यदि \(x=\sqrt{7}+3\) है तो (x-3) किस प्रकार की संख्या है?
If \(x=\sqrt{7}+3\), what type of number is (x-3)?
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A परिमेय / Rational
B अपरिमेय / Irrational
C पूर्णांक / Integer
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
B. अपरिमेय / Irrational
Step 1
Concept
\(x-3=\sqrt{7}\), which is irrational. First separate the rational terms.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. \(x-3=\sqrt{7}\), which is irrational. First separate the rational terms.
Step 3
Exam Tip
\(x-3=\sqrt{7}\) है जो अपरिमेय है। पहले परिमेय पदों को अलग करें।
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\(\sqrt{32}+\sqrt{18}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{32}+\sqrt{18}\)?
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A \(5\sqrt{2}\)
B \(7\sqrt{2}\)
C \(\sqrt{50}\)
D \(6\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
B. \(7\sqrt{2}\)
Step 1
Concept
\(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the sum is \(7\sqrt{2}\). First convert radicals into like form.
Step 2
Why this answer is correct
The correct answer is B. \(7\sqrt{2}\). \(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the sum is \(7\sqrt{2}\). First convert radicals into like form.
Step 3
Exam Tip
\(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए योग \(7\sqrt{2}\) है। पहले मूलों को समान रूप में बदलें।
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(\(\sqrt{11}\)2 -4) का मान क्या है?
What is the value of (\(\sqrt{11}\)2 -4)?
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A (7)
B \(\sqrt{7}\)
C (15)
D \(11\sqrt{4}\)
Explanation opens after your attempt
Step 1
Concept
(\(\sqrt{11}\)2 =11), so the value is (7). Squaring can remove the square root.
Step 2
Why this answer is correct
The correct answer is A. (7). (\(\sqrt{11}\)2 =11), so the value is (7). Squaring can remove the square root.
Step 3
Exam Tip
(\(\sqrt{11}\)2 =11) इसलिए मान (7) है। वर्ग करने पर वर्गमूल हट सकता है।
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\(\frac{1}{\sqrt{5}-2}\) को सरल करने पर क्या मिलता है?
What is obtained after simplifying \(\frac{1}{\sqrt{5}-2}\)?
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A \(\sqrt{5}-2\)
B \(\frac{\sqrt{5}+2}{9}\)
C \(\sqrt{5}+2\)
D \(\frac{1}{\sqrt{5}+2}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{5}+2\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (5-4=1). So the answer is \(\sqrt{5}+2\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{5}+2\). Multiplying by the conjugate makes the denominator (5-4=1). So the answer is \(\sqrt{5}+2\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (5-4=1) बनता है। इसलिए उत्तर \(\sqrt{5}+2\) है।
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यदि \(a=2\sqrt{3}+1\) और \(b=2\sqrt{3}-1\) हैं तो (a+b) क्या है?
If \(a=2\sqrt{3}+1\) and \(b=2\sqrt{3}-1\), what is (a+b)?
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A (2)
B \(4\sqrt{3}\)
C (6)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(4\sqrt{3}\)
Step 1
Concept
The constant terms cancel and \(2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\). Add like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(4\sqrt{3}\). The constant terms cancel and \(2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\). Add like radicals.
Step 3
Exam Tip
स्थिर पद कट जाते हैं और \(2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\) मिलता है। समान मूलों को जोड़ें।
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(\(4+\sqrt{7}\)\(4-\sqrt{7}\)) का मान क्या है?
What is the value of (\(4+\sqrt{7}\)\(4-\sqrt{7}\))?
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A (23)
B (9)
C \(16+\sqrt{7}\)
D \(8\sqrt{7}\)
Explanation opens after your attempt
Step 1
Concept
This is \(a^2-b^2\), so the value is (16-7=9). Conjugate multiplication removes the radical.
Step 2
Why this answer is correct
The correct answer is B. (9). This is \(a^2-b^2\), so the value is (16-7=9). Conjugate multiplication removes the radical.
Step 3
Exam Tip
यह \(a^2-b^2\) है इसलिए मान (16-7=9) है। संयुग्मी गुणन से मूल हट जाता है।
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\(\sqrt{75}-\sqrt{12}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{75}-\sqrt{12}\)?
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A \(3\sqrt{3}\)
B \(7\sqrt{3}\)
C \(5\sqrt{3}\)
D \(\sqrt{63}\)
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{3}\)
Step 1
Concept
\(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(3\sqrt{3}\). First take out perfect-square factors.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{3}\). \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(3\sqrt{3}\). First take out perfect-square factors.
