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I was \[27\] years old \[8\] years ago. How old am I now?

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint: In order to determine the age of the given problem. Some of the key points to solve the problem on ages are:
If the present age is \[y\], then \[n\] times the present age \[ = ny\].
If the present age is \[x\], then age \[n\] years later. Hence n years \[ = x + n\].
If the present age is \[x\], then age \[n\] years ago \[ = x - n\].
The ages in a ratio \[a:b\] will be \[ax\] and \[bx\].
If the current age is \[y\], then \[\dfrac{1}{n}\] of the age is \[\frac{y}{n}\].

Complete step-by-step answer:
We are given the problem, I was \[27\] years old \[8\] years ago.
So, we have to find out the present age of the person.
We can make up the equation as,
Let us consider the present age be \[x\]
With the following as per the statement in the question,
The person was \[27\] years old \[8\] years ago.
Hence while writing in the equation format,
\[x - 8 = 27\]
So, we need to find the current age of the person,
\[x = 27 + 8\]
By adding those two values we get the present age as,
\[x = 35\].
Thus, the present age of the person is \[35\] years old.
By the given question after creating the required equation,
We were able to find the present age of the person as \[x = 35\].
That is the person was \[35\] years old.
So, the correct answer is “35 Years”.

Note: To react to the questions, take the following steps:
The most important thing is to carefully read the question and gradually shape the equation that will assist you in answering it.
Basic operations such as addition, subtraction, multiplication, and division will aid a candidate in arriving at the correct answer, and no complicated calculations are needed to do so.
Arrange the values given by correctly putting them in an equation and assigning variables to the unknown values.
To find the answer, solve the equation once it has been established.
The final step is to double-check the answer by plugging it into the equation to ensure that no errors were made during the calculation.