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OK, so the test is in 5 days, and I have some PS questions.

maybe someone can help me... (I have the answers but I don't get them... so there is either a mistake in the answers or in my head...)

1. John has 7 bananas and 3 kiwis. He needs to divide the 10 in two parcels, so there is an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel.

a) 21

b) 35

c)60

d) 105

e) 120

2. From the word ZEBRA how many different letter combinations can be arranged by using the foregoing letters?

a) 20

b) 60

c)100

d) 120

e) 150

3. Five coins are tossed one after the other. what is the probability that the first three are heads?

a) 1/16

b) 5/16

c) 1/2

d) 2/3

e) 11/16

4. a monkey is at the bottom of a 30 foot well. each day he jumps up three and slips back two. at that rate, in how many days will the monkey reach the top of the well?

a) 27

b) 28

c) 29

d) 30

e) 31

5. a cook went to a merket to buy some eggs and paid 12$ for them. but since the eggs were quite small he talked the seller into adding 2 more eggs, free of charge. As the 2 eggs were added, the price per dozen went down by a dollar. How many eggs did the cook bring home from the market?

a) 12

b)16

c)15

d)18

e) 8

6. for the equation |||x|-3|-x|=3 find all the solutions

Question 1 : I dont get any of the answers you have mentioned.

7 B and 3 Ks can be arranged in 10!/ ( 7! * 3! ) = 120 ways.
Out of those 120 ways if one gets 5Bs or 3Ks then that combination is invalid.
There two caes in which invalid combinations occur.
Case 1:
Ist parcel gets 5B and second gets rest
So No of invalid ways = 5Bs in first So second parcel can be packed in
(10-5)! / ( (7-5)! * 3! ) = 10
Case 2:
2nd parcel gets 3K & 2B and second gets the rest
So No of invalid ways
(10-5)! / ( (7-5)! * 3! ) = 10

Total ways to distributes = 120-20 = 100

---------------------------------------------------------------------
Question 2
answer = 5! = 120
---------------------------------------------------------------------
Question 3
Desired events are
HHH HH = 1/32
HHH HT = 1/32
HHH TH = 1/32
HHH TT = 1/32
Total P = 1/8 which is not the answer given.
--------------------------------------------------------------------
Question 4
Mokney is effectively going up by (3-2) foot per day
so to reach 30 feet he will take 30 days.
--------------------------------------------------------------------
Question 5
n * p/12 = 12 ( n = no of eggs, p - cost of egg per dogen )
(n+2)(p-1)/12 = 12
so p = (n+2)/2

We need to only find out how many ways we can put the items in ONE parcel. The rest will be put in the second parcel:

Number of ways to put the fruits in one parcel = Number of ways so that 1 kiwis and 4 bananas + Numbers of ways so that 2 kiwis and 3 bananas

= 3C1 * 7C4 + 3C2 * 7C4 = 195

(We can not put 3 kiwis and 2 bananas in one parcel because then the other parcel will not have any kiwis)

Question 2 : Agreed with Anand

Question 3 : Agreed with Anand

Question 4: answer should be 28.

The monkey climes 1 feet per day. At the end of 27 days the monkey would have climbed 27 feet. On the 28th day, the monley would have climed 30 feet before slipping back 2 feet. Monkey does not need to slip back once he reaches the top.

Last edited by gmatblast on 04 Feb 2004, 13:48, edited 1 time in total.

2. D
3. B. (5C2/32). Or you can use a probabality formula
4. D (The reason being the monkey climbs 1 feet every day . 3-2 so
it will take 30 days to climb 30 feet wall. How can the answer be
27. Please correct me if I am wrong)

5. D . It took me a long while to figure it out. Could someone please explain a faster approach

I think there is typo in the above equation. It should be 3C1 * 7C4 + 3C2 * 7C3 = 195

Don't we need to multiply this with 2, because one can have 2K+3B in 1st basket and 1K+4B in second basket.

kpadma,

I agree that there is a typo in the equation.

As for multiplying it by 2, I do not think we need to do that. Imagin, if one parcel has 2K + 3B then only way the other parcel can have fruits is 1K + 4B.

Here, (parcel 1: 2K + 3B and parcel 2: 1K + 4B) together constitute ONE way of distribution. . It should not be considered as TWO ways of distribution. So we do not need to multiply by 2.

