कौन सा मान \( \frac{23}{6} \) से बड़ा और \( \sqrt{15} \) से छोटा है?
Which value is greater than \( \frac{23}{6} \) and less than \( \sqrt{15} \)?
#number-line
#between-values
#fraction-root
A (3.82)
B (3.90)
C (3.86)
D (4.00)
Explanation opens after your attempt
Step 1
Concept
\( \frac{23}{6}\approx3.833 \) and \( \sqrt{15}\approx3.873 \). Therefore (3.86) lies between them.
Step 2
Why this answer is correct
The correct answer is C. (3.86). \( \frac{23}{6}\approx3.833 \) and \( \sqrt{15}\approx3.873 \). Therefore (3.86) lies between them.
Step 3
Exam Tip
\( \frac{23}{6}\approx3.833 \) और \( \sqrt{15}\approx3.873 \) है। इसलिए (3.86) इनके बीच है।
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संख्या रेखा पर \( \sqrt{\frac{196}{289}} \) किसके बराबर है?
On the number line, \( \sqrt{\frac{196}{289}} \) is equal to what?
#number-line
#fraction-root
#exact-value
A \( \frac{14}{17} \)
B \( \frac{196}{17} \)
C \( \frac{17}{14} \)
D \( \frac{14}{289} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{14}{17} \)
Step 1
Concept
\( \sqrt{\frac{196}{289}}=\frac{14}{17} \). Take the positive square root of numerator and denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{14}{17} \). \( \sqrt{\frac{196}{289}}=\frac{14}{17} \). Take the positive square root of numerator and denominator.
Step 3
Exam Tip
\( \sqrt{\frac{196}{289}}=\frac{14}{17} \) है। अंश और हर का धनात्मक वर्गमूल लें।
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कौन सा मान \( \frac{17}{5} \) से बड़ा और \( \sqrt{12} \) से छोटा है?
Which value is greater than \( \frac{17}{5} \) and less than \( \sqrt{12} \)?
#number-line
#between-values
#fraction-root
A (3.30)
B (3.50)
C (3.45)
D (3.70)
Explanation opens after your attempt
Step 1
Concept
\( \frac{17}{5}=3.4 \) and \( \sqrt{12}\approx3.464 \). Therefore (3.45) lies between them.
Step 2
Why this answer is correct
The correct answer is C. (3.45). \( \frac{17}{5}=3.4 \) and \( \sqrt{12}\approx3.464 \). Therefore (3.45) lies between them.
Step 3
Exam Tip
\( \frac{17}{5}=3.4 \) और \( \sqrt{12}\approx3.464 \) है। इसलिए (3.45) इनके बीच है।
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संख्या रेखा पर \( \sqrt{\frac{169}{225}} \) किसके बराबर है?
On the number line, \( \sqrt{\frac{169}{225}} \) is equal to what?
#number-line
#fraction-root
#exact-value
A \( \frac{13}{15} \)
B \( \frac{169}{15} \)
C \( \frac{15}{13} \)
D \( \frac{13}{225} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{13}{15} \)
Step 1
Concept
\( \sqrt{\frac{169}{225}}=\frac{13}{15} \). Take the positive square root of numerator and denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{13}{15} \). \( \sqrt{\frac{169}{225}}=\frac{13}{15} \). Take the positive square root of numerator and denominator.
Step 3
Exam Tip
\( \sqrt{\frac{169}{225}}=\frac{13}{15} \) है। अंश और हर का धनात्मक वर्गमूल लें।
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कौन सा मान \( \frac{9}{4} \) से बड़ा और \( \sqrt{6} \) से छोटा है?
Which value is greater than \( \frac{9}{4} \) and less than \( \sqrt{6} \)?
#number-line
#between-values
#fraction-root
A (2.4)
B (2.2)
C (2.5)
D (3.0)
Explanation opens after your attempt
Step 1
Concept
\( \frac{9}{4}=2.25 \) and \( \sqrt{6}\approx2.449 \). Therefore (2.4) lies between them.
Step 2
Why this answer is correct
The correct answer is A. (2.4). \( \frac{9}{4}=2.25 \) and \( \sqrt{6}\approx2.449 \). Therefore (2.4) lies between them.
Step 3
Exam Tip
\( \frac{9}{4}=2.25 \) और \( \sqrt{6}\approx2.449 \) है। इसलिए (2.4) इनके बीच है।
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संख्या रेखा पर \( \sqrt{\frac{49}{121}} \) किसके बराबर है?
On the number line, \( \sqrt{\frac{49}{121}} \) is equal to what?
#number-line
#fraction-root
#exact-value
A \( \frac{7}{11} \)
B \( \frac{49}{11} \)
C \( \frac{11}{7} \)
D \( \frac{7}{121} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{7}{11} \)
Step 1
Concept
\( \sqrt{\frac{49}{121}}=\frac{7}{11} \). Take the positive square root of numerator and denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{7}{11} \). \( \sqrt{\frac{49}{121}}=\frac{7}{11} \). Take the positive square root of numerator and denominator.
