100 results found for "problem solving myth" in Class 10.
\(8x^2-14x-15=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(8x^2-14x-15=0\)?
#quadratic
#factorisation
#verification
A ((4x+3)(2x-5)=0)
B ((4x-3)(2x+5)=0)
C ((8x+5)(x-3)=0)
D ((x+5)(8x-3)=0)
Explanation opens after your attempt
Correct Answer
A. ((4x+3)(2x-5)=0)
Step 1
Concept
((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((4x+3)(2x-5)=0). ((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((4x+3)(2x-5)=8x-2 -20x+6x-15=8x-2 -14x-15), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(8x^2-23x-15=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(8x^2-23x-15=0\)?
#quadratic
#factorisation
#audit
A ((8x+5)(x-3)=0)
B ((8x-5)(x+3)=0)
C ((x+5)(8x-3)=0)
D ((8x-3)(x+5)=0)
Explanation opens after your attempt
Correct Answer
A. ((8x+5)(x-3)=0)
Step 1
Concept
((8x+5)(x-3)=8x-2 -19x-15), so it is not for the given equation. In exams, verify each option by expansion.
Step 2
Why this answer is correct
The correct answer is A. ((8x+5)(x-3)=0). ((8x+5)(x-3)=8x-2 -19x-15), so it is not for the given equation. In exams, verify each option by expansion.
Step 3
Exam Tip
((8x+5)(x-3)=8x-2 -19x-15) नहीं बल्कि यह विस्तार गलत होगा; सही गुणनखंड ((8x+5)(x-3)) से (-24x+5x=-19x) बनता है। परीक्षा में विस्तार से हर विकल्प जांचें।
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\(13x^2-52x+9=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(13x^2-52x+9=0\) by completing the square?
#quadratic
#completing-square
#expert
A ((x-2)2 =\frac{43}{13})
B ((x+2)2 =\frac{43}{13})
C ((x-4)2 =\frac{9}{13})
D ((x-2)2 =4)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =\frac{43}{13})
Step 1
Concept
First \(x^2-4x+\frac{9}{13}=0\) is obtained, then ((x-2)2 =\frac{43}{13}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. ((x-2)2 =\frac{43}{13}). First \(x^2-4x+\frac{9}{13}=0\) is obtained, then ((x-2)2 =\frac{43}{13}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-4x+\frac{9}{13}=0\) बनता है, फिर ((x-2)2 =\frac{43}{13}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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\(7x^2-19x-6=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(7x^2-19x-6=0\)?
#quadratic
#factorisation
#verification
A ((7x+2)(x-3)=0)
B ((7x-2)(x+3)=0)
C ((x+2)(7x-3)=0)
D ((7x-3)(x+2)=0)
Explanation opens after your attempt
Correct Answer
A. ((7x+2)(x-3)=0)
Step 1
Concept
((7x+2)(x-3)=7x-2 -19x-6), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((7x+2)(x-3)=0). ((7x+2)(x-3)=7x-2 -19x-6), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((7x+2)(x-3)=7x-2 -19x-6), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(11x^2-44x+7=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(11x^2-44x+7=0\) by completing the square?
#quadratic
#completing-square
#expert
A ((x-2)2 =\frac{37}{11})
B ((x+2)2 =\frac{37}{11})
C ((x-4)2 =\frac{7}{11})
D ((x-2)2 =4)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =\frac{37}{11})
Step 1
Concept
First \(x^2-4x+\frac{7}{11}=0\) is obtained, then ((x-2)2 =\frac{37}{11}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. ((x-2)2 =\frac{37}{11}). First \(x^2-4x+\frac{7}{11}=0\) is obtained, then ((x-2)2 =\frac{37}{11}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-4x+\frac{7}{11}=0\) बनता है, फिर ((x-2)2 =\frac{37}{11}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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\(6x^2-11x-10=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(6x^2-11x-10=0\)?
#quadratic
#factorisation
#verification
A ((3x+2)(2x-5)=0)
B ((3x-2)(2x+5)=0)
C ((6x+5)(x-2)=0)
D ((2x+5)(3x-2)=0)
Explanation opens after your attempt
Correct Answer
A. ((3x+2)(2x-5)=0)
Step 1
Concept
((3x+2)(2x-5)=6x-2 -11x-10), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((3x+2)(2x-5)=0). ((3x+2)(2x-5)=6x-2 -11x-10), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((3x+2)(2x-5)=6x-2 -11x-10), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(9x^2-30x+8=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(9x^2-30x+8=0\) by completing the square?
#quadratic
#completing-square
#expert
A (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9})
B (\left\(x+\frac{5}{3}\right\)2 =\frac{17}{9})
C (\left\(x-\frac{10}{3}\right\)2 =\frac{8}{9})
D (\left\(x-\frac{5}{3}\right\)2 =\frac{25}{9})
Explanation opens after your attempt
Correct Answer
A. (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9})
Step 1
Concept
First \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) is obtained, then (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9}). First \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) is obtained, then (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) बनता है, फिर (\left\(x-\frac{5}{3}\right\)2 =\frac{17}{9}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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\(5x^2-7x-6=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(5x^2-7x-6=0\)?
#quadratic
#factorisation
#verification
A ((5x+3)(x-2)=0)
B ((5x-3)(x+2)=0)
C ((x+3)(5x-2)=0)
D ((5x-2)(x+3)=0)
Explanation opens after your attempt
Correct Answer
A. ((5x+3)(x-2)=0)
Step 1
Concept
((5x+3)(x-2)=5x-2 -7x-6), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((5x+3)(x-2)=0). ((5x+3)(x-2)=5x-2 -7x-6), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((5x+3)(x-2)=5x-2 -7x-6), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(7x^2-22x+7=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(7x^2-22x+7=0\) by completing the square?
#quadratic
#completing-square
#hard
A (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49})
B (\left\(x+\frac{11}{7}\right\)2 =\frac{72}{49})
C (\left\(x-\frac{22}{7}\right\)2 =1)
D (\left\(x-\frac{11}{7}\right\)2 =\frac{121}{49})
Explanation opens after your attempt
Correct Answer
A. (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49})
Step 1
Concept
First \(x^2-\frac{22}{7}x+1=0\) is obtained, then (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49}). First \(x^2-\frac{22}{7}x+1=0\) is obtained, then (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{22}{7}x+1=0\) बनता है, फिर (\left\(x-\frac{11}{7}\right\)2 =\frac{72}{49}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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\(3x^2-5x-2=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(3x^2-5x-2=0\)?
#quadratic
#factorisation
#verification
A ((3x+1)(x-2)=0)
B ((3x-1)(x+2)=0)
C ((x+1)(3x-2)=0)
D ((3x-2)(x+1)=0)
Explanation opens after your attempt
Correct Answer
A. ((3x+1)(x-2)=0)
Step 1
Concept
((3x+1)(x-2)=3x-2 -5x-2), so it is correct. In exams, verify factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((3x+1)(x-2)=0). ((3x+1)(x-2)=3x-2 -5x-2), so it is correct. In exams, verify factorisation by expanding.
Step 3
Exam Tip
((3x+1)(x-2)=3x-2 -5x-2), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(5x^2-18x+9=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(5x^2-18x+9=0\) by completing the square?
#quadratic
#completing-square
#hard
A (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25})
B (\left\(x+\frac{9}{5}\right\)2 =\frac{36}{25})
C (\left\(x-\frac{18}{5}\right\)2 =\frac{9}{5})
D (\left\(x-\frac{9}{5}\right\)2 =\frac{81}{25})
Explanation opens after your attempt
Correct Answer
A. (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25})
Step 1
Concept
First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25}). First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) बनता है, फिर (\left\(x-\frac{9}{5}\right\)2 =\frac{36}{25}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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\(2x^2-3x-2=0\) को हल करने में कौनसा गुणनखंड रूप सही है?
Which factorised form is correct for solving \(2x^2-3x-2=0\)?
#quadratic
#factorisation
#verification
A ((2x+1)(x-2)=0)
B ((2x-1)(x+2)=0)
C ((x+1)(2x-2)=0)
D ((2x-2)(x+1)=0)
Explanation opens after your attempt
Correct Answer
A. ((2x+1)(x-2)=0)
Step 1
Concept
((2x+1)(x-2)=2x-2 -3x-2), so it is correct. In exams, verify the factorisation by expanding.
Step 2
Why this answer is correct
The correct answer is A. ((2x+1)(x-2)=0). ((2x+1)(x-2)=2x-2 -3x-2), so it is correct. In exams, verify the factorisation by expanding.
Step 3
Exam Tip
((2x+1)(x-2)=2x-2 -3x-2), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें।
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\(3x^2-10x+3=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(3x^2-10x+3=0\) by completing the square?
