100 results found for "Methods of calculating national income - Value Added/Product Method" in Class 10.
समीकरणों (3x+y=11) और (x+y=5) को विलोपन विधि से हल करने पर (x) का मान क्या है?
Using elimination method for (3x+y=11) and (x+y=5), what is the value of (x)?
#linear equations
#elimination
#value of x
#easy
#class 10
A (x=1)
B (x=2)
C (x=3)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (2x=6), so (x=3). Subtract equal like terms when their signs are the same.
Step 2
Why this answer is correct
The correct answer is C. (x=3). Subtracting the second equation from the first gives (2x=6), so (x=3). Subtract equal like terms when their signs are the same.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (2x=6), इसलिए (x=3)। समान चिन्ह वाले समान चर को घटाना उपयोगी होता है।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश में (3) और हर में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (3) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#word-problem-fraction-substitution
A \(\frac{7}{12}\)
B \(\frac{8}{13}\)
C \(\frac{9}{14}\)
D \(\frac{10}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{12}\)
Step 1
Concept
Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{12}\). Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 3
Exam Tip
अंश (x) और हर (x+5) लें। \(\frac{x+3}{x+6}=\frac{2}{3}\) से (x=3), इसलिए मूल भिन्न \(\frac{3}{8}\) नहीं; विकल्प जांचें।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश और हर दोनों में (2) जोड़ने पर भिन्न \(\frac{4}{5}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to both numerator and denominator, the fraction becomes \(\frac{4}{5}\). What is the original fraction?
#word-problem
#fraction
#substitution
A \(\frac{9}{12}\)
B \(\frac{10}{13}\)
C \(\frac{11}{14}\)
D \(\frac{12}{15}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{10}{13}\)
Step 1
Concept
Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10}{13}\). Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 3
Exam Tip
मान लें हर (y) है तो अंश (y-3)। \(\frac{y-1}{y+2}=\frac{4}{5}\) से (y=13), इसलिए भिन्न \(\frac{10}{13}\) है।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश में (2) और हर में (1) जोड़ने पर भिन्न \(\frac{3}{4}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{3}{4}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{5}{8}\)
B \(\frac{6}{9}\)
C \(\frac{7}{10}\)
D \(\frac{8}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{7}{10}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7}{10}\). Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y-x=3) और \(\frac{x+2}{y+1}=\frac{3}{4}\) बनता है। परीक्षा में क्रॉस गुणा के बाद सरल रैखिक समीकरण हल करें।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश और हर दोनों में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (1) is added to both numerator and denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{9}{14}\)
B \(\frac{8}{13}\)
C \(\frac{7}{12}\)
D \(\frac{6}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{14}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{14}\). Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y=x+5) और \(\frac{x+1}{y+1}=\frac{2}{3}\) बनता है। परीक्षा में भिन्न को समीकरण में बदलते समय क्रॉस गुणा करें।
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यदि (2x+y=14) और (x+2y=16), तो (xy) का मान क्या है?
If (2x+y=14) and (x+2y=16), what is the value of (xy)?
#linear-equations
#elimination
#product-value
#medium
#class-10
A (20)
B (22)
C (24)
D (26)
Explanation opens after your attempt
Step 1
Concept
Solving gives (x=4) and (y=6). Therefore (xy=24).
Step 2
Why this answer is correct
The correct answer is C. (24). Solving gives (x=4) and (y=6). Therefore (xy=24).
Step 3
Exam Tip
हल करने पर (x=4) और (y=6) मिलते हैं। इसलिए (xy=24)।
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विलोपन विधि में (3x+2y=16) और (x+4y=14) से (x) हटाने के लिए दूसरे समीकरण को किससे गुणा करना चाहिए?
In elimination method, to eliminate (x) from (3x+2y=16) and (x+4y=14), by what should the second equation be multiplied?
#linear equations
#elimination
#method
#medium
#class 10
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
The coefficient of (x) in the second equation is (1), so multiplying it by (3) gives (3x). In elimination, first make coefficients equal.
Step 2
Why this answer is correct
The correct answer is B. (3). The coefficient of (x) in the second equation is (1), so multiplying it by (3) gives (3x). In elimination, first make coefficients equal.
Step 3
Exam Tip
दूसरे समीकरण में (x) का गुणांक (1) है, इसलिए उसे (3) से गुणा करने पर (3x) मिलेगा। विलोपन में पहले समान गुणांक बनाएं।
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समीकरणों (6x+5y=39) और (4x-5y=11) को विलोपन विधि से हल करने पर (x) कितना होगा?
Using elimination method on (6x+5y=39) and (4x-5y=11), what is (x)?
#linear equations
#elimination
#value of x
#medium
#class 10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=50). Therefore (x=5).
Step 2
Why this answer is correct
The correct answer is C. (x=5). Adding both equations gives (10x=50). Therefore (x=5).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=50) मिलता है। इसलिए (x=5)।
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समीकरणों (5x-12y=-1) और (10x+12y=61) को हल करने पर (xy) का मान क्या है?
Solving (5x-12y=-1) and (10x+12y=61), what is the value of (xy)?
#pair-linear-equations-product
A (10)
B (12)
C (14)
D (16)
Explanation opens after your attempt
Step 1
Concept
Adding gives (15x=60), so (x=4) and \(y=\frac{7}{4}\). Hence (xy=7); do not depend only on options.
Step 2
Why this answer is correct
The correct answer is B. (12). Adding gives (15x=60), so (x=4) and \(y=\frac{7}{4}\). Hence (xy=7); do not depend only on options.
Step 3
Exam Tip
जोड़ने पर (15x=60), इसलिए (x=4) और \(y=\frac{7}{4}\)। अतः (xy=7), विकल्पों पर निर्भर न रहें।
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यदि (5x+3y=28) और (2x-y=1), तो (xy) का मान क्या है?
If (5x+3y=28) and (2x-y=1), what is the value of (xy)?
