असमानता (3x-7<11) का हल समुच्चय क्या है?
What is the solution set of the inequality (3x-7<11)?
#linear inequalities
#algebraic solution
#class 11
#hard
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A (x<6)
B (x>6)
C \(x\le 6\)
D \(x\ge 6\)
Explanation opens after your attempt
Explanation
Simple Explanation
(3x<18) से (x<6) मिलता है। परीक्षा में दोनों पक्षों पर समान क्रिया करने का ध्यान रखें। / From (3x<18), we get (x<6). In exams, keep applying the same operation on both sides.
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असमानता \(5-2x\ge 17\) को हल कीजिए।
Solve the inequality \(5-2x\ge 17\).
#linear inequalities
#negative coefficient
#sign reversal
#class 11
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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
B. \(x\le -6\)
Explanation
Simple Explanation
\(-2x\ge 12\) में ऋणात्मक संख्या से भाग देने पर चिह्न बदलता है। इसलिए \(x\le -6\) होगा। / In \(-2x\ge 12\), dividing by a negative number reverses the sign. Hence \(x\le -6\).
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असमानता \(\frac{x-3}{4}+2>\frac{x}{2}\) का हल क्या है?
What is the solution of \(\frac{x-3}{4}+2>\frac{x}{2}\)?
#linear inequalities
#fractions
#one variable
#class 11
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A (x>5)
B \(x\ge 5\)
C (x<5)
D \(x\le 5\)
Explanation opens after your attempt
Explanation
Simple Explanation
हरों को हटाने पर (x+5>2x) मिलता है। अतः (x<5) सही है। / After clearing denominators, (x+5>2x) is obtained. Therefore (x<5) is correct.
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असमानता \(0.3x-1.2\le 0.6\) का हल समुच्चय चुनिए।
Choose the solution set of \(0.3x-1.2\le 0.6\).
#linear inequalities
#decimals
#solution set
#class 11
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A (x<6)
B \(x\ge 6\)
C (x>6)
D \(x\le 6\)
Explanation opens after your attempt
Correct Answer
D. \(x\le 6\)
Explanation
Simple Explanation
\(0.3x\le 1.8\) से \(x\le 6\) मिलता है। दशमलव को भिन्न में बदलकर भी हल किया जा सकता है। / From \(0.3x\le 1.8\), we get \(x\le 6\). Decimals can also be converted into fractions.
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असमानता (-4(2x-1)<3(1-x)+5) को हल कीजिए।
Solve the inequality (-4(2x-1)<3(1-x)+5).
#linear inequalities
#brackets
#negative division
#class 11
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A \(x>-\frac{4}{5}\)
B \(x<-\frac{4}{5}\)
C \(x\ge -\frac{4}{5}\)
D \(x\le -\frac{4}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{4}{5}\)
Explanation
Simple Explanation
सरलीकरण से (-5x<4) मिलता है। ऋणात्मक से भाग देने पर उत्तर \(x>-\frac{4}{5}\) है। / Simplification gives (-5x<4). Dividing by a negative gives \(x>-\frac{4}{5}\).
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असमानता (7-(3x+2)\ge 2x-10) का सही हल कौन सा है?
Which is the correct solution of (7-(3x+2)\ge 2x-10)?
#linear inequalities
#brackets
#comparison
#class 11
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A \(x\ge 3\)
B \(x\le 3\)
C (x<3)
D (x>3)
Explanation opens after your attempt
Correct Answer
B. \(x\le 3\)
Explanation
Simple Explanation
सरलीकरण के बाद \(15\ge 5x\) आता है। इसलिए \(x\le 3\) सही है। / After simplification, \(15\ge 5x\) is obtained. Hence \(x\le 3\) is correct.
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असमानता \(\frac{2x+1}{3}\le \frac{x-4}{2}\) का हल समुच्चय क्या होगा?
What will be the solution set of \(\frac{2x+1}{3}\le \frac{x-4}{2}\)?
#linear inequalities
#fraction inequality
#hard
#class 11
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A \(x\le -14\)
B (x>-14)
C \(x\ge -14\)
D (x<-14)
Explanation opens after your attempt
Correct Answer
A. \(x\le -14\)
Explanation
Simple Explanation
(6) से गुणा करने पर \(4x+2\le 3x-12\) मिलता है। इससे \(x\le -14\) आता है। / Multiplying by (6) gives \(4x+2\le 3x-12\). This gives \(x\le -14\).
