100 results found for "fraction square root" in Class 10.
एक भिन्न में हर अंश से (5) अधिक है। यदि अंश में (3) और हर में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (3) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#word-problem-fraction-substitution
A \(\frac{7}{12}\)
B \(\frac{8}{13}\)
C \(\frac{9}{14}\)
D \(\frac{10}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{12}\)
Step 1
Concept
Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{12}\). Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 3
Exam Tip
अंश (x) और हर (x+5) लें। \(\frac{x+3}{x+6}=\frac{2}{3}\) से (x=3), इसलिए मूल भिन्न \(\frac{3}{8}\) नहीं; विकल्प जांचें।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश और हर दोनों में (2) जोड़ने पर भिन्न \(\frac{4}{5}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to both numerator and denominator, the fraction becomes \(\frac{4}{5}\). What is the original fraction?
#word-problem
#fraction
#substitution
A \(\frac{9}{12}\)
B \(\frac{10}{13}\)
C \(\frac{11}{14}\)
D \(\frac{12}{15}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{10}{13}\)
Step 1
Concept
Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10}{13}\). Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 3
Exam Tip
मान लें हर (y) है तो अंश (y-3)। \(\frac{y-1}{y+2}=\frac{4}{5}\) से (y=13), इसलिए भिन्न \(\frac{10}{13}\) है।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश में (2) और हर में (1) जोड़ने पर भिन्न \(\frac{3}{4}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{3}{4}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{5}{8}\)
B \(\frac{6}{9}\)
C \(\frac{7}{10}\)
D \(\frac{8}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{7}{10}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7}{10}\). Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y-x=3) और \(\frac{x+2}{y+1}=\frac{3}{4}\) बनता है। परीक्षा में क्रॉस गुणा के बाद सरल रैखिक समीकरण हल करें।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश और हर दोनों में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (1) is added to both numerator and denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{9}{14}\)
B \(\frac{8}{13}\)
C \(\frac{7}{12}\)
D \(\frac{6}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{14}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{14}\). Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y=x+5) और \(\frac{x+1}{y+1}=\frac{2}{3}\) बनता है। परीक्षा में भिन्न को समीकरण में बदलते समय क्रॉस गुणा करें।
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एक भिन्न का हर अंश से (3) अधिक है। यदि भिन्न और उसके व्युत्क्रम का योग \(\frac{29}{10}\) है तो अंश क्या है?
The denominator of a fraction is (3) more than its numerator. If the sum of the fraction and its reciprocal is \(\frac{29}{10}\), what is the numerator?
#quadratic equations
#fraction
#reciprocal
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 2
Why this answer is correct
The correct answer is A. (2). The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 3
Exam Tip
भिन्न \(\frac{x}{x+3}\) है। \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\) से (x=2) या (x=15) आता है और विकल्पों में (2) सही है।
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एक धनात्मक भिन्न का हर अंश से (4) अधिक है। भिन्न और उसके व्युत्क्रम का योग \(\frac{41}{20}\) है। भिन्न क्या है?
In a positive fraction, the denominator is (4) more than the numerator. The sum of the fraction and its reciprocal is \(\frac{41}{20}\). What is the fraction?
#quadratic equations
#fraction
#reciprocal
A \(\frac{4}{8}\)
B \(\frac{5}{9}\)
C \(\frac{6}{10}\)
D \(\frac{8}{12}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{5}{9}\)
Step 1
Concept
Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5}{9}\). Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 3
Exam Tip
भिन्न \(\frac{x}{x+4}\) हो, तो \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\)। इससे (x=5), इसलिए भिन्न \(\frac{5}{9}\) है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
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कौन सा वर्गमूल परिमेय है, इसलिए उस पर \(\sqrt{2}\) जैसी अपरिमेयता सिद्धि लागू नहीं होती?
Which square root is rational, so an irrationality proof like \(\sqrt{2}\) does not apply to it?
#perfect square
#rational square root
#class 10
A \(\sqrt{2}\)
B \(\sqrt{3}\)
C \(\sqrt{4}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{4}\)
Step 1
Concept
(4) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{4}=2\), which is rational.
Step 3
Exam Tip
Square roots of perfect squares are rational. चरण 1: (4) पूर्ण वर्ग है। चरण 2: \(\sqrt{4}=2\), जो परिमेय संख्या है। चरण 3: पूर्ण वर्गों के वर्गमूल परिमेय होते हैं।
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कौन सा वर्गमूल इस अध्याय की अपरिमेयता सिद्धि का उदाहरण नहीं है क्योंकि वह परिमेय है?
Which square root is not an example of irrationality proof in this chapter because it is rational?
#rational square root
#perfect square
#class 10
A \(\sqrt{9}\)
B \(\sqrt{2}\)
C \(\sqrt{3}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{9}\)
Step 1
Concept
(9) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{9}=3\), which is rational.
Step 3
Exam Tip
A square root of a perfect square does not need an irrationality proof. चरण 1: (9) पूर्ण वर्ग है। चरण 2: \(\sqrt{9}=3\), जो परिमेय है। चरण 3: पूर्ण वर्ग के वर्गमूल को अपरिमेय सिद्ध करने की जरूरत नहीं होती।
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कौन सा वर्गमूल परिमेय है, इसलिए \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) जैसी सिद्धि की जरूरत नहीं है?
Which square root is rational, so it does not need a proof like \(\sqrt{2}\), \(\sqrt{3}\), or \(\sqrt{5}\)?
#rational square root
#perfect square
#class 10
A \(\sqrt{4}\)
B \(\sqrt{2}\)
C \(\sqrt{3}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{4}\)
Step 1
Concept
(4) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{4}=2\), which is rational.
Step 3
Exam Tip
The square root of a perfect square is not proved irrational. चरण 1: (4) पूर्ण वर्ग है। चरण 2: \(\sqrt{4}=2\), जो परिमेय संख्या है। चरण 3: पूर्ण वर्ग के वर्गमूल को अपरिमेय सिद्ध नहीं करते।
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निम्न में से कौन सी संख्या पूर्ण वर्ग नहीं है और उसका वर्गमूल अपरिमेय सिद्ध किया जाता है?
Which of the following is not a perfect square and its square root is proved irrational?
#perfect square
#sqrt2
#class 10
A (2)
B (4)
C (9)
D (25)
Explanation opens after your attempt
Step 1
Concept
(4), (9), and (25) are perfect squares.
Step 2
Why this answer is correct
(2) is not a perfect square, so \(\sqrt{2}\) is proved irrational.
