Question

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A school team won 6 games this year against 4 games won last year. What is the per cent increase?​

Answer 1

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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%.

Answer 2

→

No of games won this year = 6

No of games won last year = 4

increase in number of games won = 6-4

= 2

percentage increase =

$\frac{increase \: in \: number \: of\: games \: won \: }{number \: of \: game \: won \: last \: year \: } \times 100\%$

$= \frac{2}{4} \times 100\%$

$\frac{1}{2} \times 100\%$

= 50%