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यदि (2(3x-5)-4(x+1)>8), तो (x) के लिए सही शर्त क्या है?
If (2(3x-5)-4(x+1)>8), what is the correct condition for (x)?
#linear inequalities
#expansion
#algebraic solution
#class 11
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A (x>11)
B (x<11)
C \(x\le 11\)
D \(x\ge 11\)
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Explanation
Simple Explanation
बायाँ पक्ष (2x-14) बनता है। (2x-14>8) से (x>11) मिलता है। / The left side becomes (2x-14). From (2x-14>8), we get (x>11).
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असमानता \(\frac{5x-2}{7}<\frac{3x+8}{14}\) का हल चुनिए।
Choose the solution of \(\frac{5x-2}{7}<\frac{3x+8}{14}\).
#linear inequalities
#rational coefficients
#class 11
#hard
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A \(x<\frac{12}{7}\)
B \(x>\frac{12}{7}\)
C \(x\le \frac{12}{7}\)
D \(x\ge \frac{12}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x<\frac{12}{7}\)
Explanation
Simple Explanation
(14) से गुणा करने पर (10x-4<3x+8) मिलता है। इसलिए (7x<12) और \(x<\frac{12}{7}\)। / Multiplying by (14) gives (10x-4<3x+8). So (7x<12) and \(x<\frac{12}{7}\).
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असमानता \(-\frac{x}{3}+4\le \frac{2x}{5}-1\) को हल कीजिए।
Solve the inequality \(-\frac{x}{3}+4\le \frac{2x}{5}-1\).
#linear inequalities
#fraction solution
#class 11
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A \(x\le \frac{75}{11}\)
B \(x\ge \frac{75}{11}\)
C \(x<\frac{75}{11}\)
D \(x>\frac{75}{11}\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge \frac{75}{11}\)
Explanation
Simple Explanation
हर हटाने पर \(-5x+60\le 6x-15\) मिलता है। इससे \(75\le 11x\), अतः \(x\ge \frac{75}{11}\)। / Clearing denominators gives \(-5x+60\le 6x-15\). Thus \(75\le 11x\), so \(x\ge \frac{75}{11}\).
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असमानता (9-2(4x-3)<5(x+2)) का हल क्या है?
What is the solution of (9-2(4x-3)<5(x+2))?
#linear inequalities
#bracket simplification
#class 11
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A \(x>-\frac{5}{13}\)
B \(x<-\frac{5}{13}\)
C \(x\ge -\frac{5}{13}\)
D \(x\le -\frac{5}{13}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{5}{13}\)
Explanation
Simple Explanation
सरलीकरण से (15-8x<5x+10) मिलता है। इससे (5<13x), इसलिए \(x>\frac{5}{13}\) नहीं बल्कि \(x>\frac{5}{13}\) होता है। / Simplification gives (15-8x<5x+10). This gives (5<13x), so \(x>\frac{5}{13}\).
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असमानता (4x-9\ge 2(1-x)+15) का हल समुच्चय बताइए।
Find the solution set of (4x-9\ge 2(1-x)+15).
#linear inequalities
#solution set
#class 11
#hard
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A \(x\ge \frac{13}{3}\)
B \(x\le \frac{13}{3}\)
C \(x>\frac{13}{3}\)
D \(x<\frac{13}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{13}{3}\)
Explanation
Simple Explanation
दाएँ पक्ष को सरल करने पर (17-2x) मिलता है। \(4x-9\ge 17-2x\) से \(x\ge \frac{13}{3}\)। / The right side simplifies to (17-2x). From \(4x-9\ge 17-2x\), \(x\ge \frac{13}{3}\).
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असमानता \(\frac{3x+5}{4}-\frac{x-1}{2}\ge 6\) को हल करें।
Solve the inequality \(\frac{3x+5}{4}-\frac{x-1}{2}\ge 6\).
#linear inequalities
#fraction simplification
#class 11
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A \(x\ge 15\)
B \(x\le 15\)
C (x>15)
D (x<15)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 15\)
Explanation
Simple Explanation
बायाँ पक्ष \(\frac{x+7}{4}\) बनता है। \(\frac{x+7}{4}\ge 6\) से \(x\ge 17\) होना चाहिए। / The left side becomes \(\frac{x+7}{4}\). From \(\frac{x+7}{4}\ge 6\), \(x\ge 17\) should result.
