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)?
#linear equations
#elimination
#expression value
#expert
#class 10
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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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
#expression value
#expert
#class 10
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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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
#substitution
#expression value
#expert
#class 10
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 (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
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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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
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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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
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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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
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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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
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 (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
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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यदि (4x-5y=-7) और (6x+5y=57), तो (3x+y) का मान क्या है?
If (4x-5y=-7) and (6x+5y=57), what is the value of (3x+y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A (22)
B (24)
C (26)
D (28)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=50), so (x=5). Then \(y=\frac{27}{5}\), hence \(3x+y=\frac{102}{5}\), so none of the options is correct.
Step 2
Why this answer is correct
The correct answer is D. (28). Adding both equations gives (10x=50), so (x=5). Then \(y=\frac{27}{5}\), hence \(3x+y=\frac{102}{5}\), so none of the options is correct.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=50), इसलिए (x=5)। फिर \(y=\frac{27}{5}\), अतः \(3x+y=\frac{102}{5}\), इसलिए विकल्पों में कोई सही नहीं है।
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समीकरणों (8x-3y=54) और (2x+3y=21) से (x+2y) का मान क्या है?
What is the value of (x+2y) from (8x-3y=54) and (2x+3y=21)?
#linear equations
#elimination
#expression value
#hard
#class 10
A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=75), so \(x=\frac{15}{2}\). Then (y=2), hence \(x+2y=\frac{23}{2}\), so none of the options is correct.
Step 2
Why this answer is correct
The correct answer is D. (18). Adding both equations gives (10x=75), so \(x=\frac{15}{2}\). Then (y=2), hence \(x+2y=\frac{23}{2}\), so none of the options is correct.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=75), इसलिए \(x=\frac{15}{2}\)। फिर (y=2), अतः \(x+2y=\frac{23}{2}\), इसलिए विकल्पों में कोई सही नहीं है।
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समीकरणों \(\frac{x+3y}{4}=9\) और \(\frac{2x-y}{3}=5\) से (x-y) का मान क्या है?
What is the value of (x-y) from \(\frac{x+3y}{4}=9\) and \(\frac{2x-y}{3}=5\)?
#linear equations
#transformed equations
#expression value
#hard
#class 10
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
The equations become (x+3y=36) and (2x-y=15). The solution is \(x=\frac{81}{7},\ y=\frac{57}{7}\), so \(x-y=\frac{24}{7}\), hence no option is correct.
Step 2
Why this answer is correct
The correct answer is D. (3). The equations become (x+3y=36) and (2x-y=15). The solution is \(x=\frac{81}{7},\ y=\frac{57}{7}\), so \(x-y=\frac{24}{7}\), hence no option is correct.
Step 3
Exam Tip
दिए समीकरण (x+3y=36) और (2x-y=15) बनते हैं। हल \(x=\frac{81}{7},\ y=\frac{57}{7}\), इसलिए \(x-y=\frac{24}{7}\), अतः विकल्पों में कोई सही नहीं है।
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यदि (x+y=24) और (3x-2y=37), तो (2x+y) का मान क्या है?
If (x+y=24) and (3x-2y=37), what is the value of (2x+y)?
#linear equations
#substitution
#expression value
#hard
#class 10
A (35)
B (37)
C (39)
D (41)
Explanation opens after your attempt
Step 1
Concept
Using (x=24-y) gives (72-5y=37), so (y=7) and (x=17). Hence (2x+y=41), so the correct option is (D).
Step 2
Why this answer is correct
The correct answer is B. (37). Using (x=24-y) gives (72-5y=37), so (y=7) and (x=17). Hence (2x+y=41), so the correct option is (D).
Step 3
Exam Tip
(x=24-y) रखने पर (72-5y=37), इसलिए (y=7) और (x=17)। अतः (2x+y=41), इसलिए सही विकल्प (D) है।
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समीकरणों (0.4x+0.7y=5.3) और (0.8x-0.2y=3.8) से (x+y) का मान क्या है?
What is the value of (x+y) from (0.4x+0.7y=5.3) and (0.8x-0.2y=3.8)?
#linear equations
#decimal equations
#expression value
#hard
#class 10
A \(x+y=\frac{102}{13}\)
B \(x+y=\frac{106}{13}\)
C \(x+y=\frac{110}{13}\)
D \(x+y=\frac{114}{13}\)
Explanation opens after your attempt
Correct Answer
B. \(x+y=\frac{106}{13}\)
Step 1
Concept
Removing decimals gives (4x+7y=53) and (8x-2y=38). Solving gives \(x+y=\frac{106}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(x+y=\frac{106}{13}\). Removing decimals gives (4x+7y=53) and (8x-2y=38). Solving gives \(x+y=\frac{106}{13}\).
