असमीका (3x-7>11) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of the inequality (3x-7>11).
#linear inequalities
#one variable
#algebraic solution
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A (x>6)
B (x<6)
C \(x\ge 6\)
D \(x>\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
(3x>18) so (x>6). In exams, first collect like terms on one side.
Step 2
Why this answer is correct
The correct answer is A. (x>6). (3x>18) so (x>6). In exams, first collect like terms on one side.
Step 3
Exam Tip
(3x>18) इसलिए (x>6)। परीक्षा में समान पदों को पहले एक तरफ लाएं।
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असमीका \(5-2x\le 13\) को हल कीजिए।
Solve the inequality \(5-2x\le 13\).
#linear inequalities
#negative coefficient
#sign reversal
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A \(x\le -4\)
B \(x\ge -4\)
C (x<4)
D \(x\ge 4\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge -4\)
Step 1
Concept
\(-2x\le 8\) and dividing by a negative reverses the sign so \(x\ge -4\). Watch the sign reversal in exams.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge -4\). \(-2x\le 8\) and dividing by a negative reverses the sign so \(x\ge -4\). Watch the sign reversal in exams.
Step 3
Exam Tip
\(-2x\le 8\) और ऋणात्मक संख्या से भाग देने पर चिन्ह पलटता है इसलिए \(x\ge -4\)। परीक्षा में ऋणात्मक गुणांक पर विशेष ध्यान दें।
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असमीका \(\frac{x}{3}+4<7\) का समाधान क्या है?
What is the solution of the inequality \(\frac{x}{3}+4<7\)?
#linear inequalities
#fraction
#one variable
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A (x<3)
B (x>9)
C (x<9)
D \(x\le 9\)
Explanation opens after your attempt
Step 1
Concept
\(\frac{x}{3}<3\) so (x<9). In exams, remove the constant term before clearing the denominator.
Step 2
Why this answer is correct
The correct answer is C. (x<9). \(\frac{x}{3}<3\) so (x<9). In exams, remove the constant term before clearing the denominator.
Step 3
Exam Tip
\(\frac{x}{3}<3\) इसलिए (x<9)। परीक्षा में हर हटाने से पहले स्थिर पद हटाएं।
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असमीका \(\frac{2x-5}{4}\ge 3\) का समाधान समुच्चय चुनिए।
Choose the solution set of the inequality \(\frac{2x-5}{4}\ge 3\).
#linear inequalities
#fraction
#algebraic solution
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A \(x\ge \frac{7}{2}\)
B \(x\le \frac{17}{2}\)
C \(x>\frac{17}{2}\)
D \(x\ge \frac{17}{2}\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge \frac{17}{2}\)
Step 1
Concept
From \(2x-5\ge 12\), \(2x\ge 17\), so \(x\ge \frac{17}{2}\). Multiplying by a positive denominator does not change the sign.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge \frac{17}{2}\). From \(2x-5\ge 12\), \(2x\ge 17\), so \(x\ge \frac{17}{2}\). Multiplying by a positive denominator does not change the sign.
Step 3
Exam Tip
\(2x-5\ge 12\) से \(2x\ge 17\), इसलिए \(x\ge \frac{17}{2}\)। परीक्षा में धनात्मक हर से गुणा करने पर चिन्ह नहीं बदलता।
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असमीका (4(x-2)+3\le 2x+9) को हल कीजिए।
Solve the inequality (4(x-2)+3\le 2x+9).
#linear inequalities
#brackets
#one variable
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A \(x\le 7\)
B \(x\ge 7\)
C (x<7)
D \(x\le 10\)
Explanation opens after your attempt
Correct Answer
A. \(x\le 7\)
Step 1
Concept
Simplifying gives \(4x-5\le 2x+9\), so \(2x\le 14\) and \(x\le 7\). Open brackets carefully in exams.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 7\). Simplifying gives \(4x-5\le 2x+9\), so \(2x\le 14\) and \(x\le 7\). Open brackets carefully in exams.
Step 3
Exam Tip
सरलीकरण से \(4x-5\le 2x+9\), इसलिए \(2x\le 14\) और \(x\le 7\)। परीक्षा में कोष्ठक सही खोलें।
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असमीका (7-3(x+1)>2x-6) का समाधान क्या होगा?
What will be the solution of (7-3(x+1)>2x-6)?
#linear inequalities
#brackets
#hard
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A (x>2)
B (x<2)
C \(x\le 2\)
D (x<-2)
Explanation opens after your attempt
Step 1
Concept
(4-3x>2x-6) gives (10>5x), hence (x<2). Keep variable terms on one side in exams.
Step 2
Why this answer is correct
The correct answer is B. (x<2). (4-3x>2x-6) gives (10>5x), hence (x<2). Keep variable terms on one side in exams.
Step 3
Exam Tip
(4-3x>2x-6) से (10>5x), इसलिए (x<2)। परीक्षा में चर वाले पदों को एक तरफ रखें।
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असमीका (2x+5<3x-4) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of (2x+5<3x-4).
#linear inequalities
#variable both sides
#solution set
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A (x<9)
B \(x\le 9\)
C (x>9)
D (x>-9)
Explanation opens after your attempt
Step 1
Concept
(5<x-4) gives (9<x), so (x>9). Write the final answer in standard form.
Step 2
Why this answer is correct
The correct answer is C. (x>9). (5<x-4) gives (9<x), so (x>9). Write the final answer in standard form.
Step 3
Exam Tip
(5<x-4) से (9<x), इसलिए (x>9)। परीक्षा में अंतिम उत्तर को मानक रूप में लिखें।
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असमीका (6x-1\ge 2(2x+7)) को हल कीजिए।
Solve the inequality (6x-1\ge 2(2x+7)).
#linear inequalities
#variable both sides
#brackets
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A \(x\le \frac{15}{2}\)
B \(x>\frac{15}{2}\)
C \(x\ge \frac{13}{2}\)
D \(x\ge \frac{15}{2}\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge \frac{15}{2}\)
Step 1
Concept
\(6x-1\ge 4x+14\) gives \(2x\ge 15\), so \(x\ge \frac{15}{2}\). Simplify both sides first in exams.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge \frac{15}{2}\). \(6x-1\ge 4x+14\) gives \(2x\ge 15\), so \(x\ge \frac{15}{2}\). Simplify both sides first in exams.
