Maths Olympiad Prep

Track / Stage 5 / 36 of 400 #636 of 1964

Problem 636

AIME late
Combinatorics Difficulty 5.1 Find the answer

Ayeesha had 12 guests aged 6,7,8,96,7,8,9 and 10 at her birthday party. Four of the guests were 6 years old. The most common age was 8 years old. What was the mean age of the guests?
A 6
В 6.5
C 7
D 7.5
E 8

Multiple choice: answer with the letter of the option you want.

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Official solution

Solution
D

Since there were four children aged 6 and the most common age was 8 , there were at least five children aged 8 at the party. Hence, as there were 12 children in total, there were at most three other children at the party. We are told that there were children aged 7, 9 and 10 at the event so we can deduce that there were exactly five children aged 8 , one child aged 7 , one child aged 9 and one child aged 10 at Ayeesha's party as well as the four children aged 6. Therefore the total age of the children was 4×6+7+5×8+9+10=904 \times 6+7+5 \times 8+9+10=90. Hence the mean age of the children was 90÷12=7.590 \div 12=7.5.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.