Step 3
Exam Tip
First identify perfect and non-perfect squares. चरण 1: (4), (9) और (25) पूर्ण वर्ग हैं। चरण 2: (2) पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{2}\) अपरिमेय सिद्ध किया जाता है। चरण 3: पूर्ण वर्ग और अपूर्ण वर्ग की पहचान पहले करें।
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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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\(5x^2=80\) 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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\(3x^2=12\) 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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\(x^2=169\) 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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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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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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वर्गमूल विधि से \(x^2=64\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=64\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm8\)
B (x=8)
C (x=-8)
D \(x=\pm32\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm8\)
Step 1
Concept
\(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm8\). \(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{64}=\pm8\) होता है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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किस समीकरण को वर्गमूल विधि से सीधे हल किया जा सकता है?
Which equation can be solved directly by square root method?
#quadratic
#square-root-method
#method-selection
A ((x-2)2 =9)
B \(x^2+5x+6=0\)
C \(x^2+2x+1=0\)
D \(2x^2+3x+1=0\)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =9)
Step 1
Concept
\(In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 2
Why this answer is correct
\(The correct answer is A. ((x-2)^2=9). In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 3
Exam Tip
((x-2)2 =9) में सीधे वर्गमूल लिया जा सकता है। परीक्षा में ((expression\()^2=k) रूप को पहचानें\)।
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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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यदि \(x^2=16\), तो वर्गमूल विधि से (x) का मान क्या होगा?
If \(x^2=16\), what is the value of (x) by square root method?
#quadratic
#square-root-method
#roots
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm8\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x^2=16\) से \(x=\pm\sqrt{16}=\pm4\) मिलता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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यदि \(a=7+4\sqrt{3}\), तो कौन-सा विकल्प (a) का वर्गमूल दर्शाता है?
If \(a=7+4\sqrt{3}\), which option represents a square root of (a)?
#surd square root
#identity
#class 10
A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(\sqrt{7}+2\)
D \(\sqrt{3}+1\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
(\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3).
Step 2
Why this answer is correct
This equals \(7+4\sqrt{3}\).
Step 3
Exam Tip
In such questions, identify the form \(m+n+2\sqrt{mn}\). चरण 1: (\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3)। चरण 2: यह \(7+4\sqrt{3}\) के बराबर है। चरण 3: ऐसे प्रश्नों में \(m+n+2\sqrt{mn}\) का रूप पहचानें।
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\(2^6 \times 5^2\) का वर्गमूल क्या होगा?
What is the square root of \(2^6 \times 5^2\)?
#real-numbers
#square-root
#prime-factorisation
A \(2^2 \times 5\)
B \(2^3 \times 5\)
C \(2^3 \times 5^2\)
D \(2^6 \times 5\)
Explanation opens after your attempt
Correct Answer
B. \(2^3 \times 5\)
Step 1
Concept
When taking a square root, halve the prime exponents.
Step 2
Why this answer is correct
\(2^6\) becomes \(2^3\), and \(5^2\) becomes (5).
Step 3
Exam Tip
In square roots, the base does not change; the exponent is halved. चरण 1: वर्गमूल लेते समय अभाज्य घातों को आधा करते हैं। चरण 2: \(2^6\) से \(2^3\) और \(5^2\) से (5) मिलता है। चरण 3: वर्गमूल में आधार नहीं बदलता, घात आधी होती है।
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\(2^4 \times 3^2\) का वर्गमूल क्या होगा?
What is the square root of \(2^4 \times 3^2\)?
#real-numbers
#square-root
#prime-factorisation
A \(2^2 \times 3\)
B \(2^4 \times 3\)
C \(2^2 \times 3^2\)
D \(2 \times 3\)
Explanation opens after your attempt
Correct Answer
A. \(2^2 \times 3\)
Step 1
Concept
When taking a square root, halve all prime exponents.
Step 2
Why this answer is correct
\(2^4\) becomes \(2^2\) and \(3^2\) becomes (3).
Step 3
Exam Tip
In square roots, halve the exponent, not the base. चरण 1: वर्गमूल लेते समय सभी अभाज्य घातों को आधा करते हैं। चरण 2: \(2^4\) से \(2^2\) और \(3^2\) से (3) मिलता है। चरण 3: वर्गमूल में आधार को नहीं, घात को आधा करें।
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यदि किसी संख्या का अभाज्य गुणनखंड रूप \(2^6\times3^4\times5^2\) है, तो उसका वर्गमूल क्या होगा?
If a number has prime factorisation \(2^6\times3^4\times5^2\), what is its square root?
#real numbers
#square root
#prime factorisation
#exponents
A \(2^3\times3^2\times5\)
B \(2^2\times3^2\times5\)
C \(2^3\times3\times5^2\)
D \(2^6\times3^2\times5\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5\)
Step 1
Concept
In square root, each prime exponent becomes half.
Step 2
Why this answer is correct
\(2^6\) becomes \(2^3\), \(3^4\) becomes \(3^2\), and \(5^2\) becomes (5).
Step 3
Exam Tip
This direct method works when all exponents are even. चरण 1: वर्गमूल लेते समय हर अभाज्य गुणनखंड की घात आधी हो जाती है। चरण 2: \(2^6\) से \(2^3\), \(3^4\) से \(3^2\), और \(5^2\) से (5) मिलेगा। चरण 3: यह विधि तभी सीधे लागू होती है जब सभी घातें सम हों।
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एक भिन्न का हर अंश से (4) अधिक है। यदि अंश और हर दोनों में (2) जोड़ने पर भिन्न \(\frac{3}{5}\) हो जाती है तो मूल अंश क्या है?
The denominator of a fraction is (4) more than its numerator. If (2) is added to both numerator and denominator, the fraction becomes \(\frac{3}{5}\). What is the original numerator?
#quadratic equations
#fraction word problem
#linear reduction
A (4)
B (5)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
The numerator is (x) and denominator is (x+4). From \(\frac{x+2}{x+6}=\frac{3}{5}\), (x=4).
Step 2
Why this answer is correct
The correct answer is A. (4). The numerator is (x) and denominator is (x+4). From \(\frac{x+2}{x+6}=\frac{3}{5}\), (x=4).
Step 3
Exam Tip
अंश (x) और हर (x+4) है। \(\frac{x+2}{x+6}=\frac{3}{5}\) से (x=4) मिलता है।
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यदि (4x-2 -(4h+1)x+h=0) की एक जड़ \(\frac{1}{4}\) है, तो दूसरी जड़ क्या है?
If one root of (4x-2 -(4h+1)x+h=0) is \(\frac{1}{4}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{4}\)
C (4h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{4}\) है। एक जड़ \(\frac{1}{4}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि (x-2 -(m+9)x+9m=0) की एक जड़ (9) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+9)x+9m=0) is (9), what is the other root?
#quadratic-roots
#other-root
#parameter
A (m)
B (9m)
C (m+9)
D \(\frac{m}{9}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 3
Exam Tip
जड़ों का गुणनफल (9m) है। एक जड़ (9) है, इसलिए दूसरी जड़ \(\frac{9m}{9}=m\) होगी।
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यदि (2x-2 -(3p+2)x+p(p+2)=0) की एक जड़ (p) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(3p+2)x+p(p+2)=0) is (p), what will be the other root?