#quadratic
#completing-square
#hard
A (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9})
B (\left\(x+\frac{5}{3}\right\)2 =\frac{16}{9})
C (\left\(x-\frac{10}{3}\right\)2 =1)
D (\left\(x-\frac{5}{3}\right\)2 =\frac{25}{9})
Explanation opens after your attempt
Correct Answer
A. (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9})
Step 1
Concept
First we get \(x^2-\frac{10}{3}x+1=0\), then (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9}). First we get \(x^2-\frac{10}{3}x+1=0\), then (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{10}{3}x+1=0\) बनता है और फिर (\left\(x-\frac{5}{3}\right\)2 =\frac{16}{9}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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((x-7)2 =11) को हल करने पर (x) का मान क्या होगा?
Solving ((x-7)2 =11), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=7\pm\sqrt{11}\)
B \(x=-7\pm\sqrt{11}\)
C \(x=7\pm11\)
D \(x=\sqrt{7}\pm11\)
Explanation opens after your attempt
Correct Answer
A. \(x=7\pm\sqrt{11}\)
Step 1
Concept
\(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=7\pm\sqrt{11}\). \(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x-7=\pm\sqrt{11}\), इसलिए \(x=7\pm\sqrt{11}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2+8x-33=0\) को पूर्ण वर्ग विधि से हल करने में सही चरण कौनसा है?
Which step is correct in solving \(x^2+8x-33=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x+4)2 =49)
B ((x-4)2 =49)
C ((x+8)2 =33)
D ((x+4)2 =33)
Explanation opens after your attempt
Correct Answer
A. ((x+4)2 =49)
Step 1
Concept
Adding (16) to \(x^2+8x=33\) gives ((x+4)2 =49). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x+4)2 =49). Adding (16) to \(x^2+8x=33\) gives ((x+4)2 =49). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2+8x=33\) में (16) जोड़ने पर ((x+4)2 =49) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(5x^2+16x+3=0\) को हल करने पर मूल क्या होंगे?
What will be the roots after solving \(5x^2+16x+3=0\)?
#quadratic
#factorisation
#fraction-roots
A \(x=-3,-\frac{1}{5}\)
B \(x=3,\frac{1}{5}\)
C \(x=-5,-\frac{3}{1}\)
D \(x=-1,-\frac{3}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(x=-3,-\frac{1}{5}\)
Step 1
Concept
(5x-2 +16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-3,-\frac{1}{5}\). (5x-2 +16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 3
Exam Tip
(5x-2 +16x+3=(5x+1)(x+3)), इसलिए मूल \(-\frac{1}{5}\) और (-3) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-18x+45=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-18x+45=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x-9)2 =36)
B ((x+9)2 =36)
C ((x-18)2 =45)
D ((x-9)2 =45)
Explanation opens after your attempt
Correct Answer
A. ((x-9)2 =36)
Step 1
Concept
Adding (81) to \(x^2-18x=-45\) gives ((x-9)2 =36). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x-9)2 =36). Adding (81) to \(x^2-18x=-45\) gives ((x-9)2 =36). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2-18x=-45\) में (81) जोड़ने पर ((x-9)2 =36) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(8x^2-32x=0\) को हल करते समय कौनसी गलती नहीं करनी चाहिए?
Which mistake should be avoided while solving \(8x^2-32x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ना / Missing (x=0)
B (8x) सामान्य गुणनखंड लेना / Taking (8x) common
C (8x(x-4)=0) लिखना / Writing (8x(x-4)=0)
D (x=4) को मूल मानना / Taking (x=4) as a root
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ना / Missing (x=0)
Step 1
Concept
(8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ना / Missing (x=0). (8x-2 -32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).
Step 3
Exam Tip
(8x-2 -32x=8x(x-4)), इसलिए (x=0) और (x=4) दोनों मूल हैं। परीक्षा में चर से भाग देने पर (x=0) छूट सकता है।
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\(12x^2+17x+5=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(12x^2+17x+5=0\) by factorisation?
#quadratic
#factorisation
#fraction-roots
A \(x=-1,-\frac{5}{12}\)
B \(x=1,\frac{5}{12}\)
C \(x=-\frac{12}{5},-1\)
D \(x=-5,-\frac{1}{12}\)
Explanation opens after your attempt
Correct Answer
A. \(x=-1,-\frac{5}{12}\)
Step 1
Concept
(12x-2 +17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-1,-\frac{5}{12}\). (12x-2 +17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 3
Exam Tip
(12x-2 +17x+5=(12x+5)(x+1)), इसलिए मूल \(-\frac{5}{12}\) और (-1) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-10x+24=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-10x+24=0\) by completing the square?
#quadratic
#completing-square
#steps
A ((x-5)2 =1)
B ((x+5)2 =1)
C ((x-10)2 =24)
D ((x-5)2 =24)
Explanation opens after your attempt
Correct Answer
A. ((x-5)2 =1)
Step 1
Concept
Adding (25) to \(x^2-10x=-24\) gives ((x-5)2 =1). In exams, add the same number to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x-5)2 =1). Adding (25) to \(x^2-10x=-24\) gives ((x-5)2 =1). In exams, add the same number to both sides.
Step 3
Exam Tip
\(x^2-10x=-24\) में (25) जोड़ने पर ((x-5)2 =1) मिलता है। परीक्षा में दोनों पक्षों में समान संख्या जोड़ें।
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((x+6)2 =5) को हल करने पर (x) का मान क्या होगा?
Solving ((x+6)2 =5), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=-6\pm\sqrt{5}\)
B \(x=6\pm\sqrt{5}\)
C \(x=-6\pm5\)
D \(x=\sqrt{6}\pm5\)
Explanation opens after your attempt
Correct Answer
A. \(x=-6\pm\sqrt{5}\)
Step 1
Concept
\(x+6=\pm\sqrt{5}\), so \(x=-6\pm\sqrt{5}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=-6\pm\sqrt{5}\). \(x+6=\pm\sqrt{5}\), so \(x=-6\pm\sqrt{5}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x+6=\pm\sqrt{5}\), इसलिए \(x=-6\pm\sqrt{5}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(5x^2=80\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(5x^2=80\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm16\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=16\) मिलता है, इसलिए \(x=\pm4\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2+2x-24=0\) को पूर्ण वर्ग विधि से हल करने में सही चरण कौनसा है?
Which step is correct in solving \(x^2+2x-24=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x+1)2 =25)
B ((x-1)2 =25)
C ((x+2)2 =24)
D ((x+1)2 =24)
Explanation opens after your attempt
Correct Answer
A. ((x+1)2 =25)
Step 1
Concept
Adding (1) to \(x^2+2x=24\) gives ((x+1)2 =25). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x+1)2 =25). Adding (1) to \(x^2+2x=24\) gives ((x+1)2 =25). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2+2x=24\) में (1) जोड़ने पर ((x+1)2 =25) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(3x^2+11x+10=0\) को हल करने पर मूल क्या होंगे?
What will be the roots after solving \(3x^2+11x+10=0\)?
#quadratic
#factorisation
#fraction-roots
A \(x=-2,-\frac{5}{3}\)
B \(x=2,\frac{5}{3}\)
C \(x=-3,-\frac{10}{3}\)
D (x=-1,-10)
Explanation opens after your attempt
Correct Answer
A. \(x=-2,-\frac{5}{3}\)
Step 1
Concept
(3x-2 +11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-2,-\frac{5}{3}\). (3x-2 +11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(3x-2 +11x+10=(3x+5)(x+2)), इसलिए मूल \(-\frac{5}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-16x+28=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-16x+28=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x-8)2 =36)
B ((x+8)2 =36)
C ((x-16)2 =28)
D ((x-8)2 =28)
Explanation opens after your attempt
Correct Answer
A. ((x-8)2 =36)
Step 1
Concept
Adding (64) to \(x^2-16x=-28\) gives ((x-8)2 =36). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x-8)2 =36). Adding (64) to \(x^2-16x=-28\) gives ((x-8)2 =36). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2-16x=-28\) में (64) जोड़ने पर ((x-8)2 =36) मिलता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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\(6x^2-18x=0\) को हल करते समय कौनसी गलती नहीं करनी चाहिए?
Which mistake should be avoided while solving \(6x^2-18x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ना / Missing (x=0)
B (6x) सामान्य गुणनखंड लेना / Taking (6x) common
C (6x(x-3)=0) लिखना / Writing (6x(x-3)=0)
D (x=3) को मूल मानना / Taking (x=3) as a root
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ना / Missing (x=0)
Step 1
Concept
(6x-2 -18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ना / Missing (x=0). (6x-2 -18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).
