असमानता \(3(2x-5)-4(x+1)\le 2x-19\) का हल समुच्चय क्या है?
What is the solution set of the inequality \(3(2x-5)-4(x+1)\le 2x-19\)?
#linear-inequalities
#identity-inequality
#class-11
#expert
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A \(x\ge 0\)
B \(x\in\mathbb{R}\)
C \(\varnothing\)
D \(x\le 0\)
Explanation opens after your attempt
Correct Answer
B. \(x\in\mathbb{R}\)
Step 1
Concept
Both sides become identical, so the inequality is true for every real (x). In an identity-type inequality, all real numbers are the solution.
Step 2
Why this answer is correct
The correct answer is B. \(x\in\mathbb{R}\). Both sides become identical, so the inequality is true for every real (x). In an identity-type inequality, all real numbers are the solution.
Step 3
Exam Tip
दोनों पक्ष समान बनते हैं, इसलिए असमानता हर वास्तविक (x) के लिए सत्य है। पहचान जैसी स्थिति में सभी वास्तविक संख्याएँ उत्तर होती हैं।
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असमानता \(\frac{3x-4}{5}-\frac{x+2}{3}\le \frac{1}{15}\) का हल क्या है?
What is the solution of the inequality \(\frac{3x-4}{5}-\frac{x+2}{3}\le \frac{1}{15}\)?
#linear-inequalities
#fractional-inequality
#class-11
#expert
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A \(x>\frac{23}{4}\)
B \(x\le \frac{23}{4}\)
C \(x<-\frac{23}{4}\)
D \(x\ge \frac{23}{4}\)
Explanation opens after your attempt
Correct Answer
B. \(x\le \frac{23}{4}\)
Step 1
Concept
Clearing denominators gives \(4x-22\le 1\), so \(x\le \frac{23}{4}\). In fractional inequalities, multiply by the LCM first.
Step 2
Why this answer is correct
The correct answer is B. \(x\le \frac{23}{4}\). Clearing denominators gives \(4x-22\le 1\), so \(x\le \frac{23}{4}\). In fractional inequalities, multiply by the LCM first.
Step 3
Exam Tip
हर हटाने पर \(4x-22\le 1\) मिलता है, इसलिए \(x\le \frac{23}{4}\)। भिन्नों वाली असमानता में पहले लघुत्तम समापवर्त्य से गुणा करें।
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असमानता (7-2(4x-3)>3(x+5)-5x) को हल कीजिए।
Solve the inequality (7-2(4x-3)>3(x+5)-5x).
#linear-inequalities
#bracket-simplification
#class-11
#expert
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A \(x>-\frac{1}{3}\)
B \(x<\frac{1}{3}\)
C \(x>\frac{1}{3}\)
D \(x<-\frac{1}{3}\)
Explanation opens after your attempt
Correct Answer
D. \(x<-\frac{1}{3}\)
Step 1
Concept
Simplification gives (-6x>2), so \(x<-\frac{1}{3}\). Reverse the sign when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is D. \(x<-\frac{1}{3}\). Simplification gives (-6x>2), so \(x<-\frac{1}{3}\). Reverse the sign when dividing by a negative coefficient.
Step 3
Exam Tip
सरलीकरण पर (-6x>2) मिलता है, इसलिए \(x<-\frac{1}{3}\)। ऋणात्मक गुणांक से भाग देते समय चिह्न उलटता है।
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युग्म असमानता \(-3\le \frac{7-2x}{5}<1\) का हल समुच्चय क्या है?
What is the solution set of the compound inequality \(-3\le \frac{7-2x}{5}<1\)?
#linear-inequalities
#compound-inequality
#class-11
#expert
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A \(1<x\le 11\)
B \(1\le x<11\)
C \(x\le 1\)
D (x>11)
Explanation opens after your attempt
Correct Answer
A. \(1<x\le 11\)
Step 1
Concept
Solving both parts gives \(x\le 11\) and (x>1). Therefore the combined solution is \(1<x\le 11\).
Step 2
Why this answer is correct
The correct answer is A. \(1<x\le 11\). Solving both parts gives \(x\le 11\) and (x>1). Therefore the combined solution is \(1<x\le 11\).
Step 3
Exam Tip
दोनों भाग हल करने पर \(x\le 11\) और (x>1) मिलता है। इसलिए संयुक्त हल \(1<x\le 11\) है।
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असमानता \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\) का हल क्या है?
What is the solution of the inequality \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\)?
#linear-inequalities
#fractional-inequality
#class-11
#expert
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A \(x\le \frac{26}{7}\)
B \(x\ge \frac{14}{7}\)
C \(x\ge \frac{26}{7}\)
D \(x<\frac{26}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(x\ge \frac{26}{7}\)
Step 1
Concept
Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge \frac{26}{7}\). Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.
Step 3
Exam Tip
हर हटाने पर \(7x-14\ge 12\), अतः \(x\ge \frac{26}{7}\)। भिन्नों में पहले लघुत्तम समापवर्त्य से गुणा करें।
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असमानता \(\frac{2x+7}{4}<\frac{3x-1}{6}+\frac{5}{3}\) का हल समुच्चय क्या है?
What is the solution set of the inequality \(\frac{2x+7}{4}<\frac{3x-1}{6}+\frac{5}{3}\)?
#linear-inequalities
#no-solution
#class-11
#expert
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A \(\varnothing\)
B \(x\in\mathbb{R}\)
C (x<0)
D (x>0)
Explanation opens after your attempt
Correct Answer
A. \(\varnothing\)
Step 1
Concept
After clearing denominators, (6x+21<6x+18), which is false. If the variable cancels and the statement is false, the solution is empty.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). After clearing denominators, (6x+21<6x+18), which is false. If the variable cancels and the statement is false, the solution is empty.
