Four different children have jelly beans: Aaron has 5, Bianca has 7

Hello,

Since Bianca and Callie are both within 1 jelly bean of each other and Aaron has 5, Dante must provide 3 of his 11 jelly beans so each child has no more than 1 fewer jelly bean than any other child.

Dante + Aaron = 11+5 =16/2 = 8
11-8 = 3 so Dante must provide 3 jelly beans to Aaron.

Answer (B)

Sure there's a shorter way to solve the problem but this was my method. Still early in the process of figuring out the shortcuts.

Is it B ?
If D gives 3 to A, then difference for each will be 1 or 0.

Simply use the options given,
If D gives A 3 jelly beans,
then A has=8
B has=7
C has=8
D has 8

Condition given in the question met.
Answer B

Hi All,

This question can be solved without any special knowledge and math - we just need to use a bit of 'brute force' and we can TEST THE ANSWERS....

We're told the number of jelly beans that each child has:
Aaron = 5
Bianca = 7
Callie = 8
Dante = 11

Dante is to give a certain number of jelly beans to Aaron so that every child has a number of jelly beans that is equal to (or 1 fewer than) any other child. We're asked what that number is.

Answer A: 2
If Dante gives Aaron 2 jelly beans, then the numbers will be....
Aaron = 7
Bianca = 7
Callie = 8
Dante = 9
In this scenario, Dante has 2 more jelly beans than both Aaron and Bianca. This does not match the prompt, so this is NOT the answer.

Answer B: 3
If Dante gives Aaron 3 jelly beans, then the numbers will be....
Aaron = 8
Bianca = 7
Callie = 8
Dante = 8
In this scenario, everyone is within 1 jelly bean of everyone else. This is a MATCH for what the prompt described, so this MUST be the answer.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Yep,

We have:

5 - 7 - 8 - 11 --> minus 1:
6 - 7 - 8 - 10--> still not good. minus 1:
7 - 7 - 8 - 9 --> still not good, minus 1:
8 - 7 - 8 - 8 --> this will do!

So, Dante must give 3 jelly beans to Aaaron.

Hi ,
to satisfy the conditions we have to have all four of them to 7 or 8..
it gets satisfied if dante has 8, Aaron to becomes 8.. so ans 3..B

MANHATTAN GMAT OFFICIAL SOLUTION:

Conceptually, the transfer of jelly beans from Dante to Aaron reduces the range of the number of jelly beans held by individual children. Our constraint is that no children differ in their number of jelly beans by more than 1—a condition Bianca and Callie already satisfy.

We can draw the following picture (a number line) to visualize the scenario:From the picture, we can infer that Aaron and Dante must end up with a number of jelly beans that is either 7 or 8. If either Aaron or Dante has a number of jelly beans other than 7 or 8, he will differ too much from either Bianca's or Callie's number.
A + x = 7 or 8
5 + x = 7 or 8
x = 2 or 3
D – x = 7 or 8
11 – x = 7 or 8
x = 3 or 4

The solution to both equations is x = 3. The resulting number of jelly beans is A = 8, B = 7, C = 8, and D = 8.

The correct answer is B.

A = 5
B = 7
C = 8
D = 11

How many jelly beans must Dante give to Aaron to ensure that no child has more than 1 fewer jelly beans than any other child?

i..e A and D each of them must have either 7 or 8 jelly beans at the ends of transaction because C and D are already 7 and 8 respectively and max difference between any two can be 1 only

A = 5+3 = 8
B = 7
C = 8
D = 11-3 = 8

Answer: option B

Since Bianca’s number of jelly beans and Callie’s number of jelly beans differ by exactly 1, we need to bring Aaron’s number of jelly beans to either Bianca’s or Callie’s. If Aaron’s number is brought up to Bianca’s (i.e., Dante gives Aarron 2 jelly beans), then Aaron, Bianca, Callie, and Dante have 7, 7, 8, and 9 jelly beans, respectively. However, we see that Dante still has 2 more jelly beans than either Aaron or Bianca. On the other hand, if Aaron’s number is brought up to Callie’s (i.e., Dante gives Aarron 3 jelly beans), then Aaron, Bianca, Callie, and Dante have 8, 7, 8, and 8 jelly beans, respectively. We see that in this case, no child has more than 1 fewer jelly beans than any other child.

Answer: B

Jelly beans
A 5
B 7
C 8
D 11
target : distribution of JB such that difference of JB among children is 1 or 0
possible only when D gives 3 to A
we get arrangement
A 8
B 7
C 8
D 8
now ∆ of JB is 1 & 0 among all
option B

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.