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ps #15

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ps #15 [#permalink] New post 29 May 2007, 00:55
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E

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15. In 1980 the government spent $12 billion for direct cash payments to single parents with dependent children. If this was 2,000 percent of the amount spent in 1956, what was the amount spent in 1956? (1 billion = 1,000,000,000)
(A) $6 million
(B) $24 million
(C) $60 million
(D) $240 million
(E) $600 million

not too sure how to attack this problem. do we use scientific notation?
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 [#permalink] New post 29 May 2007, 00:59
So if amount spend in 1956 = x, then 20x = 12 buillion
x = 0.6 billion = 600 million
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 [#permalink] New post 29 May 2007, 01:05
E. 600 Million

1956 = x
So 1980 = x*2000/100 = 12 billion
Solve for x
x=3/5 billion
=0.6 billion
=600 million
600,000,000
(1 million = 6 zeroes, 1 billion = 9 zeroes)
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 [#permalink] New post 29 May 2007, 10:45
$12 billion = 12*10^9
X=amount spent in 1956

then 12*10^9=(2000/100)*X
12*10^9=20*X
we can write 10^9 as 5*20*10^7 so

12*5*20*10^7=20*X (cancel out 20)

X=60*10^7 = 600,000,000 = 600 million

so Answer is E

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Re: ps #15 [#permalink] New post 16 Jun 2007, 20:53
bmwhype2 wrote:
15. In 1980 the government spent $12 billion for direct cash payments to single parents with dependent children. If this was 2,000 percent of the amount spent in 1956, what was the amount spent in 1956? (1 billion = 1,000,000,000)
(A) $6 million
(B) $24 million
(C) $60 million
(D) $240 million
(E) $600 million

not too sure how to attack this problem. do we use scientific notation?



2000 % = 2000/100 = 20
let x = 1956

20x = 12 billion
x = 12billion / 20

12 billion = 12* 10^9 = 12
20 = 2*10^1

= 12* 10^9 / (2* 10^1)
= 6* 10^8
= 600 million

therefore E
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 [#permalink] New post 17 Jun 2007, 06:38
E is the answer

---
(2000/100)X = $12 * 10^9

X = (12 * 10^9 * 100)/2000

==> X = 6 * 10^8 ==> 600* 10^6

==> $600 million
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 [#permalink] New post 18 Jun 2007, 09:44
easy proportion
12 billion ---------------- 2 000 %
x ------------------------- 100%
x=12 billion*100/2000 = 600 million :-D
  [#permalink] 18 Jun 2007, 09:44
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