असमानता \(\frac{3x-5}{4}-\frac{x+1}{6}\ge \frac{x}{3}+2\) का हल समुच्चय क्या है?
What is the solution set of the inequality \(\frac{3x-5}{4}-\frac{x+1}{6}\ge \frac{x}{3}+2\)?
#linear inequalities
#class 11
#expert
#one variable
A \(x\ge 9\)
B \(x\le 9\)
C \(x\ge \frac{9}{2}\)
D \(x\le \frac{9}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 9\)
Step 1
Concept
Multiplying by (12) and combining terms gives \(3x\ge27\). In exams, do not reverse the sign when multiplying by a positive number.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 9\). Multiplying by (12) and combining terms gives \(3x\ge27\). In exams, do not reverse the sign when multiplying by a positive number.
Step 3
Exam Tip
हर को (12) से हटाकर पदों को सावधानी से मिलाने पर \(3x\ge27\) मिलता है। परीक्षा में हर हटाने के बाद चिन्ह न बदलें जब गुणक धनात्मक हो।
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असमानता \(\frac{2-3x}{9}<\frac{x+4}{6}\) का हल क्या है?
What is the solution of \(\frac{2-3x}{9}<\frac{x+4}{6}\)?
#linear inequalities
#rational inequality
#class 11
#hard
A \(x>-\frac{8}{15}\)
B \(x<-\frac{8}{15}\)
C \(x\ge -\frac{8}{15}\)
D \(x\le -\frac{8}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{8}{15}\)
Step 1
Concept
Multiplying by (18) gives (2(2-3x)<3(x+4)). Thus (4-6x<3x+12), so \(x>-\frac{8}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{8}{15}\). Multiplying by (18) gives (2(2-3x)<3(x+4)). Thus (4-6x<3x+12), so \(x>-\frac{8}{9}\).
Step 3
Exam Tip
(18) से गुणा करने पर (2(2-3x)<3(x+4)) मिलता है। इससे (4-6x<3x+12), इसलिए \(x>-\frac{8}{9}\)।
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असमानता (4x+7>2(2x+5)) के लिए कौन सा कथन सही है?
Which statement is correct for the inequality (4x+7>2(2x+5))?
#linear inequalities
#no solution
#class 11
A सभी वास्तविक संख्याएँ / All real numbers
B कोई हल नहीं / No solution
C (x>3)
D (x<3)
Explanation opens after your attempt
Correct Answer
B. कोई हल नहीं / No solution
Step 1
Concept
The right side is (4x+10). (4x+7>4x+10) gives (7>10), which is false.
Step 2
Why this answer is correct
The correct answer is B. कोई हल नहीं / No solution. The right side is (4x+10). (4x+7>4x+10) gives (7>10), which is false.
Step 3
Exam Tip
दाएँ पक्ष (4x+10) है। (4x+7>4x+10) से (7>10) मिलता है, जो असत्य है।
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असमानता \(-6+\frac{5x}{2}\le \frac{x-3}{4}\) का हल चुनिए।
Choose the solution of \(-6+\frac{5x}{2}\le \frac{x-3}{4}\).
#linear inequalities
#fractions
#hard
#class 11
A \(x\le \frac{21}{9}\)
B \(x\le \frac{21}{9}\)
C \(x\ge \frac{21}{9}\)
D \(x<\frac{21}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{21}{9}\)
Step 1
Concept
Multiplying by (4) gives \(-24+10x\le x-3\). Thus \(9x\le 21\), so \(x\le \frac{7}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{21}{9}\). Multiplying by (4) gives \(-24+10x\le x-3\). Thus \(9x\le 21\), so \(x\le \frac{7}{3}\).
Step 3
Exam Tip
(4) से गुणा करने पर \(-24+10x\le x-3\) मिलता है। इससे \(9x\le 21\), अतः \(x\le \frac{7}{3}\)।
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यदि (9-4(x+1)<2(3-x)), तो (x) के लिए सही शर्त क्या है?
If (9-4(x+1)<2(3-x)), what is the correct condition for (x)?