Step 3
Exam Tip
(6x-2 -18x=6x(x-3)), इसलिए (x=0) और (x=3) दोनों मूल हैं। परीक्षा में चर से भाग देने पर (x=0) छूट सकता है।
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\(3x^2+8x+4=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2+8x+4=0\) by factorisation?
#quadratic
#factorisation
#fraction-roots
A \(x=-2,-\frac{2}{3}\)
B \(x=2,\frac{2}{3}\)
C (x=-3,-4)
D \(x=-\frac{3}{2},-4\)
Explanation opens after your attempt
Correct Answer
A. \(x=-2,-\frac{2}{3}\)
Step 1
Concept
(3x-2 +8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-2,-\frac{2}{3}\). (3x-2 +8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(3x-2 +8x+4=(3x+2)(x+2)), इसलिए मूल \(-\frac{2}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-8x+12=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-8x+12=0\) by completing the square?
#quadratic
#completing-square
#steps
A ((x-4)2 =4)
B ((x+4)2 =4)
C ((x-8)2 =12)
D ((x-4)2 =12)
Explanation opens after your attempt
Correct Answer
A. ((x-4)2 =4)
Step 1
Concept
Adding (16) to \(x^2-8x=-12\) gives ((x-4)2 =4). In exams, add the same number to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x-4)2 =4). Adding (16) to \(x^2-8x=-12\) gives ((x-4)2 =4). In exams, add the same number to both sides.
Step 3
Exam Tip
\(x^2-8x=-12\) में (16) जोड़ने पर ((x-4)2 =4) मिलता है। परीक्षा में दोनों पक्षों में समान संख्या जोड़ें।
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((x-5)2 =3) को हल करने पर (x) का मान क्या होगा?
Solving ((x-5)2 =3), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=5\pm\sqrt{3}\)
B \(x=-5\pm\sqrt{3}\)
C \(x=5\pm3\)
D \(x=\sqrt{5}\pm3\)
Explanation opens after your attempt
Correct Answer
A. \(x=5\pm\sqrt{3}\)
Step 1
Concept
\(x-5=\pm\sqrt{3}\), so \(x=5\pm\sqrt{3}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=5\pm\sqrt{3}\). \(x-5=\pm\sqrt{3}\), so \(x=5\pm\sqrt{3}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x-5=\pm\sqrt{3}\), इसलिए \(x=5\pm\sqrt{3}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(3x^2=12\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2=12\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=4\) मिलता है, इसलिए \(x=\pm2\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2+4x-12=0\) को पूर्ण वर्ग विधि से हल करने में सही चरण कौनसा है?
Which step is correct in solving \(x^2+4x-12=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x+2)2 =16)
B ((x-2)2 =16)
C ((x+4)2 =12)
D ((x+2)2 =12)
Explanation opens after your attempt
Correct Answer
A. ((x+2)2 =16)
Step 1
Concept
Adding (4) to \(x^2+4x=12\) gives ((x+2)2 =16). In exams, add the same term to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x+2)2 =16). Adding (4) to \(x^2+4x=12\) gives ((x+2)2 =16). In exams, add the same term to both sides.
Step 3
Exam Tip
\(x^2+4x=12\) में (4) जोड़ने पर ((x+2)2 =16) मिलता है। परीक्षा में दोनों पक्षों में समान पद जोड़ें।
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\(2x^2+7x+6=0\) को हल करने पर मूल क्या होंगे?
What will be the roots after solving \(2x^2+7x+6=0\)?
#quadratic
#factorisation
#fraction-roots
A \(x=-\frac{3}{2},-2\)
B \(x=\frac{3}{2},2\)
C \(x=-3,-\frac{1}{2}\)
D (x=-1,-6)
Explanation opens after your attempt
Correct Answer
A. \(x=-\frac{3}{2},-2\)
Step 1
Concept
(2x-2 +7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{3}{2},-2\). (2x-2 +7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(2x-2 +7x+6=(2x+3)(x+2)), इसलिए \(x=-\frac{3}{2}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
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\(x^2-12x+20=0\) को पूर्ण वर्ग विधि से हल करने में सही मध्य चरण कौनसा है?
Which middle step is correct while solving \(x^2-12x+20=0\) by completing square?
#quadratic
#completing-square
#steps
A ((x-6)2 =16)
B ((x+6)2 =16)
C ((x-12)2 =20)
D ((x-6)2 =36)
Explanation opens after your attempt
Correct Answer
A. ((x-6)2 =16)
Step 1
Concept
From \(x^2-12x+20=0\), \(x^2-12x=-20\), then adding (36) gives ((x-6)2 =16). In exams, add the same number to both sides.
Step 2
Why this answer is correct
The correct answer is A. ((x-6)2 =16). From \(x^2-12x+20=0\), \(x^2-12x=-20\), then adding (36) gives ((x-6)2 =16). In exams, add the same number to both sides.
Step 3
Exam Tip
\(x^2-12x+20=0\) से \(x^2-12x=-20\), फिर (36) जोड़कर ((x-6)2 =16) मिलता है। परीक्षा में दोनों पक्षों में समान संख्या जोड़ें।
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\(3x^2+5x-2=0\) को गुणनखंड विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2+5x-2=0\) by factorisation?
#quadratic
#factorisation
#fraction-roots
A \(x=\frac{1}{3},-2\)
B \(x=-\frac{1}{3},2\)
C (x=3,-2)
D \(x=\frac{2}{3},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{3},-2\)
Step 1
Concept
(3x-2 +5x-2=(3x-1)(x+2)), so the roots are \(\frac{1}{3}\) and (-2). In exams, solve (3x-1=0) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{3},-2\). (3x-2 +5x-2=(3x-1)(x+2)), so the roots are \(\frac{1}{3}\) and (-2). In exams, solve (3x-1=0) carefully.
Step 3
Exam Tip
(3x-2 +5x-2=(3x-1)(x+2)), इसलिए मूल \(\frac{1}{3}\) और (-2) हैं। परीक्षा में (3x-1=0) को सावधानी से हल करें।
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\(5x^2-20x=0\) को हल करते समय कौनसी सामान्य गलती हो सकती है?
What common mistake can occur while solving \(5x^2-20x=0\)?
#quadratic
#common-mistake
#zero-root
A (x=0) को छोड़ देना / Missing (x=0)
B (x=4) को लिखना / Writing (x=4)
C (5x) सामान्य गुणनखंड लेना / Taking (5x) common
D (5x(x-4)=0) लिखना / Writing (5x(x-4)=0)
Explanation opens after your attempt
Correct Answer
A. (x=0) को छोड़ देना / Missing (x=0)
Step 1
Concept
The correct form is (5x(x-4)=0), giving (x=0) and (x=4). In exams, dividing directly by the variable can miss (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x=0) को छोड़ देना / Missing (x=0). The correct form is (5x(x-4)=0), giving (x=0) and (x=4). In exams, dividing directly by the variable can miss (x=0).
Step 3
Exam Tip
सही रूप (5x(x-4)=0) है, जिससे (x=0) और (x=4) मिलते हैं। परीक्षा में चर से सीधे भाग देने से (x=0) छूट सकता है।
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\(x^2+6x+1=0\) को पूर्ण वर्ग विधि से हल करने पर कौनसा चरण सही है?
Which step is correct while solving \(x^2+6x+1=0\) by completing the square?
#quadratic
#completing-square
#steps
A ((x+3)2 =8)
B ((x+6)2 =35)
C ((x+3)2 =10)
D ((x-3)2 =8)
Explanation opens after your attempt
Correct Answer
A. ((x+3)2 =8)
Step 1
Concept
Adding (9) in \(x^2+6x+1=0\) gives ((x+3)2 =8). In exams, add the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)2 =8). Adding (9) in \(x^2+6x+1=0\) gives ((x+3)2 =8). In exams, add the square of half the coefficient.
Step 3
Exam Tip
\(x^2+6x+1=0\) में (9) जोड़ने पर ((x+3)2 =8) बनता है। परीक्षा में आधे गुणांक का वर्ग जोड़ें।
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((x-4)2 =25) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x-4)2 =25) gives which values?
#quadratic
#square-root-method
#roots
A (x=9,-1)
B (x=5,-5)
C (x=4,25)
D (x=1,9)
Explanation opens after your attempt
Correct Answer
A. (x=9,-1)
Step 1
Concept
\(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=9,-1). \(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x-4=\pm5\), इसलिए (x=9) या (x=-1) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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((x+3)2 =16) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x+3)2 =16) gives which values?
