100 results found for "value and saturation" in Class 10.
मान के अध्ययन में मध्यम मान की भूमिका क्या है?
What is the role of middle value in value study?
#middle value
#transition
#shading
A प्रकाश और छाया के बीच संक्रमण बनाना / Creating transition between light and shadow
B रंग का नाम बदलना / Changing colour name
C आकार मिटाना / Erasing shape
D कागज काटना / Cutting paper
Explanation opens after your attempt
Correct Answer
A. प्रकाश और छाया के बीच संक्रमण बनाना / Creating transition between light and shadow
Step 1
Concept
Middle value connects light and dark parts. Exam tip: connect mid value with transition.
Step 2
Why this answer is correct
The correct answer is A. प्रकाश और छाया के बीच संक्रमण बनाना / Creating transition between light and shadow. Middle value connects light and dark parts. Exam tip: connect mid value with transition.
Step 3
Exam Tip
मध्यम मान उजले और गहरे भागों को जोड़ता है। परीक्षा में mid value को transition से जोड़ें।
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एक गंभीर दृश्य में अत्यधिक संतृप्त रंगों का गलत उपयोग क्या कर सकता है?
What can wrong use of highly saturated colours do in a serious scene?
#saturation
#serious mood
#colour
A गंभीर भाव को कमजोर कर सकता है / It can weaken serious mood
B गंभीर भाव हमेशा बढ़ाएगा / It will always increase serious mood
C मान संरचना ठीक कर देगा / It will fix value structure
D स्थान को वास्तविक बना देगा / It will make space actual
Explanation opens after your attempt
Correct Answer
A. गंभीर भाव को कमजोर कर सकता है / It can weaken serious mood
Step 1
Concept
Highly intense colours can sometimes look unsuitable to subject. Exam tip: match saturation with mood.
Step 2
Why this answer is correct
The correct answer is A. गंभीर भाव को कमजोर कर सकता है / It can weaken serious mood. Highly intense colours can sometimes look unsuitable to subject. Exam tip: match saturation with mood.
Step 3
Exam Tip
अत्यधिक तीव्र रंग कभी विषय से असंगत लग सकते हैं। परीक्षा में saturation को mood से मिलाएं।
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यदि मान पट्टी में मध्य मान नहीं है तो छायांकन पर क्या असर होगा?
If a value scale has no middle values what effect will it have on shading?
#value scale
#middle value
#shading
A प्रकाश से छाया का संक्रमण कठोर लगेगा / Transition from light to shadow will look harsh
B छाया अधिक कोमल होगी / Shadow will become softer
C रंगत्व बढ़ेगा / Hue will increase
D बनावट वास्तविक होगी / Texture will become actual
Explanation opens after your attempt
Correct Answer
A. प्रकाश से छाया का संक्रमण कठोर लगेगा / Transition from light to shadow will look harsh
Step 1
Concept
Middle values create smooth transition. Exam tip: keep full range in value scale.
Step 2
Why this answer is correct
The correct answer is A. प्रकाश से छाया का संक्रमण कठोर लगेगा / Transition from light to shadow will look harsh. Middle values create smooth transition. Exam tip: keep full range in value scale.
Step 3
Exam Tip
मध्य मान smooth transition बनाते हैं। परीक्षा में value scale में full range रखें।
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छाया को अधिक गहरा दिखाने के लिए किस मान का उपयोग होगा?
Which value will be used to make shadow look darker?
#dark value
#shadow
#value
A हल्का मान / Light value
B गहरा मान / Dark value
C रंगत्व / Hue
D सकारात्मक स्थान / Positive space
Explanation opens after your attempt
Correct Answer
B. गहरा मान / Dark value
Step 1
Concept
Dark value increases the effect of shadow. Exam tip: connect shadow with dark value.
Step 2
Why this answer is correct
The correct answer is B. गहरा मान / Dark value. Dark value increases the effect of shadow. Exam tip: connect shadow with dark value.
Step 3
Exam Tip
गहरा मान छाया का प्रभाव बढ़ाता है। परीक्षा में shadow को dark value से जोड़ें।
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प्रकाश पड़ने वाले भाग को दिखाने के लिए कौन सा मान चाहिए?
