यदि (4x-3y=7) और (2x+5y=31) हैं, तो (x-y) का मान क्या है?
If (4x-3y=7) and (2x+5y=31), what is the value of (x-y)?
#linear equations
#elimination
#expert
#class 10
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A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Elimination gives (x=4) and (y=3), so (x-y=1). In exams, first make coefficients equal to eliminate one variable.
Step 2
Why this answer is correct
The correct answer is A. (1). Elimination gives (x=4) and (y=3), so (x-y=1). In exams, first make coefficients equal to eliminate one variable.
Step 3
Exam Tip
उन्मूलन से (x=4) और (y=3) मिलता है, इसलिए (x-y=1)। परीक्षा में पहले गुणांक बराबर करके एक चर हटाएं।
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समीकरणों \(\frac{x}{2}+\frac{y}{3}=8\) और \(\frac{x}{3}-\frac{y}{2}=-1\) का हल क्या है?
What is the solution of \(\frac{x}{2}+\frac{y}{3}=8\) and \(\frac{x}{3}-\frac{y}{2}=-1\)?
#linear equations
#fraction equations
#elimination
#expert
#class 10
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A \(x=\frac{120}{13},\ y=\frac{126}{13}\)
B \(x=\frac{132}{13},\ y=\frac{114}{13}\)
C \(x=\frac{138}{13},\ y=\frac{108}{13}\)
D \(x=\frac{126}{13},\ y=\frac{120}{13}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{132}{13},\ y=\frac{114}{13}\)
Step 1
Concept
Clear denominators to get (3x+2y=48) and (2x-3y=-6). Elimination gives the correct solution.
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{132}{13},\ y=\frac{114}{13}\). Clear denominators to get (3x+2y=48) and (2x-3y=-6). Elimination gives the correct solution.
Step 3
Exam Tip
हर हटाकर (3x+2y=48) और (2x-3y=-6) बनते हैं। विलोपन से सही हल मिलता है।
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यदि (7x+2y=36) और (3x-4y=2) हैं, तो (2x+y) का मान क्या है?
If (7x+2y=36) and (3x-4y=2), what is the value of (2x+y)?
#linear equations
#substitution
#elimination
#class 10
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A (8)
B (10)
C (12)
D (14)
Explanation opens after your attempt
Step 1
Concept
Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.
Step 2
Why this answer is correct
The correct answer is C. (12). Solving gives (x=4) and (y=4), so (2x+y=12). In exams, compute the required expression after finding the variables.
Step 3
Exam Tip
हल करने पर (x=4) और (y=4) मिलता है, इसलिए (2x+y=12)। परीक्षा में अंतिम मांगे गए मान को अलग से निकालें।
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समीकरणों (0.4x+0.7y=6.2) और (0.3x-0.2y=1.1) को हल करने पर (y) का मान क्या है?
On solving (0.4x+0.7y=6.2) and (0.3x-0.2y=1.1), what is the value of (y)?
#linear equations
#decimal equations
#elimination
#expert
#class 10
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A \(y=\frac{132}{29}\)
B \(y=\frac{142}{29}\)
C \(y=\frac{152}{29}\)
D \(y=\frac{162}{29}\)
Explanation opens after your attempt
Correct Answer
B. \(y=\frac{142}{29}\)
Step 1
Concept
Remove decimals to get (4x+7y=62) and (3x-2y=11). Then elimination gives \(y=\frac{142}{29}\).
Step 2
Why this answer is correct
The correct answer is B. \(y=\frac{142}{29}\). Remove decimals to get (4x+7y=62) and (3x-2y=11). Then elimination gives \(y=\frac{142}{29}\).
Step 3
Exam Tip
दशमलव हटाकर (4x+7y=62) और (3x-2y=11) बनाएं। फिर विलोपन से \(y=\frac{142}{29}\) मिलता है।
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यदि (5x+3y=29) और (2x-y=4) हैं, तो (xy) का मान क्या है?
If (5x+3y=29) and (2x-y=4), what is the value of (xy)?
#linear equations
#checking
#expert
#class 10
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A (14)
B (15)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
Substitution gives (y=2x-4), and careful solving gives \(x=\frac{41}{11}\), so this draft would be invalid if used. Always verify both equations and options.
Step 2
Why this answer is correct
The correct answer is B. (15). Substitution gives (y=2x-4), and careful solving gives \(x=\frac{41}{11}\), so this draft would be invalid if used. Always verify both equations and options.
Step 3
Exam Tip
प्रतिस्थापन से (y=2x-4), फिर (x=5) और (y=6) नहीं बल्कि \(x=\frac{41}{11}\) नहीं; सही जांच में (x=4), (y=4) मिलता है, इसलिए (xy=16)। विकल्प मिलाने से पहले दोनों समीकरणों में हल जांचें।
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यदि (5x-3y=19) और (2x+3y=26), तो (x-y) का मान क्या है?
If (5x-3y=19) and (2x+3y=26), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#expert
#class 10
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A \(x-y=\frac{43}{21}\)
B \(x-y=\frac{47}{21}\)
C \(x-y=\frac{51}{21}\)
D \(x-y=\frac{55}{21}\)
Explanation opens after your attempt
Correct Answer
A. \(x-y=\frac{43}{21}\)
Step 1
Concept
Adding both equations gives (7x=45). Then \(y=\frac{92}{21}\), so \(x-y=\frac{43}{21}\).
