Question 435682
 Five years back the aga of a father was three times the age of his son. it will be twice after the 10years.Find their present age?
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Equations:
(f-5)= 3(s-5)
(f+10) = 2(s+10)
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Rearrange:
f = 3s-10
f = 2s+10
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Equate:
3s-10 = 2s+10
s = 20 (son's age now)
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Solve for "f":
f = 3s-10
f = 3*20-10 = 50 (father's age now)
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Two persons 27km apart,starting at the same time are together in 9hours if they walk in the same direction,but 3hours if they walk in opposite direction.Find out speed of each man?
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Same Direction DATA:
Faster man: distance = d+27 km; time = 9 hrs; rate = (d+27)/9 km/h
Slower man: distance = d km ; time = 9 hrs; rate = d/9 km/h
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Opposite Direction DATA:
Faster man: rate = (d+27)/9 km/h ; time = 3 hrs ; distance = rt = (d+27)/3 km
Slower man: rate = d/9 km/h ; time = 3 hrs ; distance = rt = d/3 km
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Equation:
dist + dist = 27 km
(d+27)/3 + d/3 = 27
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d+27+d = 3*27
2d = 2*27
d = 27
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Faster man rate = (d+27)/9 = (2*27)/9 = 18 km/h
Slower man rate = d/9 = 27/9 = 3 km/h
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Cheers,
Stan H.