असमीका \(7x-9\ge 3x+15\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of the inequality \(7x-9\ge 3x+15\).
#linear inequalities
#one variable
#algebraic solution
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A \(x\ge 6\)
B \(x\le 6\)
C (x>6)
D \(x\ge -6\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 6\)
Step 1
Concept
From \(7x-3x\ge 15+9\), \(4x\ge 24\), so \(x\ge 6\). Keep variable terms and constants separate in exams.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 6\). From \(7x-3x\ge 15+9\), \(4x\ge 24\), so \(x\ge 6\). Keep variable terms and constants separate in exams.
Step 3
Exam Tip
\(7x-3x\ge 15+9\) से \(4x\ge 24\), इसलिए \(x\ge 6\)। परीक्षा में चर पदों और स्थिर पदों को अलग रखें।
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असमीका (2-5x<17) को हल कीजिए।
Solve the inequality (2-5x<17).
#linear inequalities
#negative coefficient
#sign reversal
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A (x<-3)
B (x>-3)
C \(x\ge -3\)
D (x<3)
Explanation opens after your attempt
Step 1
Concept
(-5x<15), and division by a negative reverses the sign, so (x>-3). Always reverse the sign with a negative coefficient.
Step 2
Why this answer is correct
The correct answer is B. (x>-3). (-5x<15), and division by a negative reverses the sign, so (x>-3). Always reverse the sign with a negative coefficient.
Step 3
Exam Tip
(-5x<15) और ऋणात्मक संख्या से भाग देने पर चिन्ह पलटता है, इसलिए (x>-3)। परीक्षा में ऋणात्मक गुणांक पर चिन्ह अवश्य बदलें।
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असमीका \(\frac{x+4}{3}-2\le \frac{x-1}{6}\) का समाधान क्या है?
What is the solution of \(\frac{x+4}{3}-2\le \frac{x-1}{6}\)?
#linear inequalities
#fractions
#LCM
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A \(x\ge 3\)
B (x<3)
C \(x\le 3\)
D \(x\le -3\)
Explanation opens after your attempt
Correct Answer
C. \(x\le 3\)
Step 1
Concept
Multiplying by (6) gives \(2x+8-12\le x-1\), so \(x\le 3\). After clearing fractions, combine all terms carefully.
Step 2
Why this answer is correct
The correct answer is C. \(x\le 3\). Multiplying by (6) gives \(2x+8-12\le x-1\), so \(x\le 3\). After clearing fractions, combine all terms carefully.
Step 3
Exam Tip
हर (6) से गुणा करने पर \(2x+8-12\le x-1\), इसलिए \(x\le 3\)। परीक्षा में भिन्न हटाने के बाद सभी पद ध्यान से मिलाएं।
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असमीका (4(2x-3)>5x+9) का समाधान समुच्चय चुनिए।
Choose the solution set of (4(2x-3)>5x+9).
#linear inequalities
#brackets
#hard
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A (x<7)
B \(x\ge 7\)
C (x>3)
D (x>7)
Explanation opens after your attempt
Step 1
Concept
(8x-12>5x+9) gives (3x>21), so (x>7). Multiply through brackets completely in exams.
Step 2
Why this answer is correct
The correct answer is D. (x>7). (8x-12>5x+9) gives (3x>21), so (x>7). Multiply through brackets completely in exams.
Step 3
Exam Tip
(8x-12>5x+9) से (3x>21), इसलिए (x>7)। परीक्षा में कोष्ठक खोलते समय गुणा पूरा करें।
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असमीका \(\frac{3x-5}{2}\ge \frac{x+7}{4}+3\) को हल कीजिए।
Solve the inequality \(\frac{3x-5}{2}\ge \frac{x+7}{4}+3\).
#linear inequalities
#fractions
#algebraic solution
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A \(x\ge \frac{29}{5}\)
B \(x\le \frac{29}{5}\)
C \(x>\frac{29}{5}\)
D \(x\ge \frac{5}{29}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{29}{5}\)
Step 1
Concept
Multiplying by (4) gives (2(3x-5)\ge x+7+12), so \(x\ge \frac{29}{5}\). Do not forget to multiply the entire right side.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{29}{5}\). Multiplying by (4) gives (2(3x-5)\ge x+7+12), so \(x\ge \frac{29}{5}\). Do not forget to multiply the entire right side.
Step 3
Exam Tip
हर (4) से गुणा करने पर (2(3x-5)\ge x+7+12), इसलिए \(x\ge \frac{29}{5}\)। परीक्षा में पूरे दाएं पक्ष को गुणा करना न भूलें।
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असमीका (11-2(x+5)\le 3x-4) का समाधान क्या होगा?
What will be the solution of (11-2(x+5)\le 3x-4)?
#linear inequalities
#brackets
#one variable
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A \(x\le 1\)
B \(x\ge 1\)
C (x>1)
D \(x\ge -1\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge 1\)
Step 1
Concept
Simplifying gives \(1-2x\le 3x-4\), so \(5\le 5x\) and \(x\ge 1\). Expand brackets with negative multiplication carefully.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge 1\). Simplifying gives \(1-2x\le 3x-4\), so \(5\le 5x\) and \(x\ge 1\). Expand brackets with negative multiplication carefully.
Step 3
Exam Tip
सरलीकरण से \(1-2x\le 3x-4\), इसलिए \(5\le 5x\) और \(x\ge 1\)। परीक्षा में ऋणात्मक गुणा से कोष्ठक सही खोलें।
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असमीका \(\frac{5-4x}{3}>\frac{x+2}{6}\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(\frac{5-4x}{3}>\frac{x+2}{6}\).
#linear inequalities
#fractions
#negative coefficient
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A \(x>\frac{8}{9}\)
B \(x\le \frac{8}{9}\)
C \(x<\frac{8}{9}\)
D \(x<\frac{9}{8}\)
Explanation opens after your attempt
Correct Answer
C. \(x<\frac{8}{9}\)
Step 1
Concept
Multiplying by (6) gives (10-8x>x+2), so \(x<\frac{8}{9}\). Move variables to one side before dividing.