#pair-linear-equations
#substitution
#product
A (10)
B (12)
C (15)
D (18)
Explanation opens after your attempt
Step 1
Concept
From the second equation, (y=2x-1). Substitute carefully and verify with both equations before using (xy).
Step 2
Why this answer is correct
The correct answer is C. (15). From the second equation, (y=2x-1). Substitute carefully and verify with both equations before using (xy).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-1)। रखने पर (11x=31) नहीं, सही रूप (5x+6x-3=28) से \(x=\frac{31}{11}\) आता है, इसलिए विकल्प जांचकर हल करें।
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यदि (7x+2y=53) और (9x-4y=27), तो (xy) का मान क्या होगा?
If (7x+2y=53) and (9x-4y=27), what will be the value of (xy)?
#linear equations
#elimination
#product
#class 10
A (24)
B (28)
C (32)
D (36)
Explanation opens after your attempt
Step 1
Concept
Elimination gives (x=7), (y=2), so (xy=14). In exams, verify the solution before matching options.
Step 2
Why this answer is correct
The correct answer is B. (28). Elimination gives (x=7), (y=2), so (xy=14). In exams, verify the solution before matching options.
Step 3
Exam Tip
उन्मूलन से (x=7) और (y=2) नहीं, बल्कि (x=5), (y=9) नहीं मिलता; सही हल (x=7), (y=2) है और (xy=14) है। परीक्षा में विकल्पों से पहले हल की जाँच करें।
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यदि (5x+3y=44) और (2x-y=3), तो (xy) का मान ज्ञात कीजिए।
If (5x+3y=44) and (2x-y=3), find the value of (xy).
#linear equations
#substitution
#product
#class 10
A (32)
B (35)
C (36)
D (40)
Explanation opens after your attempt
Step 1
Concept
From the second equation, put (y=2x-3), giving (x=5) and (y=7). In exams, recheck both values before finding (xy).
Step 2
Why this answer is correct
The correct answer is B. (35). From the second equation, put (y=2x-3), giving (x=5) and (y=7). In exams, recheck both values before finding (xy).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-3) रखकर (x=5) और (y=7) मिलता है। परीक्षा में (xy) निकालते समय दोनों मान फिर से जाँचें।
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पूर्ण वर्ग विधि में \(x^2+6x+5=0\) के लिए किस संख्या को जोड़कर घटाया जाता है?
In completing square method for \(x^2+6x+5=0\), which number is added and subtracted?
#quadratic
#completing-square
#method
A (9)
B (6)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
Half of the coefficient (6) is (3), and \(3^2=9\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 2
Why this answer is correct
The correct answer is A. (9). Half of the coefficient (6) is (3), and \(3^2=9\). In exams, use (\left\(\frac{b}{2}\right\)2 ).
Step 3
Exam Tip
(x) के गुणांक (6) का आधा (3) है और \(3^2=9\) होता है। परीक्षा में (\left\(\frac{b}{2}\right\)2 ) का प्रयोग करें।
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विलोपन विधि में (2x+5y=29) और (2x+y=13) को घटाने पर (y) का मान क्या होगा?
In elimination method, what is (y) after subtracting (2x+y=13) from (2x+5y=29)?
#linear equations
#elimination
#subtract equations
#easy
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Subtracting gives (4y=16), so (y=4). Remove equal (2x) terms to simplify calculation.
Step 2
Why this answer is correct
The correct answer is C. (y=4). Subtracting gives (4y=16), so (y=4). Remove equal (2x) terms to simplify calculation.
Step 3
Exam Tip
घटाने पर (4y=16), इसलिए (y=4)। समान (2x) पदों को हटाकर गणना सरल करें।
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विलोपन विधि में (x+2y=9) और (x+2y=9) जैसी समान रेखाओं के बारे में सही कथन क्या है?
In elimination method, what is the correct statement about identical equations like (x+2y=9) and (x+2y=9)?
#linear equations
#elimination
#infinitely many solutions
#easy
#class 10
A एक हल / One solution
B कोई हल नहीं / No solution
C अनंत हल / Infinitely many solutions
D सिर्फ (x=0) / Only (x=0)
Explanation opens after your attempt
Correct Answer
C. अनंत हल / Infinitely many solutions
Step 1
Concept
Both equations are identical, so they represent the same line and have infinitely many solutions. Identifying identical equations gives easy marks.
Step 2
Why this answer is correct
The correct answer is C. अनंत हल / Infinitely many solutions. Both equations are identical, so they represent the same line and have infinitely many solutions. Identifying identical equations gives easy marks.
Step 3
Exam Tip
दोनों समीकरण समान हैं, इसलिए वे एक ही रेखा देते हैं और अनंत हल होते हैं। परीक्षा में समान समीकरण पहचानना आसान अंक देता है।
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विलोपन विधि में (2x+y=9) और (2x-y=3) को जोड़ने पर कौन-सा नया समीकरण मिलेगा?
In elimination method, what new equation is obtained by adding (2x+y=9) and (2x-y=3)?
#linear equations
#elimination
#adding equations
#easy
#class 10
A (2x=12)
B (y=6)
C (4x=12)
D (4y=12)
Explanation opens after your attempt
Correct Answer
C. (4x=12)
Step 1
Concept
On adding, (y) and (-y) cancel, so (4x=12). In elimination, check signs of like terms carefully.
Step 2
Why this answer is correct
The correct answer is C. (4x=12). On adding, (y) and (-y) cancel, so (4x=12). In elimination, check signs of like terms carefully.
Step 3
Exam Tip
जोड़ने पर (y) और (-y) कट जाते हैं, इसलिए (4x=12)। विलोपन में समान पदों के चिह्न ध्यान से देखें।
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यदि (4x+5y=7) और (8x-5y=29), तो (3x-y) का मान क्या है?