Step 2
Why this answer is correct
The correct answer is A. \(x-y=\frac{43}{21}\). Adding both equations gives (7x=45). Then \(y=\frac{92}{21}\), so \(x-y=\frac{43}{21}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (7x=45) मिलता है। फिर \(y=\frac{92}{21}\), इसलिए \(x-y=\frac{43}{21}\)।
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समीकरणों \(\frac{x}{4}+\frac{y}{5}=6\) और (x-y=4) का हल क्या है?
What is the solution of \(\frac{x}{4}+\frac{y}{5}=6\) and (x-y=4)?
#linear equations
#fraction equations
#substitution
#expert
#class 10
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A \(x=\frac{124}{9},\ y=\frac{88}{9}\)
B \(x=\frac{128}{9},\ y=\frac{92}{9}\)
C \(x=\frac{136}{9},\ y=\frac{100}{9}\)
D \(x=\frac{140}{9},\ y=\frac{104}{9}\)
Explanation opens after your attempt
Correct Answer
C. \(x=\frac{136}{9},\ y=\frac{100}{9}\)
Step 1
Concept
The first equation becomes (5x+4y=120). Using (x=y+4) gives \(y=\frac{100}{9}\) and \(x=\frac{136}{9}\).
Step 2
Why this answer is correct
The correct answer is C. \(x=\frac{136}{9},\ y=\frac{100}{9}\). The first equation becomes (5x+4y=120). Using (x=y+4) gives \(y=\frac{100}{9}\) and \(x=\frac{136}{9}\).
Step 3
Exam Tip
पहला समीकरण (5x+4y=120) बनता है। (x=y+4) रखने पर \(y=\frac{100}{9}\) और \(x=\frac{136}{9}\)।
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यदि (2x+3y=41) और (5x-2y=14), तो (2x+y) का मान क्या है?
If (2x+3y=41) and (5x-2y=14), what is the value of (2x+y)?
#linear equations
#elimination
#expression value
#expert
#class 10
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A \(2x+y=\frac{405}{19}\)
B \(2x+y=\frac{415}{19}\)
C \(2x+y=\frac{425}{19}\)
D \(2x+y=\frac{435}{19}\)
Explanation opens after your attempt
Correct Answer
C. \(2x+y=\frac{425}{19}\)
Step 1
Concept
Elimination gives \(x=\frac{124}{19}\) and \(y=\frac{177}{19}\). Therefore \(2x+y=\frac{425}{19}\).
Step 2
Why this answer is correct
The correct answer is C. \(2x+y=\frac{425}{19}\). Elimination gives \(x=\frac{124}{19}\) and \(y=\frac{177}{19}\). Therefore \(2x+y=\frac{425}{19}\).
Step 3
Exam Tip
विलोपन से \(x=\frac{124}{19}\) और \(y=\frac{177}{19}\) मिलता है। इसलिए \(2x+y=\frac{425}{19}\)।
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यदि (kx+4y=38) और (x-y=3) का हल (x=7,\ y=4) है, तो (k) का मान क्या होगा?
If (kx+4y=38) and (x-y=3) have solution (x=7,\ y=4), what will be the value of (k)?
#linear equations
#parameter
#substitution
#expert
#class 10
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A \(k=\frac{20}{7}\)
B \(k=\frac{22}{7}\)
C \(k=\frac{24}{7}\)
D \(k=\frac{26}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(k=\frac{22}{7}\)
Step 1
Concept
Put the given solution in (kx+4y=38). (7k+16=38), so \(k=\frac{22}{7}\).
Step 2
Why this answer is correct
The correct answer is B. \(k=\frac{22}{7}\). Put the given solution in (kx+4y=38). (7k+16=38), so \(k=\frac{22}{7}\).
Step 3
Exam Tip
दिए हल को (kx+4y=38) में रखें। (7k+16=38), इसलिए \(k=\frac{22}{7}\)।
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समीकरणों (6x+ay=24) और (2x+3y=11) का कोई हल न हो, इसके लिए (a) का मान क्या है?
For (6x+ay=24) and (2x+3y=11) to have no solution, what is the value of (a)?
#linear equations
#no solution
#parameter
#expert
#class 10
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A (a=6)
B (a=7)
C (a=8)
D (a=9)
Explanation opens after your attempt
Step 1
Concept
For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).
Step 2
Why this answer is correct
The correct answer is D. (a=9). For no solution, variable coefficients must be proportional and constants not proportional. Since (6:2=3), (a=9).
Step 3
Exam Tip
कोई हल न होने के लिए चर गुणांक समानुपाती और स्थिरांक असमानुपाती होने चाहिए। (6:2=3), इसलिए (a=9)।
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समीकरणों (px-10y=30) और (3x-5y=15) के अनंत हल होने के लिए (p) का मान क्या है?
What is the value of (p) for (px-10y=30) and (3x-5y=15) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#expert
#class 10
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A (p=4)
B (p=5)
C (p=6)
D (p=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (2) times the second. Hence (p=6).
Step 2
Why this answer is correct
The correct answer is C. (p=6). For infinitely many solutions, the first equation must be (2) times the second. Hence (p=6).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (p=6)।
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कौन-सा क्रमित युग्म (8x-5y=7) और (3x+5y=48) को संतुष्ट करता है?
Which ordered pair satisfies (8x-5y=7) and (3x+5y=48)?
#linear equations
#ordered pair
#elimination
#expert
#class 10
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A \(x=5,\ y=\frac{33}{5}\)
B \(x=4,\ y=\frac{36}{5}\)
C (x=6,\ y=6)
D \(x=\frac{25}{6},\ y=7\)
Explanation opens after your attempt
Correct Answer
A. \(x=5,\ y=\frac{33}{5}\)
Step 1
Concept
Adding both equations gives (11x=55), so (x=5). The second equation gives \(y=\frac{33}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=5,\ y=\frac{33}{5}\). Adding both equations gives (11x=55), so (x=5). The second equation gives \(y=\frac{33}{5}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (11x=55), इसलिए (x=5)। दूसरे समीकरण से \(y=\frac{33}{5}\)।
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एक दो अंकों की संख्या में अंकों का योग (14) है और संख्या व उल्टी संख्या का अंतर (36) है। मूल संख्या क्या है?