#linear inequalities
#brackets
#class 11
A \(x>-\frac{1}{2}\)
B \(x<-\frac{1}{2}\)
C \(x\ge -\frac{1}{2}\)
D \(x\le -\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{1}{2}\)
Step 1
Concept
Simplification gives (5-4x<6-2x). Thus (-1<2x), so \(x>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{1}{2}\). Simplification gives (5-4x<6-2x). Thus (-1<2x), so \(x>-\frac{1}{2}\).
Step 3
Exam Tip
सरलीकरण से (5-4x<6-2x) मिलता है। इससे (-1<2x), अतः \(x>-\frac{1}{2}\)।
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असमानता \(\frac{x-8}{5}\ge \frac{2x+1}{3}-4\) को हल कीजिए।
Solve the inequality \(\frac{x-8}{5}\ge \frac{2x+1}{3}-4\).
#linear inequalities
#fractions
#one variable
#class 11
A \(x\le \frac{47}{7}\)
B \(x\ge \frac{47}{7}\)
C \(x<\frac{47}{7}\)
D \(x>\frac{47}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{47}{7}\)
Step 1
Concept
Multiplying by (15) gives (3x-24\ge 5(2x+1)-60). Thus \(31\ge 7x\), so \(x\le \frac{31}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{47}{7}\). Multiplying by (15) gives (3x-24\ge 5(2x+1)-60). Thus \(31\ge 7x\), so \(x\le \frac{31}{7}\).
Step 3
Exam Tip
(15) से गुणा करने पर (3x-24\ge 5(2x+1)-60) मिलता है। इससे \(31\ge 7x\), इसलिए \(x\le \frac{31}{7}\)।
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असमानता (3(2x-1)-5(x+2)\le x-20) का हल क्या है?
What is the solution of (3(2x-1)-5(x+2)\le x-20)?
#linear inequalities
#no solution
#brackets
#class 11
A \(x\ge \frac{7}{2}\)
B \(x\le \frac{7}{2}\)
C \(x>\frac{7}{2}\)
D \(x<\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{7}{2}\)
Step 1
Concept
The left side is (x-13). The inequality \(x-13\le x-20\) gives false \(-13\le -20\), so there should be no solution.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{7}{2}\). The left side is (x-13). The inequality \(x-13\le x-20\) gives false \(-13\le -20\), so there should be no solution.
Step 3
Exam Tip
बायाँ पक्ष (x-13) है। \(x-13\le x-20\) असत्य \( -13\le -20\) देता है, इसलिए कोई हल नहीं होना चाहिए।
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असमानता \(2.2x-1.1\ge 4.4-0.5x\) को हल करें।
Solve the inequality \(2.2x-1.1\ge 4.4-0.5x\).
#linear inequalities
#decimals
#hard
#class 11
A \(x\ge \frac{55}{27}\)
B \(x\le \frac{55}{27}\)
C \(x>\frac{55}{27}\)
D \(x<\frac{55}{27}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge \frac{55}{27}\)
Step 1
Concept
From \(2.7x\ge 5.5\), \(x\ge \frac{55}{27}\). Multiplying by (10) is useful for removing decimals.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge \frac{55}{27}\). From \(2.7x\ge 5.5\), \(x\ge \frac{55}{27}\). Multiplying by (10) is useful for removing decimals.
Step 3
Exam Tip
\(2.7x\ge 5.5\) से \(x\ge \frac{55}{27}\) मिलता है। दशमलव हटाने के लिए (10) से गुणा करना उपयोगी है।
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यदि \(4-\frac{3x}{2}>1+\frac{x}{6}\), तो (x) का हल क्या है?
If \(4-\frac{3x}{2}>1+\frac{x}{6}\), what is the solution for (x)?
#linear inequalities
#fractions
#class 11
#hard
A \(x<\frac{9}{5}\)
B \(x>\frac{9}{5}\)
C \(x\le \frac{9}{5}\)
D \(x\ge \frac{9}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x<\frac{9}{5}\)
Step 1
Concept
Multiplying by (6) gives (24-9x>6+x). Thus (18>10x), so \(x<\frac{9}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{9}{5}\). Multiplying by (6) gives (24-9x>6+x). Thus (18>10x), so \(x<\frac{9}{5}\).
