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Bunuel
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In 1970 there were 8,902 women stockbrokers in the United States. By 1 19 Aug 2018, 10:04
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In 1970 there were 8,902 women stockbrokers in the United States. By 1978 the number had increased to 19,947. Approximately what was the percent increase?

(A) 45%
(B) 125%
(C) 145%
(D) 150%
(E) 225%
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B
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In 1970 there were 8,902 women stockbrokers in the United States. By 1 19 Aug 2018, 10:14
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Bunuel
In 1970 there were 8,902 women stockbrokers in the United States. By 1978 the number had increased to 19,947. Approximately what was the percent increase?

(A) 45%
(B) 125%
(C) 145%
(D) 150%
(E) 225%
Show more

% increase =\(\frac{(19947-8902) * 100}{8902}\)
= \(\frac{11905}{8902 * 100}\)
~\(\frac{12000 *100}{9000}\)
< 130%
~ 125 %
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In 1970 there were 8,902 women stockbrokers in the United States. By 1 19 Aug 2018, 10:46
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11000/8900=110/89 ~110/90=11/9~1.22 ~122%~125% ans B

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In 1970 there were 8,902 women stockbrokers in the United States. By 1 27 Dec 2022, 20:05
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Bunuel
In 1970 there were 8,902 women stockbrokers in the United States. By 1978 the number had increased to 19,947. Approximately what was the percent increase?

(A) 45%
(B) 125%
(C) 145%
(D) 150%
(E) 225%

% increase =\(\frac{(19947-8902) * 100}{8902}\)
= \(\frac{11905}{8902 * 100}\)
~\(\frac{12000 *100}{9000}\)
< 130%
~ 125 %
Show more
Hey Counter Sniper
i get it that you were trying to round it up to the nearest integer for the answer but (19947-8902 doesn't equal to 11905).
thank you
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In 1970 there were 8,902 women stockbrokers in the United States. By 1 12 Feb 2023, 13:53
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I think you should try avoid calculations:

So you know that in 1970 the number of Womens were 8,902.

If the increase was 100% it should be near 18,000 (try to think 9,000x2=18,000, since 8,902 it's close to 9,000) .

If the increase was 150% (9,000x2+ 4,500) the number of Womens would be 22,500

Since 145% is realy near to 150% it can only be option B

:)
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In 1970 there were 8,902 women stockbrokers in the United States. By 1 18 Sep 2025, 19:53
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