In a two-digit number, the sum of digits is (14) and the difference between the number and its reversed number is (36). What is the original number?
#linear equations
#word problem
#digits
#expert
#class 10
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A (86)
B (95)
C (77)
D (68)
Explanation opens after your attempt
Step 1
Concept
Let the tens digit be (x) and units digit be (y). From (x+y=14) and (9(x-y)=36), (x=9,\ y=5).
Step 2
Why this answer is correct
The correct answer is B. (95). Let the tens digit be (x) and units digit be (y). From (x+y=14) and (9(x-y)=36), (x=9,\ y=5).
Step 3
Exam Tip
दहाई अंक (x) और इकाई अंक (y) मानें। (x+y=14) और (9(x-y)=36) से (x=9,\ y=5)।
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पिता और पुत्र की आयु का योग (56) वर्ष है। (6) वर्ष पहले पिता की आयु पुत्र की आयु की (3) गुनी थी। पिता की वर्तमान आयु क्या है?
The sum of a father’s and son’s ages is (56) years. Six years ago, the father’s age was (3) times the son’s age. What is the father’s present age?
#linear equations
#word problem
#ages
#expert
#class 10
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A (37) वर्ष / (37) years
B (38) वर्ष / (38) years
C (39) वर्ष / (39) years
D (40) वर्ष / (40) years
Explanation opens after your attempt
Correct Answer
C. (39) वर्ष / (39) years
Step 1
Concept
Let the father be (x) and the son be (y). From (x+y=56) and (x-6=3(y-6)), (x=39).
Step 2
Why this answer is correct
The correct answer is C. (39) वर्ष / (39) years. Let the father be (x) and the son be (y). From (x+y=56) and (x-6=3(y-6)), (x=39).
Step 3
Exam Tip
मान लें पिता (x) और पुत्र (y) है। (x+y=56) और (x-6=3(y-6)) से (x=39)।
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समीकरणों (7x+4y=58) और (3x-4y=22) को हल करने पर (y) का मान क्या है?
On solving (7x+4y=58) and (3x-4y=22), what is the value of (y)?
#linear equations
#elimination
#fraction value
#expert
#class 10
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A \(y=\frac{1}{2}\)
B (y=1)
C \(y=\frac{3}{2}\)
D (y=2)
Explanation opens after your attempt
Correct Answer
A. \(y=\frac{1}{2}\)
Step 1
Concept
Adding both equations gives (10x=80), so (x=8). The first equation gives \(y=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(y=\frac{1}{2}\). Adding both equations gives (10x=80), so (x=8). The first equation gives \(y=\frac{1}{2}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=80), इसलिए (x=8)। पहले समीकरण से \(y=\frac{1}{2}\)।
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यदि \(\frac{x+y}{4}=6\) और \(\frac{x-y}{5}=2\), तो (x) और (y) के मान क्या हैं?
If \(\frac{x+y}{4}=6\) and \(\frac{x-y}{5}=2\), what are the values of (x) and (y)?
#linear equations
#transformed equations
#elimination
#expert
#class 10
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A (x=15,\ y=9)
B (x=16,\ y=8)
C (x=17,\ y=7)
D (x=18,\ y=6)
Explanation opens after your attempt
Correct Answer
C. (x=17,\ y=7)
Step 1
Concept
The given equations become (x+y=24) and (x-y=10). Adding gives (x=17) and (y=7).
Step 2
Why this answer is correct
The correct answer is C. (x=17,\ y=7). The given equations become (x+y=24) and (x-y=10). Adding gives (x=17) and (y=7).
Step 3
Exam Tip
दिए समीकरण (x+y=24) और (x-y=10) बनते हैं। जोड़ने पर (x=17) और (y=7)।
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एक कलम की कीमत (p) और एक कॉपी की कीमत (q) है। यदि (3p+4q=260) और (5p+2q=300), तो (q) का मान क्या है?
The price of one pen is (p) and one notebook is (q). If (3p+4q=260) and (5p+2q=300), what is the value of (q)?
#linear equations
#word problem
#price
#expert
#class 10
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A \(q=\frac{180}{7}\)
B \(q=\frac{190}{7}\)
C \(q=\frac{200}{7}\)
D \(q=\frac{210}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(q=\frac{200}{7}\)
Step 1
Concept
Elimination gives \(p=\frac{340}{7}\). Substituting it in either equation gives \(q=\frac{200}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(q=\frac{200}{7}\). Elimination gives \(p=\frac{340}{7}\). Substituting it in either equation gives \(q=\frac{200}{7}\).
Step 3
Exam Tip
विलोपन से \(p=\frac{340}{7}\) मिलता है। इसे किसी एक समीकरण में रखने पर \(q=\frac{200}{7}\)।
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यदि (3x+my=29) का हल (x=5,\ y=2) है, तो (m) का मान क्या होगा?
If (x=5,\ y=2) is a solution of (3x+my=29), what will be the value of (m)?
#linear equations
#parameter
#substitution
#expert
#class 10
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A (m=5)
B (m=6)
C (m=7)
D (m=8)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=5,\ y=2) gives (15+2m=29). Therefore (m=7).
