# Word Problem - How many pieces of candy

Solve for pieces of candy

Nine people have some candy.

The person with the most candy has nine prices.

At least one person has no candy.

The average number is four pieces of candy per person.

Four pieces of candy is the median.

More people have two pieces of candy than any other number.

How many pieces of candy does each person have?

### Comments for Word Problem - How many pieces of candy

 Feb 10, 2013 Solve by: Staff AnswerPart IC = total number of pieces of candyN₁ = number of pieces of candy for person 1 N₂ = number of pieces of candy for person 2N₃ = number of pieces of candy for person 3 N₄ = number of pieces of candy for person 4 N₅ = number of pieces of candy for person 5 N₆ = number of pieces of candy for person 6 N₇ = number of pieces of candy for person 7 N₈ = number of pieces of candy for person 8 N₉ = number of pieces of candy for person 9C = N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉The average number is four pieces of candy per person. 4 = (N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉)/9The total number of pieces of candy is 36 (4 × 9 = 36) 36 = (N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉)Four pieces of candy is the median. N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉N₁ + N₂ + N₃ + N₄ + 4 + N₆ + N₇ + N₈ + N₉ = 36 The person with the most candy has nine prices.N₉ = 9At least one person has no candy.N₁ = 0N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉0 + N₂ + N₃ + N₄ + 4 + N₆ + N₇ + N₈ + 9 = 36More people have two pieces of candy than any other number. Therefore, at least two of the following three people must have 2 pieces of candy: N₂, N₃, N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉ 0 + N₂ + 2 + 2 + 4 + N₆ + N₇ + N₈ + 9 = 36N₆, N₇, and N₈ must be 5,6,7; 5,6,8; 5,7,8 or 6,7,8If these three numbers are 6,7, and 8, then the total number of pieces of candy will exceed 36If these three numbers are 5,7, and 8, then the total number of pieces of candy will exceed 36If these three numbers are 5,6, and 8, then the total number of pieces of candy does equal 36If this is the case, then N₂ = 0. ---------------------------------

 Feb 10, 2013 Solve by: Staff ---------------------------------Part IIThis is a 1st possible solution:N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉0 + 0 + 2 + 2 + 4 + 5 + 6 + 8 + 9 = 36If N₆, N₇, and N₈ are equal to 5,6,7N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉0 + N₂ + 2 + 2 + 4 + 5 + 6 + 7 + 9 = 36Since the sum of 0 + 2 + 2 + 4 + 5 + 6 + 7 + 9 = 35, N₂ must = 1N₁ + N₂ + N₃ + N₄ + N₅ + N₆ + N₇ + N₈ + N₉0 + 1 + 2 + 2 + 4 + 5 + 6 + 7 + 9 = 36The Final Solution #1 is: N₁ = 0 pieces of candyN₂ = 0 piece of candyN₃ = 2 pieces of candy N₄ = 2 pieces of candy N₅ = 4 pieces of candy N₆ = 5 pieces of candy N₇ = 6 pieces of candy N₈ = 8 pieces of candy N₉ = 9 pieces of candy The Final Solution #2 is: N₁ = 0 pieces of candyN₂ = 1 piece of candyN₃ = 2 pieces of candy N₄ = 2 pieces of candy N₅ = 4 pieces of candy N₆ = 5 pieces of candy N₇ = 6 pieces of candy N₈ = 7 pieces of candy N₉ = 9 pieces of candy Thanks for writing.Staff www.solving-math-problems.com