Question 984473
A, B and C are three groups of pupils. If group A is 1/3 of the total number of pupils in the three groups, Group B has 5 pupils more than group A and group C has 3/5 as many pupils as group B, find the original number of pupils in each group.
:
Write an equation for each statement:
" group A is 1/3 of the total number of pupils in the three groups,"
a = {{{1/3}}}(a+b+c)
3a = a + b + c (multiplied both sides by 3)
3a - a = b + c
2a = b + c
" Group B has 5 pupils more than group A"
b = a + 5
or
a = b - 5
"group C has 3/5 as many pupils as group B,"
c = {{{3/5}}}b
c = .6b (decimals will work here)
:
In the 1st equation, 2a = b + c, replace c with .6b and replace a with (b-5)
2(b-5) = b + .6b
2b - 10 = 1.6b
2b - 1.6b = 10
.4b = 10
b = 10/.4
b = 25
then
a = 25 - 5
a = 20
and
c = .6(25)
c = 15
:
find the original number of pupils in each group.
a = 20; b = 25 ; c = 15
:
;
See if that checks out in the 1st statement
 group A is 1/3 of the total number of pupils in the three groups,"
20 = {{{1/3}}}(20+25+15)
20 = {{{1/3}}}(60)