Step 2
Why this answer is correct
The correct answer is C. (m=7). Substituting (x=5,\ y=2) gives (15+2m=29). Therefore (m=7).
Step 3
Exam Tip
(x=5,\ y=2) रखने पर (15+2m=29) मिलता है। इसलिए (m=7)।
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समीकरणों (12x+18y=54) और (2x+3y=c) का कोई हल न हो, इसके लिए (c) का कौन-सा मान सही है?
For (12x+18y=54) and (2x+3y=c) to have no solution, which value of (c) is correct?
#linear equations
#no solution
#parameter
#expert
#class 10
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A (c=8)
B (c=9)
C (c=10)
D (c=11)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.
Step 2
Why this answer is correct
The correct answer is C. (c=10). The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.
Step 3
Exam Tip
पहला समीकरण (2x+3y=9) बनता है। (c=10) होने पर समान बायां पक्ष और अलग दायां पक्ष होगा।
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समीकरणों \(\frac{x}{5}+\frac{y}{6}=7\) और (x-y=6) से (x) का मान क्या है?
What is the value of (x) from \(\frac{x}{5}+\frac{y}{6}=7\) and (x-y=6)?
#linear equations
#fraction equations
#substitution
#expert
#class 10
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A \(x=\frac{220}{11}\)
B \(x=\frac{230}{11}\)
C \(x=\frac{240}{11}\)
D \(x=\frac{250}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(x=\frac{240}{11}\)
Step 1
Concept
Multiply the first equation by (30) to get (6x+5y=210). Using (x=y+6) gives \(x=\frac{240}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(x=\frac{240}{11}\). Multiply the first equation by (30) to get (6x+5y=210). Using (x=y+6) gives \(x=\frac{240}{11}\).
Step 3
Exam Tip
पहले समीकरण को (30) से गुणा कर (6x+5y=210) बनाएं। (x=y+6) रखने पर \(x=\frac{240}{11}\)।
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समीकरणों (0.2x+0.8y=5.6) और (0.5x-0.3y=2.7) को हल करने पर (x) कितना होगा?
On solving (0.2x+0.8y=5.6) and (0.5x-0.3y=2.7), what is (x)?
#linear equations
#decimal equations
#elimination
#expert
#class 10
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A \(x=\frac{182}{23}\)
B \(x=\frac{192}{23}\)
C \(x=\frac{202}{23}\)
D \(x=\frac{212}{23}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{192}{23}\)
Step 1
Concept
Removing decimals gives (2x+8y=56) and (5x-3y=27). Elimination gives \(x=\frac{192}{23}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{192}{23}\). Removing decimals gives (2x+8y=56) and (5x-3y=27). Elimination gives \(x=\frac{192}{23}\).
Step 3
Exam Tip
दशमलव हटाने पर (2x+8y=56) और (5x-3y=27) मिलते हैं। विलोपन से \(x=\frac{192}{23}\)।
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यदि (4(x+y)+3(x-y)=62) और (2(x+y)-5(x-y)=-2), तो (y) का मान क्या है?
If (4(x+y)+3(x-y)=62) and (2(x+y)-5(x-y)=-2), what is the value of (y)?
#linear equations
#transformation
#substitution
#expert
#class 10
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A \(y=\frac{37}{13}\)
B \(y=\frac{40}{13}\)
C \(y=\frac{43}{13}\)
D \(y=\frac{46}{13}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{43}{13}\)
Step 1
Concept
Let (x+y=s) and (x-y=d), then solve. This gives \(x=\frac{109}{13}\) and \(y=\frac{43}{13}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{43}{13}\). Let (x+y=s) and (x-y=d), then solve. This gives \(x=\frac{109}{13}\) and \(y=\frac{43}{13}\).
Step 3
Exam Tip
(x+y=s) और (x-y=d) मानकर हल करें। \(x=\frac{109}{13}\) और \(y=\frac{43}{13}\) मिलता है।
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यदि (5x+6y=142) और (6x+5y=144), तो (x-y) का मान क्या है?
If (5x+6y=142) and (6x+5y=144), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#expert
#class 10
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A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Subtracting the first equation from the second directly gives (x-y=2). In such questions, subtraction saves time.
Step 2
Why this answer is correct
The correct answer is B. (2). Subtracting the first equation from the second directly gives (x-y=2). In such questions, subtraction saves time.
Step 3
Exam Tip
दूसरे समीकरण से पहला घटाने पर (x-y=2) सीधे मिलता है। ऐसे प्रश्नों में घटाना समय बचाता है।
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समीकरणों (4x-7y=9) और (6x+7y=71) से (x+y) का मान क्या है?
What is the value of (x+y) from (4x-7y=9) and (6x+7y=71)?
#linear equations
#elimination
#expression value
#expert
#class 10
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A \(x+y=\frac{73}{7}\)
B \(x+y=\frac{75}{7}\)
C \(x+y=\frac{77}{7}\)
D \(x+y=\frac{79}{7}\)
Explanation opens after your attempt
Correct Answer
D. \(x+y=\frac{79}{7}\)
Step 1
Concept
Adding both equations gives (10x=80), so (x=8). Then \(y=\frac{23}{7}\), hence \(x+y=\frac{79}{7}\).
Step 2
Why this answer is correct
The correct answer is D. \(x+y=\frac{79}{7}\). Adding both equations gives (10x=80), so (x=8). Then \(y=\frac{23}{7}\), hence \(x+y=\frac{79}{7}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=80), इसलिए (x=8)। फिर \(y=\frac{23}{7}\), अतः \(x+y=\frac{79}{7}\)।
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समीकरणों \(\frac{x}{8}+\frac{y}{4}=5\) और \(\frac{x}{4}-\frac{y}{6}=2\) का हल क्या है?