Step 2
Why this answer is correct
The correct answer is C. \(x<\frac{8}{9}\). Multiplying by (6) gives (10-8x>x+2), so \(x<\frac{8}{9}\). Move variables to one side before dividing.
Step 3
Exam Tip
हर (6) से गुणा करने पर (10-8x>x+2), इसलिए \(x<\frac{8}{9}\)। परीक्षा में चर को एक तरफ लाकर फिर भाग दें।
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असमीका \(\frac{3x}{5}+2<\frac{x}{2}+7\) को हल कीजिए।
Solve the inequality \(\frac{3x}{5}+2<\frac{x}{2}+7\).
#linear inequalities
#fractions
#LCM
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A (x>50)
B \(x\le 50\)
C (x<25)
D (x<50)
Explanation opens after your attempt
Step 1
Concept
Multiplying by (10) gives (6x+20<5x+70), so (x<50). Use the LCM to clear fractions in exams.
Step 2
Why this answer is correct
The correct answer is D. (x<50). Multiplying by (10) gives (6x+20<5x+70), so (x<50). Use the LCM to clear fractions in exams.
Step 3
Exam Tip
हर (10) से गुणा करने पर (6x+20<5x+70), इसलिए (x<50)। परीक्षा में लघुत्तम समापवर्त्य से भिन्न हटाएं।
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द्वि-असमीका \(-4\le 2x-1<9\) का समाधान समुच्चय चुनिए।
Choose the solution set of the compound inequality \(-4\le 2x-1<9\).
#compound inequality
#interval
#one variable
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A \(-\frac{3}{2}\le x<5\)
B \(-\frac{3}{2}<x\le 5\)
C \(-\frac{3}{2}<x<5\)
D \(-\frac{3}{2}\le x\le 5\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{3}{2}\le x<5\)
Step 1
Concept
Adding (1) to all parts gives \(-3\le 2x<10\), so \(-\frac{3}{2}\le x<5\). Keep open and closed endpoints correct.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\le x<5\). Adding (1) to all parts gives \(-3\le 2x<10\), so \(-\frac{3}{2}\le x<5\). Keep open and closed endpoints correct.
Step 3
Exam Tip
तीनों पक्षों में (1) जोड़ने पर \(-3\le 2x<10\), इसलिए \(-\frac{3}{2}\le x<5\)। परीक्षा में खुले और बंद सिरों को सही रखें।
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असमीका \(6-\frac{3x-2}{5}\ge 1\) का समाधान क्या है?
What is the solution of \(6-\frac{3x-2}{5}\ge 1\)?
#linear inequalities
#fractions
#brackets
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A \(x\ge 9\)
B (x<9)
C \(x\le 9\)
D \(x\le -9\)
Explanation opens after your attempt
Correct Answer
C. \(x\le 9\)
Step 1
Concept
Multiplying by (5) gives (30-(3x-2)\ge 5), so \(x\le 9\). Put the full numerator in brackets after subtraction.
Step 2
Why this answer is correct
The correct answer is C. \(x\le 9\). Multiplying by (5) gives (30-(3x-2)\ge 5), so \(x\le 9\). Put the full numerator in brackets after subtraction.
Step 3
Exam Tip
हर (5) से गुणा करने पर (30-(3x-2)\ge 5), इसलिए \(x\le 9\)। परीक्षा में घटाव से पहले पूरा अंश कोष्ठक में रखें।
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असमीका (-2(3-x)+4>x+10) को हल कीजिए।
Solve the inequality (-2(3-x)+4>x+10).
#linear inequalities
#brackets
#signs
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A (x<12)
B \(x\ge 12\)
C (x>6)
D (x>12)
Explanation opens after your attempt
Step 1
Concept
(-6+2x+4>x+10) gives (x>12). Carefully handle multiplication of two negative signs.
Step 2
Why this answer is correct
The correct answer is D. (x>12). (-6+2x+4>x+10) gives (x>12). Carefully handle multiplication of two negative signs.
Step 3
Exam Tip
(-6+2x+4>x+10) से (x>12)। परीक्षा में दो ऋण चिन्हों के गुणा को सावधानी से देखें।
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असमीका \(0.25x-1.5\ge 2\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(0.25x-1.5\ge 2\).
#linear inequalities
#decimals
#one variable
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A \(x\ge 14\)
B \(x\le 14\)
C (x>14)
D \(x\ge 7\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 14\)
Step 1
Concept
\(0.25x\ge 3.5\), so \(x\ge 14\). Converting decimals to fractions can simplify calculations.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 14\). \(0.25x\ge 3.5\), so \(x\ge 14\). Converting decimals to fractions can simplify calculations.
Step 3
Exam Tip
\(0.25x\ge 3.5\), इसलिए \(x\ge 14\)। परीक्षा में दशमलव को भिन्न में बदलना गणना को सरल बना सकता है।
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असमीका \(2.4-0.6x\le -1.2\) को हल कीजिए।
Solve the inequality \(2.4-0.6x\le -1.2\).
#linear inequalities
#decimals
#sign reversal
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A \(x\le 6\)
B \(x\ge 6\)
C (x>6)
D \(x\ge -6\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge 6\)
Step 1
Concept
\(-0.6x\le -3.6\), and division by a negative gives \(x\ge 6\). The sign reversal rule applies to decimals too.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge 6\). \(-0.6x\le -3.6\), and division by a negative gives \(x\ge 6\). The sign reversal rule applies to decimals too.
Step 3
Exam Tip
\(-0.6x\le -3.6\) और ऋणात्मक संख्या से भाग देने पर \(x\ge 6\)। परीक्षा में दशमलव के साथ भी चिन्ह पलटने का नियम लागू होता है।
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असमीका \(\frac{x-8}{5}+\frac{2x+1}{10}<3\) का समाधान क्या है?
What is the solution of \(\frac{x-8}{5}+\frac{2x+1}{10}<3\)?