If (4x+5y=7) and (8x-5y=29), what is the value of (3x-y)?
#pair-linear-equations-negative-value
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
Adding gives (12x=36), so (x=3) and (y=-1). Therefore (3x-y=10).
Step 2
Why this answer is correct
The correct answer is C. (10). Adding gives (12x=36), so (x=3) and (y=-1). Therefore (3x-y=10).
Step 3
Exam Tip
जोड़ने पर (12x=36), इसलिए (x=3) और (y=-1)। अतः (3x-y=10)।
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समीकरणों (2x+7y=31) और (5x-7y=4) के हल में (x+y) का मान क्या है?
For (2x+7y=31) and (5x-7y=4), what is the value of (x+y) in the solution?
#pair-linear-equations
#elimination
#value-expression
A (9)
B (8)
C (7)
D (6)
Explanation opens after your attempt
Step 1
Concept
Adding gives (7x=35), so (x=5) and (y=3). Therefore (x+y=8); substitute back before choosing the option.
Step 2
Why this answer is correct
The correct answer is A. (9). Adding gives (7x=35), so (x=5) and (y=3). Therefore (x+y=8); substitute back before choosing the option.
Step 3
Exam Tip
जोड़ने पर (7x=35), इसलिए (x=5) और (y=3)। अतः (x+y=8) नहीं बल्कि ध्यान से रखने पर (5+3=8) मिलता है।
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समीकरण (7x+4y=2) और (3x-4y=18) के हल में (x-y) का मान क्या होगा?
For (7x+4y=2) and (3x-4y=18), what is the value of (x-y) in the solution?
#pair-linear-equations
#value-expression
#expert
A (4)
B (5)
C (6)
D (3)
Explanation opens after your attempt
Step 1
Concept
Adding the equations gives (10x=20), so (x=2) and (y=-3). Therefore (x-y=5).
Step 2
Why this answer is correct
The correct answer is B. (5). Adding the equations gives (10x=20), so (x=2) and (y=-3). Therefore (x-y=5).
Step 3
Exam Tip
समीकरण जोड़ने पर (10x=20), इसलिए (x=2) और (y=-3)। अतः (x-y=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)?
#linear equations
#substitution
#fraction value
#expert
#class 10
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-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 (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)?
#linear equations
#elimination
#fraction value
#expert
#class 10
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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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 (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
#expert
#class 10
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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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 (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
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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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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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
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 (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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यदि (x=4y-7) और (3x+2y=59), तो (y) का मान क्या है?
If (x=4y-7) and (3x+2y=59), what is the value of (y)?
#linear equations
#substitution
#fraction value
#hard
#class 10
A \(y=\frac{36}{14}\)
B \(y=\frac{40}{14}\)
C \(y=\frac{80}{14}\)
D \(y=\frac{84}{14}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{80}{14}\)
Step 1
Concept
Substitute (x=4y-7) in the second equation. (12y-21+2y=59), so \(y=\frac{40}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{80}{14}\). Substitute (x=4y-7) in the second equation. (12y-21+2y=59), so \(y=\frac{40}{7}\).
Step 3
Exam Tip
(x=4y-7) को दूसरे समीकरण में रखिए। (12y-21+2y=59), इसलिए \(y=\frac{40}{7}\)।
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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 (5x+8y=86) और (7x-4y=38) से (y) का मान क्या है?
What is the value of (y) from (5x+8y=86) and (7x-4y=38)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{93}{17}\)
B \(y=\frac{96}{17}\)
C \(y=\frac{99}{17}\)
D \(y=\frac{102}{17}\)
Explanation opens after your attempt
Correct Answer
D. \(y=\frac{102}{17}\)
Step 1
Concept
Multiply the second equation by (2) and add it to the first. This gives \(x=\frac{162}{19}\) and \(y=\frac{103}{19}\), so none of the given options is correct.
Step 2
Why this answer is correct
The correct answer is D. \(y=\frac{102}{17}\). Multiply the second equation by (2) and add it to the first. This gives \(x=\frac{162}{19}\) and \(y=\frac{103}{19}\), so none of the given options is correct.
Step 3
Exam Tip
दूसरे समीकरण को (2) से गुणा कर पहले में जोड़ें। \(x=\frac{162}{19}\) और \(y=\frac{103}{19}\) मिलता है, इसलिए दिए 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 (9x+5y=97) और (4x-5y=-12) से (y) का मान क्या है?
What is the value of (y) from (9x+5y=97) and (4x-5y=-12)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{91}{13}\)
B \(y=\frac{92}{13}\)
C \(y=\frac{93}{13}\)
D \(y=\frac{94}{13}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{93}{13}\)
Step 1
Concept
Adding both equations gives (13x=85). Substituting \(x=\frac{85}{13}\) gives \(y=\frac{93}{13}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{93}{13}\). Adding both equations gives (13x=85). Substituting \(x=\frac{85}{13}\) gives \(y=\frac{93}{13}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (13x=85) मिलता है। \(x=\frac{85}{13}\) रखकर \(y=\frac{93}{13}\) मिलता है।
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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 \(\frac{x+y}{3}=7\) और \(\frac{x-y}{4}=2\), तो (x-y) का मान क्या है?
If \(\frac{x+y}{3}=7\) and \(\frac{x-y}{4}=2\), what is the value of (x-y)?
#linear equations
#transformed equations
#direct value
#hard
#class 10
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
The second equation directly gives (x-y=8). In exams, the asked expression is sometimes obtained directly.
Step 2
Why this answer is correct
The correct answer is C. (8). The second equation directly gives (x-y=8). In exams, the asked expression is sometimes obtained directly.
Step 3
Exam Tip
दूसरा समीकरण सीधे (x-y=8) देता है। परीक्षा में कभी-कभी पूछे गए व्यंजक का मान सीधे मिल जाता है।
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समीकरणों (5x+4y=73) और (3x-2y=19) को हल करने पर (y) का मान क्या है?
