Transcript 8_6 Coin_ ticket_ and weight problemsTROUT11

8.6 Coin, Ticket, and Weight
Problems
Goal: To use a system of equations to solve
coin, ticket and weight problems
Pattern
• Similar to 8.4 problems
• Set up two equations
• One equation is a physical amount that you
can count with two different categories (ie
like the fish problem 2 kinds rock fish and
blue fish)
• The other equation will have a dollar
amount associated with it or a weight
associated with the different categories to
give a total dollar or pound amount.
Coin Problems
This coin equals how many pennies?
Nickel
5
Dime
10
Quarter
25
A vending machine only takes nickels and
dimes. There are three times as many
dimes as nickels in the machine. The face
value of the coins is $5.25. How many of
each coin are in the machine?
3n  d
315  d
45  d
5n  10d =525
5n  10 3n =525
35n =525
n =15
Try This
A jar of dimes and quarters contains
$15.25. There are 103 coins in all. How
many of each are there?
Let q = the number of quarters
Let d = the number of dimes
-10 (q  d  103 )
=
33  d  103
d  70
25q  10d  1525
10q  10d  1030
15q  495
q  33
Try This
A jar of dimes and nickels contains $2.55.
There are 30 coins in all. How many of each
are there?
5n  10d  255
-5 ( n  d  30 ) =
n  21  30
n9
5n  5d  150
5d  105
d  21
Try This
A jar of quarters and nickels contains
$3.00. There are 6 more nickels than
quarters. How many of each are there?
25 ( n  q  6 )
=
15  q  6
15  6  q
5n  25q  300
25n  25q  150
30n  450
n  15
q9
Anyone get 11 quarters and 17 nickels???
Ticket Problem
There were 166 paid admissions to a game. The price was
$2 for adults and $0.75 for children. The amount taken in
was $293.25. How many adults and children attended?
A+ C =166 and 2A + 0.75C =293.25
A = 166 –C
2(166-C) +0.75C =293.25
332- 2C +0.75C =293.25
-1.25 C = -38.75
C=31
A=135
Weight Problem
A jar contains 5 gram bolts and 10 gram
bolts. The contents of the jar weigh 3.8
kg. If there are 460 bolts, how many of
there of each kind?
X = 5 gram bolts and y =10 gram bolts
X +y = 460 and 5(x) + 10(y) =3800
You solve it
X =160 5 gram bolts and y= 170 10 gram bolts
Assignment
Page 390 #5-14 all