#quadratic-roots
#other-root
#parametric-equation
A \(\frac{p+2}{2}\)
B (p+2)
C \(\frac{p}{2}\)
D (2p+2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{p+2}{2}\)
Step 1
Concept
The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{p+2}{2}\). The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 3
Exam Tip
जड़ों का गुणनफल (\frac{p(p+2)}{2}) है। एक जड़ (p) होने पर दूसरी जड़ \(\frac{p+2}{2}\) होगी।
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यदि (3x-2 -(3h+1)x+h=0) की एक जड़ \(\frac{1}{3}\) है, तो दूसरी जड़ क्या है?
If one root of (3x-2 -(3h+1)x+h=0) is \(\frac{1}{3}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{3}\)
C (3h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{3}\) है। एक जड़ \(\frac{1}{3}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि (2x-2 -(h+1)x+h=0) की जड़ों में से एक हमेशा \(\frac{1}{2}\) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(h+1)x+h=0) is always \(\frac{1}{2}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{2}\)
C (2h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 2
Why this answer is correct
The correct answer is A. (h). The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 3
Exam Tip
गुणनफल \(\frac{h}{2}\) है। एक जड़ \(\frac{1}{2}\) है, इसलिए दूसरी जड़ \(\frac{h}{2}\div\frac{1}{2}=h\) होगी।
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यदि (x-2 -(m+2)x+3m=0) की एक जड़ (3) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+2)x+3m=0) is (3), what is the other root?
#quadratic-roots
#other-root
#parametric-equation
A (m)
B (3m)
C (m+2)
D \(\frac{m}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 2
Why this answer is correct
The correct answer is A. (m). Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 3
Exam Tip
(x=3) रखने पर समीकरण हर (m) के लिए सही हो जाता है। गुणनफल (3m) है और एक जड़ (3), इसलिए दूसरी जड़ (m) है।
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यदि (x-2 -(m-2 )x+m-6 =0) की एक जड़ (3) है, तो दूसरी जड़ क्या होगी?
If one root of (x-2 -(m-2 )x+m-6 =0) is (3), what is the other root?
#quadratic-roots
#other-root
#error-check
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 2
Why this answer is correct
The correct answer is A. (1). Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 3
Exam Tip
(x=3) रखने पर (9-3(m-2 )+m-6 =0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।
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यदि (x-2 -(m+7)x+7m=0) का एक मूल (7) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+7)x+7m=0) is (7), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (7m)
C (m+7)
D (m-7 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (7m) है और एक मूल (7) है। इसलिए दूसरा मूल \(\frac{7m}{7}=m\) है।
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यदि \(x^2+ax+54=0\) का एक मूल (6) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+54=0\) is (6), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
B दूसरा मूल (-9), (a=3) / other root (-9), (a=3)
C दूसरा मूल (9), (a=15) / other root (9), (a=15)
D दूसरा मूल (-6), (a=0) / other root (-6), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
Step 1
Concept
The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15). The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 3
Exam Tip
मूलों का गुणनफल (54) है, इसलिए दूसरा मूल (9) होगा। योग (15) है और (-a=15), इसलिए (a=-15)।
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यदि (x-2 -(m+6)x+6m=0) का एक मूल (6) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+6)x+6m=0) is (6), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (6m)
C (m+6)
D (m-6 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (6m) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{6m}{6}=m\) है।
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यदि \(x^2+ax+40=0\) का एक मूल (5) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+40=0\) is (5), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
B दूसरा मूल (-8), (a=3) / other root (-8), (a=3)
C दूसरा मूल (8), (a=13) / other root (8), (a=13)
D दूसरा मूल (-5), (a=0) / other root (-5), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
Step 1
Concept
The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13). The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 3
Exam Tip
मूलों का गुणनफल (40) है, इसलिए दूसरा मूल (8) होगा। योग (13) है और (-a=13), इसलिए (a=-13)।
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यदि (x-2 -(m+5)x+5m=0) का एक मूल (5) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+5)x+5m=0) is (5), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (5m)
C (m+5)
D (m-5 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (5m) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{5m}{5}=m\) है।
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यदि \(x^2+ax+24=0\) का एक मूल (4) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+24=0\) is (4), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
B दूसरा मूल (-6), (a=2) / other root (-6), (a=2)
C दूसरा मूल (6), (a=10) / other root (6), (a=10)
D दूसरा मूल (-4), (a=0) / other root (-4), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
Step 1
Concept
The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10). The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 3
Exam Tip
मूलों का गुणनफल (24) है, इसलिए दूसरा मूल (6) होगा। योग (10) है और (-a=10), इसलिए (a=-10)।
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यदि (x-2 -(m+4)x+4m=0) का एक मूल (4) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+4)x+4m=0) is (4), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (4m)
C (m+4)
D (m-4 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 3
Exam Tip
गुणनफल (4m) है और एक मूल (4) है। इसलिए दूसरा मूल \(\frac{4m}{4}=m\) है।
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यदि \(x^2+ax+18=0\) का एक मूल (2) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+18=0\) is (2), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
B दूसरा मूल (-9), (a=7) / other root (-9), (a=7)
C दूसरा मूल (9), (a=11) / other root (9), (a=11)
D दूसरा मूल (-2), (a=0) / other root (-2), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
Step 1
Concept
The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11). The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 3
Exam Tip
मूलों का गुणनफल (18) है, इसलिए दूसरा मूल (9) होगा। योग (11) है और (-a=11), इसलिए (a=-11)।
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यदि (x-2 -(m+2)x+2m=0) का एक मूल (2) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+2)x+2m=0) is (2), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (2m)
C (m+2)
D (m-2 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 3
Exam Tip
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है।
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यदि \(x^2+ax+12=0\) का एक मूल (3) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
B दूसरा मूल (-4), (a=1) / other root (-4), (a=1)
C दूसरा मूल (4), (a=7) / other root (4), (a=7)
D दूसरा मूल (-3), (a=0) / other root (-3), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
Step 1
Concept
The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7). The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 3
Exam Tip
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)।
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कौन-सा विकल्प पूर्ण वर्ग न होने के कारण अपरिमेय वर्गमूल का सही उदाहरण है?
Which option is a correct example of an irrational square root because it is not a perfect square?
#real-numbers
#root5
#perfect-square
#irrationality
#hard
A \(\sqrt{5}\)
B \(\sqrt{4}\)
C \(\sqrt{9}\)
D \(\sqrt{25}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
(4,9,25) are perfect squares, so their square roots are integers.
Step 2
Why this answer is correct
(5) is not a perfect square, and \(\sqrt{5}\) is irrational.