Which value is needed to show the lighted part?
#light value
#highlight
#value
A गहरा मान / Dark value
B मध्यम अंधेरा / Medium dark
C हल्का मान / Light value
D काली रेखा / Black line
Explanation opens after your attempt
Correct Answer
C. हल्का मान / Light value
Step 1
Concept
Light value shows the lit part. Exam tip: connect highlight with light value.
Step 2
Why this answer is correct
The correct answer is C. हल्का मान / Light value. Light value shows the lit part. Exam tip: connect highlight with light value.
Step 3
Exam Tip
हल्का मान प्रकाश वाले भाग को दिखाता है। परीक्षा में प्रकाशित भाग को हल्का मान से जोड़ें।
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किस स्थिति में बाहरी रेखा को हटाकर केवल रंग और मान से आकृति बनाना अधिक परिपक्व तरीका माना जा सकता है?
In which situation can removing outline and forming figure only with colour and value be considered a more mature method?
#shape
#value contrast
#naturalism
A जब आकृति को प्राकृतिक और प्रकाश आधारित दिखाना हो / When figure needs to look natural and light-based
B जब केवल बाल रेखाएं दिखानी हों / When only hair lines are needed
C जब कागज काटना हो / When paper needs cutting
D जब पाठ लिखना हो / When text must be written
Explanation opens after your attempt
Correct Answer
A. जब आकृति को प्राकृतिक और प्रकाश आधारित दिखाना हो / When figure needs to look natural and light-based
Step 1
Concept
Boundary made by colour and value can look more natural. Exam tip: understand shape formation without outline.
Step 2
Why this answer is correct
The correct answer is A. जब आकृति को प्राकृतिक और प्रकाश आधारित दिखाना हो / When figure needs to look natural and light-based. Boundary made by colour and value can look more natural. Exam tip: understand shape formation without outline.
Step 3
Exam Tip
रंग और मान से बनी सीमा अधिक प्राकृतिक दिख सकती है। परीक्षा में outline के बिना shape formation समझें।
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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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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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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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यदि (4x-y=17) और (2x+3y=19), तो (x) का मान क्या होगा?
If (4x-y=17) and (2x+3y=19), what will be the value of (x)?
#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
From the first equation use (y=4x-17). Then (2x+3(4x-17)=19) gives (x=5).
Step 2
Why this answer is correct
The correct answer is C. (x=5). From the first equation use (y=4x-17). Then (2x+3(4x-17)=19) gives (x=5).
Step 3
Exam Tip
पहले समीकरण से (y=4x-17) रखें। फिर (2x+3(4x-17)=19) से (x=5) मिलता है।
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यदि (x+2y=21) और (y=8), तो (x) का मान क्या होगा?
If (x+2y=21) and (y=8), what will be the value of (x)?
#linear equations
#substitution
#value of x
#easy
#class 10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Putting (y=8) gives (x+16=21), so (x=5). Multiply first and then subtract.
Step 2
Why this answer is correct
The correct answer is C. (x=5). Putting (y=8) gives (x+16=21), so (x=5). Multiply first and then subtract.
Step 3
Exam Tip
(y=8) रखने पर (x+16=21), इसलिए (x=5)। पहले गुणा करें फिर घटाएं।
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समीकरणों (2x+5y=31) और (2x+y=15) से (y) का मान ज्ञात कीजिए।
Find the value of (y) from (2x+5y=31) and (2x+y=15).
#linear equations
#elimination
#value of y
#easy
#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 (4y=16), so (y=4). Subtract to remove equal (2x).
Step 2
Why this answer is correct
The correct answer is B. (y=4). Subtracting the second equation from the first gives (4y=16), so (y=4). Subtract to remove equal (2x).
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (4y=16), इसलिए (y=4)। समान (2x) को हटाने के लिए घटाएं।
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समीकरणों (5x+y=31) और (3x+y=19) से (x) का मान क्या है?
What is the value of (x) from (5x+y=31) and (3x+y=19)?