#linear inequalities
#fractions
#hard
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A \(x>\frac{45}{4}\)
B \(x\le \frac{45}{4}\)
C \(x<\frac{45}{4}\)
D \(x<\frac{4}{45}\)
Explanation opens after your attempt
Correct Answer
C. \(x<\frac{45}{4}\)
Step 1
Concept
Multiplying by (10) gives (2x-16+2x+1<30), so \(x<\frac{45}{4}\). Check signs before combining like terms.
Step 2
Why this answer is correct
The correct answer is C. \(x<\frac{45}{4}\). Multiplying by (10) gives (2x-16+2x+1<30), so \(x<\frac{45}{4}\). Check signs before combining like terms.
Step 3
Exam Tip
हर (10) से गुणा करने पर (2x-16+2x+1<30), इसलिए \(x<\frac{45}{4}\)। परीक्षा में समान पदों को मिलाने से पहले चिन्ह जांचें।
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असमीका \(\frac{7x+3}{9}\ge \frac{2x-1}{3}\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{7x+3}{9}\ge \frac{2x-1}{3}\).
#linear inequalities
#fractions
#variable both sides
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A \(x\le -6\)
B (x>-6)
C \(x\ge 6\)
D \(x\ge -6\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge -6\)
Step 1
Concept
Multiplying by (9) gives \(7x+3\ge 6x-3\), so \(x\ge -6\). Multiply the complete numerator while clearing denominators.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge -6\). Multiplying by (9) gives \(7x+3\ge 6x-3\), so \(x\ge -6\). Multiply the complete numerator while clearing denominators.
Step 3
Exam Tip
हर (9) से गुणा करने पर \(7x+3\ge 6x-3\), इसलिए \(x\ge -6\)। परीक्षा में हर हटाने पर पूरे अंश को गुणा करें।
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असमीका (5x+4<31) को संतुष्ट करने वाला सबसे बड़ा पूर्णांक (x) क्या है?
What is the greatest integer (x) satisfying (5x+4<31)?
#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
(5x<27) gives \(x<\frac{27}{5}\), so the greatest integer is (5). Check integers near the boundary.
Step 2
Why this answer is correct
The correct answer is A. (5). (5x<27) gives \(x<\frac{27}{5}\), so the greatest integer is (5). Check integers near the boundary.
Step 3
Exam Tip
(5x<27) से \(x<\frac{27}{5}\), इसलिए सबसे बड़ा पूर्णांक (5) है। परीक्षा में सीमा के पास वाले पूर्णांक जांचें।
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असमीका (-4x+6<-10) को संतुष्ट करने वाला सबसे छोटा पूर्णांक (x) क्या है?
What is the least integer (x) satisfying (-4x+6<-10)?
#linear inequalities
#integer solution
#least integer
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A (4)
B (5)
C (6)
D (3)
Explanation opens after your attempt
Step 1
Concept
(-4x<-16) gives (x>4), so the least integer is (5). Do not include the boundary in a strict inequality.
Step 2
Why this answer is correct
The correct answer is B. (5). (-4x<-16) gives (x>4), so the least integer is (5). Do not include the boundary in a strict inequality.
Step 3
Exam Tip
(-4x<-16) से (x>4), इसलिए सबसे छोटा पूर्णांक (5) है। परीक्षा में सख्त असमीका में सीमा को शामिल न करें।
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असमीका \(-3\le x<4\) के कितने पूर्णांक समाधान हैं?
How many integer solutions does \(-3\le x<4\) have?
#linear inequalities
#integer count
#interval
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A (6)
B (8)
C (7)
D (5)
Explanation opens after your attempt
Step 1
Concept
The integers are (-3,-2,-1,0,1,2,3), so there are (7) solutions. Include closed endpoints and exclude open endpoints.
Step 2
Why this answer is correct
The correct answer is C. (7). The integers are (-3,-2,-1,0,1,2,3), so there are (7) solutions. Include closed endpoints and exclude open endpoints.
Step 3
Exam Tip
पूर्णांक (-3,-2,-1,0,1,2,3) हैं, इसलिए कुल (7) समाधान हैं। परीक्षा में बंद सिरा शामिल करें और खुला सिरा छोड़ें।
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असमीका \(3x-2\le 16\) के कितने प्राकृतिक संख्या समाधान हैं?
How many natural number solutions does \(3x-2\le 16\) have?
#linear inequalities
#natural numbers
#count
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A (5)
B (7)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
\(3x\le 18\) gives \(x\le 6\), so natural solutions are from (1) to (6). Count natural numbers from (1) in exams.
Step 2
Why this answer is correct
The correct answer is D. (6). \(3x\le 18\) gives \(x\le 6\), so natural solutions are from (1) to (6). Count natural numbers from (1) in exams.
Step 3
Exam Tip
\(3x\le 18\) से \(x\le 6\), इसलिए प्राकृतिक समाधान (1) से (6) तक हैं। परीक्षा में प्राकृतिक संख्याओं की गिनती (1) से करें।
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किसी संख्या (x) के दोगुने में (7) जोड़ने पर परिणाम (23) से अधिक है। (x) के लिए समाधान क्या है?
Twice a number (x) increased by (7) is greater than (23). What is the solution for (x)?
#linear inequalities
#word problem
#translation
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A (x>8)
B (x<8)
C \(x\ge 8\)
D (x>15)
Explanation opens after your attempt
Step 1
Concept
The inequality is (2x+7>23), so (x>8). Greater than means (>) in exams.
Step 2
Why this answer is correct
The correct answer is A. (x>8). The inequality is (2x+7>23), so (x>8). Greater than means (>) in exams.
Step 3
Exam Tip
असमीका (2x+7>23) बनेगी, इसलिए (x>8)। परीक्षा में अधिक है का अर्थ (>) होता है।
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टिकट का कुल खर्च (15x+40) रुपये है और यह अधिकतम (250) रुपये होना चाहिए। (x) के लिए समाधान क्या है?
The total ticket cost is (15x+40) rupees and it should be at most (250) rupees. What is the solution for (x)?
