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
Step 1
Concept
From (3x<18), we get (x<6). In exams, keep applying the same operation on both sides.
Step 2
Why this answer is correct
The correct answer is A. (x<6). From (3x<18), we get (x<6). In exams, keep applying the same operation on both sides.
Step 3
Exam Tip
(3x<18) से (x<6) मिलता है। परीक्षा में दोनों पक्षों पर समान क्रिया करने का ध्यान रखें।
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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. \(x\le -6\)
Step 1
Concept
In \(-2x\ge 12\), dividing by a negative number reverses the sign. Hence \(x\le -6\).
Step 2
Why this answer is correct
The correct answer is B. \(x\le -6\). In \(-2x\ge 12\), dividing by a negative number reverses the sign. Hence \(x\le -6\).
Step 3
Exam Tip
\(-2x\ge 12\) में ऋणात्मक संख्या से भाग देने पर चिह्न बदलता है। इसलिए \(x\le -6\) होगा।
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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
Step 1
Concept
After clearing denominators, (x+5>2x) is obtained. Therefore (x<5) is correct.
Step 2
Why this answer is correct
The correct answer is C. (x<5). After clearing denominators, (x+5>2x) is obtained. Therefore (x<5) is correct.
Step 3
Exam Tip
हरों को हटाने पर (x+5>2x) मिलता है। अतः (x<5) सही है।
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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. \(x\le 6\)
Step 1
Concept
From \(0.3x\le 1.8\), we get \(x\le 6\). Decimals can also be converted into fractions.
Step 2
Why this answer is correct
The correct answer is D. \(x\le 6\). From \(0.3x\le 1.8\), we get \(x\le 6\). Decimals can also be converted into fractions.
Step 3
Exam Tip
\(0.3x\le 1.8\) से \(x\le 6\) मिलता है। दशमलव को भिन्न में बदलकर भी हल किया जा सकता है।
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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. \(x>-\frac{4}{5}\)
Step 1
Concept
Simplification gives (-5x<4). Dividing by a negative gives \(x>-\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{4}{5}\). Simplification gives (-5x<4). Dividing by a negative gives \(x>-\frac{4}{5}\).
Step 3
Exam Tip
सरलीकरण से (-5x<4) मिलता है। ऋणात्मक से भाग देने पर उत्तर \(x>-\frac{4}{5}\) है।
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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. \(x\le 3\)
Step 1
Concept
After simplification, \(15\ge 5x\) is obtained. Hence \(x\le 3\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(x\le 3\). After simplification, \(15\ge 5x\) is obtained. Hence \(x\le 3\) is correct.
Step 3
Exam Tip
सरलीकरण के बाद \(15\ge 5x\) आता है। इसलिए \(x\le 3\) सही है।
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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. \(x\le -14\)
Step 1
Concept
Multiplying by (6) gives \(4x+2\le 3x-12\). This gives \(x\le -14\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -14\). Multiplying by (6) gives \(4x+2\le 3x-12\). This gives \(x\le -14\).
Step 3
Exam Tip
(6) से गुणा करने पर \(4x+2\le 3x-12\) मिलता है। इससे \(x\le -14\) आता है।
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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 \(x\ge 11\)
Explanation opens after your attempt
Step 1
Concept
The left side becomes (2x-14). From (2x-14>8), we get (x>11).
Step 2
Why this answer is correct
The correct answer is A. (x>11). The left side becomes (2x-14). From (2x-14>8), we get (x>11).
Step 3
Exam Tip
बायाँ पक्ष (2x-14) बनता है। (2x-14>8) से (x>11) मिलता है।
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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. \(x<\frac{12}{7}\)
Step 1
Concept
Multiplying by (14) gives (10x-4<3x+8). So (7x<12) and \(x<\frac{12}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{12}{7}\). Multiplying by (14) gives (10x-4<3x+8). So (7x<12) and \(x<\frac{12}{7}\).
