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!
_________________

Sandy

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test