Update

Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है

Subjects
0 reads0 ratings0 helpful

At least 40 marks are required to pass an examination. If \(x+12\ge 40\), what is the minimum value of \(x\)?

Answer and explanation

Correct answer: 28

Subtracting 12 from both sides of \(x+12\ge 40\) gives \(x\ge 28\). Therefore, the smallest possible value of \(x\) is 28. Option 27 is incorrect because \(27+12=39\), which is less than 40 and does not satisfy the passing condition. Exam tip: ‘At least’ means \(\ge\), so the boundary value is included.

Related tags

Linear InequalitiesMinimum ValueWord ProblemAlgebraic Solution Of Linear Inequalities In One VariableMathematicsClass 11 Mcq

Was this question useful?

No ratings yetWrite a review / Rate this question

Student Reviews

No published reviews yet.

Add Muft Shiksha to your Home Screen

In Safari, tap Share, then Add to Home Screen.