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# ps #15

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29 May 2007, 01:55
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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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29 May 2007, 01:59
So if amount spend in 1956 = x, then 20x = 12 buillion
x = 0.6 billion = 600 million
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29 May 2007, 02:05
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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29 May 2007, 11:45
$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 CEO Joined: 21 Jan 2007 Posts: 2747 Location: New York City Followers: 11 Kudos [?]: 939 [0], given: 4 Re: ps #15 [#permalink] ### Show Tags 16 Jun 2007, 21:53 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 Manager Joined: 17 May 2007 Posts: 172 Followers: 1 Kudos [?]: 12 [0], given: 0 ### Show Tags 17 Jun 2007, 07:38 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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18 Jun 2007, 10:44
easy proportion
12 billion ---------------- 2 000 %
x ------------------------- 100%
x=12 billion*100/2000 = 600 million
18 Jun 2007, 10:44
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