Question

a man is four times as old as his son after 20 years he will be twice as old as his son at that time find their present ages.

Answer 1

Hi friend
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Your answer
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LET THE AGE OF THE SON BE X AND FATHER'S AGE BE Y.

FORMING EQUATIONS
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4X = Y......... (i)

2(X+20) = Y+20....... (ii)

FROM EQ.(i), WE GET,
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Y = 4X............. (iii)

SOLVING (BY METHOD OF SUBSTITUTION)
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FROM EQ.(ii),
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2(X+20) = Y + 20

2X + 40 = Y + 20

2X + 40 = 4X + 20 [FROM (iii)]

2X - 4X = 20 - 40

-2X = -20

2X = 20

X = 20/2

X = 10

SO,
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PRESENT AGE OF SON = 10 YEARS

PRESENT AGE OF FATHER = 4X = (4×10) YEARS = 40 YEARS

HOPE IT HELPS

#ARCHITECTSETHROLLINS

✯ BRAINLY STAR ✯

Answer 2

Hey mate..
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Let,

The present age of his son be x and the man 4x

Given,

After 20 years, he will be twice as old as his son at that time.

So,

=> ( 4x + 20 ) = 2 ( x + 20 )

=> 4x + 20 = 2x + 40

=> 4x - 2x = 40 - 20

=> 2x = 20

=> x = 20 / 2

=> x = 10

Thus,

The age of his son is 10 years

And the age of the man is ( 4 × 10 ) years = 40 years

Hope it helps !!