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A number of oranges are to be distributed evenly among a [#permalink]
28 Aug 2010, 09:05

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

46% (02:21) correct
54% (01:33) wrong based on 276 sessions

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Re: Baskets and oranges [#permalink]
28 Aug 2010, 09:24

5

This post received KUDOS

Expert's post

jananijayakumar wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Given: \(\frac{20}{b}=integer=x\), basically we are told that # of baskets, \(b\), is a factor of 20, so \(b\) could be: 1, 2, 4, 5, 10, or 20.

(1) \(\frac{20}{\frac{b}{2}}=2x\) --> \(\frac{20}{b}=x\), the same info as in stem (only one possible values of \(b\) is excluded: 1, as if \(b=1\) it can not be halved). Not sufficient.

(2) \(2b>20\) --> only one value of \(b\) satisfies this: \(b=20\). Sufficient.

Re: Baskets and oranges [#permalink]
28 Aug 2010, 21:04

There could be 20 , 10, 5, 4, 2, 1 basket(s). According to the first point it could either be 20 baskets or 10 baskets initially. So it is not sufficient. Second point clearly states "If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket." So it has to be 20 and hence B

Re: Baskets and oranges [#permalink]
30 Aug 2010, 13:55

I dont really understand the problem, and the answer B. If for example, 12 basket are there originally, 20 oranges are placed in those 12 baskets (assuming initially they are not evenly distributed and task at hand is to distribute them evenly) , and doubling the basket to 24 would still satisfy statement 2 but at the same time we can't determine exact number of baskets.

I am assuming that they are NOT already evenly distributed because question or any statement doesnt say that explicitly.

Not sure what am I missing! _________________

"Learning never exhausts the mind." --Leonardo da Vinci

Re: Baskets and oranges [#permalink]
31 Aug 2010, 03:05

The question says explicitly that number of basket to be decided in such a ways that there are always even distribution. Check "distributed evenly " in the qustion

Re: Baskets and oranges [#permalink]
13 Sep 2010, 06:27

Bunuel wrote:

jananijayakumar wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Given: \(\frac{20}{b}=integer=x\), basically we are told that # of baskets, \(b\), is a factor of 20, so \(b\) could be: 1, 2, 4, 5, 10, or 20.

(1) \(\frac{20}{\frac{b}{2}}=2x\) --> \(\frac{20}{b}=x\), the same info as in stem (only one possible values of \(b\) is excluded: 1, as if \(b=1\) it can not be halved). Not sufficient.

(2) \(2b>20\) --> only one value of \(b\) satisfies this: \(b=20\). Sufficient.

Answer: B.

I am unable to understand this problem and solution!!!

Re: Baskets and oranges [#permalink]
13 Sep 2010, 08:38

2

This post received KUDOS

Expert's post

utin wrote:

Bunuel wrote:

jananijayakumar wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Given: \(\frac{20}{b}=integer=x\), basically we are told that # of baskets, \(b\), is a factor of 20, so \(b\) could be: 1, 2, 4, 5, 10, or 20.

(1) \(\frac{20}{\frac{b}{2}}=2x\) --> \(\frac{20}{b}=x\), the same info as in stem (only one possible values of \(b\) is excluded: 1, as if \(b=1\) it can not be halved). Not sufficient.

(2) \(2b>20\) --> only one value of \(b\) satisfies this: \(b=20\). Sufficient.

Answer: B.

I am unable to understand this problem and solution!!!

I agree with you that the wording is quite ambiguous.

Intended meaning of the question is as follows: 1. there are 20 oranges \(b\) baskets; 2. These 20 oranges COULD be evenly (equally) distributed among these \(b\) baskets, so the number of baskets is a factor of 20: 1, 2, 4, 5, 10, or 20;

Question: "what is the number of oranges per basket" or \(\frac{20}{b}=?\) So we should find the # of baskets.

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket --> just tells us that # of baskets is even as it can be halved, everything else we knew before: obviously if you halve the number of baskets then each basket will contain twice as many oranges as before. So number of baskets is: 2, 4, 10, or 20. Not sufficient.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket --> twice the number of baskets is more than 20 --> only \(b=20\) satisfies this, so we have 20 baskets. Sufficient.

A number of oranges are to be distributed evenly among a [#permalink]
04 Aug 2013, 11:19

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

My answer was D.