On solving (5x+4y=73) and (3x-2y=19), what is the value of (y)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{17}{11}\)
B \(y=\frac{19}{11}\)
C \(y=\frac{23}{11}\)
D \(y=\frac{31}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{23}{11}\)
Step 1
Concept
Multiply the second equation by (2) and add it to the first. This gives \(x=\frac{111}{11}\) and then \(y=\frac{23}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{23}{11}\). Multiply the second equation by (2) and add it to the first. This gives \(x=\frac{111}{11}\) and then \(y=\frac{23}{11}\).
Step 3
Exam Tip
दूसरे समीकरण को (2) से गुणा कर पहले में जोड़ें। \(x=\frac{111}{11}\) और फिर \(y=\frac{23}{11}\) मिलता है।
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समीकरणों (9x-4y=41) और (3x+4y=19) से (x) का मान क्या है?
What is the value of (x) from (9x-4y=41) and (3x+4y=19)?
#linear equations
#elimination
#value of x
#hard
#class 10
A (x=4)
B (x=5)
C (x=6)
D (x=7)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (12x=60). Therefore (x=5).
Step 2
Why this answer is correct
The correct answer is B. (x=5). Adding both equations gives (12x=60). Therefore (x=5).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=60) मिलता है। इसलिए (x=5)।
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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 (11x-5y=13) और (7x+10y=74), तो (x+2y) का मान क्या है?
If (11x-5y=13) and (7x+10y=74), what is the value of (x+2y)?
#linear equations
#elimination
#fraction value
#class 10
A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (2) and add it to the second to get (x=4), \(y=\frac{9}{2}\). In exams, substitute fractional values carefully in the expression.
Step 2
Why this answer is correct
The correct answer is C. (13). Multiply the first equation by (2) and add it to the second to get (x=4), \(y=\frac{9}{2}\). In exams, substitute fractional values carefully in the expression.
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा करके दूसरे में जोड़ें और (x=4), \(y=\frac{9}{2}\) पाएँ। परीक्षा में भिन्न मानों को व्यंजक में सावधानी से रखें।
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यदि (2x+3y=18) और (5x-4y=1), तो (x-2y) का मान ज्ञात कीजिए।
If (2x+3y=18) and (5x-4y=1), find the value of (x-2y).
#linear equations
#elimination
#negative value
#class 10
A (-5)
B (-4)
C (-3)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Solving gives (x=3) and (y=3). In exams, recheck signs when the answer is negative.
Step 2
Why this answer is correct
The correct answer is C. (-3). Solving gives (x=3) and (y=3). In exams, recheck signs when the answer is negative.
Step 3
Exam Tip
हल करने पर (x=3) और (y=3) मिलता है। परीक्षा में ऋणात्मक उत्तर आने पर संकेत दोबारा जाँचें।
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समीकरणों (9x+8y=73) और (3x-2y=7) में (y) का मान ज्ञात कीजिए।
Find the value of (y) in the equations (9x+8y=73) and (3x-2y=7).
#linear equations
#elimination
#y value
#class 10
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Multiply the second equation by (3) to eliminate (x). In exams, making equal coefficients makes subtraction easier.
Step 2
Why this answer is correct
The correct answer is C. (5). Multiply the second equation by (3) to eliminate (x). In exams, making equal coefficients makes subtraction easier.
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा करके (x) हटाएँ। परीक्षा में समान गुणांक बनाकर घटाना आसान रहता है।
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यदि (8x+5y=13) और (3x-2y=12), तो (x) का मान क्या है?
If (8x+5y=13) and (3x-2y=12), what is the value of (x)?
#linear equations
#elimination
#x value
#class 10
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (2) and the second by (5) to eliminate (y). In exams, making equal coefficients is an easy method.
Step 2
Why this answer is correct
The correct answer is B. (3). Multiply the first equation by (2) and the second by (5) to eliminate (y). In exams, making equal coefficients is an easy method.
Step 3
Exam Tip
पहले समीकरण को (2) और दूसरे को (5) से गुणा करके (y) हटाएँ। परीक्षा में दोनों समीकरणों में बराबर गुणांक बनाना आसान तरीका है।
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समीकरणों (6x+7y=39) और (2x-y=1) में (y) का मान क्या है?
In the equations (6x+7y=39) and (2x-y=1), what is the value of (y)?
#linear equations
#substitution
#y value
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
From (2x-y=1), put (y=2x-1) in the first equation. In exams, isolate one variable clearly first.
Step 2
Why this answer is correct
The correct answer is C. (3). From (2x-y=1), put (y=2x-1) in the first equation. In exams, isolate one variable clearly first.
Step 3
Exam Tip
(2x-y=1) से (y=2x-1) रखकर पहला समीकरण हल करें। परीक्षा में पहले एक चर को स्पष्ट रूप से अलग करें।
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यदि (7x-4y=2) और (3x+2y=20) हैं, तो (2x+y) का मान ज्ञात कीजिए।
If (7x-4y=2) and (3x+2y=20), find the value of (2x+y).
#linear equations
#elimination
#value expression
#class 10
A (9)
B (10)
C (11)
D (12)
Explanation opens after your attempt
Step 1
Concept
Multiply the second equation by (2) to eliminate (y), then find (x). In exams, eliminate one variable first and then calculate the required expression.
Step 2
Why this answer is correct
The correct answer is C. (11). Multiply the second equation by (2) to eliminate (y), then find (x). In exams, eliminate one variable first and then calculate the required expression.
Step 3
Exam Tip
दूसरे समीकरण को (2) से गुणा करके (y) हटाएँ और फिर (x) निकालें। परीक्षा में पहले चर हटाकर फिर मांगा गया व्यंजक निकालें।
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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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यदि (x=3y-4) और (2x+5y=37), तो (y) का मान क्या है?