Step 3
Exam Tip
(14) से गुणा करने पर (10x-4<3x+8) मिलता है। इसलिए (7x<12) और \(x<\frac{12}{7}\)।
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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. \(x\ge \frac{75}{11}\)
Step 1
Concept
Clearing denominators gives \(-5x+60\le 6x-15\). Thus \(75\le 11x\), so \(x\ge \frac{75}{11}\).
Step 2
Why this answer is correct
The correct answer is B. \(x\ge \frac{75}{11}\). Clearing denominators gives \(-5x+60\le 6x-15\). Thus \(75\le 11x\), so \(x\ge \frac{75}{11}\).
Step 3
Exam Tip
हर हटाने पर \(-5x+60\le 6x-15\) मिलता है। इससे \(75\le 11x\), अतः \(x\ge \frac{75}{11}\)।
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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. \(x>-\frac{5}{13}\)
Step 1
Concept
Simplification gives (15-8x<5x+10). This gives (5<13x), so \(x>\frac{5}{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{5}{13}\). Simplification gives (15-8x<5x+10). This gives (5<13x), so \(x>\frac{5}{13}\).
Step 3
Exam Tip
सरलीकरण से (15-8x<5x+10) मिलता है। इससे (5<13x), इसलिए \(x>\frac{5}{13}\) नहीं बल्कि \(x>\frac{5}{13}\) होता है।
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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. \(x\ge \frac{13}{3}\)
Step 1
Concept
The right side simplifies to (17-2x). From \(4x-9\ge 17-2x\), \(x\ge \frac{13}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{13}{3}\). The right side simplifies to (17-2x). From \(4x-9\ge 17-2x\), \(x\ge \frac{13}{3}\).
Step 3
Exam Tip
दाएँ पक्ष को सरल करने पर (17-2x) मिलता है। \(4x-9\ge 17-2x\) से \(x\ge \frac{13}{3}\)।
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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. \(x\ge 15\)
Step 1
Concept
The left side becomes \(\frac{x+7}{4}\). From \(\frac{x+7}{4}\ge 6\), \(x\ge 17\) should result.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 15\). The left side becomes \(\frac{x+7}{4}\). From \(\frac{x+7}{4}\ge 6\), \(x\ge 17\) should result.
Step 3
Exam Tip
बायाँ पक्ष \(\frac{x+7}{4}\) बनता है। \(\frac{x+7}{4}\ge 6\) से \(x\ge 17\) होना चाहिए।
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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 \(x\ge 0\)
Explanation opens after your attempt
Step 1
Concept
Multiplying by (6) gives (4-x>2x+4). Thus (-3x>0), so (x<0).
Step 2
Why this answer is correct
The correct answer is A. (x<0). Multiplying by (6) gives (4-x>2x+4). Thus (-3x>0), so (x<0).
Step 3
Exam Tip
(6) से गुणा करने पर (4-x>2x+4) मिलता है। इससे (-3x>0), इसलिए (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
Step 1
Concept
From (0.9x<-5.4), we get (x<-6). In decimal problems, handle place values carefully.
Step 2
Why this answer is correct
The correct answer is A. (x<-6). From (0.9x<-5.4), we get (x<-6). In decimal problems, handle place values carefully.
Step 3
Exam Tip
(0.9x<-5.4) से (x<-6) मिलता है। दशमलव वाले प्रश्नों में स्थान मान ध्यान से रखें।
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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. \(x\ge 1\)
Step 1
Concept
Multiplying by (10) gives \(20x-6\ge 5x+9\). Thus \(15x\ge 15\), so \(x\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 1\). Multiplying by (10) gives \(20x-6\ge 5x+9\). Thus \(15x\ge 15\), so \(x\ge 1\).
