Question 1034417: 1.the units digit of a two digit number is thrice the tens digit.if sum of the number and the number formed by reversing the digits is 88,then find the number?(unit digit mean?)
2.a man's age 20 years hence would be thrice his age 20 years ago.find his present age?
3.five years ago ,father's age was six times the son's age.fifteen years hence,father's age will be 2 times the son's age.find the present age of father???
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! problem 1:
a = tens digit
b = units digit
number is represented by 10a + b
number formed by reversing the digits is represented by 10b + a.
the sum of the number and the number with the digits reversed is equal to 88.
this means that 10a + b + 10b + a = 88.
combine like terms to get 11a + 11b = 88
you are given that the units digits is 3 times the tens digit.
this means that b = 3a
in the equation of 11a + 11b = 88, replace b with 3a to get:
11a + 11*3a = 88
simplify to get 11a + 33a = 88
combine like terms to get 44a = 88
divide both sides of this equation by 44 to get a = 2
since b = 3a, this means that b = 6
since b is the units digit and a is the tens digit, this means that the number is 26.
the number with the digits reversed is 62.
26 + 62 = 88
the problem is solved.
the number is 26.
problem 2:
x = the man's age today.
the man's age 20 years from now is equal to the man's age 20 years before now.
this means that (x+20) = 3 * (x-20).
simplify to get x + 20 = 3x - 60
subtract x from both sides of this equation and add 60 to both sides of this equation to get 20 + 60 = 3x - x
simplify to get 80 = 2x
divide both sides of this equation by 2 to get x = 40.
the man's age today is 40.
20 years from now, the man's age will be 60.
20 years ago, the man's age was 20.
60 / 20 = 3.
the man's age 20 years from now is 3 times the man's age 20 years ago.
the problem is solved.
the man's age today is 40.
problem 3:
x = the man's age today.
y = the son's age today.
5 years ago, the father's age was 6 times the son's age.
this means that (x-5) = 6 * (y-5).
15 years from now, the father's age will be 2 times the son's age.
this means that (x+15) = 2 * (y+15).
you have two equations that need to be solved simultaneously.
they are:
(x-5) = 6 * (y-5)
(x+15) = 2 * (y+15)
simplify these equations to get:
x - 5 = 6y - 30
x +15 = 2y + 30
subtract the second equation from the first to get -20 = 4y - 60.
add 60 to both sides of this equation to get 40 = 4y.
divide both sides of this equation by 4 to get y = 10.
the son's age today is 10.
the son's age 5 years ago was 5.
the son's age 15 years from now is 25.
you are given that, 5 years ago, the father's age is 6 times the son's age 5 years ago.
5 years ago the son's age was 5.
this means the father's age had to be 30, because 6 * 5 = 30.
this means the father's age today is 35.
the father's age today is 35.
the father's age 5 years ago was 30.
the father's age 15 years from now is 50.
5 years ago, the ratio of the father's age to the son's age was 30/5 = 6.
15 years from now, the ratio of the father's age to the son's age will be 50/25 = 2.
the problem is solved.
the father's age today is 35.
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