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12 Nov 2021, 01:26
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:50) correct
23%
(01:46)
wrong
based on 88
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A broker charged an average commission of 12 percent on two transactions: one for $600 and the other for $900. If his commission for the $900 transaction was 10 percent, what percent was his commission for the $600 transaction?
(A) 1%
(B) 12%
(C) 13%
(D) 14%
(E) 15%
(A) 1%
(B) 12%
(C) 13%
(D) 14%
(E) 15%
12 Nov 2021, 02:22
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Can use Weighted Average logic.
The 12% Average Return on the 2 investments will be closer to the 10% earned on the amount of $900 (the 10% figure is “weighted more heavily” since $900 is invested at 10% and only $600 is invested at N%)
This means N% must be greater than 12% by more than 10% is less than the overall average of 12% (12 - 10 = 2)
Therefore, N% must be greater than the overall average + 2% points or
N > 14%
Only E satisfies this possibility
Posted from my mobile device
The 12% Average Return on the 2 investments will be closer to the 10% earned on the amount of $900 (the 10% figure is “weighted more heavily” since $900 is invested at 10% and only $600 is invested at N%)
This means N% must be greater than 12% by more than 10% is less than the overall average of 12% (12 - 10 = 2)
Therefore, N% must be greater than the overall average + 2% points or
N > 14%
Only E satisfies this possibility
Posted from my mobile device
21 Nov 2021, 02:19
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Bunuel
Let the \(12\% \) commision be equal to \(x\) amt.
Hence we can write \(\frac{x}{1500}*100 =12 \%\) or \(x= 180\)
So the total commison is \($180\)
We know commision on \($900\) was \(10\%\) or \($90 \)
Hence comm. on \($600\) must be \($90\) or \(\frac{90}{600}*100 =15 \%\)
Ans- E