#quadratic
#square-root-method
#roots
A (x=1,-7)
B (x=4,-4)
C (x=7,-1)
D (x=3,16)
Explanation opens after your attempt
Correct Answer
A. (x=1,-7)
Step 1
Concept
\(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=1,-7). \(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x+3=\pm4\), इसलिए (x=1) या (x=-7) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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\(x^2=121\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=121\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm11\)
B (x=11)
C (x=-11)
D \(x=\pm121\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
\(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). \(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{121}=\pm11\) होता है। परीक्षा में केवल (11) लिखना अधूरा उत्तर है।
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((x-2)2 =9) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x-2)2 =9) gives which values?
#quadratic
#square-root-method
#roots
A (x=5,-1)
B (x=2,9)
C (x=3,-3)
D (x=5,1)
Explanation opens after your attempt
Correct Answer
A. (x=5,-1)
Step 1
Concept
\(x-2=\pm3\), so (x=5) or (x=-1). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=5,-1). \(x-2=\pm3\), so (x=5) or (x=-1). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x-2=\pm3\), इसलिए (x=5) या (x=-1) है। परीक्षा में \(\pm\) के दोनों केस लिखें।
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\(x^2=49\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=49\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm7\)
B (x=7)
C (x=-7)
D \(x=\pm49\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm7\)
Step 1
Concept
\(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm7\). \(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 3
Exam Tip
\(x=\pm\sqrt{49}=\pm7\) होता है। परीक्षा में केवल धनात्मक मूल लिखना सामान्य गलती है।
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संयुक्त राष्ट्र शरणार्थी उच्चायुक्त किस समस्या के समाधान से जुड़ा है?
The UN High Commissioner for Refugees is linked with solving which problem?
#world_history
#united_nations
#unhcr
A युद्ध और संकट से विस्थापित लोगों की सुरक्षा / Protection of people displaced by war and crisis
B सिनेमा टिकट बिक्री / Sale of cinema tickets
C फुटबॉल लीग प्रबंधन / Football league management
D निजी बैंक ऋण / Private bank loans
Explanation opens after your attempt
Correct Answer
A. युद्ध और संकट से विस्थापित लोगों की सुरक्षा / Protection of people displaced by war and crisis
Step 1
Concept
UNHCR helps refugees and displaced people. Exam tip: connect it with humanitarian assistance.
Step 2
Why this answer is correct
The correct answer is A. युद्ध और संकट से विस्थापित लोगों की सुरक्षा / Protection of people displaced by war and crisis. UNHCR helps refugees and displaced people. Exam tip: connect it with humanitarian assistance.
Step 3
Exam Tip
यू एन एच सी आर शरणार्थियों और विस्थापित लोगों की सहायता करता है। परीक्षा में इसे मानवीय सहायता से जोड़ें।
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विधवा पुनर्विवाह अधिनियम 1856 किस सामाजिक समस्या के समाधान से जुड़ा था?
The Widow Remarriage Act of 1856 was linked with solving which social problem?
#modern-indian-history
#widow-remarriage
#class-10
A विधवाओं के पुनर्विवाह पर सामाजिक रोक / Social restriction on widow remarriage
B किसानों का लगान / Peasant revenue
C रेलवे टिकट / Railway ticket
D सैनिक कारतूस / Military cartridges
Explanation opens after your attempt
Correct Answer
A. विधवाओं के पुनर्विवाह पर सामाजिक रोक / Social restriction on widow remarriage
Step 1
Concept
Ishwar Chandra Vidyasagar worked in favor of widow remarriage. Exam tip is to link it with women's reform.
Step 2
Why this answer is correct
The correct answer is A. विधवाओं के पुनर्विवाह पर सामाजिक रोक / Social restriction on widow remarriage. Ishwar Chandra Vidyasagar worked in favor of widow remarriage. Exam tip is to link it with women's reform.
Step 3
Exam Tip
ईश्वरचंद्र विद्यासागर ने विधवा पुनर्विवाह के पक्ष में प्रयास किया। परीक्षा में इसे महिला सुधार से जोड़ें।
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पीठ में छुरा घोंपने की मिथक ने जर्मन राजनीति में क्या भूमिका निभाई?
What role did the stab-in-the-back myth play in German politics?
#world history
#world wars
#stab in the back
#weimar
A उसने लोकतांत्रिक नेताओं और आंतरिक शत्रुओं पर हार का दोष डालने में मदद की / It helped blame democratic leaders and internal enemies for defeat
B उसने जर्मनी को तुरंत समृद्ध बना दिया / It immediately made Germany prosperous
C उसने जापान को जर्मनी का नेता बनाया / It made Japan the leader of Germany
D उसने राष्ट्र संघ को मजबूत किया / It strengthened the League of Nations
Explanation opens after your attempt
Correct Answer
A. उसने लोकतांत्रिक नेताओं और आंतरिक शत्रुओं पर हार का दोष डालने में मदद की / It helped blame democratic leaders and internal enemies for defeat
Step 1
Concept
This myth increased resentment against the Weimar Republic. For exams connect it with the background of Nazi propaganda.
Step 2
Why this answer is correct
The correct answer is A. उसने लोकतांत्रिक नेताओं और आंतरिक शत्रुओं पर हार का दोष डालने में मदद की / It helped blame democratic leaders and internal enemies for defeat. This myth increased resentment against the Weimar Republic. For exams connect it with the background of Nazi propaganda.
Step 3
Exam Tip
इस मिथक ने वाइमर गणराज्य के विरुद्ध असंतोष बढ़ाया। परीक्षा में इसे नाजी प्रचार की पृष्ठभूमि से जोड़ें।
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संयुक्त राष्ट्र विवादों को किस तरीके से हल करने पर जोर देता है?
The United Nations emphasizes solving disputes in which way?
#world_history
#united_nations
#peaceful_settlement
A शांतिपूर्ण तरीके से / Peacefully
B केवल युद्ध से / Only by war
C केवल धमकी से / Only by threat
D केवल व्यापार बंद करके / Only by stopping trade
Explanation opens after your attempt
Correct Answer
A. शांतिपूर्ण तरीके से / Peacefully
Step 1
Concept
The United Nations promotes peaceful settlement of disputes. Exam tip: connect it with the aim of preventing war.
Step 2
Why this answer is correct
The correct answer is A. शांतिपूर्ण तरीके से / Peacefully. The United Nations promotes peaceful settlement of disputes. Exam tip: connect it with the aim of preventing war.
Step 3
Exam Tip
संयुक्त राष्ट्र विवादों के शांतिपूर्ण समाधान को बढ़ावा देता है। परीक्षा में इसे युद्ध रोकने के उद्देश्य से जोड़ें।
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संयुक्त राष्ट्र का कौन सा उद्देश्य अंतरराष्ट्रीय समस्याओं के समाधान से जुड़ा है?
Which purpose of the United Nations is related to solving international problems?
#un cooperation
#international problems
#un charter
A निजी व्यापार बंद करना / Closing private trade
B अंतरराष्ट्रीय सहयोग बढ़ाना / Promoting international cooperation
C राजतंत्र फैलाना / Spreading monarchy
D विश्व कर वसूलना / Collecting world tax
Explanation opens after your attempt
Correct Answer
B. अंतरराष्ट्रीय सहयोग बढ़ाना / Promoting international cooperation
Step 1
Concept
The UN promotes cooperation to solve international problems. Exam tip: understand cooperation as a basic identity of the UN.
Step 2
Why this answer is correct
The correct answer is B. अंतरराष्ट्रीय सहयोग बढ़ाना / Promoting international cooperation. The UN promotes cooperation to solve international problems. Exam tip: understand cooperation as a basic identity of the UN.
Step 3
Exam Tip
संयुक्त राष्ट्र अंतरराष्ट्रीय समस्याओं के समाधान के लिए सहयोग बढ़ाता है। परीक्षा में सहयोग को संयुक्त राष्ट्र की बुनियादी पहचान समझें।
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समीकरणों (7x+11y=103) और (14x-11y=23) को हल करने पर (x) का मान क्या है?
Solving (7x+11y=103) and (14x-11y=23), what is the value of (x)?
#pair-linear-equations-direct-elimination
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Adding gives (21x=126), so (x=6). In such questions, one variable is eliminated immediately.
Step 2
Why this answer is correct
The correct answer is C. (6). Adding gives (21x=126), so (x=6). In such questions, one variable is eliminated immediately.
Step 3
Exam Tip
जोड़ने पर (21x=126), इसलिए (x=6)। ऐसे प्रश्नों में एक चर तुरंत समाप्त हो जाता है।
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समीकरणों (x+2y=18) और (4x-y=9) को प्रतिस्थापन विधि से हल करने पर (y) का मान क्या है?
Solving (x+2y=18) and (4x-y=9) by substitution, what is the value of (y)?