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए अंतर \(3\sqrt{3}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें।
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दशमलव \(2.303003000300003\ldots\) किस प्रकार की संख्या है?
What type of number is the decimal \(2.303003000300003\ldots\)?
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A आवर्ती परिमेय / Repeating rational
B अपरिमेय / Irrational
C सांत परिमेय / Terminating rational
D पूर्णांक / Integer
Explanation opens after your attempt
Correct Answer
B. अपरिमेय / Irrational
Step 1
Concept
It has no fixed repeating block, so it is irrational. A non-terminating non-repeating decimal is irrational.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. It has no fixed repeating block, so it is irrational. A non-terminating non-repeating decimal is irrational.
Step 3
Exam Tip
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव अपरिमेय होता है।
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\(\frac{\sqrt{75}}{\sqrt{3}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{75}}{\sqrt{3}}\)?
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A (5)
B \(\sqrt{72}\)
C (25)
D \(5\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\). Combine radicals in division and simplify.
Step 2
Why this answer is correct
The correct answer is A. (5). \(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\). Combine radicals in division and simplify.
Step 3
Exam Tip
\(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\) है। भाग में मूलों को एक साथ लिखकर सरल करें।
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\(\sqrt{80}+\sqrt{125}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{80}+\sqrt{125}\)?
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A \(9\sqrt{5}\)
B \(4\sqrt{5}\)
C \(5\sqrt{5}\)
D \(7\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\) and \(\sqrt{125}=5\sqrt{5}\), so the sum is \(9\sqrt{5}\). Like radicals should be added.
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{5}\). \(\sqrt{80}=4\sqrt{5}\) and \(\sqrt{125}=5\sqrt{5}\), so the sum is \(9\sqrt{5}\). Like radicals should be added.
Step 3
Exam Tip
\(\sqrt{80}=4\sqrt{5}\) और \(\sqrt{125}=5\sqrt{5}\), इसलिए योग \(9\sqrt{5}\) है। समान मूलों को जोड़ना चाहिए।
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\(\sqrt{15}\) और \(\sqrt{20}\) के बीच कौन-सी संख्या है?
Which number lies between \(\sqrt{15}\) and \(\sqrt{20}\)?
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A (3)
B \(\sqrt{14}\)
C \(\sqrt{17}\)
D (5)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{17}\)
Step 1
Concept
Since (15<17<20), \(\sqrt{17}\) lies between them. Compare square roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{17}\). Since (15<17<20), \(\sqrt{17}\) lies between them. Compare square roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (15<17<20), इसलिए \(\sqrt{17}\) इनके बीच है। वर्गमूल की तुलना अंदर की संख्याओं से करें।
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यदि \(y=\sqrt{13}\) है तो \(y^2+6\) का मान क्या है?
If \(y=\sqrt{13}\), what is the value of \(y^2+6\)?
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A (19)
B \(\sqrt{19}\)
C \(13\sqrt{6}\)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(y^2=13\), so \(y^2+6=19\). First square the given radical.
Step 2
Why this answer is correct
The correct answer is A. (19). \(y^2=13\), so \(y^2+6=19\). First square the given radical.
Step 3
Exam Tip
\(y^2=13\) इसलिए \(y^2+6=19\) है। पहले दिए गए मूल का वर्ग करें।
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कौन-सा विकल्प अपरिमेय संख्या का उदाहरण है?
Which option is an example of an irrational number?
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A \(0.828282\ldots\)
B (9 / 17)
C \(0.4141141114\ldots\)
D (3.625)
Explanation opens after your attempt
Correct Answer
C. \(0.4141141114\ldots\)
Step 1
Concept
The third decimal is non-terminating and non-repeating, so it is irrational. Repeating and terminating decimals are rational.
Step 2
Why this answer is correct
The correct answer is C. \(0.4141141114\ldots\). The third decimal is non-terminating and non-repeating, so it is irrational. Repeating and terminating decimals are rational.
Step 3
Exam Tip
तीसरा दशमलव असांत और अनावर्ती है इसलिए अपरिमेय है। आवर्ती और सांत दशमलव परिमेय होते हैं।
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\(\sqrt{147}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{147}\)?
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A \(7\sqrt{3}\)
B \(3\sqrt{7}\)
C (21)
D \(49\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{3}\)
Step 1
Concept
\(\sqrt{147}=\sqrt{49\times3}=7\sqrt{3}\). Choose the largest perfect-square factor.
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\). \(\sqrt{147}=\sqrt{49\times3}=7\sqrt{3}\). Choose the largest perfect-square factor.