Step 3
Exam Tip
(10) से गुणा करने पर \(20x-6\ge 5x+9\) मिलता है। इससे \(15x\ge 15\), इसलिए \(x\ge 1\)।
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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. \(x\ge 5\)
Step 1
Concept
Simplification gives \(16-5x\le 6-3x\). Thus \(10\le 2x\), hence \(x\ge 5\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 5\). Simplification gives \(16-5x\le 6-3x\). Thus \(10\le 2x\), hence \(x\ge 5\).
Step 3
Exam Tip
सरलीकरण से \(16-5x\le 6-3x\) मिलता है। इससे \(10\le 2x\), अतः \(x\ge 5\)।
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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
Step 1
Concept
The left side is (5x-10), and \(5x-10\le 5x-7\) is always true. Hence the answer should be all real numbers.
Step 2
Why this answer is correct
The correct answer is B. कोई हल नहीं / No solution. The left side is (5x-10), and \(5x-10\le 5x-7\) is always true. Hence the answer should be all real numbers.
Step 3
Exam Tip
बायाँ पक्ष (5x-10) है और असमानता \(5x-10\le 5x-7\) हमेशा सत्य है। इसलिए सही उत्तर सभी वास्तविक संख्याएँ होना चाहिए।
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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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⏭ Skip Next question
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? Hint Small clue
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
Step 1
Concept
Simplification gives (8x+12>8x+15), i.e. (12>15). This is false, so there is no solution.
Step 2
Why this answer is correct
The correct answer is C. कोई हल नहीं / No solution. Simplification gives (8x+12>8x+15), i.e. (12>15). This is false, so there is no solution.
Step 3
Exam Tip
सरलीकरण से (8x+12>8x+15) अर्थात (12>15) मिलता है। यह असत्य है, इसलिए कोई हल नहीं।
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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. \(x\ge \frac{13}{7}\)
Step 1
Concept
Multiplying by (10) gives \(14-4x\le 3x+1\). Thus \(13\le 7x\), 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 (10) gives \(14-4x\le 3x+1\). Thus \(13\le 7x\), so \(x\ge \frac{13}{7}\).
Step 3
Exam Tip
(10) से गुणा करने पर \(14-4x\le 3x+1\) मिलता है। इससे \(13\le 7x\), अतः \(x\ge \frac{13}{7}\)।
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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. \(x\le -2\)
Step 1
Concept
Simplification gives \(1-3x\ge 3-2x\). Thus \(-2\ge x\), i.e. \(x\le -2\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -2\). Simplification gives \(1-3x\ge 3-2x\). Thus \(-2\ge x\), i.e. \(x\le -2\).
Step 3
Exam Tip
सरलीकरण से \(1-3x\ge 3-2x\) मिलता है। इससे \(-2\ge x\), यानी \(x\le -2\)।
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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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A (x>-2)
B (x<-2)
C \(x\ge -2\)
D \(x\le -2\)
Explanation opens after your attempt
Step 1
Concept
The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
Step 2
Why this answer is correct
The correct answer is A. (x>-2). The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
Step 3
Exam Tip
बायाँ पक्ष \(\frac{10-x}{8}\) बनता है। \(\frac{10-x}{8}<1\) से (x>2) मिलता है।
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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
Step 1
Concept
From (10>5x), (x<2) is obtained. In interval form, it is (\(-\infty,2\)).
Step 2
Why this answer is correct
The correct answer is A. (x<2). From (10>5x), (x<2) is obtained. In interval form, it is (\(-\infty,2\)).
Step 3
Exam Tip
(10>5x) से (x<2) मिलता है। अंतराल में यह (\(-\infty,2\)) होगा।
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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. \(x>-\frac{14}{5}\)
Step 1
Concept
We get (-5x<14). Dividing by a negative gives \(x>-\frac{14}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{14}{5}\). We get (-5x<14). Dividing by a negative gives \(x>-\frac{14}{5}\).