Step 3
Exam Tip
\(6x-1\ge 4x+14\) से \(2x\ge 15\), इसलिए \(x\ge \frac{15}{2}\)। परीक्षा में दोनों पक्षों को पहले सरल करें।
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असमीका \(\frac{x-2}{5}\le \frac{x+4}{3}\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{x-2}{5}\le \frac{x+4}{3}\).
#linear inequalities
#fractions
#LCM
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A \(x\ge -13\)
B \(x\le -13\)
C \(x\le 13\)
D (x>-13)
Explanation opens after your attempt
Correct Answer
A. \(x\ge -13\)
Step 1
Concept
(3(x-2)\le 5(x+4)) gives \(3x-6\le 5x+20\), hence \(x\ge -13\). Use the LCM to clear denominators.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -13\). (3(x-2)\le 5(x+4)) gives \(3x-6\le 5x+20\), hence \(x\ge -13\). Use the LCM to clear denominators.
Step 3
Exam Tip
(3(x-2)\le 5(x+4)) से \(3x-6\le 5x+20\), इसलिए \(x\ge -13\)। परीक्षा में लघुत्तम समापवर्त्य से हर हटाएं।
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असमीका \(\frac{3x+1}{2}>\frac{x-5}{4}\) को हल कीजिए।
Solve the inequality \(\frac{3x+1}{2}>\frac{x-5}{4}\).
#linear inequalities
#fractions
#hard
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A \(x<-\frac{7}{5}\)
B \(x>-\frac{7}{5}\)
C \(x>\frac{7}{5}\)
D \(x<\frac{7}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(x>-\frac{7}{5}\)
Step 1
Concept
(2(3x+1)>x-5) gives (6x+2>x-5), so \(x>-\frac{7}{5}\). Multiply both sides correctly while clearing denominators.
Step 2
Why this answer is correct
The correct answer is B. \(x>-\frac{7}{5}\). (2(3x+1)>x-5) gives (6x+2>x-5), so \(x>-\frac{7}{5}\). Multiply both sides correctly while clearing denominators.
Step 3
Exam Tip
(2(3x+1)>x-5) से (6x+2>x-5), इसलिए \(x>-\frac{7}{5}\)। परीक्षा में हर हटाते समय हर पक्ष पर सही गुणा करें।
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असमीका \(0.4x+1.2\le 3.6\) का समाधान क्या है?
What is the solution of \(0.4x+1.2\le 3.6\)?
#linear inequalities
#decimals
#one variable
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A \(x\ge 6\)
B (x<6)
C \(x\le 6\)
D \(x\le 12\)
Explanation opens after your attempt
Correct Answer
C. \(x\le 6\)
Step 1
Concept
\(0.4x\le 2.4\), so \(x\le 6\). Converting decimals to fractions can simplify exam calculations.
Step 2
Why this answer is correct
The correct answer is C. \(x\le 6\). \(0.4x\le 2.4\), so \(x\le 6\). Converting decimals to fractions can simplify exam calculations.
Step 3
Exam Tip
\(0.4x\le 2.4\), इसलिए \(x\le 6\)। परीक्षा में दशमलव को भिन्न में बदलना गणना आसान कर सकता है।
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असमीका (1.5-0.3x<0.6) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of (1.5-0.3x<0.6).
#linear inequalities
#decimals
#sign reversal
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A (x<3)
B (x>-3)
C \(x\le 3\)
D (x>3)
Explanation opens after your attempt
Step 1
Concept
(-0.3x<-0.9), and division by a negative gives (x>3). The sign reversal rule also applies with decimals.
Step 2
Why this answer is correct
The correct answer is D. (x>3). (-0.3x<-0.9), and division by a negative gives (x>3). The sign reversal rule also applies with decimals.
Step 3
Exam Tip
(-0.3x<-0.9) और ऋणात्मक संख्या से भाग देने पर (x>3)। परीक्षा में दशमलव के साथ भी चिन्ह पलटने का नियम वही रहता है।
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असमीका \(-4x+9\ge 25\) को हल कीजिए।
Solve the inequality \(-4x+9\ge 25\).
#linear inequalities
#negative coefficient
#one variable
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A \(x\le -4\)
B \(x\ge -4\)
C (x<-4)
D \(x\le 4\)
Explanation opens after your attempt
Correct Answer
A. \(x\le -4\)
Step 1
Concept
\(-4x\ge 16\), and dividing by a negative gives \(x\le -4\). Check the inequality sign in the final step.
Step 2
Why this answer is correct
The correct answer is A. \(x\le -4\). \(-4x\ge 16\), and dividing by a negative gives \(x\le -4\). Check the inequality sign in the final step.
Step 3
Exam Tip
\(-4x\ge 16\) और ऋणात्मक से भाग देने पर \(x\le -4\)। परीक्षा में अंतिम चरण में असमीका चिन्ह जांचें।
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असमीका (12-5x<2) का समाधान क्या है?
What is the solution of (12-5x<2)?
#linear inequalities
#sign reversal
#hard
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A (x<2)
B (x>2)
C \(x\le 2\)
D (x>-2)
Explanation opens after your attempt
Step 1
Concept
(-5x<-10), so reversing the sign gives (x>2). Always reverse the sign when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is B. (x>2). (-5x<-10), so reversing the sign gives (x>2). Always reverse the sign when dividing by a negative.
Step 3
Exam Tip
(-5x<-10), इसलिए चिन्ह पलटकर (x>2) मिलता है। परीक्षा में ऋणात्मक भाग देने पर चिन्ह अवश्य पलटें।
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असमीका (2(3x-1)-5\ge 4x+7) का समाधान समुच्चय चुनिए।
Choose the solution set of (2(3x-1)-5\ge 4x+7).
