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

Question

The ten children in a certain group contributed a total of 28 pieces of clothing to a charity. 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
ll. 5
lll. 9

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

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Official Answer

D

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chetan2u · Accepted Answer

If I were to do this question, I would look at the average of all, which would be 28/10 or 2.8
Thus, the average is closer to 3.

Possible ranges: 1-3, 2-4. No other range is possible.

Options:
1) 0
Only range is 2-4. This means few are 2 and remaining are 4 and this should make the average 2.8
Weighted average: 2.......2.8.......4 => 2.....(0.8)....2.8.....(1.2)......4
Number of 2s to total = 10*1.2/(0.8+1.2) = 12/2 = 6. Thus, six of 2s and four of 4s.

2) 5
1-3 range is not possible. Only possibility is 2-4
So total of remaining 5 is 28-(5*3) = 13. 
We have to use five of 2s and 4s to get 13, that is 2x+4(5-x)=13. But 2x+4(5-x) will always be even. 
Not possible

3) 9
1-3 range is a possibility 
The 10th number would be 28-3*9 or 1. Possible.

Thus, 0 and 9 are possible

sachi-in · Answer

This can be solved in under 3 min if we simply look at the options from down below :

A 0
B 5
C 9

C. Well 9*3 + 1*1 = 27 + 1 = 28 and we have 28 shirts exactly hence 3 shirts contributed by 9 boys possible

B. Well 5*3 = 15, so 28-15 = 13 contributions are left, 5 boys have already contributed, so rest 5 must contribute 13
 since range is 2; only 2,4 or 4,5 or 1,2 are possible contributions in this range
 Using weighted average :
 2a + 4b = 13 not possible ( 2*5 = 10 and 4*5 = 20 weighted average possible but sum must be even and it is not )
 4a + 5b = 13 not possible ( 4*5 = 20 and 5*5 = 25 in no way the weighted avg can be below 20 )
 1a + 2b = 13 not possible ( 1*5 = 5 and 2*5 = 10 in no way weighted avg can be above 10 )

A. Since contributors with 3 shirts are not present,
 Simply by observation we can notice in above case (B) 2*10 = 20 can be contributed by 10 people, now rest 8 we can divide among 4 contributors who contributes 4 shirts each.
 2a + 4b = 28 and a + b = 10 can have Integer solutions as a = 6, b = 4

So A, C possible

gmatophobia · Answer

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

jm2k5 · Answer

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.

mmdfl · 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.

sayan640 · Answer

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.

chetan2u · Answer

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

Posted from my mobile device

KarishmaB · Answer

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. Screenshot 2024-06-22 at 12.33.43 PM.png 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. Screenshot 2024-06-22 at 12.33.53 PM.png 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

Kinshook · Answer

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

Aboyhasnoname · Answer

I went the algebraic way......

Range is 2 ..Total is 28....Keep these constraints......

With Range 2...
0 to 2 Possible? No..Coz maximum will be 20 then 
1 to 3 Possible? Yes.... 
2 to 4 Possible? Yes... 
3 to 5 Possible? No..Coz minimum will be 30 then...

Now...Either 1,2,3 (Case 1)
or 2,3,4 ( Case 2) 
Looking at it ...0 is not possible in 1,2,3....because then maximum will be 20(If all give 2 each)

Coming to Case 2... 
2,3,4..if 0 gave 3...means all the other 10 gave either 2 or 4.... Let those who gave 2 be = x ...so those who gave 4 should be 10-x 
Equation....2x + 4(10-x) = 28 ...2x + 40 -4x = 28.....-2x = -12..x = 6...Its possible.....

II. 5.... 
Case 1...1,2,3.... 
if 5 Gave 3....means 15.....remaining is 13....(The other 5 will give a total of 13..) Is it possible? No...because the maximum they can give is 2 each....so max it can go to 10....

Case 2...2,3,4 
5 gave 3..means 15...remaining 13....(The other 5 will give a total of 13...)....so 2(x) +4(5-x) = 13...Need not to solve...Something x 2 + Something x 4 should give us even number whereas we are getting odd number..... 
So not possible......

III.... 9...
Means 9 x 3 is 27...the remaining 1 will give 1...hence 28 total..possible.... 
So I and III possible not II....answer D

Puranjaysapra · Answer

I got the answer right but I wanted to check this, there is no where mentioned that each student contributed at least 1, so the possible range could go anywhere till 28,29,30 right? I know that it is not very feasible, I just wanted to know

chetan2u · Answer

The only restriction is that range is 2, and this negates all other possibilities except range 2 to 4 or 1 to 3. Otherwise it would have been possible that only one gave all 28 and rest 0.

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

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

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