Question

A school team won 6 games this year against 4 games won last year. What is the percent increase

Answer 1

Given that the school won 6 games this year against 4 won last year
Increase in number of wins = 6 – 4 = 2
Hence per cent increase = (2/4) × 100
                                       = (1/2) × 100
                                       = 50%
Thus the percentage increase in winning the games this year is 50%.
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Answer 2

$\huge\underline\textbf{Answer:-}$

The increase in the number of wins or the amount of change =6 - 4 = 2

$Percentage \: increase = \: \frac{amount \: of \: change \: }{orginal \:amount \: or \: base } \times 100$

$= \frac{increase \: in \: the \: number \: of \: wins}{orginal \: number \: of \: wins} \times 100 = \frac{2}{4} \times 100 = 50$

Threrefore, the answer is 50