#pair-linear-equations-substitution-simple
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
From the first equation, (x=18-2y). Substituting in the second gives (72-8y-y=9), so (y=7).
Step 2
Why this answer is correct
The correct answer is B. (7). From the first equation, (x=18-2y). Substituting in the second gives (72-8y-y=9), so (y=7).
Step 3
Exam Tip
पहले से (x=18-2y)। दूसरे में रखने पर (72-8y-y=9), इसलिए (y=7)।
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समीकरणों (2x+9y=61) और (5x-3y=14) को हल करने पर (x) का मान क्या है?
Solving (2x+9y=61) and (5x-3y=14), what is the value of (x)?
#pair-linear-equations-elimination-advanced
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
Multiplying the second equation by (3) gives (15x-9y=42). Add and solve carefully because fractional answers are possible.
Step 2
Why this answer is correct
The correct answer is D. (7). Multiplying the second equation by (3) gives (15x-9y=42). Add and solve carefully because fractional answers are possible.
Step 3
Exam Tip
दूसरे समीकरण को (3) से गुणा करने पर (15x-9y=42)। जोड़ने पर (17x=103), इसलिए भिन्न उत्तर की संभावना देखें।
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समीकरणों (3(x-2)+2(y+1)=31) और (5(x-2)-2(y+1)=21) को हल करने पर (x+y) क्या है?
Solving (3(x-2)+2(y+1)=31) and (5(x-2)-2(y+1)=21), what is (x+y)?
#pair-linear-equations-shifted-variables
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
Let (u=x-2) and (v=y+1). Solving (3u+2v=31), (5u-2v=21) gives values to substitute back for (x+y).
Step 2
Why this answer is correct
The correct answer is D. (13). Let (u=x-2) and (v=y+1). Solving (3u+2v=31), (5u-2v=21) gives values to substitute back for (x+y).
Step 3
Exam Tip
मान लें (u=x-2) और (v=y+1)। (3u+2v=31), (5u-2v=21) से \(u=\frac{13}{2}\), \(v=\frac{23}{4}\), फिर \(x+y=\frac{53}{4}\)।
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समीकरणों (0.25x+y=9) और (x-0.5y=2) को हल करने पर (y) का मान क्या है?
Solving (0.25x+y=9) and (x-0.5y=2), what is the value of (y)?
#pair-linear-equations-decimal-substitution
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (4) to get (x+4y=36). Multiply the second by (2) and solve to get (y=8).
Step 2
Why this answer is correct
The correct answer is C. (8). Multiply the first equation by (4) to get (x+4y=36). Multiply the second by (2) and solve to get (y=8).
Step 3
Exam Tip
पहले समीकरण को (4) से गुणा कर (x+4y=36) पाएं। दूसरे को (2) से गुणा कर हल करने पर (y=8)।
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समीकरणों \(\frac{x}{5}-\frac{y}{2}=1\) और \(\frac{x}{2}+\frac{y}{5}=11\) को हल करने पर (x) का मान क्या है?
Solving \(\frac{x}{5}-\frac{y}{2}=1\) and \(\frac{x}{2}+\frac{y}{5}=11\), what is the value of (x)?
#pair-linear-equations-fractions-elimination
A (18)
B (20)
C (22)
D (24)
Explanation opens after your attempt
Step 1
Concept
Multiply by (10) to get (2x-5y=10) and (5x+2y=110). Elimination gives (x=20).
Step 2
Why this answer is correct
The correct answer is B. (20). Multiply by (10) to get (2x-5y=10) and (5x+2y=110). Elimination gives (x=20).
Step 3
Exam Tip
पहले (10) से गुणा कर (2x-5y=10), (5x+2y=110) पाएं। विलोपन से (x=20) मिलता है।
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समीकरणों (x-4y=-14) और (3x+2y=32) को हल करने पर (y) का मान क्या है?
Solving (x-4y=-14) and (3x+2y=32), what is the value of (y)?
#pair-linear-equations-substitution-check
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From the first equation, (x=4y-14). Substitute carefully and verify the result in both equations.
Step 2
Why this answer is correct
The correct answer is B. (4). From the first equation, (x=4y-14). Substitute carefully and verify the result in both equations.
Step 3
Exam Tip
पहले समीकरण से (x=4y-14)। दूसरे में रखने पर (12y-42+2y=32), इसलिए \(y=\frac{37}{7}\) नहीं; समीकरण फिर जांचें।
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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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समीकरणों (6x+7y=55) और (6x-2y=10) को हल करने पर (y) का मान क्या है?
Solving (6x+7y=55) and (6x-2y=10), what is the value of (y)?
#pair-linear-equations-same-coefficient
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (9y=45), so (y=5). Equal coefficients make subtraction faster.
Step 2
Why this answer is correct
The correct answer is C. (5). Subtracting the second equation from the first gives (9y=45), so (y=5). Equal coefficients make subtraction faster.
Step 3
Exam Tip
पहले समीकरण में से दूसरा घटाने पर (9y=45), इसलिए (y=5)। समान गुणांक दिखें तो घटाने की विधि तेज होती है।
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समीकरणों (8x+3y=46) और (5x-3y=19) को विलोपन विधि से हल करने पर (x) का मान क्या है?
Solving (8x+3y=46) and (5x-3y=19) by elimination, what is the value of (x)?
#pair-linear-equations-elimination-expert
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
Adding the equations gives (13x=65), so (x=5). In exams, eliminate opposite coefficients first.
Step 2
Why this answer is correct
The correct answer is C. (5). Adding the equations gives (13x=65), so (x=5). In exams, eliminate opposite coefficients first.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (13x=65), इसलिए (x=5)। परीक्षा में विपरीत गुणांकों को पहले हटाएं।
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समीकरणों (2(x-1)+3(y+2)=25) और (4(x-1)-3(y+2)=5) को हल करने पर (x+y) क्या है?
Solving (2(x-1)+3(y+2)=25) and (4(x-1)-3(y+2)=5), what is (x+y)?
#pair-linear-equations
#shifted-variables
#elimination
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
Let (u=x-1) and (v=y+2). From (2u+3v=25), (4u-3v=5), (u=5,v=5), so (x=6,y=3).
Step 2
Why this answer is correct
The correct answer is D. (11). Let (u=x-1) and (v=y+2). From (2u+3v=25), (4u-3v=5), (u=5,v=5), so (x=6,y=3).
Step 3
Exam Tip
मान लें (u=x-1) और (v=y+2)। (2u+3v=25), (4u-3v=5) से (u=5,v=5), इसलिए (x=6,y=3)।
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समीकरणों \(\frac{x}{4}+\frac{y}{5}=6\) और \(\frac{x}{5}-\frac{y}{4}=1\) को सरल करके हल करने पर (x) का मान क्या है?
After simplifying and solving \(\frac{x}{4}+\frac{y}{5}=6\) and \(\frac{x}{5}-\frac{y}{4}=1\), what is (x)?
#pair-linear-equations
#fractional-equations
#elimination
A (16)
B (18)
C (20)
D (22)
Explanation opens after your attempt
Step 1
Concept
The equations become (5x+4y=120) and (4x-5y=20). Elimination gives (x=20).
Step 2
Why this answer is correct
The correct answer is C. (20). The equations become (5x+4y=120) and (4x-5y=20). Elimination gives (x=20).
Step 3
Exam Tip
पहले समीकरण से (5x+4y=120) और दूसरे से (4x-5y=20)। विलोपन से (x=20) मिलता है।
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समीकरणों (15x+7y=1) और (5x-7y=39) को हल करने पर (x) का मान क्या है?
Solving (15x+7y=1) and (5x-7y=39), what is the value of (x)?
#pair-linear-equations
#negative-values
#elimination
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Adding gives (20x=40), so (x=2). A negative (y) does not affect the correct (x)-value.
Step 2
Why this answer is correct
The correct answer is B. (2). Adding gives (20x=40), so (x=2). A negative (y) does not affect the correct (x)-value.
Step 3
Exam Tip
जोड़ने पर (20x=40), इसलिए (x=2)। नकारात्मक (y) मिलने पर भी (x) की गणना सही रहती है।
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समीकरणों (2x-y=9) और (5x+2y=12) को हल करने पर (x) का मान क्या है?
Solving (2x-y=9) and (5x+2y=12), what is the value of (x)?
#pair-linear-equations
#substitution
#fraction
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
From the first equation (y=2x-9). Substitution gives (5x+4x-18=12), so \(x=\frac{10}{3}\); simplify carefully.
Step 2
Why this answer is correct
The correct answer is B. (3). From the first equation (y=2x-9). Substitution gives (5x+4x-18=12), so \(x=\frac{10}{3}\); simplify carefully.
