Question 1120860: Ann goes swimming regularly she wants to improve her fitness, so she
decides to swim 10 lengths in the first session and increase the number of
lengths she swims by 2 every session. When she reaches 50 lengths in a
session, she will not increase the number any further. If Ann asks her
friend Joy to come swimming with her, Joy starts coming at Sue's 8th
session, Joy starts to swim 15 lengths and increases the number of
lengths by 5 each time. After how many of Joy's sessions does she swim
the same number of lengths as Ann?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I will assume that "Ann" and "Sue" are supposed to be the same person....
Joy starts swimming at Ann's 8th session.
Ann starts doing 10 lengths and increases the number by 2 each session. So in her 8th session, the number of lengths she swims is 10+7(2) = 24.
In Joy's first session, she swims 15 lengths, and she increases that number by 5 each session.
Let x be the number of Joy's sessions after her first (i.e., after Ann's 8th).
The number of lengths Ann swims x sessions after Joy's first is 24+2x; the number Joy swims that same session is 15+5x.
The number of lengths they swim will be the same when 24+2x = 15+5x.
24+2x = 15+5x
3x = 9
x = 3
It will be 3 sessions after her first (i.e., her 4th session) when Joy swims the same number of lengths as Ann.
Ann's sessions:
session number 1 2 3 4 5 6 7 8 9 10 11 12...
lengths 10 12 14 16 18 20 22 24 26 28 30 32...
Joy's sessions:
session number 1 2 3 4 5...
lengths 15 20 25 30 35...
Of course, you could solve the problem without formal algebra, simply by writing out a table like the one shown.
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