MATH 020
Fall 2004
Lesson 15 - Translating Sentences into Equations Word Problems - Section 4.1
page 153-4: 30,37,40 and 42
- (30)
- Call number of Teaching Assistants T
and number of Research Assistants R
then, T+R=600
"there are 3 times as many teaching assistants as research assistants"
can be translated to 3R=T
so the 2 equations are T+R=600 and 3R=T
start with T+R=600 and substitute 3R in place of T. thus,
(T)+R=600 is the same as (3R)+R=600.
Now just solve for R
3R+R=600
4R=600
[4R/4]=[600/4], so R=150
So there are 150 Reasearch Assistants.
- (37)
- Fee is $5.50 and each ticket costs $37.50. Total charge is $343.00
so the equation is 5.50+37.50x=343
Solve for x, where x represents the number of tickets purchaced.
5.50+37.50x=343
(5.50+37.50x)-5.5=343-5.5
37.50x=337.5
[37.50/37.50]x=[337.5/37.5]
x=9
Thus, a total of 9 tickets were purchased.
- (40)
- A 20 foot board is cut into 2 pieces of unknown and unequal length. Call the shorter length a, and the longer
b. Since the sum of these lengths must be equal to 20, then our first equation is a+b=20. "Twice the length of
the shorter piece is 4ft more than the length of the longer piece" can be translated into 2a=b+4.
So our two equations are a+b=20 and 2a=b+4
By subtracting a from both sides of a+b=20, we find that
(a+b)-a=20-a
b=20-a.
Now substitute "20-a" in place of b as follows:
2a=(b)+4
2a=(20-a)+4 then solve for a. This will give us the length of the shorter piece.
2a=20-a+4
2a=24-a
(2a)+a=(24-a)+a
3a=24
[3a/3]=[24/3]
a=8
So the length of the shorter board is 8ft, and the other is 20-8=12ft.
- (42)
- This one is just like #37. Monthly cell phone charge is $35.00. Add $.40 for each minute of phone use.
The total bill is $99.80
Thus, the equation is 35+.40x=99.80, where x stands for the number of minutes used. There is only on unknown here
that we wish to solve for, thus one equation will do.
35+.40x=99.80 solve for x
(35+.40x)-35=(99.80)-35
.40x=64.80
[.40x/.40]=[64.80/.40]
x=162
Thus, the cell phone was used for 162 minutes.
Lesson 16 - Perimeter Problems - Section 4.3
page 170: 5,7,9,13,15
- (5)
- 2L+2W=42
L=2W-3
2(2W-3)+2W=42
4W-6+2W=42
6W-6=42
6W=48
W=8 and L=2(8)-3=13, L=13
- (7)
- 2L+2W=120
L=2W
2(2W)+2W=120
4W+2W=120
6W=120
W=20 and L=2(20)=40
- (9)
- Perimeter of triangle is 110cm
1st side: call it a
2nd side: call it b
3rd side: call it c
a=2b and c=b+30
a+b+c=110
(2b)+b+(b+30)=110
4b+30=110
4b=80
b=20
if b=20, then a=2(20)=40 and c=(20)+30=50
- (11)
- Perimeter of rectangle is 64
Length L=20
find width:
P=2L+2W
(64)=2(20)+2W
64=40+2W
24=2W
W=12
- (13)
- Square picture frame:
perimeter P=48
all four sides are equal.
Call each side a.
thus, 4a=48 or a=12
each side's length is 12.
Lesson 17 -Mixture Problems - Section 4.6
page 193: 3,5,7,9,17,25,27 and 32
- (3)
- High-protein dog food costs: $6.75 per lb.Vitamin supplement $3.25 per lb,
contains p lbs of the high protein stuff, and v lbs. of the vitamin supplement.
Find v and p such that 5 lb. of the mixture costs $4.65 per lb.
first, we see that v+p=5, that is, the sum of the weights of each type of dog food is equal to 5 lbs.
the we have the mixture equation: 6.75p+3.25v=4.65(5)
So the two equations are v+p=5 and 6.75p+3.25v=23.25
since v+p=5, then v=5-p
just replace v with 5-p in the equation 6.75p+3.25v=23.25
6.75p+3.25(v)=23.25
6.75p+3.25(5-p)=23.25 and solve for p
6.75p+3.25(5-p)=23.25
6.75p+16.25-3.25p=23.25
3.50p+16.25=23.25
3.50p+16.25-16.25=23.25-16.25
3.50p=7
[3.5p/3.5]=[7/3.5]
p=2
if p=2, then
thus, 2 lbs. of the high protein dog food is used in the mixture, and 3 lbs of the vitamin supplement.
- (5)
- The entire coffee mix weighs 8 pounds plus 12 pounds, or 20 lb total. The unknown is
the total cost per pound, call it x.
(9.20)8+(5.50)12=20(total cost per pound)
(9.20)8+(5.50)12=20x just solve for x
73.6+66=20x
139.6=20x
[139.6/20]=[20x/20]
x=6.98
the total cost per pound is $6.98
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