#linear inequalities
#brackets
#solution set
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A \(x\le 7\)
B \(x\ge 9\)
C \(x\ge 7\)
D (x>7)
Explanation opens after your attempt
Correct Answer
C. \(x\ge 7\)
Step 1
Concept
\(6x-7\ge 4x+7\) gives \(2x\ge 14\), so \(x\ge 7\). Simplify brackets and constants carefully.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge 7\). \(6x-7\ge 4x+7\) gives \(2x\ge 14\), so \(x\ge 7\). Simplify brackets and constants carefully.
Step 3
Exam Tip
\(6x-7\ge 4x+7\) से \(2x\ge 14\), इसलिए \(x\ge 7\)। परीक्षा में कोष्ठक और स्थिर पद दोनों ध्यान से सरल करें।
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असमीका (9-2(4x+1)\le 3x-4) को हल कीजिए।
Solve (9-2(4x+1)\le 3x-4).
#linear inequalities
#brackets
#negative multiplication
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A \(x\le 1\)
B \(x\ge 3\)
C \(x\le 3\)
D \(x\ge 1\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge 1\)
Step 1
Concept
\(7-8x\le 3x-4\) gives \(11\le 11x\), hence \(x\ge 1\). Be careful with signs while expanding a negative multiple.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge 1\). \(7-8x\le 3x-4\) gives \(11\le 11x\), hence \(x\ge 1\). Be careful with signs while expanding a negative multiple.
Step 3
Exam Tip
\(7-8x\le 3x-4\) से \(11\le 11x\), इसलिए \(x\ge 1\)। परीक्षा में ऋणात्मक गुणा से कोष्ठक खोलते समय चिन्ह ध्यान रखें।
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द्वि-असमीका \(-3<2x+1\le 9\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of the compound inequality \(-3<2x+1\le 9\).
#compound inequality
#one variable
#solution set
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A \(-2<x\le 4\)
B (-1<x<4)
C \(-2\le x\le 4\)
D (-2<x<4)
Explanation opens after your attempt
Correct Answer
A. \(-2<x\le 4\)
Step 1
Concept
Subtracting (1) from all parts gives \(-4<2x\le 8\), so \(-2<x\le 4\). Apply the same operation to all three parts.
Step 2
Why this answer is correct
The correct answer is A. \(-2<x\le 4\). Subtracting (1) from all parts gives \(-4<2x\le 8\), so \(-2<x\le 4\). Apply the same operation to all three parts.
Step 3
Exam Tip
सभी पक्षों से (1) घटाने पर \(-4<2x\le 8\), इसलिए \(-2<x\le 4\)। परीक्षा में द्वि-असमीका में तीनों पक्षों पर समान क्रिया करें।
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द्वि-असमीका \(4\le 3x-2<13\) को हल कीजिए।
Solve the compound inequality \(4\le 3x-2<13\).
#compound inequality
#interval
#algebraic solution
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A \(2<x\le 5\)
B \(2\le x<5\)
C \(x\le 2\) या (x>5)
D \(2\le x\le 5\)
Explanation opens after your attempt
Correct Answer
B. \(2\le x<5\)
Step 1
Concept
Adding (2) to all parts gives \(6\le 3x<15\), so \(2\le x<5\). Keep open and closed endpoints correct.
Step 2
Why this answer is correct
The correct answer is B. \(2\le x<5\). Adding (2) to all parts gives \(6\le 3x<15\), so \(2\le x<5\). Keep open and closed endpoints correct.
Step 3
Exam Tip
तीनों पक्षों में (2) जोड़ने पर \(6\le 3x<15\), इसलिए \(2\le x<5\)। परीक्षा में खुले और बंद सिरों को सही रखें।
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असमीका \(\frac{x+3}{2}-\frac{x-1}{3}<4\) का समाधान क्या है?
What is the solution of \(\frac{x+3}{2}-\frac{x-1}{3}<4\)?
#linear inequalities
#fractions
#LCM
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A (x<15)
B (x>13)
C (x<13)
D \(x\le 13\)
Explanation opens after your attempt
Step 1
Concept
Multiplying by (6) gives (3(x+3)-2(x-1)<24), hence (x<13). After clearing fractions, expand brackets carefully.
Step 2
Why this answer is correct
The correct answer is C. (x<13). Multiplying by (6) gives (3(x+3)-2(x-1)<24), hence (x<13). After clearing fractions, expand brackets carefully.
Step 3
Exam Tip
हर (6) से गुणा करने पर (3(x+3)-2(x-1)<24), इसलिए (x<13)। परीक्षा में भिन्नों को हटाने के बाद कोष्ठक जरूर खोलें।
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असमीका \(\frac{2x-1}{3}+\frac{x+5}{6}\ge 2\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{2x-1}{3}+\frac{x+5}{6}\ge 2\).
#linear inequalities
#fractions
#hard
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A \(x\le \frac{9}{5}\)
B \(x\ge 1\)
C \(x> \frac{9}{5}\)
D \(x\ge \frac{9}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge \frac{9}{5}\)
Step 1
Concept
Multiplying by (6) gives (2(2x-1)+(x+5)\ge 12), so \(5x+3\ge 12\) and \(x\ge \frac{9}{5}\). Use the LCM to clear denominators.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge \frac{9}{5}\). Multiplying by (6) gives (2(2x-1)+(x+5)\ge 12), so \(5x+3\ge 12\) and \(x\ge \frac{9}{5}\). Use the LCM to clear denominators.
Step 3
Exam Tip
हर (6) से गुणा करने पर (2(2x-1)+(x+5)\ge 12), इसलिए \(5x+3\ge 12\) और \(x\ge \frac{9}{5}\)। परीक्षा में लघुत्तम समापवर्त्य से हर हटाएं।
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असमीका \(\frac{5x+2}{4}\le \frac{3x-1}{2}+1\) को हल कीजिए।
Solve the inequality \(\frac{5x+2}{4}\le \frac{3x-1}{2}+1\).
#linear inequalities
#fractions
#variable both sides
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A \(x\ge 0\)
B \(x\le 0\)
C (x>0)
D \(x\ge 2\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 0\)
Step 1
Concept
Multiplying by (4) gives (5x+2\le 2(3x-1)+4), hence \(x\ge 0\). Multiply the entire right side while clearing denominators.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 0\). Multiplying by (4) gives (5x+2\le 2(3x-1)+4), hence \(x\ge 0\). Multiply the entire right side while clearing denominators.
