Transcript File - Miss Greiner

Coin Problems
Solving Coin Problems
To solve coin problems, organize
the facts in the problem using
verbal sentences or a table.
Number of coins × Value of each = Total value of the
coin in cents
coins in cents
Helen has $1.35 in her bank in nickels and dimes. There
are 9 more nickels than dimes. Find the number she has
of each kind.
Who is the “x man” in the story?
value
nickels
.05
dimes
.10
˟
quantity
x+9
x
=
total
.05(x + 9)
.10x
$1.35
Let x = the number of dimes = 6
x + 9 = the number of nickels = 15
.10x + .05(x + 9) = 1.35
.10x + .05x + .45 = 1.35
.15x + .45 = 1.35
-.45 -.45
.15x = .90
Check:
.15
.15
6 dimes = 6(.10) = $0.60
x=6
15 nickels = 15(.05) = $0.75
Total: $0.60 + 0.75 = $1.35
$1.35 = $1.35 √
A boy has $3.20 in his bank made up of nickels, dimes, and
quarters. There are three times as many quarters as nickels, and 5
more dimes than nickels. How many coins of each kind are there?
value
×
quantity
=
total
nickels
.05
x
.05x
dimes
.10
x+5
.10(x + 5)
quarters
.25
3x
.25(3x)
$3.20
Let x = number of nickels = 3
x + 5 = number of dimes = 8
3x = number of quarters = 9
.05x + .10(x + 5) + .25(3x) = 3.20
.05x + .10x + .50 + .75x = 3.20
.90x +.50 = 3.20
-.50 -.50
.90x = 2.70
Check:
.90
.90
3 nickels = 3(.05) = $0.15
x=3
8 dimes = 8(.10) = $0.80
9 quarters = 9(.25) = $2.25
Total: $0.15 + 0.80 + 2.25 = $3.20
$3.20 = $3.20 √
Viki has $2.80 in quarters and dimes. The number of
dimes is 7 less than the number of quarters. Find the
number she has of each kind.
value
˟
quantity
dimes
.10
x-7
quarters
.25
x
=
total
.10(x – 7)
.25x
2.80
Let x = number of quarters = 10
x – 7 = number of dimes = 3
.25x + .10(x – 7) = 2.80
.25x + .10x - .70 = 2.80
.35x - .70 = 2.80
+ .70 + .70
.35x = 3.50
.35
.35
x = 10
Check:
10 quarters = 10(.25) = $2.50
3 dimes = 3(.10) = $0.30
Total: $2.50 + 0.30 = $2.80
$2.80 = $2.80 √
Total 11/29/11
A purse contains $1.35 in nickels and dimes. In all there
are 15 coins. How many coins of each kind are there?
value
X quantity
=
total
.05(15 – x)
nickels
.05
15 - x
dimes
.10
x
.10x
15
$1.35
For example: If the sum of two number
is 15 and one of the numbers is 6, then
the other number is 15 – 6 or 9.
Let
x = the number of dimes = 12
15 – x = the number of nickels = 3
.10x + .05(15 – x) = 1.35
.10x + .75 - .05x = 1.35
.05x + .75 = 1.35
-.75 -.75
.05x = .60
Check:
12 dimes = .10(12) = $1.20
.05 .05
3 nickels = .05(3) = $0.15
x = 12
Total: $1.20 + 0.15 = $1.35
$1.35 = $1.35√
A purse that contains $3.20 in quarters and dimes has,
in all, 20 coins. Find the number of each kind of coin.
value
×
quantity
=
total
dimes
.10
x
.10x
quarters
.25
20 - x
.25(20 - x)
20
$3.20
Let x = # of dimes = 12
20 – x = # of quarters = 8
.10x + .25(20 – x) = $3.20
.10x + 5.00 - .25x = 3.20
5.00 - .15x = 3.20
-5.00
-5.00
-.15x = -1.80
-.15
-.15
x = 12
Check:
12 dimes: .10(12) = $1.20
8 quarters: .25(8) = $2.00
Total: $1.20 + 2.00 = $3.20
$3.20 = $3.20 √
Mildred bought 1-cent stamps, 33-cent stamps, and
34-cent stamps for $22.45. The number of 1-cent
stamps exceeded the number of 34-cent stamps by 50.
The number of 33-cent stamps was 10 less than twice
the number of 34-cent stamps. How many of each
stamp did she buy?
value
total
1-cent
× quantity =
.01
x + 50
.01(x + 50)
33-cent
.33
2x - 10
.33(2x – 10)
34-cent
.34
x
.34x
$22.45
Let x = # of 34-cent stamps = 25
2x – 10 = # of 33-cent stamps = 40
x + 50 = # of 1-cent stamps = 75
.01(x + 50) + .33(2x – 10) + .34x = $22.45
.01x + .50 + .66x – 3.30 + .34x = 22.45
1.01x – 2.80 = 22.45
+ 2.80 +2.80
1.01x = 25.25
Check:
1.01
1.01
(25) 34-cent stamps: .34(25) =$8.50
(40) 33-cent stamps: .33(40) = $13.20
x = 25
(75) 1- cent stamps: .01(75) = $0.75
Total: $8.50 + 13.20 + 0.75 = $22.45
$22.45 = $22.45 √