SOLUTION: Aunt Alice gave each of her three nieces a number of silver dollars equal to their ages. The youngest felt that this was unfair. They agreed to redistribute the money. The youngest

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: Aunt Alice gave each of her three nieces a number of silver dollars equal to their ages. The youngest felt that this was unfair. They agreed to redistribute the money. The youngest      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 12331: Aunt Alice gave each of her three nieces a number of silver dollars equal to their ages. The youngest felt that this was unfair. They agreed to redistribute the money. The youngest would split half of her silver dollars evenly with the other two sisters. The middle would then give each of the others 4 silver dollars. Finally, the oldest was to split half of her dollars equally between the two younger sisters. After exchanging the money, each girl had 16 silver dollars. How old are the sisters? (Their ages = 48)
MUST HAVE PROCESS
Answer by bonster(299) About Me  (Show Source):
You can put this solution on YOUR website!
youngest=x
middle=y
oldest=z

The youngest kept half and gave 1/4 each to her sisters
x:x%2F2
y:Y+x%2F4
z:Z+x%2F4

The middle gave the oldest and youngest 4 each
x:x%2F2+4
y:Y+x%2F4-8
z:Z+x%2F4+4


the oldest kept half z so (Z+x%2F4+4) becomes:z%2F2+x%2F8+2
she gives half of that to each sister so they receive: (z%2F4+x%2F16+1)
x:x%2F2+4+(z%2F4+x%2F16+1)=16
y:Y+x%2F4-8+(z%2F4+x%2F16+1)=16
z:z%2F2+x%2F8+2=16


x%2F2+4+(z%2F4+x%2F16+1)=16 <--subtract x%2F2 and 4 from both sides
(z%2F4+x%2F16+1)=16-4-x%2F2
(z%2F4+x%2F16+1)=12-x%2F2

y+x%2F4-8+(z%2F4+x%2F16+1)=16 <--subtract y and x%2F4 and add 8 to both sides
(z%2F4+x%2F16+1)=16+8-y-x%2F4
(z%2F4+x%2F16+1)=24-y-x%2F4


since 12-x%2F2 and 24-y-x%2F4 both equal (z%2F4+x%2F16+1), you can equate them and solve for y:
12-x%2F2=24-y-x%2F4
12-2x%2F4=24-y-x%2F4 <--add 2x%2F4 to both sides
12=24-y+x%2F4 <--add y to both sides
12+y=24+x%2F4 <--subtract 12 from both sides
y=12+x%2F4

to solve for z:
z%2F2+x%2F8+2=16 <--subtract 2 from both sides
z%2F2+x%2F8=14 <--subtract x%2F8 from both sides
z%2F2=14-x%2F8 <--multiply both sides by 2
z=28-x%2F4


plug it back to x+y+z=48
x+(12+x%2F4)+(28-x%2F4)=48
x+x%2F4-x%2F4+12+28=48
x+40=48
x=8


since the youngest=x, she is 8 years old
middle y=(12+x%2F4)=12+8%2F4=12+2=14
oldest z=28-x%2F4=28-8%2F4=28-2=26


They are 8, 14, and 26 years old.