A group of children have a number of pens,

Question

A group of children have a number of pens, such that each child has at least one pen. If one of the children, Ann, takes 1 pen from each of the other, the number of pens with her would be thrice the number of children in the group. If the total number of pens among the children is 42, which of the following could be the number of children in the group, so that it can be ensured that Ann has the greatest number of pens?

I. 5
II. 9
III. 15

(A) Only I
(B) Only II
(C) Only III
(D) Both I and II
(E) Both II and III

I. 5
II. 9
III. 15

(A) Only I
(B) Only II
(C) Only III
(D) Both I and II
(E) Both II and III

rajtare · Answer

let no. of pens with ann be x, and number of children be n

x+(n-1)=3n
or x= 2n+1

Also, sum of pens =42

let's consider I) x= 10+1=11, so balance pen left are 42-11=31, and student left are 4. so one of the possible solution is 1,1,1,28. here clearly 28 is the largest, hence I is out
II) x=2*9+1 =19 so balance pen left are 42-19= 23, and number of children left are 8, since each student must have at least 1 pen, so to build a scenario in which we have to beat ann's number we must allocate 1 pen to every student except the last one i.e 1,1,1,1,1,1,1,16. Clearly Ann has the highest number of pen. hence 2) will hold true

3) x=30+1=31 so balance pen left are 42-31 =11, and number of children left are 14. Since each student must have atleast 1 pen, we won't be able to distribute 11 pens among 14 students. hence n=15 is not the acceptable solution

Hence the correct answer is option B)

justinioane · Answer

For this one it was easy for me use the answer choice to solve for each. Ann having the greatest number of pens is a key condition. This is important!
"the number of pens with her would be thrice the number of children in the group."

Let’s start with the first option of 5 students:
Ann’s pens = 5 × 3 = 15
The other 4 students will have: 42 – 15 = 27 pens divided among them.
If each of the 3 students has the minimum (1 pen), one student could have: 27 – 3 = 24 pens — which is more than Ann’s 15.
Therefore, Option 1 is not valid.

Option 2:
Ann’s pens = 9 × 3 = 27
The remaining students have: 42 – 27 = 15 pens
With 8 students sharing 15 pens, no one can have more than Ann.
Option 2 is valid.

Option 3:
Ann’s pens = 15 × 3 = 45
But the total number of pens is only 42, which is less than 45.
Option 3 is not valid.

Answer: Only Option 2

A group of children have a number of pens,

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A group of children have a number of pens,

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