Source: Manhattan 5 lb 13.74
Problem statement:
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
My Concern:
The solution provided seems to make an assumption that the gains/losses in each trade is fixed. Is this a valid assumption to make, or did I miss something in question/solution?
Solution:
(B). The stockbroker has made a profit on 80% of his 40 trades this year, so (0.80)(40) = 32 of his trades so farhave been profitable.Quantity B asks for the maximum number of additional losses in a row he can have without dropping below 50%. If hestays at 32 profitable trades and 8 non-profitable trades, and all future trades are losses, then he canât go above 32 Ă 2= 64 trades without dropping below a 50% success rate. At 64 trades exactly, he would have 32 profitable trades and32 non-profitable trades, for a âsuccessâ percentage of 50% profitable trades. 64 - 40 = 24 trades, so he can have 24losses in a row without dropping below 50%. Quantity B is larger.
Alternatively, you can use algebra. Set the number of additional losing trades (above 40) to x. Then, the number ofwinning trades will remain constant at 32, the total number of trades will increase to 40 + x, and the total number oflosing trades will be 8 + x. Quantity B asks for the maximum number of additional losses in a row he can have withoutdropping below 50% of profitable trades, so set up an inequality. The percentage of profitable trades must be greaterthan or equal to 50:
Cross multiply (note: you know the variable represents a positive number, so you donât need to do anything to theinequality sign):64 Ă 40 + x24 Ă xTherefore, the stockbroker can lose money on 24 trades in a row and still have 50% of trades be profitable, soQuantity B is 24. Quantity B is larger.