Step 3
Exam Tip
\(\sqrt{147}=\sqrt{49\times3}=7\sqrt{3}\) है। सबसे बड़ा पूर्ण वर्ग गुणनखंड चुनें।
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\(\frac{3}{\sqrt{7}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{3}{\sqrt{7}}\)?
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A \(\frac{3\sqrt{7}}{7}\)
B \(3\sqrt{7}\)
C \(\frac{\sqrt{7}}{3}\)
D \(\frac{7}{3\sqrt{7}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3\sqrt{7}}{7}\)
Step 1
Concept
Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{3\sqrt{7}}{7}\). Rationalisation keeps the value same.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3\sqrt{7}}{7}\). Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{3\sqrt{7}}{7}\). Rationalisation keeps the value same.
Step 3
Exam Tip
अंश और हर को \(\sqrt{7}\) से गुणा करने पर \(\frac{3\sqrt{7}}{7}\) मिलता है। परिमेयकरण में मान समान रहता है।
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(\(\sqrt{8}+\sqrt{2}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{8}+\sqrt{2}\)2 )?
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A (10)
B (18)
C (12)
D \(8+2\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), so (\(3\sqrt{2}\)2 =18). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is B. (18). \(\sqrt{8}=2\sqrt{2}\), so (\(3\sqrt{2}\)2 =18). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए (\(3\sqrt{2}\)2 =18) है। पहले कोष्ठक को सरल करें।
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\(6-\sqrt{10}\) में अपरिमेय भाग कौन-सा है?
What is the irrational part in \(6-\sqrt{10}\)?
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A (6)
B \(-\sqrt{10}\)
C (10)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
B. \(-\sqrt{10}\)
Step 1
Concept
(6) is rational and \(-\sqrt{10}\) is the irrational part. Identify the radical term in mixed form.
Step 2
Why this answer is correct
The correct answer is B. \(-\sqrt{10}\). (6) is rational and \(-\sqrt{10}\) is the irrational part. Identify the radical term in mixed form.
Step 3
Exam Tip
(6) परिमेय है और \(-\sqrt{10}\) अपरिमेय भाग है। मिश्रित रूप में मूल वाला पद पहचानें।
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\(\sqrt{45}+\sqrt{125}-\sqrt{20}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{45}+\sqrt{125}-\sqrt{20}\)?
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A \(4\sqrt{5}\)
B \(6\sqrt{5}\)
C \(8\sqrt{5}\)
D \(10\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(6\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(6\sqrt{5}\). Add and subtract coefficients of like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(6\sqrt{5}\). \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(6\sqrt{5}\). Add and subtract coefficients of like radicals.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए उत्तर \(6\sqrt{5}\) है। समान मूलों के गुणांक जोड़ें और घटाएं।
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यदि \(\sqrt{m}=9\) है तो (m) का मान क्या है?
If \(\sqrt{m}=9\), what is the value of (m)?
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A (18)
B (9)
C (81)
D \(\sqrt{9}\)
Explanation opens after your attempt
Step 1
Concept
Squaring both sides gives (m=81). Remembering perfect squares is useful.
Step 2
Why this answer is correct
The correct answer is C. (81). Squaring both sides gives (m=81). Remembering perfect squares is useful.
Step 3
Exam Tip
दोनों पक्षों का वर्ग करने पर (m=81) मिलता है। पूर्ण वर्गों को याद रखना उपयोगी है।
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यदि (n) धनात्मक पूर्ण संख्या है और (n) पूर्ण वर्ग है तो \(\sqrt{n}\) कैसी संख्या होगी?
If (n) is a positive integer and (n) is a perfect square, what type of number will \(\sqrt{n}\) be?
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A अपरिमेय / Irrational
B परिमेय / Rational
C न वास्तविक / Not real
D अनावर्ती दशमलव / Non-repeating decimal
Explanation opens after your attempt
Correct Answer
B. परिमेय / Rational
Step 1
Concept
The square root of a perfect square is an integer, so it is rational. First check whether the number is a perfect square.
Step 2
Why this answer is correct
The correct answer is B. परिमेय / Rational. The square root of a perfect square is an integer, so it is rational. First check whether the number is a perfect square.
Step 3
Exam Tip
पूर्ण वर्ग का वर्गमूल पूर्णांक होता है इसलिए परिमेय होता है। पहले यह देखें कि संख्या पूर्ण वर्ग है या नहीं।
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\(\sqrt{18}\times\sqrt{32}\) का मान क्या है?
What is the value of \(\sqrt{18}\times\sqrt{32}\)?
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A (24)
B \(\sqrt{50}\)
C \(12\sqrt{2}\)
D (576)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{18}\times\sqrt{32}=\sqrt{576}=24\). The product of two irrationals can be rational.
