Question: A mother is twice as old as her son. If 20 years ago, the age of the mother was 10 times the age of the son, what is the present age of the mother?
A. 38 years
B. 40 years
C. 43 years
D. 45 years
Answer: 45 years
Solution:
Let the mother’s age be $x$
Let the son’s age be $y$.
According to the first statement “A mother is twice as old as her son”. Mathematically we can write,
$$x=2y \quad …(i)$$
According to the second statement “If 20 years ago, the age of the mother was 10 times the age of the son”. Mathematically we can write,
$$x-20=10(y-20) \quad …(ii)$$
From equation $(i)$ and $(ii)$ we get,
$$2y-20=10y-200$$$$8y=180$$ $$y=\frac{45}{2} \quad …(iii)$$
So, the son’s age is $\frac{45}{2}$ years.
Therefore from equation $(i)$, the mother’s age is $45$ years.
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