I have 4 word problems that I can't solve

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rfxw

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1) A rectangle is 6cm long and 5 cm wide. When each dimension is increased by x cm, the area is tripled. Find the value of x.

2)Two positive real numbers have a sum of 5 and product of 5. Find the numbers.

3)A rectangular animal pen with area 1200 m2 (m squared) has one side along a barn. The other three sides are enclosed by 100 m of fencing. Find the dimensions of the pen.

4)A 5 in. by 7in. photo is surrounded by a frame of uniform width. The area of the frame equals the area of the photo. Find the width of the frame.
 
These are good problems.
Are you asking us to do them for you?
That is not what this site is about.
You have to show some effort on your part.
What have you done on these problem?
 
We just started this section so I don't know to much about these problems but I know for numbers 1,3,and 4 the area= 1/2 base x height but I don't know what to do beyond that point and for number 2 I know that sum is addition and product is multiplication and I was thinking maybe the answer is 3 and 2 but that isn't the product. Thats what I know about these problems.
 
rfxw said:
1) A rectangle is 6cm long and 5 cm wide. When each dimension is increased by x cm, the area is tripled. Find the value of x.

2)Two positive real numbers have a sum of 5 and product of 5. Find the numbers.

3)A rectangular animal pen with area 1200 m2 (m squared) has one side along a barn. The other three sides are enclosed by 100 m of fencing. Find the dimensions of the pen.

4)A 5 in. by 7in. photo is surrounded by a frame of uniform width. The area of the frame equals the area of the photo. Find the width of the frame.
Click to expand...

I'll help you set up the first problem....if you TRULY don't have a clue on the rest of them, then you MUST talk to your teacher.

For problem 1....

IF the rectangle has a length of 6 cm, and a width of 5 cm, and both the length and width are increased by x cm, then

new length = 6 + x cm
and
new width = 5 + x cm

The area of the original figure is 6 cm * 5 cm, or 30 cm^2

If the area of the new figure is TRIPLE the area of the original, then

length of new figure * width of new figure = 3 * area of original

(6 + x)(5 + x) = 3*30

Ok...now, can you solve that for x?

If you're still having difficulty, please repost, showing ALL of the work you've done to try to solve this problem.

If we can see your work, we can determine how best to help you.
 
rfxw said:
1) A rectangle is 6cm long and 5 cm wide. When each dimension is increased by x cm, the area is tripled. Find the value of x.

2)Two positive real numbers have a sum of 5 and product of 5. Find the numbers.

3)A rectangular animal pen with area 1200 m2 (m squared) has one side along a barn. The other three sides are enclosed by 100 m of fencing. Find the dimensions of the pen.

4)A 5 in. by 7in. photo is surrounded by a frame of uniform width. The area of the frame equals the area of the photo. Find the width of the frame.
Click to expand...
1)

What is the area of the old rectangle?

\(\displaystyle A_{old} \, = \, 5 \cdot 6 \, = \, 30\) ......................................................................(1)

When each dimension is increased by x cm - What is the area of the new rectangle?

\(\displaystyle A_{new} \, = \, (5\, + \, x) \cdot (6 \, + \, x)\)........................................................(2)

the area is tripled

\(\displaystyle A_{new} \, = \, 3\cdot A_{old}\)................................................................................(3)

Now use (1) and (2) in (3) - and solve for 'x'

2)

Let the numbers be 'x' & 'y'

Then

x + y = 5.............................................................................(1)

xy = 5 ---> x = 5/y.............................................................(2)(assuming y is not equal to zero)

then using (2) in(1)

5/y + y = 5

y^2 - 5y + 5 = 0

Above is a quadratic equation - solve for 'y'. Then solve for 'x' from (1)

These look like some type of take-home test - are you sure you are supposed to get help on these?

two unknowns
 
I also need help with #3. this is what I have done but I can't find the answer.
A = x(100 - x)
1200 = 100x - x^2
-x^2 +100x - 1200 = 0
then i used the quadratic formula
-100 +- sqrt(10000-4800)/2
-100 +- 20sqrt(13)/-2
50 +- -10sqrt(13)
That is all I've gotten to, but when I check it with the answer my teacher gave me, it's wrong.
The answer should be either 20m x 16m or 30m x 40m.
Please show me how to do this
 
nguyenvanessa21 said:
I also need help with #3. this is what I have done but I can't find the answer.
A = x(100 - x)
1200 = 100x - x^2
-x^2 +100x - 1200 = 0
then i used the quadratic formula
-100 +- sqrt(10000-4800)/2
-100 +- 20sqrt(13)/-2
50 +- -10sqrt(13)
That is all I've gotten to, but when I check it with the answer my teacher gave me, it's wrong.
The answer should be either 20m x 16m or 30m x 40m.
Please show me how to do this
Click to expand...

You need to start by NAMING things. What does "x" stand for in your problem?

You also might find it very helpful to draw a diagram for the problem.

Here's how I understand problem 3.....ONE side of the pen is along the barn, so you don't need fence for that side.

You have 100 m of fence to use.

Suppose you let x = the side of the pen which is parallel to the barn (length), and
let y = the sides of the fence that are perpendicular to the barn (width)

PLEASE draw a diagram!!! You'll see that you have ONE side of length x, and TWO sides of length y.

The area of the pen is 1200 m^2.....so,

length * width = 1200
x * y = 1200

Since you have 100 m of fencing to use to make two sides of y m, and one side of x m (refer to your drawing, please). So....

x + y + y = 100
x + 2y = 100

Now...you have two equations:

x*y = 1200
x + 2y = 100

Solve this system of equations. If you're having trouble with this, please repost, showing ALL of the work you've done, so we can see where you need further help.
 
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