Math Help: Percentages?

Question

Mark made a poster for his room by enlarging a photo of his favourite hockey player, which he took himself. The original photo had the dimensions 6 by 9cm. Mark enlarged the picture by 180% but found it still wasn't big enough. He then had the enlarged photo enlarged again by 190%. What are the dimensions of his final poster?

Please don't just give me the answer, but how you got it as well. I've tried many times but I just can't get it right.

Aqua · Accepted Answer

ok , the initial dimensions of the photo are 6 cm by 9 cm.

To obtain the size after 180 % enlargement , you take 180% of 6 cm and add it to 6 cm and same with 9 cm.

 180% of 6 cm is (180/100) * 6 cm thats  (1.8)*6cm = 10.8cm  and 180% of 9 cm is (1.9)*9cm = 16.2 cm.

you get the 180% enlargement + original size as:
(10.8 + 6) cm = 16.8 cm  by (16.2 + 9) cm = 25.2 cm
the dimensions now of the enlarged poster are 16.8cm by 25.2 cm

Again , now the already enlarged photo has to be enlarged by 190% , so you take 190 % of 16.8 cm and add 16.8 to it and same with 25.2 cm. <remember the original photo dimensions are not to be considered here>

you get, 190% enlargement of already enlarged pic:
 (31.92 + 16.8)cm = 48.72 cm by (47.88 + 25.2)cm = 73.08cm

the dimensions you get now are 48.72 cm by 73.08 cm which is the solution to your problem.

or the final dimension of the poster are 0.4872 m by 0.7308 m.

I hope this helps you , good luck !
:)

Anonymous · Answer

190% is just the original dimensions times 1.9 -- that's it.  100% is the original times 1, 50% is the original times .5, etc.

So 6cm times 1.9 = 11.4cm, and 9cm times 1.9 = 17.1cm.  Those are the final dimensions.

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Math Help: Percentages?

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