Step 3
Exam Tip
In options, identify perfect squares first. चरण 1: (4,9,25) पूर्ण वर्ग हैं, इसलिए उनके वर्गमूल पूर्णांक हैं। चरण 2: (5) पूर्ण वर्ग नहीं है और \(\sqrt{5}\) अपरिमेय है। चरण 3: विकल्पों में पहले पूर्ण वर्ग पहचानें।
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किसी संख्या में (9) जोड़ने पर प्राप्त संख्या का वर्ग (676) है। धनात्मक मूल के अनुसार मूल संख्या क्या है?
When (9) is added to a number, the square of the result is (676). According to the positive root, what is the original number?
#quadratic equations
#square
#number problem
A (15)
B (17)
C (19)
D (26)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =676) and (x+9=26), so (x=17). If positive root is stated, take (26).
Step 2
Why this answer is correct
The correct answer is B. (17). ((x+9)2 =676) and (x+9=26), so (x=17). If positive root is stated, take (26).
Step 3
Exam Tip
((x+9)2 =676) और (x+9=26), इसलिए (x=17) है। धनात्मक मूल लिखा हो तो (26) लें।
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किसी संख्या में (5) जोड़ने पर प्राप्त संख्या का वर्ग (225) है। धनात्मक मूल के अनुसार मूल संख्या क्या है?
When (5) is added to a number, the square of the result is (225). According to the positive root, what is the original number?
#quadratic equations
#square
#number problem
A (8)
B (10)
C (12)
D (15)
Explanation opens after your attempt
Step 1
Concept
((x+5)2 =225) and (x+5=15), so (x=10). If the positive root is given, do not take the negative root.
Step 2
Why this answer is correct
The correct answer is B. (10). ((x+5)2 =225) and (x+5=15), so (x=10). If the positive root is given, do not take the negative root.
Step 3
Exam Tip
((x+5)2 =225) और (x+5=15), इसलिए (x=10) है। धनात्मक मूल दिया हो तो ऋणात्मक मूल नहीं लेना है।
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किसी संख्या में (3) जोड़ने पर प्राप्त संख्या का वर्ग (100) है। धनात्मक मूल के अनुसार संख्या क्या है?
When (3) is added to a number, the square of the result is (100). According to the positive root, what is the number?
#quadratic equations
#number problem
#square
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
((x+3)2 =100) gives (x+3=10), so (x=7). If the question says positive root, take (10).
Step 2
Why this answer is correct
The correct answer is C. (7). ((x+3)2 =100) gives (x+3=10), so (x=7). If the question says positive root, take (10).
Step 3
Exam Tip
((x+3)2 =100) से (x+3=10), इसलिए (x=7) है। प्रश्न में धनात्मक मूल कहा हो तो (10) लें।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (15)
B (10)
C (25)
D (-15)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (15). The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (25) है, इसलिए दूसरा मूल (25-10=15) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (14)
B (9)
C (23)
D (-14)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (14). The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (23) है, इसलिए दूसरा मूल (23-9=14) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (13)
B (8)
C (21)
D (-13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (13). The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (21) है, इसलिए दूसरा मूल (21-8=13) होगा। परीक्षा में एक मूल दिया हो तो योग का उपयोग करें।
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यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (9)
B (6)
C (15)
D (-9)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (9). The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (15) है, इसलिए दूसरा मूल (15-6=9) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (5)
B (4)
C (9)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (5). The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (9) है, इसलिए दूसरा मूल (9-4=5) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (3)
B (2)
C (5)
D (-3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (3). The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए दूसरा मूल (5-2=3) होगा। परीक्षा में एक मूल दिया हो तो योग या गुणनफल का प्रयोग करें।
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यदि (x=0), \(ax^2+bx+c=0\) की जड़ है, तो कौन-सी शर्त निश्चित रूप से सही है?
If (x=0) is a root of \(ax^2+bx+c=0\), which condition must be true?
#quadratic-roots
#zero-root
#root-verification
A (c=0)
B (b=0)
C (a=0)
D (a+b=0)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य जड़ होने की सीधी शर्त (c=0) है।
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यदि \(x^2+ax+12=0\) की एक जड़ दूसरी जड़ से (3) अधिक है, तो (a) के संभव मान क्या हैं?
If one root of \(x^2+ax+12=0\) is (3) more than the other root, what are the possible values of (a)?
#quadratic-roots
#difference-of-roots
#parameter
A \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
B \(\sqrt{21}\) और \(-\sqrt{21}\) / \(\sqrt{21}\) and \(-\sqrt{21}\)
C (3) और (-3) / (3) and (-3)
D (12) और (-12) / (12) and (-12)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
Step 1
Concept
Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\). Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 3
Exam Tip
जड़ें (r) और (r+3) मानने पर (r(r+3)=12) मिलता है। इससे जड़ों का योग \(\pm\sqrt{57}\) होता है, इसलिए \(a=\mp\sqrt{57}\)।
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यदि (x-2 +(k-3)x+k=0) की एक जड़ दूसरी जड़ की दुगुनी है, तो (k) का मान क्या होगा?
If one root of (x-2 +(k-3)x+k=0) is twice the other root, what is the value of (k)?
#quadratic-roots
#roots-ratio
#parameter
A \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
B \(\frac{3+\sqrt{33}}{4}\) या \(\frac{3-\sqrt{33}}{4}\) / \(\frac{3+\sqrt{33}}{4}\) or \(\frac{3-\sqrt{33}}{4}\)
C (6) या (3) / (6) or (3)
D (9) या (2) / (9) or (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
Step 1
Concept
Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\). Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 3
Exam Tip
जड़ें (r) और (2r) मानने पर (3r=3-k) और \(2r^2=k\) मिलता है। हल करने पर \(2k^2-21k+18=0\), इसलिए दिए गए दोनों मान मिलते हैं।
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यदि \(x^2-13x+42=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-13x+42=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (13)
C (42)
D (6)
Explanation opens after your attempt
Step 1
Concept
The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (6) और (7) हैं। इसलिए छोटा मूल (6) और बड़ा मूल (7) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (2) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (2), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{2}{3}\)
B \(\frac{2}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{2}{3}\)
Step 1
Concept
Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{2}{3}\). Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 3
Exam Tip
(x=2) रखने पर (4+3a=0) से \(a=-\frac{4}{3}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{2}{3}\) होगा।
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यदि \(x^2-11x+30=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-11x+30=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (11)
C (30)
D (5)
Explanation opens after your attempt
Step 1
Concept
The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (5) और (6) हैं। इसलिए छोटा मूल (5) और बड़ा मूल (6) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (1), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{1}{2}\)
B \(\frac{1}{2}\)
C (2)
D (-2)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{2}\)
Step 1
Concept
Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 3
Exam Tip
(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।
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समीकरण \(x^2-17x+70=0\) का एक मूल (7) है तो दूसरा मूल क्या है?
If one root of \(x^2-17x+70=0\) is (7), what is the other root?
