Question 503132: 1.a father is now 5 times as old as his son. 21 years hence he will be twice as old as his son . what are their ages.
2.find a number such that when 58 is added to it , the result is equal to 3 x the number.
3.in 8 year time time mart age will be 4 times her age 7 years ago. if Mary is x year now. what is Marys age in 8 year time .
Marys age 7 years ago
solve x
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! Q1.
A father is now 5 times as old as his son. 21 years hence he will be twice as old as his son . what are their ages.
Let
present age of father = f
present age of son = s
Father is 5 times as old as son
f=5s ..............(1)
After 21 years means we have to add 21 in the present ages of son and father
f+21=s+21
then
father will be twice as old as his son
f+21=2(s+21)
f+21=2s+42
f=2s+42-21
f=2s+21 ...........(2)
Put the value of from (1) to (2), we have
f=2s+21
5s=2s+21
5s-2s=21
3s=21
s=21/3
s=7
Put the value of s in (1), we have
f=5s
f=5*7
f=35
present age of father = 35
present age of son = 7
Q2.
find a number such that when 58 is added to it , the result is equal to 3 x the number.
Let
Number = x
3x=x+58
3x-x=58
2x=58
x=58/2
x=29
Number = 29
Q3.
In 8 year time time Mary age will be 4 times her age 7 years ago. if Mary is x year now. what is Marys age in 8 year time .
Marys age 7 years ago
solve x
Mary's age now = x
7 years ago = x-7
After 8 years time = x+8
After 8 years from now, Mary age will be 4 times her age 7 years ago
4(x-7)=x+8
4x-28=x+8
4x-x=8+28
3x=36
x=36/3
x=12
Mary is 12 years old now.
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