Step 3
Exam Tip
हर हटाने पर (6x+21<6x+18), जो असत्य है। यदि चर हट जाए और कथन असत्य हो, तो हल रिक्त होता है।
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असमानता (4(x-2)-3(2-x)\ge 5x+1) को हल कीजिए।
Solve the inequality (4(x-2)-3(2-x)\ge 5x+1).
#linear-inequalities
#brackets
#class-11
#expert
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A \(x\le \frac{15}{2}\)
B \(x\ge \frac{15}{2}\)
C \(x> \frac{15}{2}\)
D \(x< \frac{15}{2}\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge \frac{15}{2}\)
Step 1
Concept
Simplification gives \(2x\ge 15\), so \(x\ge \frac{15}{2}\). Apply the negative sign carefully while opening brackets.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge \frac{15}{2}\). Simplification gives \(2x\ge 15\), so \(x\ge \frac{15}{2}\). Apply the negative sign carefully while opening brackets.
Step 3
Exam Tip
सरलीकरण से \(2x\ge 15\) मिलता है, इसलिए \(x\ge \frac{15}{2}\)। कोष्ठक खोलते समय ऋण चिह्न ध्यान से लगाएँ।
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असमानता (\frac{2}{3}(3x-6)-\frac{1}{5}(10x+5)\le x-8) का हल क्या है?
What is the solution of (\frac{2}{3}(3x-6)-\frac{1}{5}(10x+5)\le x-8)?
#linear-inequalities
#constant-reduction
#class-11
#expert
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A \(x\le 3\)
B (x<3)
C \(x\ge 3\)
D (x>3)
Explanation opens after your attempt
Correct Answer
C. \(x\ge 3\)
Step 1
Concept
The left side becomes (-5), so \(-5\le x-8\) and \(x\ge 3\). When variables cancel, place the remaining constant correctly.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge 3\). The left side becomes (-5), so \(-5\le x-8\) and \(x\ge 3\). When variables cancel, place the remaining constant correctly.
Step 3
Exam Tip
बायाँ पक्ष (-5) बनता है, इसलिए \(-5\le x-8\) और \(x\ge 3\)। चर कटने पर बचे स्थिर पद को सही तरफ रखें।
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असमानता (-3(2x+1)+5\ge 4(1-x)-2x) का हल समुच्चय क्या है?
What is the solution set of (-3(2x+1)+5\ge 4(1-x)-2x)?
#linear-inequalities
#false-statement
#class-11
#expert
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A \(x\le -1\)
B \(x\ge -1\)
C \(x\in\mathbb{R}\)
D \(\varnothing\)
Explanation opens after your attempt
Correct Answer
D. \(\varnothing\)
Step 1
Concept
After simplification, \(2\ge 4\), which is false. A false constant inequality has an empty solution set.
Step 2
Why this answer is correct
The correct answer is D. \(\varnothing\). After simplification, \(2\ge 4\), which is false. A false constant inequality has an empty solution set.
Step 3
Exam Tip
सरलीकरण के बाद \(2\ge 4\) मिलता है, जो असत्य है। असत्य स्थिर असमानता का हल रिक्त समुच्चय होता है।
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असमानता \(\frac{x-2}{7}+\frac{x+3}{5}>1\) को हल कीजिए।
Solve the inequality \(\frac{x-2}{7}+\frac{x+3}{5}>1\).
#linear-inequalities
#fraction-addition
#class-11
#expert
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A (x>2)
B (x<2)
C \(x\ge 2\)
D \(x\le 2\)
Explanation opens after your attempt
Step 1
Concept
Clearing denominators gives (12x+11>35), so (x>2). Multiplying by a positive LCM does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. (x>2). Clearing denominators gives (12x+11>35), so (x>2). Multiplying by a positive LCM does not change the sign.
Step 3
Exam Tip
हर हटाने पर (12x+11>35), इसलिए (x>2)। धनात्मक लघुत्तम समापवर्त्य से गुणा करने पर चिह्न नहीं बदलता।
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असमानता \(6-\frac{3x-4}{2}\le \frac{5-x}{3}\) का हल क्या है?
What is the solution of \(6-\frac{3x-4}{2}\le \frac{5-x}{3}\)?
#linear-inequalities
#mixed-fraction
#class-11
#expert
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A \(x\le \frac{38}{7}\)
B \(x\ge \frac{38}{7}\)
C \(x> \frac{38}{7}\)
D \(x< \frac{38}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge \frac{38}{7}\)
Step 1
Concept
After clearing denominators and simplifying, \(38\le 7x\). Therefore \(x\ge \frac{38}{7}\) is the correct solution.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge \frac{38}{7}\). After clearing denominators and simplifying, \(38\le 7x\). Therefore \(x\ge \frac{38}{7}\) is the correct solution.
Step 3
Exam Tip
हर हटाने और सरलीकरण पर \(38\le 7x\) मिलता है। इसलिए \(x\ge \frac{38}{7}\) सही हल है।
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असमानता (\frac{2}{5}(5x-3)-\frac{1}{4}(8x+4)<\frac{5}{2}) का हल समुच्चय क्या है?
What is the solution set of (\frac{2}{5}(5x-3)-\frac{1}{4}(8x+4)<\frac{5}{2})?
#linear-inequalities
#true-statement
#class-11
#expert
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A (x<0)
B \(\varnothing\)
C \(x\in\mathbb{R}\)
D (x>0)
Explanation opens after your attempt
Correct Answer
C. \(x\in\mathbb{R}\)
Step 1
Concept
The left side becomes \(-\frac{11}{5}\), and \(-\frac{11}{5}<\frac{5}{2}\) is true. Therefore every real (x) is a solution.