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यदि \(\frac{4-x}{6}>\frac{x+2}{3}\), तो (x) का सही अंतराल क्या है?
If \(\frac{4-x}{6}>\frac{x+2}{3}\), what is the correct interval for (x)?
#linear inequalities
#interval
#sign reversal
#class 11
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A (x<0)
B (x>0)
C \(x\le 0\)
D \(x\ge 0\)
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Explanation
Simple Explanation
(6) से गुणा करने पर (4-x>2x+4) मिलता है। इससे (-3x>0), इसलिए (x<0)। / Multiplying by (6) gives (4-x>2x+4). Thus (-3x>0), so (x<0).
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असमानता (1.5x+2.4<0.6x-3) का हल क्या है?
What is the solution of (1.5x+2.4<0.6x-3)?
#linear inequalities
#decimals
#class 11
#hard
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A (x<-6)
B (x>-6)
C \(x\le -6\)
D \(x\ge -6\)
Explanation opens after your attempt
Explanation
Simple Explanation
(0.9x<-5.4) से (x<-6) मिलता है। दशमलव वाले प्रश्नों में स्थान मान ध्यान से रखें। / From (0.9x<-5.4), we get (x<-6). In decimal problems, handle place values carefully.
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असमानता \(2x-\frac{3}{5}\ge \frac{x}{2}+\frac{9}{10}\) के लिए (x) की न्यूनतम सीमा क्या है?
What is the lower bound for (x) in \(2x-\frac{3}{5}\ge \frac{x}{2}+\frac{9}{10}\)?
#linear inequalities
#fractions
#lower bound
#class 11
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A \(x\ge 1\)
B \(x\le 1\)
C (x>1)
D (x<1)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 1\)
Explanation
Simple Explanation
(10) से गुणा करने पर \(20x-6\ge 5x+9\) मिलता है। इससे \(15x\ge 15\), इसलिए \(x\ge 1\)। / Multiplying by (10) gives \(20x-6\ge 5x+9\). Thus \(15x\ge 15\), so \(x\ge 1\).
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असमानता (6-5(x-2)\le 3(2-x)) को हल कीजिए।
Solve the inequality (6-5(x-2)\le 3(2-x)).
#linear inequalities
#brackets
#algebra
#class 11
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A \(x\ge 5\)
B \(x\le 5\)
C (x<5)
D (x>5)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 5\)
Explanation
Simple Explanation
सरलीकरण से \(16-5x\le 6-3x\) मिलता है। इससे \(10\le 2x\), अतः \(x\ge 5\)। / Simplification gives \(16-5x\le 6-3x\). Thus \(10\le 2x\), hence \(x\ge 5\).
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यदि (3(x-4)+2(x+1)\le 5x-7), तो हल क्या होगा?
If (3(x-4)+2(x+1)\le 5x-7), what will be the solution?
#linear inequalities
#identity case
#all real numbers
#class 11
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A सभी वास्तविक संख्याएँ / All real numbers
B कोई हल नहीं / No solution
C \(x\le 7\)
D \(x\ge 7\)
Explanation opens after your attempt
Correct Answer
B. कोई हल नहीं / No solution
Explanation
Simple Explanation
बायाँ पक्ष (5x-10) है और असमानता \(5x-10\le 5x-7\) हमेशा सत्य है। इसलिए सही उत्तर सभी वास्तविक संख्याएँ होना चाहिए। / The left side is (5x-10), and \(5x-10\le 5x-7\) is always true. Hence the answer should be all real numbers.
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असमानता (4(2x+3)>8x+15) के लिए सही निष्कर्ष क्या है?
What is the correct conclusion for (4(2x+3)>8x+15)?
#linear inequalities
#no solution
#class 11
#hard
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A सभी वास्तविक संख्याएँ / All real numbers
B (x>3)
C कोई हल नहीं / No solution
D (x<3)
Explanation opens after your attempt
Correct Answer
C. कोई हल नहीं / No solution
Explanation
Simple Explanation
सरलीकरण से (8x+12>8x+15) अर्थात (12>15) मिलता है। यह असत्य है, इसलिए कोई हल नहीं। / Simplification gives (8x+12>8x+15), i.e. (12>15). This is false, so there is no solution.
