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A stockbroker has made a profit on 80% of his 40 trades this
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Updated on: 17 Jun 2018, 02:01
Updated on: 17 Jun 2018, 02:01
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Question Stats:
31% (01:56) correct
68% (01:25) wrong based on 77 sessions
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A stockbroker has made a profit on 80% of his 40 trades this 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.
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.
Originally posted by Shrija Roy on 16 Jun 2018, 09:02.
Last edited by Carcass on 17 Jun 2018, 02:01, edited 2 times in total.
Last edited by Carcass on 17 Jun 2018, 02:01, edited 2 times in total.
Formatted
Re: Percent
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16 Jun 2018, 10:34
16 Jun 2018, 10:34
Expert Reply
Are you sure this is a question from the OG 1st edition ??
Moreover, it is QCQ not a multiple answer choice question
Moreover, it is QCQ not a multiple answer choice question
Re: Percent
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16 Jun 2018, 13:43
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
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!
