The ten children in a certain group contributed a total of 28 pieces

Answer choice elimination

I. 0

The group consists of 10 children. Let's assume each child initially donates two clothes

G = {2, 2, 2, 2, 2, 2, 2, 2, 2, 2}

The number of clothes donated = 2*10 = 20

The remaining 8 clothes can be distributed as follows

G = {2, 2, 2, 2, 2, 2, 4, 4, 4, 4}

Hence, we can have a scenario in which no children donates 3 clothes.

II. 5

Let's assume that 5 children donate 3 clothes each.

Number of clothes donated = 5 * 13 = 15

Remaining = 28 - 15 = 13

There are two possible cases from here

Case 1 : 3 is the maximum number of clothes donated

G = {_, _, _, _, _, 3, 3, 3, 3, 3}

As the range is 2, the minimum number of clothes donated = 1

G = {1, _, _, _, _, 3, 3, 3, 3, 3}

The remaining 12 clothes are therefore donated by the remaining children. However, the maximum number of clothes that they can donate = is 4*2 = 8. Hence, this case is not possible.

Case 2 : 3 is not the maximum number of clothes donated

G = {.... 3, 3, 3, 3, 3 ...}

Let's assume 4 is the maximum number of clothes donated.

G = { _, _, _, _, 3, 3, 3, 3, 3, 4}

Remaining = 28 - 19 = 9. However the maximum number of clothes that the remaining students can donate = 4*2 = 8. Hence, this case is not possible.

Therefore II is not possible.

III. 9

Let's assume 9 students donate three pieces of clothes

G = { _, 3, 3, 3, 3, 3, 3, 3, 3, 3}

The remaining 1 cloth can be accommodated as follows

G = { 1, 3, 3, 3, 3, 3, 3, 3, 3, 3}

Hence, this is a possible case.

Option D

Thanks you gmatophobia, This is the method I also used to solve this problem. But it is quite time consuming and I was wondering if there is another quicker way to approach this question.

Let:
x students donate (n-1) clothes.
y students donate n clothes.
z students donate (n+1) clothes.

From the problem statement we can write these two equations:

x+y+z = 10 ........ (I)
(n-1)x + ny + (n+1)z = 28 ...........(II)

Simplify equation (II)

(n-1)x + ny + (n+1)z = 28
nx-x+ny+nz+z = 28
n(x+y+z)+z-x = 28

Put x+y+z = 10 from equation (I)

10n+z-x = 28
or, z = x+28-10n

Put n=1:
z = x+18
This is not possible since z has to be less than or equal to 10 (because there are only 10 students)

Put n=2:
z= x+8
If n=2, then number of students donating 3 clothes will be z.

Put n=3:
z = x-2
If n=3, then the number of students donating 3 clothes will be y.
y = 12-2x .......(Use the equation x+y+z=10)

Put n=4:
x = z+12
This is not possible since x<=10.

So we get 2 expressions for number of students donating 3 clothes = (x+8) and (12-2x)

Notice that:

0 can be obtained by putting x=6 in (12-2x).
9 can be obtained by putting x=1 in (x+8).

5 can never be obtained since x is a non-negative integer. So, the minimum value of (x+8) is 8.
(12-2x) is an even number so it can never be 5.

So, option D is the correct answer.

Pre-thinking
Since the range =2, the list can have two formats:
a) 2 _ _ _ _ _ _ _ _ 4
b) 1 _ _ _ _ _ _ _ _ 3

I. To have zero 3's, given that the range = 2, we have a list that begun with 2 and ends with 4 (if it begin with 1, it would necessarily finish with 3, with contradicts the initial assumption of zero 3's). By trying an error, you will quickly combine 2's (5) and 4's (3) that sum up to 28.

III. If you have 9 3's, it sums up to 27 and the last available slot in the list must necessarily be filled with 1 so the sum of the 10 elements can be 28, which makes sense with the range = 2 said in the question.

II. Let's see each arrangement
a) we need to fill five consecultive blanks 3's and three blanks (consecultive or not) with 2's or 4's. After fill with the 3's, the sum left is 28-(5*3+2+4)=7. It is impossible combine an natural amount of 2's and 4's to sum up 7.
b) we need to fill FOUR (because we already have a 3 in the list) consecultive blanks with 3's and four blanks with 2's or 1's.
1 _ _ _ _ 3 3 3 3 3
After fill with the 3's, the sum left is 28-(5*3+1)=12. The maximum that we can fill is 4 2's = 8. So this arrange is also impossible.­

chetan2u , requesting to provide an easier solution ..

chetan2u,
I faced problem in understanding the highlighted part. 
Which test do you perform to say at the outset that "1-3 range is not possible. " or "1-3 range is  possible." Please help.
­

First we know that only possibilities are 1-3 and 2-4.

