For integer (x), how many solutions are there for (\frac{2x-3}{5}<x-1\le\frac{x+8}{2})?पूर्णांक (x) के लिए (\frac{2x-3}{5}<x-1\le\frac{x+8}{2}) में कितने हल हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (11)(11)
Solving both parts gives (x>-\frac{2}{3}) and (x\le10). Thus integers are (x=0,1,\ldots,10), totaling (11).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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