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
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 \(\frac{4x+9}{5}\le \frac{x-6}{2}\) 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
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 (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
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 (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
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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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
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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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
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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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
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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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
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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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
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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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
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 \(\frac{x-2}{7}+\frac{x+3}{5}>1\) को हल कीजिए।
Solve the inequality \(\frac{x-2}{7}+\frac{x+3}{5}>1\).
#linear-inequalities
#fraction-addition
#class-11
#expert
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 (-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
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{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
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 (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
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 \(\frac{2x+7}{4}<\frac{3x-1}{6}+\frac{5}{3}\) 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
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 \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\) 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
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 (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
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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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
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 (3(2x-1)-2(x+5)\le x+4) का सही हल चुनिए।
Choose the correct solution of (3(2x-1)-2(x+5)\le x+4).
#linear inequalities
#class 11
#expert
#one variable
A \(x\le17\)
B \(x\ge\frac{17}{3}\)
C \(x\le\frac{17}{3}\)
D \(x\ge17\)
Explanation opens after your attempt
Correct Answer
C. \(x\le\frac{17}{3}\)
Step 1
Concept
Simplification gives \(4x-13\le x+4\). Hence \(3x\le17\), and the final solution is \(x\le\frac{17}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(x\le\frac{17}{3}\). Simplification gives \(4x-13\le x+4\). Hence \(3x\le17\), and the final solution is \(x\le\frac{17}{3}\).
Step 3
Exam Tip
सरलीकरण से \(4x-13\le x+4\) मिलता है। इसलिए \(3x\le17\) और अंतिम हल \(x\le\frac{17}{3}\) है।
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किस (x) के लिए (3(2x-1)-2(x+5)\le x+4) सत्य है?
For which (x) is (3(2x-1)-2(x+5)\le x+4) true?
#linear inequalities
#class 11
#expert
#one variable
A \(x\le17\)
B \(x\ge17\)
C \(x\le\frac{17}{3}\)
D \(x\ge\frac{17}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le17\)
Step 1
Concept
The left side is (6x-3-2x-10=4x-13). From \(4x-13\le x+4\), \(3x\le17\), so \(x\le\frac{17}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le17\). The left side is (6x-3-2x-10=4x-13). From \(4x-13\le x+4\), \(3x\le17\), so \(x\le\frac{17}{3}\).
Step 3
Exam Tip
बाईं ओर (6x-3-2x-10=4x-13) है। \(4x-13\le x+4\) से \(3x\le17\), इसलिए \(x\le\frac{17}{3}\)।
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यदि (x) पूर्णांक है और \(-10<4x+2\le18\), तो (x) का सबसे बड़ा मान क्या है?
If (x) is an integer and \(-10<4x+2\le18\), what is the greatest value of (x)?
#linear inequalities
#class 11
#expert
#one variable
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Subtracting gives \(-12<4x\le16\), so \(-3<x\le4\). The greatest integer solution is (4).
Step 2
Why this answer is correct
The correct answer is B. (4). Subtracting gives \(-12<4x\le16\), so \(-3<x\le4\). The greatest integer solution is (4).
Step 3
Exam Tip
घटाने पर \(-12<4x\le16\), इसलिए \(-3<x\le4\) मिलता है। पूर्णांक हलों में सबसे बड़ा (4) है।
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असमानता \(4-\frac{2x+1}{3}\ge\frac{1-x}{2}\) का हल क्या है?
What is the solution of \(4-\frac{2x+1}{3}\ge\frac{1-x}{2}\)?
#linear inequalities
#class 11
#expert
#one variable
A \(x\le19\)
B \(x\ge19\)
C \(x\le\frac{19}{2}\)
D \(x\ge\frac{19}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le19\)
Step 1
Concept
Multiplying by positive (6) gives (24-2(2x+1)\ge3(1-x)). Thus \(22-4x\ge3-3x\), so \(x\le19\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le19\). Multiplying by positive (6) gives (24-2(2x+1)\ge3(1-x)). Thus \(22-4x\ge3-3x\), so \(x\le19\).
