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Worked Examples · Example 1

Q.Solve 30x<20030x < 200 when

(i) xx is a natural number,
(ii) xx is an integer.
Uttar Pradesh UpmspTextbookSubjective· 2mImportance★★★★★
29% · 27/94 Questions
✓ Free question

Dividing both sides by 3030 gives x<203≈6.67x < \frac{20}{3} \approx 6.67; the solution set depends on whether we restrict xx to natural numbers ({1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}) or all integers ({…,−2,−1,0,1,2,3,4,5,6}\{\ldots, -2, -1, 0, 1, 2, 3, 4, 5, 6\}).

The heart of solving a linear inequality is isolating the variable, just as with equations. The crucial difference: when you multiply or divide both sides by a negative number, the inequality sign flips. Here we divide by a positive number, so the direction stays the same.

Once we have the inequality in the form x<some valuex < \text{some value}, the solution set is determined by what kind of numbers xx is allowed to be. Natural numbers start from 11 (in most Indian syllabi) and go upward; integers include zero and all negative whole numbers too.

Step-by-step solution:

  1. Isolate xx by dividing both sides by 3030:

30x<200  ⟹  x<20030=20330x < 200 \implies x < \frac{200}{30} = \frac{20}{3}

Simplifying the fraction: 203=6.6‾≈6.67\frac{20}{3} = 6.\overline{6} \approx 6.67.

  1. Interpret x<203x < \frac{20}{3} for natural numbers:

    Natural numbers are {1,2,3,4,5,…}\{1, 2, 3, 4, 5, \ldots\}. We need all natural numbers strictly less than 6.676.67.

    The largest natural number less than 6.676.67 is 66. So the solution set is:

x∈{1,2,3,4,5,6}x \in \{1, 2, 3, 4, 5, 6\}

  1. Interpret x<203x < \frac{20}{3} for integers:

    Integers are {…,−3,−2,−1,0,1,2,3,…}\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}. We need all integers strictly less than 6.676.67.

    The largest integer less than 6.676.67 is still 66, but now we include zero, all negative integers, and so on. The solution set is:

x∈{…,−2,−1,0,1,2,3,4,5,6}x \in \{\ldots, -2, -1, 0, 1, 2, 3, 4, 5, 6\}

In interval notation (though typically we list integers explicitly for finite ranges), this is all integers in (−∞,6](-\infty, 6] intersected with Z\mathbb{Z}.

Watch out

A common mistake is to round 203≈6.67\frac{20}{3} \approx 6.67 up to 77 and include 77 in the solution. Remember: the inequality is strict (<<, not ≤\leq), so xx must be less than 6.676.67, which means x≤6x \leq 6 for integers.

Tip

When the boundary is a fraction or decimal, always check the largest whole number (or natural number) below it. Here, 6<6.67<76 < 6.67 < 7, so 66 is in, 77 is out.

✓Final answer

  1. When xx is a natural number, the solution is x∈{1,2,3,4,5,6}x \in \{1, 2, 3, 4, 5, 6\}.
  2. When xx is an integer, the solution is x∈{…,−2,−1,0,1,2,3,4,5,6}x \in \{\ldots, -2, -1, 0, 1, 2, 3, 4, 5, 6\}.

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