#roots
#other_root
#product
A (10)
B (7)
C (17)
D (70)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 2
Why this answer is correct
The correct answer is A. (10). The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 3
Exam Tip
मूलों का गुणनफल (70) है और एक मूल (7) है। दूसरा मूल \(\frac{70}{7}=10\) होगा।
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समीकरण \(x^2-15x+54=0\) का एक मूल (6) है तो दूसरा मूल क्या है?
If one root of \(x^2-15x+54=0\) is (6), what is the other root?
#roots
#other_root
#product
A (9)
B (6)
C (15)
D (54)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 2
Why this answer is correct
The correct answer is A. (9). The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 3
Exam Tip
मूलों का गुणनफल (54) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{54}{6}=9\) है।
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समीकरण \(x^2-13x+40=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-13x+40=0\) is (5), what is the other root?
#roots
#other_root
#product
A (8)
B (5)
C (13)
D (40)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 2
Why this answer is correct
The correct answer is A. (8). The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 3
Exam Tip
मूलों का गुणनफल (40) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{40}{5}=8\) है।
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यदि मूलों का योग (12) है और एक मूल (5) है तो दूसरा मूल क्या है?
If the sum of roots is (12) and one root is (5), what is the other root?
#roots
#other_root
#sum
A (5)
B (7)
C (12)
D (17)
Explanation opens after your attempt
Step 1
Concept
The other root is (12-5=7). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is (12-5=7). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (12-5=7) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (6) है और मूलों का गुणनफल (48) है तो दूसरा मूल क्या होगा?
If one root is (6) and the product of roots is (48), what is the other root?
#roots
#other_root
#product
A (6)
B (8)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (8). The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{48}{6}=8\) होगा। गुणनफल को दिए हुए मूल से भाग करें।
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समीकरण \(x^2-11x+30=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-11x+30=0\) is (5), what is the other root?
#roots
#other_root
#factorisation
A (5)
B (6)
C (11)
D (30)
Explanation opens after your attempt
Step 1
Concept
(x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -11x+30=(x-5)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (8) है और एक मूल (3) है तो दूसरा मूल क्या है?
If the sum of roots is (8) and one root is (3), what is the other root?
#roots
#other_root
#sum
A (3)
B (5)
C (8)
D (11)
Explanation opens after your attempt
Step 1
Concept
The other root is (8-3=5). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (5). The other root is (8-3=5). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (8-3=5) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (5) है और मूलों का गुणनफल (35) है तो दूसरा मूल क्या होगा?
If one root is (5) and the product of roots is (35), what is the other root?
#roots
#other_root
#product
A (5)
B (7)
C (30)
D (35)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{35}{5}=7\) होगा। गुणनफल में दिए हुए मूल से भाग करें।
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समीकरण \(x^2-9x+18=0\) का एक मूल (3) है तो दूसरा मूल क्या है?
If one root of \(x^2-9x+18=0\) is (3), what is the other root?
#roots
#other_root
#factorisation
A (3)
B (6)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
(x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -9x+18=(x-3)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (5) है और एक मूल (2) है तो दूसरा मूल क्या है?
If the sum of roots is (5) and one root is (2) then what is the other root?
#roots
#other_root
#sum
A (2)
B (3)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
The other root is (5-2=3). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (3). The other root is (5-2=3). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (5-2=3) है। योग में से दिए हुए मूल को घटाएं।
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यदि एक मूल (3) है और मूलों का गुणनफल (12) है तो दूसरा मूल क्या होगा?
If one root is (3) and the product of roots is (12) then what is the other root?
#roots
#other_root
#product
A (3)
B (4)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 2
Why this answer is correct
The correct answer is B. (4). The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{12}{3}=4\) होगा। गुणनफल वाले प्रश्न में दिए मूल से भाग दें।
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समीकरण \(x^2-6x+8=0\) का एक मूल (4) है तो दूसरा मूल क्या है?
If one root of \(x^2-6x+8=0\) is (4) then what is the other root?
#roots
#other_root
#factorisation
A (1)
B (2)
C (-2)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 2
Why this answer is correct
The correct answer is B. (2). (x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 3
Exam Tip
(x-2 -6x+8=(x-4)(x-2)) इसलिए दूसरा मूल (2) है। दिए गए एक मूल से दूसरा गुणनखंड खोजें।
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किस समीकरण में (x=-2) मूल नहीं है?
In which equation is (x=-2) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2+2x=0\)
B \(x^2+5x+6=0\)
C \(2x^2+3x-2=0\)
D \(x^2-2x+4=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2-2x+4=0\)
Step 1
Concept
Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2-2x+4=0\). Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 3
Exam Tip
(x=-2) रखने पर \(4+4+4=12\neq0\) मिलता है। मूल न होने की जांच भी प्रतिस्थापन से करें।
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किस समीकरण में (x=1) मूल नहीं है?
In which equation is (x=1) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2-1=0\)
B \(x^2-3x+2=0\)
C \(2x^2+x-3=0\)
D \(x^2+x+1=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2+x+1=0\)
Step 1
Concept
Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2+x+1=0\). Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 3
Exam Tip
(x=1) रखने पर \(1+1+1=3\neq 0\) मिलता है। किसी विकल्प में मूल न होने की जांच भी प्रतिस्थापन से करें।
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\(\sqrt{13}\) का वर्ग किसके बराबर है?
The square of \(\sqrt{13}\) is equal to what?
#real-numbers
#square-root
#square
A (169)
B \(\sqrt{13}\)
C (13)
D (26)
Explanation opens after your attempt
Step 1
Concept
Squaring a square root gives the number inside it.
Step 2
Why this answer is correct
(\(\sqrt{13}\)2 =13).
Step 3
Exam Tip
Apply (\(\sqrt{a}\)2 =a) directly. चरण 1: वर्गमूल का वर्ग करने पर अंदर की संख्या मिलती है। चरण 2: (\(\sqrt{13}\)2 =13)। चरण 3: (\(\sqrt{a}\)2 =a) को सीधे लागू करें।
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यदि ग्राफ में प्रतिच्छेद बिंदु (\left\(3.75,-2.5\right\)) पढ़ा गया है, तो भिन्न रूप क्या होगा?
If the intersection point is read as (\left\(3.75,-2.5\right\)) on a graph, what is its fraction form?
#decimal coordinates
#fraction form
#graph reading
A (\left\(\frac{15}{4},-\frac{5}{2}\right\))
B (\left\(\frac{5}{2},-\frac{15}{4}\right\))
C (\left\(\frac{375}{10},-\frac{25}{10}\right\))
D (\left\(-\frac{15}{4},\frac{5}{2}\right\))
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{15}{4},-\frac{5}{2}\right\))
Step 1
Concept
\(3.75=\frac{15}{4}\) and \(-2.5=-\frac{5}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{15}{4},-\frac{5}{2}\right\)). \(3.75=\frac{15}{4}\) and \(-2.5=-\frac{5}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 3
Exam Tip
\(3.75=\frac{15}{4}\) और \(-2.5=-\frac{5}{2}\)। दशमलव निर्देशांक को सरल भिन्न में बदलना बेहतर रहता है।
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यदि ग्राफ में प्रतिच्छेद बिंदु (\left\(2.25,-1.5\right\)) पढ़ा गया है, तो भिन्न रूप क्या होगा?
