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# M27-16

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:27
5
22
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Difficulty:

95% (hard)

Question Stats:

36% (01:52) correct 64% (01:54) wrong based on 205 sessions

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A certain fruit stand sold a total of 76 oranges to 19 customers. How many of them bought only one orange?

(1) None of the customers bought more than 4 oranges.

(2) The difference between the number of oranges bought by any two customers is even.

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16 Sep 2014, 01:27
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4
Official Solution:

(1) None of the customers bought more than 4 oranges. This basically means that all customers bought exactly 4 oranges ($$\frac{76}{19}=4$$), because if even one customer bought less than 4, the sum would be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

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01 Dec 2014, 19:08
Bunuel wrote:
Official Solution:

(1) None of the customers bought more than 4 oranges. This basically means that all customers bought exactly 4 oranges ($$\frac{76}{19}=4$$), because if even one customer bought less than 4, the sum would be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

Hi,

i had a question regarding the 1st statement.
" None of the customers bought more than 4 oranges."
why would this mean everyone bought EXACTLY 4 oranges. What if someone buys 1,2 or 3 oranges (coz it is still less than 4)
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02 Dec 2014, 00:44
arunpkumar wrote:
Bunuel wrote:
Official Solution:

(1) None of the customers bought more than 4 oranges. This basically means that all customers bought exactly 4 oranges ($$\frac{76}{19}=4$$), because if even one customer bought less than 4, the sum would be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

Hi,

i had a question regarding the 1st statement.
" None of the customers bought more than 4 oranges."
why would this mean everyone bought EXACTLY 4 oranges. What if someone buys 1,2 or 3 oranges (coz it is still less than 4)

There are 19 customers who bought total of 76 oranges and none of the customers bought more than 4 oranges. Could any of them bought less than 4 oranges?
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17 Nov 2015, 16:56
The explanation for (2) is great Bunnel. I was frustrated with trying pairs of (1,5), (1,7)....
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25 Nov 2015, 04:13
Statement two was a hard one. Great explanation!
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25 Dec 2015, 23:01
1
I thought that 0 is an even number? If it's so then in case 2, two or more people can buy the same odd amount of fruit ( 1, 3, 3 etc or 3, 5, 5 etc.) as long as the sum is 76. Therefore, one may or may not buy exactly 1 orange
--> option 2 is insufficient

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26 Dec 2015, 00:04
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notMD wrote:
I thought that 0 is an even number? If it's so then in case 2, two or more people can buy the same odd amount of fruit ( 1, 3, 3 etc or 3, 5, 5 etc.) as long as the sum is 76. Therefore, one may or may not buy exactly 1 orange
--> option 2 is insufficient

Hi,
You rae correct that 0 is an even number..
and you are also correct soeonecan buy 1,1,3 or 3,5,5...
but you have to look at two points
1)the total number of oranges..it is an even number
2) since the difference is even.. either all numbers are even or all numbers will be odd,,..
why? say you have someone with 4 and other with 3.. the difference will be odd...

having seen these two points
we have to ask ourselves " Is it possible to have odd number of people having odd number of oranges and still having total number of oranges as even?"
The answer is NO... Because Sum of odd quantity of numbers having odd values will always be odd
So any odd value gets discarded and along with it goes out the value of '1'..
Thus suff
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30 Jul 2016, 05:28
I think this is a high-quality question and I agree with explanation.
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28 Nov 2016, 20:36
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please explain how S2 is sufficient on its own. Did not understand from official solution
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28 Nov 2016, 22:22
pratyushk1 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please explain how S2 is sufficient on its own. Did not understand from official solution

Can you please what part of the following is unclear:

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

Thank you.
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08 Jun 2017, 06:14
I think this is a high-quality question and I agree with explanation. Great Question +100 Kudos
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08 Jun 2017, 06:42
Hi Bunuel I got the answer alright... but what level question would this be... thanks

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24 Dec 2017, 06:24
I think this is a high-quality question and I agree with explanation.
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08 May 2018, 03:38
I think this is a high-quality question and I agree with explanation.
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17 May 2018, 00:09
I think this is a high-quality question and I agree with explanation. Statement 2
Best ODD-EVEN Concept so far
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03 Aug 2018, 04:02
1
Bunuel wrote:
A certain fruit stand sold a total of 76 oranges to 19 customers. How many of them bought only one orange?

(1) None of the customers bought more than 4 oranges.

(2) The difference between the number of oranges bought by any two customers is even.

Statement 1: None bought < 4- Clearly sufficient since then everyone bought 4 Oranges
Statement 2: Tricky and I got wrong at first However, given a thought
Everyone bought something so there is no customer who has 0 orange
Moving ahead, if 18 customers each bought 1 and 19th customer bought (76-18)= 58 now if I take difference of any 2 numbers from it such as 1-1=0 or 58-1 Not valid
Giving 2 Oranges to each of customers , hence 18 customers have 2 and 18th one has 76-36=40 oranges, now difference of any number in set is even . It means none of them had 1 orange- Sufficient

Ans: D
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23 Mar 2019, 02:59
I think this is a high-quality question and I agree with explanation.
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15 Apr 2019, 01:16
Bunuel wrote:
Official Solution:

(1) None of the customers bought more than 4 oranges. This basically means that all customers bought exactly 4 oranges ($$\frac{76}{19}=4$$), because if even one customer bought less than 4, the sum would be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

Hi Bunuel

Can you please give an example for the following portion:-
But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange.

Thanks
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15 Apr 2019, 01:18
JIAA wrote:
Bunuel wrote:
Official Solution:

(1) None of the customers bought more than 4 oranges. This basically means that all customers bought exactly 4 oranges ($$\frac{76}{19}=4$$), because if even one customer bought less than 4, the sum would be less than 76. Hence, no one bought only one orange. Sufficient.

(2) The difference between the number of oranges bought by any two customers is even. In order for the difference between ANY number of oranges bought to be even, either all customers must have bought an odd number of oranges or all customers must have bought an even number of oranges. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange. Sufficient.

Hi Bunuel

Can you please give an example for the following portion:-
But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one (i.e. odd number) orange.

Thanks

Well the sum of odd number of odd numbers is always odd: odd + odd + odd = odd. For example, 1 + 1 + 1 = 3= odd.
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# M27-16

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