Step 3
Exam Tip
हर (4) से गुणा करने पर (5x+2\le 2(3x-1)+4), इसलिए \(x\ge 0\)। परीक्षा में पूरे दाएं पक्ष पर हर का गुणा करें।
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असमीका \(\frac{3x}{4}-2>\frac{x}{2}+5\) का समाधान क्या है?
What is the solution of \(\frac{3x}{4}-2>\frac{x}{2}+5\)?
#linear inequalities
#fractions
#algebraic solution
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A (x<28)
B (x>28)
C \(x>\frac{7}{4}\)
D \(x\le 28\)
Explanation opens after your attempt
Step 1
Concept
We get \(\frac{x}{4}>7\), so (x>28). Keep variable terms on one side and constants on the other.
Step 2
Why this answer is correct
The correct answer is B. (x>28). We get \(\frac{x}{4}>7\), so (x>28). Keep variable terms on one side and constants on the other.
Step 3
Exam Tip
\(\frac{x}{4}>7\) मिलता है, इसलिए (x>28)। परीक्षा में चर पदों को एक तरफ और स्थिर पदों को दूसरी तरफ रखें।
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असमीका (5(2-x)\ge 3(4+x)+1) का समाधान समुच्चय चुनिए।
Choose the solution set of (5(2-x)\ge 3(4+x)+1).
#linear inequalities
#brackets
#negative coefficient
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A \(x\ge -\frac{3}{8}\)
B \(x<-\frac{3}{8}\)
C \(x\le -\frac{3}{8}\)
D \(x\le \frac{3}{8}\)
Explanation opens after your attempt
Correct Answer
C. \(x\le -\frac{3}{8}\)
Step 1
Concept
\(10-5x\ge 13+3x\) gives \(-8x\ge 3\), so \(x\le -\frac{3}{8}\). Reverse the sign when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is C. \(x\le -\frac{3}{8}\). \(10-5x\ge 13+3x\) gives \(-8x\ge 3\), so \(x\le -\frac{3}{8}\). Reverse the sign when dividing by a negative coefficient.
Step 3
Exam Tip
\(10-5x\ge 13+3x\) से \(-8x\ge 3\), इसलिए \(x\le -\frac{3}{8}\)। परीक्षा में ऋणात्मक गुणांक से भाग देने पर चिन्ह पलटें।
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असमीका (-2(x-4)+5<3(1-x)) को हल कीजिए।
Solve the inequality (-2(x-4)+5<3(1-x)).
#linear inequalities
#brackets
#hard
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A (x>-10)
B \(x\le -10\)
C (x<-5)
D (x<-10)
Explanation opens after your attempt
Correct Answer
D. (x<-10)
Step 1
Concept
(-2x+13<3-3x) gives (x+13<3), hence (x<-10). Add variable terms to both sides to simplify.
Step 2
Why this answer is correct
The correct answer is D. (x<-10). (-2x+13<3-3x) gives (x+13<3), hence (x<-10). Add variable terms to both sides to simplify.
Step 3
Exam Tip
(-2x+13<3-3x) से (x+13<3), इसलिए (x<-10)। परीक्षा में दोनों पक्षों में चर जोड़कर सरल करें।
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असमीका \(\frac{x+1}{2}+\frac{x+2}{3}\le \frac{x+3}{4}\) का समाधान क्या है?
What is the solution of \(\frac{x+1}{2}+\frac{x+2}{3}\le \frac{x+3}{4}\)?
#linear inequalities
#fractions
#LCM
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A \(x\le -\frac{5}{7}\)
B \(x\ge -\frac{5}{7}\)
C \(x<-\frac{5}{7}\)
D \(x\le \frac{5}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le -\frac{5}{7}\)
Step 1
Concept
Multiplying by (12) gives \(6x+6+4x+8\le 3x+9\), so \(x\le -\frac{5}{7}\). Clear denominators and combine linear terms.
Step 2
Why this answer is correct
The correct answer is A. \(x\le -\frac{5}{7}\). Multiplying by (12) gives \(6x+6+4x+8\le 3x+9\), so \(x\le -\frac{5}{7}\). Clear denominators and combine linear terms.
Step 3
Exam Tip
हर (12) से गुणा करने पर \(6x+6+4x+8\le 3x+9\), इसलिए \(x\le -\frac{5}{7}\)। परीक्षा में समान हर हटाकर रैखिक पद मिलाएं।
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असमीका \(\frac{4x-7}{5}>\frac{x+2}{10}\) को हल कीजिए।
Solve the inequality \(\frac{4x-7}{5}>\frac{x+2}{10}\).
#linear inequalities
#fractions
#hard
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A \(x<\frac{16}{7}\)
B \(x>\frac{16}{7}\)
C \(x\ge \frac{16}{7}\)
D \(x>\frac{7}{16}\)
Explanation opens after your attempt
Correct Answer
B. \(x>\frac{16}{7}\)
Step 1
Concept
Multiplying by (10) gives (8x-14>x+2), hence \(x>\frac{16}{7}\). Multiply the complete numerator while clearing fractions.
Step 2
Why this answer is correct
The correct answer is B. \(x>\frac{16}{7}\). Multiplying by (10) gives (8x-14>x+2), hence \(x>\frac{16}{7}\). Multiply the complete numerator while clearing fractions.
Step 3
Exam Tip
हर (10) से गुणा करने पर (8x-14>x+2), इसलिए \(x>\frac{16}{7}\)। परीक्षा में भिन्न हटाते समय पूरा अंश गुणा करें।
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असमीका \(\frac{2x}{3}-\frac{5}{6}\ge \frac{x}{2}+1\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(\frac{2x}{3}-\frac{5}{6}\ge \frac{x}{2}+1\).
#linear inequalities
#fractions
#one variable
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A \(x\le 11\)
B (x>11)
C \(x\ge 11\)
D \(x\ge \frac{11}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x\ge 11\)
Step 1
Concept
Multiplying by (6) gives \(4x-5\ge 3x+6\), so \(x\ge 11\). Quickly identify the LCM of small denominators.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge 11\). Multiplying by (6) gives \(4x-5\ge 3x+6\), so \(x\ge 11\). Quickly identify the LCM of small denominators.