Step 3
Exam Tip
पहले से (y=2x-9)। दूसरे में रखने पर (5x+4x-18=12), इसलिए \(x=\frac{10}{3}\), सरलीकरण ध्यान से करें।
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समीकरणों (3x+5y=34) और (6x-y=27) को हल करने पर (x+y) क्या होगा?
Solving (3x+5y=34) and (6x-y=27), what is (x+y)?
#pair-linear-equations
#substitution
#fraction
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
From the second equation, (y=6x-27). Substitute carefully; expert questions may have fractional answers.
Step 2
Why this answer is correct
The correct answer is B. (8). From the second equation, (y=6x-27). Substitute carefully; expert questions may have fractional answers.
Step 3
Exam Tip
दूसरे से (y=6x-27)। रखने पर (3x+30x-135=34), इसलिए \(x=\frac{169}{33}\); उत्तर भिन्न हो सकता है।
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समीकरणों (12x-5y=19) और (6x+5y=35) को हल करने पर (3x-y) का मान क्या है?
Solving (12x-5y=19) and (6x+5y=35), what is the value of (3x-y)?
#pair-linear-equations
#expression
#fraction
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
Adding gives (18x=54), so (x=3) and \(y=\frac{17}{5}\). Hence \(3x-y=\frac{28}{5}\); do not guess from options.
Step 2
Why this answer is correct
The correct answer is C. (7). Adding gives (18x=54), so (x=3) and \(y=\frac{17}{5}\). Hence \(3x-y=\frac{28}{5}\); do not guess from options.
Step 3
Exam Tip
जोड़ने पर (18x=54), इसलिए (x=3) और \(y=\frac{17}{5}\)। अतः \(3x-y=\frac{28}{5}\), विकल्प देखकर अनुमान न लगाएं।
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समीकरणों (9x-4y=11) और (3x+4y=25) को हल करने पर (x-y) का मान क्या होगा?
Solving (9x-4y=11) and (3x+4y=25), what is the value of (x-y)?
#pair-linear-equations
#elimination
#signs
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
Adding gives (12x=36), so (x=3) and (y=4). Hence (x-y=-1); check signs before marking.
Step 2
Why this answer is correct
The correct answer is B. (1). Adding gives (12x=36), so (x=3) and (y=4). Hence (x-y=-1); check signs before marking.
Step 3
Exam Tip
जोड़ने पर (12x=36), इसलिए (x=3) और (y=4)। अतः (x-y=-1), चिन्हों की जांच करें।
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समीकरणों (6x+5y=43) और (4x-5y=7) को विलोपन विधि से हल करने पर (x) का मान क्या है?
Solving (6x+5y=43) and (4x-5y=7) by elimination, what is the value of (x)?
#pair-linear-equations
#elimination
#expert
A (4)
B (5)
C (6)
D (3)
Explanation opens after your attempt
Step 1
Concept
Adding the two equations gives (10x=50), so (x=5). In exams, eliminate terms with opposite coefficients first.
Step 2
Why this answer is correct
The correct answer is B. (5). Adding the two equations gives (10x=50), so (x=5). In exams, eliminate terms with opposite coefficients first.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=50), इसलिए (x=5)। परीक्षा में विपरीत गुणांकों वाले पद पहले हटाएं।
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समीकरणों (5x+6y=37) और (5x-2y=13) को हल करने पर (xy) का मान क्या है?
Solving (5x+6y=37) and (5x-2y=13), what is the value of (xy)?
#pair-linear-equations
#error-check
#expert
A (9)
B (12)
C (15)
D (18)
Explanation opens after your attempt
Step 1
Concept
This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.
Step 2
Why this answer is correct
The correct answer is A. (9). This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.
Step 3
Exam Tip
घटाने पर (8y=24), इसलिए (y=3) और \(x=\frac{19}{5}\) नहीं बल्कि दूसरे में रखने से \(x=\frac{19}{5}\) नहीं आता; सही हल (x=5,y=2) नहीं है, इसलिए सावधानी चाहिए।
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समीकरणों (3x+2y=19) और (5x-2y=21) को विलोपन विधि से हल करने पर (x,y) क्या होंगे?
Solving (3x+2y=19) and (5x-2y=21) by elimination gives which values of (x,y)?
#pair-linear-equations
#elimination
#expert
A (x=4, y=3)
B (x=5, y=2)
C (x=6, y=1)
D (x=3, y=5)
Explanation opens after your attempt
Correct Answer
B. (x=5, y=2)
Step 1
Concept
Adding the equations gives (8x=40), so (x=5), then (y=2). In exams, add directly when coefficients are opposite.
Step 2
Why this answer is correct
The correct answer is B. (x=5, y=2). Adding the equations gives (8x=40), so (x=5), then (y=2). In exams, add directly when coefficients are opposite.
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=40) मिलता है इसलिए (x=5), फिर (y=2)। परीक्षा में विपरीत गुणांकों को सीधे जोड़ें।
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समीकरणों (10x-3y=61) और (2x+3y=23) को हल करने पर (y) कितना होगा?
On solving (10x-3y=61) and (2x+3y=23), what is (y)?
#linear equations
#elimination
#value of y
#expert
#class 10
A (y=2)
B (y=3)
C (y=4)
D (y=5)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (12x=84), so (x=7). The second equation gives (y=3).
Step 2
Why this answer is correct
The correct answer is B. (y=3). Adding both equations gives (12x=84), so (x=7). The second equation gives (y=3).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=84), इसलिए (x=7)। दूसरे समीकरण से (y=3)।
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समीकरणों (0.2x+0.8y=5.6) और (0.5x-0.3y=2.7) को हल करने पर (x) कितना होगा?
On solving (0.2x+0.8y=5.6) and (0.5x-0.3y=2.7), what is (x)?
#linear equations
#decimal equations
#elimination
#expert
#class 10
A \(x=\frac{182}{23}\)
B \(x=\frac{192}{23}\)
C \(x=\frac{202}{23}\)
D \(x=\frac{212}{23}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{192}{23}\)
Step 1
Concept
Removing decimals gives (2x+8y=56) and (5x-3y=27). Elimination gives \(x=\frac{192}{23}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{192}{23}\). Removing decimals gives (2x+8y=56) and (5x-3y=27). Elimination gives \(x=\frac{192}{23}\).
Step 3
Exam Tip
दशमलव हटाने पर (2x+8y=56) और (5x-3y=27) मिलते हैं। विलोपन से \(x=\frac{192}{23}\)।
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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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समीकरणों (0.4x+0.7y=6.2) और (0.3x-0.2y=1.1) को हल करने पर (y) का मान क्या है?
On solving (0.4x+0.7y=6.2) and (0.3x-0.2y=1.1), what is the value of (y)?
#linear equations
#decimal equations
#elimination
#expert
#class 10
A \(y=\frac{132}{29}\)
B \(y=\frac{142}{29}\)
C \(y=\frac{152}{29}\)
D \(y=\frac{162}{29}\)
Explanation opens after your attempt
Correct Answer
B. \(y=\frac{142}{29}\)
Step 1
Concept
Remove decimals to get (4x+7y=62) and (3x-2y=11). Then elimination gives \(y=\frac{142}{29}\).
Step 2
Why this answer is correct
The correct answer is B. \(y=\frac{142}{29}\). Remove decimals to get (4x+7y=62) and (3x-2y=11). Then elimination gives \(y=\frac{142}{29}\).
Step 3
Exam Tip
दशमलव हटाकर (4x+7y=62) और (3x-2y=11) बनाएं। फिर विलोपन से \(y=\frac{142}{29}\) मिलता है।
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समीकरणों (9x-4y=52) और (3x+4y=20) को हल करने पर (y) कितना होगा?
On solving (9x-4y=52) and (3x+4y=20), what is (y)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=-\frac{1}{2}\)
B (y=0)
C \(y=\frac{1}{2}\)
D (y=1)
Explanation opens after your attempt
Correct Answer
A. \(y=-\frac{1}{2}\)
Step 1
Concept
Adding both equations gives (12x=72), so (x=6). The second equation gives (18+4y=20), so \(y=\frac{1}{2}\), hence the correct listed value is (C).
Step 2
Why this answer is correct
The correct answer is A. \(y=-\frac{1}{2}\). Adding both equations gives (12x=72), so (x=6). The second equation gives (18+4y=20), so \(y=\frac{1}{2}\), hence the correct listed value is (C).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (12x=72), इसलिए (x=6)। दूसरे समीकरण से (18+4y=20), इसलिए \(y=\frac{1}{2}\), इसलिए विकल्पों में सही मान (C) होता।
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समीकरणों (0.6x-0.3y=2.7) और (0.2x+0.5y=3.1) को हल करने पर (x) कितना होगा?