If the intersection point is read as (\left\(2.25,-1.5\right\)) on a graph, what is its fraction form?
#decimal coordinates
#fraction form
#graph reading
A (\left\(\frac{9}{4},-\frac{3}{2}\right\))
B (\left\(\frac{3}{2},-\frac{9}{4}\right\))
C (\left\(\frac{225}{10},-\frac{15}{10}\right\))
D (\left\(-\frac{9}{4},\frac{3}{2}\right\))
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{9}{4},-\frac{3}{2}\right\))
Step 1
Concept
\(2.25=\frac{9}{4}\) and \(-1.5=-\frac{3}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{9}{4},-\frac{3}{2}\right\)). \(2.25=\frac{9}{4}\) and \(-1.5=-\frac{3}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 3
Exam Tip
\(2.25=\frac{9}{4}\) और \(-1.5=-\frac{3}{2}\)। दशमलव निर्देशांक को सरल भिन्न में बदलना बेहतर रहता है।
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संख्या रेखा पर (-1.25) के समान बिंदु कौन सा भिन्न दिखाता है?
Which fraction shows the same point as (-1.25) on the number line?
#number-line
#decimal-to-fraction
#negative-rational
#equivalent
A \(-\frac{3}{4}\)
B \(-\frac{5}{4}\)
C \(\frac{5}{4}\)
D \(-\frac{7}{4}\)
Explanation opens after your attempt
Correct Answer
B. \(-\frac{5}{4}\)
Step 1
Concept
\(-1.25=-\frac{125}{100}=-\frac{5}{4}\). Convert the decimal into a simplified fraction.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{5}{4}\). \(-1.25=-\frac{125}{100}=-\frac{5}{4}\). Convert the decimal into a simplified fraction.
Step 3
Exam Tip
\(-1.25=-\frac{125}{100}=-\frac{5}{4}\) है। दशमलव को सरल भिन्न में बदलें।
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संख्या रेखा पर (2.125) के समान बिंदु को कौन सा भिन्न दिखाता है?
Which fraction represents the same point as (2.125) on the number line?
#number-line
#decimal-to-fraction
#rational-numbers
#equivalent
A \(\frac{15}{8}\)
B \(\frac{16}{8}\)
C \(\frac{17}{8}\)
D \(\frac{19}{8}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{17}{8}\)
Step 1
Concept
\(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\). Convert decimals to fractions to identify the same point.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{17}{8}\). \(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\). Convert decimals to fractions to identify the same point.
Step 3
Exam Tip
\(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\) है। दशमलव को भिन्न में बदलकर समान बिंदु पहचानें।
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संख्या रेखा पर (-1.125) किस भिन्न के बराबर है?
Which fraction is equal to (-1.125) on the number line?
#polynomials
#number-line
#negative-decimal-fraction
#medium
A \(-\frac{9}{8}\)
B \(-\frac{8}{9}\)
C \(\frac{9}{8}\)
D \(-\frac{11}{25}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{9}{8}\)
Step 1
Concept
\(-1.125=-\frac{1125}{1000}=-\frac{9}{8}\). In exams, keep the negative sign while simplifying.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{9}{8}\). \(-1.125=-\frac{1125}{1000}=-\frac{9}{8}\). In exams, keep the negative sign while simplifying.
Step 3
Exam Tip
\(-1.125=-\frac{1125}{1000}=-\frac{9}{8}\) है। परीक्षा में ऋण चिह्न को सरल करते समय साथ रखें।
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संख्या रेखा पर (0.375) को किस भिन्न से सही दर्शाया जाएगा?
Which fraction correctly represents (0.375) on the number line?
#polynomials
#number-line
#decimal-fraction
#medium
A \(\frac{3}{8}\)
B \(\frac{37}{5}\)
C \(\frac{5}{8}\)
D \(\frac{8}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{8}\)
Step 1
Concept
\(0.375=\frac{375}{1000}=\frac{3}{8}\). In exams, convert the decimal into a fraction and simplify.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{8}\). \(0.375=\frac{375}{1000}=\frac{3}{8}\). In exams, convert the decimal into a fraction and simplify.
Step 3
Exam Tip
\(0.375=\frac{375}{1000}=\frac{3}{8}\) है। परीक्षा में दशमलव को भिन्न में बदलकर सरल करें।
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संख्या रेखा पर \(0.333\ldots\) किस भिन्न के बराबर है?
On the number line, \(0.333\ldots\) is equal to which fraction?
#number-line
#recurring-decimal
#fraction
#rational-numbers
A \(\frac{1}{2}\)
B \(\frac{1}{3}\)
C \(\frac{3}{10}\)
D \(\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{1}{3}\)
Step 1
Concept
\(0.333\ldots=\frac{1}{3}\), so both are at the same point. Connect recurring decimals with fractions.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{3}\). \(0.333\ldots=\frac{1}{3}\), so both are at the same point. Connect recurring decimals with fractions.
Step 3
Exam Tip
\(0.333\ldots=\frac{1}{3}\), इसलिए दोनों एक ही बिंदु पर होंगे। आवर्ती दशमलव को भिन्न से जोड़कर याद रखें।
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संख्या रेखा पर (0.6) को किस भिन्न से दिखाया जा सकता है?
Which fraction can show (0.6) on the number line?
#polynomials
#number-line
#decimal-to-fraction
#class-10
A \(\frac{3}{5}\)
B \(\frac{5}{3}\)
C \(\frac{6}{1}\)
D \(\frac{1}{6}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{5}\)
Step 1
Concept
\(0.6=\frac{6}{10}=\frac{3}{5}\). In exams, use place value to form the fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5}\). \(0.6=\frac{6}{10}=\frac{3}{5}\). In exams, use place value to form the fraction.
Step 3
Exam Tip
\(0.6=\frac{6}{10}=\frac{3}{5}\) है। परीक्षा में दशमलव के स्थान मान से भिन्न बनाएं।
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संख्या रेखा पर (3.75) को भिन्न रूप में किस संख्या से दर्शाया जा सकता है?
Which fraction can represent (3.75) on the number line?
#polynomials
#number-line
#decimal-to-fraction
#class-10
A \(\frac{15}{4}\)
B \(\frac{7}{4}\)
C \(\frac{3}{4}\)
D \(\frac{75}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{15}{4}\)
Step 1
Concept
\(3.75=\frac{375}{100}=\frac{15}{4}\). In exams, converting a decimal into a simple fraction is useful.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{15}{4}\). \(3.75=\frac{375}{100}=\frac{15}{4}\). In exams, converting a decimal into a simple fraction is useful.