Step 2
Why this answer is correct
The correct answer is C. \(x\in\mathbb{R}\). The left side becomes \(-\frac{11}{5}\), and \(-\frac{11}{5}<\frac{5}{2}\) is true. Therefore every real (x) is a solution.
Step 3
Exam Tip
बायाँ पक्ष \(-\frac{11}{5}\) बनता है और \(-\frac{11}{5}<\frac{5}{2}\) सत्य है। इसलिए हर वास्तविक (x) हल है।
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युग्म असमानता \(4\le 2x+6<14\) को हल कीजिए।
Solve the compound inequality \(4\le 2x+6<14\).
#linear-inequalities
#interval-solution
#class-11
#expert
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A \(-1\le x<4\)
B (x<-1)
C \(x\ge 4\)
D \(-1<x\le 4\)
Explanation opens after your attempt
Correct Answer
A. \(-1\le x<4\)
Step 1
Concept
Subtracting (6) from all parts gives \(-2\le 2x<8\). Dividing by (2) gives \(-1\le x<4\).
Step 2
Why this answer is correct
The correct answer is A. \(-1\le x<4\). Subtracting (6) from all parts gives \(-2\le 2x<8\). Dividing by (2) gives \(-1\le x<4\).
Step 3
Exam Tip
सभी भागों से (6) घटाकर \(-2\le 2x<8\) मिलता है। फिर (2) से भाग देने पर \(-1\le x<4\) आता है।
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युग्म असमानता \(-7<\frac{x+2}{3}\le 1\) का हल समुच्चय क्या है?
What is the solution set of \(-7<\frac{x+2}{3}\le 1\)?
#linear-inequalities
#compound-fraction
#class-11
#expert
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A \(x\le -23\)
B (x>1)
C \(-23<x\le 1\)
D \(-23\le x<1\)
Explanation opens after your attempt
Correct Answer
C. \(-23<x\le 1\)
Step 1
Concept
Multiplying by (3) gives \(-21<x+2\le 3\). Therefore \(-23<x\le 1\) is the solution.
Step 2
Why this answer is correct
The correct answer is C. \(-23<x\le 1\). Multiplying by (3) gives \(-21<x+2\le 3\). Therefore \(-23<x\le 1\) is the solution.
Step 3
Exam Tip
(3) से गुणा करने पर \(-21<x+2\le 3\) मिलता है। इसलिए \(-23<x\le 1\) हल है।
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यदि \(\frac{3x-1}{2}\ge x+4\) और (2x-7<5), तो (x) का संयुक्त हल क्या है?
If \(\frac{3x-1}{2}\ge x+4\) and (2x-7<5), what is the combined solution for (x)?
#linear-inequalities
#and-condition
#class-11
#expert
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A \(x\ge 9\)
B \(\varnothing\)
C (x<6)
D (6<x<9)
Explanation opens after your attempt
Correct Answer
B. \(\varnothing\)
Step 1
Concept
The first inequality gives \(x\ge 9\), and the second gives (x<6). Their intersection is empty.
Step 2
Why this answer is correct
The correct answer is B. \(\varnothing\). The first inequality gives \(x\ge 9\), and the second gives (x<6). Their intersection is empty.
Step 3
Exam Tip
पहली असमानता से \(x\ge 9\) और दूसरी से (x<6) मिलता है। दोनों का प्रतिच्छेद रिक्त है।
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यदि \(\frac{x-5}{4}<2\) या \(3x+1\le -8\), तो संयुक्त हल क्या है?
If \(\frac{x-5}{4}<2\) or \(3x+1\le -8\), what is the combined solution?
#linear-inequalities
#or-condition
#class-11
#expert
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A (x<13)
B \(x\le -3\)
C (x>13)
D (-3<x<13)
Explanation opens after your attempt
Step 1
Concept
The first inequality gives (x<13), and the second gives \(x\le -3\). The second set is contained in the first, so the answer is (x<13).
Step 2
Why this answer is correct
The correct answer is A. (x<13). The first inequality gives (x<13), and the second gives \(x\le -3\). The second set is contained in the first, so the answer is (x<13).
Step 3
Exam Tip
पहली असमानता से (x<13) और दूसरी से \(x\le -3\) मिलता है। दूसरी शर्त पहले हल में शामिल है, इसलिए उत्तर (x<13) है।
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असमानता (5(1-2x)\le 3-4(2x+1)) को हल कीजिए।
Solve the inequality (5(1-2x)\le 3-4(2x+1)).
#linear-inequalities
#sign-handling
#class-11
#expert
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A \(x\le 3\)
B (x<3)
C (x>3)
D \(x\ge 3\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge 3\)
Step 1
Concept
Simplification gives \(6\le 2x\). Hence \(x\ge 3\) is the correct solution.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge 3\). Simplification gives \(6\le 2x\). Hence \(x\ge 3\) is the correct solution.
Step 3
Exam Tip
सरलीकरण पर \(6\le 2x\) मिलता है। अतः \(x\ge 3\) सही हल है।
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असमानता (2(x+3)-\frac{3x-1}{2}>7) का हल क्या है?
What is the solution of (2(x+3)-\frac{3x-1}{2}>7)?
#linear-inequalities
#linear-fraction
#class-11
#expert
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A (x<1)
B (x>1)
C \(x\ge 1\)
D \(x\le 1\)
Explanation opens after your attempt
Step 1
Concept
After clearing the denominator, (x+13>14). Therefore (x>1) is the solution.
Step 2
Why this answer is correct
The correct answer is B. (x>1). After clearing the denominator, (x+13>14). Therefore (x>1) is the solution.
Step 3
Exam Tip
हर हटाने पर (x+13>14) बनता है। इसलिए (x>1) हल है।
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असमानता \(\frac{4x+9}{5}\le \frac{x-6}{2}\) का हल समुच्चय क्या है?