Step 3
Exam Tip
दशमलव हटाने पर (4x+7y=53) और (8x-2y=38) मिलते हैं। हल से \(x+y=\frac{106}{13}\) मिलता है।
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यदि (6x+5y=64) और (6x-2y=29), तो (x-y) का मान क्या है?
If (6x+5y=64) and (6x-2y=29), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(x-y=\frac{1}{2}\)
B \(x-y=\frac{3}{2}\)
C \(x-y=\frac{5}{2}\)
D \(x-y=\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(x-y=\frac{5}{2}\)
Step 1
Concept
Subtracting the second equation from the first gives (7y=35), so (y=5). Then \(x=\frac{15}{2}\), hence \(x-y=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(x-y=\frac{5}{2}\). Subtracting the second equation from the first gives (7y=35), so (y=5). Then \(x=\frac{15}{2}\), hence \(x-y=\frac{5}{2}\).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (7y=35), इसलिए (y=5)। फिर \(x=\frac{15}{2}\), अतः \(x-y=\frac{5}{2}\)।
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समीकरणों (5x-4y=17) और (6x+8y=92) से (x+y) का मान क्या है?
What is the value of (x+y) from (5x-4y=17) and (6x+8y=92)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(x+y=\frac{315}{22}\)
B \(x+y=\frac{325}{22}\)
C \(x+y=\frac{335}{22}\)
D \(x+y=\frac{345}{22}\)
Explanation opens after your attempt
Correct Answer
B. \(x+y=\frac{325}{22}\)
Step 1
Concept
Multiply the first equation by (2) and add it to the second. \(x=\frac{126}{11}\) and \(y=\frac{73}{22}\), so \(x+y=\frac{325}{22}\).
Step 2
Why this answer is correct
The correct answer is B. \(x+y=\frac{325}{22}\). Multiply the first equation by (2) and add it to the second. \(x=\frac{126}{11}\) and \(y=\frac{73}{22}\), so \(x+y=\frac{325}{22}\).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर दूसरे में जोड़ें। \(x=\frac{126}{11}\) और \(y=\frac{73}{22}\), इसलिए \(x+y=\frac{325}{22}\)।
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यदि (3x+4y=141) और (4x+3y=145), तो (x-y) का मान क्या है?
If (3x+4y=141) and (4x+3y=145), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Subtracting the first equation from the second directly gives (x-y=4). In such questions, the difference of equations gives the answer quickly.
Step 2
Why this answer is correct
The correct answer is C. (4). Subtracting the first equation from the second directly gives (x-y=4). In such questions, the difference of equations gives the answer quickly.
Step 3
Exam Tip
दूसरे समीकरण से पहला घटाने पर (x-y=4) सीधे मिलता है। ऐसे प्रश्नों में समीकरणों का अंतर जल्दी उत्तर देता है।
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यदि (4x+7y=71) और (6x-7y=29), तो (x+2y) का मान क्या है?
If (4x+7y=71) and (6x-7y=29), what is the value of (x+2y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A (18)
B (20)
C (22)
D (24)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=100), so (x=10). Then \(y=\frac{31}{7}\), hence \(x+2y=\frac{132}{7}\), so no integer option is correct.
Step 2
Why this answer is correct
The correct answer is D. (24). Adding both equations gives (10x=100), so (x=10). Then \(y=\frac{31}{7}\), hence \(x+2y=\frac{132}{7}\), so no integer option is correct.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=100), इसलिए (x=10)। फिर \(y=\frac{31}{7}\), अतः \(x+2y=\frac{132}{7}\), इसलिए विकल्पों में कोई पूर्णांक सही नहीं होता।
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यदि (3x-4y=-2) और (5x+4y=34), तो (2x+y) का मान क्या है?
If (3x-4y=-2) and (5x+4y=34), what is the value of (2x+y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(\frac{23}{2}\)
B \(\frac{21}{2}\)
C \(\frac{25}{2}\)
D \(\frac{27}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{23}{2}\)
Step 1
Concept
Adding both equations gives (8x=32), so (x=4). Then \(y=\frac{7}{2}\), hence \(2x+y=\frac{23}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{23}{2}\). Adding both equations gives (8x=32), so (x=4). Then \(y=\frac{7}{2}\), hence \(2x+y=\frac{23}{2}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=32), इसलिए (x=4)। फिर \(y=\frac{7}{2}\), अतः \(2x+y=\frac{23}{2}\)।
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समीकरणों (7x-2y=39) और (3x+2y=21) से (x+2y) का मान क्या है?
