Question

In September 2009, the sales of a product were \(\frac{2}{3}\)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%

Hint:

Take July as 100, find September as two-thirds of it, raise September by 5% to get November, then compare with the July base of 100.

Answer 1

The Correct Option is C

This can also be solved by carrying the sales figures as fractions of the July value instead of picking a base number.

1. Let the July 2009 sales be $J$.
2. September sales are two-thirds of July sales: $S = \frac{2}{3}J$.
3. November sales are 5% higher than September sales, so $N = S \times \frac{21}{20} = \frac{2}{3}J \times \frac{21}{20}$.
4. Multiply the fractions: $N = \frac{2 \times 21}{3 \times 20}J = \frac{42}{60}J = \frac{7}{10}J = 0.7J$.
5. So November sales equal 70% of the July figure, which means they are 30% below the July figure.
6. The percentage change from July to November is $\frac{N-J}{J}\times 100 = \frac{0.7J - J}{J}\times 100 = -30\%$.

This confirms the sales fell by 30% compared to the July base.

\[\boxed{-30\%}\]