Habs 2010 - Questions 23 and 25

[earleymom]

Hi,

Can someone help me with the following questions from Habs 2010 paper:

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 _________

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?

Next Question 25:

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?

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?

[IronMikeTyson]

Its 2 xenox in a row to get the bonus

So
xenox 2500
Zoid 50000
Yondo 40000
xieno 2500

total 95000

[earleymom]

Just like my DS! Didn't read the question properly . Thanks for pointing it out magi22.

[Okanagan]

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?

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?

Yes

[earleymom]

Thanks Okanagan

[IronMikeTyson]

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)

[Okanagan]

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)

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?".

[daveg]

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)

That's the expected number, not the minimum number. You can't use probability like that.

[IronMikeTyson]

Thanks !