CA Foundation · Quantitative Aptitude · Linear Inequalities
If x > y and both x and y are positive real numbers, which of the following must be true?
1/x < 1/y must be true. For positive numbers, taking reciprocals reverses the inequality, so if x > y then 1/x < 1/y. For instance, 4 > 2 but 1/4 < 1/2. The other options reverse or misapply the inequality rules.
- A1/x > 1/y
- Bx² < y²
- C1/x < 1/yCorrect
- D−x > −y
Explanation
For positive numbers, taking reciprocals reverses the inequality, so x > y gives 1/x < 1/y. Example: 4 > 2 but 1/4 < 1/2. Squaring preserves the order for positives, so x² > y², and negating reverses it to −x < −y. Hence the other options are false.
Did you get it right without looking?
One question tells you little. A timed set on Linear Inequalities shows your real accuracy, how long you take and where you lose marks.
More Linear Inequalities questions
- Solve the inequality (x − 2)/3 − (x + 1)/2 < 1 for real x.
- The solution of the double inequality −7 < 3 − 2x ≤ 9, where x is real, is:
- A firm makes x units of A and y units of B. Constraints: 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0. Profit is Z = 30x + 40y (in ₹). What is the…
- The set of all real values of x satisfying |2x − 5| ≤ 7 is:
- Which of the following points lies in the solution region of the system x + y ≤ 8, x − y ≥ 2, x ≥ 0, y ≥ 0?
- The solution set of 2x − 3 ≥ 5 together with 3x + 1 < 22, where x is a real number, is: