Pals need help on few gre sample questions.

[abhishek1787]

1.Tom, brik and Harry agree to chip in to buy a television. Tom agrees to spend twice as much as Harry, and brik agrees to spend $40 more than Tom. If they had all chipped in the same amount, Harry would have had to put in $60 more than he does. How much does Harry spend on the television?

A) $50

B) $60

C) $70

D) $80

E) $100

 

2.Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least one of the people chosen is a man?

A) 1/6B)1/3 C)2/3 D) 4/5E) 5/6

 

[apo]

1. h is amount that Harry spent, then 2h+h+2h+40 = 3(h+60)

5h+40=3h+180

2h=140

h=70

2. the prob. that no man selected is 6/10*5/9*4/8=120/720

the prob at least one man selected is 1-120/720=600/720=5/6

[abhishek1787]

Thanks apo!!

 

Need some light on the below prob as well.Any shortcut to solve?

 

a merchant sells 3 diff sizes of canned tOMATOES. A large can costs as much as 5 medium cans or 7 small cans. If a customer buys an equal number of small and large cans of tomatoes for th eexact amount of money that would buy 200 medium cans, how many small cans will she buy??

a. 35

b. 45

c. 72

d. 199

e. 208

Edited June 27, 2012 by abhishek1787
correction

[abhishek1787]

little algebra helped me to get the answer.

 

Let L be the cost of large can

M=number of medium sized can.

S=number of small sized can.

 

then 5M=7S=L

 

200 medium sized can costs 5*200=$1000.(where 5 is cost of each medium sized can)

 

deriving the equation which says -"customer buys an equal number of small and large cans of tomatoes for th eexact amount of money that would buy 200 medium cans".

 

5M+7S=1000;solve for M and S by plugging in the given options for S.

 

S=35 is the answer.Let me know if i am correct.

[apo]

a is the number of large cans

b is the number of medium cans

c is the number of small cans

pa is the price of a large can

pb is the price of a medium can

pc is the price of a small can

Then,

 

pa=5pb

pa=7pc

pc=5pb/7

a*pa+a*pc=200*pb

then,

a*5pb+a*5pb/7=200*pb

a*5+a*5/7=200

40*a/7=200

a/7=5

a=35

[abhishek1787]

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

 

(A) 31

(B) 51

© 56

(D) 62

(E) 93

[thepillow]

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

 

(A) 31

(B) 51

© 56

(D) 62

(E) 93

 

Wow, this is a tough question. I got 56. Is that right?

 

This was my method:

 

Just to get them out of the way: 5, 50, 500, and 5000 = 4 possibilities.

 

Then:

 

2 digits: Use the pairs {2,3} and {1,4}. Each has 2 orders = 4 possibilites

 

3 digits: Use the sets:

{2,3,0} = 4 possibilities (don't count 0 as first digit)

{4,1,0} = 4 possibilities (don't count 0 as first digit)

{1,1,3} = 3 possibilities (don't count the two 1s as different)

{2,2,1} = 3 possibilities (don't count the two 2s as different)

= 14 possibilites

4 digits: Use the sets:

Use the sets:

{2,3,0,0} = 6 possibilities (no 0 as first digit and don't count two 0s as different)

{4,1,0,0} = 6 possibilities (no 0 as first digit and don't count two 0s as different)

{1,1,3,0} = 9 possibilities (no 0 as first digit and don't count two 1s as different)

{2,1,1,1} = 4 possibilities (don't count three 1s as different)

{2,2,0,1} = 9 possibilites (no 0 as first digit and don't count two 2s as different)

= 34 possibilites

 

So adding these all up that's 4 + 4 + 14 + 34 = 56 total possibilities

 

Did I forget any?

Edited August 3, 2012 by thepillow

[thepillow]

Actually, I think I just wasted a whole lot of time. After seeing that the answer is 56 it occurred to me that 56 is the same as 8c3, which makes sense:

 

___ ___ ___ ___ ___ ___ ___ ___

 

There are 8 empty slots above and we're looking for a number (4 digits or less) where all the digits together add to 5. If we consider each of the 8 slots to have a value of 1 then all we need to do is choose 3 out of those 8 to cross out and we'll be left with 5. We can do that in 8c3 = 56 ways. Depending on which slots we cross out, the left over slots will be grouped together in various ways that will all be equivalent to 4 digits or less and sum to 5.

 

For example:

 

___ ___ ___ ___ X X X ___ = 41

 

X ___ X ___ X ___ ___ ___ = 113

 

___ ___ X ___ ___ ___ X X = 23

 

___ ___ ___ ___ ___ X X X = 5000

 

etc.

 

 

To me this seems like a much much faster solution but also much harder to conceptualize initially.

[mxplusb]

That's beautiful. I never would have looked at it that way.

 

Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the digits are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56.

 

In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed.

 

And I would have gotten the correct answer very quickly, if I hadn't forgotten about the 0122 combinations. Drat.

[thepillow]

 

Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the digits are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56.

 

In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed.

 

 

Hi mxplusb,

 

You're right, that's much faster. Thanks!!

[drdevis]

good well leaned

[shadoWizard]

That's beautiful. I never would have looked at it that way.

 

Anyway, thepillow, there's a way that's slightly faster than your first approach, I think. You take all the digits combinations -- 0005, 0014, 0023, 0122, 0113, and 1112. For the first and last, the ways to arrange the are 4!/3! (the 3! is because the three 0's or three 1's are interchangeable). For the others, it's 4!/2!. Adds up to 56.