What is the solution set of \(\frac{4x+9}{5}\le \frac{x-6}{2}\)?
#linear-inequalities
#fraction-comparison
#class-11
#expert
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A \(x\ge -16\)
B (x<-16)
C \(x\le -16\)
D (x>-16)
Explanation opens after your attempt
Correct Answer
C. \(x\le -16\)
Step 1
Concept
Clearing denominators gives \(8x+18\le 5x-30\), so \(3x\le -48\). This gives \(x\le -16\).
Step 2
Why this answer is correct
The correct answer is C. \(x\le -16\). Clearing denominators gives \(8x+18\le 5x-30\), so \(3x\le -48\). This gives \(x\le -16\).
Step 3
Exam Tip
हर हटाने पर \(8x+18\le 5x-30\), इसलिए \(3x\le -48\)। इससे \(x\le -16\) मिलता है।
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असमानता (9-3(x-4)\ge 2(7-x)+x) को हल कीजिए।
Solve the inequality (9-3(x-4)\ge 2(7-x)+x).
#linear-inequalities
#bracket-expansion
#class-11
#expert
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A \(x\le \frac{7}{2}\)
B \(x\ge \frac{7}{2}\)
C \(x< \frac{7}{2}\)
D \(x> \frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{7}{2}\)
Step 1
Concept
Simplification gives \(7\ge 2x\). Therefore \(x\le \frac{7}{2}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{7}{2}\). Simplification gives \(7\ge 2x\). Therefore \(x\le \frac{7}{2}\) is correct.
Step 3
Exam Tip
सरलीकरण पर \(7\ge 2x\) मिलता है। इसलिए \(x\le \frac{7}{2}\) सही है।
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असमानता \(\frac{2x-5}{3}+\frac{x+1}{6}\le \frac{x}{2}\) का हल क्या है?
What is the solution of \(\frac{2x-5}{3}+\frac{x+1}{6}\le \frac{x}{2}\)?
#linear-inequalities
#fraction-sum
#class-11
#expert
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A \(x\ge \frac{9}{2}\)
B \(x<\frac{9}{2}\)
C \(x>\frac{9}{2}\)
D \(x\le \frac{9}{2}\)
Explanation opens after your attempt
Correct Answer
D. \(x\le \frac{9}{2}\)
Step 1
Concept
Clearing denominators gives \(5x-9\le 3x\). This gives \(2x\le 9\), so \(x\le \frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is D. \(x\le \frac{9}{2}\). Clearing denominators gives \(5x-9\le 3x\). This gives \(2x\le 9\), so \(x\le \frac{9}{2}\).
Step 3
Exam Tip
हर हटाने पर \(5x-9\le 3x\) मिलता है। इससे \(2x\le 9\) और \(x\le \frac{9}{2}\) आता है।
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युग्म असमानता \(3x-8<2x+1\le 5x-11\) का हल क्या है?
What is the solution of the compound inequality \(3x-8<2x+1\le 5x-11\)?
#linear-inequalities
#double-inequality
#class-11
#expert
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A (x<4)
B \(4\le x<9\)
C \(x\ge 9\)
D \(4<x\le 9\)
Explanation opens after your attempt
Correct Answer
B. \(4\le x<9\)
Step 1
Concept
The first part gives (x<9), and the second gives \(x\ge 4\). Combining both gives \(4\le x<9\).
Step 2
Why this answer is correct
The correct answer is B. \(4\le x<9\). The first part gives (x<9), and the second gives \(x\ge 4\). Combining both gives \(4\le x<9\).
Step 3
Exam Tip
पहले भाग से (x<9) और दूसरे भाग से \(x\ge 4\) मिलता है। दोनों को मिलाकर \(4\le x<9\) मिलता है।
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युग्म असमानता \(-4\le 5-3x<11\) को हल कीजिए।
Solve the compound inequality \(-4\le 5-3x<11\).
#linear-inequalities
#negative-coefficient
#class-11
#expert
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A \(x\le -2\)
B (x>3)
C \(-2<x\le 3\)
D \(-2\le x<3\)
Explanation opens after your attempt
Correct Answer
C. \(-2<x\le 3\)
Step 1
Concept
Solving both sides separately gives \(x\le 3\) and (x>-2). Therefore \(-2<x\le 3\) is the correct interval.
Step 2
Why this answer is correct
The correct answer is C. \(-2<x\le 3\). Solving both sides separately gives \(x\le 3\) and (x>-2). Therefore \(-2<x\le 3\) is the correct interval.
Step 3
Exam Tip
दोनों तरफ अलग-अलग हल करने पर \(x\le 3\) और (x>-2) मिलता है। इसलिए \(-2<x\le 3\) सही अंतराल है।
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यदि (7x-2(3x+5)\ge 4) और \(\frac{x-1}{2}>3\), तो संयुक्त हल क्या है?
If (7x-2(3x+5)\ge 4) and \(\frac{x-1}{2}>3\), what is the combined solution?
#linear-inequalities
#intersection
#class-11
#expert
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A \(x\ge 14\)
B (x>7)
C (7<x<14)
D \(x\le 14\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 14\)
Step 1
Concept
The first inequality gives \(x\ge 14\), and the second gives (x>7). Their intersection is \(x\ge 14\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 14\). The first inequality gives \(x\ge 14\), and the second gives (x>7). Their intersection is \(x\ge 14\).
Step 3
Exam Tip
पहली असमानता से \(x\ge 14\) और दूसरी से (x>7) मिलता है। प्रतिच्छेद \(x\ge 14\) है।
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यदि \(2x-9\le x-4\) या (5x+3>18), तो संयुक्त हल समुच्चय क्या है?
