युग्म \(3x+2y=\lambda\) और (18x+12y=48) के अनंत हलों के लिए \(\lambda\) क्या होगा?
For infinitely many solutions of \(3x+2y=\lambda\) and (18x+12y=48), what is \(\lambda\)?
#class10
#linear-equations
#solvability
A \(\lambda=6\)
B \(\lambda=8\)
C \(\lambda=10\)
D \(\lambda=12\)
Explanation opens after your attempt
Correct Answer
B. \(\lambda=8\)
Step 1
Concept
The coefficient ratio is \(\frac{1}{6}\). For infinitely many solutions, \(\frac{\lambda}{48}=\frac{1}{6}\), so \(\lambda=8\).
Step 2
Why this answer is correct
The correct answer is B. \(\lambda=8\). The coefficient ratio is \(\frac{1}{6}\). For infinitely many solutions, \(\frac{\lambda}{48}=\frac{1}{6}\), so \(\lambda=8\).
Step 3
Exam Tip
गुणांक अनुपात \(\frac{1}{6}\) है। अनंत हलों के लिए \(\frac{\lambda}{48}=\frac{1}{6}\) इसलिए \(\lambda=8\)।
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युग्म (wx-6y=4) और (16x-24y=12) का अद्वितीय हल कब होगा?
When will (wx-6y=4) and (16x-24y=12) have a unique solution?
#class10
#linear-equations
#solvability
A (w=4)
B \(w\neq4\)
C (w=16)
D \(w\neq16\)
Explanation opens after your attempt
Correct Answer
B. \(w\neq4\)
Step 1
Concept
For a unique solution, \(\frac{w}{16}\neq\frac{-6}{-24}\) must hold. Therefore, \(w\neq4\).
Step 2
Why this answer is correct
The correct answer is B. \(w\neq4\). For a unique solution, \(\frac{w}{16}\neq\frac{-6}{-24}\) must hold. Therefore, \(w\neq4\).
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{w}{16}\neq\frac{-6}{-24}\) होना चाहिए। इसलिए \(w\neq4\) होगा।
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युग्म (5x+(v-2)y=3) और (15x+9y=14) में कोई हल न होने के लिए (v) का मान क्या है?
What is the value of (v) for no solution in (5x+(v-2)y=3) and (15x+9y=14)?
#class10
#linear-equations
#solvability
A (v=3)
B (v=4)
C (v=5)
D (v=6)
Explanation opens after your attempt
Step 1
Concept
Equating coefficient ratios gives (v-2=3). Thus (v=5), and the different constant ratio gives no solution.
Step 2
Why this answer is correct
The correct answer is C. (v=5). Equating coefficient ratios gives (v-2=3). Thus (v=5), and the different constant ratio gives no solution.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर (v-2=3) मिलता है। इसलिए (v=5) और स्थिर अनुपात अलग होने से कोई हल नहीं।
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युग्म ((u+3)x+8y=16) और (10x+20y=40) के अनंत हलों के लिए (u) क्या होगा?
For infinitely many solutions of ((u+3)x+8y=16) and (10x+20y=40), what is (u)?
#class10
#linear-equations
#solvability
A (u=1)
B (u=2)
C (u=3)
D (u=4)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, \(\frac{u+3}{10}=\frac{8}{20}=\frac{16}{40}\) is needed. This gives (u=1).
Step 2
Why this answer is correct
The correct answer is A. (u=1). For infinitely many solutions, \(\frac{u+3}{10}=\frac{8}{20}=\frac{16}{40}\) is needed. This gives (u=1).
Step 3
Exam Tip
अनंत हलों के लिए \(\frac{u+3}{10}=\frac{8}{20}=\frac{16}{40}\) चाहिए। इससे (u=1) मिलता है।
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युग्म (12x-7y=5) और (18x-11y=8) की हल-संख्या क्या है?
What is the number of solutions of (12x-7y=5) and (18x-11y=8)?
#class10
#linear-equations
#solvability
A कोई हल नहीं / No solution
B अनंत हल / Infinitely many solutions
C अद्वितीय हल / Unique solution
D शून्य या अनंत / Zero or infinite
Explanation opens after your attempt
Correct Answer
C. अद्वितीय हल / Unique solution
Step 1
Concept
Since \(\frac{12}{18}\neq\frac{-7}{-11}\), the lines intersect. Hence a unique solution is obtained.
Step 2
Why this answer is correct
The correct answer is C. अद्वितीय हल / Unique solution. Since \(\frac{12}{18}\neq\frac{-7}{-11}\), the lines intersect. Hence a unique solution is obtained.
Step 3
Exam Tip
\(\frac{12}{18}\neq\frac{-7}{-11}\) होने से रेखाएँ काटती हैं। इसलिए अद्वितीय हल मिलेगा।
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युग्म (13x+4y=6) और (26x+8y=12) के लिए सही विकल्प चुनिए।
Choose the correct option for (13x+4y=6) and (26x+8y=12).
#class10
#linear-equations
#solvability
A अद्वितीय हल / Unique solution
B कोई हल नहीं / No solution
C अनंत हल / Infinitely many solutions
D केवल (x=0) हल / Only (x=0) solution
Explanation opens after your attempt
Correct Answer
C. अनंत हल / Infinitely many solutions
Step 1
Concept
All three ratios are equal. Thus the lines are coincident and the pair has infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. अनंत हल / Infinitely many solutions. All three ratios are equal. Thus the lines are coincident and the pair has infinitely many solutions.
Step 3
Exam Tip
तीनों अनुपात समान हैं। अतः रेखाएँ संपाती हैं और युग्म के अनंत हल हैं।
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युग्म (4x+9y=2) और (8x+18y=7) की हल-स्थिति क्या है?
What is the solution status of (4x+9y=2) and (8x+18y=7)?
#class10
#linear-equations
#solvability
A अद्वितीय हल / Unique solution
B कोई हल नहीं / No solution
C अनंत हल / Infinitely many solutions
D सभी बिंदु हल / All points are solutions
Explanation opens after your attempt
Correct Answer
B. कोई हल नहीं / No solution
Step 1
Concept
The coefficient ratio is equal but \(\frac{2}{7}\) is different. Hence both are distinct parallel lines.
Step 2
Why this answer is correct
The correct answer is B. कोई हल नहीं / No solution. The coefficient ratio is equal but \(\frac{2}{7}\) is different. Hence both are distinct parallel lines.
Step 3
Exam Tip
गुणांक अनुपात समान है लेकिन \(\frac{2}{7}\) अलग है। इसलिए दोनों अलग समांतर रेखाएँ हैं।
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युग्म (10x+3y=17) और (20x+9y=31) में कितने हल होंगे?
How many solutions will the pair (10x+3y=17) and (20x+9y=31) have?
#class10
#linear-equations
#solvability
A कोई हल नहीं / No solution
B अनंत हल / Infinitely many solutions
C अद्वितीय हल / Unique solution
D निर्धारित नहीं / Cannot be decided
Explanation opens after your attempt
Correct Answer
C. अद्वितीय हल / Unique solution
Step 1
Concept
Here \(\frac{10}{20}\neq\frac{3}{9}\). Therefore, the lines meet at one point.
Step 2
Why this answer is correct
The correct answer is C. अद्वितीय हल / Unique solution. Here \(\frac{10}{20}\neq\frac{3}{9}\). Therefore, the lines meet at one point.
Step 3
Exam Tip
यहाँ \(\frac{10}{20}\neq\frac{3}{9}\) है। इसलिए रेखाएँ एक बिंदु पर मिलती हैं।
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रेखाएँ (9x-5y=13) और (18x-10y=26) के लिए सही निष्कर्ष क्या है?
What is the correct conclusion for the lines (9x-5y=13) and (18x-10y=26)?
#class10
#linear-equations
#solvability
A कोई हल नहीं / No solution
B अद्वितीय हल / Unique solution
C अनंत हल / Infinitely many solutions
D ठीक दो हल / Exactly two solutions
Explanation opens after your attempt
Correct Answer
C. अनंत हल / Infinitely many solutions
Step 1
Concept
The second equation is (2) times the first. So both lines are coincident and give infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. अनंत हल / Infinitely many solutions. The second equation is (2) times the first. So both lines are coincident and give infinitely many solutions.
Step 3
Exam Tip
दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों रेखाएँ संपाती हैं और अनंत हल देती हैं।
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रेखाएँ (7x+2y=9) और (21x+6y=25) किस प्रकार का युग्म बनाती हैं?
What type of pair is formed by the lines (7x+2y=9) and (21x+6y=25)?
#class10
#linear-equations
#solvability
A संगत और स्वतंत्र / Consistent and independent
B संगत और आश्रित / Consistent and dependent
C असंगत / Inconsistent
D सदैव अद्वितीय / Always unique
Explanation opens after your attempt
Correct Answer
C. असंगत / Inconsistent
Step 1
Concept
Coefficient ratios are equal but the constant ratio is different. Hence the lines are parallel and the pair is inconsistent.
Step 2
Why this answer is correct
The correct answer is C. असंगत / Inconsistent. Coefficient ratios are equal but the constant ratio is different. Hence the lines are parallel and the pair is inconsistent.
Step 3
Exam Tip
गुणांक अनुपात समान है पर स्थिर पद का अनुपात अलग है। इसलिए रेखाएँ समांतर हैं और युग्म असंगत है।
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युग्म (6x-ty=18) और (14x-21y=42) के अनंत हलों के लिए (t) का मान क्या होगा?
What is the value of (t) for infinitely many solutions of (6x-ty=18) and (14x-21y=42)?
#class10
#linear-equations
#solvability
A (t=6)
B (t=8)
C (t=9)
D (t=12)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, \(\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}\) must hold. This gives (t=9).
Step 2
Why this answer is correct
The correct answer is C. (t=9). For infinitely many solutions, \(\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}\) must hold. This gives (t=9).
Step 3
Exam Tip
अनंत हलों के लिए \(\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}\) होना चाहिए। इससे (t=9) मिलता है।
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युग्म (11x+3y=4) और (22x+sy=10) का अद्वितीय हल कब होगा?
When will (11x+3y=4) and (22x+sy=10) have a unique solution?
#class10
#linear-equations
#solvability
A (s=6)
B \(s\neq6\)
C (s=3)
D \(s\neq3\)
Explanation opens after your attempt
Correct Answer
B. \(s\neq6\)
Step 1
Concept
For a unique solution, \(\frac{11}{22}\neq\frac{3}{s}\) must hold. Hence \(s\neq6\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(s\neq6\). For a unique solution, \(\frac{11}{22}\neq\frac{3}{s}\) must hold. Hence \(s\neq6\) is correct.
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{11}{22}\neq\frac{3}{s}\) होना चाहिए। अतः \(s\neq6\) सही है।
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युग्म (2x+ny=7) और (6x+15y=19) में कोई हल न होने के लिए (n) का मान क्या है?
What is the value of (n) for no solution in (2x+ny=7) and (6x+15y=19)?
#class10
#linear-equations
#solvability
A (n=3)
B (n=4)
C (n=5)
D (n=6)
Explanation opens after your attempt
Step 1
Concept
Equating coefficient ratios, \(\frac{2}{6}=\frac{n}{15}\) gives (n=5). The constant ratio is different, so the pair is inconsistent.
Step 2
Why this answer is correct
The correct answer is C. (n=5). Equating coefficient ratios, \(\frac{2}{6}=\frac{n}{15}\) gives (n=5). The constant ratio is different, so the pair is inconsistent.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर \(\frac{2}{6}=\frac{n}{15}\) से (n=5) मिलता है। स्थिर अनुपात अलग है इसलिए युग्म असंगत है।
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युग्म (5x+(r+1)y=11) और (15x+18y=33) के अनंत हलों के लिए (r) क्या होगा?
For infinitely many solutions of (5x+(r+1)y=11) and (15x+18y=33), what is (r)?
#class10
#linear-equations
#solvability
A (r=4)
B (r=5)
C (r=6)
D (r=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, \(\frac{5}{15}=\frac{r+1}{18}=\frac{11}{33}\) is needed. This gives (r=5).
Step 2
Why this answer is correct
The correct answer is B. (r=5). For infinitely many solutions, \(\frac{5}{15}=\frac{r+1}{18}=\frac{11}{33}\) is needed. This gives (r=5).
Step 3
Exam Tip
अनंत हलों के लिए \(\frac{5}{15}=\frac{r+1}{18}=\frac{11}{33}\) चाहिए। इससे (r=5) मिलता है।
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युग्म (qx+4y=9) और (15x+10y=18) का अद्वितीय हल कब होगा?
When will (qx+4y=9) and (15x+10y=18) have a unique solution?
#class10
#linear-equations
#solvability
A \(q\neq6\)
B (q=6)
C \(q\neq15\)
D (q=15)
Explanation opens after your attempt
Correct Answer
A. \(q\neq6\)
Step 1
Concept
For a unique solution, \(\frac{q}{15}\neq\frac{4}{10}\) must hold. Hence \(q\neq6\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(q\neq6\). For a unique solution, \(\frac{q}{15}\neq\frac{4}{10}\) must hold. Hence \(q\neq6\) is correct.
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{q}{15}\neq\frac{4}{10}\) होना चाहिए। इसलिए \(q\neq6\) सही शर्त है।
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युग्म (3x-2y=5) और (12x-8y=c) में कोई हल न हो इसके लिए (c) पर कौन-सी शर्त होगी?
For (3x-2y=5) and (12x-8y=c) to have no solution, what condition is required on (c)?
#class10
#linear-equations
#solvability
A (c=20)
B \(c\neq20\)
C (c=15)
D \(c\neq15\)
Explanation opens after your attempt
Correct Answer
B. \(c\neq20\)
Step 1
Concept
The coefficient ratio is \(\frac{1}{4}\). For no solution, \(\frac{5}{c}\neq\frac{1}{4}\), so \(c\neq20\).
Step 2
Why this answer is correct
The correct answer is B. \(c\neq20\). The coefficient ratio is \(\frac{1}{4}\). For no solution, \(\frac{5}{c}\neq\frac{1}{4}\), so \(c\neq20\).
Step 3
Exam Tip
गुणांक अनुपात \(\frac{1}{4}\) है। कोई हल नहीं के लिए \(\frac{5}{c}\neq\frac{1}{4}\) इसलिए \(c\neq20\)।
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युग्म (8x+ay=12) और (20x+10y=30) के अनंत हलों के लिए (a) का मान बताइए।
Find the value of (a) for infinitely many solutions of (8x+ay=12) and (20x+10y=30).
#class10
#linear-equations
#solvability
A (a=3)
B (a=4)
C (a=5)
D (a=6)
Explanation opens after your attempt
Step 1
Concept
Here \(\frac{8}{20}=\frac{12}{30}\). Thus \(\frac{a}{10}=\frac{2}{5}\) gives (a=4).
Step 2
Why this answer is correct
The correct answer is B. (a=4). Here \(\frac{8}{20}=\frac{12}{30}\). Thus \(\frac{a}{10}=\frac{2}{5}\) gives (a=4).
Step 3
Exam Tip
यहाँ \(\frac{8}{20}=\frac{12}{30}\) है। अतः \(\frac{a}{10}=\frac{2}{5}\) से (a=4) होगा।
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युग्म ((p-2)x+7y=14) और (9x+21y=42) के अनंत हलों के लिए (p) क्या होगा?
For infinitely many solutions of ((p-2)x+7y=14) and (9x+21y=42), what is (p)?
#class10
#linear-equations
#solvability
A (p=3)
B (p=4)
C (p=5)
D (p=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, all three ratios must be equal. From \(\frac{p-2}{9}=\frac{7}{21}\), (p=5).
Step 2
Why this answer is correct
The correct answer is C. (p=5). For infinitely many solutions, all three ratios must be equal. From \(\frac{p-2}{9}=\frac{7}{21}\), (p=5).
Step 3
Exam Tip
अनंत हलों के लिए तीनों अनुपात समान होने चाहिए। \(\frac{p-2}{9}=\frac{7}{21}\) से (p=5) मिलता है।
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युग्म (4x+my=8) और (10x+15y=21) में कोई हल न होने के लिए (m) का मान क्या है?
What is the value of (m) for no solution in (4x+my=8) and (10x+15y=21)?
#class10
#linear-equations
#solvability
A (m=4)
B (m=5)
C (m=6)
D (m=8)
Explanation opens after your attempt
Step 1
Concept
Equating coefficient ratios gives (m=6). The constant ratio is different, so there is no solution.
Step 2
Why this answer is correct
The correct answer is C. (m=6). Equating coefficient ratios gives (m=6). The constant ratio is different, so there is no solution.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर (m=6) मिलता है। स्थिर पद का अनुपात अलग है इसलिए कोई हल नहीं होगा।
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युग्म (kx+5y=10) और (6x+15y=18) का अद्वितीय हल कब होगा?
When will the pair (kx+5y=10) and (6x+15y=18) have a unique solution?
#class10
#linear-equations
#solvability
A (k=2)
B \(k\neq2\)
C (k=6)
D \(k\neq6\)
Explanation opens after your attempt
Correct Answer
B. \(k\neq2\)
Step 1
Concept
For a unique solution, \(\frac{k}{6}\neq\frac{5}{15}\) must hold. Therefore, \(k\neq2\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(k\neq2\). For a unique solution, \(\frac{k}{6}\neq\frac{5}{15}\) must hold. Therefore, \(k\neq2\) is correct.
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{k}{6}\neq\frac{5}{15}\) होना चाहिए। इसलिए \(k\neq2\) सही है।
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युग्म ((s+2)x+3y=1) और (5x+(s-1 )y=4) के अद्वितीय हल की सही शर्त कौन-सी है?
Which condition gives a unique solution for ((s+2)x+3y=1) and (5x+(s-1 )y=4)?
#class10
#linear-equations
#solvability
A \(s^2+s-17=0\)
B \(s^2+s-17\neq0\)
C \(s^2-s-17=0\)
D \(s^2-s-17\neq0\)
Explanation opens after your attempt
Correct Answer
B. \(s^2+s-17\neq0\)
Step 1
Concept
The determinant is (D=(s+2)(s-1 )-15=s-2 +s-17 ). For a unique solution, \(D\neq0\) is required.
Step 2
Why this answer is correct
The correct answer is B. \(s^2+s-17\neq0\). The determinant is (D=(s+2)(s-1 )-15=s-2 +s-17 ). For a unique solution, \(D\neq0\) is required.
Step 3
Exam Tip
सारणिक (D=(s+2)(s-1 )-15=s-2 +s-17 ) है। अद्वितीय हल के लिए \(D\neq0\) चाहिए।
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युग्म \(5x-2y=\gamma\) और (20x-8y=36) के अनंत हलों के लिए \(\gamma\) का मान क्या है?
What is \(\gamma\) for infinitely many solutions of \(5x-2y=\gamma\) and (20x-8y=36)?
#class10
#linear-equations
#solvability
A \(\gamma=6\)
B \(\gamma=8\)
C \(\gamma=9\)
D \(\gamma=12\)
Explanation opens after your attempt
Correct Answer
C. \(\gamma=9\)
Step 1
Concept
The coefficient ratio is \(\frac{1}{4}\). For infinitely many solutions, \(\frac{\gamma}{36}=\frac{1}{4}\), so \(\gamma=9\).
Step 2
Why this answer is correct
The correct answer is C. \(\gamma=9\). The coefficient ratio is \(\frac{1}{4}\). For infinitely many solutions, \(\frac{\gamma}{36}=\frac{1}{4}\), so \(\gamma=9\).
Step 3
Exam Tip
गुणांक अनुपात \(\frac{1}{4}\) है। अनंत हलों के लिए \(\frac{\gamma}{36}=\frac{1}{4}\), इसलिए \(\gamma=9\)।
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युग्म (3x+\(\beta-1\)y=10) और (6x+4y=18) में कोई हल न होने के लिए \(\beta\) क्या होगा?
For no solution in (3x+\(\beta-1\)y=10) and (6x+4y=18), what is \(\beta\)?
#class10
#linear-equations
#solvability
A \(\beta=1\)
B \(\beta=2\)
C \(\beta=3\)
D \(\beta=4\)
Explanation opens after your attempt
Correct Answer
C. \(\beta=3\)
Step 1
Concept
Equating coefficient ratios gives \(\beta=3\). The constant ratio \(\frac{5}{9}\) is different, so there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(\beta=3\). Equating coefficient ratios gives \(\beta=3\). The constant ratio \(\frac{5}{9}\) is different, so there is no solution.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर \(\beta=3\) मिलता है। स्थिर अनुपात \(\frac{5}{9}\) अलग है, इसलिए कोई हल नहीं।
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युग्म \(\alpha x+4y=7\) और (9x+12y=21) का अद्वितीय हल कब होगा?
When will \(\alpha x+4y=7\) and (9x+12y=21) have a unique solution?
#class10
#linear-equations
#solvability
A \(\alpha=3\)
B \(\alpha\neq3\)
C \(\alpha=9\)
D \(\alpha\neq9\)
Explanation opens after your attempt
Correct Answer
B. \(\alpha\neq3\)
Step 1
Concept
For a unique solution, \(\frac{\alpha}{9}\neq\frac{4}{12}\) must hold. Thus \(\alpha\neq3\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(\alpha\neq3\). For a unique solution, \(\frac{\alpha}{9}\neq\frac{4}{12}\) must hold. Thus \(\alpha\neq3\) is correct.
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{\alpha}{9}\neq\frac{4}{12}\) होना चाहिए। इसलिए \(\alpha\neq3\) सही शर्त है।
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युग्म \(rx+\frac{3}{2}y=6\) और (8x+6y=24) के अनंत हलों के लिए (r) का मान क्या है?
What is (r) for infinitely many solutions of \(rx+\frac{3}{2}y=6\) and (8x+6y=24)?
#class10
#linear-equations
#solvability
A (r=1)
B (r=2)
C (r=3)
D (r=4)
Explanation opens after your attempt
Step 1
Concept
All three ratios must be \(\frac{1}{4}\). Therefore, (r=2) is the correct value.
Step 2
Why this answer is correct
The correct answer is B. (r=2). All three ratios must be \(\frac{1}{4}\). Therefore, (r=2) is the correct value.
Step 3
Exam Tip
तीनों अनुपात \(\frac{1}{4}\) होने चाहिए। इसलिए (r=2) सही मान है।
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युग्म \(\frac{1}{2}x+qy=3\) और (2x+8y=5) के लिए कोई हल न होने पर (q) क्या होगा?
For no solution in \(\frac{1}{2}x+qy=3\) and (2x+8y=5), what is (q)?
#class10
#linear-equations
#solvability
A (q=1)
B (q=2)
C (q=4)
D (q=8)
Explanation opens after your attempt
Step 1
Concept
Equating coefficient ratios gives (q=2). The constant ratio is not equal, so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. (q=2). Equating coefficient ratios gives (q=2). The constant ratio is not equal, so there is no solution.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर (q=2) मिलता है। स्थिर पद अनुपात समान नहीं है, इसलिए कोई हल नहीं।
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युग्म ((p+4)x-9y=2) और (5x-15y=6) का अद्वितीय हल कब होगा?
When will ((p+4)x-9y=2) and (5x-15y=6) have a unique solution?
#class10
#linear-equations
#solvability
A (p=-1)
B (p=1)
C \(p\neq-1\)
D \(p\neq1\)
Explanation opens after your attempt
Correct Answer
C. \(p\neq-1\)
Step 1
Concept
For a unique solution, \(\frac{p+4}{5}\neq\frac{-9}{-15}\) must hold. Hence \(p\neq-1\).
Step 2
Why this answer is correct
The correct answer is C. \(p\neq-1\). For a unique solution, \(\frac{p+4}{5}\neq\frac{-9}{-15}\) must hold. Hence \(p\neq-1\).
Step 3
Exam Tip
अद्वितीय हल के लिए \(\frac{p+4}{5}\neq\frac{-9}{-15}\) होना चाहिए। इसलिए \(p\neq-1\)।
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युग्म (4x+(m-5 )y=9) और (12x+6y=27) के अनंत हलों के लिए (m) कितना है?
What is (m) for infinitely many solutions of (4x+(m-5 )y=9) and (12x+6y=27)?
#class10
#linear-equations
#solvability
A (m=5)
B (m=6)
C (m=7)
D (m=9)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, \(\frac{4}{12}=\frac{m-5}{6}=\frac{9}{27}\) is required. This gives (m=7).
Step 2
Why this answer is correct
The correct answer is C. (m=7). For infinitely many solutions, \(\frac{4}{12}=\frac{m-5}{6}=\frac{9}{27}\) is required. This gives (m=7).
Step 3
Exam Tip
अनंत हलों के लिए \(\frac{4}{12}=\frac{m-5}{6}=\frac{9}{27}\) चाहिए। इससे (m=7) मिलता है।
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युग्म ((k+1)x+6y=3) और (2x+12y=8) के लिए कोई हल न होने पर (k) का मान क्या होगा?
For no solution in ((k+1)x+6y=3) and (2x+12y=8), what is (k)?
#class10
#linear-equations
#solvability
A (k=-1)
B (k=0)
C (k=1)
D (k=2)
Explanation opens after your attempt
Step 1
Concept
Equating coefficient ratios gives (k=0). The constant ratio \(\frac{3}{8}\) is different, so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. (k=0). Equating coefficient ratios gives (k=0). The constant ratio \(\frac{3}{8}\) is different, so there is no solution.
Step 3
Exam Tip
गुणांक अनुपात समान करने पर (k=0) मिलता है। स्थिर अनुपात \(\frac{3}{8}\) अलग है, इसलिए कोई हल नहीं।
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युग्म ((k-3)x+2y=5) और (4x+ky=11) अद्वितीय न हो, इसके लिए (k) क्या हो सकता है?
For ((k-3)x+2y=5) and (4x+ky=11) to be non-unique, what can (k) be?
#class10
#linear-equations
#solvability
A \(k=\frac{3\pm\sqrt{41}}{2}\)
B \(k=\frac{-3\pm\sqrt{41}}{2}\)
C (k=3)
D (k=4)
Explanation opens after your attempt
Correct Answer
A. \(k=\frac{3\pm\sqrt{41}}{2}\)
Step 1
Concept
For non-unique solutions, the determinant must be zero. From ((k-3)k-8=0), \(k=\frac{3\pm\sqrt{41}}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{3\pm\sqrt{41}}{2}\). For non-unique solutions, the determinant must be zero. From ((k-3)k-8=0), \(k=\frac{3\pm\sqrt{41}}{2}\).
Step 3
Exam Tip
अद्वितीय न होने के लिए सारणिक शून्य होना चाहिए। ((k-3)k-8=0) से \(k=\frac{3\pm\sqrt{41}}{2}\) मिलता है।
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