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असमानता \(\frac{7-2x}{5}\le \frac{3x+1}{10}\) का हल चुनिए।
Choose the solution of \(\frac{7-2x}{5}\le \frac{3x+1}{10}\).
#linear inequalities
#fraction inequality
#class 11
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A \(x\ge \frac{13}{7}\)
B \(x\le \frac{13}{7}\)
C \(x>\frac{13}{7}\)
D \(x<\frac{13}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{13}{7}\)
Explanation
Simple Explanation
(10) से गुणा करने पर \(14-4x\le 3x+1\) मिलता है। इससे \(13\le 7x\), अतः \(x\ge \frac{13}{7}\)। / Multiplying by (10) gives \(14-4x\le 3x+1\). Thus \(13\le 7x\), so \(x\ge \frac{13}{7}\).
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असमानता (-3(x+2)+7\ge 2(4-x)-5) का हल क्या है?
What is the solution of (-3(x+2)+7\ge 2(4-x)-5)?
#linear inequalities
#sign handling
#class 11
#hard
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A \(x\le -2\)
B \(x\ge -2\)
C (x<-2)
D (x>-2)
Explanation opens after your attempt
Correct Answer
A. \(x\le -2\)
Explanation
Simple Explanation
सरलीकरण से \(1-3x\ge 3-2x\) मिलता है। इससे \(-2\ge x\), यानी \(x\le -2\)। / Simplification gives \(1-3x\ge 3-2x\). Thus \(-2\ge x\), i.e. \(x\le -2\).
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असमानता \(\frac{x+6}{8}-\frac{x-2}{4}<1\) को हल करें।
Solve the inequality \(\frac{x+6}{8}-\frac{x-2}{4}<1\).
#linear inequalities
#fraction subtraction
#class 11
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? Hint Small clue
A (x>-2)
B (x<-2)
C \(x\ge -2\)
D \(x\le -2\)
Explanation opens after your attempt
Explanation
Simple Explanation
बायाँ पक्ष \(\frac{10-x}{8}\) बनता है। \(\frac{10-x}{8}<1\) से (x>2) मिलता है। / The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
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असमानता (11-3x>2x+1) का हल अंतराल रूप में क्या है?
What is the interval-form solution of (11-3x>2x+1)?
#linear inequalities
#interval form
#class 11
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A (x<2)
B (x>2)
C \(x\le 2\)
D \(x\ge 2\)
Explanation opens after your attempt
Explanation
Simple Explanation
(10>5x) से (x<2) मिलता है। अंतराल में यह (\(-\infty,2\)) होगा। / From (10>5x), (x<2) is obtained. In interval form, it is (\(-\infty,2\)).
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यदि (-7x+4<18-2x), तो (x) किससे बड़ा होगा?
If (-7x+4<18-2x), then (x) will be greater than what?
#linear inequalities
#negative division
#class 11
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A \(x>-\frac{14}{5}\)
B \(x<-\frac{14}{5}\)
C \(x\ge -\frac{14}{5}\)
D \(x\le -\frac{14}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{14}{5}\)
Explanation
Simple Explanation
(-5x<14) प्राप्त होता है। ऋणात्मक से भाग देने पर \(x>-\frac{14}{5}\) होगा। / We get (-5x<14). Dividing by a negative gives \(x>-\frac{14}{5}\).
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असमानता (2.5(2x-1)\ge 1.5(x+3)) का हल क्या है?
What is the solution of (2.5(2x-1)\ge 1.5(x+3))?
#linear inequalities
#decimal coefficients
#class 11
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? Hint Small clue
A \(x\ge 2\)
B \(x\le 2\)
C (x>2)
D (x<2)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 2\)
Explanation
Simple Explanation
सरलीकरण से \(5x-2.5\ge 1.5x+4.5\) मिलता है। इससे \(3.5x\ge 7\), अतः \(x\ge 2\)। / Simplification gives \(5x-2.5\ge 1.5x+4.5\). Thus \(3.5x\ge 7\), so \(x\ge 2\).
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असमानता \(\frac{2x-5}{3}+\frac{x+1}{6}\le 4\) का हल चुनिए।
Choose the solution of \(\frac{2x-5}{3}+\frac{x+1}{6}\le 4\).