Step 3
Exam Tip
\( \sqrt{\frac{49}{121}}=\frac{7}{11} \) है। अंश और हर दोनों का धनात्मक वर्गमूल लें।
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समीकरण \(25x^2-10x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(25x^2-10x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{5}\)
B -\(\frac{1}{5}\)
C (5)
D (-5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
(25x-2 -10x+1=(5x-1)2 ). Therefore the repeated root is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). (25x-2 -10x+1=(5x-1)2 ). Therefore the repeated root is \(\frac{1}{5}\).
Step 3
Exam Tip
(25x-2 -10x+1=(5x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{5}\) है।
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यदि \(x=\frac{5}{2}\) समीकरण \(2x^2-9x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{5}{2}\) is a root of \(2x^2-9x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (10)
B (-10)
C \(\frac{5}{2}\)
D -\(\frac{5}{2}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{5}{2}\) gives \(\frac{25}{2}-\frac{45}{2}+c=0\), so (c=10). Calculate each term carefully with a fractional root.
Step 2
Why this answer is correct
The correct answer is A. (10). Putting \(x=\frac{5}{2}\) gives \(\frac{25}{2}-\frac{45}{2}+c=0\), so (c=10). Calculate each term carefully with a fractional root.
Step 3
Exam Tip
\(x=\frac{5}{2}\) रखने पर \(\frac{25}{2}-\frac{45}{2}+c=0\) इसलिए (c=10) है। भिन्न मूल में हर पद सावधानी से निकालें।
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समीकरण \(4x^2-17x+4=0\) के मूल कौन से हैं?
What are the roots of \(4x^2-17x+4=0\)?
#roots
#factorisation
#fraction_root
A (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
B (-4) और \(-\frac{1}{4}\) / (-4) and \(-\frac{1}{4}\)
C (2) और (4) / (2) and (4)
D \(\frac{4}{3}\) और (1) / \(\frac{4}{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\)
Step 1
Concept
(4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और \(\frac{1}{4}\) / (4) and \(\frac{1}{4}\). (4x-2 -17x+4=(4x-1)(x-4)). Therefore the roots are \(\frac{1}{4}\) and (4).
Step 3
Exam Tip
(4x-2 -17x+4=(4x-1)(x-4)) है। इसलिए मूल \(\frac{1}{4}\) और (4) हैं।
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समीकरण \(16x^2-8x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(16x^2-8x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{4}\)
B -\(\frac{1}{4}\)
C (4)
D (-4)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{4}\)
Step 1
Concept
(16x-2 -8x+1=(4x-1)2 ). Therefore the repeated root is \(\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{4}\). (16x-2 -8x+1=(4x-1)2 ). Therefore the repeated root is \(\frac{1}{4}\).
Step 3
Exam Tip
(16x-2 -8x+1=(4x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{4}\) है।
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यदि \(x=\frac{4}{3}\) समीकरण \(3x^2-7x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{4}{3}\) is a root of \(3x^2-7x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (4)
B (-4)
C \(\frac{4}{3}\)
D -\(\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{4}{3}\) gives \(\frac{16}{3}-\frac{28}{3}+c=0\), so (c=4). For fractional roots calculate each term separately.
Step 2
Why this answer is correct
The correct answer is A. (4). Putting \(x=\frac{4}{3}\) gives \(\frac{16}{3}-\frac{28}{3}+c=0\), so (c=4). For fractional roots calculate each term separately.
Step 3
Exam Tip
\(x=\frac{4}{3}\) रखने पर \(\frac{16}{3}-\frac{28}{3}+c=0\) इसलिए (c=4) है। भिन्न मूल में हर पद अलग निकालें।
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समीकरण \(3x^2-13x+4=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-13x+4=0\)?
#roots
#factorisation
#fraction_root
A (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\)
B (-4) और \(-\frac{1}{3}\) / (-4) and \(-\frac{1}{3}\)
C (3) और (4) / (3) and (4)
D \(\frac{4}{3}\) और (1) / \(\frac{4}{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\)
Step 1
Concept
(3x-2 -13x+4=(3x-1)(x-4)). Therefore the roots are \(\frac{1}{3}\) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और \(\frac{1}{3}\) / (4) and \(\frac{1}{3}\). (3x-2 -13x+4=(3x-1)(x-4)). Therefore the roots are \(\frac{1}{3}\) and (4).
Step 3
Exam Tip
(3x-2 -13x+4=(3x-1)(x-4)) है। इसलिए मूल \(\frac{1}{3}\) और (4) हैं।
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समीकरण \(9x^2-6x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(9x^2-6x+1=0\)?
#roots
#repeated_root
#fraction_root
A \(\frac{1}{3}\)
B -\(\frac{1}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{3}\)
Step 1
Concept
(9x-2 -6x+1=(3x-1)2 ). Therefore the repeated root is \(\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{3}\). (9x-2 -6x+1=(3x-1)2 ). Therefore the repeated root is \(\frac{1}{3}\).
