Asked by Anonymous
Max, Ruby and Steve shared a packet of sweets. The ratio of the total number of sweets Ruby and Steve received to the number of sweets Max received was 2:3. After Max gave 9 sweets to Ruby and 23 sweets to Steve, the 3 children had the same number of sweets in the end. How many sweets did Ruby have in the end?
Answers
Answered by
mathhelper
At the start:
Max has m sweets
Ruby has r sweets
Steve has s sweets
(r+s)/m = 2/3
2m = 3r + 3s ***
after exchange:
Max has m-32 , he gave 9 to Ruby and 23 to Steve
Ruby has r+9
Steve has s+23 , but these are equal, so
m-32 = r + 9
r = m - 41 ***
m-32 = s + 23
s = m - 55 ****
sub *** and **** into **
2m = 3r + 3s
2m = 3(m-41) + 3(m-55)
2m = 3m - 123 + 3m - 165
288 = 4m
m = 72 , r = 31 , s = 17 <---- before the exchange
so.... after exchange each will have 40
Max has m sweets
Ruby has r sweets
Steve has s sweets
(r+s)/m = 2/3
2m = 3r + 3s ***
after exchange:
Max has m-32 , he gave 9 to Ruby and 23 to Steve
Ruby has r+9
Steve has s+23 , but these are equal, so
m-32 = r + 9
r = m - 41 ***
m-32 = s + 23
s = m - 55 ****
sub *** and **** into **
2m = 3r + 3s
2m = 3(m-41) + 3(m-55)
2m = 3m - 123 + 3m - 165
288 = 4m
m = 72 , r = 31 , s = 17 <---- before the exchange
so.... after exchange each will have 40
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