 

In this case it's okay for the numbers to start with a zero because that just represents a less-than-four-digit number, which is allowed.

 

And I would have gotten the correct answer very quickly, if I hadn't forgotten about the 0122 combinations. Drat.

 

mxplus. my perm and comb skill are not too good. So would you mind helping me by elaborating a little more.

 

Since order does matter, so its a permutation.

For first and last, I get it that out of four slots, we can arrange 5 (for 1st) and 2 (for last) on 4 each ways.

For 2nd and 3rd, I understand that 1 and 4 (for 2nd) and 2 and 3 (for 3rd) can be placed in 4P2 ways. (I interpret the problem as "there are 4 rooms and two paintings. in how many ways we can hang the two paintings".

Now this is where I get confused. (0122 and 0113). We have 4 rooms and 3 paintings. The permutation we have is 4P3 = 4.

 

Kindly explain.

[thepillow]

Now this is where I get confused. (0122 and 0113). We have 4 rooms and 3 paintings. The permutation we have is 4P3 = 4.

 

Kindly explain.

 

Hi Fahad, mxplusb might have a different explanation, but here is one way of looking at it:

 

Using 0122 as an example, if we ignore the fact that the two 2s are identical then there are 4! = 24 ways to arrange the numbers. Of course we now need to eliminate all of the orders in which we have counted the 2s as different. To see why we can use 4!/2!, I think it's helpful to look at it visually by changing the color of the identical 2s to make them look different:

 

One of our possible numbers is 2210, but we have actually counted it as 2 orders: 2210 2210.

 

Similarly:

 

•1022 is counted as 1022 and 1022

•2102 is counted as 2102 and 2102

etcetera...

In fact, for every one of the possible orders there are 2! ways to arrange the two 2s. So we divide by 2! to make sure we don't count these extra orders. In other words, each order that we want has an identical twin that we don't want.

 

Does that help?

[shadoWizard]

thank you thepillow. your explanations always help.... without a doubt. And mxplus, this is by far the best (fastest and easiest) solution for this type of question that i've ever come across.

[thepillow]

thank you thepillow. your explanations always help.... without a doubt. And mxplus, this is by far the best (fastest and easiest) solution for this type of question that i've ever come across.

 

Hi Fahad, I'm glad I could help.

 

I realized, though, that I forgot to specifically answer your question about why the "4 rooms and 3 paintings" strategy doesn't quite work here. It's pretty tricky, but in this case we don't actually have exactly 3 paintings. I think it's easiest to see this visually:

 

If we label the 4 rooms A, B, C and D, we can show them like this:

 

a___ b___ c___ d___

 

 

The 3 paintings would be:

 

[ 0 ] [ 1 ] and [ 22 ]

 

If we now think about the ways to choose which of the 4 rooms will have which painting then, you're right, we can do 4p3 as our permutation. However, in this situation the two 2s must always be next to each other because they are part of the same painting. It would be impossible to make the number 2102, for example.

 

So, we don't want to count 2102 and 2102 as different numbers, but we definitely want to be able to put the 2s next to each or apart from each other.

 

I hope that helps!

[mxplusb]

thepillow's explanation looks good to me. Thank you for that.

 

I find that probability and combination/permutation problems are the most confusing problems on this sort of test. There are many different rules that can be applied, and it is often hard to figure out which one fits best.

 

Personally I hate having to rely on an answer sheet to tell if I got the right or wrong answer -- I'd much rather be able to double-check myself. Of course on standardized tests there isn't always time. But when you're studying there is -- either you can allow yourself all the time you need to get the problem right (remembering that practicing doing a problem wrong is worse than not practicing it at all) or you can go through a practice test once while timing yourself, and then go back to the problems you're not sure of and take your time on the second untimed round, before looking up the answers.

 

Permutation and combination formulas are shortcuts. Many problems you can test out long-hand, by actually writing down all the possibilities, and on the ones you can't you can often make a smaller version of the same problem and test that one. This way you don't have to take the strategies on trust. You can use them because you know they work, you've tested it yourself.

 

I was thinking about this, because I've noticed the way I do test problems is often different than the way a textbook would say to do them. Textbooks tend to focus on memorizing formulas and strategies, then figuring out which one to use. I tend to not lean on formulas unless I'm very sure about them, either because I've used them so much that they're second nature or because I know how to derive them. I also rely on estimation and common sense a lot -- not relevant for this problem, but some problems will have one or a few answers that are clearly impossible. So I can cross those off without even checking.

[The Rain]

Can anyone tell me from which book or test these problems are taken ? I wanted to solve more like this...

[shadoWizard]

Can anyone tell me from which book or test these problems are taken ? I wanted to solve more like this...

I guess I saw this question in the Math Review of GRE Official Guide 11th Ed.

[Ibnea Sina]

lets try with plug in...

 

if the Large can is cost $35

the medium can is cost 35/5=$7

and small can is cost 35/7= $5

 

Now the cost of 200 medium can is 200*7=1400 which gives us the exact cost of buying same amount of Large and small sized cans.

lets the number is x.

 

x*35+x*5= 1400

=> x= 1400/40

=> x=35

 

its 35 cans.

[shadoWizard]

lets try with plug in...

 

if the Large can is cost $35

the medium can is cost 35/5=$7

and small can is cost 35/7= $5

 

Now the cost of 200 medium can is 200*7=1400 which gives us the exact cost of buying same amount of Large and small sized cans.

lets the number is x.

 

x*35+x*5= 1400

=> x= 1400/40

=> x=35

 

its 35 cans.

thats good