Step 3
Exam Tip
(6) से गुणा करने पर (24-9x>6+x) मिलता है। इससे (18>10x), अतः \(x<\frac{9}{5}\)।
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असमानता (12-5x\ge 3(4-2x)+x) का हल क्या है?
What is the solution of (12-5x\ge 3(4-2x)+x)?
#linear inequalities
#always true
#equality boundary
#class 11
A सभी वास्तविक संख्याएँ / All real numbers
B \(x\le 0\)
C कोई हल नहीं / No solution
D \(x\ge 0\)
Explanation opens after your attempt
Correct Answer
A. सभी वास्तविक संख्याएँ / All real numbers
Step 1
Concept
The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
Step 3
Exam Tip
दाएँ पक्ष (12-5x) बनता है। समानता \(12-5x\ge 12-5x\) हमेशा सत्य है।
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असमानता \(7x-4\le 2x+16\) को हल कीजिए।
Solve the inequality \(7x-4\le 2x+16\).
#linear inequalities
#basic hard
#class 11
A \(x\le 4\)
B \(x\ge 4\)
C (x<4)
D (x>4)
Explanation opens after your attempt
Correct Answer
A. \(x\le 4\)
Step 1
Concept
From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 4\). From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
Step 3
Exam Tip
\(5x\le 20\) से \(x\le 4\) मिलता है। सरल रैखिक असमानता में पहले (x) वाले पद एक ओर लाएँ।
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यदि (6x+2<3(2x+1)), तो सही निष्कर्ष क्या है?
If (6x+2<3(2x+1)), what is the correct conclusion?
#linear inequalities
#always true
#class 11
A सभी वास्तविक संख्याएँ / All real numbers
B कोई हल नहीं / No solution
C (x<1)
D (x>1)
Explanation opens after your attempt
Correct Answer
A. सभी वास्तविक संख्याएँ / All real numbers
Step 1
Concept
The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक संख्याएँ / All real numbers. The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
Step 3
Exam Tip
दाएँ पक्ष (6x+3) है। (6x+2<6x+3) हमेशा सत्य है, इसलिए सभी वास्तविक संख्याएँ हल हैं।
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असमानता \(5-\frac{x-3}{2}\ge \frac{3x+1}{4}\) को हल करें।
Solve the inequality \(5-\frac{x-3}{2}\ge \frac{3x+1}{4}\).
#linear inequalities
#fraction with brackets
#class 11
A \(x\le \frac{21}{5}\)
B \(x\ge \frac{21}{5}\)
C \(x<\frac{21}{5}\)
D \(x>\frac{21}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{21}{5}\)
Step 1
Concept
Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{21}{5}\). Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
Step 3
Exam Tip
(4) से गुणा करने पर (20-2(x-3)\ge 3x+1) मिलता है। इससे \(25\ge 5x\), इसलिए \(x\le 5\)।
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असमानता (2(1-3x)<4-7x) का हल अंतराल कौन सा है?
Which interval is the solution of (2(1-3x)<4-7x)?
#linear inequalities
#interval
#algebra
#class 11
A (x<2)
B (x>2)
C \(x\le 2\)
D \(x\ge 2\)
Explanation opens after your attempt
Step 1
Concept
Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
Step 2
Why this answer is correct
The correct answer is A. (x<2). Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
Step 3
Exam Tip
(2-6x<4-7x) में (7x) जोड़ने पर (2+x<4) मिलता है। इसलिए (x<2)।
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असमानता \(-\frac{4x-1}{3}\le 5-x\) का हल क्या है?
What is the solution of \(-\frac{4x-1}{3}\le 5-x\)?
#linear inequalities
#negative numerator
#class 11
A \(x\ge -14\)
B \(x\le -14\)
C (x>-14)
D (x<-14)
Explanation opens after your attempt
Correct Answer
A. \(x\ge -14\)
Step 1
Concept
Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -14\). Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
Step 3
Exam Tip
(3) से गुणा करने पर \(-4x+1\le 15-3x\) मिलता है। इससे \(x\ge -14\)।
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यदि \(\frac{x}{3}-\frac{x-5}{6}>2\), तो (x) के लिए सही शर्त क्या है?