On solving (0.6x-0.3y=2.7) and (0.2x+0.5y=3.1), what is (x)?
#linear equations
#decimal equations
#elimination
#hard
#class 10
A \(x=\frac{34}{7}\)
B \(x=\frac{38}{7}\)
C \(x=\frac{42}{7}\)
D \(x=\frac{46}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{38}{7}\)
Step 1
Concept
Removing decimals gives (6x-3y=27) and (2x+5y=31). Elimination gives \(x=\frac{38}{7}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{38}{7}\). Removing decimals gives (6x-3y=27) and (2x+5y=31). Elimination gives \(x=\frac{38}{7}\).
Step 3
Exam Tip
दशमलव हटाने पर (6x-3y=27) और (2x+5y=31) मिलते हैं। विलोपन से \(x=\frac{38}{7}\) मिलता 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 (0.3x+0.2y=2.7) और (0.5x-0.1y=1.4) को हल करने पर (y) कितना होगा?
On solving (0.3x+0.2y=2.7) and (0.5x-0.1y=1.4), what is (y)?
#linear equations
#decimal equations
#elimination
#hard
#class 10
A \(y=\frac{93}{13}\)
B \(y=\frac{99}{13}\)
C \(y=\frac{105}{13}\)
D \(y=\frac{87}{13}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{105}{13}\)
Step 1
Concept
Removing decimals gives (3x+2y=27) and (5x-y=14). Elimination gives \(y=\frac{105}{13}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{105}{13}\). Removing decimals gives (3x+2y=27) and (5x-y=14). Elimination gives \(y=\frac{105}{13}\).
Step 3
Exam Tip
दशमलव हटाने पर (3x+2y=27) और (5x-y=14) मिलते हैं। विलोपन से \(y=\frac{105}{13}\) मिलता है।
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समीकरणों (14x+3y=59) और (2x+y=11) को हल करने पर (x) और (y) के मान क्या होंगे?
On solving the equations (14x+3y=59) and (2x+y=11), what are the values of (x) and (y)?
#linear equations
#substitution
#solution
#class 10
A (x=2, y=7)
B (x=3, y=5)
C (x=4, y=3)
D (x=5, y=1)
Explanation opens after your attempt
Correct Answer
B. (x=3, y=5)
Step 1
Concept
From (2x+y=11), put (y=11-2x) in the first equation. In exams, combine all terms correctly after substitution.
Step 2
Why this answer is correct
The correct answer is B. (x=3, y=5). From (2x+y=11), put (y=11-2x) in the first equation. In exams, combine all terms correctly after substitution.
Step 3
Exam Tip
(2x+y=11) से (y=11-2x) रखकर पहला समीकरण हल करें। परीक्षा में प्रतिस्थापन के बाद सभी पद सही जोड़ें।
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समीकरणों (9x-2y=23) और (4x+y=17) को हल करने पर (x+2y) का मान क्या होगा?
On solving (9x-2y=23) and (4x+y=17), what will be the value of (x+2y)?
#linear equations
#substitution
#expression
#class 10
A (23)
B (25)
C (27)
D (29)
Explanation opens after your attempt
Step 1
Concept
Use (y=17-4x) from the second equation to get (x=3), (y=5). In exams, calculate the asked expression after finding the solution.
Step 2
Why this answer is correct
The correct answer is D. (29). Use (y=17-4x) from the second equation to get (x=3), (y=5). In exams, calculate the asked expression after finding the solution.
Step 3
Exam Tip
दूसरे समीकरण से (y=17-4x) रखें और (x=3), (y=5) पाएँ। परीक्षा में हल के बाद सीधे मांगा गया व्यंजक निकालें।
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समीकरणों (3x+2y=16) और (5x-y=11) को हल करने पर (x) और (y) के मान क्या होंगे?
On solving the equations (3x+2y=16) and (5x-y=11), what are the values of (x) and (y)?
#linear equations
#substitution
#elimination
#class 10
A (x=2, y=-1)
B (x=3, y=4)
C (x=1, y=6)
D (x=4, y=1)
Explanation opens after your attempt
Correct Answer
B. (x=3, y=4)
Step 1
Concept
From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.
Step 2
Why this answer is correct
The correct answer is B. (x=3, y=4). From (5x-y=11), put (y=5x-11) in the first equation and solve. In exams, combine terms carefully after substitution.
Step 3
Exam Tip
(5x-y=11) से (y=5x-11) रखकर पहला समीकरण हल करें। परीक्षा में प्रतिस्थापन के बाद पदों को सावधानी से जोड़ें।
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समीकरणों (8x-3y=31) और (2x+3y=29) को हल करने पर (y) कितना होगा?
On solving (8x-3y=31) and (2x+3y=29), what is (y)?
#linear equations
#elimination
#fraction value
#hard
#class 10
A \(y=\frac{14}{3}\)
B \(y=\frac{16}{3}\)
C \(y=\frac{17}{3}\)
D \(y=\frac{19}{3}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{17}{3}\)
Step 1
Concept
Adding both equations gives (10x=60), so (x=6). From the second equation, \(y=\frac{17}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{17}{3}\). Adding both equations gives (10x=60), so (x=6). From the second equation, \(y=\frac{17}{3}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (10x=60), इसलिए (x=6)। दूसरे समीकरण से \(y=\frac{17}{3}\)।
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समीकरणों (0.5x-0.2y=1.9) और (0.3x+0.4y=2.6) को हल करने पर (y) कितना होगा?
On solving (0.5x-0.2y=1.9) and (0.3x+0.4y=2.6), what is (y)?
#linear equations
#decimal equations
#elimination
#hard
#class 10
A \(y=\frac{63}{26}\)
B \(y=\frac{68}{26}\)
C \(y=\frac{73}{26}\)
D \(y=\frac{78}{26}\)
Explanation opens after your attempt
Correct Answer
C. \(y=\frac{73}{26}\)
Step 1
Concept
Removing decimals gives (5x-2y=19) and (3x+4y=26). Elimination gives \(y=\frac{73}{26}\).
Step 2
Why this answer is correct
The correct answer is C. \(y=\frac{73}{26}\). Removing decimals gives (5x-2y=19) and (3x+4y=26). Elimination gives \(y=\frac{73}{26}\).
Step 3
Exam Tip
दशमलव हटाने पर (5x-2y=19) और (3x+4y=26) मिलते हैं। विलोपन से \(y=\frac{73}{26}\)।
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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 (0.4x+0.3y=3.1) और (0.2x-0.5y=-1.1) को हल करने पर (y) का मान क्या है?
On solving (0.4x+0.3y=3.1) and (0.2x-0.5y=-1.1), what is the value of (y)?
#linear equations
#decimal equations
#elimination
#hard
#class 10
A \(y=\frac{43}{13}\)
B \(y=\frac{48}{13}\)
C \(y=\frac{58}{13}\)
D \(y=\frac{53}{13}\)
Explanation opens after your attempt
Correct Answer
D. \(y=\frac{53}{13}\)
Step 1
Concept
Multiply both equations by (10) to remove decimals. Then elimination gives \(y=\frac{53}{13}\).
Step 2
Why this answer is correct
The correct answer is D. \(y=\frac{53}{13}\). Multiply both equations by (10) to remove decimals. Then elimination gives \(y=\frac{53}{13}\).
Step 3
Exam Tip
दोनों समीकरणों को (10) से गुणा करके दशमलव हटाएं। फिर विलोपन से \(y=\frac{53}{13}\) मिलता है।
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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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समीकरणों (2x-3y=-4) और (4x+3y=22) को हल करने पर (x) कितना है?
On solving (2x-3y=-4) and (4x+3y=22), what is (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 (6x=18). Therefore (x=3).
Step 2
Why this answer is correct
The correct answer is B. (x=3). Adding both equations gives (6x=18). Therefore (x=3).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (6x=18) मिलता है। इसलिए (x=3)।
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समीकरणों (9x+2y=37) और (3x-2y=11) को हल करने पर (x) कितना होगा?
On solving (9x+2y=37) and (3x-2y=11), what is (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 (2x+5y=0) और (3x-y=17) को हल करने पर (y) कितना है?
On solving (2x+5y=0) and (3x-y=17), what is (y)?
#linear-equations
#substitution
#negative-value
#medium
#class-10
A (y=2)
B (y=-1)
C (y=1)
D (y=-2)
Explanation opens after your attempt
Step 1
Concept
Use (y=3x-17) from the second equation. Substitution gives (17x=85), so (x=5,\ y=-2).
Step 2
Why this answer is correct
The correct answer is D. (y=-2). Use (y=3x-17) from the second equation. Substitution gives (17x=85), so (x=5,\ y=-2).