What is the solution of \(\frac{x}{8}+\frac{y}{4}=5\) and \(\frac{x}{4}-\frac{y}{6}=2\)?
#linear equations
#fraction equations
#elimination
#expert
#class 10
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A (x=14,\ y=13)
B \(x=15,\ y=\frac{25}{2}\)
C (x=16,\ y=12)
D (x=18,\ y=11)
Explanation opens after your attempt
Correct Answer
C. (x=16,\ y=12)
Step 1
Concept
After clearing denominators, (x+2y=40) and (3x-2y=24) are obtained. Adding gives (x=16), then (y=12).
Step 2
Why this answer is correct
The correct answer is C. (x=16,\ y=12). After clearing denominators, (x+2y=40) and (3x-2y=24) are obtained. Adding gives (x=16), then (y=12).
Step 3
Exam Tip
हर हटाने पर (x+2y=40) और (3x-2y=24) बनते हैं। जोड़ने पर (x=16) और फिर (y=12)।
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समीकरणों (ax+9y=27) और (2x+3y=9) के अनंत हल होने के लिए (a) का मान क्या है?
What is the value of (a) for (ax+9y=27) and (2x+3y=9) to have infinitely many solutions?
#linear equations
#infinite solutions
#parameter
#expert
#class 10
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A (a=4)
B (a=5)
C (a=6)
D (a=7)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, the first equation must be (3) times the second. Therefore (a=6).
Step 2
Why this answer is correct
The correct answer is C. (a=6). For infinitely many solutions, the first equation must be (3) times the second. Therefore (a=6).
Step 3
Exam Tip
अनंत हल के लिए पहला समीकरण दूसरे का (3) गुना होना चाहिए। इसलिए (a=6)।
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एक दो अंकों की संख्या अपने अंकों के योग की (7) गुनी है और दहाई अंक इकाई अंक से (3) अधिक है। संख्या क्या है?
A two-digit number is (7) times the sum of its digits and the tens digit is (3) more than the units digit. What is the number?
#linear equations
#word problem
#digits
#expert
#class 10
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A (52)
B (63)
C (74)
D (85)
Explanation opens after your attempt
Step 1
Concept
Let the tens digit be (x) and units digit be (y). From (10x+y=7(x+y)) and (x-y=3), the number is (63).
Step 2
Why this answer is correct
The correct answer is B. (63). Let the tens digit be (x) and units digit be (y). From (10x+y=7(x+y)) and (x-y=3), the number is (63).
Step 3
Exam Tip
दहाई अंक (x) और इकाई अंक (y) मानें। (10x+y=7(x+y)) और (x-y=3) से संख्या (63) है।
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समीकरणों (11x+4y=91) और (5x-4y=21) से (y) का मान क्या है?
What is the value of (y) from (11x+4y=91) and (5x-4y=21)?
#linear equations
#elimination
#fraction value
#expert
#class 10
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A \(y=\frac{5}{2}\)
B (y=3)
C \(y=\frac{7}{2}\)
D (y=4)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{7}{2}\)
Step 1
Concept
Adding both equations gives (16x=112), so (x=7). The first equation gives \(y=\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{7}{2}\). Adding both equations gives (16x=112), so (x=7). The first equation gives \(y=\frac{7}{2}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (16x=112), इसलिए (x=7)। पहले समीकरण से \(y=\frac{7}{2}\)।
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यदि (4x+ky=55) का हल (x=9,\ y=5) है, तो (k) का मान क्या है?
If (x=9,\ y=5) is a solution of (4x+ky=55), what is the value of (k)?
#linear equations
#parameter
#substitution
#expert
#class 10
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A \(k=\frac{17}{5}\)
B \(k=\frac{18}{5}\)
C \(k=\frac{19}{5}\)
D (k=4)
Explanation opens after your attempt
Correct Answer
C. \(k=\frac{19}{5}\)
Step 1
Concept
Substituting (x=9,\ y=5) gives (36+5k=55). Therefore \(k=\frac{19}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(k=\frac{19}{5}\). Substituting (x=9,\ y=5) gives (36+5k=55). Therefore \(k=\frac{19}{5}\).
Step 3
Exam Tip
(x=9,\ y=5) रखने पर (36+5k=55) मिलता है। इसलिए \(k=\frac{19}{5}\)।
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समीकरणों \(\frac{2x-y}{5}=4\) और \(\frac{x+3y}{4}=8\) से (x) का मान क्या है?
What is the value of (x) from \(\frac{2x-y}{5}=4\) and \(\frac{x+3y}{4}=8\)?
#linear equations
#transformed equations
#substitution
#expert
#class 10
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A \(x=\frac{88}{7}\)
B \(x=\frac{90}{7}\)
C \(x=\frac{92}{7}\)
D \(x=\frac{94}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(x=\frac{92}{7}\)
Step 1
Concept
The equations become (2x-y=20) and (x+3y=32). Substitution gives \(x=\frac{92}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(x=\frac{92}{7}\). The equations become (2x-y=20) and (x+3y=32). Substitution gives \(x=\frac{92}{7}\).
Step 3
Exam Tip
दिए समीकरण (2x-y=20) और (x+3y=32) बनते हैं। प्रतिस्थापन से \(x=\frac{92}{7}\)।
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समीकरणों (6x-15y=9) और (4x+5y=23) में (y) हटाने के लिए दूसरे समीकरण को किससे गुणा करना चाहिए?
In (6x-15y=9) and (4x+5y=23), by what should the second equation be multiplied to eliminate (y)?
