Algebra II: Summary of Word Problems for 4.8, 5.8, 5.9
Number Relationships:
Ex 1: two numbers differ by 5 and have a product of 84. Find the numbers
Define variables:
1st # = x
2nd # = x + 5
If you were to subtract x + 5 – x, you would get 5
Their product is 84, so x(x+5) = 84. Distribute, set = to zero and factor.
(x +12)(x - 7) = 0
x2 + 5x -84=0
so, x = -12 or 7
If x = -12, then x + 5 = -7
If x = 7, then x + 5 =12
There are two answers: -12 and -7; 7 and 12 – both differ by 5 and have a product of 84
Ex. 2: The sum of two numbers is 9. The sum of their squares is 101.
Define variables: 1st # = x
2nd # = 9 – x
If you add these numbers, x + 9-x, you get 9
The sum of their squares is 101 results in the equation x2 + (9-x)2 = 101. Foil, combine like terms and set
= to zero.
x2 + 81 -18x + x2 – 100 = 0
2x2 -18x -20 =0 All numbers are even, so divide both sides by 2 to simplify
X2 – 9x -10 =0
factors into (x-10)(x +1) =0,
If x = 10, then 9-10 = -1
so, x = 10 0r -1
If x = -1, then 9 – (-1) = 10
My answers are -1 and 10
Ex. 3: The sum of squares of two consecutive even numbers is 100. Find the numbers.
Define variables: if the first number is x, then the second number is x+2
*with consecutive even or odd, you use x and x + 2; for consecutive you use x and x+1
x2 +(x+2)2 = 100 reflects that the sum of the squares is 100
x2 + x2 + 4x + 4 – 100 = 0
Foil, combine like terms and set = to zero
2x2 + 4x – 96 = 0
Divide both sides by 2
x2 +2x – 48 = o factors into (x+8)(x-6) = 0, so x = -8 or 6
If x = -8, x+2 = -6;
if x = 6, then x+2= 8
There are two answers: -8, -6 and 6, 8 both are consecutive even integers; the sum of the squares is 100.
Area: Posters, walkways, photographs Area Problems involving area and framing: Always draw a picture
Total area = area of frame + area of picture
Area of frame = total area – area of picture
Area of picture = total area – area of frame
Ex. P. 198
A graphic artist is designing a poster that consists of a rectangular print with a uniform border. The print
is to be twice as tall as it is wide, and the border is to be 3 inches wide. If the area of the poster is to be
680 in2, find the dimensions of the print.
2x+6
Draw and label:
3
I know the area of the area of the whole poster is
2x
3
680, so (2x+6)(x+6) = 680.
3
x+6
If I solve for x, I can find the length and width.
x
3
(2x+6)(x+6)=680
Foil
2x2 + 18x+36= 680
Divide both sides by 2 since it will make the problem sassier
X2 + 9x + 18 = 340
Set = to zero and factor
X2 + 9x -322 =0
So factor (x – 14)(x + 23)=0
X= 14 or -23
Drop the negative answer since we are dealing with length
X=14, so 2x = 28
Plug the numbers back into the expressions for length and width of the poster
Dimensions are 14in x 28 in
Area: Working with area and perimeter
Ex 1: The length of a rectangle is 1 m less than twice the width. If the area is 55 m2, find the perimeter.
Draw a picture with the given information:
You know that the area is 55, so x(2x-1)=55
x
2x-1
Distribute, set = to zero, and factor to solve.