If \(2x-9\le x-4\) or (5x+3>18), what is the combined solution set?
#linear-inequalities
#union
#class-11
#expert
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A \(x\le 5\)
B (x>3)
C \(\varnothing\)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
D. \(x\in\mathbb{R}\)
Step 1
Concept
The first condition gives \(x\le 5\), and the second gives (x>3). Their union is all real numbers.
Step 2
Why this answer is correct
The correct answer is D. \(x\in\mathbb{R}\). The first condition gives \(x\le 5\), and the second gives (x>3). Their union is all real numbers.
Step 3
Exam Tip
पहली शर्त \(x\le 5\) देती है और दूसरी (x>3) देती है। दोनों का संघ सभी वास्तविक संख्याएँ है।
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असमानता (\frac{1}{3}(6x-9)-\frac{1}{2}(4x+2)\ge -5) का हल समुच्चय क्या है?
What is the solution set of (\frac{1}{3}(6x-9)-\frac{1}{2}(4x+2)\ge -5)?
#linear-inequalities
#always-true
#class-11
#expert
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A \(\varnothing\)
B \(x\in\mathbb{R}\)
C \(x\ge -5\)
D \(x\le -5\)
Explanation opens after your attempt
Correct Answer
B. \(x\in\mathbb{R}\)
Step 1
Concept
The left side becomes (-4), and \(-4\ge -5\) is true. Therefore every real (x) is a solution.
Step 2
Why this answer is correct
The correct answer is B. \(x\in\mathbb{R}\). The left side becomes (-4), and \(-4\ge -5\) is true. Therefore every real (x) is a solution.
Step 3
Exam Tip
बायाँ पक्ष (-4) बनता है और \(-4\ge -5\) सत्य है। इसलिए हर वास्तविक (x) हल है।
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असमानता (\frac{3}{4}(x-8)<\frac{1}{2}(x+2)-5) को हल कीजिए।
Solve the inequality (\frac{3}{4}(x-8)<\frac{1}{2}(x+2)-5).
#linear-inequalities
#coefficient-fraction
#class-11
#expert
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A (x>8)
B \(x\le 8\)
C (x<8)
D \(x\ge 8\)
Explanation opens after your attempt
Step 1
Concept
Simplification gives \(\frac{x}{4}<2\). Therefore (x<8) is the solution.
Step 2
Why this answer is correct
The correct answer is C. (x<8). Simplification gives \(\frac{x}{4}<2\). Therefore (x<8) is the solution.
Step 3
Exam Tip
सरलीकरण पर \(\frac{x}{4}<2\) मिलता है। इसलिए (x<8) हल है।
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असमानता (11-2(5-x)>3x+4) का हल क्या है?
What is the solution of (11-2(5-x)>3x+4)?
#linear-inequalities
#negative-division
#class-11
#expert
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A (x<-3)
B (x>-3)
C \(x\le -3\)
D \(x\ge -3\)
Explanation opens after your attempt
Step 1
Concept
Simplification gives (-x>3). Therefore (x<-3) is correct.
Step 2
Why this answer is correct
The correct answer is A. (x<-3). Simplification gives (-x>3). Therefore (x<-3) is correct.
Step 3
Exam Tip
सरलीकरण से (-x>3) मिलता है। इसलिए (x<-3) सही है।
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असमानता \(\frac{x+7}{3}-\frac{2x-1}{4}\ge 0\) को हल कीजिए।
Solve the inequality \(\frac{x+7}{3}-\frac{2x-1}{4}\ge 0\).
#linear-inequalities
#fraction-difference
#class-11
#expert
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A \(x\ge \frac{31}{2}\)
B \(x<\frac{31}{2}\)
C \(x>\frac{31}{2}\)
D \(x\le \frac{31}{2}\)
Explanation opens after your attempt
Correct Answer
D. \(x\le \frac{31}{2}\)
Step 1
Concept
Clearing denominators gives \(31-2x\ge 0\). Hence \(x\le \frac{31}{2}\) is correct.
Step 2
Why this answer is correct
The correct answer is D. \(x\le \frac{31}{2}\). Clearing denominators gives \(31-2x\ge 0\). Hence \(x\le \frac{31}{2}\) is correct.
Step 3
Exam Tip
हर हटाने पर \(31-2x\ge 0\) मिलता है। अतः \(x\le \frac{31}{2}\) सही है।
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असमानता (6x-5<2(3x-4)) का हल समुच्चय क्या है?
What is the solution set of (6x-5<2(3x-4))?
#linear-inequalities
#no-real-solution
#class-11
#expert
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A (x<3)
B (x>3)
C \(\varnothing\)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
C. \(\varnothing\)
Step 1
Concept
Simplification gives (-5<-8), which is false. Therefore there is no real solution.
Step 2
Why this answer is correct
The correct answer is C. \(\varnothing\). Simplification gives (-5<-8), which is false. Therefore there is no real solution.
Step 3
Exam Tip
सरलीकरण पर (-5<-8) मिलता है, जो असत्य है। इसलिए कोई वास्तविक हल नहीं है।
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असमानता (2(4-x)\le \frac{x+9}{3}) को हल कीजिए।
Solve the inequality (2(4-x)\le \frac{x+9}{3}).
#linear-inequalities
#mixed-linear
#class-11
#expert
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A \(x\le \frac{15}{7}\)
B \(x\ge \frac{15}{7}\)
C \(x>\frac{15}{7}\)
D \(x<\frac{15}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge \frac{15}{7}\)
Step 1
Concept
Clearing the denominator gives \(24-6x\le x+9\). Hence \(15\le 7x\), so \(x\ge \frac{15}{7}\).