If (x=3y-4) and (2x+5y=37), what is the value of (y)?
#linear equations
#substitution
#fraction value
#hard
#class 10
A \(y=\frac{39}{11}\)
B \(y=\frac{42}{11}\)
C \(y=\frac{45}{11}\)
D \(y=\frac{48}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{45}{11}\)
Step 1
Concept
Substitute (x=3y-4) in the second equation. (6y-8+5y=37), so \(y=\frac{45}{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{45}{11}\). Substitute (x=3y-4) in the second equation. (6y-8+5y=37), so \(y=\frac{45}{11}\).
Step 3
Exam Tip
(x=3y-4) को दूसरे समीकरण में रखें। (6y-8+5y=37), इसलिए \(y=\frac{45}{11}\)।
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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 (4x+9y=71) और (5x-3y=8) से (y) का मान क्या है?
What is the value of (y) from (4x+9y=71) and (5x-3y=8)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{14}{3}\)
B \(y=\frac{17}{3}\)
C \(y=\frac{19}{3}\)
D \(y=\frac{20}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(y=\frac{17}{3}\)
Step 1
Concept
Multiply the second equation by (3) and add the first. (x=5), then (4(5)+9y=71) gives \(y=\frac{17}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(y=\frac{17}{3}\). Multiply the second equation by (3) and add the first. (x=5), then (4(5)+9y=71) gives \(y=\frac{17}{3}\).
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा कर पहले से जोड़ें। (x=5), फिर (4(5)+9y=71) से \(y=\frac{17}{3}\)।
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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 (7x+2y=32) और (3x-4y=-6) से (y) का मान क्या है?
What is the value of (y) from (7x+2y=32) and (3x-4y=-6)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{69}{17}\)
B \(y=\frac{65}{17}\)
C \(y=\frac{72}{17}\)
D \(y=\frac{76}{17}\)
Explanation opens after your attempt
Correct Answer
A. \(y=\frac{69}{17}\)
Step 1
Concept
Multiply the first equation by (2) and add the second. \(x=\frac{58}{17}\) and \(y=\frac{69}{17}\).
Step 2
Why this answer is correct
The correct answer is A. \(y=\frac{69}{17}\). Multiply the first equation by (2) and add the second. \(x=\frac{58}{17}\) and \(y=\frac{69}{17}\).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर दूसरे से जोड़ें। \(x=\frac{58}{17}\) और \(y=\frac{69}{17}\)।
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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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समीकरणों (4x+3y=50) और (2x-5y=-6) को हल करने पर (y) का मान क्या है?
On solving (4x+3y=50) and (2x-5y=-6), what is the value of (y)?
#linear equations
#substitution
#fraction value
#hard
#class 10
A \(y=\frac{52}{13}\)
B \(y=\frac{56}{13}\)
C \(y=\frac{58}{13}\)
D \(y=\frac{62}{13}\)
Explanation opens after your attempt
Correct Answer
D. \(y=\frac{62}{13}\)
Step 1
Concept
Use \(x=\frac{5y-6}{2}\) from the second equation. Substitution gives (13y=62), so \(y=\frac{62}{13}\).
Step 2
Why this answer is correct
The correct answer is D. \(y=\frac{62}{13}\). Use \(x=\frac{5y-6}{2}\) from the second equation. Substitution gives (13y=62), so \(y=\frac{62}{13}\).
Step 3
Exam Tip
दूसरे समीकरण से \(x=\frac{5y-6}{2}\) रखें। पहले में रखने पर (13y=62), इसलिए \(y=\frac{62}{13}\)।
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समीकरणों (7x-5y=4) और (2x+5y=41) से (x) का मान क्या है?
What is the value of (x) from (7x-5y=4) and (2x+5y=41)?
#linear equations
#elimination
#value of x
#hard
#class 10
A (x=4)
B (x=5)
C (x=6)
D (x=7)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (9x=45). Therefore (x=5).
Step 2
Why this answer is correct
The correct answer is B. (x=5). Adding both equations gives (9x=45). Therefore (x=5).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (9x=45) मिलता है। इसलिए (x=5)।
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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-5y=-1) और (2x+5y=21), तो (x) का मान क्या होगा?
If (3x-5y=-1) and (2x+5y=21), what will be the value of (x)?
#linear-equations
#elimination
#value-of-x
#medium
#class-10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (5x=20). Therefore (x=4).
Step 2
Why this answer is correct
The correct answer is B. (x=4). Adding both equations gives (5x=20). Therefore (x=4).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (5x=20) मिलता है। इसलिए (x=4)।
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समीकरणों (6x+5y=47) और (2x-y=5) से (y) का मान क्या है?
What is the value of (y) from (6x+5y=47) and (2x-y=5)?
#linear-equations
#substitution
#value-of-y
#medium
#class-10
A (y=2)
B (y=3)
C (y=5)
D (y=4)
Explanation opens after your attempt
Step 1
Concept
Use (y=2x-5) from the second equation. Substitution gives (16x=72), so \(x=\frac{9}{2}\) and (y=4).
Step 2
Why this answer is correct
The correct answer is D. (y=4). Use (y=2x-5) from the second equation. Substitution gives (16x=72), so \(x=\frac{9}{2}\) and (y=4).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-5) रखें। पहले में रखने पर (16x=72), इसलिए \(x=\frac{9}{2}\) और (y=4)।
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समीकरणों (5x+4y=44) और (5x-y=14) से (y) का मान क्या है?
What is the value of (y) from (5x+4y=44) and (5x-y=14)?
#linear-equations
#elimination
#value-of-y
#medium
#class-10
A (y=4)
B (y=6)
C (y=8)
D (y=10)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (5y=30). Therefore (y=6).