Step 2
Why this answer is correct
The correct answer is A. (24). \(\sqrt{18}\times\sqrt{32}=\sqrt{576}=24\). The product of two irrationals can be rational.
Step 3
Exam Tip
\(\sqrt{18}\times\sqrt{32}=\sqrt{576}=24\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है।
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\(\sqrt{5}+\sqrt{7}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{5}+\sqrt{7}\)?
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A यह \(\sqrt{12}\) के बराबर है / It is equal to \(\sqrt{12}\)
B यह परिमेय है / It is rational
C यह अपरिमेय है / It is irrational
D यह (12) है / It is (12)
Explanation opens after your attempt
Correct Answer
C. यह अपरिमेय है / It is irrational
Step 1
Concept
\(\sqrt{5}\) and \(\sqrt{7}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly as \(\sqrt{12}\).
Step 2
Why this answer is correct
The correct answer is C. यह अपरिमेय है / It is irrational. \(\sqrt{5}\) and \(\sqrt{7}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly as \(\sqrt{12}\).
Step 3
Exam Tip
\(\sqrt{5}\) और \(\sqrt{7}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। अलग मूलों को सीधे जोड़कर \(\sqrt{12}\) नहीं बनाते।
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\(\frac{\sqrt{112}}{\sqrt{7}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{112}}{\sqrt{7}}\)?
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A (8)
B (4)
C \(\sqrt{119}\)
D (16)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{16}=4\). In division of roots take the quotient inside.
Step 2
Why this answer is correct
The correct answer is B. (4). \(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{16}=4\). In division of roots take the quotient inside.
Step 3
Exam Tip
\(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{16}=4\) है। मूलों के भाग में अंदर का भागफल लें।
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\(8\sqrt{2}-5\sqrt{2}\) का परिणाम क्या है?
What is the result of \(8\sqrt{2}-5\sqrt{2}\)?
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A \(13\sqrt{2}\)
B \(3\sqrt{2}\)
C (3)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
B. \(3\sqrt{2}\)
Step 1
Concept
Subtracting coefficients of like radicals gives \(3\sqrt{2}\). It is irrational because \(\sqrt{2}\) remains.
Step 2
Why this answer is correct
The correct answer is B. \(3\sqrt{2}\). Subtracting coefficients of like radicals gives \(3\sqrt{2}\). It is irrational because \(\sqrt{2}\) remains.
Step 3
Exam Tip
समान मूलों के गुणांक घटाने पर \(3\sqrt{2}\) मिलता है। यह अपरिमेय है क्योंकि \(\sqrt{2}\) बचता है।
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(\sqrt{2}\(\sqrt{8}+\sqrt{18}\)) का मान क्या है?
What is the value of (\sqrt{2}\(\sqrt{8}+\sqrt{18}\))?
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A (10)
B \(5\sqrt{2}\)
C (20)
D \(2\sqrt{26}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(5\sqrt{2}\) and the product is (10). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. (10). \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(5\sqrt{2}\) and the product is (10). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए कोष्ठक \(5\sqrt{2}\) और गुणनफल (10) है। पहले कोष्ठक सरल करें।
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\(\sqrt{29}\) और \(\sqrt{31}\) की तुलना में कौन-सा कथन सही है?
Which statement is correct when comparing \(\sqrt{29}\) and \(\sqrt{31}\)?
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A \(\sqrt{29}>\sqrt{31}\)
B \(\sqrt{29}=\sqrt{31}\)
C \(\sqrt{29}<\sqrt{31}\)
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{29}<\sqrt{31}\)
Step 1
Concept
Since (29<31), \(\sqrt{29}<\sqrt{31}\). For positive roots compare the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{29}<\sqrt{31}\). Since (29<31), \(\sqrt{29}<\sqrt{31}\). For positive roots compare the numbers inside.
Step 3
Exam Tip
क्योंकि (29<31), इसलिए \(\sqrt{29}<\sqrt{31}\)। धनात्मक मूलों में अंदर की संख्याओं की तुलना करें।
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(\sqrt{3}\times\(2\sqrt{3}+5\)) का सरल रूप क्या है?
What is the simplified form of (\sqrt{3}\times\(2\sqrt{3}+5\))?
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A \(6+5\sqrt{3}\)
B \(2+5\sqrt{3}\)
C (6+5)
D \(10\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(6+5\sqrt{3}\)
Step 1
Concept
Distributing gives \(2\times3+5\sqrt{3}=6+5\sqrt{3}\). Keep radical and rational terms separate.
Step 2
Why this answer is correct
The correct answer is A. \(6+5\sqrt{3}\). Distributing gives \(2\times3+5\sqrt{3}=6+5\sqrt{3}\). Keep radical and rational terms separate.