#linear inequalities
#word problem
#cost
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A \(x\ge 14\)
B \(x\le 14\)
C (x<14)
D \(x\le 15\)
Explanation opens after your attempt
Correct Answer
B. \(x\le 14\)
Step 1
Concept
\(15x+40\le 250\) gives \(15x\le 210\), so \(x\le 14\). At most means \(\le\) in exams.
Step 2
Why this answer is correct
The correct answer is B. \(x\le 14\). \(15x+40\le 250\) gives \(15x\le 210\), so \(x\le 14\). At most means \(\le\) in exams.
Step 3
Exam Tip
\(15x+40\le 250\) से \(15x\le 210\), इसलिए \(x\le 14\)। परीक्षा में अधिकतम का अर्थ \(\le\) होता है।
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एक आयत की चौड़ाई ((x-2)) सेमी और लंबाई (8) सेमी है। यदि परिमाप कम से कम (40) सेमी है, तो (x) के लिए समाधान क्या है?
The breadth of a rectangle is ((x-2)) cm and length is (8) cm. If the perimeter is at least (40) cm, what is the solution for (x)?
#linear inequalities
#word problem
#perimeter
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A \(x\le 14\)
B (x>14)
C \(x\ge 14\)
D \(x\ge 10\)
Explanation opens after your attempt
Correct Answer
C. \(x\ge 14\)
Step 1
Concept
(2((x-2)+8)\ge 40) gives \(2x+12\ge 40\), so \(x\ge 14\). At least means \(\ge\) in exams.
Step 2
Why this answer is correct
The correct answer is C. \(x\ge 14\). (2((x-2)+8)\ge 40) gives \(2x+12\ge 40\), so \(x\ge 14\). At least means \(\ge\) in exams.
Step 3
Exam Tip
(2((x-2)+8)\ge 40) से \(2x+12\ge 40\), इसलिए \(x\ge 14\)। परीक्षा में कम से कम का अर्थ \(\ge\) होता है।
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किसी खाते में (500) रुपये हैं और हर दिन (25x) रुपये खर्च होते हैं। यदि बची राशि (200) रुपये से अधिक रहनी चाहिए, तो (x) के लिए समाधान क्या है?
An account has (500) rupees and (25x) rupees are spent. If the remaining amount must be more than (200) rupees, what is the solution for (x)?
#linear inequalities
#word problem
#sign reversal
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A (x>12)
B \(x\le 12\)
C (x<8)
D (x<12)
Explanation opens after your attempt
Step 1
Concept
(500-25x>200) gives (-25x>-300), so (x<12). Reverse the sign when dividing by a negative coefficient.
Step 2
Why this answer is correct
The correct answer is D. (x<12). (500-25x>200) gives (-25x>-300), so (x<12). Reverse the sign when dividing by a negative coefficient.
Step 3
Exam Tip
(500-25x>200) से (-25x>-300), इसलिए (x<12)। परीक्षा में ऋणात्मक गुणांक से भाग देने पर चिन्ह पलटें।
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द्वि-असमीका \(-1\le \frac{2x+5}{3}<7\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(-1\le \frac{2x+5}{3}<7\).
#compound inequality
#interval notation
#fraction
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A ([-4,8))
B ((-4,8])
C ((-4,8))
D ([-4,8])
Explanation opens after your attempt
Correct Answer
A. ([-4,8))
Step 1
Concept
\(-3\le 2x+5<21\) gives \(-8\le 2x<16\), so ([-4,8)). Write open and closed endpoints correctly in interval form.
Step 2
Why this answer is correct
The correct answer is A. ([-4,8)). \(-3\le 2x+5<21\) gives \(-8\le 2x<16\), so ([-4,8)). Write open and closed endpoints correctly in interval form.
Step 3
Exam Tip
\(-3\le 2x+5<21\) से \(-8\le 2x<16\), इसलिए ([-4,8))। परीक्षा में अंतराल रूप में खुले और बंद सिरों को सही लिखें।
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द्वि-असमीका \(-6<4-2x\le 10\) को हल कीजिए।
Solve the compound inequality \(-6<4-2x\le 10\).
#compound inequality
#negative coefficient
#interval
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A (x<-3) या \(x\ge 5\)
B \(-3\le x<5\)
C \(-3<x\le 5\)
D (-3<x<5)
Explanation opens after your attempt
Correct Answer
B. \(-3\le x<5\)
Step 1
Concept
We get \(-10<-2x\le 6\), and division by (-2) reverses direction, so \(-3\le x<5\). Reverse both signs when dividing by a negative.
Step 2
Why this answer is correct
The correct answer is B. \(-3\le x<5\). We get \(-10<-2x\le 6\), and division by (-2) reverses direction, so \(-3\le x<5\). Reverse both signs when dividing by a negative.
Step 3
Exam Tip
\(-10<-2x\le 6\) मिलता है और (-2) से भाग देने पर दिशा बदलती है, इसलिए \(-3\le x<5\)। परीक्षा में ऋणात्मक भाग से दोनों चिन्हों की दिशा बदलें।
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द्वि-असमीका \(0<\frac{9-3x}{6}\le 2\) का समाधान क्या है?
What is the solution of the compound inequality \(0<\frac{9-3x}{6}\le 2\)?
#compound inequality
#fraction
#sign reversal
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A (x<-1) या \(x\ge 3\)
B \(-1<x\le 3\)
C \(-1\le x<3\)
D (-1<x<3)
Explanation opens after your attempt
Correct Answer
C. \(-1\le x<3\)
Step 1
Concept
\(0<9-3x\le 12\) gives (x<3) and \(x\ge -1\), so \(-1\le x<3\). Check both bounds separately.
Step 2
Why this answer is correct
The correct answer is C. \(-1\le x<3\). \(0<9-3x\le 12\) gives (x<3) and \(x\ge -1\), so \(-1\le x<3\). Check both bounds separately.
Step 3
Exam Tip
\(0<9-3x\le 12\) से (x<3) और \(x\ge -1\), इसलिए \(-1\le x<3\)। परीक्षा में दोनों सीमाओं को अलग-अलग जांचें।
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द्वि-असमीका \(-2\le \frac{5x-1}{4}<3\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(-2\le \frac{5x-1}{4}<3\).