Step 3
Exam Tip
(-5x<14) प्राप्त होता है। ऋणात्मक से भाग देने पर \(x>-\frac{14}{5}\) होगा।
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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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A \(x\ge 2\)
B \(x\le 2\)
C (x>2)
D (x<2)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 2\)
Step 1
Concept
Simplification gives \(5x-2.5\ge 1.5x+4.5\). Thus \(3.5x\ge 7\), so \(x\ge 2\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 2\). Simplification gives \(5x-2.5\ge 1.5x+4.5\). Thus \(3.5x\ge 7\), so \(x\ge 2\).
Step 3
Exam Tip
सरलीकरण से \(5x-2.5\ge 1.5x+4.5\) मिलता है। इससे \(3.5x\ge 7\), अतः \(x\ge 2\)।
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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. \(x\le \frac{17}{5}\)
Step 1
Concept
Clearing denominators gives \(4x-10+x+1\le 24\). Hence \(5x\le 33\), so \(x\le \frac{33}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{17}{5}\). Clearing denominators gives \(4x-10+x+1\le 24\). Hence \(5x\le 33\), so \(x\le \frac{33}{5}\).
Step 3
Exam Tip
हर हटाने पर \(4x-10+x+1\le 24\) मिलता है। अतः \(5x\le 33\), इसलिए \(x\le \frac{33}{5}\)।
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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)
Step 1
Concept
Simplification gives (-x-13<9). Thus (-x<22), so (x>-22) should result.
Step 2
Why this answer is correct
The correct answer is B. (x<-22). Simplification gives (-x-13<9). Thus (-x<22), so (x>-22) should result.
Step 3
Exam Tip
सरलीकरण से (-x-13<9) मिलता है। इससे (-x<22), इसलिए (x>-22) होना चाहिए।
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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. \(x\le -\frac{21}{10}\)
Step 1
Concept
Multiplying by (6) gives \(9-12x\ge 30-2x\). Thus \(-21\ge 10x\), so \(x\le -\frac{21}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -\frac{21}{10}\). Multiplying by (6) gives \(9-12x\ge 30-2x\). Thus \(-21\ge 10x\), so \(x\le -\frac{21}{10}\).
Step 3
Exam Tip
(6) से गुणा करने पर \(9-12x\ge 30-2x\) मिलता है। इससे \(-21\ge 10x\), अतः \(x\le -\frac{21}{10}\)।
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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. \(x>-\frac{5}{7}\)
Step 1
Concept
Multiplying by (4) gives (8-(5x-1)<2x+14). Thus (9-5x<2x+14), so \(x>-\frac{5}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{5}{7}\). Multiplying by (4) gives (8-(5x-1)<2x+14). Thus (9-5x<2x+14), so \(x>-\frac{5}{7}\).
Step 3
Exam Tip
(4) से गुणा करने पर (8-(5x-1)<2x+14) मिलता है। इससे (9-5x<2x+14), इसलिए \(x>-\frac{5}{7}\)।
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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. \(x\ge 11\)
Step 1
Concept
The right side is (6x-9). From \(13+4x\le 6x-9\), \(22\le 2x\), so \(x\ge 11\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 11\). The right side is (6x-9). From \(13+4x\le 6x-9\), \(22\le 2x\), so \(x\ge 11\).
Step 3
Exam Tip
दाएँ पक्ष (6x-9) है। \(13+4x\le 6x-9\) से \(22\le 2x\), अतः \(x\ge 11\)।
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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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A (x<-23)
B (x>-23)
C \(x\le -23\)
D \(x\ge -23\)
Explanation opens after your attempt
Correct Answer
A. (x<-23)
Step 1
Concept
The left side becomes (5x-22). From (5x-22>6x+1), we get (x<-23).
Step 2
Why this answer is correct
The correct answer is A. (x<-23). The left side becomes (5x-22). From (5x-22>6x+1), we get (x<-23).
