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

What is the value of k for no solution in 2x + 3y = 4 and kx + 6y = 11?

Advertisement

Answer and explanation

Correct answer: k = 4

A pair of linear equations has no solution when the coefficient ratios are equal but the constant ratio is different: a1/a2 = b1/b2 ≠ c1/c2. Here, 3/6 = 1/2. Therefore, 2/k must also equal 1/2, which gives k = 4. However, the constant ratio is 4/11, and 4/11 is not equal to 1/2. Thus the lines have equal slopes but different intercepts; they are parallel and never intersect. Therefore, there is no solution and option C is correct. Other listed values do not make the lines parallel.

Related tags

Class10Linear-EquationsNo-SolutionConditions For SolvabilityPair Of Linear Equations In Two VariablesMathematicsClass 10 Mcq

Frequently asked questions

What is the correct answer to this question?

k = 4

Why is this the correct answer?

A pair of linear equations has no solution when the coefficient ratios are equal but the constant ratio is different: a1/a2 = b1/b2 ≠ c1/c2. Here, 3/6 = 1/2. Therefore, 2/k must also equal 1/2, which gives k = 4. However, the constant ratio is 4/11, and 4/11 is not equal to 1/2. Thus the lines have equal slopes but different intercepts; they are parallel and never intersect. Therefore, there is no solution and option C is correct. Other listed values do not make the lines parallel.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Pair of Linear Equations in Two Variables. Topic: Conditions for solvability.

Was this question useful?

No ratings yetWrite a review / Rate this question

Student Reviews

No published reviews yet.

Advertisement