#compound inequality
#interval
#fraction
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A \(x\le -\frac{7}{5}\) या \(x>\frac{13}{5}\)
B \(-\frac{7}{5}<x\le \frac{13}{5}\)
C \(-\frac{7}{5}<x<\frac{13}{5}\)
D \(-\frac{7}{5}\le x<\frac{13}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(-\frac{7}{5}\le x<\frac{13}{5}\)
Step 1
Concept
\(-8\le 5x-1<12\) gives \(-7\le 5x<13\), so \(-\frac{7}{5}\le x<\frac{13}{5}\). Preserve boundary signs at every step.
Step 2
Why this answer is correct
The correct answer is D. \(-\frac{7}{5}\le x<\frac{13}{5}\). \(-8\le 5x-1<12\) gives \(-7\le 5x<13\), so \(-\frac{7}{5}\le x<\frac{13}{5}\). Preserve boundary signs at every step.
Step 3
Exam Tip
\(-8\le 5x-1<12\) से \(-7\le 5x<13\), इसलिए \(-\frac{7}{5}\le x<\frac{13}{5}\)। परीक्षा में हर चरण में सीमा चिन्ह बनाए रखें।
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असमीका \(\frac{x+2}{3}+\frac{x-5}{4}\ge 1\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(\frac{x+2}{3}+\frac{x-5}{4}\ge 1\).
#linear inequalities
#fractions
#LCM
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A \(x\ge \frac{19}{7}\)
B \(x\le \frac{19}{7}\)
C \(x>\frac{19}{7}\)
D \(x\ge \frac{7}{19}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{19}{7}\)
Step 1
Concept
Multiplying by (12) gives \(4x+8+3x-15\ge 12\), so \(x\ge \frac{19}{7}\). Use the LCM to clear denominators.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{19}{7}\). Multiplying by (12) gives \(4x+8+3x-15\ge 12\), so \(x\ge \frac{19}{7}\). Use the LCM to clear denominators.
Step 3
Exam Tip
हर (12) से गुणा करने पर \(4x+8+3x-15\ge 12\), इसलिए \(x\ge \frac{19}{7}\)। परीक्षा में लघुत्तम समापवर्त्य से हर हटाएं।
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असमीका \(\frac{2x-3}{5}-\frac{x+1}{2}\le -1\) को हल कीजिए।
Solve the inequality \(\frac{2x-3}{5}-\frac{x+1}{2}\le -1\).
#linear inequalities
#fractions
#subtraction
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A \(x\le -1\)
B \(x\ge -1\)
C (x>-1)
D \(x\ge 1\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge -1\)
Step 1
Concept
Multiplying by (10) gives \(4x-6-5x-5\le -10\), so \(x\ge -1\). Change signs of both terms after a subtracted fraction.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge -1\). Multiplying by (10) gives \(4x-6-5x-5\le -10\), so \(x\ge -1\). Change signs of both terms after a subtracted fraction.
Step 3
Exam Tip
हर (10) से गुणा करने पर \(4x-6-5x-5\le -10\), इसलिए \(x\ge -1\)। परीक्षा में घटाव वाले भिन्न के दोनों पदों के चिन्ह बदलें।
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असमीका \(\frac{3-2x}{4}+\frac{x-6}{3}>0\) का समाधान क्या है?
What is the solution of \(\frac{3-2x}{4}+\frac{x-6}{3}>0\)?
#linear inequalities
#fractions
#negative coefficient
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A \(x>-\frac{15}{2}\)
B \(x\le -\frac{15}{2}\)
C \(x<-\frac{15}{2}\)
D \(x<\frac{15}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x<-\frac{15}{2}\)
Step 1
Concept
Multiplying by (12) gives (9-6x+4x-24>0), so \(x<-\frac{15}{2}\). Reverse the sign when a negative coefficient remains.
Step 2
Why this answer is correct
The correct answer is C. \(x<-\frac{15}{2}\). Multiplying by (12) gives (9-6x+4x-24>0), so \(x<-\frac{15}{2}\). Reverse the sign when a negative coefficient remains.
Step 3
Exam Tip
हर (12) से गुणा करने पर (9-6x+4x-24>0), इसलिए \(x<-\frac{15}{2}\)। परीक्षा में ऋणात्मक गुणांक मिलने पर चिन्ह पलटें।
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असमीका \(\frac{5x+4}{8}<\frac{x-2}{2}+3\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{5x+4}{8}<\frac{x-2}{2}+3\).
#linear inequalities
#fractions
#variable both sides
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A (x>12)
B \(x\le 12\)
C (x<6)
D (x<12)
Explanation opens after your attempt
Step 1
Concept
Multiplying by (8) gives (5x+4<4x-8+24), so (x<12). Combine constants correctly after clearing denominators.
Step 2
Why this answer is correct
The correct answer is D. (x<12). Multiplying by (8) gives (5x+4<4x-8+24), so (x<12). Combine constants correctly after clearing denominators.
Step 3
Exam Tip
हर (8) से गुणा करने पर (5x+4<4x-8+24), इसलिए (x<12)। परीक्षा में हर हटाने के बाद स्थिर पदों को सही मिलाएं।
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असमीका (2x+5>2x+9) का समाधान समुच्चय क्या है?
What is the solution set of the inequality (2x+5>2x+9)?
#linear inequalities
#no solution
#identity check
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A \(\varnothing\)
B \(x\in \mathbb{R}\)
C (x>0)
D (x<0)
Explanation opens after your attempt
Correct Answer
A. \(\varnothing\)
Step 1
Concept
Subtracting (2x) from both sides gives (5>9), which is false. In such a case, there is no solution.
Step 2
Why this answer is correct
The correct answer is A. \(\varnothing\). Subtracting (2x) from both sides gives (5>9), which is false. In such a case, there is no solution.
Step 3
Exam Tip
दोनों पक्षों से (2x) हटाने पर (5>9) मिलता है, जो असत्य है। परीक्षा में ऐसी स्थिति में कोई समाधान नहीं होता।
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असमीका \(3x-4\le 3x+1\) का समाधान समुच्चय क्या है?