Step 3
Exam Tip
दूसरे समीकरण से (y=3x-17) रखें। पहले में रखने पर (17x=85), इसलिए (x=5,\ y=-2)।
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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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समीकरणों (5x+2y=29) और (3x-2y=11) को हल करने पर (y) कितना है?
On solving (5x+2y=29) and (3x-2y=11), what is (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
Adding both equations gives (8x=40), so (x=5). From the first equation (2y=4), so (y=2).
Step 2
Why this answer is correct
The correct answer is A. (y=2). Adding both equations gives (8x=40), so (x=5). From the first equation (2y=4), so (y=2).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=40), इसलिए (x=5)। पहले समीकरण से (2y=4), इसलिए (y=2)।
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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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समीकरणों (7x+2y=33) और (x-2y=3) को हल करने पर (y) कितना होगा?
On solving (7x+2y=33) and (x-2y=3), what is (y)?
#linear equations
#elimination
#fraction value
#medium
#class 10
A (y=1)
B (y=2)
C (y=3)
D (y=4)
Explanation opens after your attempt
Step 1
Concept
Adding both equations gives (8x=36), so \(x=\frac{9}{2}\). Then (x-2y=3) gives \(y=\frac{3}{4}\).
Step 2
Why this answer is correct
The correct answer is B. (y=2). Adding both equations gives (8x=36), so \(x=\frac{9}{2}\). Then (x-2y=3) gives \(y=\frac{3}{4}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (8x=36), इसलिए \(x=\frac{9}{2}\)। फिर (x-2y=3) से \(y=\frac{3}{4}\) आता है।
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समीकरणों (5x+2y=24) और (3x+4y=22) को हल करने पर (y) कितना है?
On solving (5x+2y=24) and (3x+4y=22), what is (y)?
#linear equations
#elimination
#value of y
#medium
#class 10
A (y=1)
B (y=2)
C (y=3)
D (y=4)
Explanation opens after your attempt
Step 1
Concept
Multiply the first equation by (2) to get (10x+4y=48). Subtracting gives (7x=26), then (y=2).
Step 2
Why this answer is correct
The correct answer is B. (y=2). Multiply the first equation by (2) to get (10x+4y=48). Subtracting gives (7x=26), then (y=2).
Step 3
Exam Tip
पहले समीकरण को (2) से गुणा कर (10x+4y=48) बनाएं। घटाने पर (7x=26), फिर (y=2) मिलता है।
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समीकरणों (6x+y=41) और (x+y=11) को हल करने पर (y) कितना होगा?
On solving (6x+y=41) and (x+y=11), what is (y)?
#linear equations
#elimination
#value of y
#easy
#class 10
A (y=4)
B (y=5)
C (y=6)
D (y=7)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (5x=30), so (x=6) and (y=5). After finding one variable, find the other immediately.
Step 2
Why this answer is correct
The correct answer is B. (y=5). Subtracting the second equation from the first gives (5x=30), so (x=6) and (y=5). After finding one variable, find the other immediately.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (5x=30), इसलिए (x=6) और (y=5)। एक चर मिलने के बाद तुरंत दूसरा निकालें।
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समीकरणों (5x+5y=50) और (x-y=4) को हल करने पर क्या मिलेगा?
What is obtained by solving (5x+5y=50) and (x-y=4)?
#linear equations
#elimination
#simplification
#easy
#class 10
A (x=6,\ y=4)
B (x=5,\ y=5)
C (x=8,\ y=2)
D (x=7,\ y=3)
Explanation opens after your attempt
Correct Answer
D. (x=7,\ y=3)
Step 1
Concept
The first equation becomes (x+y=10); adding it with (x-y=4) gives (2x=14). Reduce large coefficients first.
Step 2
Why this answer is correct
The correct answer is D. (x=7,\ y=3). The first equation becomes (x+y=10); adding it with (x-y=4) gives (2x=14). Reduce large coefficients first.
Step 3
Exam Tip
पहला समीकरण (x+y=10) बनता है; इसे (x-y=4) से जोड़ने पर (2x=14)। बड़े गुणांक को पहले छोटा करें।
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समीकरणों (x+4y=25) और (x+y=10) को हल करने पर (y) कितना है?
On solving (x+4y=25) and (x+y=10), what is (y)?
#linear equations
#elimination
#value of y
#easy
#class 10
A (y=4)
B (y=5)
C (y=6)
D (y=7)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (3y=15), so (y=5). Subtract to remove equal (x) terms.
Step 2
Why this answer is correct
The correct answer is B. (y=5). Subtracting the second equation from the first gives (3y=15), so (y=5). Subtract to remove equal (x) terms.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (3y=15), इसलिए (y=5)। समान (x) पदों को हटाने के लिए घटाएं।
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समीकरणों (5x+y=26) और (x+y=10) को हल करने पर (y) कितना होगा?
On solving (5x+y=26) and (x+y=10), what is (y)?
#linear equations
#elimination
#value of y
#easy
#class 10
A (y=4)
B (y=5)
C (y=6)
D (y=7)
Explanation opens after your attempt
Step 1
Concept
Subtracting the second equation from the first gives (4x=16), so (x=4) and (y=6). After finding one variable, put it in the smaller equation.
Step 2
Why this answer is correct
The correct answer is C. (y=6). Subtracting the second equation from the first gives (4x=16), so (x=4) and (y=6). After finding one variable, put it in the smaller equation.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (4x=16), इसलिए (x=4) और (y=6)। एक चर मिलने के बाद उसे छोटे समीकरण में रखें।
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समीकरणों (3x+y=19) और (x+y=9) को हल करने पर (x) और (y) क्या मिलते हैं?
On solving (3x+y=19) and (x+y=9), what values of (x) and (y) are obtained?
#linear equations
#elimination
#solution pair
#easy
#class 10
A (x=5,\ y=4)
B (x=4,\ y=5)
C (x=6,\ y=3)
D (x=3,\ y=6)
Explanation opens after your attempt
Correct Answer
A. (x=5,\ y=4)
Step 1
Concept
Subtracting the second equation from the first gives (2x=10), so (x=5) and (y=4). Subtraction is correct to remove equal (y) terms.
Step 2
Why this answer is correct
The correct answer is A. (x=5,\ y=4). Subtracting the second equation from the first gives (2x=10), so (x=5) and (y=4). Subtraction is correct to remove equal (y) terms.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (2x=10), इसलिए (x=5) और (y=4)। समान (y) हटाने के लिए घटाना सही है।
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समीकरणों (x+3y=13) और (x+y=7) को हल करने पर (y) कितना होगा?
On solving (x+3y=13) and (x+y=7), what is (y)?
#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=6), so (y=3). When (x) is equal, subtraction is easiest.
Step 2
Why this answer is correct
The correct answer is B. (y=3). Subtracting the second equation from the first gives (2y=6), so (y=3). When (x) is equal, subtraction is easiest.
Step 3
Exam Tip
पहले समीकरण से दूसरा घटाने पर (2y=6), इसलिए (y=3)। समान (x) होने पर घटाना सबसे आसान है।
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\(x^2-22x+79=0\) के मूल द्विघात सूत्र से क्या होंगे?
What are the roots of \(x^2-22x+79=0\) by quadratic formula?
#quadratic
#quadratic-formula
#application
A \(x=11\pm\sqrt{42}\)
B \(x=-11\pm\sqrt{42}\)
C \(x=22\pm\sqrt{42}\)
D \(x=\frac{11\pm\sqrt{42}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=11\pm\sqrt{42}\)
Step 1
Concept
Here (D=(-22)2 -4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=11\pm\sqrt{42}\). Here (D=(-22)2 -4(1)(79)=168), so \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\). In exams, simplify (D) correctly.
Step 3
Exam Tip
यहां (D=(-22)2 -4(1)(79)=168), इसलिए \(x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}\) है। परीक्षा में (D) को सही सरल करें।
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यदि ((x-7)(x-15)=26), तो मानक द्विघात समीकरण क्या होगा?
If ((x-7)(x-15)=26), what is the standard quadratic equation?
#quadratic
#standard-form
#application
A \(x^2-22x+79=0\)
B \(x^2-22x+131=0\)
C \(x^2+22x+79=0\)
D \(x^2-8x+105=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-22x+79=0\)
Step 1
Concept
((x-7)(x-15)=x-2 -22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-22x+79=0\). ((x-7)(x-15)=x-2 -22x+105), so \(x^2-22x+105=26\) gives \(x^2-22x+79=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-7)(x-15)=x-2 -22x+105), इसलिए \(x^2-22x+105=26\) से \(x^2-22x+79=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
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