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1)The price of the tickets went up from $48 to $58: 56. What percent of a change does this represent?
(Hint: the change is what percent of the original price?)

2) We have 73 coins, all dimes and quarters, in the value of $13. How many quarters, how many
dimes?

2006-11-05 17:45:02 · 7 answers · asked by Agentj100 4 in Science & Mathematics Mathematics

7 answers

1) Percentage change = [(new price - old price) / old price] x 100
so:
[($58.56 - $48) / $48] x 100 = 22%

2)
a + b = 73
where a = number of dimes
where b = number of quarters
[ie. some number of dimes plus some number of quarters equals 73 coins]

First, write it in terms of dollar value.
($0.10 x a) + ($0.25 x b) = $13
[this means: some number of dimes multiplied by their dollar value, plus some number of quarters multiplied by their dollar value, equals $13]

Remember that "a + b = 73"?
It can be written as "a = 73 - b"

Put this relationship into: ($0.10 x a) + ($0.25 x b) = $13
($0.10 x [73 - b]) + ($0.25 x b) = $13

Which gives:
$7.3 - 0.1b + 0.25b = $13

Organise the "b"s into one side, the integers into the other:
0.15b = 5.7
b = 38

This means you need 38 quarters.

Since you had a total of 73 coins to play with, and 38 are already known to be quarters, 35 must be dimes.

2006-11-05 17:56:33 · answer #1 · answered by Carinna C 2 · 0 0

1) You want the percentage of the second value compared to the first value which is 58.56?48=1.22=122%

2) you want to right two equations representing how many dimes and quarters you have and there values.

25q+10d=1300
q+d=73
the second equation gives you q=73-d
input that value for q in the first equation you get
25(73-d)+10d=1300
1825-15d=1300
subtract 1825 from both sides
-15d=-525
divide by -15
d=35
then use our equation from earlier
to get 73-35=38=q
So we have 35 dimes and 38 quarters

2006-11-05 18:01:32 · answer #2 · answered by Anonymous · 0 0

% Change = Change / Original Price x 100

% = (48 - 58)/48 x 100
you do the math.

D = Dime
Q = Quarter

0.10D + 0.25Q = 13
D + Q = 73

10D + 25Q = 1300
10D + 10Q = 730

Subtract the equations

15Q = 570
Q = 38
D = 35

2006-11-05 18:02:13 · answer #3 · answered by gizmeaux1 2 · 0 0

1) 48 + 48*p/100 = 58.56
48p/100 = 10.56
p=10.56*100/48 = 22%

2) x number of dimes and y number of quarters

x+y = 73 => x=73 - y
0.1 x + 0.25 y = 13 => 0.1(73 - y) + 0.25y = 13
7.3 - 0.1y + 0.25y = 13
0.15 y = 5.7
so y=38
and x=35

2006-11-05 18:42:37 · answer #4 · answered by Anonymous · 0 0

2)

d + q = 73

10d + 25q = 1300
..................
d = 73 - q

10(73 - q) + 25q = 1300
730 - 10q + 25q = 1300
15q = 1300 - 730 = 570
q = 570/15 = 38
d = 73 - 38 = 35

.... check

10d + 25q = 1300
10(35) + 25(38) = 350 + 950 = 1300

(you made a typo in other problem and don't get it)

2006-11-05 18:01:54 · answer #5 · answered by atheistforthebirthofjesus 6 · 0 0

1) 58.56 / 48 = 1.22. Therefore it increased by 22%.


2)
0.25x + 0.1 * (73 - x) = 13

0.15x + 7.3 = 13

0.15 * x = 5.7

x = 38 quarters
73 - x = 35 dimes

2006-11-05 18:00:01 · answer #6 · answered by Jason 2 · 0 0

1.) (48-58)/48
2.) use 2 equations,
where x=dimes
y=quarters
1st equation x+y=73
2nd equation 0.1*x+0.25*y=13

you do the computations, i dont have a calculator n_n

2006-11-05 17:56:04 · answer #7 · answered by jamezu 2 · 0 0

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