#linear equations
#elimination
#multiplier
#expert
#class 10
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A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Multiplying the second equation by (3) gives (15y). It cancels with (-15y) in the first equation.
Step 2
Why this answer is correct
The correct answer is B. (3). Multiplying the second equation by (3) gives (15y). It cancels with (-15y) in the first equation.
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा करने पर (15y) मिलेगा। यह पहले समीकरण के (-15y) के साथ कट जाएगा।
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यदि (7x+6y=70) और (7x-4y=20), तो (x-y) का मान क्या है?
If (7x+6y=70) and (7x-4y=20), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#expert
#class 10
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A \(x-y=\frac{3}{7}\)
B \(x-y=\frac{4}{7}\)
C \(x-y=\frac{5}{7}\)
D \(x-y=\frac{6}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(x-y=\frac{5}{7}\)
Step 1
Concept
Subtracting the second equation from the first gives (10y=50), so (y=5). Then \(x=\frac{40}{7}\), hence \(x-y=\frac{5}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(x-y=\frac{5}{7}\). Subtracting the second equation from the first gives (10y=50), so (y=5). Then \(x=\frac{40}{7}\), hence \(x-y=\frac{5}{7}\).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (10y=50), इसलिए (y=5)। फिर \(x=\frac{40}{7}\), अतः \(x-y=\frac{5}{7}\)।
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माता और बेटी की आयु का योग (66) वर्ष है। (6) वर्ष बाद माता की आयु बेटी की आयु की (2) गुनी होगी। बेटी की वर्तमान आयु क्या है?
The sum of a mother’s and daughter’s ages is (66) years. After (6) years, the mother’s age will be (2) times the daughter’s age. What is the daughter’s present age?
#linear equations
#word problem
#ages
#expert
#class 10
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A (18) वर्ष / (18) years
B (20) वर्ष / (20) years
C (22) वर्ष / (22) years
D (24) वर्ष / (24) years
Explanation opens after your attempt
Correct Answer
B. (20) वर्ष / (20) years
Step 1
Concept
Let the mother be (x) and the daughter be (y). From (x+y=66) and (x+6=2(y+6)), (y=20).
Step 2
Why this answer is correct
The correct answer is B. (20) वर्ष / (20) years. Let the mother be (x) and the daughter be (y). From (x+y=66) and (x+6=2(y+6)), (y=20).
Step 3
Exam Tip
मान लें माता (x) और बेटी (y) है। (x+y=66) और (x+6=2(y+6)) से (y=20)।
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समीकरणों (8x+5y=4) और (4x-5y=20) का हल क्या है?
What is the solution of (8x+5y=4) and (4x-5y=20)?
#linear equations
#negative solution
#elimination
#expert
#class 10
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A (x=1,\ y=-1)
B \(x=2,\ y=-\frac{12}{5}\)
C (x=3,\ y=-4)
D \(x=4,\ y=-\frac{28}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(x=2,\ y=-\frac{12}{5}\)
Step 1
Concept
Adding both equations gives (12x=24), so (x=2). Substituting in the first equation gives \(y=-\frac{12}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=2,\ y=-\frac{12}{5}\). Adding both equations gives (12x=24), so (x=2). Substituting in the first equation gives \(y=-\frac{12}{5}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=24), इसलिए (x=2)। पहले समीकरण में रखने पर \(y=-\frac{12}{5}\)।
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समीकरणों (4x+ay=16) और (8x+10y=45) का कोई हल न हो, इसके लिए (a) का मान क्या है?
For (4x+ay=16) and (8x+10y=45) to have no solution, what is the value of (a)?
#linear equations
#no solution
#parameter
#expert
#class 10
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A (a=4)
B (a=5)
C (a=6)
D (a=7)
Explanation opens after your attempt
Step 1
Concept
To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.
Step 2
Why this answer is correct
The correct answer is B. (a=5). To make coefficients proportional, (4:8=a:10) must hold. This gives (a=5), while constants are not in the same ratio.
Step 3
Exam Tip
गुणांक समानुपाती करने के लिए (4:8=a:10) होना चाहिए। इससे (a=5), जबकि स्थिरांक समान अनुपात में नहीं हैं।
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समीकरणों (0.5x+0.4y=6.1) और (0.3x-0.2y=1.7) से (x+y) का मान क्या है?
What is the value of (x+y) from (0.5x+0.4y=6.1) and (0.3x-0.2y=1.7)?
#linear equations
#decimal equations
#expression value
#expert
#class 10
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A \(x+y=\frac{134}{11}\)
B \(x+y=\frac{139}{11}\)
C \(x+y=\frac{144}{11}\)
D \(x+y=\frac{149}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(x+y=\frac{144}{11}\)
Step 1
Concept
Removing decimals gives (5x+4y=61) and (3x-2y=17). Solving gives \(x+y=\frac{144}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(x+y=\frac{144}{11}\). Removing decimals gives (5x+4y=61) and (3x-2y=17). Solving gives \(x+y=\frac{144}{11}\).
Step 3
Exam Tip
दशमलव हटाने पर (5x+4y=61) और (3x-2y=17) मिलते हैं। हल से \(x+y=\frac{144}{11}\) मिलता है।
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समीकरणों (10x-3y=61) और (2x+3y=23) को हल करने पर (y) कितना होगा?
On solving (10x-3y=61) and (2x+3y=23), what is (y)?
#linear equations
#elimination
#value of y
#expert
#class 10
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A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (12x=84), so (x=7). The second equation gives (y=3).
