barn and the garden shed."

H-20 = B-20 + S-20

Now solve for B, S, and H

TchrWill
06-22-2006, 04:51 PM
Hi :)
I was wondering if some one can help me with this problem I'm in trouble from the set up :(

The problem says: The barn on Sue's small farm was built 10 years before the garden shed and twenty years after the house. Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed. What is the present age of each building?

Barn Shed house

if the shed is 'x' and the barn is x-10 what is the relationship with the shed and the house, if the barn and the house are (x-10) + 20? or is all that wrong? I don't know what I'm doing
1--The barn on Sue's small farm was built 10 years before the garden shed or B = S + 10
2--and twenty years after the house or B = H - 20.
3--Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed or H - 20 = (B - 20)0 + (S - 20)

Can you take it from here?

jonboy
06-22-2006, 08:14 PM
Hello vonsmiley :D

I was wondering if some one can help me with this problem I'm in trouble from the set up

The problem says: The barn on Sue's small farm was built 10 years before the garden shed and twenty years after the house. Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed. What is the present age of each building?

Barn Shed house

if the shed is 'x' and the barn is x-10 what is the relationship with the shed and the house, if the barn and the house are (x-10) + 20? or is all that wrong? I don't know what I'm doing!

Alright we will call the shed \L\bold x, the barn \L\bold y and the house \L\bold z.

1)The barn on Sue's small farm was built 10 years before the garden shed: \L\bold y=x-10

2)The barn on Sue's small farm was built twenty years after the house: \L\bold y=20+z

3) Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed: \L\bold z=y+x

Notice that my first 2 equations have the same variable except \L\bold z. So that means we should use that as our system to start off. Now we need to find out what the z is in terms of x and y. Well I worked it out and this is what I get.

\L\bold y=20+z
\L\bold y=20+y+x
\L\bold 0=20+x

Since filling \L\bold x+y in for \L\bold z yields \L\bold y, we can eradicate that. But we do have another way to solve this. We can change the \L\bold x in terms of \L\bold y and \L\bold z. So we use our third equation: \L\bold z=y+x to find what \L\bold x equals in our third equation \L\bold y=x-10 since the former did not work.

\L\bold x=z-y

So fill that into our original first equation:\L\bold y=z-y-10

Now we use are other equation that is equal to y and has the terms of \L\bold y and \L\bold z to solve a system.

\L\bold y=z-y-10
\L\bold y=20+z

So since both are equal to \L\bold y, set them equal to each other making:

\L\bold z-y-10=20+z

Simplify this to get to \L\bold y (barn age). If you get a negative answer, make it positive since time cannot be negative. The reason you might get one is because people can interpret this statement conversely "the barn on Sue's small farm was built 10 years before the garden shed" by making the garden shed age=barn age+10yrs, thus making a positive answer.

Once you get \L\bold z you can get the \L\bold x and \L\bold y by filling them back into our equations (if you get get \L\bold z=-1, which it is not, then put \L\bold 1 in for \L\bold z not \L\bold-1).

Can you finish :?:

Denis
06-22-2006, 11:02 PM
\L\bold z-7-10=20+z

z - z = 20 + 7 + 10
0 = 37 ??

Try and be a little more careful, jonboy.

Also, if you go back over your other equations, you'll find
most are not accurate; as example y = x - 10 should be y = x + 10.

Anyhow, I'm stopping here since I'm not a moderator...

vonsmiley
06-23-2006, 11:57 AM
First let me thank everyone for the help they have given. Now, let me state just how really dumb I am, because even though now see this:

B = S + 10
B = H - 20

"Twenty years ago"

That's

B-20 = Age of Barn
S-20 = Age of Shed
H-20 = Age of House

And, I understand there must be a way to substitute, I DON'T SEE IT :oops: :cry:

so, if I tried to slove for (H) using (B)

B = H-20 and I know
H - 20 = (B-20) + (S - 20)
[u] +20

H = B -20 + S :shock:

I told you dumb as a box of rocks :roll:

stapel
06-23-2006, 12:43 PM
Twenty years ago, the age of the house was the same as the combined ages of the barn and the garden shed. What is the present age of each building?
This means:

. . . . .(house, 20 years ago) = (barn back then) + (shed back then)

Insert the appropriate "back then" expressions. Note that "B = S + 10" means "S = B - 10", and that "B = H - 20" means "H = B + 20". Substitute for two of the variables in the equation above, to get an equation in only one variable. Solve for that variable. Back-solve for the requested information.

Eliz.

tkhunny
06-23-2006, 01:29 PM
B = S + 10
B = H - 20
H-20 = B-20 + S-20

This is one of the joys of the notation. It doesn't care where it came from. Just solve it without confusing yourself with the source.

In this case, it looks simplest to solve the first two for NOT-B

We have
B = S + 10
B = H - 20

We want
S = B - 10
H = B + 20

These substitute nicely into the third equation.

H-20 = B-20 + S-20

becomes

(B + 20)-20 = B-20 + (B - 10)-20

Can you now solve this for 'B'?

Don't forget to go back and reread the problem statement. Sometimes, one wnders off and answers some question other than that which was asked.

vonsmiley
06-23-2006, 03:21 PM
I am so close. I thought I understood everything, but one of my attempts to check my work came up wrong.

B = S +10 -->so, S = B - 10
B = H - 20 -->so, H = B + 20
H - 20 = ( B -20 ) + ( S -20 )

H - 20 = (B -20) + (S -20)
(B + 20) - 20 = (B - 20) + (B – 10) - 20
1B = 2B- 30 - 20
-2B -2B
-1B = -50

B = 50

B = S +10
30 = S+ 10
-10 -10

20 = S

H = B + 20
H = 30 + 20

H = 50

CHECK:

WRONG:
S = B - 10
20 = 50 -10
20= 40

RIGHT:
H - 20 = (B -20) + (S -20)
50 -20 = (50-20) + 20 -20
30 = 30

Yvonne (less smart than even she knew) :oops:

Denis
06-23-2006, 04:49 PM
B = 50

B = S +10
30 = S+ 10
-10 -10

20 = S

Look at that CAREFULLY :idea:
Btw Yvonne, you don't seem dumb to me...just absent minded :wink:

tkhunny
06-23-2006, 04:51 PM
You have
B = 50
S = B - 10
H = B + 20

Substitute the value for B

S = (50) - 10 = 40
H = (50) + 20 = 70

What were you doing?!

vonsmiley
06-23-2006, 05:07 PM
Thank you everyone for putting up with my slowness and as dennis said absentmindedness/"blindness" :oops: Now my face really is red!

Denis
06-23-2006, 05:17 PM
Isaac Newton and Albert Einstein were both absent minded... :idea:

vonsmiley
06-23-2006, 05:20 PM
My two favorite dead men :)

tkhunny
06-24-2006, 11:28 PM
The color of the barn?