2x2 – x – 55 = 0 factors into (2x - 11)(x + 5) = 0
x = 5.5 or -5
Drop negative answer since we are working with length
if x = 5.5, then 2(5.5)-1 = 10
Perimeter = 2(5.5) + 2(10) = 31 m
Ex 2: The perimeter of a rectangle is 18 cm. The area is 14 cm2. Find the dimensions.
x
Start by labeling with x and y. Then set up formulas for area and
y
perimeter. 2x + 2y = 18 and xy = 14
If you take the perimeter formula and divide both sides by 2, it is simpler.
x + y = 9 and xy = 14
x=9–y
Solve the perimeter formula for x and substitute it into the area formula
(9-y)y=14
Distribute, set = to zero, factor and solve
0 = y2 – 9y + 14 Factors into 0 = (y-7)(y-2)
so, 9y – y2 = 14
so, Y = 7 or 2
Using the original equation xy = 14, you can see that if y = 2, the x = 7, or if y = 7, then x = 2
So the dimensions are 2 cm by 7 cm
Triangles: # 10, p. 200
The height of a triangle is 7 cm greater than the length of the base, and its area is 15 cm2. Find the
height.
Area for a triangle: A = ½(bh), so 15 = (1/2)(x)(x+7).
Start by multiplying both sides by 2 and distributing.
Draw:
X +7
x
30 = x2 + 7x
0 = x2 + 7x – 30
X = -10 or 3
Factors into 0=(x + 10)(x-3)
Drop negative answer since working with length
Base = 3, so x + 7 = 10 Height = 10 cm.
#6 p. 200: A rectangular garden has a perimeter of 66 ft and an area of 216 ft2. Find the dimensions.
Y
P = 2x + 2y so 66 = 2x + 2y or 33 = x + y if you divide both sides by 2
X
A = xy or 216 = xy so y = 216/x
To get your equation substitute y = 216/x into x + y = 33
𝑥+
216
𝑥
= 33
#14 p. 200: A rectangle is twice as long as it is wide. If its length is increased by 4 cm and its width is
decreased by 3 cm, then the new rectangle formed has an area of 100 cm2. Find the dimensions of the
original rectangle.
W
w–3
2W
2W + 4
Equation: (w-3)(2w+4) =100
You can finish! Answer should be 8 cm by 16 cm
Vertical Motion
There are two formulas given, one if the problem is given in meters and another if the problem is given
in feet. In either case, h is height, v = velocity (or speed), and t = time
If given in meters, use h = vt – 4.9t2
If given feet, use h = vt – 16t2
Page 199, Example 2: A batter hits a baseball directly upwards at 96 ft/s. How long is the ball in the air
before being caught by the catcher and how high does the ball go?
We are working in feet, so use h = vt – 16t2 and plug in the information given. IF the ball returns to the
ground (the pitcher), h = 0 if the a ball goes up and comes back down.
0 = 96t – 16t2
factor out a gcf of 16t
0 = 16t(6 – t)
so t – 0 or 6
This means that when the ball is at the ground at the beginning t= o and when it returns to the ground, 6
seconds have passed. At the highest point, t= 3 (just cut time in half.
To find max height, plug in t=3 into h = 96t – 16t2, so h = 96(3) – 16(9) = 144 feet
#16 p. 200
A ball is thrown directly upward from ground level with an initial speed of 80 ft/s. How high will it go?
When will it return to the ground?
The equation for vertical motion in feet is h = vt-16t2
The height is zero at the beginning and end, so entering zero for height and 80 for v will give me
the beginning time and ending time.
0 = 80t – 16t2
0 = 16t(t – 5)
When h = 0, we will factor using gcf
16t = 0 t-5=0
T = 0 and t = 5
At the beginning, time is zero,
at the end, 5 seconds have passed
At the top, half of the time has elapsed, so I plug in 2.5 for t and 80 for v to find the max height
H = 80(2.5) – 16(2.5)2 so h = 100 ft
# 17 p. 200
A signal flare is fired upward from ground level with an initial speed of 294 m/s. A balloonist cruising at
a height of 2450 m sees it pass on the way up. How long will it be before the balloonist sees it pass on
the way down?
We want to know the time at height 2450, so that is what we will use for h. We use 294 for v.
We use h = vt – 4.9t2 since we are working in meters rather than feet
Once we solve the equation, we will have two values for t; the smaller one is when the flare
passes on the way up and the bigger one is when it passes on the way down. If we subtract
these two times, we will have the time that has elapsed between seeing it for the first time and
second time.