Step 3
Exam Tip
\(3.75=\frac{375}{100}=\frac{15}{4}\) है। परीक्षा में दशमलव को सरल भिन्न में बदलना उपयोगी है।
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संख्या रेखा पर (2.25) किस भिन्न के बराबर है?
Which fraction is equal to (2.25) on the number line?
#decimal
#fraction
#number-line
A \(\frac{9}{4}\)
B \(\frac{4}{9}\)
C \(\frac{5}{4}\)
D \(\frac{11}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{4}\)
Step 1
Concept
\(2.25=\frac{225}{100}=\frac{9}{4}\). Converting a decimal to simplest fraction gives the correct point.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{4}\). \(2.25=\frac{225}{100}=\frac{9}{4}\). Converting a decimal to simplest fraction gives the correct point.
Step 3
Exam Tip
\(2.25=\frac{225}{100}=\frac{9}{4}\)। दशमलव को सरल भिन्न में बदलना सही बिंदु देता है।
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संख्या रेखा पर (3.5) किस भिन्न के बराबर है?
Which fraction is equal to (3.5) on the number line?
#decimal
#fraction
#number-line
A \(\frac{7}{2}\)
B \(\frac{5}{3}\)
C \(\frac{3}{5}\)
D \(\frac{2}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{2}\)
Step 1
Concept
\(3.5=\frac{35}{10}=\frac{7}{2}\). Convert the decimal to a simplest fraction to fix its position.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{2}\). \(3.5=\frac{35}{10}=\frac{7}{2}\). Convert the decimal to a simplest fraction to fix its position.
Step 3
Exam Tip
\(3.5=\frac{35}{10}=\frac{7}{2}\)। दशमलव को सरल भिन्न में बदलकर सही स्थान तय करें।
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संख्या रेखा पर (0.2) किस भिन्न के बराबर है?
Which fraction is equal to (0.2) on the number line?
#decimal
#fraction
#number-line
A \(\frac{1}{5}\)
B \(\frac{1}{2}\)
C \(\frac{2}{5}\)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
\(0.2=\frac{2}{10}=\frac{1}{5}\). Use decimal place value to form the simplest fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). \(0.2=\frac{2}{10}=\frac{1}{5}\). Use decimal place value to form the simplest fraction.
Step 3
Exam Tip
\(0.2=\frac{2}{10}=\frac{1}{5}\)। दशमलव स्थान देखकर सरल भिन्न बनाइए।
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संख्या रेखा पर (1.25) किस भिन्न के बराबर है?
Which fraction is equal to (1.25) on the number line?
#decimal
#fraction
#number-line
A \(\frac{5}{4}\)
B \(\frac{4}{5}\)
C \(\frac{3}{4}\)
D \(\frac{7}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{4}\)
Step 1
Concept
\(1.25=\frac{125}{100}=\frac{5}{4}\). Convert a decimal to a fraction to locate the point correctly.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{4}\). \(1.25=\frac{125}{100}=\frac{5}{4}\). Convert a decimal to a fraction to locate the point correctly.
Step 3
Exam Tip
\(1.25=\frac{125}{100}=\frac{5}{4}\)। दशमलव को भिन्न में बदलकर सही बिंदु पहचानें।
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(0.015625) को सरलतम भिन्न में लिखने पर हर क्या होगा?
What is the denominator when (0.015625) is written in lowest fraction form?
#decimal-to-fraction
#lowest-denominator
#terminating-decimal
#class-10
A (32)
B (64)
C (128)
D (256)
Explanation opens after your attempt
Step 1
Concept
\(0.015625=\frac{15625}{1000000}=\frac{1}{64}\). Convert a terminating decimal to a fraction and reduce the denominator.
Step 2
Why this answer is correct
The correct answer is B. (64). \(0.015625=\frac{15625}{1000000}=\frac{1}{64}\). Convert a terminating decimal to a fraction and reduce the denominator.
Step 3
Exam Tip
\(0.015625=\frac{15625}{1000000}=\frac{1}{64}\) है। सांत दशमलव को भिन्न में बदलकर हर को सरलतम रूप में देखें।
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(0.046875) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of (0.046875)?
#lowest-fraction
#decimal-conversion
#terminating-decimal
#real-numbers
A \(\frac{3}{64}\)
B \(\frac{15}{320}\)
C \(\frac{75}{1600}\)
D \(\frac{1}{64}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{64}\)
Step 1
Concept
\(0.046875=\frac{46875}{1000000}\), and reducing gives \(\frac{3}{64}\). Convert the decimal to a fraction and reduce fully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{64}\). \(0.046875=\frac{46875}{1000000}\), and reducing gives \(\frac{3}{64}\). Convert the decimal to a fraction and reduce fully.
Step 3
Exam Tip
\(0.046875=\frac{46875}{1000000}\) है और सरल करने पर \(\frac{3}{64}\) मिलता है। दशमलव से भिन्न बनाकर अंतिम रूप तक सरल करें।
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\(0.00\overline{54}\) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of \(0.00\overline{54}\)?
#recurring-decimal
#fraction-form
#lowest-form
#expert
A \(\frac{3}{550}\)
B \(\frac{54}{990}\)
C \(\frac{6}{1100}\)
D \(\frac{1}{550}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{550}\)
Step 1
Concept
Two non-repeating zeros and two repeating digits give \(\frac{54}{9900}\). Reducing it gives \(\frac{3}{550}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{550}\). Two non-repeating zeros and two repeating digits give \(\frac{54}{9900}\). Reducing it gives \(\frac{3}{550}\).
Step 3
Exam Tip
दो अनावर्ती शून्य और दो आवर्ती अंकों से \(\frac{54}{9900}\) बनता है। इसे सरल करने पर \(\frac{3}{550}\) मिलता है।
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(0.00084) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of (0.00084)?
#decimal-to-fraction
#lowest-form
#terminating-decimal
#expert
A \(\frac{21}{25000}\)
B \(\frac{84}{10000}\)
C \(\frac{42}{50000}\)
D \(\frac{7}{2500}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21}{25000}\)
Step 1
Concept
\(0.00084=\frac{84}{100000}\), and reducing by (4) gives \(\frac{21}{25000}\). Even for small decimals, check the greatest common factor carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21}{25000}\). \(0.00084=\frac{84}{100000}\), and reducing by (4) gives \(\frac{21}{25000}\). Even for small decimals, check the greatest common factor carefully.
Step 3
Exam Tip
\(0.00084=\frac{84}{100000}\) है और (4) से सरल करने पर \(\frac{21}{25000}\) मिलता है। छोटे दशमलव में भी महत्तम सामान्य गुणनखंड ध्यान से देखें।
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(0.03125) को सरलतम भिन्न में लिखने पर हर क्या होगा?
