Lesson Progress
0% Complete

# Understand the Problem

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day? $\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$

#### Number Theory in Median

Easy
##### Suggested Book
Mathematical Circles

Do you really need a hint? Try it first!

Recall the concept of Median . https://en.wikipedia.org/wiki/Median
As here 252 cans of soda sold to  even no of customers i.e.  to 100 customers , so median will be the average of the 50th and 51st largest amount of cans per  person .
First they each have one can atleast one can . Now think for the rest 152 cans and 50th and 51st customers . Then in order to proceed divide 100 customers in two groups as first 49 customers and second 51 customers .
To minimize the first 49, they will  be sold no more cans. Now 152 cans left to divide among next 51 customers . Taking $$\frac {152}{51}$$ gives us 2 and a remainder of 50. Seeing this, the largest number of cans the 50th person could have is 3 ( including the can in 1st distribution), which leaves 4( including the can in 1st distribution)  to the rest of the people. The average of 3 and 4 is 3.5  . So the maximum possible median is 3.5  .

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Colour Problem | PRMO-2018 | Problem No-27

Try this beautiful Problem on Combinatorics from PRMO -2018.You may use sequential hints to solve the problem.

## Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on triangle from AMC 10A, 2018. Problem-23. You may use sequential hints to solve the problem.

## Area of region | AMC 10B, 2016| Problem No 21

Try this beautiful Problem on Geometry on Circle from AMC 10B, 2016. Problem-20. You may use sequential hints to solve the problem.

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.