Jan and 5 of his friends sold candy bars to raise money for their scho

Question

Jan and 5 of his friends sold candy bars to raise money for their school trip. Jan sold 20 bars and each of his 5 friends sold at least one bar. Did Jan sell more candy bars than each of at least 3 of his friends?

(1) The median of the number of bars sold by Jan's 5 friends is 18. 
(2) The average (arithmetic mean) of the number of bars sold by Jan's 5 friends is 12.

Bunuel · Accepted Answer

Jan and 5 of his friends sold candy bars to raise money for their school trip. Jan sold 20 bars and  each of his 5 friends sold at least one bar . Did Jan sell more candy bars than each of at least 3 of his friends?

(1) The median of the number of bars sold by Jan's 5 friends is 18.

{a, b, 18, c, d}, so at least 3 friends sold less than 20 bars (a, b and 18). Sufficient.

(2) The average (arithmetic mean) of the number of bars sold by Jan's 5 friends is 12.

The total number of bars sold by Jan's 5 friends is 5*12=60. Can there be a case where at least 3 of Jan's friends sold 20 or more bars? No, since in this case the total number of bars sold by his friends would be more than 60 (20+20+20+1+1=62). Sufficient.

Answer: D.

KarishmaB · Answer

Jan sold 20 bars. So the question is whether at least 3 friends sold less than 20 bars? Or is it possible that 3 friends sold at least 20 bars each? If not, then we know that Jan did sell more than at least 3 friends. (1) The median of the number of bars sold by Jan's 5 friends is 18. Since there are 5 friends, this means that at least 3 friends sold 18 or less than 18 bars. So Jan did sell more than at least 3 of his friends. Sufficient alone (2) The average (arithmetic mean) of the number of bars sold by Jan's 5 friends is 12. This means Jan's 5 friends sold total 60 bars. Then could 3 friends have sold at least 20 bars each? No. If 3 friends were to sell 20 bars each, the leftover 2 friends would have sold 0 bars but we know that each friend sold at least 1 bar. Hence Jan did sell more bars than at least 3 friends. Answer (D) Also check this question on mean, median, range: https://anaprep.com/sets-statistics-mea ... -question/

Amobnc · Answer

Hi Bunuel,
The question asked is if Jan sold more candy bars than each of at least 3 of his friends:
In statement 2.  I analyzed the following situations:
12+12+12+12+12=60 which would render a NO answer; and
1+1+12+23+23=60 which would render a YES answer.

Why is this wrong?
How come you negated the question asked to get to the right answer?

Bunuel · Answer

{12, 12, 12, 12, 12} also gives a YES answer to the question: Jan sold more candy bars than each of  at least 3  of his friend.

Amobnc · Answer

Thanks @Bunuel. 
For some reason I interpreted 'than each of at least 3 of his friends' as the sum of the number of candy bars that each of at least 3 of his friends had..
Twist in the brain..!

av722 · Answer

we're trying to see if he's sold more candy bars than  at least 3 of his friends .

1) if the median of his 5 friends' candy bars is 18, then the set would look something like this
A B 18 D E
where A and B would have to be <= 18. Since 18 is less than 20, then Jan sold more candy bars than friend A, B and Mr. "Sold 18 bars".

2) We should think about how to break the scenario. If the average candy bars sold of his five friends is 12, then the avg equation would look like this

(sum of bars sold by his 5 friends) / 5 = 12.

So sum of bars sold by his 5 friends = 60. In order to  not  have Jan sell more candy bars than at least 3 friends, then those 3 friends would have to sell  at least  21 candy bars. However, 3(21) = 63, and 63>60.

Thus statements 1 and 2 are sufficient alone. D.

esseeveniet · Answer

Hi, q-  Did Jan sell more candy bars than each of at least 3 of his friends?

would need your help to clear my reasoning? (i) Median is 18, therefore 3rd person have 18, if i take the lowest possible scenario, {1,1,18,morethan 18, more than 18}} , in this case Jan and atleast 3 friends have equal candy's i.e 20. Which is not clear to answer.

Average of the of 5=12, then total candy's by friends can be 60. and possible scenario to counter this is {1,1,18,20,20}

Bunuel · Answer

It's not clear what you're doing there.

From (1), we know Jan's friends sold {a, b, 18, c, d} candies and Jan sold 20. So, the answer to the question is definitely YES. Friends who sold a, b, and 18 bars sold fewer than Jan, who sold 20, meaning at least 3 friends sold fewer.

For (2), since the 5 friends sold a total of 60 candies, it's impossible for 3 or more friends to have sold 20 or more, because 3 * 20 is already 60.

rashminuligonda · Answer

hi bunuel, for
 1) The median of the number of bars sold by Jan's 5 friends is 18.

consider { 1, 1, 18, 18,18} considering first 3 friends. how is this condidition sufficient?

similarly for 
 (2) The average (arithmetic mean) of the number of bars sold by Jan's 5 friends is 12.

consider { 1, 1, 18, 20,20} considering first 3 friends. how is this condidition sufficient?

please help

Bunuel · Answer

The question asks:  Did Jan sell more candy bars than each of  at least 3  of his friends?

Your examples do not show a problem because they still give the same answer the question asks for, which is YES. So the only thing we care about is  whether Jan sold more than at least 3 friends.

In your median example {1, 1, 18, 18, 18}, Jan’s 20 is greater than all 5 friends, so the answer to the question is definitely YES.
In your average example {1, 1, 18, 20, 20}, Jan’s 20 is greater than 3 friends (1, 1, and 18), so again the answer to the question is YES.

So those examples do not make the statements insufficient. They are consistent with the guaranteed YES.

Borrat · Answer

statement 1)

scenario 1: 1 1 18 _ _ yes 20 is greater than 1st 2d 3rd 
 scenario 2: 18 18 18 _ _ yes 20 is greater than 1st 2d 3rd ---- 1 alone is sufficient

statement 2)
total/5 = 12 , so total is 60 , since theres a myriad of ways to get 5 numbers to sum to 60, ill find 2 cases that contradict each other :

scenario 1: 12 12 12 12 12 --- yes 20 is greater than 1st 2nd 3rd

now for scenario 2: lets force it to  not be  .. how? : 20 20 20 _ _. here we are pushing it to be wrong... but can we actually do this based on the fact that total sum of 5 numbers must be 60? .... check again: 20 + 20 + 20 + _ + _ =60 ... this means the blanks have to be zero, but the constraint specifically says at least 1. so fill the blanks with 1 and take away that amount from the others. : 20 20 18 1 1. ... so the way this problem is created forces sufficiency
20 20 18 1 1 = 60 ,

Jan and 5 of his friends sold candy bars to raise money for their scho

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Jan and 5 of his friends sold candy bars to raise money for their scho

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