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.
Answers
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:
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