Step 3
Exam Tip
हर (6) से गुणा करने पर \(4x-5\ge 3x+6\), इसलिए \(x\ge 11\)। परीक्षा में छोटे हरों का लघुत्तम समापवर्त्य जल्दी लिखें।
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असमीका \(\frac{7-2x}{3}\le \frac{x+1}{6}\) को हल कीजिए।
Solve the inequality \(\frac{7-2x}{3}\le \frac{x+1}{6}\).
#linear inequalities
#fractions
#negative x
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A \(x\le \frac{13}{5}\)
B \(x>\frac{13}{5}\)
C \(x\ge \frac{5}{13}\)
D \(x\ge \frac{13}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge \frac{13}{5}\)
Step 1
Concept
Multiplying by (6) gives \(14-4x\le x+1\), hence \(x\ge \frac{13}{5}\). Moving the variable to the right can sometimes simplify the work.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge \frac{13}{5}\). Multiplying by (6) gives \(14-4x\le x+1\), hence \(x\ge \frac{13}{5}\). Moving the variable to the right can sometimes simplify the work.
Step 3
Exam Tip
हर (6) से गुणा करने पर \(14-4x\le x+1\), इसलिए \(x\ge \frac{13}{5}\)। परीक्षा में चर को दाईं तरफ ले जाना कभी-कभी आसान होता है।
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द्वि-असमीका \(1<\frac{x-2}{3}\le 4\) का समाधान समुच्चय क्या है?
What is the solution set of the compound inequality \(1<\frac{x-2}{3}\le 4\)?
#compound inequality
#fraction
#interval
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A \(5<x\le 14\)
B \(5\le x<14\)
C (x>5) या \(x\le 14\)
D (5<x<14)
Explanation opens after your attempt
Correct Answer
A. \(5<x\le 14\)
Step 1
Concept
\(3<x-2\le 12\) gives \(5<x\le 14\). Carry open and closed signs correctly throughout.
Step 2
Why this answer is correct
The correct answer is A. \(5<x\le 14\). \(3<x-2\le 12\) gives \(5<x\le 14\). Carry open and closed signs correctly throughout.
Step 3
Exam Tip
\(3<x-2\le 12\) से \(5<x\le 14\)। परीक्षा में खुले और बंद चिन्हों को वैसा ही आगे बढ़ाएं।
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द्वि-असमीका \(-2\le \frac{3x+1}{5}<4\) को हल कीजिए।
Solve the compound inequality \(-2\le \frac{3x+1}{5}<4\).
#compound inequality
#fraction
#hard
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A \( -\frac{11}{3}<x\le \frac{19}{3}\)
B \(-\frac{11}{3}\le x<\frac{19}{3}\)
C \(x\le -\frac{11}{3}\) या \(x>\frac{19}{3}\)
D \(-\frac{11}{3}<x<\frac{19}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(-\frac{11}{3}\le x<\frac{19}{3}\)
Step 1
Concept
Multiplying by (5) gives \(-10\le 3x+1<20\), so \(-\frac{11}{3}\le x<\frac{19}{3}\). A positive multiplier does not change the signs.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{11}{3}\le x<\frac{19}{3}\). Multiplying by (5) gives \(-10\le 3x+1<20\), so \(-\frac{11}{3}\le x<\frac{19}{3}\). A positive multiplier does not change the signs.
Step 3
Exam Tip
पहले (5) से गुणा करके \(-10\le 3x+1<20\) मिलता है, इसलिए \(-\frac{11}{3}\le x<\frac{19}{3}\)। परीक्षा में धनात्मक गुणा से चिन्ह नहीं बदलता।
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असमीका (4x-3<18) को संतुष्ट करने वाला सबसे बड़ा पूर्णांक (x) क्या है?
What is the greatest integer (x) satisfying (4x-3<18)?
#linear inequalities
#integer solution
#greatest integer
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A (5)
B (6)
C (4)
D (7)
Explanation opens after your attempt
Step 1
Concept
(4x<21) gives \(x<\frac{21}{4}\), so the greatest integer is (5). When an integer answer is asked, check values near the boundary.
Step 2
Why this answer is correct
The correct answer is A. (5). (4x<21) gives \(x<\frac{21}{4}\), so the greatest integer is (5). When an integer answer is asked, check values near the boundary.
Step 3
Exam Tip
(4x<21) से \(x<\frac{21}{4}\), इसलिए सबसे बड़ा पूर्णांक (5) है। परीक्षा में पूर्णांक उत्तर मांगने पर सीमा के पास जांच करें।
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असमीका \(-3x+10\le 1\) को संतुष्ट करने वाला सबसे छोटा पूर्णांक (x) क्या है?
What is the least integer (x) satisfying \(-3x+10\le 1\)?
#linear inequalities
#integer solution
#least integer
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A (2)
B (3)
C (4)
D (-3)
Explanation opens after your attempt
Step 1
Concept
\(-3x\le -9\) gives \(x\ge 3\), so the least integer is (3). Check the boundary after division by a negative.
Step 2
Why this answer is correct
The correct answer is B. (3). \(-3x\le -9\) gives \(x\ge 3\), so the least integer is (3). Check the boundary after division by a negative.
Step 3
Exam Tip
\(-3x\le -9\) से \(x\ge 3\), इसलिए सबसे छोटा पूर्णांक (3) है। परीक्षा में ऋणात्मक भाग देने के बाद सीमा जांचें।
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असमीका \(-2<x\le 4\) के कितने पूर्णांक समाधान हैं?
How many integer solutions does \(-2<x\le 4\) have?
#linear inequalities
#integer count
#interval
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A (5)
B (7)
C (6)
D (4)
Explanation opens after your attempt
Step 1
Concept
The integers are (-1,0,1,2,3,4), so there are (6) solutions. Do not include an open endpoint in exams.
Step 2
Why this answer is correct
The correct answer is C. (6). The integers are (-1,0,1,2,3,4), so there are (6) solutions. Do not include an open endpoint in exams.