Step 3
Exam Tip
वितरण करने पर \(2\times3+5\sqrt{3}=6+5\sqrt{3}\) मिलता है। मूल वाले पद और परिमेय पद अलग रखें।
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\(\frac{2}{\sqrt{3}+1}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{2}{\sqrt{3}+1}\)?
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A \(\sqrt{3}+1\)
B \(\sqrt{3}-1\)
C \(2\sqrt{3}-2\)
D \(\frac{2}{\sqrt{3}-1}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{3}-1\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{2\(\sqrt{3}-1\)}{3-1}=\sqrt{3}-1). Make the denominator rational.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{3}-1\). Multiplying by the conjugate gives (\frac{2\(\sqrt{3}-1\)}{3-1}=\sqrt{3}-1). Make the denominator rational.
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{2\(\sqrt{3}-1\)}{3-1}=\sqrt{3}-1) मिलता है। हर को परिमेय बनाएं।
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\(8+\sqrt{17}\) और \(8-\sqrt{17}\) का योग क्या है?
What is the sum of \(8+\sqrt{17}\) and \(8-\sqrt{17}\)?
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A (16)
B \(2\sqrt{17}\)
C (8)
D (17)
Explanation opens after your attempt
Step 1
Concept
The irrational terms cancel and the sum is (16). The sum of conjugate numbers is rational.
Step 2
Why this answer is correct
The correct answer is A. (16). The irrational terms cancel and the sum is (16). The sum of conjugate numbers is rational.
Step 3
Exam Tip
अपरिमेय पद कट जाते हैं और योग (16) है। संयुग्मी संख्याओं का योग परिमेय होता है।
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\(8+\sqrt{17}\) और \(8-\sqrt{17}\) का गुणनफल क्या है?
What is the product of \(8+\sqrt{17}\) and \(8-\sqrt{17}\)?
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A (81)
B (47)
C \(64+\sqrt{17}\)
D \(16\sqrt{17}\)
Explanation opens after your attempt
Step 1
Concept
The product is (82 -\(\sqrt{17}\)2 =64-17=47). Use \(a^2-b^2\) in conjugate multiplication.
Step 2
Why this answer is correct
The correct answer is B. (47). The product is (82 -\(\sqrt{17}\)2 =64-17=47). Use \(a^2-b^2\) in conjugate multiplication.
Step 3
Exam Tip
गुणनफल (82 -\(\sqrt{17}\)2 =64-17=47) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं।
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यदि \(q=5-\sqrt{2}\) है तो (q) किस प्रकार की संख्या है?
If \(q=5-\sqrt{2}\), what type of number is (q)?
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A परिमेय / Rational
B पूर्णांक / Integer
C अपरिमेय / Irrational
D सांत दशमलव / Terminating decimal
Explanation opens after your attempt
Correct Answer
C. अपरिमेय / Irrational
Step 1
Concept
Subtracting irrational \(\sqrt{2}\) from rational (5) gives an irrational number. The irrational part remains.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. Subtracting irrational \(\sqrt{2}\) from rational (5) gives an irrational number. The irrational part remains.
Step 3
Exam Tip
परिमेय (5) में से अपरिमेय \(\sqrt{2}\) घटाने पर अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है।
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संख्या रेखा पर \(\sqrt{5}\) बनाने के लिए किस समकोण त्रिभुज का कर्ण उपयोग हो सकता है?
To construct \(\sqrt{5}\) on the number line, which right triangle hypotenuse can be used?
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A भुजाएँ (1) और (1) / Legs (1) and (1)
B भुजाएँ (1) और (2) / Legs (1) and (2)
C भुजाएँ (2) और (2) / Legs (2) and (2)
D भुजाएँ (3) और (1) / Legs (3) and (1)
Explanation opens after your attempt
Correct Answer
B. भुजाएँ (1) और (2) / Legs (1) and (2)
Step 1
Concept
In a right triangle, the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Pythagoras theorem is used in construction.
Step 2
Why this answer is correct
The correct answer is B. भुजाएँ (1) और (2) / Legs (1) and (2). In a right triangle, the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Pythagoras theorem is used in construction.
Step 3
Exam Tip
समकोण त्रिभुज में कर्ण \(\sqrt{1^2+2^2}=\sqrt{5}\) होगा। निर्माण में पाइथागोरस प्रमेय लगती है।
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\(\sqrt{54}\div\sqrt{6}\) का मान क्या है?
What is the value of \(\sqrt{54}\div\sqrt{6}\)?