2450 = 294t – 4.9t2
Make it equal zero and then divide it all by 4.9.
It will then factor more easily.
Work
#5 P. 245: Pump A can unload the Lunar Petro in 30 h and pump B can unload it in 24 h. Because of an
approaching storm, both pumps were used. How long did it take to empty the ship.
Work rate is how much they can do per hour
They work the same amount of time
They complete one job, so add their work done to equal one job
Pump A
Pump B
Hourly work rate x
1/30
1/24
Hours worked =
T
T
𝑡
𝑡
+
=1
30 24
Work done
t/30
t/24
P. 250 #6: The intake pipe can fill a tank in 6 h when the outlet pipe is closed. With the outlet pipe open,
it takes 9 h. How long would it take the outlet pipe to empty a full tank?
What is being asked for is the outlet pipes time to empty the tank alone (we will say it is x)
That means that it empties it in 1/x per hours
The tank will fill in 9 h if both the outlet and intake pipes are open so they both work 9 hours to
complete 1 job. We will subtract since the tank is filling while it is emptying
Hourly work rate x
1/6
1/x
In
out
9
6
9
−𝑥 =1
Hours worked =
9
9
Work done
9/6
9/x
Finish; Answer is 18 hrs
Mixture: These three tiers because you are mixing two things to get a final mixture
#8 P. 245: How much water must be evaporated from a 300 L tank of 2% salt solution to obtain a 5%
solution?
In this one, we subtract since we are evaporating, water out. We usually add if we put to things
together. In this case, water that evaporates has 0% salt. (If something is pure, it is 100%)
Liters x
%=
Original
300
0.02
Add (Evaporate)
X
0
Final
300 – x
0.05
Equation: 6 – 0 = 0.05(300 – x) Finish- Answer should be 180 L
Salt
6
0
0.05(300-x)
Distance
#10 from p. 246: Pam jogged up a hill at 6 km/h and then jogged back down at 10 km/h. How many km
di she travel in all if her total jogging time was 1 hr 20 min.
Since you know total time, use d/r = t to fill in the middle column of your chart.
Add the time up to the time down and make it equal to total time. Once you get time, you will
figure out total distance. The distance up = distance down.
1 hr 20 min = 1 20/60 = 1 2/3 = 5/3
rate x
6
10
Up
Down
𝑑
𝑑
5
Time =
d/6
d/10
Equation: 6 + 10 = 3; Finish, answer is 10 km
distance
d
d
#10 p. 251: Tim paddled his kayak 12 km upstream against a 3 km/h current and back again in 5 h 20
min. What is the kayak’s rate in still water (Changed problem slightly).
Again, you know total time, so use d/r=t to fill in the middle column of the chart.
Add the distance up to the distance down to get total distance
Total time = 5 + 20/60 = 5 1/3 = 16/3
Up
rate x
r–3
Down
r+3
Equation:
12
12
+
𝑟−3
𝑟+3
=
16
;
3
Time =
12
𝑟−3
12
𝑟+3
distance
12
12
Finish, answer is 32 km
Cost Per person
#12 P. 251 Members of the ski club contributes equally to obtain $1800 for a trip. When 6 members
couldn’t go, the cost per person increased by $10.
Cost per person equals the total amount divided by the number of people
Original cost per person + 10 new cost per person
Original
# of people
x
New
x-6
Equation:
1800
+
𝑥
10 =
1800
;
𝑥−6
Cost per person
1800
𝑥
1800
𝑥−6
Total cost
1800
1800
Finish, answer is 30
Investment
#11 P. 246: The computer club invested $2200, part at 4.5% and the rest at 7%. The total earnings were
$144. How much was invested at each rate?
If the total invested is 2200, and you invest x amount at one interest rate, you incest 2200-x at
the other
You add the two amounts earned to get the total amount earned
Amount invested
At 4.5%
X
At 7%
2200 – x
Equation: 0.045x + 0.07(2200 – x) = 144
%=
0.045
0.07
Amount earned
0.045x
0.07(2200-x)