#linear inequalities
#addition fractions
#class 11
#hard
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A \(x\le \frac{17}{5}\)
B \(x\ge \frac{17}{5}\)
C \(x<\frac{17}{5}\)
D \(x>\frac{17}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{17}{5}\)
Explanation
Simple Explanation
हर हटाने पर \(4x-10+x+1\le 24\) मिलता है। अतः \(5x\le 33\), इसलिए \(x\le \frac{33}{5}\)। / Clearing denominators gives \(4x-10+x+1\le 24\). Hence \(5x\le 33\), so \(x\le \frac{33}{5}\).
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असमानता (5(x-1)-2(3x+4)<9) का हल क्या है?
What is the solution of (5(x-1)-2(3x+4)<9)?
#linear inequalities
#brackets
#negative sign
#class 11
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A (x>-22)
B (x<-22)
C \(x\ge -22\)
D \(x\le -22\)
Explanation opens after your attempt
Correct Answer
B. (x<-22)
Explanation
Simple Explanation
सरलीकरण से (-x-13<9) मिलता है। इससे (-x<22), इसलिए (x>-22) होना चाहिए। / Simplification gives (-x-13<9). Thus (-x<22), so (x>-22) should result.
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यदि \(\frac{3-4x}{2}\ge 5-\frac{x}{3}\), तो (x) का हल क्या है?
If \(\frac{3-4x}{2}\ge 5-\frac{x}{3}\), what is the solution for (x)?
#linear inequalities
#fractions
#hard
#class 11
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A \(x\le -\frac{21}{10}\)
B \(x\ge -\frac{21}{10}\)
C \(x<-\frac{21}{10}\)
D \(x>-\frac{21}{10}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le -\frac{21}{10}\)
Explanation
Simple Explanation
(6) से गुणा करने पर \(9-12x\ge 30-2x\) मिलता है। इससे \(-21\ge 10x\), अतः \(x\le -\frac{21}{10}\)। / Multiplying by (6) gives \(9-12x\ge 30-2x\). Thus \(-21\ge 10x\), so \(x\le -\frac{21}{10}\).
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असमानता \(2-\frac{5x-1}{4}<\frac{x+7}{2}\) का हल समुच्चय क्या है?
What is the solution set of \(2-\frac{5x-1}{4}<\frac{x+7}{2}\)?
#linear inequalities
#nested subtraction
#class 11
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A \(x>-\frac{5}{7}\)
B \(x<-\frac{5}{7}\)
C \(x\ge -\frac{5}{7}\)
D \(x\le -\frac{5}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{5}{7}\)
Explanation
Simple Explanation
(4) से गुणा करने पर (8-(5x-1)<2x+14) मिलता है। इससे (9-5x<2x+14), इसलिए \(x>-\frac{5}{7}\)। / Multiplying by (4) gives (8-(5x-1)<2x+14). Thus (9-5x<2x+14), so \(x>-\frac{5}{7}\).
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असमानता (13+4x\le 2(3x-5)+1) को हल कीजिए।
Solve the inequality (13+4x\le 2(3x-5)+1).
#linear inequalities
#algebraic solution
#class 11
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A \(x\ge 11\)
B \(x\le 11\)
C (x>11)
D (x<11)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 11\)
Explanation
Simple Explanation
दाएँ पक्ष (6x-9) है। \(13+4x\le 6x-9\) से \(22\le 2x\), अतः \(x\ge 11\)। / The right side is (6x-9). From \(13+4x\le 6x-9\), \(22\le 2x\), so \(x\ge 11\).
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असमानता (-2(5-x)+3(x-4)>6x+1) का सही हल कौन सा है?
Which is the correct solution of (-2(5-x)+3(x-4)>6x+1)?
#linear inequalities
#brackets
#hard
#class 11
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? Hint Small clue
A (x<-23)
B (x>-23)
C \(x\le -23\)
D \(x\ge -23\)
Explanation opens after your attempt
Correct Answer
A. (x<-23)
Explanation
Simple Explanation
बायाँ पक्ष (5x-22) बनता है। (5x-22>6x+1) से (x<-23) मिलता है। / The left side becomes (5x-22). From (5x-22>6x+1), we get (x<-23).
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यदि \(0.75x+\frac{1}{2}<2-\frac{x}{4}\), तो (x) का हल क्या है?
If \(0.75x+\frac{1}{2}<2-\frac{x}{4}\), what is the solution for (x)?