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A \(\sqrt{48}\)
B (3)
C \(\sqrt{9}\)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{54}\div\sqrt{6}=\sqrt{9}=3\). In division of roots take the quotient inside.
Step 2
Why this answer is correct
The correct answer is B. (3). \(\sqrt{54}\div\sqrt{6}=\sqrt{9}=3\). In division of roots take the quotient inside.
Step 3
Exam Tip
\(\sqrt{54}\div\sqrt{6}=\sqrt{9}=3\) है। मूलों के भाग में अंदर का भागफल लें।
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यदि एक वर्ग का क्षेत्रफल (50) वर्ग इकाई है तो उसकी भुजा का सरल रूप क्या होगा?
If the area of a square is (50) square units, what will be the simplified form of its side?
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A \(25\sqrt{2}\)
B (10)
C \(5\sqrt{2}\)
D \(2\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
C. \(5\sqrt{2}\)
Step 1
Concept
The side will be \(\sqrt{50}=5\sqrt{2}\). In a square, side equals the square root of area.
Step 2
Why this answer is correct
The correct answer is C. \(5\sqrt{2}\). The side will be \(\sqrt{50}=5\sqrt{2}\). In a square, side equals the square root of area.
Step 3
Exam Tip
भुजा \(\sqrt{50}=5\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है।
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\(\frac{1}{4-\sqrt{7}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{1}{4-\sqrt{7}}\)?
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A \(\frac{4+\sqrt{7}}{9}\)
B \(\frac{4-\sqrt{7}}{9}\)
C \(4+\sqrt{7}\)
D \(\frac{1}{4+\sqrt{7}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4+\sqrt{7}}{9}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (16-7=9). So the answer is \(\frac{4+\sqrt{7}}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4+\sqrt{7}}{9}\). Multiplying by the conjugate makes the denominator (16-7=9). So the answer is \(\frac{4+\sqrt{7}}{9}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (16-7=9) बनता है। इसलिए उत्तर \(\frac{4+\sqrt{7}}{9}\) है।
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\(\sqrt{2}+\sqrt{50}\) और \(6\sqrt{2}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{2}+\sqrt{50}\) and \(6\sqrt{2}\)?
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A पहला बड़ा है / The first is greater
B दूसरा बड़ा है / The second is greater
C दोनों बराबर हैं / Both are equal
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
C. दोनों बराबर हैं / Both are equal
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\), so \(\sqrt{2}+\sqrt{50}=6\sqrt{2}\). Simplify before comparing.
Step 2
Why this answer is correct
The correct answer is C. दोनों बराबर हैं / Both are equal. \(\sqrt{50}=5\sqrt{2}\), so \(\sqrt{2}+\sqrt{50}=6\sqrt{2}\). Simplify before comparing.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\), इसलिए \(\sqrt{2}+\sqrt{50}=6\sqrt{2}\) है। तुलना से पहले सरल करें।
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\(3\sqrt{7}+2\sqrt{28}\) का सरल रूप क्या है?
What is the simplified form of \(3\sqrt{7}+2\sqrt{28}\)?
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A \(5\sqrt{7}\)
B \(7\sqrt{7}\)
C \(11\sqrt{7}\)
D \(14\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
B. \(7\sqrt{7}\)
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\), so \(3\sqrt{7}+4\sqrt{7}=7\sqrt{7}\). Watch both coefficients and radicals carefully.
Step 2
Why this answer is correct
The correct answer is B. \(7\sqrt{7}\). \(\sqrt{28}=2\sqrt{7}\), so \(3\sqrt{7}+4\sqrt{7}=7\sqrt{7}\). Watch both coefficients and radicals carefully.
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\), इसलिए \(3\sqrt{7}+4\sqrt{7}=7\sqrt{7}\) है। गुणांक और मूल दोनों ध्यान से देखें।
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किस विकल्प का दशमलव प्रसार असांत अनावर्ती होगा?
Which option will have a non-terminating non-repeating decimal expansion?
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A \(\frac{3}{8}\)
B \(\frac{5}{6}\)
C \(\sqrt{37}\)
D \(0.121212\ldots\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{37}\)
Step 1
Concept
\(\sqrt{37}\) is irrational because (37) is not a perfect square. An irrational number has a non-terminating non-repeating decimal.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{37}\). \(\sqrt{37}\) is irrational because (37) is not a perfect square. An irrational number has a non-terminating non-repeating decimal.
Step 3
Exam Tip
\(\sqrt{37}\) अपरिमेय है क्योंकि (37) पूर्ण वर्ग नहीं है। अपरिमेय का दशमलव असांत अनावर्ती होता है।
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यदि \(a=\sqrt{3}+2\) और \(b=\sqrt{3}-2\) हैं तो (ab) क्या है?