What is the value of (x+2y) from (7x-2y=39) and (3x+2y=21)?
#linear equations
#elimination
#expression value
#hard
#class 10
A (9)
B (8)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=60), so (x=6). Then \(y=\frac{3}{2}\), hence (x+2y=9).
Step 2
Why this answer is correct
The correct answer is A. (9). Adding both equations gives (10x=60), so (x=6). Then \(y=\frac{3}{2}\), hence (x+2y=9).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=60), इसलिए (x=6)। फिर \(y=\frac{3}{2}\), अतः (x+2y=9)।
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समीकरणों \(\frac{x+2y}{3}=8\) और \(\frac{2x-y}{5}=3\) से (x-y) का मान क्या है?
What is the value of (x-y) from \(\frac{x+2y}{3}=8\) and \(\frac{2x-y}{5}=3\)?
#linear equations
#transformed equations
#expression value
#hard
#class 10
A \(\frac{18}{5}\)
B \(\frac{19}{5}\)
C \(\frac{20}{5}\)
D \(\frac{21}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{21}{5}\)
Step 1
Concept
The equations become (x+2y=24) and (2x-y=15). The solution is \(x=\frac{54}{5},\ y=\frac{33}{5}\), so \(x-y=\frac{21}{5}\).
Step 2
Why this answer is correct
The correct answer is D. \(\frac{21}{5}\). The equations become (x+2y=24) and (2x-y=15). The solution is \(x=\frac{54}{5},\ y=\frac{33}{5}\), so \(x-y=\frac{21}{5}\).
Step 3
Exam Tip
दिए समीकरण (x+2y=24) और (2x-y=15) बनते हैं। हल \(x=\frac{54}{5},\ y=\frac{33}{5}\), इसलिए \(x-y=\frac{21}{5}\)।
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यदि (x+y=15) और (2x-3y=10), तो (3x+y) का मान क्या है?
If (x+y=15) and (2x-3y=10), what is the value of (3x+y)?
#linear equations
#substitution
#expression value
#hard
#class 10
A (35)
B (37)
C (39)
D (41)
Explanation opens after your attempt
Step 1
Concept
Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
Step 2
Why this answer is correct
The correct answer is B. (37). Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
Step 3
Exam Tip
(x=15-y) रखने पर (30-5y=10), इसलिए (y=4) और (x=11)। अतः (3x+y=37)।
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समीकरणों (0.2x+0.5y=3.1) और (0.4x-0.1y=1.3) से (x+y) का मान क्या है?
What is the value of (x+y) from (0.2x+0.5y=3.1) and (0.4x-0.1y=1.3)?
#linear equations
#decimal equations
#expression value
#hard
#class 10
A \(x+y=\frac{97}{11}\)
B \(x+y=\frac{89}{11}\)
C \(x+y=\frac{101}{11}\)
D \(x+y=\frac{105}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(x+y=\frac{97}{11}\)
Step 1
Concept
Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(x+y=\frac{97}{11}\). Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).
Step 3
Exam Tip
दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)।
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यदि (4x+7y=53) और (4x-3y=13), तो (x-y) का मान क्या है?
If (4x+7y=53) and (4x-3y=13), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(x-y=\frac{9}{4}\)
B \(x-y=\frac{7}{4}\)
C \(x-y=\frac{11}{4}\)
D \(x-y=\frac{13}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(x-y=\frac{9}{4}\)
Step 1
Concept
Subtracting the equations gives (10y=40), so (y=4). Then \(x=\frac{25}{4}\), hence \(x-y=\frac{9}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(x-y=\frac{9}{4}\). Subtracting the equations gives (10y=40), so (y=4). Then \(x=\frac{25}{4}\), hence \(x-y=\frac{9}{4}\).
Step 3
Exam Tip
दोनों समीकरण घटाने पर (10y=40), इसलिए (y=4)। फिर \(x=\frac{25}{4}\), अतः \(x-y=\frac{9}{4}\)।
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समीकरणों (4x-3y=7) और (5x+6y=44) से (x+y) का मान क्या है?