What is the solution set of \(3x-4\le 3x+1\)?
#linear inequalities
#all real numbers
#identity
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A \(\varnothing\)
B \(x\in \mathbb{R}\)
C \(x\le 1\)
D \(x\ge -4\)
Explanation opens after your attempt
Correct Answer
B. \(x\in \mathbb{R}\)
Step 1
Concept
Subtracting (3x) from both sides gives \(-4\le 1\), which is always true. Such an inequality has all real numbers as solution.
Step 2
Why this answer is correct
The correct answer is B. \(x\in \mathbb{R}\). Subtracting (3x) from both sides gives \(-4\le 1\), which is always true. Such an inequality has all real numbers as solution.
Step 3
Exam Tip
दोनों पक्षों से (3x) हटाने पर \(-4\le 1\) मिलता है, जो सदैव सत्य है। परीक्षा में ऐसी असमीका का समाधान सभी वास्तविक संख्याएं होता है।
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यदि (4x-1>7) और \(2x+3\le 17\), तो संयुक्त समाधान समुच्चय क्या है?
If (4x-1>7) and \(2x+3\le 17\), what is the common solution set?
#linear inequalities
#intersection
#compound condition
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A \(x\le 2\) या (x>7)
B \(2\le x<7\)
C \(2<x\le 7\)
D (2<x<7)
Explanation opens after your attempt
Correct Answer
C. \(2<x\le 7\)
Step 1
Concept
The first inequality gives (x>2), and the second gives \(x\le 7\). Their intersection is \(2<x\le 7\).
Step 2
Why this answer is correct
The correct answer is C. \(2<x\le 7\). The first inequality gives (x>2), and the second gives \(x\le 7\). Their intersection is \(2<x\le 7\).
Step 3
Exam Tip
पहली असमीका से (x>2) और दूसरी से \(x\le 7\) मिलता है। दोनों का प्रतिच्छेद \(2<x\le 7\) है।
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यदि \(-3x+5\ge -7\) और (x+2>0), तो संयुक्त समाधान समुच्चय ज्ञात कीजिए।
If \(-3x+5\ge -7\) and (x+2>0), find the common solution set.
#linear inequalities
#intersection
#one variable
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A \(x\le -2\) या (x>4)
B \(-2\le x<4\)
C (-2<x<4)
D \(-2<x\le 4\)
Explanation opens after your attempt
Correct Answer
D. \(-2<x\le 4\)
Step 1
Concept
The first inequality gives \(x\le 4\), and the second gives (x>-2). The common solution is \(-2<x\le 4\).
Step 2
Why this answer is correct
The correct answer is D. \(-2<x\le 4\). The first inequality gives \(x\le 4\), and the second gives (x>-2). The common solution is \(-2<x\le 4\).
Step 3
Exam Tip
पहली असमीका से \(x\le 4\) और दूसरी से (x>-2) मिलता है। संयुक्त समाधान \(-2<x\le 4\) है।
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असमीका \(\frac{3x+7}{2}\ge -4\) को अंतराल रूप में लिखिए।
Write the inequality \(\frac{3x+7}{2}\ge -4\) in interval form.
#linear inequalities
#interval notation
#solution set
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A \([-5,\infty\))
B (\(-\infty,-5]\)
C (\(-5,\infty\))
D (\(-\infty,-5\))
Explanation opens after your attempt
Correct Answer
A. \([-5,\infty\))
Step 1
Concept
\(3x+7\ge -8\) gives \(x\ge -5\), so the interval is \([-5,\infty\)). Use a closed endpoint for \(\ge\).
Step 2
Why this answer is correct
The correct answer is A. \([-5,\infty\)). \(3x+7\ge -8\) gives \(x\ge -5\), so the interval is \([-5,\infty\)). Use a closed endpoint for \(\ge\).
Step 3
Exam Tip
\(3x+7\ge -8\) से \(x\ge -5\), इसलिए अंतराल \([-5,\infty\)) है। परीक्षा में \(\ge\) के लिए बंद सिरा लगाएं।
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असमीका \(\frac{5-2x}{7}<-1\) का अंतराल रूप क्या है?
What is the interval form of \(\frac{5-2x}{7}<-1\)?
#linear inequalities
#interval notation
#sign reversal
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A (\(-\infty,6\))
B (\(6,\infty\))
C \([6,\infty\))
D (\(-\infty,6]\)
Explanation opens after your attempt
Correct Answer
B. (\(6,\infty\))
Step 1
Concept
(5-2x<-7) gives (-2x<-12), so (x>6). Use an open endpoint for a strict inequality.
Step 2
Why this answer is correct
The correct answer is B. (\(6,\infty\)). (5-2x<-7) gives (-2x<-12), so (x>6). Use an open endpoint for a strict inequality.
Step 3
Exam Tip
(5-2x<-7) से (-2x<-12), इसलिए (x>6)। परीक्षा में सख्त असमीका के लिए खुला सिरा लगाएं।
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निम्न में से कौन-सा मान असमीका \(2x-3\le 5\) को संतुष्ट नहीं करता है?
Which of the following values does not satisfy the inequality \(2x-3\le 5\)?
#linear inequalities
#checking solution
#one variable
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A (x=1)
B (x=4)
C (x=5)
D (x=0)
Explanation opens after your attempt
Step 1
Concept
The inequality gives \(2x\le 8\) and \(x\le 4\). Hence (x=5) is not a solution.
Step 2
Why this answer is correct
The correct answer is C. (x=5). The inequality gives \(2x\le 8\) and \(x\le 4\). Hence (x=5) is not a solution.
Step 3
Exam Tip
असमीका से \(2x\le 8\) और \(x\le 4\) मिलता है। इसलिए (x=5) समाधान नहीं है।
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यदि (x) पूर्णांक है, तो \(-5<3x+1\le 10\) के कितने समाधान हैं?
If (x) is an integer, how many solutions does \(-5<3x+1\le 10\) have?
