Solving Inequality Word Questions
(You might like to read Introduction to Inequalities and Solving Inequalities first.)
In Algebra we have “ inequality ” questions like :
Sam and Alex play in the same soccer team.
Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.
What are the possible number of goals Alex scored?
How do we solve them ?
Reading: Solving Inequality Word Questions
The flim-flam is to break the solution into two parts :
Turn the English into Algebra .
then practice Algebra to solve .
Turning English into Algebra
To turn the English into Algebra it helps to :
- Read the whole thing first
- Do a sketch if needed
- Assign letters for the values
- Find or work out formulas
We should besides write down what is actually being asked for, so we know where we are going and when we have arrived !
The best manner to learn this is by example, so let ‘s try our first case :
Sam and Alex play in the same soccer team.
Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.
What are the possible number of goals Alex scored?
arrogate Letters :
- the number of goals Alex scored: A
- the number of goals Sam scored: S
We know that Alex scored 3 more goals than Sam did, so : A = S + 3
And we know that in concert they scored less than 9 goals : S + A < 9
We are being asked for how many goals Alex might have scored : A
Solve:
Start with:
S + A < 9
A = S + 3, so:
S +
(S + 3)
< 9
Simplify:
2S + 3 < 9
Subtract 3 from both sides:
2S < 9 − 3
Simplify:
2S < 6
Divide both sides by 2:
S < 3
Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals .
Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .
arrest :
- When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
- When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
- When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
- (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)
Lots More Examples!
Example: Of 8 pups, there are more girls than boys.
How many girl pups could there be?
assign Letters :
- the number of girls: g
- the number of boys: b
We know that there are 8 pups, so : g + b = 8, which can be rearranged to
b = 8 − g
We besides know there are more girls than boys, so :
g > barn
We are being asked for the number of girl pups : g
clear :
Start with:
g > b
b = 8 − g, so:
g >
8 − g
Add g to both sides:
g + g > 8
Simplify:
2g > 8
Divide both sides by 2:
g > 4
so there could be 5, 6, 7 or 8 daughter pups .
Could there be 8 daughter pups ? then there would be no boys at all, and the interview is n’t authorize on that point ( sometimes questions are like that ) .
Check
- When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
- When g = 7, then b = 1 and g > b is correct
- When g = 6, then b = 2 and g > b is correct
- When g = 5, then b = 3 and g > b is correct
- (But if g = 4, then b = 4 and g > b is incorrect)
A rapid example :
Example: Joe enters a race where he has to cycle and run.
He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed.
Joe completes the race in less than 2½ hours, what can we say about his average speeds?
assign Letters :
- Average running speed: s
- So average cycling speed: 2s
Formulas :
- Speed = distance Time
- Which can be rearranged to: Time = distance Speed
We are being asked for his average speeds : s and 2s
The race is divided into two parts :
1. Cycling
- Distance = 25 km
- Average speed = 2s km/h
- So Time = distance Average Speed = 25 2s hours
2. Running
- Distance = 20 km
- Average speed = s km/h
- So Time = outdistance Average Speed = 20 s hours
Joe completes the slipstream in less than 2½ hours
- The total time < 2½
- 25 2s + 20 s < 2½
resolve :
Start with:
252s
+
20s
< 2½
Multiply all terms by 2s:
25 + 40 < 5s
Simplify:
65 < 5s
Divide both sides by 5:
13 < s
Swap sides:
s > 13
So his median rush run is greater than 13 kilometers per hour and his average focal ratio cycling is greater than 26 km/h
In this example we get to use two inequalities at once :
Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t, where t is the time in seconds.
At what times will the velocity be between 10 m/s and 15 m/s?
Letters :
- velocity in m/s: v
- the time in seconds: t
formula :
- v = 20 − 10t
We are being asked for the prison term t when v is between 5 and 15 m/s :
10 < five < 15
10 < 20 − 10t < 15
clear :
Start with:
10 < 20 − 10t < 15
Subtract 20 from each:
10
− 20
< 20 − 10t
− 20
< 15
− 20
Simplify:
−10 < −10t < −5
Divide each by 10:
−1 < −t < −0.5
Change signs and reverse inequalities:
1
>
t
>
0.5
It is neater to show the smaller
number first,
so
swap over:
0.5 < t < 1
So the speed is between 10 m/s and 15 m/s between 0.5 and 1 second after .
And a sanely hard exercise to finish with :
Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area.
The perimeter of the room is 16 m.
What could the width and length of the room be?
Make a sketch : we do n’t know the size of the tables, lone their area, they may fit absolutely or not !
assign Letters :
- the length of the room: L
- the width of the room: W
The formula for the perimeter is 2(W + L), and we know it is 16 thousand
- 2(W + L) = 16
- W + L = 8
- L = 8 − W
We besides know the area of a rectangle is the width times the length : Area = W × L
And the area must be greater than or equal to 7 :
- W × L ≥ 7
We are being asked for the potential values of W and L
Let ‘s clear :
Start with:
W × L ≥ 7
Substitute L = 8 − W:
W ×
(8 − W)
≥ 7
Expand:
8W − W2 ≥ 7
Bring all terms to left hand side:
W2 − 8W + 7 ≤ 0
This is a quadratic inequality. It can be solved many way, here we will solve it by completing the squarely :
Move the number term −7 to the right side of the inequality:
W2 − 8W ≤ −7
Complete the square on the left side of the inequality and balance this by adding the same value to the
right side of the inequality:
W2 − 8W + 16 ≤ −7 + 16
Simplify:
(W − 4)2 ≤ 9
Take the square root on both sides of the inequality:
−3 ≤ W − 4 ≤ 3
Yes we have two inequalities, because 32 = 9 AND (−3)2 = 9
Add 4 to both sides of each inequality:
1 ≤ W ≤ 7
So the width must be between 1 m and 7 m ( inclusive ) and the length is 8−width .
check :
- Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m2 (fits exactly 7 tables)
- Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m2 (7 won’t fit)
- Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m2 (7 fit easily)
- Likewise for W around 7 m