Step 3
Exam Tip
पूर्णांक (-1,0,1,2,3,4) हैं, इसलिए कुल (6) समाधान हैं। परीक्षा में खुले सिरे को शामिल न करें।
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असमीका (2x+1<12) के कितने प्राकृतिक संख्या समाधान हैं?
How many natural number solutions does (2x+1<12) have?
#linear inequalities
#natural numbers
#count
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A (4)
B (6)
C (7)
D (5)
Explanation opens after your attempt
Step 1
Concept
\(x<\frac{11}{2}\), so the natural numbers are (1,2,3,4,5). For natural numbers, start counting from (1).
Step 2
Why this answer is correct
The correct answer is D. (5). \(x<\frac{11}{2}\), so the natural numbers are (1,2,3,4,5). For natural numbers, start counting from (1).
Step 3
Exam Tip
\(x<\frac{11}{2}\), इसलिए प्राकृतिक संख्याएं (1,2,3,4,5) हैं। परीक्षा में प्राकृतिक संख्या के लिए (1) से गिनना याद रखें।
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यदि (x) पूर्णांक है, तो \(3x+2\ge -4\) और (x<5) का समाधान समुच्चय क्या है?
If (x) is an integer, what is the solution set of \(3x+2\ge -4\) and (x<5)?
#linear inequalities
#integer set
#intersection
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A \(x\in{-2,-1,0,1,2,3,4}\)
B \(x\in{-1,0,1,2,3,4,5}\)
C \(x\in{-2,-1,0,1,2,3,4,5}\)
D \(x\in{-3,-2,-1,0,1,2,3}\)
Explanation opens after your attempt
Correct Answer
A. \(x\in{-2,-1,0,1,2,3,4}\)
Step 1
Concept
The first inequality gives \(x\ge -2\) and the second gives (x<5). The integer solutions run from (-2) to (4).
Step 2
Why this answer is correct
The correct answer is A. \(x\in{-2,-1,0,1,2,3,4}\). The first inequality gives \(x\ge -2\) and the second gives (x<5). The integer solutions run from (-2) to (4).
Step 3
Exam Tip
पहली असमीका से \(x\ge -2\) और दूसरी से (x<5)। पूर्णांक समाधान (-2) से (4) तक हैं।
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असमीका \(2x-5\le 7\) को अंतराल रूप में लिखिए।
Write the inequality \(2x-5\le 7\) in interval form.
#linear inequalities
#interval notation
#solution set
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A (\(-\infty,6\))
B (\(-\infty,6]\)
C \([6,\infty\))
D (\(6,\infty\))
Explanation opens after your attempt
Correct Answer
B. (\(-\infty,6]\)
Step 1
Concept
\(2x\le 12\) gives \(x\le 6\), so the interval is (\(-\infty,6]\). Use a closed endpoint for \(\le\).
Step 2
Why this answer is correct
The correct answer is B. (\(-\infty,6]\). \(2x\le 12\) gives \(x\le 6\), so the interval is (\(-\infty,6]\). Use a closed endpoint for \(\le\).
Step 3
Exam Tip
\(2x\le 12\) से \(x\le 6\), इसलिए अंतराल (\(-\infty,6]\) है। परीक्षा में \(\le\) के लिए बंद सिरा लगाएं।
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असमीका \(\frac{x}{2}-\frac{3}{4}>\frac{x}{3}+\frac{1}{6}\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{x}{2}-\frac{3}{4}>\frac{x}{3}+\frac{1}{6}\).
#linear inequalities
#fractions
#LCM
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A \(x<\frac{11}{2}\)
B \(x>\frac{5}{2}\)
C \(x>\frac{11}{2}\)
D \(x\ge \frac{11}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x>\frac{11}{2}\)
Step 1
Concept
Multiplying by (12) gives (6x-9>4x+2), hence \(x>\frac{11}{2}\). Use the LCM to remove small fractions.
Step 2
Why this answer is correct
The correct answer is C. \(x>\frac{11}{2}\). Multiplying by (12) gives (6x-9>4x+2), hence \(x>\frac{11}{2}\). Use the LCM to remove small fractions.
Step 3
Exam Tip
हर (12) से गुणा करने पर (6x-9>4x+2), इसलिए \(x>\frac{11}{2}\)। परीक्षा में छोटे भिन्नों को हटाने के लिए लघुत्तम समापवर्त्य लें।
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असमीका \(\frac{5-3x}{2}\ge \frac{x+1}{4}\) को हल कीजिए।
Solve the inequality \(\frac{5-3x}{2}\ge \frac{x+1}{4}\).
#linear inequalities
#fractions
#negative coefficient
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A \(x\ge \frac{9}{7}\)
B \(x<\frac{9}{7}\)
C \(x\le \frac{7}{9}\)
D \(x\le \frac{9}{7}\)
Explanation opens after your attempt
Correct Answer
D. \(x\le \frac{9}{7}\)
Step 1
Concept
Multiplying by (4) gives \(10-6x\ge x+1\), so \(x\le \frac{9}{7}\). Combine variable terms before dividing.
Step 2
Why this answer is correct
The correct answer is D. \(x\le \frac{9}{7}\). Multiplying by (4) gives \(10-6x\ge x+1\), so \(x\le \frac{9}{7}\). Combine variable terms before dividing.
Step 3
Exam Tip
हर (4) से गुणा करने पर \(10-6x\ge x+1\), इसलिए \(x\le \frac{9}{7}\)। परीक्षा में चर पदों को मिलाने के बाद ही भाग दें।
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एक विद्यार्थी को बोनस (5) अंक जोड़ने के बाद कम से कम (75) अंक चाहिए। यदि मूल अंक (x) हैं, तो (x) का न्यूनतम मान क्या होगा?
A student needs at least (75) marks after adding a bonus of (5) marks. If the original marks are (x), what is the minimum value of (x)?
#linear inequalities
#word problem
#minimum value
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A (70)
B (75)
C (80)
D (65)
Explanation opens after your attempt
Step 1
Concept
The inequality is \(x+5\ge 75\), so \(x\ge 70\). In exams, at least means \(\ge\).
Step 2
Why this answer is correct
The correct answer is A. (70). The inequality is \(x+5\ge 75\), so \(x\ge 70\). In exams, at least means \(\ge\).