#linear inequalities
#integer count
#compound inequality
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A (4)
B (6)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
\(-6<3x\le 9\) gives \(-2<x\le 3\), so the integers are (-1,0,1,2,3). There are (5) solutions.
Step 2
Why this answer is correct
The correct answer is D. (5). \(-6<3x\le 9\) gives \(-2<x\le 3\), so the integers are (-1,0,1,2,3). There are (5) solutions.
Step 3
Exam Tip
\(-6<3x\le 9\) से \(-2<x\le 3\), इसलिए पूर्णांक (-1,0,1,2,3) हैं। कुल (5) समाधान मिलते हैं।
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असमीका \(\frac{x}{4}+\frac{x}{6}\le 5\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(\frac{x}{4}+\frac{x}{6}\le 5\).
#linear inequalities
#fractions
#one variable
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A \(x\le 12\)
B \(x\ge 12\)
C (x<12)
D \(x\le 10\)
Explanation opens after your attempt
Correct Answer
A. \(x\le 12\)
Step 1
Concept
Multiplying by (12) gives \(3x+2x\le 60\), so \(x\le 12\). Add like variable terms to simplify.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 12\). Multiplying by (12) gives \(3x+2x\le 60\), so \(x\le 12\). Add like variable terms to simplify.
Step 3
Exam Tip
हर (12) से गुणा करने पर \(3x+2x\le 60\), इसलिए \(x\le 12\)। परीक्षा में समान चर पदों को जोड़कर सरल करें।
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असमीका \(\frac{3x}{2}-\frac{5}{3}>\frac{1}{6}\) को हल कीजिए।
Solve the inequality \(\frac{3x}{2}-\frac{5}{3}>\frac{1}{6}\).
#linear inequalities
#fractions
#LCM
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A \(x<\frac{11}{9}\)
B \(x>\frac{11}{9}\)
C \(x\ge \frac{11}{9}\)
D \(x>\frac{9}{11}\)
Explanation opens after your attempt
Correct Answer
B. \(x>\frac{11}{9}\)
Step 1
Concept
Multiplying by (6) gives (9x-10>1), so \(x>\frac{11}{9}\). Take the LCM of all fractions together.
Step 2
Why this answer is correct
The correct answer is B. \(x>\frac{11}{9}\). Multiplying by (6) gives (9x-10>1), so \(x>\frac{11}{9}\). Take the LCM of all fractions together.
Step 3
Exam Tip
हर (6) से गुणा करने पर (9x-10>1), इसलिए \(x>\frac{11}{9}\)। परीक्षा में सभी भिन्नों का एक साथ लघुत्तम समापवर्त्य लें।
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असमीका \(\frac{2x-7}{3}\le \frac{4-x}{6}\) का समाधान क्या है?
What is the solution of \(\frac{2x-7}{3}\le \frac{4-x}{6}\)?
#linear inequalities
#fractions
#variable both sides
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A \(x\ge \frac{18}{5}\)
B \(x<\frac{18}{5}\)
C \(x\le \frac{18}{5}\)
D \(x\le \frac{5}{18}\)
Explanation opens after your attempt
Correct Answer
C. \(x\le \frac{18}{5}\)
Step 1
Concept
Multiplying by (6) gives \(4x-14\le 4-x\), so \(x\le \frac{18}{5}\). Combine variable terms from both sides carefully.
Step 2
Why this answer is correct
The correct answer is C. \(x\le \frac{18}{5}\). Multiplying by (6) gives \(4x-14\le 4-x\), so \(x\le \frac{18}{5}\). Combine variable terms from both sides carefully.
Step 3
Exam Tip
हर (6) से गुणा करने पर \(4x-14\le 4-x\), इसलिए \(x\le \frac{18}{5}\)। परीक्षा में दोनों पक्षों के चर पदों को सावधानी से मिलाएं।
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असमीका \(\frac{x+9}{-3}<2\) का समाधान समुच्चय चुनिए।
Choose the solution set of \(\frac{x+9}{-3}<2\).
#linear inequalities
#negative denominator
#sign reversal
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A (x<-15)
B \(x\le -15\)
C (x>-3)
D (x>-15)
Explanation opens after your attempt
Correct Answer
D. (x>-15)
Step 1
Concept
Multiplying by (-3) reverses the inequality sign and gives (x+9>-6). Hence (x>-15).
Step 2
Why this answer is correct
The correct answer is D. (x>-15). Multiplying by (-3) reverses the inequality sign and gives (x+9>-6). Hence (x>-15).
Step 3
Exam Tip
(-3) से गुणा करने पर असमीका का चिन्ह पलटता है और (x+9>-6) मिलता है। इसलिए (x>-15)।
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असमीका \(-\frac{2x-1}{5}\ge 3\) को हल कीजिए।
Solve the inequality \(-\frac{2x-1}{5}\ge 3\).
#linear inequalities
#negative numerator
#signs
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A \(x\le -7\)
B \(x\ge -7\)
C (x<-7)
D \(x\le 7\)
Explanation opens after your attempt
Correct Answer
A. \(x\le -7\)
Step 1
Concept
Multiplying by (5) gives (-(2x-1)\ge 15), so \(x\le -7\). Apply the negative sign to the whole numerator.
Step 2
Why this answer is correct
The correct answer is A. \(x\le -7\). Multiplying by (5) gives (-(2x-1)\ge 15), so \(x\le -7\). Apply the negative sign to the whole numerator.
Step 3
Exam Tip
हर (5) से गुणा करने पर (-(2x-1)\ge 15), इसलिए \(x\le -7\)। परीक्षा में ऋण चिन्ह को पूरे अंश पर लागू करें।
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असमीका \(8-\frac{x-3}{2}\le 1\) का समाधान क्या है?
What is the solution of \(8-\frac{x-3}{2}\le 1\)?
#linear inequalities
#fractions
#brackets
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A \(x\le 17\)
B \(x\ge 17\)
C (x>17)
D \(x\ge 5\)
Explanation opens after your attempt
Correct Answer
B. \(x\ge 17\)
Step 1
Concept
Multiplying by (2) gives (16-(x-3)\le 2), so \(x\ge 17\). Change signs properly when opening brackets after subtraction.