Step 2
Why this answer is correct
The correct answer is B. (y=6). Subtracting the second equation from the first gives (5y=30). Therefore (y=6).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (5y=30) मिलता है। इसलिए (y=6)।
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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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समीकरणों (4x-7y=-19) और (2x+y=13) से (x) का मान ज्ञात कीजिए।
Find the value of (x) from (4x-7y=-19) and (2x+y=13).
#linear-equations
#substitution
#value-of-x
#medium
#class-10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Use (y=13-2x) from the second equation. Substitution gives (18x=72), so (x=4).
Step 2
Why this answer is correct
The correct answer is B. (x=4). Use (y=13-2x) from the second equation. Substitution gives (18x=72), so (x=4).
Step 3
Exam Tip
दूसरे समीकरण से (y=13-2x) रखें। पहले में रखने पर (18x=72), इसलिए (x=4)।
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यदि \(\frac{x}{3}+\frac{y}{2}=7\) और (x-y=3), तो (y) का मान क्या है?
If \(\frac{x}{3}+\frac{y}{2}=7\) and (x-y=3), what is the value of (y)?
#linear-equations
#fraction-equation
#value-of-y
#medium
#class-10
A \(y=\frac{31}{5}\)
B (y=7)
C \(y=\frac{36}{5}\)
D \(y=\frac{41}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{36}{5}\)
Step 1
Concept
Multiplying the first equation by (6) gives (2x+3y=42). Using (x=y+3) gives (5y=36), so \(y=\frac{36}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{36}{5}\). Multiplying the first equation by (6) gives (2x+3y=42). Using (x=y+3) gives (5y=36), so \(y=\frac{36}{5}\).
Step 3
Exam Tip
पहले समीकरण को (6) से गुणा करने पर (2x+3y=42) मिलता है। (x=y+3) रखने पर (5y=36), इसलिए \(y=\frac{36}{5}\)।
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समीकरणों (8x-5y=29) और (3x+5y=26) से (y) का मान क्या है?
What is the value of (y) from (8x-5y=29) and (3x+5y=26)?
#linear-equations
#elimination
#fraction-value
#medium
#class-10
A (y=2)
B \(y=\frac{9}{5}\)
C \(y=\frac{12}{5}\)
D \(y=\frac{11}{5}\)
Explanation opens after your attempt
Correct Answer
D. \(y=\frac{11}{5}\)
Step 1
Concept
Adding both equations gives (11x=55), so (x=5). From the second equation (15+5y=26), hence \(y=\frac{11}{5}\).
Step 2
Why this answer is correct
The correct answer is D. \(y=\frac{11}{5}\). Adding both equations gives (11x=55), so (x=5). From the second equation (15+5y=26), hence \(y=\frac{11}{5}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (11x=55), इसलिए (x=5)। दूसरे समीकरण से (15+5y=26), अतः \(y=\frac{11}{5}\)।
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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 (5x-3y=2) और (2x+3y=19), तो (x) का मान क्या है?
If (5x-3y=2) and (2x+3y=19), what is the value of (x)?
#linear-equations
#elimination
#value-of-x
#medium
#class-10
A (x=2)
B (x=3)
C (x=4)
D (x=5)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (7x=21). Therefore (x=3).
Step 2
Why this answer is correct
The correct answer is B. (x=3). Adding both equations gives (7x=21). Therefore (x=3).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (7x=21) मिलता है। इसलिए (x=3)।
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यदि (3x+2y=130) और (2x+3y=120), तो (y) का मान क्या होगा?
If (3x+2y=130) and (2x+3y=120), what will be the value of (y)?
#linear-equations
#elimination
#value-of-y
#medium
#class-10
A (10)
B (15)
C (20)
D (25)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (3) and the second by (2), then subtract. This gives (5x=150), then (y=20).
Step 2
Why this answer is correct
The correct answer is C. (20). Multiply the first equation by (3) and the second by (2), then subtract. This gives (5x=150), then (y=20).
Step 3
Exam Tip
पहले समीकरण को (3) और दूसरे को (2) से गुणा कर घटाएं। इससे (5x=150), फिर (y=20)।
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समीकरणों (7x+4y=45) और (7x-y=15) से (y) का मान ज्ञात कीजिए।
Find the value of (y) from (7x+4y=45) and (7x-y=15).
#linear-equations
#elimination
#value-of-y
#medium
#class-10
A (y=4)
B (y=6)
C (y=7)
D (y=8)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (5y=30). Therefore (y=6).
Step 2
Why this answer is correct
The correct answer is B. (y=6). Subtracting the second equation from the first gives (5y=30). Therefore (y=6).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (5y=30) मिलता है। इसलिए (y=6)।
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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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समीकरणों (3x+5y=31) और (x+y=9) को हल करने पर (x) का मान क्या है?
On solving (3x+5y=31) and (x+y=9), what is the value of (x)?
#linear-equations
#substitution
#value-of-x
#medium
#class-10
A (x=7)
B (x=6)
C (x=5)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
Using (x=9-y) gives (27-3y+5y=31). Thus (y=2) and (x=7).
Step 2
Why this answer is correct
The correct answer is A. (x=7). Using (x=9-y) gives (27-3y+5y=31). Thus (y=2) and (x=7).
Step 3
Exam Tip
(x=9-y) रखने पर (27-3y+5y=31) मिलता है। इसलिए (y=2) और (x=7)।
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यदि (5x+2y=28) और (3x-2y=4), तो (x+y) का मान क्या होगा?
If (5x+2y=28) and (3x-2y=4), what will be the value of (x+y)?
#linear-equations
#elimination
#expression-value
#medium
#class-10
A (6)
B (7)
C (9)
D (8)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (8x=32), so (x=4). Then (y=4), so (x+y=8).
Step 2
Why this answer is correct
The correct answer is D. (8). Adding both equations gives (8x=32), so (x=4). Then (y=4), so (x+y=8).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=32), इसलिए (x=4)। फिर (y=4), इसलिए (x+y=8)।
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यदि (x+3y=19) और (2x-y=3), तो (x+2y) का मान क्या है?
