Asked by Anonymous
Alice had 2/3 as many sweets as Jason. The ratio of Jason’s sweets to Bob’s sweets is 1:4. If Bob had 60 more sweets than Alice, find the total number of sweets that was shared among the three children at first.
Answers
Answered by
Anonymous
A/J = 2/3 so 3 A = 2 J so 6 A = 4 J
J/B = 1/4 so 4 J = B = 6 A
and
B - A = 60
but we already know that B = 6 A
so 5 A = 60
A = 12 , (well there is hope)
then B= 6 A = 72
and J= B/4 = 18
12 + 72 + 18 = 102
==================
Check
==================
J/B = 4 ?
18 * 4 = 72 yes
A/J = 2/3 = ? 12/18 yes
B-A = 72-12 = 60 yes
Whew !
J/B = 1/4 so 4 J = B = 6 A
and
B - A = 60
but we already know that B = 6 A
so 5 A = 60
A = 12 , (well there is hope)
then B= 6 A = 72
and J= B/4 = 18
12 + 72 + 18 = 102
==================
Check
==================
J/B = 4 ?
18 * 4 = 72 yes
A/J = 2/3 = ? 12/18 yes
B-A = 72-12 = 60 yes
Whew !
Answered by
mathhelper
number of sweets for Jason --- x
number for Alice = (2/3)x or 2x/3
Jason : Bob = 1 : 4
x/Bob = 1/4
Bob = 4x
number for Bob = 4x
Bob - Alice = 60
4x - 2x/3 = 60
times 3
12x - 2x = 180
10x = 180
x = 18
Jason had 18, Alice had 12 and Bob had 72
for a total of 102
Check:
did Bob have 60 more than Alice ? YES
ratio of Jason : Bob = x : 4x = 1:4, as required
All looks good
number for Alice = (2/3)x or 2x/3
Jason : Bob = 1 : 4
x/Bob = 1/4
Bob = 4x
number for Bob = 4x
Bob - Alice = 60
4x - 2x/3 = 60
times 3
12x - 2x = 180
10x = 180
x = 18
Jason had 18, Alice had 12 and Bob had 72
for a total of 102
Check:
did Bob have 60 more than Alice ? YES
ratio of Jason : Bob = x : 4x = 1:4, as required
All looks good
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