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
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 \(\frac{7-2x}{5}\le \frac{3x+1}{10}\) का हल चुनिए।
Choose the solution of \(\frac{7-2x}{5}\le \frac{3x+1}{10}\).
#linear inequalities
#fraction inequality
#class 11
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 (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
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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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
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 (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
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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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
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 (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
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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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
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 \(\frac{3x+5}{4}-\frac{x-1}{2}\ge 6\) को हल करें।
Solve the inequality \(\frac{3x+5}{4}-\frac{x-1}{2}\ge 6\).
#linear inequalities
#fraction simplification
#class 11
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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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
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 (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
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 \(-\frac{x}{3}+4\le \frac{2x}{5}-1\) को हल कीजिए।
Solve the inequality \(-\frac{x}{3}+4\le \frac{2x}{5}-1\).
#linear inequalities
#fraction solution
#class 11
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 \(\frac{5x-2}{7}<\frac{3x+8}{14}\) का हल चुनिए।
Choose the solution of \(\frac{5x-2}{7}<\frac{3x+8}{14}\).
#linear inequalities
#rational coefficients
#class 11
#hard
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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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
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 \(\frac{2x+1}{3}\le \frac{x-4}{2}\) 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
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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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
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 (-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
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 \(0.3x-1.2\le 0.6\) का हल समुच्चय चुनिए।
Choose the solution set of \(0.3x-1.2\le 0.6\).
#linear inequalities
#decimals
#solution set
#class 11
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 \(\frac{x-3}{4}+2>\frac{x}{2}\) का हल क्या है?
What is the solution of \(\frac{x-3}{4}+2>\frac{x}{2}\)?
#linear inequalities
#fractions
#one variable
#class 11
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 \(5-2x\ge 17\) को हल कीजिए।
Solve the inequality \(5-2x\ge 17\).
#linear inequalities
#negative coefficient
#sign reversal
#class 11
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 (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
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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प्रतिबंध एंजाइम किस कार्य के लिए प्रसिद्ध हैं?
Restriction enzymes are known for which function?
#restriction enzyme
#dna cutting
#class 11
A आरएनए बनाना / Making RNA
B डीएनए को विशिष्ट स्थानों पर काटना / Cutting DNA at specific sites
C लिपिड बनाना / Making lipids
D प्रकाश संश्लेषण कराना / Performing photosynthesis
Explanation opens after your attempt
Correct Answer
B. डीएनए को विशिष्ट स्थानों पर काटना / Cutting DNA at specific sites
Step 1
Concept
Restriction enzymes cut DNA at specific recognition sites. For exams remember them as molecular scissors.
Step 2
Why this answer is correct
The correct answer is B. डीएनए को विशिष्ट स्थानों पर काटना / Cutting DNA at specific sites. Restriction enzymes cut DNA at specific recognition sites. For exams remember them as molecular scissors.
Step 3
Exam Tip
प्रतिबंध एंजाइम डीएनए को विशेष पहचान स्थलों पर काटते हैं। परीक्षा में इन्हें आणविक कैंची याद रखें।
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असमानता \(9-2x\le 1\) का हल क्या है?
What is the solution of the inequality \(9-2x\le 1\)?
#linear-inequalities
#sign-change
#class-11
#easy
A \(x\ge 4\)
B \(x\le 4\)
C \(x\ge -4\)
D \(x\le -4\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 4\)
Step 1
Concept
Subtracting (9) from \(9-2x\le 1\) gives \(-2x\le -8\), and dividing by (-2) gives \(x\ge 4\). The sign reverses when dividing by a negative number.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 4\). Subtracting (9) from \(9-2x\le 1\) gives \(-2x\le -8\), and dividing by (-2) gives \(x\ge 4\). The sign reverses when dividing by a negative number.
Step 3
Exam Tip
\(9-2x\le 1\) में (9) घटाने पर \(-2x\le -8\) मिलता है और (-2) से भाग देने पर \(x\ge 4\) मिलता है। ऋणात्मक संख्या से भाग देने पर चिन्ह उलटता है।
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किसी संख्या में (3) जोड़ने पर परिणाम (16) से अधिक नहीं है। संख्या का हल क्या है?
When (3) is added to a number, the result is not more than (16). What is the solution for the number?
#linear-inequalities
#word-problem
#class-11
#easy
A \(x\le 13\)
B \(x\ge 13\)
C (x<13)
D (x>13)
Explanation opens after your attempt
Correct Answer
A. \(x\le 13\)
Step 1
Concept
The statement gives \(x+3\le 16\), and subtracting (3) gives \(x\le 13\). Not more than means \(\le\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le 13\). The statement gives \(x+3\le 16\), and subtracting (3) gives \(x\le 13\). Not more than means \(\le\).
Step 3
Exam Tip
वाक्य से \(x+3\le 16\) बनता है और (3) घटाने पर \(x\le 13\) मिलता है। अधिक नहीं का अर्थ \(\le\) होता है।
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असमानता \(2x+1\le 5x-8\) का हल ज्ञात कीजिए।
Find the solution of \(2x+1\le 5x-8\).