If \(\frac{x}{3}-\frac{x-5}{6}>2\), what is the correct condition for (x)?
#linear inequalities
#fraction subtraction
#class 11
A (x>7)
B (x<7)
C \(x\ge 7\)
D \(x\le 7\)
Explanation opens after your attempt
Step 1
Concept
The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
Step 2
Why this answer is correct
The correct answer is A. (x>7). The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
Step 3
Exam Tip
बायाँ पक्ष \(\frac{x+5}{6}\) बनता है। \(\frac{x+5}{6}>2\) से (x>7) मिलता है।
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असमानता (8-3(2-x)\ge 4x-1) का हल समुच्चय चुनिए।
Choose the solution set of (8-3(2-x)\ge 4x-1).
#linear inequalities
#bracket simplification
#class 11
A \(x\le 3\)
B \(x\ge 3\)
C (x<3)
D (x>3)
Explanation opens after your attempt
Correct Answer
A. \(x\le 3\)
Step 1
Concept
The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le 3\). The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
Step 3
Exam Tip
बायाँ पक्ष (2+3x) है। \(2+3x\ge 4x-1\) से \(x\le 3\) मिलता है।
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असमानता \(3-\frac{2x+5}{7}\le \frac{1-x}{2}\) को हल करें।
Solve the inequality \(3-\frac{2x+5}{7}\le \frac{1-x}{2}\).
#linear inequalities
#complex fractions
#class 11
A \(x\ge -\frac{27}{3}\)
B \(x\le -9\)
C \(x\ge -9\)
D (x<-9)
Explanation opens after your attempt
Correct Answer
C. \(x\ge -9\)
Step 1
Concept
Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
Step 2
Why this answer is correct
The correct answer is C. \(x\ge -9\). Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
Step 3
Exam Tip
(14) से गुणा करने पर (42-2(2x+5)\le 7(1-x)) मिलता है। इससे \(32-4x\le 7-7x\), अतः \(x\le -\frac{25}{3}\) नहीं बल्कि \(3x\le -25\) है।
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असमानता \(\frac{9x+4}{5}\ge 2x-3\) का हल क्या है?
What is the solution of \(\frac{9x+4}{5}\ge 2x-3\)?
#linear inequalities
#fraction
#class 11
#hard
A \(x\le 19\)
B \(x\ge 19\)
C (x<19)
D (x>19)
Explanation opens after your attempt
Correct Answer
A. \(x\le 19\)
Step 1
Concept
Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le 19\). Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
Step 3
Exam Tip
(5) से गुणा करने पर \(9x+4\ge 10x-15\) मिलता है। अतः \(x\le 19\)।
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यदि \(0.75x+\frac{1}{2}<2-\frac{x}{4}\), तो (x) का हल क्या है?
If \(0.75x+\frac{1}{2}<2-\frac{x}{4}\), what is the solution for (x)?
#linear inequalities
#decimal fraction
#class 11
A \(x<\frac{3}{2}\)
B \(x>\frac{3}{2}\)
C \(x\le \frac{3}{2}\)
D \(x\ge \frac{3}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x<\frac{3}{2}\)
Step 1
Concept
Treat (0.75x) as \(\frac{3x}{4}\). From \(x+\frac{1}{2}<2\), \(x<\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x<\frac{3}{2}\). Treat (0.75x) as \(\frac{3x}{4}\). From \(x+\frac{1}{2}<2\), \(x<\frac{3}{2}\).
Step 3
Exam Tip
\(0.75x=\frac{3x}{4}\) मानकर हल करें। \(x+\frac{1}{2}<2\) से \(x<\frac{3}{2}\) मिलता है।
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असमानता (-2(5-x)+3(x-4)>6x+1) का सही हल कौन सा है?
Which is the correct solution of (-2(5-x)+3(x-4)>6x+1)?