Step 3
Exam Tip
बायाँ पक्ष (5x-22) बनता है। (5x-22>6x+1) से (x<-23) मिलता है।
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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. \(x<\frac{3}{2}\)
Step 1
Concept
Treat (0.75x) as \(\frac{3x}{4}\). From \(x+\frac{1}{2}<2\), \(x<\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{3}{2}\). Treat (0.75x) as \(\frac{3x}{4}\). From \(x+\frac{1}{2}<2\), \(x<\frac{3}{2}\).
Step 3
Exam Tip
\(0.75x=\frac{3x}{4}\) मानकर हल करें। \(x+\frac{1}{2}<2\) से \(x<\frac{3}{2}\) मिलता है।
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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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A \(x\le 19\)
B \(x\ge 19\)
C (x<19)
D (x>19)
Explanation opens after your attempt
Correct Answer
A. \(x\le 19\)
Step 1
Concept
Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le 19\). Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
Step 3
Exam Tip
(5) से गुणा करने पर \(9x+4\ge 10x-15\) मिलता है। अतः \(x\le 19\)।
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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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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. \(x\ge -9\)
Step 1
Concept
Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
Step 2
Why this answer is correct
The correct answer is C. \(x\ge -9\). Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
Step 3
Exam Tip
(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।
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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. \(x\le 3\)
Step 1
Concept
The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le 3\). The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
Step 3
Exam Tip
बायाँ पक्ष (2+3x) है। \(2+3x\ge 4x-1\) से \(x\le 3\) मिलता है।
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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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A (x>7)
B (x<7)
C \(x\ge 7\)
D \(x\le 7\)
Explanation opens after your attempt
Step 1
Concept
The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
Step 2
Why this answer is correct
The correct answer is A. (x>7). The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
Step 3
Exam Tip
बायाँ पक्ष \(\frac{x+5}{6}\) बनता है। \(\frac{x+5}{6}>2\) से (x>7) मिलता है।
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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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A \(x\ge -14\)
B \(x\le -14\)
C (x>-14)
D (x<-14)
Explanation opens after your attempt
Correct Answer
A. \(x\ge -14\)
Step 1
Concept
Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -14\). Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
Step 3
Exam Tip
(3) से गुणा करने पर \(-4x+1\le 15-3x\) मिलता है। इससे \(x\ge -14\)।
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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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A (x<2)
B (x>2)
C \(x\le 2\)
D \(x\ge 2\)
Explanation opens after your attempt
Step 1
Concept
Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
Step 2
Why this answer is correct
The correct answer is A. (x<2). Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
Step 3
Exam Tip
(2-6x<4-7x) में (7x) जोड़ने पर (2+x<4) मिलता है। इसलिए (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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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. \(x\le \frac{21}{5}\)
Step 1
Concept
Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{21}{5}\). Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
Step 3
Exam Tip
(4) से गुणा करने पर (20-2(x-3)\ge 3x+1) मिलता है। इससे \(25\ge 5x\), इसलिए \(x\le 5\)।
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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
Step 1
Concept
The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
Step 3
Exam Tip
दाएँ पक्ष (6x+3) है। (6x+2<6x+3) हमेशा सत्य है, इसलिए सभी वास्तविक संख्याएँ हल हैं।
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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. \(x\le 4\)
Step 1
Concept
From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 4\). From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
Step 3
Exam Tip
\(5x\le 20\) से \(x\le 4\) मिलता है। सरल रैखिक असमानता में पहले (x) वाले पद एक ओर लाएँ।
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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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⏭ Skip Next question
+10 Time+ 10 sec extra
? 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
Step 1
Concept
The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
Step 3
Exam Tip
दाएँ पक्ष (12-5x) बनता है। समानता \(12-5x\ge 12-5x\) हमेशा सत्य है।
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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. \(x<\frac{9}{5}\)
Step 1
Concept
Multiplying by (6) gives (24-9x>6+x). Thus (18>10x), so \(x<\frac{9}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{9}{5}\). Multiplying by (6) gives (24-9x>6+x). Thus (18>10x), so \(x<\frac{9}{5}\).
