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

Step 1: Assume a base value for July sales.
Let the sales in July 2009 be 100 units. Using 100 as the base makes it easy to read off the final percentage change directly.
Step 2: Work out the September sales.
September sales were \(\frac{2}{3}\) of July sales, so September sales = \(\frac{2}{3} \times 100 = 66.67\) units.
Step 3: Work out the November sales.
November sales were 5% higher than September sales, so November sales = \(66.67 \times 1.05 = 70\) units.
Step 4: Compare November sales with the July base.
July sales are 100 units and November sales are 70 units. The difference is \(70 - 100 = -30\) units on a base of 100, which is a change of -30%.
Step 5: Match this with the given options.
A drop of 30% from the July base matches option (3). Options (1) and (4) show a rise in sales, which is wrong because sales actually fell over this period. Option (2), a drop of 20%, does not match the -30% figure worked out above, so it is also ruled out.