#linear inequalities
#brackets
#hard
#class 11
A (x<-23)
B (x>-23)
C \(x\le -23\)
D \(x\ge -23\)
Explanation opens after your attempt
Correct Answer
A. (x<-23)
Step 1
Concept
The left side becomes (5x-22). From (5x-22>6x+1), we get (x<-23).
Step 2
Why this answer is correct
The correct answer is A. (x<-23). The left side becomes (5x-22). From (5x-22>6x+1), we get (x<-23).
Step 3
Exam Tip
बायाँ पक्ष (5x-22) बनता है। (5x-22>6x+1) से (x<-23) मिलता है।
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असमानता (13+4x\le 2(3x-5)+1) को हल कीजिए।
Solve the inequality (13+4x\le 2(3x-5)+1).
#linear inequalities
#algebraic solution
#class 11
A \(x\ge 11\)
B \(x\le 11\)
C (x>11)
D (x<11)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 11\)
Step 1
Concept
The right side is (6x-9). From \(13+4x\le 6x-9\), \(22\le 2x\), so \(x\ge 11\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 11\). The right side is (6x-9). From \(13+4x\le 6x-9\), \(22\le 2x\), so \(x\ge 11\).
Step 3
Exam Tip
दाएँ पक्ष (6x-9) है। \(13+4x\le 6x-9\) से \(22\le 2x\), अतः \(x\ge 11\)।
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असमानता \(2-\frac{5x-1}{4}<\frac{x+7}{2}\) का हल समुच्चय क्या है?
What is the solution set of \(2-\frac{5x-1}{4}<\frac{x+7}{2}\)?
#linear inequalities
#nested subtraction
#class 11
A \(x>-\frac{5}{7}\)
B \(x<-\frac{5}{7}\)
C \(x\ge -\frac{5}{7}\)
D \(x\le -\frac{5}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{5}{7}\)
Step 1
Concept
Multiplying by (4) gives (8-(5x-1)<2x+14). Thus (9-5x<2x+14), so \(x>-\frac{5}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{5}{7}\). Multiplying by (4) gives (8-(5x-1)<2x+14). Thus (9-5x<2x+14), so \(x>-\frac{5}{7}\).
Step 3
Exam Tip
(4) से गुणा करने पर (8-(5x-1)<2x+14) मिलता है। इससे (9-5x<2x+14), इसलिए \(x>-\frac{5}{7}\)।
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यदि \(\frac{3-4x}{2}\ge 5-\frac{x}{3}\), तो (x) का हल क्या है?
If \(\frac{3-4x}{2}\ge 5-\frac{x}{3}\), what is the solution for (x)?
#linear inequalities
#fractions
#hard
#class 11
A \(x\le -\frac{21}{10}\)
B \(x\ge -\frac{21}{10}\)
C \(x<-\frac{21}{10}\)
D \(x>-\frac{21}{10}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le -\frac{21}{10}\)
Step 1
Concept
Multiplying by (6) gives \(9-12x\ge 30-2x\). Thus \(-21\ge 10x\), so \(x\le -\frac{21}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le -\frac{21}{10}\). Multiplying by (6) gives \(9-12x\ge 30-2x\). Thus \(-21\ge 10x\), so \(x\le -\frac{21}{10}\).
Step 3
Exam Tip
(6) से गुणा करने पर \(9-12x\ge 30-2x\) मिलता है। इससे \(-21\ge 10x\), अतः \(x\le -\frac{21}{10}\)।
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असमानता (5(x-1)-2(3x+4)<9) का हल क्या है?
What is the solution of (5(x-1)-2(3x+4)<9)?
#linear inequalities
#brackets
#negative sign
#class 11
A (x>-22)
B (x<-22)
C \(x\ge -22\)
D \(x\le -22\)
Explanation opens after your attempt
Correct Answer
B. (x<-22)
Step 1
Concept
Simplification gives (-x-13<9). Thus (-x<22), so (x>-22) should result.
Step 2
Why this answer is correct
The correct answer is B. (x<-22). Simplification gives (-x-13<9). Thus (-x<22), so (x>-22) should result.
