Two years ago, Dilip was three times as old as his son and two years
hence, twice his age will be equal to five times that of his son. Find
their present ages.
please answer this question if you know the answer with showing all the workings
Answers
Answered by
1
Answer:
Step-by-step explanation:
Let us take dilip's current age to be x and his son's current age to be y
Two years ago,
x - 2 = 3(y - 2)
x - 2 = 3y - 6
x = 3y - 4 - (a)
Two years hence,
2(x + 2) = 5(y + 2)
2x + 4 = 5y + 10
2x = 5y + 6
x = (5y + 6)/2 - (b)
Equate (a) and (b)
3y - 4 = 5y/2 + 6/2
3y - 4 = 5y/2 + 3
3y - 5y/2 = 7
(6y - 5y)/2 = 7
y/2 = 7
y = 14
Now, from (a)
x = 3y - 4
x = 3(14) - 4
x = 42 - 4
x = 38
Answered by
5
2 years ago,
Son's age = x
Dilip's age = 3x
Present age of Son = x+2
Present age of Dilip = 3x+2
After 2 years,
Son's age = x+2+2 = x+4
Dilip's age = 3x+2+2 = 3x+4
A/Q 2(3x+4) = 5(x+4)
6x+8 = 5x+20
x = 12
∴ ᴘʀᴇsᴇɴᴛ ᴀɢᴇ ᴏғ sᴏɴ = x+2 = 12+2 = 14
ᴘʀᴇsᴇɴᴛ ᴀɢᴇ ᴏғ ᴅɪʟɪᴘ = 3x+2 = 3x12+2 = 38
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