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04 Jul 2024, 01:30
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E
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Difficulty:
55%
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Question Stats:
61% (01:29) correct
39%
(01:13)
wrong
based on 114
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Company X was profitable in both 2010 and 2011. What was the percent change in Company X’s profits from 2010 to 2011?
(1) Company X’s revenue increased by 10% from 2010 to 2011.
(2) Company X’s expenses decreased by 5% from 2010 to 2011.
(1) Company X’s revenue increased by 10% from 2010 to 2011.
(2) Company X’s expenses decreased by 5% from 2010 to 2011.
04 Jul 2024, 02:16
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I think E,
Revenue increased from 2010 to 2011 by 10% and Expenses decreased by 5% from 2010 to 2011. %profit change is +ve.
But from either of the statements we are not clear about relation between expenses and revenue hence we cant calculate percent change in Company X’s profits.
Revenue increased from 2010 to 2011 by 10% and Expenses decreased by 5% from 2010 to 2011. %profit change is +ve.
But from either of the statements we are not clear about relation between expenses and revenue hence we cant calculate percent change in Company X’s profits.
04 Jul 2024, 06:14
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As we are comparing profits of, let the revenue for 2010 = \(r\), the expenses for 2010 = \(e\) and the profit for 2010 = \(p\).
There for the equation for the profit for 2010 will be \(p = r - e\)
(1) Company X’s revenue increased by 10% from 2010 to 2011.
With this we know that the revenue for 2011 will be \(1.1r\). This alone does not provide enough information.
INSUFFICIENT
(2) Company X’s expenses decreased by 5% from 2010 to 2011.
With this we know that the expenses for 2011 will be \(0.95e\). This alone does not provide enough information.
INSUFFICIENT
(1+2)
Putting this together, we know that the profit for 2011 will be \(1.1r - 0.95e\). As the increase/decrease for expenses and revenues is not uniform, and we are dealing with algebraic terms, it is impossible to solve for the percentage change in profits.
To illustrate this: the percentage change formula in this case will be 100*[(Profit for 2011 - Profit 2010)/Profit 2010]
\(100*\frac{(1.1r - 0.95e) - (r - e)}{(r - e)}\)
\(100*\frac{(0.1r + 0.5e)}{(r - e)}\)
\(\frac{10r + 50e}{(r - e)}\)
As the values in the numerator are not the same and the sign between the values is not the same, it is impossible to take out a common factor which would isolate and cancel out the algebraic. ie. had it been \(\frac{10r - 10e}{(r - e)}\)
Then \(\frac{10(r - e)}{(r - e)}\)
\(10*1\)
\(10%\) there would have been a 10% increase in profit.
INSUFFICIENT
ANSWER E
There for the equation for the profit for 2010 will be \(p = r - e\)
(1) Company X’s revenue increased by 10% from 2010 to 2011.
With this we know that the revenue for 2011 will be \(1.1r\). This alone does not provide enough information.
INSUFFICIENT
(2) Company X’s expenses decreased by 5% from 2010 to 2011.
With this we know that the expenses for 2011 will be \(0.95e\). This alone does not provide enough information.
INSUFFICIENT
(1+2)
Putting this together, we know that the profit for 2011 will be \(1.1r - 0.95e\). As the increase/decrease for expenses and revenues is not uniform, and we are dealing with algebraic terms, it is impossible to solve for the percentage change in profits.
To illustrate this: the percentage change formula in this case will be 100*[(Profit for 2011 - Profit 2010)/Profit 2010]
\(100*\frac{(1.1r - 0.95e) - (r - e)}{(r - e)}\)
\(100*\frac{(0.1r + 0.5e)}{(r - e)}\)
\(\frac{10r + 50e}{(r - e)}\)
As the values in the numerator are not the same and the sign between the values is not the same, it is impossible to take out a common factor which would isolate and cancel out the algebraic. ie. had it been \(\frac{10r - 10e}{(r - e)}\)
Then \(\frac{10(r - e)}{(r - e)}\)
\(10*1\)
\(10%\) there would have been a 10% increase in profit.
INSUFFICIENT
ANSWER E