Step 2
Why this answer is correct
The correct answer is B. \(x\ge 17\). Multiplying by (2) gives (16-(x-3)\le 2), so \(x\ge 17\). Change signs properly when opening brackets after subtraction.
Step 3
Exam Tip
हर (2) से गुणा करने पर (16-(x-3)\le 2), इसलिए \(x\ge 17\)। परीक्षा में घटाव के बाद कोष्ठक खोलते समय चिन्ह बदलें।
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असमीका \(3-\frac{2x+1}{4}>\frac{1}{2}\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of \(3-\frac{2x+1}{4}>\frac{1}{2}\).
#linear inequalities
#fractions
#negative coefficient
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A \(x>\frac{9}{2}\)
B \(x\le \frac{9}{2}\)
C \(x<\frac{9}{2}\)
D \(x<\frac{2}{9}\)
Explanation opens after your attempt
Correct Answer
C. \(x<\frac{9}{2}\)
Step 1
Concept
Multiplying by (4) gives (12-(2x+1)>2), so \(x<\frac{9}{2}\). Remember sign reversal with a negative variable coefficient.
Step 2
Why this answer is correct
The correct answer is C. \(x<\frac{9}{2}\). Multiplying by (4) gives (12-(2x+1)>2), so \(x<\frac{9}{2}\). Remember sign reversal with a negative variable coefficient.
Step 3
Exam Tip
हर (4) से गुणा करने पर (12-(2x+1)>2), इसलिए \(x<\frac{9}{2}\)। परीक्षा में ऋणात्मक चर गुणांक पर चिन्ह पलटना याद रखें।
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असमीका \(\frac{5x-2}{3}-\frac{2x+1}{6}\ge \frac{x}{2}\) को हल कीजिए।
Solve the inequality \(\frac{5x-2}{3}-\frac{2x+1}{6}\ge \frac{x}{2}\).
#linear inequalities
#fractions
#subtraction
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A \(x\le 1\)
B (x>1)
C \(x\ge 2\)
D \(x\ge 1\)
Explanation opens after your attempt
Correct Answer
D. \(x\ge 1\)
Step 1
Concept
Multiplying by (6) gives (2(5x-2)-(2x+1)\ge 3x), so \(x\ge 1\). Put brackets around the subtracted numerator.
Step 2
Why this answer is correct
The correct answer is D. \(x\ge 1\). Multiplying by (6) gives (2(5x-2)-(2x+1)\ge 3x), so \(x\ge 1\). Put brackets around the subtracted numerator.
Step 3
Exam Tip
हर (6) से गुणा करने पर (2(5x-2)-(2x+1)\ge 3x), इसलिए \(x\ge 1\)। परीक्षा में घटाए गए अंश पर कोष्ठक जरूर लगाएं।
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किसी व्यक्ति की आयु से संबंधित शर्त \(4x+3\le 35\) है। (x) का अधिकतम मान क्या हो सकता है?
A person's age-related condition is \(4x+3\le 35\). What can be the maximum value of (x)?
#linear inequalities
#word problem
#maximum value
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A (8)
B (7)
C (9)
D (6)
Explanation opens after your attempt
Step 1
Concept
\(4x\le 32\) gives \(x\le 8\), so the maximum value is (8). For maximum value, check the upper bound.
Step 2
Why this answer is correct
The correct answer is A. (8). \(4x\le 32\) gives \(x\le 8\), so the maximum value is (8). For maximum value, check the upper bound.
Step 3
Exam Tip
\(4x\le 32\) से \(x\le 8\), इसलिए अधिकतम मान (8) है। परीक्षा में अधिकतम मान के लिए ऊपरी सीमा देखें।
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मोबाइल डेटा (120-8x) एमबी बचा है। यदि कम से कम (40) एमबी बचना चाहिए, तो (x) के लिए समाधान क्या है?
Mobile data remaining is (120-8x) MB. If at least (40) MB should remain, what is the solution for (x)?
#linear inequalities
#word problem
#sign reversal
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A \(x\ge 10\)
B \(x\le 10\)
C (x<10)
D \(x\le 20\)
Explanation opens after your attempt
Correct Answer
B. \(x\le 10\)
Step 1
Concept
\(120-8x\ge 40\) gives \(-8x\ge -80\), so \(x\le 10\). Dividing by a negative reverses the sign.
Step 2
Why this answer is correct
The correct answer is B. \(x\le 10\). \(120-8x\ge 40\) gives \(-8x\ge -80\), so \(x\le 10\). Dividing by a negative reverses the sign.
Step 3
Exam Tip
\(120-8x\ge 40\) से \(-8x\ge -80\), इसलिए \(x\le 10\)। परीक्षा में ऋणात्मक से भाग देने पर चिन्ह पलटता है।
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असमीका \(\frac{4x-5}{6}+\frac{x+2}{9}<\frac{3x+1}{2}\) का समाधान समुच्चय ज्ञात कीजिए।
Find the solution set of the inequality \(\frac{4x-5}{6}+\frac{x+2}{9}<\frac{3x+1}{2}\).
#linear inequalities
#fractions
#algebraic solution
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A \(x>\frac{1}{20}\)
B \(x<\frac{1}{20}\)
C \(x\ge \frac{1}{20}\)
D \(x>\frac{20}{1}\)
Explanation opens after your attempt
Correct Answer
A. \(x>\frac{1}{20}\)
Step 1
Concept
Multiplying by (18) gives (3(4x-5)+2(x+2)<9(3x+1)), so \(x>\frac{1}{20}\). After clearing denominators, simplify both sides carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x>\frac{1}{20}\). Multiplying by (18) gives (3(4x-5)+2(x+2)<9(3x+1)), so \(x>\frac{1}{20}\). After clearing denominators, simplify both sides carefully.
Step 3
Exam Tip
हर (18) से गुणा करने पर (3(4x-5)+2(x+2)<9(3x+1)), इसलिए \(x>\frac{1}{20}\)। परीक्षा में हर हटाने के बाद दोनों पक्षों को ध्यान से सरल करें।
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