Step 3
Exam Tip
सरलीकरण से (-x-13<9) मिलता है। इससे (-x<22), इसलिए (x>-22) होना चाहिए।
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असमानता \(\frac{2x-5}{3}+\frac{x+1}{6}\le 4\) का हल चुनिए।
Choose the solution of \(\frac{2x-5}{3}+\frac{x+1}{6}\le 4\).
#linear inequalities
#addition fractions
#class 11
#hard
A \(x\le \frac{17}{5}\)
B \(x\ge \frac{17}{5}\)
C \(x<\frac{17}{5}\)
D \(x>\frac{17}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le \frac{17}{5}\)
Step 1
Concept
Clearing denominators gives \(4x-10+x+1\le 24\). Hence \(5x\le 33\), so \(x\le \frac{33}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\le \frac{17}{5}\). Clearing denominators gives \(4x-10+x+1\le 24\). Hence \(5x\le 33\), so \(x\le \frac{33}{5}\).
Step 3
Exam Tip
हर हटाने पर \(4x-10+x+1\le 24\) मिलता है। अतः \(5x\le 33\), इसलिए \(x\le \frac{33}{5}\)।
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असमानता (2.5(2x-1)\ge 1.5(x+3)) का हल क्या है?
What is the solution of (2.5(2x-1)\ge 1.5(x+3))?
#linear inequalities
#decimal coefficients
#class 11
A \(x\ge 2\)
B \(x\le 2\)
C (x>2)
D (x<2)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 2\)
Step 1
Concept
Simplification gives \(5x-2.5\ge 1.5x+4.5\). Thus \(3.5x\ge 7\), so \(x\ge 2\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 2\). Simplification gives \(5x-2.5\ge 1.5x+4.5\). Thus \(3.5x\ge 7\), so \(x\ge 2\).
Step 3
Exam Tip
सरलीकरण से \(5x-2.5\ge 1.5x+4.5\) मिलता है। इससे \(3.5x\ge 7\), अतः \(x\ge 2\)।
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यदि (-7x+4<18-2x), तो (x) किससे बड़ा होगा?
If (-7x+4<18-2x), then (x) will be greater than what?
#linear inequalities
#negative division
#class 11
A \(x>-\frac{14}{5}\)
B \(x<-\frac{14}{5}\)
C \(x\ge -\frac{14}{5}\)
D \(x\le -\frac{14}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x>-\frac{14}{5}\)
Step 1
Concept
We get (-5x<14). Dividing by a negative gives \(x>-\frac{14}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x>-\frac{14}{5}\). We get (-5x<14). Dividing by a negative gives \(x>-\frac{14}{5}\).
Step 3
Exam Tip
(-5x<14) प्राप्त होता है। ऋणात्मक से भाग देने पर \(x>-\frac{14}{5}\) होगा।
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असमानता (11-3x>2x+1) का हल अंतराल रूप में क्या है?
What is the interval-form solution of (11-3x>2x+1)?
#linear inequalities
#interval form
#class 11
A (x<2)
B (x>2)
C \(x\le 2\)
D \(x\ge 2\)
Explanation opens after your attempt
Step 1
Concept
From (10>5x), (x<2) is obtained. In interval form, it is (\(-\infty,2\)).
Step 2
Why this answer is correct
The correct answer is A. (x<2). From (10>5x), (x<2) is obtained. In interval form, it is (\(-\infty,2\)).
Step 3
Exam Tip
(10>5x) से (x<2) मिलता है। अंतराल में यह (\(-\infty,2\)) होगा।
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असमानता \(\frac{x+6}{8}-\frac{x-2}{4}<1\) को हल करें।
Solve the inequality \(\frac{x+6}{8}-\frac{x-2}{4}<1\).
#linear inequalities
#fraction subtraction
#class 11
A (x>-2)
B (x<-2)
C \(x\ge -2\)
D \(x\le -2\)
Explanation opens after your attempt
Step 1
Concept
The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
Step 2
Why this answer is correct
The correct answer is A. (x>-2). The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
Step 3
Exam Tip
बायाँ पक्ष \(\frac{10-x}{8}\) बनता है। \(\frac{10-x}{8}<1\) से (x>2) मिलता है।
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