Step 2
Why this answer is correct
The correct answer is B. \(x\ge \frac{15}{7}\). Clearing the denominator gives \(24-6x\le x+9\). Hence \(15\le 7x\), so \(x\ge \frac{15}{7}\).
Step 3
Exam Tip
हर हटाने पर \(24-6x\le x+9\) मिलता है। इसलिए \(15\le 7x\) और \(x\ge \frac{15}{7}\) आता है।
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युग्म असमानता \(-6\le \frac{4-x}{2}<3\) को हल कीजिए।
Solve the compound inequality \(-6\le \frac{4-x}{2}<3\).
#linear-inequalities
#compound-negative-x
#class-11
#expert
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A \(x\le -2\)
B (x>16)
C \(-2\le x<16\)
D \(-2<x\le 16\)
Explanation opens after your attempt
Correct Answer
D. \(-2<x\le 16\)
Step 1
Concept
Solving both parts gives \(x\le 16\) and (x>-2). Therefore \(-2<x\le 16\) is the solution.
Step 2
Why this answer is correct
The correct answer is D. \(-2<x\le 16\). Solving both parts gives \(x\le 16\) and (x>-2). Therefore \(-2<x\le 16\) is the solution.
Step 3
Exam Tip
दोनों भाग हल करने पर \(x\le 16\) और (x>-2) मिलता है। इसलिए \(-2<x\le 16\) हल है।
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यदि \(\frac{x-2}{3}\ge \frac{2x+1}{5}\) और \(x+4\le 10\), तो संयुक्त हल क्या है?
If \(\frac{x-2}{3}\ge \frac{2x+1}{5}\) and \(x+4\le 10\), what is the combined solution?
#linear-inequalities
#intersection-fraction
#class-11
#expert
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A \(x\le 6\)
B \(x\le -13\)
C \(-13<x\le 6\)
D \(x\ge -13\)
Explanation opens after your attempt
Correct Answer
B. \(x\le -13\)
Step 1
Concept
The first inequality gives \(x\le -13\), and the second gives \(x\le 6\). Their intersection is \(x\le -13\).
Step 2
Why this answer is correct
The correct answer is B. \(x\le -13\). The first inequality gives \(x\le -13\), and the second gives \(x\le 6\). Their intersection is \(x\le -13\).
Step 3
Exam Tip
पहली असमानता से \(x\le -13\) और दूसरी से \(x\le 6\) मिलता है। दोनों का प्रतिच्छेद \(x\le -13\) है।
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असमानता \(\frac{5x+2}{6}<\frac{x-3}{2}+4\) को हल कीजिए।
Solve the inequality \(\frac{5x+2}{6}<\frac{x-3}{2}+4\).
#linear-inequalities
#fraction-linear
#class-11
#expert
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A \(x>\frac{13}{2}\)
B \(x\le \frac{13}{2}\)
C \(x<\frac{13}{2}\)
D \(x\ge \frac{13}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x<\frac{13}{2}\)
Step 1
Concept
Clearing denominators gives (5x+2<3x+15). Thus (2x<13), so \(x<\frac{13}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(x<\frac{13}{2}\). Clearing denominators gives (5x+2<3x+15). Thus (2x<13), so \(x<\frac{13}{2}\).
Step 3
Exam Tip
हर हटाने पर (5x+2<3x+15) मिलता है। इससे (2x<13), अतः \(x<\frac{13}{2}\) है।
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असमानता (8-3(2-x)\le 5x-10) का हल क्या है?
What is the solution of (8-3(2-x)\le 5x-10)?
#linear-inequalities
#bracket-linear
#class-11
#expert
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A \(x\ge 6\)
B \(x\le 6\)
C (x>6)
D (x<6)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 6\)
Step 1
Concept
Simplification gives \(12\le 2x\). Therefore \(x\ge 6\) is the correct solution.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 6\). Simplification gives \(12\le 2x\). Therefore \(x\ge 6\) is the correct solution.
Step 3
Exam Tip
सरलीकरण पर \(12\le 2x\) मिलता है। इसलिए \(x\ge 6\) सही हल है।
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यदि (4x+1>2x+9) या (x-5<-12), तो संयुक्त हल क्या है?
If (4x+1>2x+9) or (x-5<-12), what is the combined solution?
#linear-inequalities
#disjoint-union
#class-11
#expert
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A (x>4)
B (x<-7)
C (-7<x<4)
D (x<-7) या (x>4) / (x<-7) or (x>4)
Explanation opens after your attempt
Correct Answer
D. (x<-7) या (x>4) / (x<-7) or (x>4)
Step 1
Concept
The first inequality gives (x>4), and the second gives (x<-7). For an OR condition, take the union of both solution sets.
Step 2
Why this answer is correct
The correct answer is D. (x<-7) या (x>4) / (x<-7) or (x>4). The first inequality gives (x>4), and the second gives (x<-7). For an OR condition, take the union of both solution sets.
Step 3
Exam Tip
पहली असमानता (x>4) देती है और दूसरी (x<-7) देती है। या वाली स्थिति में दोनों हलों का संघ लिया जाता है।
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असमानता \(\frac{7-2x}{3}\le \frac{x+5}{6}\) को हल कीजिए।
Solve the inequality \(\frac{7-2x}{3}\le \frac{x+5}{6}\).
#linear-inequalities
#fraction-comparison
#class-11
#expert
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A \(x\le \frac{9}{5}\)
B \(x\ge \frac{9}{5}\)
C \(x>\frac{9}{5}\)
D \(x<\frac{9}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge \frac{9}{5}\)
Step 1
Concept
Clearing denominators gives \(14-4x\le x+5\). Therefore \(9\le 5x\), so \(x\ge \frac{9}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(x\ge \frac{9}{5}\). Clearing denominators gives \(14-4x\le x+5\). Therefore \(9\le 5x\), so \(x\ge \frac{9}{5}\).
