hksfton 2019-10-14 01:03 AM

a) Express the number of questions Joseph answers wrongly in terms of x.

b) If Joseph obtains 130 marks, find the value of x.

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[[i] 本帖最後由 hksfton 於 2019-10-14 01:20 AM 編輯 [/i]]

biggoldentooth 2019-10-14 10:47 AM

[quote]原帖由 [i]hksfton[/i] 於 2019-10-14 01:03 AM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=508369607&ptid=28600331][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

Joseph participates in a competition. It is known that each participant has to answer x question, where 30 marks will be obtained for each correct answer, 20 marks will be deducted for each wrong ans ... [/quote]

(a)

Let the number of questions Joseph answered wrongly be Y.

Y/X + 8/X + 3/8 = 1

Y/X + 8/X = 5/8

Y + 8 = 5X/8

8Y + 64 = 5X

Y = (5X - 64)/8

Answer: The number of questions Joseph answered wrongly is (5X - 64)/8

(b)

3X/8 * 30 - 20Y = 130

90X/8 - 20(5X-64)/8 = 130

90X -100X + 1280 = 130 * 8

-10X = -240

X = 24

*********************

Joseph had to answer totally X questions = 24 questions.

He answered 3/8 of the questions correctly = 3X/8 = 9 questions correctly.

He answered Y = (5X-64)/8 questions wrongly = 7 questions wrongly.

He did not answer 8 questions (as stated on 1#)

Joseph participates in a competition. It is known that each participant has to answer x question, where 30 marks will be obtained for each correct answer, 20 marks will be deducted for each wrong ans ... [/quote]

(a)

Let the number of questions Joseph answered wrongly be Y.

Y/X + 8/X + 3/8 = 1

Y/X + 8/X = 5/8

Y + 8 = 5X/8

8Y + 64 = 5X

Y = (5X - 64)/8

Answer: The number of questions Joseph answered wrongly is (5X - 64)/8

(b)

3X/8 * 30 - 20Y = 130

90X/8 - 20(5X-64)/8 = 130

90X -100X + 1280 = 130 * 8

-10X = -240

X = 24

*********************

Joseph had to answer totally X questions = 24 questions.

He answered 3/8 of the questions correctly = 3X/8 = 9 questions correctly.

He answered Y = (5X-64)/8 questions wrongly = 7 questions wrongly.

He did not answer 8 questions (as stated on 1#)

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