3. Rene is 6 years older than her younger sister. After 10 years, the sum of
their ages will be 50 years. Find their present ages
Answers
Answer:
Step-by-step explanation:
Let the present age of Rene's younger sister be x
The present age of Rene be x + 6
After 10 years the sum of their ages = 50 years
A/q,
=> [(x + 6) + 10] + [x + 10] = 50
=> [(x + 16) + (x + 10)] = 50
=> 2x + 26 = 50
=> 2(x + 13) = 50
=> x + 13 = 50/2
=> x + 13 = 25
=> x = 25 - 13
=> x = 12
The age of Rene's younger sister = x = 12 years
The age of Rene = x + 6 = 18 years
Given :
Rene is 6 years older than her younger sister.
After 10 years the sum of there ages will be 50 years.
To find :
Their present ages =?
Step-by-step explanation :
The present age of Rene's younger Sister be, x.
Then, the present age of Rene's be, x + 6.
It is Given that,
After 10 years the sum of there ages will be 50 years .
So,
After 10 years Rene's age will be = x + 10
After 10 years Rene's younger sister age will be = x + 6 + 10.
According to the question,
[(x + 6) + 10] + [x + 10] = 50
[(x + 16) + (x + 10)] = 50
2x + 26 = 50
2(x + 13) = 50
x + 13 = 50/2
x + 13 = 25
x = 25 - 13
x = 12
Therefore, We got the value of, x = 12.
Hence,
The age of Rene's younger sister = x = 12 years
The age of Rene = x + 6 = 18 years
Step-by-step explanation: