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1329 questions tagged with expert.
If (cx+18y=72) and (24x+48y=145) have no solution, what will (c) be?
C. (9)
कोई हल नहीं के लिए (c/24=18/48) और (72/145) अलग होना चाहिए। इसलिए (c=9)। / For no solution, (c/24=18/48) and (72/145) must be different. Therefore, (c=9).
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What is the correct conclusion for the equations (15x+8y-47=0) and (30x+17y-94=0)?
C. एक अद्वितीय हलOne unique solution
यहां \(15/30 \ne 8/17\), इसलिए रेखाएं एक बिंदु पर कटती हैं। ऐसी स्थिति में एक अद्वितीय हल मिलता है। / Here \(15/30 \ne 8/17\), so the lines intersect at one point. In this case, one unique solution is obtained.
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What is the correct ratio relation for the equations (16x-9y+55=0) and (32x-18y+113=0)?
A. (1632=(-9) / (-18) \ne 55 / 113)
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा। / The first two ratios are equal and the third is different. Therefore, there will be no solution.
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Which ratio relation is correct for the equations (10x+13y-71=0) and (30x+39y-213=0)?
C. (1030=13 / 39=(-71) / (-213))
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं। / Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
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What is the correct description of the graph of the equations (5x+17y=69) and (12x+41y=166)?
C. एक बिंदु पर कटती रेखाएंLines intersecting at one point
यहां \(5/12 \ne 17/41\), इसलिए रेखाएं एक बिंदु पर कटेंगी। इससे एक अद्वितीय हल मिलेगा। / Here \(5/12 \ne 17/41\), so the lines will intersect at one point. This gives one unique solution.
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What will be shown in the graph of the equations (30x+18y=126) and (5x+3y=21)?
A. एक ही रेखाSame line
पहला समीकरण दूसरे का (6) गुना है। इसलिए दोनों समीकरण एक ही रेखा दिखाते हैं। / The first equation is (6) times the second. Therefore, both equations show the same line.
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Which statement is correct about the graph of the equations (22x+33y=99) and (2x+3y=12)?
C. रेखाएं अलग समानांतर हैंLines are distinct parallel
यहां (22/2=33/3) लेकिन (99/12) अलग है। इसलिए ग्राफ में अलग समानांतर रेखाएं बनेंगी। / Here (22/2=33/3) but (99/12) is different. Therefore, the graph will show distinct parallel lines.
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What is the value of (b) for the equations (bx+16y=64) and (14x+28y=131) to have no solution?
C. (8)
कोई हल नहीं के लिए (b/14=16/28) और (64/131) अलग होना चाहिए। इसलिए (b=8)। / For no solution, (b/14=16/28) and (64/131) must be different. Hence, (b=8).
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What will (a) be for the equations (8x+ay=72) and (24x+30y=216) to have infinitely many solutions?
C. (10)
अनंत हल के लिए (8/24=a/30=72/216) होना चाहिए। इससे (a=10) मिलता है। / For infinitely many solutions, (8/24=a/30=72/216) must hold. This gives (a=10).
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Which condition is correct for the equations (12x+py=60) and (3x+5y=16) to have a unique solution?
B. \(p\ne20\)
अद्वितीय हल के लिए \(12/3 \ne p/5\) होना चाहिए। इसलिए \(p\ne20\) सही शर्त है। / For a unique solution, \(12/3 \ne p/5\) must hold. Therefore, \(p\ne20\) is the correct condition.
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C. (9)
दूसरा समीकरण पहले का (2) गुना है, इसलिए (q/18=1/2) होना चाहिए। अतः (q=9)। / The second equation is (2) times the first, so (q/18=1/2) must hold. Hence, (q=9).
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What will (k) be for the equations (kx+14y=42) and (18x+21y=63) to have infinitely many solutions?
C. (12)
अनंत हल के लिए (k/18=14/21=42/63) होना चाहिए। इसलिए (k=12) सही है। / For infinitely many solutions, (k/18=14/21=42/63) must hold. Therefore, (k=12) is correct.
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What is the correct solution status for the equations (11x+14y=53) and (17x+22y=83)?
B. एक अद्वितीय हलOne unique solution
यहां \(11/17 \ne 14/22\), इसलिए रेखाएं एक बिंदु पर कटती हैं। पहले दो अनुपात अलग हों तो एक अद्वितीय हल मिलता है। / Here \(11/17 \ne 14/22\), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.
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How many solutions will the equations (9x+5y=41) and (18x+10y=85) have?
C. कोई हल नहींNo solution
यहां (9/18=5/10) लेकिन (41/85) अलग है। इसलिए रेखाएं समानांतर और अलग हैं। / Here (9/18=5/10) but (41/85) is different. Therefore, the lines are parallel and distinct.
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Which conclusion is correct for the equations (7x-4y=29) and (21x-12y=87)?
C. अनंत हलInfinitely many solutions
दूसरा समीकरण पहले का (3) गुना है। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं। / The second equation is (3) times the first. Therefore, both lines are coincident and have infinitely many solutions.
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If (5x+8y=37) and (15x+24y=m) form an inconsistent pair, what is the correct condition for (m)?
B. \(m\ne111\)
पहले दो अनुपात बराबर हैं। असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए। / The first two ratios are equal. For inconsistency, the constant ratio must be different.
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What is the value of (p) for the equations (px+9y=45) and (20x+30y=103) to have no solution?
B. (6)
कोई हल नहीं के लिए (p/20=9/30) और (45/103) अलग होना चाहिए। इससे (p=6) मिलता है। / For no solution, (p/20=9/30) and (45/103) must be different. This gives (p=6).
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C. (7)
अनंत हल के लिए (4/12=a/21=32/96) होना चाहिए। इसलिए (a=7) सही है। / For infinitely many solutions, (4/12=a/21=32/96) must hold. Therefore, (a=7) is correct.
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If (3(x+y)+2(x-y)=41) and (2(x+y)-3(x-y)=-1), what is the value of (x)?
D. (7)
मान लें (u=x+y) और (v=x-y)। (3u+2v=41), (2u-3v=-1) से (u=7,v=10), इसलिए \(x=\frac{17}{2}\)। / Let (u=x+y) and (v=x-y). Solving (3u+2v=41), (2u-3v=-1) gives (u=7,v=10), so \(x=\frac{17}{2}\).
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If (3x-2y=4) and (x+y=11), what is the value of (y) by substitution?
D. (7)
दूसरे समीकरण से (x=11-y) रखकर (33-5y=4) नहीं बल्कि (33-3y-2y=4) मिलता है, इसलिए \(y=\frac{29}{5}\) नहीं होगा; सही जांच जरूरी है। / Substitute (x=11-y) carefully in the first equation; incorrect simplification changes the answer. Always check the obtained values in both equations.
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Solving (6x+5y=43) and (4x-5y=7) by elimination, what is the value of (x)?
B. (5)
दोनों समीकरण जोड़ने पर (10x=50), इसलिए (x=5)। परीक्षा में विपरीत गुणांकों वाले पद पहले हटाएं। / Adding the two equations gives (10x=50), so (x=5). In exams, eliminate terms with opposite coefficients first.
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Solving (5x+6y=37) and (5x-2y=13), what is the value of (xy)?
A. (9)
घटाने पर (8y=24), इसलिए (y=3) और \(x=\frac{19}{5}\) नहीं बल्कि दूसरे में रखने से \(x=\frac{19}{5}\) नहीं आता; सही हल (x=5,y=2) नहीं है, इसलिए सावधानी चाहिए। / This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.
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For (7x+4y=2) and (3x-4y=18), what is the value of (x-y) in the solution?
B. (5)
समीकरण जोड़ने पर (10x=20), इसलिए (x=2) और (y=-3)। अतः (x-y=5)। / Adding the equations gives (10x=20), so (x=2) and (y=-3). Therefore (x-y=5).
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If (2x-3y=-7) and (4x+y=19), what is the value of (y) by substitution?
A. (y=3)
दूसरे समीकरण से (y=19-4x) रखकर हल करने पर (x=4) और (y=3) मिलता है। परीक्षा में पहले सरल चर को अलग करें। / From the second equation, put (y=19-4x), giving (x=4) and (y=3). In exams, isolate the easier variable first.
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Solving (3x+2y=19) and (5x-2y=21) by elimination gives which values of (x,y)?
B. (x=5, y=2)
दोनों समीकरण जोड़ने पर (8x=40) मिलता है इसलिए (x=5), फिर (y=2)। परीक्षा में विपरीत गुणांकों को सीधे जोड़ें। / Adding the equations gives (8x=40), so (x=5), then (y=2). In exams, add directly when coefficients are opposite.
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For (9x+15y=45) and (kx+5y=18) to have no solution, what is the value of (k)?
B. (k=3)
पहला समीकरण (3x+5y=15) बनता है। (k=3) पर दूसरा (3x+5y=18) होगा, इसलिए कोई हल नहीं। / The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
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If (x=5y-8) and (4x+3y=61), what is the value of (y)?
C. \(y=\frac{93}{23}\)
(x=5y-8) को दूसरे समीकरण में रखें। (20y-32+3y=61), इसलिए \(y=\frac{93}{23}\)। / Substitute (x=5y-8) in the second equation. (20y-32+3y=61), so \(y=\frac{93}{23}\).
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What is the value of (x+2y) from (9x-5y=42) and (3x+5y=30)?
C. \(x+2y=\frac{54}{5}\)
दोनों समीकरण जोड़ने पर (12x=72), इसलिए (x=6)। फिर \(y=\frac{12}{5}\), अतः \(x+2y=\frac{54}{5}\)। / Adding both equations gives (12x=72), so (x=6). Then \(y=\frac{12}{5}\), hence \(x+2y=\frac{54}{5}\).
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If (6a+5b=460) and (4a+7b=444), what is the value of (b)?
C. \(b=\frac{412}{11}\)
पहले समीकरण को (2) और दूसरे को (3) से गुणा कर घटाएं। इससे \(b=\frac{412}{11}\) मिलता है। / Multiply the first equation by (2) and the second by (3), then subtract. This gives \(b=\frac{412}{11}\).
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What is the value of (y) from (6x+9y=117) and (8x-3y=37)?
C. \(y=\frac{119}{15}\)
दूसरे समीकरण को (3) से गुणा कर पहले में जोड़ें। हल करने पर \(y=\frac{119}{15}\) मिलता है। / Multiply the second equation by (3) and add it to the first. Solving gives \(y=\frac{119}{15}\).
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