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A stockbroker has made a profit on 80% of his 40 trades this

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A stockbroker has made a profit on 80% of his 40 trades this [#permalink] New post 16 Jun 2018, 09:02
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A stockbroker has made a profit on 80% of his 40 trades this year.

Quantity A
Quantity B
23
The maximum number of consecutive trades that the stockbroker can lose before his profitable trades drop below 50% for the year


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by Carcass on 17 Jun 2018, 02:01, edited 2 times in total.
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Re: Percent [#permalink] New post 16 Jun 2018, 10:34
Expert's post
Are you sure this is a question from the OG 1st edition ??

Moreover, it is QCQ not a multiple answer choice question
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Re: Percent [#permalink] New post 16 Jun 2018, 13:43
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Shrija Roy wrote:
A stockbroker has made a profit on 80% of his 40 trades this year.

A: 23
B: The maximum number of consecutive trades that the stockbroker can lose before his profitable trades drop below 50% for the year



The boker has already made 40 trades and 80% of them are profitable.

Let us suppose that the broker makes \(x\) more trades and looses them all.

Now this value of x is such that his percent of profitable trades become 50%.


Initially the broker had made: 40 trades of which 80% were profitable or \(\frac{80}{100} \times 40=32\) profitable and 8 loss making.

Now with the new \(x\) number of trades he is 50% loss making so \(\frac{50}{100} \times (40+x)=8+x\).

Remember that he already had 8 trades which lost money previously. Solve for x and you shall get the value \(x=24\).

So if he makes 24 more trades and losses them all he will have 50% profitability.


You can cross check by putting the values

40 trades ---------------- 32 profit ---------------- 8 loss

40 +24 trades ------------ 32 profit ---------------- 8 loss +24 loss =32 loss.

After 24 the profitability is 50% not BELOW 50%.

He needs to loose 25 trades to go below 50%. Hence option B is correct!
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Re: A stockbroker has made a profit on 80% of his 40 trades this [#permalink] New post 30 Dec 2018, 21:08
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Lets use the average formula here.

total profitable trades/total trades= average profitable trades.

we dont yet know the total profitable trades, lets call it p
average profit= .8
total trades = 40

P/40=.8
P = 32
there were 32 profitable trades

once we have this we can use the mixture ratio
we need to find out how many negative trades he needs to get in order for the average
profitable trade to drop to .5
the negative trades will be x
32/(40+x) = .5
32=20+.5x
12=.5x
x=24
Thus, the answer is B
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Re: A stockbroker has made a profit on 80% of his 40 trades this [#permalink] New post 31 Dec 2018, 03:22
Expert's post
Shrija Roy wrote:
A stockbroker has made a profit on 80% of his 40 trades this year.

Quantity A
Quantity B
23
The maximum number of consecutive trades that the stockbroker can lose before his profitable trades drop below 50% for the year


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Let us solve for B..
Now, 80% of 40 = \(\frac{80}{100}*40=32\)..
Now there are no wins and these 32 wins become less than 50% of total..
So let the total be X, so \(32\geq\frac{50}{100}*X...X\leq{64}\)..
Thus max losses are 64-40=24..
Thus B>A
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Re: A stockbroker has made a profit on 80% of his 40 trades this   [#permalink] 31 Dec 2018, 03:22
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