Next, for number of 3s to be NINE.
Can the range be 2-4?
No, as we have only one number to choose, so for range to be 2, the number has to be two away from 3, that is 1-3 or 3-5. Now, 3-5 is not possible as the total then will be more than 28.
Only possibility 1-3.

Next, for number of 3s to be FIVE.
Can the range be 1-3? The possible maximum sum will be 1*1+4*2+5*3 or 24.
Thus, 2-4 is the only possibility

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­10 children contributed 28 pieces which means each child contributed 2.8 pieces on average. 
Since range of the number of pieces contributed is 2, and average is 2.8, the number of pieces contributed can range from either 1 to 3 or 2 to 4.

1-3 pieces contributed
For average to be 2.8 which is very close to 3, most people would have contributed 3 pieces. Using the scale method, we can see that if 1 person contributed 1 piece, 9 must have contributed 3 pieces.  So III is possible. 
Now, 1 person must have contributed 1 piece for range to be 1 - 3 so no other case is possible here. No one must have contributed 2 pieces.



2-4 pieces contributed
For average to be 2.8, using the scale method we see that number of people who contributed 2 pieces and number of people who contributed 4 pieces could be in the ratio 3:2. 
If 6 people contributed 2 pieces and 4 people contributed 4 pieces, 0 people would have contributed 3 pieces. So 0 is possible. 


Could 5 people have contributed 3 pieces? No. Then 15 pieces would be contributed by 5 people and rest of the 13 pieces would have been contributed by the other people such that each person gave either 2 pieces or 4 pieces. But 2s and 4s cannot add up to make an odd integer 13. So 5 is not possible

Answer (D)
 ­
Check Scale method here:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
 

nowhere in the question does it say that everyone donated at least one so the for II.5, it can be 5,4,0,0,3,3,3,3,3 Whar am I missing?

­The range of clothes donates is 2. So you cant have 5 and 0. That makes the range 5.

ShortcutoSpeed
Given: The ten children in a certain group contributed a total of 28 pieces of clothing to a charity.
Asked: If the range of the numbers of pieces of clothing contributed by the ten children was 2, which of the following could be the number of children in the group who contributed 3 pieces of clothing each?

l. 0; 2*6+ 4*4 = 28; Range = 4-2=2; Possible
ll. 5; 5*3=15; 28-15=13; 2x + 4(5-x) = 13; x=7/2 = 3.5: Not Possible
lll. 9: 3*9 = 27; 28-27=1; Range = 3-1=2; Possible

A) l only
B) lll only
C) l and ll only
D) l and lll only
E) l, ll, and lll

IMO D
 ­

Why is the range for option 1 not 4 (4-0)? Why is zero not considered to be part of the range.

The range is given to be 2. If you have 4 & 0 as 2 numbers the range changes to 4 and you violate the given condition.

Step by step inferences

Sum = 28
Number = 10

Range = 2
H-L = 2 [H= Highest number, L=Lowest Number]

Now
A few ways to construct this
Combination of 0 & 2
Combination of 1 & 3
Combination of 2 & 4

0 & 2 cannot give us sum 28
2*9+0 = 18 that's max sum we can have while keeping the range constraint

1&3
3x+y = 28
x+y=10
Where x and y are the number of times 3 and 1 recurr respectively

So 3 comes 9 times and 1 comes 1 time. Hence this helps........(1)

2&4
2x+4y=28
x+y=10
So 4 appears 4 times and 2 appears 6 times.............(2)

So Already using 1 and 2 we can solve for 2 options
That is 3 appearing 9 times from (1) and 3 appearing 0 times from (2)

Now for 3 appearing 5 times
3*5 = Sum is 15
So rest of it has to be either made by 1s and 2s
OR
2s and 4s to keep the range condition fulfill

No matter what I do, I ccanot make up for the deficit of (28-15 = 13 with these 2 combinations by 5 remaining numbers (10-5)
Hence not possible

Bunuel, could you please help with your take on this question, I tried listing out no.s, using the options, but its taking too long..