What is the denominator when (0.03125) is written in lowest fraction form?
#decimal-to-fraction
#lowest-denominator
#terminating-decimal
#class-10
A (16)
B (32)
C (64)
D (128)
Explanation opens after your attempt
Step 1
Concept
\(0.03125=\frac{3125}{100000}=\frac{1}{32}\). Convert a terminating decimal to a fraction and reduce the denominator.
Step 2
Why this answer is correct
The correct answer is B. (32). \(0.03125=\frac{3125}{100000}=\frac{1}{32}\). Convert a terminating decimal to a fraction and reduce the denominator.
Step 3
Exam Tip
\(0.03125=\frac{3125}{100000}=\frac{1}{32}\) है। सांत दशमलव को भिन्न में बदलकर हर को सरलतम रूप में देखें।
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(0.01875) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of (0.01875)?
#lowest-fraction
#decimal-conversion
#terminating-decimal
#real-numbers
A \(\frac{3}{160}\)
B \(\frac{15}{800}\)
C \(\frac{1875}{10000}\)
D \(\frac{1}{160}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{160}\)
Step 1
Concept
\(0.01875=\frac{1875}{100000}\), and dividing by (625) gives \(\frac{3}{160}\). Convert the decimal to a fraction and reduce fully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{160}\). \(0.01875=\frac{1875}{100000}\), and dividing by (625) gives \(\frac{3}{160}\). Convert the decimal to a fraction and reduce fully.
Step 3
Exam Tip
\(0.01875=\frac{1875}{100000}\) है और (625) से भाग देने पर \(\frac{3}{160}\) मिलता है। दशमलव से भिन्न बनाकर अंतिम रूप तक सरल करें।
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(0.01875) को सरलतम भिन्न में लिखने पर हर का अभाज्य गुणनखंडन क्या होगा?
When (0.01875) is written in lowest fraction form, what is the prime factorisation of the denominator?
#decimal-to-fraction
#prime-factorisation
#option-audit
#real-numbers
A \(2^4\cdot 5\)
B \(2^5\cdot 5\)
C \(2^4\cdot 5^2\)
D \(2^6\cdot 5\)
Explanation opens after your attempt
Correct Answer
A. \(2^4\cdot 5\)
Step 1
Concept
\(0.01875=\frac{1875}{100000}=\frac{3}{160}\), and \(160=2^5\cdot 5\). The correct prime factorisation is \(2^5\cdot 5\), so complete the calculation before choosing.
Step 2
Why this answer is correct
The correct answer is A. \(2^4\cdot 5\). \(0.01875=\frac{1875}{100000}=\frac{3}{160}\), and \(160=2^5\cdot 5\). The correct prime factorisation is \(2^5\cdot 5\), so complete the calculation before choosing.
Step 3
Exam Tip
\(0.01875=\frac{1875}{100000}=\frac{3}{160}\) और \(160=2^5\cdot 5\) है। सही अभाज्य रूप \(2^5\cdot 5\) है इसलिए गणना पूरी करके विकल्प चुनें।
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\(0.\overline{045}\) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of \(0.\overline{045}\)?
#recurring-decimal
#lowest-form
#fraction-conversion
#expert
A \(\frac{5}{111}\)
B \(\frac{15}{333}\)
C \(\frac{45}{99}\)
D \(\frac{1}{37}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{111}\)
Step 1
Concept
\(0.\overline{045}=\frac{45}{999}\), and reducing by (9) gives \(\frac{5}{111}\). First form the denominator with (9)'s according to the repeating digits.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{111}\). \(0.\overline{045}=\frac{45}{999}\), and reducing by (9) gives \(\frac{5}{111}\). First form the denominator with (9)'s according to the repeating digits.
Step 3
Exam Tip
\(0.\overline{045}=\frac{45}{999}\) और (9) से सरल करने पर \(\frac{5}{111}\) मिलता है। आवर्ती अंकों की संख्या के अनुसार पहले (9) वाला हर बनाएं।
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\(0.00\overline{63}\) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of \(0.00\overline{63}\)?
#recurring-decimal
#fraction-form
#lowest-form
#expert
A \(\frac{7}{1100}\)
B \(\frac{63}{990}\)
C \(\frac{9}{1100}\)
D \(\frac{21}{3300}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{1100}\)
Step 1
Concept
Two non-repeating zeros and two repeating digits give \(\frac{63}{9900}\). Reducing it gives \(\frac{7}{1100}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{1100}\). Two non-repeating zeros and two repeating digits give \(\frac{63}{9900}\). Reducing it gives \(\frac{7}{1100}\).
Step 3
Exam Tip
दो अनावर्ती शून्य और दो आवर्ती अंकों से \(\frac{63}{9900}\) बनता है। इसे सरल करने पर \(\frac{7}{1100}\) मिलता है।
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(0.00096) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of (0.00096)?
#decimal-to-fraction
#lowest-form
#terminating-decimal
#expert
A \(\frac{3}{3125}\)
B \(\frac{6}{6250}\)
C \(\frac{12}{12500}\)
D \(\frac{96}{100000}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{3125}\)
Step 1
Concept
\(0.00096=\frac{96}{100000}\), and reducing by (32) gives \(\frac{3}{3125}\). Even for small decimals, check the greatest common factor carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{3125}\). \(0.00096=\frac{96}{100000}\), and reducing by (32) gives \(\frac{3}{3125}\). Even for small decimals, check the greatest common factor carefully.
Step 3
Exam Tip
\(0.00096=\frac{96}{100000}\) है और (32) से सरल करने पर \(\frac{3}{3125}\) मिलता है। छोटे दशमलव में भी महत्तम सामान्य गुणनखंड ध्यान से देखें।
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\(0.2\overline{54}\) का सरलतम भिन्न रूप कौन-सा है?
Which is the lowest fraction form of \(0.2\overline{54}\)?
#recurring-decimal
#lowest-fraction
#decimal-conversion
#real-numbers
A \(\frac{14}{55}\)
B \(\frac{28}{110}\)
C \(\frac{252}{990}\)
D \(\frac{254}{999}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{14}{55}\)
Step 1
Concept
The non-repeating part (2) and repeating part (54) give \(\frac{252}{990}\), which reduces to \(\frac{14}{55}\). In exams, identify repeating and non-repeating digits separately.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{14}{55}\). The non-repeating part (2) and repeating part (54) give \(\frac{252}{990}\), which reduces to \(\frac{14}{55}\). In exams, identify repeating and non-repeating digits separately.
Step 3
Exam Tip
सांत भाग (2) और आवर्ती भाग (54) से भिन्न \(\frac{252}{990}\) बनती है जो \(\frac{14}{55}\) तक सरल होती है। परीक्षा में आवर्ती और अनावर्ती अंकों को अलग पहचानें।
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