The sum of the length, width, and height of a box is 80 cm. The length is 10 cm less than twice the sum of the width and the height, and twice the width exceeds the height by 6 cm. Find the height of the box.
Let's assume that the width of the box is W cm, the height is H cm, and the length is L cm.
We are given the following information:
1. The sum of the length, width, and height of the box is 80 cm, so we can write the equation: L + W + H = 80.
2. The length is 10 cm less than twice the sum of the width and the height, which can be written as L = 2(W + H) - 10.
3. Twice the width exceeds the height by 6 cm, so we can write the equation: 2W = H + 6.
Now, let's solve these equations step-by-step.
Step 1: Solve the equation 2W = H + 6 for H.
Subtract 6 from both sides: 2W - 6 = H.
Step 2: Substitute this value of H in the first equation.
L + W + (2W - 6) = 80.
Simplify: L + 3W - 6 = 80.
Rearrange the equation: L + 3W = 86.
Step 3: Substitute the value of L from equation 2 into equation 3.
(2W + 10) + 3W = 86.
Simplify: 5W + 10 = 86.
Subtract 10 from both sides: 5W = 76.
Divide both sides by 5: W = 15.2.
Step 4: Substitute the value of W back into equation 2 to find H.
2(15.2) = H + 6.
Simplify: 30.4 - 6 = H.
H = 24.4.
So, the height of the box is 24.4 cm.
To solve this problem, we need to set up equations based on the given information and then apply algebraic techniques to find the solution.
Let's start by assigning variables to the unknowns in the problem. Let:
- L be the length of the box
- W be the width of the box
- H be the height of the box
Now let's translate the given information into equations:
1) The sum of the length, width, and height of a box is 80 cm:
L + W + H = 80
2) The length is 10 cm less than twice the sum of the width and the height:
L = 2(W + H) - 10
3) Twice the width exceeds the height by 6 cm:
2W = H + 6
Now we have a system of three equations with three variables. We can solve this system by substitution or elimination.
Let's use substitution to solve this system:
From equation 3, we can express H in terms of W:
H = 2W - 6
Substitute this expression for H into equations 1 and 2:
L + W + (2W - 6) = 80 (equation 1a, with substituted value of H)
L = 2(2W - 6) - 10 (equation 2a, with substituted value of H)
Now simplify equations 1a and 2a:
L + 3W - 6 = 80 (equation 1b)
L = 4W - 22 (equation 2b)
Now we have a system of two equations with two variables (L and W). Let's solve this system using elimination:
Multiply equation 1b by 4 to match the coefficients of W:
4L + 12W - 24 = 320 (equation 1c)
Subtract equation 2b from equation 1c to eliminate L:
4L - L + 12W - 4W = 320 - (4W - 22)
3L + 8W - 4W = 342
Combine like terms:
3L + 4W = 342 (equation 3a)
Now we have a system of two equations with two variables (L and W). We can now solve this system using elimination:
Multiply equation 2b by 3 to match the coefficients of L:
3L = 12W - 66 (equation 2c)
Substitute the value of 12W - 66 for 3L in equation 3a:
12W - 66 + 4W = 342
Combine like terms:
16W - 66 = 342
Add 66 to both sides:
16W = 408
Divide both sides by 16:
W = 25.5
Now substitute the value of W back into equation 2b to find L:
L = 4(25.5) - 22
L = 102 - 22
L = 80
Finally, substitute the value of W back into equation 3 to find H:
2W = H + 6
2(25.5) = H + 6
51 = H + 6
H = 45
Therefore, the height of the box is 45 cm.
If the dimensions are x,y,z then we have
x+y+z = 80
x = 2(y+z)-10
2y = z+6
Now just solve for z.