Step 3
Exam Tip
(9x-2 -6x+1=(3x-1)2 ) है। इसलिए दोहराया मूल \(\frac{1}{3}\) है।
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यदि \(x=\frac{3}{2}\) समीकरण \(2x^2-5x+c=0\) का मूल है तो (c) का मान क्या है?
If \(x=\frac{3}{2}\) is a root of \(2x^2-5x+c=0\), what is the value of (c)?
#roots
#parameter
#fraction_root
A (3)
B (-3)
C \(\frac{3}{2}\)
D \(-\frac{3}{2}\)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{3}{2}\) gives \(\frac{9}{2}-\frac{15}{2}+c=0\), so (c=3). For fractional roots calculate each term separately.
Step 2
Why this answer is correct
The correct answer is A. (3). Putting \(x=\frac{3}{2}\) gives \(\frac{9}{2}-\frac{15}{2}+c=0\), so (c=3). For fractional roots calculate each term separately.
Step 3
Exam Tip
\(x=\frac{3}{2}\) रखने पर \(\frac{9}{2}-\frac{15}{2}+c=0\) इसलिए (c=3) है। भिन्न मूल में हर पद अलग निकालें।
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समीकरण \(2x^2-7x+3=0\) के मूल कौन से हैं?
What are the roots of \(2x^2-7x+3=0\)?
#roots
#factorisation
#fraction_root
A (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\)
B (-3) और \(-\frac{1}{2}\) / (-3) and \(-\frac{1}{2}\)
C (2) और (3) / (2) and (3)
D \(\frac{3}{2}\) और (1) / \(\frac{3}{2}\) and (1)
Explanation opens after your attempt
Correct Answer
A. (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\)
Step 1
Concept
(2x-2 -7x+3=(2x-1)(x-3)). Therefore the roots are \(\frac{1}{2}\) and (3).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\frac{1}{2}\) / (3) and \(\frac{1}{2}\). (2x-2 -7x+3=(2x-1)(x-3)). Therefore the roots are \(\frac{1}{2}\) and (3).
Step 3
Exam Tip
(2x-2 -7x+3=(2x-1)(x-3)) है। इसलिए मूल \(\frac{1}{2}\) और (3) हैं।
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समीकरण \(4x^2-12x+9=0\) के मूल कौन से हैं?
What are the roots of \(4x^2-12x+9=0\)?
#roots
#factorisation
#fraction_root
A \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \)
B \(-\frac{3}{2}\) और \(-\frac{3}{2}\) / \(-\frac{3}{2}\) and \(-\frac{3}{2}\)
C (2) और (3) / (2) and (3)
D \(\frac{2}{3}\) और \(\frac{2}{3}\) / \(\frac{2}{3}\) and \(\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \)
Step 1
Concept
(4x-2 -12x+9=(2x-3)2 ). Therefore the repeated root is \(\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{3}{2} \) और \( \frac{3}{2} \) / \( \frac{3}{2} \) and \( \frac{3}{2} \). (4x-2 -12x+9=(2x-3)2 ). Therefore the repeated root is \(\frac{3}{2}\).
Step 3
Exam Tip
(4x-2 -12x+9=(2x-3)2 ) है। इसलिए दोहराया मूल \(\frac{3}{2}\) है।
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यदि \(x=\frac{1}{2}\) समीकरण \(4x^2+px-3=0\) का मूल है तो (p) का मान क्या है?
If \(x=\frac{1}{2}\) is a root of \(4x^2+px-3=0\), what is the value of (p)?
#roots
#parameter
#fraction_root
A (2)
B (4)
C (-4)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Putting \(x=\frac{1}{2}\) gives \(1+\frac{p}{2}-3=0\), so (p=4). When substituting a fractional root, solve each term carefully.
Step 2
Why this answer is correct
The correct answer is B. (4). Putting \(x=\frac{1}{2}\) gives \(1+\frac{p}{2}-3=0\), so (p=4). When substituting a fractional root, solve each term carefully.
Step 3
Exam Tip
\(x=\frac{1}{2}\) रखने पर \(1+\frac{p}{2}-3=0\) इसलिए (p=4)। भिन्न मूल रखते समय हर पद सावधानी से हल करें।
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समीकरण \(3x^2-10x+3=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-10x+3=0\)?
#roots
#factorisation
#fraction_root
A (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\)
B (1) और (3) / (1) and (3)
C (-3) और \(-\frac{1}{3}\) / (-3) and \(-\frac{1}{3}\)
D \(\frac{3}{2}\) और (2) / \(\frac{3}{2}\) and (2)
Explanation opens after your attempt
Correct Answer
A. (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\)
Step 1
Concept
(3x-2 -10x+3=(3x-1)(x-3)). Therefore the roots are \(\frac{1}{3}\) and (3).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\frac{1}{3}\) / (3) and \(\frac{1}{3}\). (3x-2 -10x+3=(3x-1)(x-3)). Therefore the roots are \(\frac{1}{3}\) and (3).
Step 3
Exam Tip
(3x-2 -10x+3=(3x-1)(x-3)) है। इसलिए मूल \(\frac{1}{3}\) और (3) हैं।
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