If \(a=\sqrt{3}+2\) and \(b=\sqrt{3}-2\), what is (ab)?
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A (-1)
B (1)
C (7)
D \(4\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{3}\)2 -22 =3-4=-1). A conjugate product can be rational.
Step 2
Why this answer is correct
The correct answer is A. (-1). (ab=\(\sqrt{3}\)2 -22 =3-4=-1). A conjugate product can be rational.
Step 3
Exam Tip
(ab=\(\sqrt{3}\)2 -22 =3-4=-1) है। संयुग्मी गुणनफल परिमेय हो सकता है।
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\(\sqrt{7}+\sqrt{112}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{7}+\sqrt{112}\)?
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A \(3\sqrt{7}\)
B \(5\sqrt{7}\)
C \(17\sqrt{7}\)
D \(\sqrt{119}\)
Explanation opens after your attempt
Correct Answer
B. \(5\sqrt{7}\)
Step 1
Concept
\(\sqrt{112}=4\sqrt{7}\), so the sum is \(5\sqrt{7}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(5\sqrt{7}\). \(\sqrt{112}=4\sqrt{7}\), so the sum is \(5\sqrt{7}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{112}=4\sqrt{7}\), इसलिए योग \(5\sqrt{7}\) है। समान मूल जोड़ने से पहले सरल करें।
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(\(3\sqrt{2}\)2 ) का मान क्या है?
What is the value of (\(3\sqrt{2}\)2 )?
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A (6)
B (12)
C (18)
D \(9\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(\(3\sqrt{2}\)2 =9\times2=18). Square both the coefficient and the radical.
Step 2
Why this answer is correct
The correct answer is C. (18). (\(3\sqrt{2}\)2 =9\times2=18). Square both the coefficient and the radical.
Step 3
Exam Tip
(\(3\sqrt{2}\)2 =9\times2=18) है। गुणांक और मूल दोनों का वर्ग करें।
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यदि \(u=\sqrt{23}+4\) और \(v=\sqrt{23}-4\) हैं तो (uv) का मान क्या है?
If \(u=\sqrt{23}+4\) and \(v=\sqrt{23}-4\), what is the value of (uv)?
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A (7)
B (39)
C \(\sqrt{23}\)
D \(8\sqrt{23}\)
Explanation opens after your attempt
Step 1
Concept
In conjugate multiplication (uv=\(\sqrt{23}\)2 -42 =7). Use \(a^2-b^2\) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (7). In conjugate multiplication (uv=\(\sqrt{23}\)2 -42 =7). Use \(a^2-b^2\) in such questions.
Step 3
Exam Tip
संयुग्मी गुणन में (uv=\(\sqrt{23}\)2 -42 =7) होता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएं।
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\(\frac{\sqrt{150}+\sqrt{54}}{\sqrt{6}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{150}+\sqrt{54}}{\sqrt{6}}\)?
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A (6)
B (8)
C (14)
D \(\sqrt{204}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{150}=5\sqrt{6}\) and \(\sqrt{54}=3\sqrt{6}\), so division gives (8). First convert the numerator into like radicals.
Step 2
Why this answer is correct
The correct answer is B. (8). \(\sqrt{150}=5\sqrt{6}\) and \(\sqrt{54}=3\sqrt{6}\), so division gives (8). First convert the numerator into like radicals.
Step 3
Exam Tip
\(\sqrt{150}=5\sqrt{6}\) और \(\sqrt{54}=3\sqrt{6}\), इसलिए भाग देने पर (8) मिलता है। पहले अंश को समान मूल में बदलें।
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यदि \(\sqrt{k}\) संख्या (7) और (8) के बीच है, तो (k) के लिए कौन-सा मान संभव है?
If \(\sqrt{k}\) lies between (7) and (8), which value of (k) is possible?
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A (48)
B (64)
C (57)
D (81)
Explanation opens after your attempt
Step 1
Concept
Since (49<57<64), \(7<\sqrt{57}<8\). Decide square-root bounds using squares.
Step 2
Why this answer is correct
The correct answer is C. (57). Since (49<57<64), \(7<\sqrt{57}<8\). Decide square-root bounds using squares.
Step 3
Exam Tip
क्योंकि (49<57<64), इसलिए \(7<\sqrt{57}<8\)। वर्गमूल की सीमा वर्गों से तय करें।
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(\sqrt{10}\(3\sqrt{10}-2\sqrt{40}\)) का सरल रूप क्या है?
What is the simplified form of (\sqrt{10}\(3\sqrt{10}-2\sqrt{40}\))?