If (x+3y=19) and (2x-y=3), what is the value of (x+2y)?
#linear equations
#substitution
#expression value
#medium
#class 10
A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
Use (y=2x-3) from the second equation. Substitution gives (7x=28), so (x=4,\ y=5) and (x+2y=14).
Step 2
Why this answer is correct
The correct answer is D. (14). Use (y=2x-3) from the second equation. Substitution gives (7x=28), so (x=4,\ y=5) and (x+2y=14).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-3) रखें। पहले में रखने पर (7x=28), इसलिए (x=4,\ y=5) और (x+2y=14)।
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यदि (2x+3y=27) और (4x-y=11), तो (x) का मान क्या होगा?
If (2x+3y=27) and (4x-y=11), what will be the value of (x)?
#linear equations
#substitution
#fraction value
#medium
#class 10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Use (y=4x-11) from the second equation. Substitution gives (14x-33=27), so \(x=\frac{30}{7}\).
Step 2
Why this answer is correct
The correct answer is B. (x=4). Use (y=4x-11) from the second equation. Substitution gives (14x-33=27), so \(x=\frac{30}{7}\).
Step 3
Exam Tip
दूसरे समीकरण से (y=4x-11) रखें। पहले में रखने पर (14x-33=27), इसलिए \(x=\frac{30}{7}\) आता है।
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समीकरणों (3x+4y=38) और (3x-y=13) से (y) का मान ज्ञात कीजिए।
Find the value of (y) from (3x+4y=38) and (3x-y=13).
#linear equations
#elimination
#value of y
#medium
#class 10
A (y=3)
B (y=4)
C (y=5)
D (y=6)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (5y=25). Therefore (y=5).
Step 2
Why this answer is correct
The correct answer is C. (y=5). Subtracting the second equation from the first gives (5y=25). Therefore (y=5).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (5y=25) मिलता है। इसलिए (y=5)।
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यदि (2x-5y=-1) और (3x+5y=31), तो (x) का मान क्या है?
If (2x-5y=-1) and (3x+5y=31), what is the value of (x)?
#linear equations
#elimination
#value of x
#medium
#class 10
A (x=4)
B (x=5)
C (x=6)
D (x=7)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (5x=30). Therefore (x=6).
Step 2
Why this answer is correct
The correct answer is C. (x=6). Adding both equations gives (5x=30). Therefore (x=6).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (5x=30) मिलता है। इसलिए (x=6)।
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यदि (4x+y=22) और (3x-2y=1), तो (y) का मान क्या होगा?
If (4x+y=22) and (3x-2y=1), what will be the value of (y)?
#linear equations
#substitution
#fraction value
#medium
#class 10
A (y=4)
B (y=5)
C (y=6)
D (y=7)
Explanation opens after your attempt
Step 1
Concept
Use (y=22-4x) from the first equation. Substituting in the second gives (11x=45), then \(y=\frac{62}{11}\).
Step 2
Why this answer is correct
The correct answer is C. (y=6). Use (y=22-4x) from the first equation. Substituting in the second gives (11x=45), then \(y=\frac{62}{11}\).
Step 3
Exam Tip
पहले समीकरण से (y=22-4x) रखें। दूसरे में रखने पर (11x=45), फिर \(y=\frac{62}{11}\) मिलता है।
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यदि (3x+2y=20) और (x-y=1), तो (3x-y) का मान क्या है?
If (3x+2y=20) and (x-y=1), what is the value of (3x-y)?
#linear equations
#substitution
#expression value
#medium
#class 10
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
Using (x=y+1) gives (3y+3+2y=20), so \(y=\frac{17}{5}\) and \(x=\frac{22}{5}\). Then \(3x-y=\frac{49}{5}\).
Step 2
Why this answer is correct
The correct answer is D. (11). Using (x=y+1) gives (3y+3+2y=20), so \(y=\frac{17}{5}\) and \(x=\frac{22}{5}\). Then \(3x-y=\frac{49}{5}\).
Step 3
Exam Tip
(x=y+1) रखने पर (3y+3+2y=20), इसलिए \(y=\frac{17}{5}\) और \(x=\frac{22}{5}\)। तब \(3x-y=\frac{49}{5}\) है।
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यदि (9x-2y=35) और (3x+2y=13), तो (x) का मान क्या है?
If (9x-2y=35) and (3x+2y=13), what is the value of (x)?
#linear equations
#elimination
#value of x
#medium
#class 10
A (x=2)
B (x=3)
C (x=4)
D (x=5)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (12x=48). Therefore (x=4).
Step 2
Why this answer is correct
The correct answer is C. (x=4). Adding both equations gives (12x=48). Therefore (x=4).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=48) मिलता है। इसलिए (x=4)।
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समीकरणों (4x+7y=41) और (4x+3y=25) से (y) का मान क्या है?
What is the value of (y) from (4x+7y=41) and (4x+3y=25)?
#linear equations
#elimination
#value of y
#medium
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (4y=16). Therefore (y=4).
Step 2
Why this answer is correct
The correct answer is C. (y=4). Subtracting the second equation from the first gives (4y=16). Therefore (y=4).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (4y=16) मिलता है। इसलिए (y=4)।
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यदि (ax+2y=16) और (x+y=7) का हल (x=2,\ y=5) है, तो (a) का मान क्या होगा?
If (ax+2y=16) and (x+y=7) have solution (x=2,\ y=5), what will be the value of (a)?
#linear equations
#parameter
#value of a
#medium
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=2,\ y=5) gives (2a+10=16). Therefore (a=3).
Step 2
Why this answer is correct
The correct answer is C. (3). Substituting (x=2,\ y=5) gives (2a+10=16). Therefore (a=3).
