Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 541-550 of 600 questions.
B. (₹25)
मूल कीमत (x) हो, तो \(\frac{300}{x-5}-\frac{300}{x}=3\)। इससे \(x^2-5x-500=0\), इसलिए (x=25)। / Let original price be (x), then \(\frac{300}{x-5}-\frac{300}{x}=3\). This gives \(x^2-5x-500=0\), so (x=25).
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B. (27 m)
चौड़ाई (x) हो, तो वृद्धि ((x+5)(x+15)-x(x+10)=350)। इससे (10x+75=350), इसलिए (x=27.5), अतः दिए विकल्पों में कोई सही नहीं। / Let breadth be (x), then increase is ((x+5)(x+15)-x(x+10)=350). This gives (10x+75=350), so (x=27.5), hence no listed option is exact.
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A. (4)
प्रति कमरा (x) छात्र हों, तो \(\frac{60}{x}-\frac{60}{x+2}=5\)। इससे \(5x^2+10x-120=0\), इसलिए (x=4)। / Let students per room be (x), then \(\frac{60}{x}-\frac{60}{x+2}=5\). This gives \(5x^2+10x-120=0\), so (x=4).
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The sum of a positive number and its square is (156). What is the number?
C. (12)
संख्या (x) हो, तो \(x^2+x=156\)। इससे \(x^2+x-156=0\), इसलिए धनात्मक मूल (12) है। / Let the number be (x), then \(x^2+x=156\). This gives \(x^2+x-156=0\), so the positive root is (12).
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B. (25 m)
चौड़ाई (x) हो, तो लंबाई (2x-5) और (x(2x-5)=875)। इससे \(2x^2-5x-875=0\), इसलिए (x=25)। / Let breadth be (x), then length is (2x-5), and (x(2x-5)=875). This gives \(2x^2-5x-875=0\), so (x=25).
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A. \((2\text{ km\)h})
धारा की चाल (x) हो, तो \(\frac{48}{10+x}+\frac{32}{10-x}=10\)। इससे \(x^2+4x-12=0\), इसलिए (x=2)। / Let the stream speed be (x), then \(\frac{48}{10+x}+\frac{32}{10-x}=10\). This gives \(x^2+4x-12=0\), so (x=2).
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B. (12 cm)
मूल चौड़ाई (x) हो, तो डिब्बे का आधार ((x-4)(x+4)) और ऊँचाई (2) है। (2(x-4)(x+4)=240) से \(x^2=136\), इसलिए विकल्पों में कोई सही नहीं। / Let original breadth be (x), then the box base is ((x-4)(x+4)) and height is (2). From (2(x-4)(x+4)=240), \(x^2=136\), so none of the options is exact.
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B. \((50\text{ km\)h})
वास्तविक चाल (x km/h) मानें, तब \(\frac{300}{x}-\frac{300}{x+10}=1\)। इससे \(x^2+10x-3000=0\), इसलिए धनात्मक मूल (x=50) है। / Let the actual speed be (x km/h\(), then (\frac{300}{x}-\frac{300}{x+10}=1). This gives (x^2+10x-3000=0), so the positive root is (x=50).\)
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C. (68 m)
चौड़ाई (x m) मानें, तब (x(x+8)=273)। \(इससे (x=13) और लंबाई (21) है, इसलिए परिधि (2(13+21)=68\) m) है। / Let the breadth be (x m\(), then (x(x+8)=273). This gives (x=13) and length (21), so the perimeter is (2(13+21)=68\) m).
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A. (10 days)
पहले मजदूर का समय (x days) मानें, तब \(\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\)। इससे \(x^2-7x-30=0\), इसलिए (x=10) है। / Let the first worker's time be (x days\(), then (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}). This gives (x^2-7x-30=0), so (x=10).\)
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