HOT Competition 28 Aug/8PM: If 12 children purchased 34 candies and th

Given: If 12 children purchased 34 candies and the range of the number of candies that they purchased is 2, which of the following can be the number of children who purchased 3 candies?
Candies children purchased ={2,3,4} or {1,2,3}

I. 2
2×3=6 candies
Remaining candies =34-6=28
Let us assume that children who purchased 2 candies=x
2x + 4(10-x) = 40-2x =28
x=6
6 Children purchased 2 candies, 2 children purchased 3 candies and 4 children purchased 4 candies.
Possible

II. 5
3×5=15
Remaining candies =34-15=19
Let us assume that x children purchased 2 candies
2x + 4(7-x) = 19
28-2x =19
x=4.5
Not possible

Let us assume that x children purchased 1 candies
x+2(7-x)=19
14-x=19
x=-5
Not possible

III. 11
11×3=33
Remaining 1 candy is purchased by remaining 1 child.
Possible

IMO D

Posted from my mobile device

If 12 children purchased 34 candies and the range of the number of candies that they purchased is 2, which of the following can be the number of children who purchased 3 candies?

I. 2
II. 5
III. 11


A. I only
B. II only
C. I and II
D. I and III
E. I, II and III

- III. 11 X 3 = 33; 1 child gets 1 candy; total 12 children; satisfies the condition.- 11 is possible.

-II. 5 X 3 = 15 candies, 19 candy remain.
now all other children must receive 3 candies and no less than that, because the range has to be 2.
7 X 3 = 21 ------- exceeds the total; NOT possible.

Looking at options, III. is possible, but II. is not

Answer has to be D.
Answer D.

using the options
if 11 children can have 3 then we can have
1 + 11(3) = 34 ,
1 child 1 and 11 children 3 , range is 2 , all conds. satisfied. Possible

If 2 children have 3 then
6*2 + 2*3 +4*4=34
Six children have 2 , 2 children have 3 and 4 children have 4 .
Range is 2 . All conditions satisfied : Possible

If 5 children have 3 then then 15 candies are used
7 children and 19 candies cannot be distributed such that the range is 4 and only 5 children have 3 candies

IMO D

The question gives that
- the range of the number of candies that children purchased is 2
- the number of children is 12
- the number of candies is 34

The question asks the number of children who purchased 3 candies.

Then I will test the answer.
I. 2
If there are 2 children who purchased 3 candies, the remaining number of candies is 28 candies and the remaining number of children is 10.
x = the number of children who purchased 4 candies.
y = the number of children who purchased 2 candies.
4x+2y = 28
x + y = 10
I find that x = 4 , y = 6.
Keep it.

II. 5
If there are 5 children who purchased 3 candies, the remaining number of candies is 19 candies and the remaining number of children is 7.
1) x+y = 7
4x+2y = 19
x = 2.5 y = 4.5 Incorrect.
x = the number of children who purchased 4 candies.
z = the number of children who purchased 5 candies.
2) x+z = 7
4x+5z = 19
x = 16 , z= -9 Incorrect
Out.

III. 11
If there are 11 children who purchased 3 candies, the remaining number of candies is 1 candy and the remaining number of children is 1.
So the remaining number of children who purchased 1 candy = 1.
Keep it.

So the answer is I+III = D.

Posted from my mobile device

Start by listing possible distributions with Range=2 that includes 3
1) 1 | 2 | 3
2) 2 | 3 | 4
3) 3 | 4 | 5

Refer to the possible choices and

Start with 11 (as it's the most limiting)
11/12 children x 3 candies = 33/34 candies
1/12 child x 1 candy = 1/34 candy
-> total 12/12 children & 34/34 candies
(III - 11 is possible)

Eliminate answer choices A, B & C

Now try with 5 (as it will eliminate either D or E)
5/12 children x 3 candies = 15/34 candies
7/12 children MUST get 19/34 candies ->
not possible with 1 & 2 pairs (too less, even if the 7 children got 2 candies each)
not possible with 2 & 4 pairs (whatever distribution will be an even number, cannot get 19 candies)
not possible with 4 & 5 pairs (too much, even if the 7 children got 4 candies only each)
(II - 5 is not possible)

Eliminate answer choice E
IMO the answer should be D

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