I did this one correct. But it took me a lot of time.

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I did this one correct. But it took me a lot of time. [#permalink]

03 Sep 2007, 01:34

I did this one correct. But it took me a lot of time.
Please, share the approach you would use.

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03 Sep 2007, 02:30

4 tests -> total of 320 questions.
# of correct question = 256 questions.
5 test -> total if 400 questions. 82% = 328 questions. So he needs 72 questions right on the last test.

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Re: average, percent, have better approach? [#permalink]

03 Sep 2007, 06:44

IrinaOK wrote:

I did this one correct. But it took me a lot of time.
Please, share the approach you would use.

nice one:

he has taken 4 tests or 320 questions and 80% of which he answered correct. 320*80%=256 correct

now he is taking test 5 or another 80 questions. this test makes total question 320+80=400

put it this way: 256+x=400*0.82
x=72

it should not take u more than 1.5 mins.

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03 Sep 2007, 07:47

72 questions

80% of 4X 80 = 256
82% of 400 = 328

he needs 72 more questions to be at 82%

[#permalink] 03 Sep 2007, 07:47

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I did this one correct. But it took me a lot of time.

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I did this one correct. But it took me a lot of time.

I did this one correct. But it took me a lot of time.

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