#linear-inequalities
#variables-both-sides
#class-11
#easy
A \(x\ge 3\)
B \(x\le 3\)
C \(x\ge -3\)
D \(x\le -3\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 3\)
Step 1
Concept
Subtracting (2x) from both sides gives \(1\le 3x-8\), then \(9\le 3x\). Hence \(x\ge 3\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 3\). Subtracting (2x) from both sides gives \(1\le 3x-8\), then \(9\le 3x\). Hence \(x\ge 3\) is correct.
Step 3
Exam Tip
दोनों पक्षों से (2x) जोड़ने के बजाय (2x) घटाने पर \(1\le 3x-8\) और फिर \(9\le 3x\) मिलता है। इसलिए \(x\ge 3\) सही है।
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असमानता \(\frac{x+3}{4}<2\) का हल क्या है?
What is the solution of \(\frac{x+3}{4}<2\)?
#linear-inequalities
#fractions
#class-11
#easy
A (x<5)
B (x>5)
C (x<8)
D (x>8)
Explanation opens after your attempt
Step 1
Concept
Multiplying by (4) gives (x+3<8), and subtracting (3) gives (x<5). The sign stays the same when clearing a positive denominator.
Step 2
Why this answer is correct
The correct answer is A. (x<5). Multiplying by (4) gives (x+3<8), and subtracting (3) gives (x<5). The sign stays the same when clearing a positive denominator.
Step 3
Exam Tip
(4) से गुणा करने पर (x+3<8) और (3) घटाने पर (x<5) मिलता है। धनात्मक हर हटाते समय चिन्ह वही रहता है।
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असमानता (-4x+5<17) का सही हल चुनिए।
Choose the correct solution of (-4x+5<17).
#linear-inequalities
#sign-change
#class-11
#easy
A (x>-3)
B (x<-3)
C (x>3)
D (x<3)
Explanation opens after your attempt
Step 1
Concept
Subtracting (5) gives (-4x<12), and dividing by (-4) gives (x>-3). The inequality sign reverses when dividing by a negative number.
Step 2
Why this answer is correct
The correct answer is A. (x>-3). Subtracting (5) gives (-4x<12), and dividing by (-4) gives (x>-3). The inequality sign reverses when dividing by a negative number.
Step 3
Exam Tip
(5) घटाने पर (-4x<12) मिलता है और (-4) से भाग देने पर (x>-3) मिलता है। ऋणात्मक संख्या से भाग देने पर असमानता का चिन्ह उलटता है।
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असमानता \(7x+2\le 23\) का हल क्या है?
What is the solution of the inequality \(7x+2\le 23\)?
#linear-inequalities
#algebraic-solution
#class-11
#easy
A \(x\le 3\)
B \(x\ge 3\)
C \(x\le 21\)
D \(x\ge 21\)
Explanation opens after your attempt
Correct Answer
A. \(x\le 3\)
Step 1
Concept
Subtracting (2) from \(7x+2\le 23\) gives \(7x\le 21\), and dividing by (7) gives \(x\le 3\). Division by a positive number does not change the sign.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 3\). Subtracting (2) from \(7x+2\le 23\) gives \(7x\le 21\), and dividing by (7) gives \(x\le 3\). Division by a positive number does not change the sign.
Step 3
Exam Tip
\(7x+2\le 23\) में (2) घटाने पर \(7x\le 21\) और (7) से भाग देने पर \(x\le 3\) मिलता है। धनात्मक संख्या से भाग देने पर चिन्ह नहीं बदलता।
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असमानता \(x\le -3\) के लिए कौन सा मान हल है?
Which value is a solution of \(x\le -3\)?
#linear-inequalities
#verification
#class-11
#easy
A (-4)
B (0)
C (-2)
D (1)
Explanation opens after your attempt
Step 1
Concept
\(-4\le -3\) is true, so (-4) is a solution. Compare negative numbers using the number line.
Step 2
Why this answer is correct
The correct answer is A. (-4). \(-4\le -3\) is true, so (-4) is a solution. Compare negative numbers using the number line.
Step 3
Exam Tip
\(-4\le -3\) सत्य है इसलिए (-4) हल है। ऋणात्मक संख्याओं की तुलना संख्या रेखा से करें।
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असमानता \(3x+4\le 13\) में (x=3) रखने पर क्या निष्कर्ष है?
What is the conclusion after putting (x=3) in \(3x+4\le 13\)?
#linear-inequalities
#verification
#class-11
#easy
A (x=3) हल है / (x=3) is a solution
B (x=3) हल नहीं है / (x=3) is not a solution
C (x=3) से (13<13) मिलता है / (x=3) gives (13<13)
D (x=3) से (13>13) मिलता है / (x=3) gives (13>13)
Explanation opens after your attempt
Correct Answer
A. (x=3) हल है / (x=3) is a solution
Step 1
Concept
Putting (x=3), (3x+4=13), and \(13\le 13\) is true. Equality is included in \(\le\).
Step 2
Why this answer is correct
The correct answer is A. (x=3) हल है / (x=3) is a solution. Putting (x=3), (3x+4=13), and \(13\le 13\) is true. Equality is included in \(\le\).
Step 3
Exam Tip
(x=3) रखने पर (3x+4=13) और \(13\le 13\) सत्य है। \(\le\) में बराबरी भी शामिल होती है।
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