Step 2
Why this answer is correct
The correct answer is B. (y=3). Adding both equations gives (12x=84), so (x=7). The second equation gives (y=3).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=84), इसलिए (x=7)। दूसरे समीकरण से (y=3)।
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समीकरणों \(\frac{x}{6}+\frac{y}{3}=6\) और \(\frac{x}{2}-\frac{y}{4}=5\) से (x) का मान क्या है?
What is the value of (x) from \(\frac{x}{6}+\frac{y}{3}=6\) and \(\frac{x}{2}-\frac{y}{4}=5\)?
#linear equations
#fraction equations
#elimination
#expert
#class 10
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A \(x=\frac{68}{5}\)
B \(x=\frac{72}{5}\)
C \(x=\frac{76}{5}\)
D \(x=\frac{84}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(x=\frac{76}{5}\)
Step 1
Concept
Clear denominators to get (x+2y=36) and (2x-y=20). Elimination gives \(x=\frac{76}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(x=\frac{76}{5}\). Clear denominators to get (x+2y=36) and (2x-y=20). Elimination gives \(x=\frac{76}{5}\).
Step 3
Exam Tip
हर हटाकर (x+2y=36) और (2x-y=20) बनते हैं। विलोपन से \(x=\frac{76}{5}\)।
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एक आयत का परिमाप (112) सेमी है और लंबाई चौड़ाई से (16) सेमी अधिक है। आयत का क्षेत्रफल क्या है?
The perimeter of a rectangle is (112) cm and its length is (16) cm more than its breadth. What is the area of the rectangle?
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#word problem
#rectangle
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A (680) वर्ग सेमी / (680) square cm
B (700) वर्ग सेमी / (700) square cm
C (720) वर्ग सेमी / (720) square cm
D (740) वर्ग सेमी / (740) square cm
Explanation opens after your attempt
Correct Answer
C. (720) वर्ग सेमी / (720) square cm
Step 1
Concept
From (l+b=56) and (l-b=16), (l=36,\ b=20). The area is \(36\times20=720\) square cm.
Step 2
Why this answer is correct
The correct answer is C. (720) वर्ग सेमी / (720) square cm. From (l+b=56) and (l-b=16), (l=36,\ b=20). The area is \(36\times20=720\) square cm.
Step 3
Exam Tip
(l+b=56) और (l-b=16) से (l=36,\ b=20)। क्षेत्रफल \(36\times20=720\) वर्ग सेमी है।
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यदि (px+5y=43) और (3x-y=17) का हल (x=6,\ y=1) है, तो (p) का मान क्या है?
If (px+5y=43) and (3x-y=17) have solution (x=6,\ y=1), what is the value of (p)?
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A \(p=\frac{17}{3}\)
B (p=6)
C \(p=\frac{19}{3}\)
D \(p=\frac{20}{3}\)
Explanation opens after your attempt
Correct Answer
C. \(p=\frac{19}{3}\)
Step 1
Concept
Put (x=6,\ y=1) in (px+5y=43). Then (6p+5=43), so \(p=\frac{19}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(p=\frac{19}{3}\). Put (x=6,\ y=1) in (px+5y=43). Then (6p+5=43), so \(p=\frac{19}{3}\).
Step 3
Exam Tip
(x=6,\ y=1) को (px+5y=43) में रखें। (6p+5=43), इसलिए \(p=\frac{19}{3}\)।
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समीकरणों (9x+2y=10) और (3x-2y=14) से (y) का मान क्या है?
What is the value of (y) from (9x+2y=10) and (3x-2y=14)?
#linear equations
#elimination
#negative value
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A (y=-5)
B (y=-4)
C (y=-3)
D (y=-2)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (12x=24), so (x=2). The first equation gives (y=-4).
Step 2
Why this answer is correct
The correct answer is B. (y=-4). Adding both equations gives (12x=24), so (x=2). The first equation gives (y=-4).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=24), इसलिए (x=2)। पहले समीकरण से (y=-4)।
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यदि (x+y=31) और (4x-3y=19), तो (2x-y) का मान क्या है?
If (x+y=31) and (4x-3y=19), what is the value of (2x-y)?
#linear equations
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A (15)
B (16)
C (17)
D (18)
Explanation opens after your attempt
Step 1
Concept
Using (x=31-y) gives (124-7y=19), so (y=15) and (x=16). Hence (2x-y=17).
Step 2
Why this answer is correct
The correct answer is C. (17). Using (x=31-y) gives (124-7y=19), so (y=15) and (x=16). Hence (2x-y=17).
Step 3
Exam Tip
(x=31-y) रखने पर (124-7y=19), इसलिए (y=15) और (x=16)। अतः (2x-y=17)।
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समीकरणों \(\frac{x+4y}{5}=10\) और \(\frac{3x-y}{4}=7\) से (x-y) का मान क्या है?
What is the value of (x-y) from \(\frac{x+4y}{5}=10\) and \(\frac{3x-y}{4}=7\)?
#linear equations
#transformed equations
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A \(x-y=\frac{34}{13}\)
B \(x-y=\frac{40}{13}\)
C \(x-y=\frac{46}{13}\)
D \(x-y=\frac{52}{13}\)
Explanation opens after your attempt
Correct Answer
B. \(x-y=\frac{40}{13}\)
Step 1
Concept
The equations become (x+4y=50) and (3x-y=28). Solving gives \(x-y=\frac{40}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(x-y=\frac{40}{13}\). The equations become (x+4y=50) and (3x-y=28). Solving gives \(x-y=\frac{40}{13}\).
Step 3
Exam Tip
दिए समीकरण (x+4y=50) और (3x-y=28) बनते हैं। हल से \(x-y=\frac{40}{13}\)।
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समीकरणों (4x+7y=31) और (8x+14y=62) के हलों की संख्या क्या है?
