Peter and Paul are two friends. The sum of their ages is 35 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter?
Answers
Answered by
6
Let peter's age be " x "
Let Paul's present age be "y"
Let z be the no of years indicated as "when" in the question
(x - z) = peter's age z years ago
(y - z) = paul's age z years ago
(x + y) = 35
z years ago peter was ( x - z ) and that is the age of Paul is now, so
x - y = z
but Paul was (y - z)
As peter is now twice the age of paul's age then
x = 2(y - z)
x= 2y - 2z
so, the equations we have are
(x + y) = 35
x - z = y
x = 2y - 2z
lining them all,
x + y = 35
x - y - z = 0
x - 2y - 2z = 0
so, solving this we get
x = 20
y = 15
z = 5
so peter's age = 20 years
Let Paul's present age be "y"
Let z be the no of years indicated as "when" in the question
(x - z) = peter's age z years ago
(y - z) = paul's age z years ago
(x + y) = 35
z years ago peter was ( x - z ) and that is the age of Paul is now, so
x - y = z
but Paul was (y - z)
As peter is now twice the age of paul's age then
x = 2(y - z)
x= 2y - 2z
so, the equations we have are
(x + y) = 35
x - z = y
x = 2y - 2z
lining them all,
x + y = 35
x - y - z = 0
x - 2y - 2z = 0
so, solving this we get
x = 20
y = 15
z = 5
so peter's age = 20 years
kvnmurty:
good
Answered by
4
let present age of peter = x
paul = y
x+y = 35 and
x= 2(y-(x-y))
⇒x = 4y - 2x
⇒3x =4y
⇒x = 4y/3
7y/3 = 35
⇒y = 15 yrs (paul's age)
x = 4×15/3 = 20 (peter's age)
paul = y
x+y = 35 and
x= 2(y-(x-y))
⇒x = 4y - 2x
⇒3x =4y
⇒x = 4y/3
7y/3 = 35
⇒y = 15 yrs (paul's age)
x = 4×15/3 = 20 (peter's age)
Similar questions