What is the value of (x+y) from (4x-3y=7) and (5x+6y=44)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(x+y=\frac{105}{13}\)
B \(x+y=\frac{99}{13}\)
C \(x+y=\frac{111}{13}\)
D \(x+y=\frac{117}{13}\)
Explanation opens after your attempt
Correct Answer
A. \(x+y=\frac{105}{13}\)
Step 1
Concept
Multiply the first equation by (2) and add the second. \(x=\frac{58}{13}\) and \(y=\frac{47}{13}\), so \(x+y=\frac{105}{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(x+y=\frac{105}{13}\). Multiply the first equation by (2) and add the second. \(x=\frac{58}{13}\) and \(y=\frac{47}{13}\), so \(x+y=\frac{105}{13}\).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर दूसरे से जोड़ें। \(x=\frac{58}{13}\) और \(y=\frac{47}{13}\), अतः \(x+y=\frac{105}{13}\)।
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यदि (3x+2y=28) और (5x-4y=8), तो (x-y) का मान क्या है?
If (3x+2y=28) and (5x-4y=8), what is the value of (x-y)?
#linear equations
#elimination
#expression value
#hard
#class 10
A \(\frac{6}{11}\)
B \(\frac{8}{11}\)
C \(\frac{10}{11}\)
D \(\frac{12}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{6}{11}\)
Step 1
Concept
Multiply the first equation by (2) and eliminate (y). \(x=\frac{64}{11}\) and \(y=\frac{58}{11}\), so \(x-y=\frac{6}{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{6}{11}\). Multiply the first equation by (2) and eliminate (y). \(x=\frac{64}{11}\) and \(y=\frac{58}{11}\), so \(x-y=\frac{6}{11}\).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर (y) हटाएं। \(x=\frac{64}{11}\) और \(y=\frac{58}{11}\), इसलिए \(x-y=\frac{6}{11}\)।
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समीकरणों (x+3y=21) और (3x-y=11) को हल करने पर (2x+y) का मान क्या है?
On solving (x+3y=21) and (3x-y=11), what is the value of (2x+y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (14)
B (13)
C (12)
D (11)
Explanation opens after your attempt
Step 1
Concept
Use (y=3x-11) from the second equation. Substitution gives \(x=\frac{27}{5},\ y=\frac{16}{5}\), so (2x+y=14).
Step 2
Why this answer is correct
The correct answer is A. (14). Use (y=3x-11) from the second equation. Substitution gives \(x=\frac{27}{5},\ y=\frac{16}{5}\), so (2x+y=14).
Step 3
Exam Tip
दूसरे समीकरण से (y=3x-11) रखें। पहले में रखने पर \(x=\frac{27}{5},\ y=\frac{16}{5}\), इसलिए (2x+y=14)।
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यदि (3x+y=22) और (x+2y=19), तो (x-y) का मान क्या है?
If (3x+y=22) and (x+2y=19), what is the value of (x-y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (2)
B (0)
C (-1)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Use (y=22-3x) from the first equation. Substitution gives (x=5,\ y=7), so (x-y=-2).
Step 2
Why this answer is correct
The correct answer is D. (-2). Use (y=22-3x) from the first equation. Substitution gives (x=5,\ y=7), so (x-y=-2).
Step 3
Exam Tip
पहले समीकरण से (y=22-3x) रखें। दूसरे में रखने पर (x=5,\ y=7), इसलिए (x-y=-2)।
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यदि (2x+y=23) और (x+3y=19), तो (x-2y) का मान क्या है?
If (2x+y=23) and (x+3y=19), what is the value of (x-2y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (2)
B (3)
C (5)
D (4)
Explanation opens after your attempt
Step 1
Concept
Use (y=23-2x) from the first equation. Substitution gives (x=10,\ y=3), so (x-2y=4).
Step 2
Why this answer is correct
The correct answer is D. (4). Use (y=23-2x) from the first equation. Substitution gives (x=10,\ y=3), so (x-2y=4).
Step 3
Exam Tip
पहले समीकरण से (y=23-2x) रखें। दूसरे में रखने पर (x=10,\ y=3), इसलिए (x-2y=4)।
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यदि (3x+2y=25) और (x-y=1), तो (x+y) का मान क्या है?
If (3x+2y=25) and (x-y=1), what is the value of (x+y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (8)
B (9)
C \(\frac{49}{5}\)
D \(\frac{51}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{49}{5}\)
Step 1
Concept
Using (x=y+1) gives (5y+3=25), so \(y=\frac{22}{5}\) and \(x=\frac{27}{5}\). Hence \(x+y=\frac{49}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{49}{5}\). Using (x=y+1) gives (5y+3=25), so \(y=\frac{22}{5}\) and \(x=\frac{27}{5}\). Hence \(x+y=\frac{49}{5}\).
Step 3
Exam Tip
(x=y+1) रखने पर (5y+3=25), इसलिए \(y=\frac{22}{5}\) और \(x=\frac{27}{5}\)। अतः \(x+y=\frac{49}{5}\)।
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