कितने पूर्णांक (x) असमानता \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\) को संतुष्ट करते हैं?

How many integers (x) satisfy \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?

Explanation opens after your attempt
Correct Answer

D. अनंतinfinitely many

Step 1

Concept

The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).

Step 2

Why this answer is correct

The correct answer is D. अनंत / infinitely many. The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).

Step 3

Exam Tip

पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)।

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Mathematics Answer, Explanation and Revision Hints

कितने पूर्णांक (x) असमानता \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\) को संतुष्ट करते हैं? / How many integers (x) satisfy \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?

Correct Answer: D. अनंत / infinitely many. Explanation: पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)। / The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).

Which concept should I revise for this Mathematics MCQ?

The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).

What exam hint can help solve this Mathematics question?

पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)।