Word Problems

Question

The sum of the length, width and height of a rectangular box is 75cm. The length is twice the sum of the width and height, and twice the width is 5cm greater than the height. Find the dimensions of the box.

Answer 1

Lets create a system of equations. Let l = length, w = width, and h = height.

Based on the information given, we can create these three equations

l + w + h = 75
l = 2 (w + h)
2w = h + 5

Just to make life easier later, lets distribute the "2" in the second equation

l = 2w + 2h

Notice how the first two equations have "l", "w", and "h" in all of them. Because of the way the second equation is written, we could substitute "2w + 2h" in place of "l" in the first equation.

l + w + h = 75
l = 2w + 2h
2w + 2h + w + h = 75
3w + 3h = 75
w + h = 25

Now let's focus on these two equations:
2w = h + 5
w + h = 25

It looks like if we multiply the second equation by (-1) and rearrange the order, we could eliminate the variable "h" when adding these two equations together.

- w - h = -25 ---------> -25 = - h - w

2w = h + 5
-25 = - h - w
___________
2w - 25 = 5 - w
+w + w
3w - 25 = 5
+25 +25
3w = 30
w = 10

Now, lets substitute "10" for "w" in ANY of the equations that have two variables in it:


w + h = 25
10 + h = 25
h = 15


Now, lets substitute "10" for "w" and "15" for "h" in ANY of the equations that have three variables in it.

l + w + h = 75
l + 10 + 15 = 75
l + 25 = 75
l = 50

The dimensions of the box are:
length = 50 cm
width = 10 cm
height = 15 cm

Answer 2

Hii, Monisola - 

 

To do this problem, we need to set up 3 algebraic equations with 3 unknowns. As long as the number of equations is equal to the number of unknowns, you should be able to solve for the unknown values.

 

Let us say that length is L, width is W and height is H.

 

Then from the first sentence of the problem, you can state that:

 

L + W + H = 75

 

It further states that the length is twice the sum of the width and height. You can show this as:

 

L = 2(W+H)

 

It next says that twice the width is 5 cm greater than the height. This can be shown as:

 

2W = H + 5

 

So the 3 equations look like this:

 

L + W + H = 75

L = 2(W+H)

2W = H + 5

 

We can start by taking the value of L from the second equation and plugging it into L for the first one. That gives:

 

2(W + H) + W + H = 75

Multiplying this out gives:

2W + 2H +W + H = 75

Combining like terms, we get:

3W + 3H = 75

 

Now we can take the last of the 3 equations, and solve for H and plug that value into the equation immediately above.

 

2W = H + 5

H = 2W - 5

 

Plugging this value for H into the equation above, (3W + 3H = 75) gives the value of W as follows:

3W + 3 (2W - 5) = 75

3W + 6W -15 = 75

9W - 15 = 75

9W = 90

W = 10

 

From the equation above, H = 2W - 5. Therefore we can solve for H as follows:

H = (2)(10) - 5

H = 15

 

Because we also know that L + W+ H = 75, we can now solve for L:

 

L + 10 + 15 = 75

L + 25 = 75

L = 50

 

So the answer is length is 50cm, width is 10cm, and height is 15cm.