Step 3
Exam Tip
हर हटाने पर \(14-4x\le x+5\) मिलता है। इसलिए \(9\le 5x\) और \(x\ge \frac{9}{5}\) है।
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युग्म असमानता \(5\le 3x-1<2x+11\) का हल क्या है?
What is the solution of the compound inequality \(5\le 3x-1<2x+11\)?
#linear-inequalities
#compound-variable-bound
#class-11
#expert
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A (x<2)
B \(x\ge 12\)
C \(2\le x<12\)
D \(2<x\le 12\)
Explanation opens after your attempt
Correct Answer
C. \(2\le x<12\)
Step 1
Concept
The first part gives \(x\ge 2\), and the second gives (x<12). Together they give \(2\le x<12\).
Step 2
Why this answer is correct
The correct answer is C. \(2\le x<12\). The first part gives \(x\ge 2\), and the second gives (x<12). Together they give \(2\le x<12\).
Step 3
Exam Tip
पहले भाग से \(x\ge 2\) और दूसरे भाग से (x<12) मिलता है। दोनों मिलकर \(2\le x<12\) देते हैं।
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असमानता (-2(3x-4)+7\ge x-6) को हल कीजिए।
Solve the inequality (-2(3x-4)+7\ge x-6).
#linear-inequalities
#negative-bracket
#class-11
#expert
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A \(x\le 3\)
B \(x\ge 3\)
C (x<3)
D (x>3)
Explanation opens after your attempt
Correct Answer
A. \(x\le 3\)
Step 1
Concept
Simplification gives \(21\ge 7x\). Therefore \(x\le 3\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 3\). Simplification gives \(21\ge 7x\). Therefore \(x\le 3\) is correct.
Step 3
Exam Tip
सरलीकरण पर \(21\ge 7x\) मिलता है। इसलिए \(x\le 3\) सही है।
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असमानता \(\frac{3x+8}{4}-\frac{x-6}{2}>5\) का हल क्या है?
What is the solution of \(\frac{3x+8}{4}-\frac{x-6}{2}>5\)?
#linear-inequalities
#fraction-subtraction
#class-11
#expert
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A (x<0)
B (x>0)
C \(x\le 0\)
D \(x\ge 0\)
Explanation opens after your attempt
Step 1
Concept
Clearing denominators gives (x+20>20). Therefore (x>0) is the solution.
Step 2
Why this answer is correct
The correct answer is B. (x>0). Clearing denominators gives (x+20>20). Therefore (x>0) is the solution.
Step 3
Exam Tip
हर हटाने पर (x+20>20) मिलता है। इसलिए (x>0) हल है।
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असमानता (9x-4\le 3(3x-1)-2) का हल समुच्चय क्या है?
What is the solution set of (9x-4\le 3(3x-1)-2)?
#linear-inequalities
#contradiction
#class-11
#expert
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A \(x\le 1\)
B \(x\ge 1\)
C \(x\in\mathbb{R}\)
D \(\varnothing\)
Explanation opens after your attempt
Correct Answer
D. \(\varnothing\)
Step 1
Concept
Simplification gives \(-4\le -5\), which is false. Therefore the solution set is empty.
Step 2
Why this answer is correct
The correct answer is D. \(\varnothing\). Simplification gives \(-4\le -5\), which is false. Therefore the solution set is empty.
Step 3
Exam Tip
सरलीकरण पर \(-4\le -5\) मिलता है, जो असत्य है। इसलिए हल समुच्चय रिक्त है।
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असमानता \(\frac{x}{2}-\frac{x-3}{6}\le 4\) को हल कीजिए।
Solve the inequality \(\frac{x}{2}-\frac{x-3}{6}\le 4\).
#linear-inequalities
#simple-fraction
#class-11
#expert
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A \(x\ge \frac{21}{2}\)
B \(x<\frac{21}{2}\)
C \(x\le \frac{21}{2}\)
D \(x>\frac{21}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x\le \frac{21}{2}\)
Step 1
Concept
Clearing denominators gives \(2x+3\le 24\). Therefore \(x\le \frac{21}{2}\) is correct.
Step 2
Why this answer is correct
The correct answer is C. \(x\le \frac{21}{2}\). Clearing denominators gives \(2x+3\le 24\). Therefore \(x\le \frac{21}{2}\) is correct.
Step 3
Exam Tip
हर हटाने पर \(2x+3\le 24\) मिलता है। इसलिए \(x\le \frac{21}{2}\) सही है।
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यदि \(2x+5\ge 13\) और \(-x+4>1\), तो संयुक्त हल क्या है?
If \(2x+5\ge 13\) and (-x+4>1), what is the combined solution?
#linear-inequalities
#empty-intersection
#class-11
#expert
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A \(\varnothing\)
B \(x\ge 4\)
C (x<3)
D (3<x<4)
Explanation opens after your attempt
Correct Answer
A. \(\varnothing\)
Step 1
Concept
The first inequality gives \(x\ge 4\), and the second gives (x<3). Both conditions cannot hold together.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). The first inequality gives \(x\ge 4\), and the second gives (x<3). Both conditions cannot hold together.
Step 3
Exam Tip
पहली असमानता से \(x\ge 4\) और दूसरी से (x<3) मिलता है। दोनों शर्तें साथ में पूरी नहीं हो सकतीं।
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युग्म असमानता \(7-4x<3(2-x)\le 12\) को हल कीजिए।
Solve the compound inequality \(7-4x<3(2-x)\le 12\).