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A (10)
B (-10)
C \(30-4\sqrt{10}\)
D \(-2\sqrt{40}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{40}=2\sqrt{10}\), so the bracket is \(-\sqrt{10}\) and the product is (-10). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is B. (-10). \(\sqrt{40}=2\sqrt{10}\), so the bracket is \(-\sqrt{10}\) and the product is (-10). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{40}=2\sqrt{10}\), इसलिए कोष्ठक \(-\sqrt{10}\) और गुणनफल (-10) है। पहले कोष्ठक को सरल करें।
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कौन-सा विकल्प \(\frac{6}{\sqrt{19}}\) का सही परिमेयकृत रूप है?
Which option is the correct rationalised form of \(\frac{6}{\sqrt{19}}\)?
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A \(\frac{6}{19}\)
B \(\frac{\sqrt{19}}{6}\)
C \(\frac{6\sqrt{19}}{19}\)
D \(6\sqrt{19}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{6\sqrt{19}}{19}\)
Step 1
Concept
Multiplying numerator and denominator by \(\sqrt{19}\) gives \(\frac{6\sqrt{19}}{19}\). Rationalisation removes the radical from the denominator.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{6\sqrt{19}}{19}\). Multiplying numerator and denominator by \(\sqrt{19}\) gives \(\frac{6\sqrt{19}}{19}\). Rationalisation removes the radical from the denominator.
Step 3
Exam Tip
अंश और हर को \(\sqrt{19}\) से गुणा करने पर \(\frac{6\sqrt{19}}{19}\) मिलता है। परिमेयकरण में हर से मूल हटता है।
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\(\sqrt{12}+\sqrt{108}\) और \(8\sqrt{3}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{12}+\sqrt{108}\) and \(8\sqrt{3}\)?
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A पहला बड़ा है / The first is greater
B दूसरा बड़ा है / The second is greater
C दोनों बराबर हैं / Both are equal
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
C. दोनों बराबर हैं / Both are equal
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the sum is \(8\sqrt{3}\). Find simplified forms before comparing.
Step 2
Why this answer is correct
The correct answer is C. दोनों बराबर हैं / Both are equal. \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the sum is \(8\sqrt{3}\). Find simplified forms before comparing.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{108}=6\sqrt{3}\), इसलिए योग \(8\sqrt{3}\) है। तुलना से पहले सरल रूप निकालें।
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यदि किसी वर्ग की भुजा \(2\sqrt{7}\) इकाई है तो उसका क्षेत्रफल क्या होगा?
If the side of a square is \(2\sqrt{7}\) units, what will be its area?
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A (14) वर्ग इकाई / (14) square units
B (21) वर्ग इकाई / (21) square units
C (28) वर्ग इकाई / (28) square units
D (49) वर्ग इकाई / (49) square units
Explanation opens after your attempt
Correct Answer
C. (28) वर्ग इकाई / (28) square units
Step 1
Concept
The area is (\(2\sqrt{7}\)2 =28) square units. A square with irrational side can have rational area.
Step 2
Why this answer is correct
The correct answer is C. (28) वर्ग इकाई / (28) square units. The area is (\(2\sqrt{7}\)2 =28) square units. A square with irrational side can have rational area.
Step 3
Exam Tip
क्षेत्रफल (\(2\sqrt{7}\)2 =28) वर्ग इकाई है। अपरिमेय भुजा का क्षेत्रफल परिमेय हो सकता है।
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दशमलव \(5.12012001200012\ldots\) किस प्रकार की संख्या है?
What type of number is the decimal \(5.12012001200012\ldots\)?
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A सांत परिमेय / Terminating rational
B आवर्ती परिमेय / Repeating rational
C अपरिमेय / Irrational
D पूर्णांक / Integer
Explanation opens after your attempt
Correct Answer
C. अपरिमेय / Irrational
Step 1
Concept
It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.
Step 3
Exam Tip
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव को पहचानना जरूरी है।
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\(\frac{1}{5+\sqrt{6}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{1}{5+\sqrt{6}}\)?
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A \(\frac{5-\sqrt{6}}{19}\)
B \(\frac{5+\sqrt{6}}{19}\)
C \(5-\sqrt{6}\)
D \(\frac{1}{5-\sqrt{6}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5-\sqrt{6}}{19}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (25-6=19). So the rationalised form is \(\frac{5-\sqrt{6}}{19}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5-\sqrt{6}}{19}\). Multiplying by the conjugate makes the denominator (25-6=19). So the rationalised form is \(\frac{5-\sqrt{6}}{19}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (25-6=19) बनता है। इसलिए परिमेयकृत रूप \(\frac{5-\sqrt{6}}{19}\) है।
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