Habs 2010 - Questions 23 and 25
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Habs 2010 - Questions 23 and 25
Hi,
Can someone help me with the following questions from Habs 2010 paper:
Next Question 25:
Can someone help me with the following questions from Habs 2010 paper:
Not sure how to get 95000 from combination of 50,000, 40,000 and 2,500(and double xenox being 5500). The only possibility for score in ninety thousands(that I can think of) is one Zoid, one Yondo and two Xenox, but that will be 95,500 isn't?23. In the computer game “Fizz Darkweek” a player scores points by hitting
certain targets:
“Zoid” scores fifty thousand,
“Yondo” scores forty thousand,
“Xenox” scores two thousand five hundred.
Also, if you hit two Xenox in a row you get an extra five hundred
bonus points.
It is Debbie’s turn. She has four shots on target with a total score of
ninety-five thousand points . Write down one possible set of targets that
she might have hit:
First Shot _________
Second Shot _________
Third Shot _________
Fourth Shot _________
Next Question 25:
I think the answer is five. Because they are asking the "least number", assuming 5 left-handed boys can wear glasses, the least combination of right-handed boys wearing glasses could be 10 - 5 = 5. Am I correct?25. Of the 26 boys in a class, 21 are right-handed. If 10 of the boys in the
class wear glasses, what is the least number of boys in the class who
are both right-handed and wear glasses?
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Re: Habs 2010 - Questions 23 and 25
Its 2 xenox in a row to get the bonus
So
xenox 2500
Zoid 50000
Yondo 40000
xieno 2500
total 95000
So
xenox 2500
Zoid 50000
Yondo 40000
xieno 2500
total 95000
"To err is human;to forgive ,divine"
Re: Habs 2010 - Questions 23 and 25
Just like my DS! Didn't read the question properly . Thanks for pointing it out magi22.
Re: Habs 2010 - Questions 23 and 25
Yesearleymom wrote:I think the answer is five. Because they are asking the "least number", assuming 5 left-handed boys can wear glasses, the least combination of right-handed boys wearing glasses could be 10 - 5 = 5. Am I correct?25. Of the 26 boys in a class, 21 are right-handed. If 10 of the boys in the
class wear glasses, what is the least number of boys in the class who
are both right-handed and wear glasses?
Re: Habs 2010 - Questions 23 and 25
Thanks Okanagan
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Re: Habs 2010 - Questions 23 and 25
you may be right but I am not sure.
Boys could be ambidextrous
So prob being right handed and wearing glasses = 21/26 *10/26
number = 26 * 21/26*10/26= 8 (rounded)
Boys could be ambidextrous
So prob being right handed and wearing glasses = 21/26 *10/26
number = 26 * 21/26*10/26= 8 (rounded)
"To err is human;to forgive ,divine"
Re: Habs 2010 - Questions 23 and 25
The question isn't the probability though - it's "what is the least number of boys in the class who are both right-handed and wear glasses?".magi22 wrote:you may be right but I am not sure.
Boys could be ambidextrous
So prob being right handed and wearing glasses = 21/26 *10/26
number = 26 * 21/26*10/26= 8 (rounded)
Re: Habs 2010 - Questions 23 and 25
That's the expected number, not the minimum number. You can't use probability like that.magi22 wrote:you may be right but I am not sure.
Boys could be ambidextrous
So prob being right handed and wearing glasses = 21/26 *10/26
number = 26 * 21/26*10/26= 8 (rounded)
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