Statement 1:

No of baskets halved = N /2

twice as many oranges in every remaining basket:

I interpreted it as

N / 2 2 (N / 20) ------- = 20

40/N = N/10 N^2 = 400 N = 20

So sufficient.

Statement 2: Factors of 20: 1,2,4,5,10,20.

So out of the above factors, only doubling 20 would not be able to accommodate 20 oranges. so Sufficient.

My answer was D. But MGMAT answer is different. Could someone explain whats wrong with this?

Last edited by Zarrolou on 04 Aug 2013, 11:22, edited 1 time in total.

Re: A number of oranges are to be distributed evenly among a [#permalink]
04 Aug 2013, 11:28

sudharsansuski wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

My answer was D.

Statement 1:

No of baskets halved = N /2

twice as many oranges in every remaining basket:

I interpreted it as

N / 2 2 (N / 20) ------- = 20

40/N = N/10 N^2 = 400 N = 20

So sufficient.

Statement 2: Factors of 20: 1,2,4,5,10,20.

So out of the above factors, only doubling 20 would not be able to accommodate 20 oranges. so Sufficient.

My answer was D. But MGMAT answer is different. Could someone explain whats wrong with this?

let number of basket = x number of oranges in each basket = y therefore x*y = 20

now statement 1: If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket no. of basket = x/2 then no. of oranges in each basket = 2y therefore x/2)*2y = 20 again same comes out x*y = 20 hence in sufficient.

statement 2: If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket no. of basket = 2x then 2x>20 x>10 now for evenly distributing 20 oranges with number of baskets greater than 10 number of baskets must be = 20 hence suffiecient

hope it helps _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: A number of oranges are to be distributed evenly among a [#permalink]
01 Sep 2013, 12:05

To me it seems a slightly ambiguous question. I agree with Bunuel, no of baskets could be either of 1, 2, 4, 5, 10, 20 and thus no of oranges per basket could be 20, 10, 5, 4, 2, 1 respectively..

(1) there are more than one cases which satisfy the given condition, viz., no of baskets could be 2, 4, 10 or 20 for this statement.. so not sufficient

(2) if no of baskets is doubled, it isnt possible to place at least one orange per basket.. according to me it can happen for TWO cases: one where the no of baskets is '20' and two, where no of baskets is '4'.. because for this second case too, if no of baskets is doubled (made 8), How can we place one orange per basket so that we keep no of oranges per basket constant (remember we have to keep the question in mind which says oranges are to be distributed Evenly among the baskets.. which essentially means same no of oranges per basket)... so i think it cannot be 'b'.. this statement is not sufficient

IF we now combine the two statements, we still have both these cases satisfying the given conditions: no of baskets as '4' and no of baskets as '20'.. i think answer should be 'E'..

Re: A number of oranges are to be distributed evenly among a [#permalink]
02 Sep 2013, 01:08

Expert's post

amanvermagmat wrote:

To me it seems a slightly ambiguous question. I agree with Bunuel, no of baskets could be either of 1, 2, 4, 5, 10, 20 and thus no of oranges per basket could be 20, 10, 5, 4, 2, 1 respectively..

(1) there are more than one cases which satisfy the given condition, viz., no of baskets could be 2, 4, 10 or 20 for this statement.. so not sufficient

(2) if no of baskets is doubled, it isnt possible to place at least one orange per basket.. according to me it can happen for TWO cases: one where the no of baskets is '20' and two, where no of baskets is '4'.. because for this second case too, if no of baskets is doubled (made 8), How can we place one orange per basket so that we keep no of oranges per basket constant (remember we have to keep the question in mind which says oranges are to be distributed Evenly among the baskets.. which essentially means same no of oranges per basket)... so i think it cannot be 'b'.. this statement is not sufficient

IF we now combine the two statements, we still have both these cases satisfying the given conditions: no of baskets as '4' and no of baskets as '20'.. i think answer should be 'E'..

I see your point but still you are not correct. The point is that the condition about even distribution applies only to the original # of baskets. _________________

Re: Baskets and oranges [#permalink]
04 Nov 2013, 04:39

1

This post received KUDOS

Bunuel wrote:

I agree with you that the wording is quite ambiguous.

