Question

3 children were competing to see how many puzzles they could compete in 1 hour. Child A completed 5 times as many puzzles as Child B. Child C completed 4 more puzzles than Child B. How many puzzles did Child A solve if the average number of puzzles they solved was 13?

Answer 1

child A solved 25 puzzles

Explanation:

child A completed 5 times as many as child B , that is 5B

child C completed 4 more than child B , that is B + 4

then number of puzzles completed by each child is

child A = 5B , child B = B , child C = B + 4

given average number solved is 13 ,then create an equation and solve for B

average is calculated as

average =
`(sum)/(count)` =
`(5B+B+B+4)/(3)` =
`(7B+4)/(3)`

equate to 13

`(7B+4)/(3)` = 13 ( multiply both sides by 3 to clear the fraction )

3 ×
`(7B+4)/(3)` = 3 × 13 ( simplify )

7B + 4 = 39 ( subtract 4 from both sides )

7B + 4 - 4 = 39 - 4 , that is

7B = 35 ( divide both sides by 7 )

`(7)/(7)` B =
`(35)/(7)`

B = 5

then A = 5B = 5 × 5 = 25

child A solved 25 puzzles