Case 1: x E [0, +inf] ||x-3|-x| =3
Case 1a. x E [3, +inf] |x-3 -x| = 3, valid for all x in the aforesaid interval.
Case 1b. x E [0,3] x = 0

Case 2: x E (-inf, 0] ||x+3|-x| = 3
Case 2a: x E [-3,0] |x+3 -x| =3, valid for all x in the interval mentioned hitherto.
Case 2b: x E (-inf, -3), x = 0, no solution

Here, (parcel 1: 2K + 3B and parcel 2: 1K + 4B) together constitute ONE way of distribution. . It should not be considered as TWO ways of distribution. So we do not need to multiply by 2.

Yes. You are correct !
It is a two level combination problem. Good Question.

Does this statement: "He needs to divide the 10 in two parcels, so there is an equal total number of fruit in either parcel" mean that each parcel is to have exactly five fruits in it? Is banana number one any different than banana number two?

ETS never uses wording that's this confusing.

The way I read it, the answer is 2-- Box a will have 1 kiwi and 4 bananas, box 2 will have 2 kiwi and 3 bananas. And you can trade the boxes.

So I'm sure that approach is wrong.

So let's try a different way.

Box 1 will contain either
Kiwi A
Kiwi B
Kiwi C
Kiwi A and B
Kiwi A and C
Kiwi B and C

Taking those,
Kiwi A can have 7 c 4 bananas--35
Kiwi B can have 7 c 4 bananas--35
Kiwi C can have 7 c 4 bananas--35
Kiwi A and B can have can have 7 c 3 bananas--35
Kiwi A and C can have can have 7 c 3 bananas--35
Kiwi B and C can have can have 7 c 3 bananas--35

Giving an answer of 210. Now we may have to divide this answer by two, because 105 of these are mirror images of the other 105. The question is unclear in this regard, as well.

For Q1 I assumed all B and all K are same and each box should have 5 fruits and both kinds of fruits must be present in each box.

I got similar answer and I am posting it to support your method, using an alternative method.

No of ways in which one box can be filled by 5 fruits = 10C5 = 252
Out of these invalid combinations are 5 Bs, 3K and 2 Bs
One can chose 5Bs in 7C5 ways = 21
One can choose 3Ks and 2Bs in 3C3 * 7C2 = 21 ways
Total ways in which fruits can be distributed = 252-(21+21) = 210

As stoolfi said half of this set is mirror image of the other set
so I get 105.

Re: some ps questions [#permalink]
05 Feb 2004, 02:14

dalina wrote:

1. John has 7 bananas and 3 kiwis. He needs to divide the 10 in two parcels, so there is an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel.

a) 21

b) 35

c)60

d) 105

e) 120

This question is very poorly worded (in fact, there is no question). You need to specify whether the fruits and boxes are distinguishable for one thing.

Let's assume the question is: How many ways can ghe fruits be packaged? An argument can be made that the answer is ONE.

here goes. Let's assume that the both fruits and boxes are indistiguisheable. Since we must have at least one kiwi in each box, one must have 1 and the other 2. Hence the boxes have 4 and 3 bananas respectively. If the boxes are interchangable, then this is the only way we can pack them!

As for the answer choices, the way the question that makes sense is if the fruits are distinguishable but the boxes are not. There are 3 ways to put one kiwi in box 1 and 7C4 or 35 ways to choose 4 of 7 bananna. Hence there are 105 ways to put 1 kiwi in box 1.

There are 3 ways to put two kiwis in box 1 and 7C3 or 35 ways to choose 3 of 7 banannas. Hence there are 105 ways to put 2 kiwis in box 2. If the boxes are indistiguishable, each way to put 1 kiwi in box one corresponds to a way to put 2 kiwis in box 2 so there are the same. hence, there are 105 ways to package the fruits. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

3. Five coins are tossed one after the other. what is the probability that the first three are heads?

a) 1/16

b) 5/16

c) 1/2

d) 2/3

e) 11/16

This only makes sense if the question is:
Five coins are tossed one after the other. what is the probability that exactly three are heads? _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

2. From the word ZEBRA how many different letter combinations can be arranged by using the foregoing letters?

a) 20

b) 60

c)100

d) 120

e) 150

This is another poorly worded and hence ambiguous question. Better is:

The five letters in the word ZEBRA can be arranged into how many different 5-letter words? _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993