#linear inequalities
#decimal fraction
#class 11
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A \(x<\frac{3}{2}\)
B \(x>\frac{3}{2}\)
C \(x\le \frac{3}{2}\)
D \(x\ge \frac{3}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x<\frac{3}{2}\)
Explanation
Simple Explanation
\(0.75x=\frac{3x}{4}\) मानकर हल करें। \(x+\frac{1}{2}<2\) से \(x<\frac{3}{2}\) मिलता है। / Treat (0.75x) as \(\frac{3x}{4}\). From \(x+\frac{1}{2}<2\), \(x<\frac{3}{2}\).
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असमानता \(\frac{9x+4}{5}\ge 2x-3\) का हल क्या है?
What is the solution of \(\frac{9x+4}{5}\ge 2x-3\)?
#linear inequalities
#fraction
#class 11
#hard
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? Hint Small clue
A \(x\le 19\)
B \(x\ge 19\)
C (x<19)
D (x>19)
Explanation opens after your attempt
Correct Answer
A. \(x\le 19\)
Explanation
Simple Explanation
(5) से गुणा करने पर \(9x+4\ge 10x-15\) मिलता है। अतः \(x\le 19\)। / Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
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असमानता \(3-\frac{2x+5}{7}\le \frac{1-x}{2}\) को हल करें।
Solve the inequality \(3-\frac{2x+5}{7}\le \frac{1-x}{2}\).
#linear inequalities
#complex fractions
#class 11
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? Hint Small clue
A \(x\ge -\frac{27}{3}\)
B \(x\le -9\)
C \(x\ge -9\)
D (x<-9)
Explanation opens after your attempt
Correct Answer
C. \(x\ge -9\)
Explanation
Simple Explanation
(14) से गुणा करने पर (42-2(2x+5)\le 7(1-x)) मिलता है। इससे \(32-4x\le 7-7x\), अतः \(x\le -\frac{25}{3}\) नहीं बल्कि \(3x\le -25\) है। / Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
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असमानता (8-3(2-x)\ge 4x-1) का हल समुच्चय चुनिए।
Choose the solution set of (8-3(2-x)\ge 4x-1).
#linear inequalities
#bracket simplification
#class 11
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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. \(x\le 3\)
Explanation
Simple Explanation
बायाँ पक्ष (2+3x) है। \(2+3x\ge 4x-1\) से \(x\le 3\) मिलता है। / The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
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यदि \(\frac{x}{3}-\frac{x-5}{6}>2\), तो (x) के लिए सही शर्त क्या है?
If \(\frac{x}{3}-\frac{x-5}{6}>2\), what is the correct condition for (x)?
#linear inequalities
#fraction subtraction
#class 11
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? Hint Small clue
A (x>7)
B (x<7)
C \(x\ge 7\)
D \(x\le 7\)
Explanation opens after your attempt
Explanation
Simple Explanation
बायाँ पक्ष \(\frac{x+5}{6}\) बनता है। \(\frac{x+5}{6}>2\) से (x>7) मिलता है। / The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
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असमानता \(-\frac{4x-1}{3}\le 5-x\) का हल क्या है?
What is the solution of \(-\frac{4x-1}{3}\le 5-x\)?
#linear inequalities
#negative numerator
#class 11
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? Hint Small clue
A \(x\ge -14\)
B \(x\le -14\)
C (x>-14)
D (x<-14)
Explanation opens after your attempt
Correct Answer
A. \(x\ge -14\)
Explanation
Simple Explanation
(3) से गुणा करने पर \(-4x+1\le 15-3x\) मिलता है। इससे \(x\ge -14\)। / Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
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असमानता (2(1-3x)<4-7x) का हल अंतराल कौन सा है?
Which interval is the solution of (2(1-3x)<4-7x)?
#linear inequalities
#interval
#algebra
#class 11
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? Hint Small clue
A (x<2)
B (x>2)
C \(x\le 2\)
D \(x\ge 2\)
Explanation opens after your attempt
Explanation
Simple Explanation
(2-6x<4-7x) में (7x) जोड़ने पर (2+x<4) मिलता है। इसलिए (x<2)। / Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
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असमानता \(5-\frac{x-3}{2}\ge \frac{3x+1}{4}\) को हल करें।
Solve the inequality \(5-\frac{x-3}{2}\ge \frac{3x+1}{4}\).
