पूर्णांक (x) के लिए \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\) में कितने हल हैं?
For integer (x), how many solutions are there for \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?
Explanation opens after your attempt
Correct Answer
C. (11)
Concept
Solving both parts gives \(x>-\frac{2}{3}\) and \(x\le10\). Thus integers are \(x=0,1,\ldots,10\), totaling (11).
Why this answer is correct
The correct answer is C. (11). Solving both parts gives \(x>-\frac{2}{3}\) and \(x\le10\). Thus integers are \(x=0,1,\ldots,10\), totaling (11).
Exam Tip
दोनों भाग हल करने पर \(x>-\frac{2}{3}\) और \(x\le10\) मिलता है। अतः पूर्णांक \(x=0,1,\ldots,10\) हैं, कुल (11)।
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