Question

Find three smallest consecutive whole numbers such that the difference between one-fourth of the largest and one-fifth of the smallest is atleast 3.

Answer 1

Let the required numbers be $x,x+1,x+2$.

The given statement can be written in inequations form as follows;

$\frac{x+2}{4}-\frac{x}{5}\ge 3$

$\Rightarrow \frac{5x+10-4x}{20}\ge 3$

$\Rightarrow \frac{x+10}{20}\ge 3$

Rule: If both the sides of an inequation are multiplied or divided by the same positive number, then the sign of the inequality will remain the same.
On multiplying both sides by 20 in both the inequations;

$\Rightarrow x+10\ge 60$

Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the following inequation.

$\Rightarrow x\ge 60-10$
$\Rightarrow x\ge 50$

Therefore, $x=50$ is the smallest value of $x$ that satisfies the inequation $x\ge 50$.

$\therefore$ required smallest consecutive whole numbers are: 50, 51, 52.