Step 3
Exam Tip
असमीका \(x+5\ge 75\) बनेगी, इसलिए \(x\ge 70\)। परीक्षा में कम से कम का अर्थ \(\ge\) होता है।
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एक वस्तु की कीमत (12x+30) रुपये है और यह (150) रुपये से कम होनी चाहिए। (x) के लिए समाधान क्या है?
The cost of an item is (12x+30) rupees and it must be less than (150) rupees. What is the solution for (x)?
#linear inequalities
#word problem
#cost
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A (x<12)
B (x<10)
C \(x\le 10\)
D (x>10)
Explanation opens after your attempt
Step 1
Concept
(12x+30<150) gives (12x<120), so (x<10). Distinguish less than from at most in word problems.
Step 2
Why this answer is correct
The correct answer is B. (x<10). (12x+30<150) gives (12x<120), so (x<10). Distinguish less than from at most in word problems.
Step 3
Exam Tip
(12x+30<150) से (12x<120), इसलिए (x<10)। परीक्षा में कम से कम और कम से कम नहीं जैसे शब्दों में फर्क करें।
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एक आयत की लंबाई ((x+3)) सेमी और चौड़ाई (5) सेमी है। यदि परिमाप अधिकतम (30) सेमी है, तो (x) के लिए समाधान क्या है?
The length of a rectangle is ((x+3)) cm and breadth is (5) cm. If the perimeter is at most (30) cm, what is the solution for (x)?
#linear inequalities
#word problem
#perimeter
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A \(x\ge 7\)
B (x<7)
C \(x\le 7\)
D \(x\le 14\)
Explanation opens after your attempt
Correct Answer
C. \(x\le 7\)
Step 1
Concept
(2((x+3)+5)\le 30) gives \(2x+16\le 30\), so \(x\le 7\). At most means \(\le\) in exams.
Step 2
Why this answer is correct
The correct answer is C. \(x\le 7\). (2((x+3)+5)\le 30) gives \(2x+16\le 30\), so \(x\le 7\). At most means \(\le\) in exams.
Step 3
Exam Tip
(2((x+3)+5)\le 30) से \(2x+16\le 30\), इसलिए \(x\le 7\)। परीक्षा में अधिकतम का अर्थ \(\le\) होता है।
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तापमान (40) से घटकर (40-3x) हो जाता है। यदि नया तापमान (25) से कम है, तो (x) के लिए समाधान क्या है?
The temperature decreases from (40) to (40-3x). If the new temperature is less than (25), what is the solution for (x)?
#linear inequalities
#word problem
#sign reversal
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A (x<5)
B \(x\le 5\)
C (x>3)
D (x>5)
Explanation opens after your attempt
Step 1
Concept
(40-3x<25) gives (-3x<-15), so (x>5). Do not forget to reverse the sign when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is D. (x>5). (40-3x<25) gives (-3x<-15), so (x>5). Do not forget to reverse the sign when dividing by a negative.
Step 3
Exam Tip
(40-3x<25) से (-3x<-15), इसलिए (x>5)। परीक्षा में ऋणात्मक से भाग देने पर चिन्ह पलटना न भूलें।
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असमीका \(\frac{2x+3}{5}-\frac{x-4}{2}\le 1\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(\frac{2x+3}{5}-\frac{x-4}{2}\le 1\).
#linear inequalities
#fractions
#brackets
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A \(x\ge 16\)
B \(x\le 16\)
C (x>16)
D \(x\ge 8\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 16\)
Step 1
Concept
Multiplying by (10) gives (2(2x+3)-5(x-4)\le 10), so \(x\ge 16\). Watch the signs after a subtraction before a bracket.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 16\). Multiplying by (10) gives (2(2x+3)-5(x-4)\le 10), so \(x\ge 16\). Watch the signs after a subtraction before a bracket.
Step 3
Exam Tip
हर (10) से गुणा करने पर (2(2x+3)-5(x-4)\le 10), इसलिए \(x\ge 16\)। परीक्षा में घटाव के बाद आने वाले कोष्ठक के चिन्ह ध्यान रखें।
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असमीका \(\frac{3x-2}{7}+\frac{x+1}{14}>2\) को हल कीजिए।
Solve the inequality \(\frac{3x-2}{7}+\frac{x+1}{14}>2\).
#linear inequalities
#fractions
#hard
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A \(x<\frac{31}{7}\)
B \(x>\frac{31}{7}\)
C \(x\ge \frac{31}{7}\)
D \(x>\frac{7}{31}\)
Explanation opens after your attempt
Correct Answer
B. \(x>\frac{31}{7}\)
Step 1
Concept
Multiplying by (14) gives (2(3x-2)+(x+1)>28), hence \(x>\frac{31}{7}\). Use the same simplification process as equations while tracking the sign.
Step 2
Why this answer is correct
The correct answer is B. \(x>\frac{31}{7}\). Multiplying by (14) gives (2(3x-2)+(x+1)>28), hence \(x>\frac{31}{7}\). Use the same simplification process as equations while tracking the sign.
Step 3
Exam Tip
हर (14) से गुणा करने पर (2(3x-2)+(x+1)>28), इसलिए \(x>\frac{31}{7}\)। परीक्षा में असमीका में भी सामान्य समीकरण जैसी सरलीकरण प्रक्रिया अपनाएं।
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असमीका \(\frac{x-5}{4}\ge \frac{2x+1}{3}-2\) का समाधान क्या है?
What is the solution of \(\frac{x-5}{4}\ge \frac{2x+1}{3}-2\)?
#linear inequalities
#fractions
#variable both sides
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A \(x\ge 1\)
B (x<1)
C \(x\le 1\)
D \(x\le -1\)
Explanation opens after your attempt
Correct Answer
C. \(x\le 1\)
Step 1
Concept
Multiplying by (12) gives (3x-15\ge 4(2x+1)-24), so \(x\le 1\). Combine constants carefully after clearing denominators.
Step 2
Why this answer is correct
The correct answer is C. \(x\le 1\). Multiplying by (12) gives (3x-15\ge 4(2x+1)-24), so \(x\le 1\). Combine constants carefully after clearing denominators.