Step 3
Exam Tip
धनात्मक (6) से गुणा करने पर (24-2(2x+1)\ge3(1-x)) मिलता है। इससे \(22-4x\ge3-3x\), इसलिए \(x\le19\)।
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यदि \(x\in\mathbb{R}\), तो (5x-2<3x+10) और \(x+4\ge2\) का संयुक्त हल क्या है?
If \(x\in\mathbb{R}\), what is the combined solution of (5x-2<3x+10) and \(x+4\ge2\)?
#linear inequalities
#class 11
#expert
#one variable
A \(-2\le x<6\)
B \(-2<x\le6\)
C (x<6)
D \(x\ge-2\)
Explanation opens after your attempt
Correct Answer
A. \(-2\le x<6\)
Step 1
Concept
The first inequality gives (x<6), and the second gives \(x\ge-2\). Their intersection is \(-2\le x<6\).
Step 2
Why this answer is correct
The correct answer is A. \(-2\le x<6\). The first inequality gives (x<6), and the second gives \(x\ge-2\). Their intersection is \(-2\le x<6\).
Step 3
Exam Tip
पहली असमानता (x<6) देती है और दूसरी \(x\ge-2\) देती है। दोनों का प्रतिच्छेद \(-2\le x<6\) है।
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किस (x) के लिए \(-1<\frac{x-4}{2}\le 6\) है?
For which (x) is \(-1<\frac{x-4}{2}\le 6\)?
#linear inequalities
#class 11
#expert
#one variable
A \(2<x\le16\)
B \(2\le x<16\)
C (x<2) या \(x\ge16\) / (x<2) or \(x\ge16\)
D \(-2<x\le12\)
Explanation opens after your attempt
Correct Answer
A. \(2<x\le16\)
Step 1
Concept
Multiplying by positive (2) gives \(-2<x-4\le12\). Adding (4) gives \(2<x\le16\).
Step 2
Why this answer is correct
The correct answer is A. \(2<x\le16\). Multiplying by positive (2) gives \(-2<x-4\le12\). Adding (4) gives \(2<x\le16\).
Step 3
Exam Tip
धनात्मक (2) से गुणा करने पर \(-2<x-4\le12\) मिलता है। (4) जोड़ने पर \(2<x\le16\)।
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असमानता \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\) का सही हल समुच्चय क्या है?
What is the correct solution set of \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\)?
#linear inequalities
#class 11
#expert
#one variable
A कोई हल नहीं / no solution
B (x>1)
C सभी वास्तविक (x) / all real (x)
D (x<1)
Explanation opens after your attempt
Correct Answer
C. सभी वास्तविक (x) / all real (x)
Step 1
Concept
After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.
Step 2
Why this answer is correct
The correct answer is C. सभी वास्तविक (x) / all real (x). After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.
Step 3
Exam Tip
सरलीकरण के बाद (1>0) जैसा सत्य कथन मिलता है। इसलिए हर वास्तविक (x) इस असमानता को संतुष्ट करता है।
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यदि \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\), तो हल क्या है?
If \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\), what is the solution?
#linear inequalities
#class 11
#expert
#one variable
A (x<1)
B (x>1)
C (x<-1)
D (x>-1)
Explanation opens after your attempt
Step 1
Concept
Multiplying by positive (8) gives (4x-2(x-3)>2x+5). This reduces to (2x+6>2x+5), always true.
Step 2
Why this answer is correct
The correct answer is A. (x<1). Multiplying by positive (8) gives (4x-2(x-3)>2x+5). This reduces to (2x+6>2x+5), always true.
Step 3
Exam Tip
धनात्मक (8) से गुणा करने पर (4x-2(x-3)>2x+5) मिलता है। इससे (2x+6>2x+5), जो सदैव सत्य है।
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असमानता (2(1-3x)\le 5-4(2x+1)) का हल क्या है?
