Question

A man's age is three times the sum of the ages of his
two sons, one of whom is twice as old as the other.
In 4 years, the sum of the sons' ages will be half of
their father's age. Calculate their present ages.​

Answer 1

Answer:

Step-by-step explanation:

Step-by-step explanation:

let age of man be x and his son's ages be y

and z

according to the question

x = 3(y+z) .......... ......(1)

and

y = 2z................... ....(2)

again

(x+4)/2 = y+z+8........(3)

from (1) and (2)

x= 9z..........................(4)

from (2) and (3)

(x+4)/2 = 3z+8

=> x+4 = 6z+16

=> x = 6z+12.............(5)

from (4) and (5)

9z = 6z+12

=> 3z = 12

=> z = 4 years

substituting the value of z in (2)

y = 2×4 = 8

=> y = 8 years

from (1)

=> x = 3(4+8)

=> x = 36 years

therefore age of man = 36 years

age of younger son = 4 years

age of elder son = 8 years

Hope this helps you

Answer 2

Answer:

45 years

Step-by-step explanation:

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8th

Maths

Linear Equations in One Variable

Solving Linear Equations with Variable on Both Sides

The age of a man is three t...

MATHS

The age of a man is three times the sum of the ages of his two sons. Five years hence, his age will be double of the sum of the ages of his sons. The father's present age is?

MEDIUM

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ANSWER

Let the sum of ages of two sons be x years

Age of man = 3xyears

After 5 years age of the man =(3x+5)years

Sum of ages of two sons =(x+10)years

Given, (3x+5)=2(x+10)

⇒ (3x+5)=2x+20

⇒ x=15

Hence 3x=3(15)=45

∴ Present age of father is 45years.

HOPE IT HELPS YOU