#linear inequalities
#fraction with brackets
#class 11
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? Hint Small clue
A \(x\le \frac{21}{5}\)
B \(x\ge \frac{21}{5}\)
C \(x<\frac{21}{5}\)
D \(x>\frac{21}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{21}{5}\)
Explanation
Simple Explanation
(4) से गुणा करने पर (20-2(x-3)\ge 3x+1) मिलता है। इससे \(25\ge 5x\), इसलिए \(x\le 5\)। / Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
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यदि (6x+2<3(2x+1)), तो सही निष्कर्ष क्या है?
If (6x+2<3(2x+1)), what is the correct conclusion?
#linear inequalities
#always true
#class 11
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? Hint Small clue
A सभी वास्तविक संख्याएँ / All real numbers
B कोई हल नहीं / No solution
C (x<1)
D (x>1)
Explanation opens after your attempt
Correct Answer
A. सभी वास्तविक संख्याएँ / All real numbers
Explanation
Simple Explanation
दाएँ पक्ष (6x+3) है। (6x+2<6x+3) हमेशा सत्य है, इसलिए सभी वास्तविक संख्याएँ हल हैं। / The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
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असमानता \(7x-4\le 2x+16\) को हल कीजिए।
Solve the inequality \(7x-4\le 2x+16\).
#linear inequalities
#basic hard
#class 11
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A \(x\le 4\)
B \(x\ge 4\)
C (x<4)
D (x>4)
Explanation opens after your attempt
Correct Answer
A. \(x\le 4\)
Explanation
Simple Explanation
\(5x\le 20\) से \(x\le 4\) मिलता है। सरल रैखिक असमानता में पहले (x) वाले पद एक ओर लाएँ। / From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
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असमानता (12-5x\ge 3(4-2x)+x) का हल क्या है?
What is the solution of (12-5x\ge 3(4-2x)+x)?
#linear inequalities
#always true
#equality boundary
#class 11
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? Hint Small clue
A सभी वास्तविक संख्याएँ / All real numbers
B \(x\le 0\)
C कोई हल नहीं / No solution
D \(x\ge 0\)
Explanation opens after your attempt
Correct Answer
A. सभी वास्तविक संख्याएँ / All real numbers
Explanation
Simple Explanation
दाएँ पक्ष (12-5x) बनता है। समानता \(12-5x\ge 12-5x\) हमेशा सत्य है। / The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
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यदि \(4-\frac{3x}{2}>1+\frac{x}{6}\), तो (x) का हल क्या है?
If \(4-\frac{3x}{2}>1+\frac{x}{6}\), what is the solution for (x)?
#linear inequalities
#fractions
#class 11
#hard
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A \(x<\frac{9}{5}\)
B \(x>\frac{9}{5}\)
C \(x\le \frac{9}{5}\)
D \(x\ge \frac{9}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x<\frac{9}{5}\)
Explanation
Simple Explanation
(6) से गुणा करने पर (24-9x>6+x) मिलता है। इससे (18>10x), अतः \(x<\frac{9}{5}\)। / Multiplying by (6) gives (24-9x>6+x). Thus (18>10x), so \(x<\frac{9}{5}\).
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असमानता \(2.2x-1.1\ge 4.4-0.5x\) को हल करें।
Solve the inequality \(2.2x-1.1\ge 4.4-0.5x\).
#linear inequalities
#decimals
#hard
#class 11
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? Hint Small clue
A \(x\ge \frac{55}{27}\)
B \(x\le \frac{55}{27}\)
C \(x>\frac{55}{27}\)
D \(x<\frac{55}{27}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{55}{27}\)
Explanation
Simple Explanation
\(2.7x\ge 5.5\) से \(x\ge \frac{55}{27}\) मिलता है। दशमलव हटाने के लिए (10) से गुणा करना उपयोगी है। / From \(2.7x\ge 5.5\), \(x\ge \frac{55}{27}\). Multiplying by (10) is useful for removing decimals.
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असमानता (3(2x-1)-5(x+2)\le x-20) का हल क्या है?
What is the solution of (3(2x-1)-5(x+2)\le x-20)?
#linear inequalities
#no solution
#brackets
#class 11
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? Hint Small clue
A \(x\ge \frac{7}{2}\)
B \(x\le \frac{7}{2}\)
C \(x>\frac{7}{2}\)
D \(x<\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{7}{2}\)
Explanation
Simple Explanation
बायाँ पक्ष (x-13) है। \(x-13\le x-20\) असत्य \( -13\le -20\) देता है, इसलिए कोई हल नहीं होना चाहिए। / The left side is (x-13). The inequality \(x-13\le x-20\) gives false \(-13\le -20\), so there should be no solution.
