Download Real Life Examples of Algebra

Real Life Examples of Algebra
Sport Announcement
Turning on the radio you hear the following commentary,
“What a match, 9 goals scored already and the home side are trailing by 1”
What’s the score for each team?
Identify the variable
We don’t know the individual scores so we assign them variables. Lets
say the away team’s score is a.
We also know that the home side’s score is 1 less so we can write the
expression a-1 for their score.
Derive an Equation
We need to form an equation so we can work out what a is; from the
commentary we know the two scores added together is 9
(a-1)+a = 9
Simplifying (grouping like terms)
2a – 1 = 9
Solving (getting all numbers on one side and letters on the other)
2a = 9 +1
2a = 10
a = 10/2
a=5
So the away teams score is 5 and the home teams score is one less, 4
Check
A quick check, the total number of goals scored is 4+5 = 9
Wedding
I have a budget set aside for my wedding, I can only spend €4000 on the hotel and meal, given the
information below how many people can I invite?
 Band and DJ €800
 Dinner €40 a head
 Sandwiches and sausages €8 a head, but only order for half the people
Identify the variable
We need to know the number of guests, let’s use the variable g for the
number of guests.
Derive an equation
The cost will be equal to the constant 800 plus 40 times g plus 8 times
half of g
4000 = 800 + 40g + 8(g/2)
Simplify
4000-800 = 40g + 4g
3200 = 44g
Solve
g = 3200/44
g = 72.7
So I can only invite 72 guests without going over the budget
Check
800 + 40(72) + 8(36) = €3,968
Plugs
You have bought a 2kW heater, but there is no fuse in the plug. If I told you that the power (P) of a
heater is equal to the voltage (V) multiplied by current (I), what size fuse would you need.
Other information,
 The voltage in the sockets is 230 volts
 Standard size fuses are 3, 5 and 13 amps and will blow if the current exceeds this number.
 Remember to convert kW to W
Identify the variables
Power (P), Current (I) and Voltage (V) are all variables
Derive the equation (write the equation down)
P = VI
Substitute the numbers we know, remembering to convert kW to W
2000 = 230I
Solve
2000/230 = I
I = 8.70 amps
So we must use a 13 amp fuse because the heater would blow a 5 amp fuse
in normal use.
Size of a house
When building a new house it was decided that the front of the house would be 1.4 times bigger
than the side. The rural renewal scheme limits the size of the floor area to 210m2. What is the
maximum size of the house?
Identify the variable
One side of the house is dependent on the other side so we only need one
variable, we’ll call the shorter side x
We can write the expression 1.4x for the length of the longer side.
Derive an Equation
We know that the area of a rectangle is one side multiplied by the other
A = (1.4x)(x)
210 = 1.4x2
Solving (getting all numbers on one side and letters on the other)
x2 = 210 / 1.4
x2 = 150
x = √150
x = 12.247m
The longer side is
1.4x = 17.146m
we wouldn’t need that accuracy when building, so let’s round it to
12.25m and 17.15m.
Check
A quick check, 12.25m and 17.15m = 210.09 (this is a little bit bigger
because we rounded up)
Finding the volume of a oil tank
It is much easier to measure the circumference of an oil tank accurately than it is to measure the
radius (we can just wrap the measuring tape around!) Measuring the height is also easy enough. But
how can we find the volume of a tank if we don’t know the radius.
 Create a new formula to calculate the volume when you know the radius.
 Solve the equation for a oil tank 450cm in circumference and 1400cm high, and for a gas
tank 80cm in circumference and 1200cm high (remember 1000cm3 = 1 litre)
Identify the Variables
Circumference (C) Volume (V) Radius (r) height (h)
Note: π is not a variable, it is a constant.
Write down the equations we know
V = πr2h
C = 2πr
We want volume but we only have C, but notice that r is present in both.
We can rearrange the formula for C to find r then, substitute this back in
to the first.
Rearrange
C = 2r
r = C / (2π)
We could use this formula every time and simply plug the value of r
found into the equation for V and solve.
More Advanced Way
But if we have many cylinders to measure it would be handier to create a
whole new formula to relate C to V so we’d only have one sum to do.
Substitute
We know that
V = πr2h
And from above
r = C / (2π)
 C2
C 
V  
 h    2
 2 
 4
2
V 
Solving

h

C 2h
4
450 2  140 202,500  140

 2,257,166cm 3  2,260l
43.14
12.56
2
80  120 768000
gas cylinder V 

 61,146cm 3  61l
43.14
12.56
Oil Cylinder V 