Step 3
Exam Tip
(6) से गुणा करने पर (24-9x>6+x) मिलता है। इससे (18>10x), अतः \(x<\frac{9}{5}\)।
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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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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. \(x\ge \frac{55}{27}\)
Step 1
Concept
From \(2.7x\ge 5.5\), \(x\ge \frac{55}{27}\). Multiplying by (10) is useful for removing decimals.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{55}{27}\). From \(2.7x\ge 5.5\), \(x\ge \frac{55}{27}\). Multiplying by (10) is useful for removing decimals.
Step 3
Exam Tip
\(2.7x\ge 5.5\) से \(x\ge \frac{55}{27}\) मिलता है। दशमलव हटाने के लिए (10) से गुणा करना उपयोगी है।
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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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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. \(x\ge \frac{7}{2}\)
Step 1
Concept
The left side is (x-13). The inequality \(x-13\le x-20\) gives false \(-13\le -20\), so there should be no solution.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{7}{2}\). The left side is (x-13). The inequality \(x-13\le x-20\) gives false \(-13\le -20\), so there should be no solution.
Step 3
Exam Tip
बायाँ पक्ष (x-13) है। \(x-13\le x-20\) असत्य \( -13\le -20\) देता है, इसलिए कोई हल नहीं होना चाहिए।
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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
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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. \(x\le \frac{47}{7}\)
Step 1
Concept
Multiplying by (15) gives (3x-24\ge 5(2x+1)-60). Thus \(31\ge 7x\), so \(x\le \frac{31}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{47}{7}\). Multiplying by (15) gives (3x-24\ge 5(2x+1)-60). Thus \(31\ge 7x\), so \(x\le \frac{31}{7}\).
Step 3
Exam Tip
(15) से गुणा करने पर (3x-24\ge 5(2x+1)-60) मिलता है। इससे \(31\ge 7x\), इसलिए \(x\le \frac{31}{7}\)।
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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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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. \(x>-\frac{1}{2}\)
Step 1
Concept
Simplification gives (5-4x<6-2x). Thus (-1<2x), so \(x>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{1}{2}\). Simplification gives (5-4x<6-2x). Thus (-1<2x), so \(x>-\frac{1}{2}\).
Step 3
Exam Tip
सरलीकरण से (5-4x<6-2x) मिलता है। इससे (-1<2x), अतः \(x>-\frac{1}{2}\)।
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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
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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. \(x\le \frac{21}{9}\)
Step 1
Concept
Multiplying by (4) gives \(-24+10x\le x-3\). Thus \(9x\le 21\), so \(x\le \frac{7}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{21}{9}\). Multiplying by (4) gives \(-24+10x\le x-3\). Thus \(9x\le 21\), so \(x\le \frac{7}{3}\).
Step 3
Exam Tip
(4) से गुणा करने पर \(-24+10x\le x-3\) मिलता है। इससे \(9x\le 21\), अतः \(x\le \frac{7}{3}\)।
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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
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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
Step 1
Concept
The right side is (4x+10). (4x+7>4x+10) gives (7>10), which is false.
Step 2
Why this answer is correct
The correct answer is B. कोई हल नहीं / No solution. The right side is (4x+10). (4x+7>4x+10) gives (7>10), which is false.
Step 3
Exam Tip
दाएँ पक्ष (4x+10) है। (4x+7>4x+10) से (7>10) मिलता है, जो असत्य है।
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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
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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. \(x>-\frac{8}{15}\)
Step 1
Concept
Multiplying by (18) gives (2(2-3x)<3(x+4)). Thus (4-6x<3x+12), so \(x>-\frac{8}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{8}{15}\). Multiplying by (18) gives (2(2-3x)<3(x+4)). Thus (4-6x<3x+12), so \(x>-\frac{8}{9}\).
Step 3
Exam Tip
(18) से गुणा करने पर (2(2-3x)<3(x+4)) मिलता है। इससे (4-6x<3x+12), इसलिए \(x>-\frac{8}{9}\)।
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