A number of oranges are to be distributed evenly among a

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!

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.

Answer: B.

Hope it's clear.

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.

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

I am not clear with the sufficiency of statement B.It says that no of baskets doubles such that number of oranges per basket is <1.

in this case 2b>20, which means that b can be any value >10 , b could be 18,19,20, and so on as 20 divided by that number of baskets will continue to be <1. For e.g 20/36=0.555 ,where b=18.
When we have various values to get less than 1,then how is it sufficient

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.

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