Step 3
Exam Tip
हर (12) से गुणा करने पर (3x-15\ge 4(2x+1)-24), इसलिए \(x\le 1\)। परीक्षा में हर हटाने के बाद स्थिर पदों को सावधानी से मिलाएं।
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असमीका \(\frac{6-x}{3}<\frac{x-2}{2}\) को हल कीजिए।
Solve the inequality \(\frac{6-x}{3}<\frac{x-2}{2}\).
#linear inequalities
#fractions
#one variable
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A \(x<\frac{18}{5}\)
B \(x>\frac{5}{18}\)
C \(x\ge \frac{18}{5}\)
D \(x>\frac{18}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(x>\frac{18}{5}\)
Step 1
Concept
Multiplying by (6) gives (12-2x<3x-6), hence \(x>\frac{18}{5}\). Move variable terms in the correct direction.
Step 2
Why this answer is correct
The correct answer is D. \(x>\frac{18}{5}\). Multiplying by (6) gives (12-2x<3x-6), hence \(x>\frac{18}{5}\). Move variable terms in the correct direction.
Step 3
Exam Tip
हर (6) से गुणा करने पर (12-2x<3x-6), इसलिए \(x>\frac{18}{5}\)। परीक्षा में दोनों तरफ के चर पदों को सही दिशा में ले जाएं।
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द्वि-असमीका \(-5\le 2-3x<11\) का समाधान समुच्चय चुनिए।
Choose the solution set of the compound inequality \(-5\le 2-3x<11\).
#compound inequality
#negative coefficient
#interval
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A \(-3<x\le \frac{7}{3}\)
B \(-3\le x<\frac{7}{3}\)
C \(x\le -3\) या \(x>\frac{7}{3}\)
D \(-3<x<\frac{7}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(-3<x\le \frac{7}{3}\)
Step 1
Concept
We get \(-7\le -3x<9\), and dividing by (-3) reverses the order, so \(-3<x\le \frac{7}{3}\). Division by a negative changes the direction of a compound inequality.
Step 2
Why this answer is correct
The correct answer is A. \(-3<x\le \frac{7}{3}\). We get \(-7\le -3x<9\), and dividing by (-3) reverses the order, so \(-3<x\le \frac{7}{3}\). Division by a negative changes the direction of a compound inequality.
Step 3
Exam Tip
\(-7\le -3x<9\) मिलता है और (-3) से भाग देने पर क्रम पलटता है, इसलिए \(-3<x\le \frac{7}{3}\)। परीक्षा में ऋणात्मक भाग से द्वि-असमीका की दिशा बदलती है।
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द्वि-असमीका \(-4<\frac{x+2}{2}\le 1\) को हल कीजिए।
Solve the compound inequality \(-4<\frac{x+2}{2}\le 1\).
#compound inequality
#fraction
#interval
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A \(-10\le x<0\)
B \(-10<x\le 0\)
C (-10<x<0)
D \(x\le -10\) या (x>0)
Explanation opens after your attempt
Correct Answer
B. \(-10<x\le 0\)
Step 1
Concept
Multiplying first by (2) gives \(-8<x+2\le 2\), so \(-10<x\le 0\). A positive multiplier does not change the inequality sign.
Step 2
Why this answer is correct
The correct answer is B. \(-10<x\le 0\). Multiplying first by (2) gives \(-8<x+2\le 2\), so \(-10<x\le 0\). A positive multiplier does not change the inequality sign.
Step 3
Exam Tip
पहले (2) से गुणा करने पर \(-8<x+2\le 2\), इसलिए \(-10<x\le 0\)। परीक्षा में धनात्मक गुणा से असमीका का चिन्ह नहीं बदलता।
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असमीका \(\frac{4x+3}{6}-\frac{x-2}{3}\ge \frac{x+5}{2}\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of the inequality \(\frac{4x+3}{6}-\frac{x-2}{3}\ge \frac{x+5}{2}\).
#linear inequalities
#fractions
#algebraic solution
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A \(x\ge -8\)
B (x<-8)
C \(x\le -8\)
D (x> -8)
Explanation opens after your attempt
Correct Answer
C. \(x\le -8\)
Step 1
Concept
Multiplying by (6) gives (4x+3-2(x-2)\ge 3(x+5)), so \(x\le -8\). After clearing denominators, simplify brackets and signs carefully.
Step 2
Why this answer is correct
The correct answer is C. \(x\le -8\). Multiplying by (6) gives (4x+3-2(x-2)\ge 3(x+5)), so \(x\le -8\). After clearing denominators, simplify brackets and signs carefully.
Step 3
Exam Tip
हर (6) से गुणा करने पर (4x+3-2(x-2)\ge 3(x+5)), इसलिए \(x\le -8\)। परीक्षा में हर हटाने के बाद कोष्ठक और चिन्ह ध्यान से सरल करें।
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द्वि-असमीका \(-2<\frac{5-2x}{3}\le 3\) को हल कीजिए।
Solve the compound inequality \(-2<\frac{5-2x}{3}\le 3\).
#compound inequality
#sign reversal
#interval
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A (x<-2) या \(x\ge \frac{11}{2}\)
B \(-2\le x<\frac{11}{2}\)
C \(-2<x\le \frac{11}{2}\)
D \(-2\le x\le \frac{11}{2}\)
Explanation opens after your attempt
Correct Answer
B. \(-2\le x<\frac{11}{2}\)
Step 1
Concept
From \(-6<5-2x\le 9\), we get \(-11<-2x\le 4\), so \(-2\le x<\frac{11}{2}\). In compound inequalities, division by a negative reverses the direction.
Step 2
Why this answer is correct
The correct answer is B. \(-2\le x<\frac{11}{2}\). From \(-6<5-2x\le 9\), we get \(-11<-2x\le 4\), so \(-2\le x<\frac{11}{2}\). In compound inequalities, division by a negative reverses the direction.
Step 3
Exam Tip
\(-6<5-2x\le 9\) से \(-11<-2x\le 4\), इसलिए \(-2\le x<\frac{11}{2}\)। परीक्षा में ऋणात्मक संख्या से भाग देने पर द्वि-असमीका की दिशा बदलती है।
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