the ages of two persons differ by 16 years if 6 years ago the elder one be three times as old as the younger one find their present ages
Answers
The ages of two persons differ by 16 years and 6 years ago, the elder one is three times as old as the younger one.
The present ages of both persons.
Let the age of elder and the younger person be x and y years respectively.
Then,
According to the question-
6 years ago-
Put y = 14 in equation (i) -
Hence, the ages of elder and younger person is 30 years and 14 years respectively.
Answer :
The present ages of younger one and elder one are 14 years and 30 years respectively.
Explanation :
Given : Ages of two persons differ by 16 years. If 6 years ago the elder one will be three times as older as the younger one.
To find : Present ages of them.
Soultion :
Let the age of elder one be x years
Let the age of the younger one be y years
Difference of age between them = 16 years
=> x - y = 16
x = 16 + y ------- eq(1)
6 years ago:
The age of elder one = (x - 6) years
The age of younger one = (y - 6) years
According to the question:
(x - 6) = 3(y - 6)
16 + y - 6 = 3(y - 6)
10 + y = 3y - 18
3y - y = 10 + 18
2y = 28
y = 28/2
y = 14
Subtitute y = 14 in eq(1)
x = 16 + 14
x = 30
So the present ages of younger one and elder one are 14 years and 30 years respectively.