#linear-inequalities
#compound-with-brackets
#class-11
#expert
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A \(x\ge -2\)
B \(x\le 1\)
C \(-2\le x<1\)
D (x>1)
Explanation opens after your attempt
Step 1
Concept
The first part gives (x>1), and the second gives \(x\ge -2\). Their combined solution is (x>1).
Step 2
Why this answer is correct
The correct answer is D. (x>1). The first part gives (x>1), and the second gives \(x\ge -2\). Their combined solution is (x>1).
Step 3
Exam Tip
पहले भाग से (x>1) और दूसरे भाग से \(x\ge -2\) मिलता है। दोनों का संयुक्त हल (x>1) है।
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असमानता \(\frac{2x+1}{7}\ge \frac{x-5}{3}\) का हल क्या है?
What is the solution of \(\frac{2x+1}{7}\ge \frac{x-5}{3}\)?
#linear-inequalities
#cross-multiplication
#class-11
#expert
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A \(x\ge 38\)
B \(x\le 38\)
C (x<38)
D (x>38)
Explanation opens after your attempt
Correct Answer
B. \(x\le 38\)
Step 1
Concept
Clearing denominators gives \(6x+3\ge 7x-35\). Therefore \(x\le 38\) is the solution.
Step 2
Why this answer is correct
The correct answer is B. \(x\le 38\). Clearing denominators gives \(6x+3\ge 7x-35\). Therefore \(x\le 38\) is the solution.
Step 3
Exam Tip
हर हटाने पर \(6x+3\ge 7x-35\) मिलता है। इसलिए \(x\le 38\) हल है।
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असमानता (3(1-x)-2(4-3x)>5x-10) को हल कीजिए।
Solve the inequality (3(1-x)-2(4-3x)>5x-10).
#linear-inequalities
#bracket-simplification
#class-11
#expert
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A \(x>\frac{5}{2}\)
B \(x\le \frac{5}{2}\)
C \(x<\frac{5}{2}\)
D \(x\ge \frac{5}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x<\frac{5}{2}\)
Step 1
Concept
Simplification gives (5>2x). Therefore \(x<\frac{5}{2}\) is correct.
Step 2
Why this answer is correct
The correct answer is C. \(x<\frac{5}{2}\). Simplification gives (5>2x). Therefore \(x<\frac{5}{2}\) is correct.
Step 3
Exam Tip
सरलीकरण पर (5>2x) मिलता है। इसलिए \(x<\frac{5}{2}\) सही है।
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असमानता \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{6}\ge 4\) का हल क्या है?
What is the solution of \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{6}\ge 4\)?
#linear-inequalities
#multiple-fractions
#class-11
#expert
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A \(x\ge \frac{17}{3}\)
B \(x\le \frac{17}{3}\)
C \(x>\frac{17}{3}\)
D \(x<\frac{17}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{17}{3}\)
Step 1
Concept
Clearing denominators gives \(6x-10\ge 24\). Hence \(x\ge \frac{17}{3}\) is the solution.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{17}{3}\). Clearing denominators gives \(6x-10\ge 24\). Hence \(x\ge \frac{17}{3}\) is the solution.
Step 3
Exam Tip
हर हटाने पर \(6x-10\ge 24\) मिलता है। अतः \(x\ge \frac{17}{3}\) हल है।
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युग्म असमानता \(-5<2-\frac{x}{3}\le 4\) को हल कीजिए।
Solve the compound inequality \(-5<2-\frac{x}{3}\le 4\).
#linear-inequalities
#compound-negative-fraction
#class-11
#expert
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A (x<-6)
B \(x\ge 21\)
C \(-6<x\le 21\)
D \(-6\le x<21\)
Explanation opens after your attempt
Correct Answer
D. \(-6\le x<21\)
Step 1
Concept
Solving both parts gives (x<21) and \(x\ge -6\). Therefore \(-6\le x<21\) is correct.
Step 2
Why this answer is correct
The correct answer is D. \(-6\le x<21\). Solving both parts gives (x<21) and \(x\ge -6\). Therefore \(-6\le x<21\) is correct.
Step 3
Exam Tip
दोनों भाग हल करने पर (x<21) और \(x\ge -6\) मिलता है। इसलिए \(-6\le x<21\) सही है।
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असमानता \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\) का हल क्या है?
What is the solution of \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\)?
#linear-inequalities
#negative-coefficient
#class-11
#expert
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A \(x\ge -\frac{1}{5}\)
B \(x<-\frac{1}{5}\)
C \(x\le -\frac{1}{5}\)
D \(x>-\frac{1}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(x\le -\frac{1}{5}\)
Step 1
Concept
Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(x\le -\frac{1}{5}\). Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).
Step 3
Exam Tip
हर हटाने पर \(9-3x\ge 2x+10\) मिलता है। इसलिए \(-5x\ge 1\) और \(x\le -\frac{1}{5}\) है।
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असमानता \(4(x+1)-\frac{2x-3}{3}<\frac{5x+6}{2}\) को हल कीजिए।
Solve the inequality \(4(x+1)-\frac{2x-3}{3}<\frac{5x+6}{2}\).
#linear-inequalities
#final-algebraic-solution
#class-11
#expert
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A \(x>-\frac{12}{5}\)
B \(x<-\frac{12}{5}\)
C \(x\ge -\frac{12}{5}\)
D \(x\le -\frac{12}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(x<-\frac{12}{5}\)
Step 1
Concept
Clearing denominators gives \(20x+30<15x+18\). Thus \(5x<-12\), so \(x<-\frac{12}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(x<-\frac{12}{5}\). Clearing denominators gives \(20x+30<15x+18\). Thus \(5x<-12\), so \(x<-\frac{12}{5}\).
Step 3
Exam Tip
हर हटाने पर \(20x+30<15x+18\) मिलता है। इससे \(5x<-12\), इसलिए \(x<-\frac{12}{5}\) है।
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