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If Leah is 6 years older than her sister, Sue, and John is 5 [#permalink]
20 Nov 2006, 18:05

1. If Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?

8 answer. HOW?

2. Two cyclists start biking from a trailâ€™s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

4 Â½ hours answer. HOW?

what are the solutions?
could you recommend a website that lays out problems for each kind?

Sue's Age = x
Leah's Age = x+6
John's Age = (x+6) + 5
x+(x+6)+(x+6+5) = 41
3x+17=41
This gives x=(41-17)/3=8
---------
Suppose A and B are cyclists.
A starts first (@6 miles/hr) and bikes for 3 hrs. Distance travelled = 6x3 = 18 miles
Distance between A and B = 18 miles.
Difference between A's and B's speed = 10-6 = 4 m/h
Time taken = 18/4 = 4 1/2 hrs.

1. If Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?

8 answer. HOW?

(1)L = S + 6
(2) J = L + 5
Add (1) and (2)
L+J= S+L+11
Cancel out L
J = S+11
We know
L+S+J = 41
Substitute L=S+6 and J = S+11
S+6 + S + S+11=41
3S=41-11-6=24
S=8

2. Two cyclists start biking from a trailâ€™s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

4 Â½ hours answer. HOW?

In 3 hour 1st cyclist has lead of 6*3=18 miles

2nd has to overcome this lead to catch up

Diff of speed b/w 1st and nd cyclist = 10 miles - 6 miles = 4 miles

5nd cyclist gains 4 miles in 1 hour
so he gains 18 miles in 18/4=4.5 hours

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