Question

In September 2009, the sales of a product were \((\frac 23)\)^rd of that in July 2009. In November 2009, the sales of the product were higher by 5% as compared to September 2009. How much is the percentage of increase in sales in November 2009 with respect to the base figure in July 2009?

• A. +40%
• B. -20%
• C. -30%
• D. +25%

Answer 1

The Correct Option is C

Let's denote the sales in July 2009 as \( S_J \). 

According to the problem, sales in September 2009 are \(\left(\frac{2}{3}\right)S_J\).

Sales in November 2009 are higher by 5% compared to September 2009. Therefore, sales in November 2009 are:

\( S_N = \left(\frac{2}{3}S_J\right) \times 1.05 \)

Simplifying this, we get:

\( S_N = \frac{2}{3} \times 1.05 \times S_J \)

\( S_N = \frac{2.1}{3}S_J \)

Now, we need to calculate the percentage increase or decrease in sales in November 2009 with respect to July 2009. This can be calculated using the formula:

\(\text{Percentage Change} = \left(\frac{S_N - S_J}{S_J}\right) \times 100\%\)

Substituting the values we have:

\(\text{Percentage Change} = \left(\frac{\frac{2.1}{3}S_J - S_J}{S_J}\right) \times 100\%\)

Simplifying the expression:

\(\text{Percentage Change} = \left(\frac{\frac{2.1}{3} - 1}{1}\right) \times 100\%\)

\(\text{Percentage Change} = \left(\frac{2.1 - 3}{3}\right) \times 100\%\)

\(\text{Percentage Change} = \left(\frac{-0.9}{3}\right) \times 100\%\)

\(\text{Percentage Change} = -0.3 \times 100\%\)

\(\text{Percentage Change} = -30\%\)

Therefore, the percentage of increase in sales in November 2009 with respect to July 2009 is -30%.