A teacher distributes n apples to some students. If she giv

[Max@Math Revolution]

[GMAT math practice question]

A teacher distributes n apples to some students. If she gives 4 apples to each student, 7 apples remain. If she tried to give 5 apples to each student, 3 students would not get anything. What is the range of possible values of n?

A. 14
B. 16
C. 18
D. 20
E. 22

[Max@Math Revolution]

=>

Assume s is the number of students.
We have n = 4s + 7 and 5(s - 4) < n ≤ 5(s - 3).
Then we have 5s - 20 < 4s + 7 ≤ 5s - 15.
5s - 20 < 4s + 7 is equivalent to s < 27 and 4s + 7 ≤ 5s - 15 is equivalent to 22 ≤ s.
Therefore we have 22 ≤ s < 27 or 22 ≤ s ≤ 26.
We have 88 ≤ 4s ≤ 104 and 95 ≤ 4s + 7 ≤ 111.
The range is 111 - 95 = 16.

Therefore, B is the answer.
Answer: B

[Scott@TargetTestPrep]

[GMAT math practice question]

A teacher distributes n apples to some students. If she gives 4 apples to each student, 7 apples remain. If she tried to give 5 apples to each student, 3 students would not get anything. What is the range of possible values of n?

A. 14
B. 16
C. 18
D. 20
E. 22

Let k be the number of students. Since 7 apples remain when each student is given 4 apples, we have:

n = 4k + 7

Suppose all but three of the students get 5 apples when the teacher tries to give each student 5 apples. Then, we have:

n = 5(k - 3)

Substituting n = 5(k - 3) into n = 4k + 7, we get:

5(k - 3) = 4k + 7

5k - 15 = 4k + 7

k = 22

In this scenario, there are 22 students and 5(22 - 3) = 5(19) = 95 apples.

To find an upper bound for the number of apples, suppose all but four students get 5 apples, one student gets only one apple and three students get no apples. Thus, we have:

n = 5(k - 4) + 1

Again substituting n = 4k + 7 in n = 5(k - 4) + 1, we get:

4k + 7 = 5(k - 4) + 1

4k + 7 = 5k - 20 + 1

k = 26

In this scenario, there are 26 students and 5(26 - 4) + 1 = 111 apples.

So, the number of apples can be any number between 95 and 111, inclusive. The range for the number of apples is 111 - 95 = 16.

Answer: B