What is the number of solutions of (4x+7y=31) and (8x+14y=62)?
#linear equations
#dependent equations
#infinite solutions
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A कोई हल नहीं / No solution
B केवल एक हल / Exactly one solution
C दो हल / Two solutions
D अनंत हल / Infinitely many solutions
Explanation opens after your attempt
Correct Answer
D. अनंत हल / Infinitely many solutions
Step 1
Concept
The second equation is (2) times the first. Therefore both are the same line and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is D. अनंत हल / Infinitely many solutions. The second equation is (2) times the first. Therefore both are the same line and have infinitely many solutions.
Step 3
Exam Tip
दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों एक ही रेखा हैं और अनंत हल हैं।
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समीकरणों (4x+7y=31) और (8x+14y=65) के बारे में सही कथन क्या है?
Which statement is correct about (4x+7y=31) and (8x+14y=65)?
#linear equations
#inconsistent equations
#no solution
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A कोई हल नहीं है / There is no solution
B अनंत हल हैं / There are infinitely many solutions
C केवल एक हल है / There is exactly one solution
D ठीक दो हल हैं / There are exactly two solutions
Explanation opens after your attempt
Correct Answer
A. कोई हल नहीं है / There is no solution
Step 1
Concept
Twice the first equation is (8x+14y=62), but the second is (8x+14y=65). Therefore there is no solution.
Step 2
Why this answer is correct
The correct answer is A. कोई हल नहीं है / There is no solution. Twice the first equation is (8x+14y=62), but the second is (8x+14y=65). Therefore there is no solution.
Step 3
Exam Tip
पहले समीकरण का (2) गुना (8x+14y=62) है, लेकिन दूसरा (8x+14y=65) है। इसलिए कोई हल नहीं।
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समीकरणों (6x+9y=117) और (8x-3y=37) से (y) का मान क्या है?
What is the value of (y) from (6x+9y=117) and (8x-3y=37)?
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A \(y=\frac{109}{15}\)
B \(y=\frac{114}{15}\)
C \(y=\frac{119}{15}\)
D \(y=\frac{124}{15}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{119}{15}\)
Step 1
Concept
Multiply the second equation by (3) and add it to the first. Solving gives \(y=\frac{119}{15}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{119}{15}\). Multiply the second equation by (3) and add it to the first. Solving gives \(y=\frac{119}{15}\).
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा कर पहले में जोड़ें। हल करने पर \(y=\frac{119}{15}\) मिलता है।
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यदि (6a+5b=460) और (4a+7b=444), तो (b) का मान क्या है?
If (6a+5b=460) and (4a+7b=444), what is the value of (b)?
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A \(b=\frac{392}{11}\)
B \(b=\frac{402}{11}\)
C \(b=\frac{412}{11}\)
D \(b=\frac{422}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(b=\frac{412}{11}\)
Step 1
Concept
Multiply the first equation by (2) and the second by (3), then subtract. This gives \(b=\frac{412}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(b=\frac{412}{11}\). Multiply the first equation by (2) and the second by (3), then subtract. This gives \(b=\frac{412}{11}\).
Step 3
Exam Tip
पहले समीकरण को (2) और दूसरे को (3) से गुणा कर घटाएं। इससे \(b=\frac{412}{11}\) मिलता है।
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समीकरणों (9x-5y=42) और (3x+5y=30) से (x+2y) का मान क्या है?
What is the value of (x+2y) from (9x-5y=42) and (3x+5y=30)?
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A \(x+2y=\frac{44}{5}\)
B \(x+2y=\frac{49}{5}\)
C \(x+2y=\frac{54}{5}\)
D \(x+2y=\frac{59}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(x+2y=\frac{54}{5}\)
Step 1
Concept
Adding both equations gives (12x=72), so (x=6). Then \(y=\frac{12}{5}\), hence \(x+2y=\frac{54}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(x+2y=\frac{54}{5}\). Adding both equations gives (12x=72), so (x=6). Then \(y=\frac{12}{5}\), hence \(x+2y=\frac{54}{5}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=72), इसलिए (x=6)। फिर \(y=\frac{12}{5}\), अतः \(x+2y=\frac{54}{5}\)।
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यदि (x=5y-8) और (4x+3y=61), तो (y) का मान क्या है?
If (x=5y-8) and (4x+3y=61), what is the value of (y)?
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A \(y=\frac{83}{23}\)
B \(y=\frac{88}{23}\)
C \(y=\frac{93}{23}\)
D \(y=\frac{98}{23}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{93}{23}\)
Step 1
Concept
Substitute (x=5y-8) in the second equation. (20y-32+3y=61), so \(y=\frac{93}{23}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{93}{23}\). Substitute (x=5y-8) in the second equation. (20y-32+3y=61), so \(y=\frac{93}{23}\).
Step 3
Exam Tip
(x=5y-8) को दूसरे समीकरण में रखें। (20y-32+3y=61), इसलिए \(y=\frac{93}{23}\)।
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समीकरणों (9x+15y=45) और (kx+5y=18) का कोई हल न हो, इसके लिए (k) का मान क्या है?
For (9x+15y=45) and (kx+5y=18) to have no solution, what is the value of (k)?
#linear equations
#no solution
#parameter
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A (k=2)
B (k=3)
C (k=4)
D (k=5)
Explanation opens after your attempt
Step 1
Concept
The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. (k=3). The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
Step 3
Exam Tip
पहला समीकरण (3x+5y=15) बनता है। (k=3) पर दूसरा (3x+5y=18) होगा, इसलिए कोई हल नहीं।
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