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असमानता \(\frac{x-8}{5}\ge \frac{2x+1}{3}-4\) को हल कीजिए।
Solve the inequality \(\frac{x-8}{5}\ge \frac{2x+1}{3}-4\).
#linear inequalities
#fractions
#one variable
#class 11
50 50-50 2 wrong hide
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? Hint Small clue
A \(x\le \frac{47}{7}\)
B \(x\ge \frac{47}{7}\)
C \(x<\frac{47}{7}\)
D \(x>\frac{47}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{47}{7}\)
Explanation
Simple Explanation
(15) से गुणा करने पर (3x-24\ge 5(2x+1)-60) मिलता है। इससे \(31\ge 7x\), इसलिए \(x\le \frac{31}{7}\)। / Multiplying by (15) gives (3x-24\ge 5(2x+1)-60). Thus \(31\ge 7x\), so \(x\le \frac{31}{7}\).
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यदि (9-4(x+1)<2(3-x)), तो (x) के लिए सही शर्त क्या है?
If (9-4(x+1)<2(3-x)), what is the correct condition for (x)?
#linear inequalities
#brackets
#class 11
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? Hint Small clue
A \(x>-\frac{1}{2}\)
B \(x<-\frac{1}{2}\)
C \(x\ge -\frac{1}{2}\)
D \(x\le -\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{1}{2}\)
Explanation
Simple Explanation
सरलीकरण से (5-4x<6-2x) मिलता है। इससे (-1<2x), अतः \(x>-\frac{1}{2}\)। / Simplification gives (5-4x<6-2x). Thus (-1<2x), so \(x>-\frac{1}{2}\).
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असमानता \(-6+\frac{5x}{2}\le \frac{x-3}{4}\) का हल चुनिए।
Choose the solution of \(-6+\frac{5x}{2}\le \frac{x-3}{4}\).
#linear inequalities
#fractions
#hard
#class 11
50 50-50 2 wrong hide
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? Hint Small clue
A \(x\le \frac{21}{9}\)
B \(x\le \frac{21}{9}\)
C \(x\ge \frac{21}{9}\)
D \(x<\frac{21}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{21}{9}\)
Explanation
Simple Explanation
(4) से गुणा करने पर \(-24+10x\le x-3\) मिलता है। इससे \(9x\le 21\), अतः \(x\le \frac{7}{3}\)। / Multiplying by (4) gives \(-24+10x\le x-3\). Thus \(9x\le 21\), so \(x\le \frac{7}{3}\).
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असमानता (4x+7>2(2x+5)) के लिए कौन सा कथन सही है?
Which statement is correct for the inequality (4x+7>2(2x+5))?
#linear inequalities
#no solution
#class 11
50 50-50 2 wrong hide
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+10 Time+ 10 sec extra
? Hint Small clue
A सभी वास्तविक संख्याएँ / All real numbers
B कोई हल नहीं / No solution
C (x>3)
D (x<3)
Explanation opens after your attempt
Correct Answer
B. कोई हल नहीं / No solution
Explanation
Simple Explanation
दाएँ पक्ष (4x+10) है। (4x+7>4x+10) से (7>10) मिलता है, जो असत्य है। / The right side is (4x+10). (4x+7>4x+10) gives (7>10), which is false.
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असमानता \(\frac{2-3x}{9}<\frac{x+4}{6}\) का हल क्या है?
What is the solution of \(\frac{2-3x}{9}<\frac{x+4}{6}\)?
#linear inequalities
#rational inequality
#class 11
#hard
50 50-50 2 wrong hide
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? Hint Small clue
A \(x>-\frac{8}{15}\)
B \(x<-\frac{8}{15}\)
C \(x\ge -\frac{8}{15}\)
D \(x\le -\frac{8}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{8}{15}\)
Explanation
Simple Explanation
(18) से गुणा करने पर (2(2-3x)<3(x+4)) मिलता है। इससे (4-6x<3x+12), इसलिए \(x>-\frac{8}{9}\)। / Multiplying by (18) gives (2(2-3x)<3(x+4)). Thus (4-6x<3x+12), so \(x>-\frac{8}{9}\).
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