Nuber of 18 yrs old students = x1
Nuber of 19 yrs old students = x2
Nuber of 20 yrs old students = x3
The number of 18 year old was six more than the combined number of 19 and 20 year olds.
This mean :
x1 = x2 + x3 + 6
x1 = x2 + x3 + 6 Subtract 6 to both sides
x1 - 6 = x2 + x3 + 6 - 6
x1 - 6 = x2 + x3
x2 + x3 = x1 - 6
A class have 32 students
This mean :
x1 + x2 + x3 = 32
x1 + ( x2 + x3 ) = 32
x1 + x1 - 6 = 32
2 x1 - 6 = 32 Add 6 to both sides
2 x1 - 6 + 6 = 32 + 6
2 x1 = 38 Divide both sides by 2
x1 = 19
x2 + x3 = x1 - 6
x2 + x3 = 19 - 6
x2 + x3 = 13 Subtract x3 to both sides
x2 + x3 - x3 = 13 - x3
x2 = 13 - x3
The average of their ages was 18.5
This mean :
( 18 x1 + 19 x2 + 20 x3 ) / 32 = 18.5
( 18 x1 + 19 x2 + 20 x3 ) / 32 = 18.5
( 18 * 19 + 19 x2 + 20 x3 ) / 32 = 18.5
( 342 + 19 x2 + 20 x3 ) / 32 = 18.5
[ 342 + 19 * ( 13 - x3 ) + 20 x3 ] / 32 = 18.5
( 342 + 247 - 19 x3 + 20 x3 ) / 32 = 18.5
( 589 + x3 ) / 32 = 18.5 Multiply both sides by 32
589 + x3 = 18.5 * 32
589 + x3 = 592 Subtract 589 to both sides
589 + x3 - 589 = 592 - 589
x3 = 3
x2 = 13 - x3
x2 = 13 - 3 = 10
Nuber of 18 yrs old students = x1 = 19
Nuber of 19 yrs old students = x2 = 10
Nuber of 20 yrs old students = x3 = 3
Proof :
( 18 * x1 + 19 * x2 + 20 * x3 ) / 32 =
( 18 * 19 + 19 * 10 + 20 * 3 ) / 32 =
( 342 + 190 + 60 ) / 32 =
592 / 32 =
18.5
I need help with this problem.
A class of 32 students was made up of people who were all 18, 19, and 20 year olds. The average of their ages was 18.5. How many of each age were in the class if the number of 18 year old was six more than the combined number of 19 and 20 year olds?
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