Q&R questions

[Wicked]

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Can someone explain these questions to me please?

1. A fly sits at the edge of a 12 inch-diameter phonograph record that is playing at 33 1/3 (100/3) revolutions per minute. apporximately how fast, in feet per minute, is the fly moving?

2. one pump can fill a pool in 10 min. another pump can fill the pool in 15 min. how many min. are required to fill the pool if both pumps are operating at the same time?


these are questions from Barron's DAT.
thank you!

 

[littlemissy]

starting with the second one

1/10 + 1/15= 1/x

solve for x

first one

100/3 revs/1 min x 12pi/1 rev x 1 ft/ 12 in

simply and multiply that gives you ft/min

 

[Wicked]

starting with the second one

1/10 + 1/15= 1/x

solve for x

first one

100/3 revs/1 min x 12pi/1 rev x 1 ft/ 12 in

simply and multiply that gives you ft/min

wow..cool
thank you..i feel stupid..😳
i really appreciate it!

 

[Wicked]

one more question..

Kate has 6 different books. assuming the order of selection doesn't matter, how many ways can Kate select, from these 6 books, 2 different books?

im sure this is an easy, basic problem..but i always had trouble understanding probability problems..
thanks for your help!

 

[CyborgNinjaX]

one more question..

Kate has 6 different books. assuming the order of selection doesn't matter, how many ways can Kate select, from these 6 books, 2 different books?

im sure this is an easy, basic problem..but i always had trouble understanding probability problems..
thanks for your help!

I'm guessing its 6*5 because first selection she has 6 books to select, second selection she has 5 books to select.

 

[Wicked]

I'm guessing its 6*5 because first selection she has 6 books to select, second selection she has 5 books to select.

that is what i thought at first, but it is wrong.
the equation is

nCr=n!/[r!(n-r)!]

n being the larger # r being the smaller #.

so it should be 6!/[2! 4!] therefore, 15.

someone explained this to me well..
hope this helps!🙂

 

[domonas]

Can someone explain these questions to me please?

2. one pump can fill a pool in 10 min. another pump can fill the pool in 15 min. how many min. are required to fill the pool if both pumps are operating at the same time?


these are questions from Barron's DAT.
thank you!

1 fills in 10 min
1 fills in 15 min

setup

1/10 + 1/15 = 1/x

15/150 + 10/150 = 1/x

25/150 = 1/x

150 = 25x

x=6

it'll take both pumps together 6 min to fill the pool.

edit: oops, looks like someone already solved it.

 

[scottyhoop]

1 fills in 10 min
1 fills in 15 min

setup

1/10 + 1/15 = 1/x

15/150 + 10/150 = 1/x

25/150 = 1/x

150 = 25x

x=6

it'll take both pumps together 6 min to fill the pool.

edit: oops, looks like someone already solved it.

I like the way Destroyer does the first question using:

(Time1)(Time2)/Time1+Time2=Time to fill
(10X15)/(10+15)=150/25=6 mins
Just another option

 

[littlemissy]

that is what i thought at first, but it is wrong.
the equation is

nCr=n!/[r!(n-r)!]

n being the larger # r being the smaller #.

so it should be 6!/[2! 4!] therefore, 15.

someone explained this to me well..
hope this helps!🙂

yeah it is a permutation its taking n things r different ways....i know that is vague pick up probablity demistified it helps. i know how and when to use it just can't explain it well

 

[fsapien]

I think that's actually a combination.
The permutation equation should be: n!/[(n-r)!]

 

[scottyhoop]

that is what i thought at first, but it is wrong.
the equation is

nCr=n!/[r!(n-r)!]

n being the larger # r being the smaller #.

so it should be 6!/[2! 4!] therefore, 15.

someone explained this to me well..
hope this helps!🙂

Why would you use a combo equation for this?? I'm not good at probability at all! I thought since order doesn't matter you would use n!/(n-r)! Any help would be great!
Scott

 

[Streetwolf]

yeah it is a permutation its taking n things r different ways....i know that is vague pick up probablity demistified it helps. i know how and when to use it just can't explain it well

You need the 2! because there are 2! ways of taking 2 objects and ordering them.

If you had 5 people and wanted to line them up, you have 5 for the first person, 4 for the second, etc = 5*4*3*2*1 = 5! ways to order them.

So going back to the 2!...

We can list the objects you choose. Let's say you have A B C D E F.

AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF
BA, CA, DA, EA, FA, CB, DB, EB, FB, DC, EC, FC, ED, FD, FE

That's 30 total. But see how only 15 of them are unique. That's because if you have 2 objects, you can order them either AB or BA. Thus, half of the selections are unique.

Similarly with 3 objects chosen from a group, only 1/6 of the selections are unique. This is because you can order 3 objects in 3! or 6 ways - ABC, ACB, BAC, BCA, CAB, and CBA. It doesn't matter which of these ways you select the three objects, because you don't care about the order they were chosen.

 

[littlemissy]

sorry guys you are right trying to think on my feet in a meeting

 

[HeatingHomer]

If the order does matter it's a permutation (nPr [n!/(n-r)!]). If the order does not matter it's a combination (nCr [n!/r!(n-r)!]).