Question

If Neptune’s average distance from the sun is $4.503 \times {10^9}km$. and Mercury's average distance is $5.791 \times {10^7}km$, how many times is Neptune's distance from Sun as compared to that of Mercury?

Answer 1

Hint: In order to solve this question, first we will assume Neptune’s average distance from the sun and Mercury's average distance from the sun to be certain variables.Then on dividing the two variables and simplifying them, we will arrive at the correct answer.

Complete step by step answer:
Given, Neptune’s average distance from the sun is $4.503 \times {10^9}\,km$.Mercury's average distance from the sun is $5.791 \times {10^7}\,km$. Let us assume Neptune’s average distance from the sun to be $x$, and Mercury's average distance from the sun to be $y$. So, $x = 4.503 \times {10^9}\,km$ and $y = 5.791 \times {10^7}\,km$. So, the number of times Neptune's distance from the Sun as compared to that of Mercury will be given by the $\dfrac{x}{y}$.

On putting the value of $x$ and $y$ in the above expression, we get,
$\dfrac{x}{y} = \dfrac{{4.503 \times {{10}^9}}}{{5.791 \times {{10}^7}}}$
This can be written as,
$\dfrac{x}{y} = \dfrac{{4.503 \times {{10}^{9 - 7}}}}{{5.791}}$
$\dfrac{x}{y} = 0.7776 \times {10^2}$
On further solving, we get,
$\therefore \dfrac{x}{y} = 77.76$

Therefore, Neptune's distance from the Sun is $77.76$ times as compared to Mercury’s distance from the sun.

Note: It is important to note that while solving such questions in which we have to find the ratio of two terms, we should always have the same unit of the term in the numerator and the term in the denominator. If the ratio is solved with different units of the numerator and the denominator, then the answer of the question will be wrong. So, the units should be converted so that the unit of both denominator and the numerator becomes the same.