What is the solution of (2(1-3x)\le 5-4(2x+1))?
#linear inequalities
#class 11
#expert
#one variable
A \(x\le-\frac{1}{2}\)
B \(x\ge-\frac{1}{2}\)
C \(x\le\frac{1}{2}\)
D \(x\ge\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le-\frac{1}{2}\)
Step 1
Concept
Simplification gives \(2-6x\le1-8x\). Thus \(2x\le-1\), so \(x\le-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le-\frac{1}{2}\). Simplification gives \(2-6x\le1-8x\). Thus \(2x\le-1\), so \(x\le-\frac{1}{2}\).
Step 3
Exam Tip
सरलीकरण से \(2-6x\le1-8x\) मिलता है। इससे \(2x\le-1\), इसलिए \(x\le-\frac{1}{2}\)।
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यदि \(x\in\mathbb{Z}\) और \(1\le\frac{5x-2}{3}<8\), तो (x) के कितने मान हैं?
If \(x\in\mathbb{Z}\) and \(1\le\frac{5x-2}{3}<8\), how many values of (x) are there?
#linear inequalities
#class 11
#expert
#one variable
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
This gives \(3\le5x-2<24\), i.e. \(5\le5x<26\). Hence \(1\le x<\frac{26}{5}\), so (x=1,2,3,4,5).
Step 2
Why this answer is correct
The correct answer is B. (5). This gives \(3\le5x-2<24\), i.e. \(5\le5x<26\). Hence \(1\le x<\frac{26}{5}\), so (x=1,2,3,4,5).
Step 3
Exam Tip
इससे \(3\le5x-2<24\), यानी \(5\le5x<26\) मिलता है। इसलिए \(1\le x<\frac{26}{5}\), अतः (x=1,2,3,4,5)।
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असमानता \(\frac{7-2x}{5}\le\frac{3x+1}{10}\) का हल क्या है?
What is the solution of \(\frac{7-2x}{5}\le\frac{3x+1}{10}\)?
#linear inequalities
#class 11
#expert
#one variable
A \(x\ge\frac{13}{7}\)
B \(x\le\frac{13}{7}\)
C \(x\ge\frac{7}{13}\)
D \(x\le-\frac{13}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge\frac{13}{7}\)
Step 1
Concept
Multiplying by positive (10) gives \(14-4x\le3x+1\). Thus \(13\le7x\), so \(x\ge\frac{13}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge\frac{13}{7}\). Multiplying by positive (10) gives \(14-4x\le3x+1\). Thus \(13\le7x\), so \(x\ge\frac{13}{7}\).
Step 3
Exam Tip
धनात्मक (10) से गुणा करने पर \(14-4x\le3x+1\) मिलता है। इससे \(13\le7x\), अतः \(x\ge\frac{13}{7}\)।
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यदि \(\frac{4x-1}{3}-\frac{x+2}{9}\ge x+1\), तो (x) का हल क्या है?
If \(\frac{4x-1}{3}-\frac{x+2}{9}\ge x+1\), what is the solution for (x)?
#linear inequalities
#class 11
#expert
#one variable
A \(x\ge7\)
B \(x\le7\)
C \(x\ge\frac{7}{2}\)
D \(x\le\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge7\)
Step 1
Concept
Multiplying by positive (9) gives (3(4x-1)-(x+2)\ge9x+9). Thus \(2x\ge14\), so \(x\ge7\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge7\). Multiplying by positive (9) gives (3(4x-1)-(x+2)\ge9x+9). Thus \(2x\ge14\), so \(x\ge7\).
Step 3
Exam Tip
धनात्मक (9) से गुणा करने पर (3(4x-1)-(x+2)\ge9x+9) मिलता है। इससे \(2x\ge14\), अतः \(x\ge7\)।
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