Question 1/11
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक आयत की लंबाई (x+10) और चौड़ाई (x-7) है। क्षेत्रफल (160) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+10) and breadth (x-7). If its area is (160), which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-230=0\)
Step 1
Concept
The area is ((x+10)(x-7)=160). Expanding gives \(x^2+3x-70=160\), so \(x^2+3x-230=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-230=0\). The area is ((x+10)(x-7)=160). Expanding gives \(x^2+3x-70=160\), so \(x^2+3x-230=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+10)(x-7)=160) होगा। विस्तार से \(x^2+3x-70=160\), इसलिए \(x^2+3x-230=0\)।
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Question 2/11
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
एक आयत की लंबाई (x+9) और चौड़ाई (x-6) है। क्षेत्रफल (130) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+9) and breadth (x-6). If its area is (130), which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-184=0\)
Step 1
Concept
The area is ((x+9)(x-6)=130). Expanding gives \(x^2+3x-54=130\), so \(x^2+3x-184=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-184=0\). The area is ((x+9)(x-6)=130). Expanding gives \(x^2+3x-54=130\), so \(x^2+3x-184=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+9)(x-6)=130) होगा। विस्तार से \(x^2+3x-54=130\), इसलिए \(x^2+3x-184=0\)।
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Question 3/11
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
एक आयत की लंबाई (2x+1) और चौड़ाई (x-4) है। यदि क्षेत्रफल (45) है, तो सही द्विघात समीकरण कौन-सा है?
A rectangle has length (2x+1) and breadth (x-4). If the area is (45), which quadratic equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(2x^2-7x-49=0\)
Step 1
Concept
The area is ((2x+1)(x-4)=45). Expanding gives \(2x^2-7x-4=45\), so \(2x^2-7x-49=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-7x-49=0\). The area is ((2x+1)(x-4)=45). Expanding gives \(2x^2-7x-4=45\), so \(2x^2-7x-49=0\).
Step 3
Exam Tip
क्षेत्रफल ((2x+1)(x-4)=45) होगा। विस्तार करने पर \(2x^2-7x-4=45\), इसलिए \(2x^2-7x-49=0\)।
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Question 4/11
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
एक आयत की लंबाई (x+8) और चौड़ाई (x-5) है। क्षेत्रफल (104) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+8) and breadth (x-5). If its area is (104), which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-144=0\)
Step 1
Concept
The area is ((x+8)(x-5)=104). Expanding gives \(x^2+3x-40=104\), so \(x^2+3x-144=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-144=0\). The area is ((x+8)(x-5)=104). Expanding gives \(x^2+3x-40=104\), so \(x^2+3x-144=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+8)(x-5)=104) होगा। विस्तार से \(x^2+3x-40=104\), इसलिए \(x^2+3x-144=0\)।
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Question 5/11
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक आयत की लंबाई (x+7) और चौड़ाई (x-3) है। क्षेत्रफल (90) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+7) and breadth (x-3). If its area is (90), which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-111=0\)
Step 1
Concept
The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-111=0\). The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+7)(x-3)=90) होगा। विस्तार से \(x^2+4x-21=90\) और \(x^2+4x-111=0\) मिलता है।
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Question 6/11
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
एक आयत की लंबाई (x+5) और चौड़ाई (x-2) है। क्षेत्रफल (84) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+5) and breadth (x-2). If its area is (84), which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-94=0\)
Step 1
Concept
The area is ((x+5)(x-2)=84). Expanding gives \(x^2+3x-10=84\) and \(x^2+3x-94=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-94=0\). The area is ((x+5)(x-2)=84). Expanding gives \(x^2+3x-10=84\) and \(x^2+3x-94=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+5)(x-2)=84) होगा। विस्तार से \(x^2+3x-10=84\) और \(x^2+3x-94=0\) मिलता है।
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Question 7/11
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक आयत की लंबाई चौड़ाई से (7) अधिक है और क्षेत्रफल (120) है। यदि चौड़ाई (x) है, तो समीकरण क्या होगा?
A rectangle has length (7) more than its breadth and area (120). If the breadth is (x), what is the equation?
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x-120=0\)
Step 1
Concept
The length is (x+7), and the area is (x(x+7)=120). Therefore \(x^2+7x-120=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x-120=0\). The length is (x+7), and the area is (x(x+7)=120). Therefore \(x^2+7x-120=0\).
Step 3
Exam Tip
लंबाई (x+7) होगी और क्षेत्रफल (x(x+7)=120) होगा। इसलिए \(x^2+7x-120=0\) है।
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Question 8/11
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
एक आयत की लंबाई चौड़ाई से (6) अधिक है और क्षेत्रफल (72) है। यदि चौड़ाई (x) है, तो समीकरण क्या होगा?
A rectangle has length (6) more than its breadth and area (72). If the breadth is (x), what is the equation?
Explanation opens after your attempt
Correct Answer
A. \(x^2+6x-72=0\)
Step 1
Concept
The length is (x+6) and the area is (x(x+6)=72). Therefore \(x^2+6x-72=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+6x-72=0\). The length is (x+6) and the area is (x(x+6)=72). Therefore \(x^2+6x-72=0\).
Step 3
Exam Tip
लंबाई (x+6) होगी और क्षेत्रफल (x(x+6)=72) होगा। इसलिए \(x^2+6x-72=0\) है।
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Question 9/11
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
एक आयत की लंबाई चौड़ाई से (4) अधिक है और क्षेत्रफल (45) है। यदि चौड़ाई (x) है, तो समीकरण क्या होगा?
A rectangle has length (4) more than its breadth and area (45). If the breadth is (x), what is the equation?
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-45=0\)
Step 1
Concept
The length is (x+4), so (x(x+4)=45). This forms \(x^2+4x-45=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-45=0\). The length is (x+4), so (x(x+4)=45). This forms \(x^2+4x-45=0\).
Step 3
Exam Tip
लंबाई (x+4) होगी, इसलिए (x(x+4)=45)। इससे \(x^2+4x-45=0\) बनता है।
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Question 10/11
Easy Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि एक वर्ग की भुजा (x+2) है और क्षेत्रफल (49) है तो समीकरण क्या बनेगा?
If the side of a square is (x+2) and the area is (49), what equation is formed?
Explanation opens after your attempt
Correct Answer
A. ((x+2)2=49)
Step 1
Concept
The area of a square is the square of its side. So the equation is ((x+2)2=49).
Step 2
Why this answer is correct
The correct answer is A. ((x+2)2=49). The area of a square is the square of its side. So the equation is ((x+2)2=49).
Step 3
Exam Tip
वर्ग का क्षेत्रफल भुजा का वर्ग होता है। इसलिए समीकरण ((x+2)2=49) बनेगा।
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Question 11/11
Easy Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि एक आयत की लंबाई (x+3) और चौड़ाई (x) है तथा क्षेत्रफल (10) है तो समीकरण क्या बनेगा?
If a rectangle has length (x+3) and width (x), and area (10), what equation is formed?
Explanation opens after your attempt
Correct Answer
A. (x(x+3)=10)
Step 1
Concept
Area is the product of length and width. So the equation is (x(x+3)=10).
Step 2
Why this answer is correct
The correct answer is A. (x(x+3)=10). Area is the product of length and width. So the equation is (x(x+3)=10).
Step 3
Exam Tip
क्षेत्रफल लंबाई और चौड़ाई का गुणनफल होता है। इसलिए समीकरण (x(x+3)=10) बनेगा।
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