Intended meaning of the question is as follows: 1. there are 20 oranges \(b\) baskets; 2. These 20 oranges COULD be evenly (equally) distributed among these \(b\) baskets, so the number of baskets is a factor of 20: 1, 2, 4, 5, 10, or 20;

Question: "what is the number of oranges per basket" or \(\frac{20}{b}=?\) So we should find the # of baskets.

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket --> just tells us that # of baskets is even as it can be halved, everything else we knew before: obviously if you halve the number of baskets then each basket will contain twice as many oranges as before. So number of baskets is: 2, 4, 5, 10, or 20. Not sufficient.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket --> twice the number of baskets is more than 20 --> only \(b=20\) satisfies this, so we have 20 baskets. Sufficient.

Answer: B.

Hope it's clear.

I think according to statement 1, the number of baskets CANNOT be 5 as we cannot have a decimal number of baskets when 5 is divided by 2. Correct me if I'm wrong.

Re: Baskets and oranges [#permalink]
04 Nov 2013, 04:45

Expert's post

emailmkarthik wrote:

Bunuel wrote:

I agree with you that the wording is quite ambiguous.

Intended meaning of the question is as follows: 1. there are 20 oranges \(b\) baskets; 2. These 20 oranges COULD be evenly (equally) distributed among these \(b\) baskets, so the number of baskets is a factor of 20: 1, 2, 4, 5, 10, or 20;

Question: "what is the number of oranges per basket" or \(\frac{20}{b}=?\) So we should find the # of baskets.

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket --> just tells us that # of baskets is even as it can be halved, everything else we knew before: obviously if you halve the number of baskets then each basket will contain twice as many oranges as before. So number of baskets is: 2, 4, 5, 10, or 20. Not sufficient.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket --> twice the number of baskets is more than 20 --> only \(b=20\) satisfies this, so we have 20 baskets. Sufficient.

Answer: B.

Hope it's clear.

I think according to statement 1, the number of baskets CANNOT be 5 as we cannot have a decimal number of baskets when 5 is divided by 2. Correct me if I'm wrong.

Re: Baskets and oranges [#permalink]
08 Nov 2013, 03:18

Bunuel wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Given: \(\frac{20}{b}=integer=x\), basically we are told that # of baskets, \(b\), is a factor of 20, so \(b\) could be: 1, 2, 4, 5, 10, or 20.

(1) \(\frac{20}{\frac{b}{2}}=2x\) --> \(\frac{20}{b}=x\), the same info as in stem (only one possible values of \(b\) is excluded: 1, as if \(b=1\) it can not be halved). Not sufficient.

(2) \(2b>20\) --> only one value of \(b\) satisfies this: \(b=20\). Sufficient.

Re: A number of oranges are to be distributed evenly among a [#permalink]
13 Jan 2014, 05:27

jananijayakumar wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Statement 1

Many possibilities, basically all combinations of factors of 20, namely 4-5, 20-1, 2-10 etc..

So 2 cases for instance

If baskets were 10 and then halved to 5 then oranges per basket increase from 2 to 4

If baskets were 20 and then halved to 10 then oranges per basket increase from 1 to 2

Insuff

Statement 2

One has to be careful here, it says that if double it won't be possible to place AT LEAST 1 orange. This doesn't mean that we won't be able to divide evenly, this means that the number of oranges will be less than 1 if we double. So only possible answer is 20 since 20/40=.5 oranges per basket, not possible

Re: A number of oranges are to be distributed evenly among a [#permalink]
06 Sep 2014, 04:33

jananijayakumar wrote:

A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?

(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.

(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

I always used to afraid of word problem but i got this .

info from question: A number of oranges are to be distributed evenly among a number of baskets . Each basket will contain at least one orange.

so total 20 oranges - when it is evenly distributed then basket count would be , factors of 20 i.e. 1, 2,4,5,10,20.

1. if there are 20 basket then each have 1 orange & if baskets are reduced to half then 10 baskets then 2 oranges in each.. but there are other possibility as well if there were 4 basket at the begin & its reduced to 2 baskets then 10 oranges in each.. so we could get more than 2 values for count of oranges in each basket. Not suff.

2. only one possible condition would suffice this statement . if there are 20 basket & each has one then when basket count is doubled then some baskets will not get oranges. hence there should be one orange per basket. Suff.

Answer B

gmatclubot

Re: A number of oranges are to be distributed evenly among a
[#permalink]
06 Sep 2014, 04:33

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