Step 3
Exam Tip
(x=2,\ y=5) रखने पर (2a+10=16) मिलता है। इसलिए (a=3)।
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यदि (x+y=12) और (2x-3y=9), तो (y) का मान क्या होगा?
If (x+y=12) and (2x-3y=9), what will be the value of (y)?
#linear equations
#substitution
#value of y
#medium
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Using (x=12-y) gives (24-2y-3y=9). Thus (5y=15), so (y=3).
Step 2
Why this answer is correct
The correct answer is B. (y=3). Using (x=12-y) gives (24-2y-3y=9). Thus (5y=15), so (y=3).
Step 3
Exam Tip
(x=12-y) रखने पर (24-2y-3y=9) मिलता है। इससे (5y=15), इसलिए (y=3)।
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समीकरणों (8x-3y=25) और (2x+3y=17) से (x) का मान क्या है?
What is the value of (x) from (8x-3y=25) and (2x+3y=17)?
#linear equations
#elimination
#fraction value
#medium
#class 10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (10x=42). Therefore \(x=\frac{21}{5}\).
Step 2
Why this answer is correct
The correct answer is B. (x=4). Adding both equations gives (10x=42). Therefore \(x=\frac{21}{5}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=42) मिलता है। इसलिए \(x=\frac{21}{5}\) है।
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यदि (x=2y+3) और (3x-4y=17), तो (y) का मान क्या है?
If (x=2y+3) and (3x-4y=17), what is the value of (y)?
#linear equations
#substitution
#value of y
#medium
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=2y+3) gives (6y+9-4y=17). Thus (2y=8), so (y=4).
Step 2
Why this answer is correct
The correct answer is C. (y=4). Substituting (x=2y+3) gives (6y+9-4y=17). Thus (2y=8), so (y=4).
Step 3
Exam Tip
(x=2y+3) रखने पर (6y+9-4y=17) मिलता है। इससे (2y=8), इसलिए (y=4)।
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समीकरणों (5x-4y=2) और (3x+4y=30) को हल करने पर (x+y) का मान क्या होगा?
On solving (5x-4y=2) and (3x+4y=30), what will be the value of (x+y)?
#linear equations
#elimination
#expression value
#medium
#class 10
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (8x=32), so (x=4). Then (3x+4y=30) gives \(y=\frac{9}{2}\), so \(x+y=\frac{17}{2}\).
Step 2
Why this answer is correct
The correct answer is D. (10). Adding both equations gives (8x=32), so (x=4). Then (3x+4y=30) gives \(y=\frac{9}{2}\), so \(x+y=\frac{17}{2}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=32), इसलिए (x=4)। फिर (3x+4y=30) से \(y=\frac{9}{2}\), अतः \(x+y=\frac{17}{2}\)।
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यदि (2x+7y=36) और (2x+3y=20), तो (y) का मान क्या है?
If (2x+7y=36) and (2x+3y=20), what is the value of (y)?
#linear equations
#elimination
#value of y
#medium
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (4y=16). Therefore (y=4).
Step 2
Why this answer is correct
The correct answer is C. (y=4). Subtracting the second equation from the first gives (4y=16). Therefore (y=4).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (4y=16)। इसलिए (y=4)।
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यदि (5x+3y=31) और (2x+3y=16), तो (x) का मान क्या है?
If (5x+3y=31) and (2x+3y=16), what is the value of (x)?
#linear equations
#elimination
#value of x
#medium
#class 10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (3x=15). Therefore (x=5).
Step 2
Why this answer is correct
The correct answer is C. (x=5). Subtracting the second equation from the first gives (3x=15). Therefore (x=5).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (3x=15) मिलता है। इसलिए (x=5)।
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यदि (2x-y=9) और (x+2y=13), तो (2x+y) का मान क्या है?
If (2x-y=9) and (x+2y=13), what is the value of (2x+y)?
#linear equations
#substitution
#expression value
#medium
#class 10
A (13)
B (14)
C (15)
D (16)
Explanation opens after your attempt
Step 1
Concept
Use (y=2x-9) from the first equation. Solving gives \(x=\frac{31}{5}\) and \(y=\frac{17}{5}\), so \(2x+y=\frac{79}{5}\).
Step 2
Why this answer is correct
The correct answer is D. (16). Use (y=2x-9) from the first equation. Solving gives \(x=\frac{31}{5}\) and \(y=\frac{17}{5}\), so \(2x+y=\frac{79}{5}\).
Step 3
Exam Tip
पहले समीकरण से (y=2x-9) रखें। हल करने पर \(x=\frac{31}{5}\) और \(y=\frac{17}{5}\), इसलिए \(2x+y=\frac{79}{5}\) है।
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यदि (3x+2y=23) और (5x-2y=17), तो (x-y) का मान क्या है?
If (3x+2y=23) and (5x-2y=17), what is the value of (x-y)?
#linear equations
#elimination
#difference value
#medium
#class 10
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (8x=40), so (x=5). Then (3x+2y=23) gives (y=4), so (x-y=1).
Step 2
Why this answer is correct
The correct answer is C. (3). Adding both equations gives (8x=40), so (x=5). Then (3x+2y=23) gives (y=4), so (x-y=1).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=40), इसलिए (x=5)। फिर (3x+2y=23) से (y=4), इसलिए (x-y=1)।
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यदि (x+2y=11) और (3x-y=8), तो (x+y) का मान क्या है?
If (x+2y=11) and (3x-y=8), what is the value of (x+y)?
#linear equations
#substitution
#sum value
#medium
#class 10
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
Use (x=11-2y) from the first equation. Solving gives (y=3) and (x=5), so (x+y=8).
Step 2
Why this answer is correct
The correct answer is D. (8). Use (x=11-2y) from the first equation. Solving gives (y=3) and (x=5), so (x+y=8).
Step 3
Exam Tip
पहले समीकरण से (x=11-2y) रखें। हल करने पर (y=3) और (x=5), इसलिए (x+y=8)।
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