#linear equations
#elimination
#value of x
#easy
#class 10
A (x=6)
B (x=5)
C (x=4)
D (x=3)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (2x=12), so (x=6). Use elimination immediately on equal (y) terms.
Step 2
Why this answer is correct
The correct answer is A. (x=6). Subtracting the second equation from the first gives (2x=12), so (x=6). Use elimination immediately on equal (y) terms.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (2x=12), इसलिए (x=6)। समान (y) पदों पर विलोपन तुरंत करें।
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समीकरणों (4x-y=22) और (x-y=4) से (x) का मान ज्ञात कीजिए।
Find the value of (x) from (4x-y=22) and (x-y=4).
#linear equations
#elimination
#value of x
#easy
#class 10
A (x=6)
B (x=5)
C (x=4)
D (x=3)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (3x=18), so (x=6). Subtract to remove equal (-y) terms.
Step 2
Why this answer is correct
The correct answer is A. (x=6). Subtracting the second equation from the first gives (3x=18), so (x=6). Subtract to remove equal (-y) terms.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (3x=18), इसलिए (x=6)। समान (-y) हटाने के लिए घटाएं।
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यदि (x=14-3y) और (x+y=8), तो (y) का मान क्या है?
If (x=14-3y) and (x+y=8), what is the value of (y)?
#linear equations
#substitution
#value of y
#easy
#class 10
A (y=1)
B (y=2)
C (y=4)
D (y=3)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=14-3y) gives (14-2y=8), so (y=3). After substitution, keep variable terms on one side.
Step 2
Why this answer is correct
The correct answer is D. (y=3). Substituting (x=14-3y) gives (14-2y=8), so (y=3). After substitution, keep variable terms on one side.
Step 3
Exam Tip
(x=14-3y) रखने पर (14-2y=8), इसलिए (y=3)। प्रतिस्थापन के बाद चर वाले पद एक तरफ रखें।
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यदि (x-y=8) और (y=5), तो (x) का मान क्या है?
If (x-y=8) and (y=5), what is the value of (x)?
#linear equations
#substitution
#value of x
#easy
#class 10
A (x=13)
B (x=12)
C (x=11)
D (x=10)
Explanation opens after your attempt
Step 1
Concept
Putting (y=5) gives (x-5=8), so (x=13). Substitute the given value in the original equation.
Step 2
Why this answer is correct
The correct answer is A. (x=13). Putting (y=5) gives (x-5=8), so (x=13). Substitute the given value in the original equation.
Step 3
Exam Tip
(y=5) रखने पर (x-5=8), इसलिए (x=13)। दिए हुए मान को मूल समीकरण में रखें।
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समीकरणों (2x+3y=22) और (2x+y=12) से (y) का मान ज्ञात कीजिए।
Find the value of (y) from (2x+3y=22) and (2x+y=12).
#linear equations
#elimination
#value of y
#easy
#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 (2y=10), so (y=5). Eliminate the equal (2x) terms.
Step 2
Why this answer is correct
The correct answer is D. (y=5). Subtracting the second equation from the first gives (2y=10), so (y=5). Eliminate the equal (2x) terms.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (2y=10), इसलिए (y=5)। समान (2x) पदों को विलोपित करें।
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यदि (y=3x-2) और (x+y=18), तो (y) का मान क्या होगा?
If (y=3x-2) and (x+y=18), what will be the value of (y)?
#linear equations
#substitution
#value of y
#easy
#class 10
A (y=10)
B (y=11)
C (y=13)
D (y=15)
Explanation opens after your attempt
Step 1
Concept
Substituting (y=3x-2) gives (4x-2=18), so (x=5) and (y=13). Find (x) first, then use the expression for (y).
Step 2
Why this answer is correct
The correct answer is C. (y=13). Substituting (y=3x-2) gives (4x-2=18), so (x=5) and (y=13). Find (x) first, then use the expression for (y).
Step 3
Exam Tip
(y=3x-2) रखने पर (4x-2=18), इसलिए (x=5) और (y=13)। पहले (x) निकालें फिर (y) के रूप में रखें।
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