After some digits, (45) repeats, so it is a recurring decimal. A recurring decimal is rational.
Step 2
Why this answer is correct
The correct answer is A. परिमेय संख्या / Rational number. After some digits, (45) repeats, so it is a recurring decimal. A recurring decimal is rational.
Step 3
Exam Tip
कुछ अंकों के बाद (45) दोहरता है, इसलिए यह आवर्ती दशमलव है। आवर्ती दशमलव परिमेय होता है।
\(0.\overline{216}=\frac{216}{999}=\frac{8}{37}\). For a purely recurring decimal, first use a denominator of (9)'s and then reduce fully.
Step 2
Why this answer is correct
The correct answer is A. (37). \(0.\overline{216}=\frac{216}{999}=\frac{8}{37}\). For a purely recurring decimal, first use a denominator of (9)'s and then reduce fully.
Step 3
Exam Tip
\(0.\overline{216}=\frac{216}{999}=\frac{8}{37}\) है। पूर्ण आवर्ती दशमलव में पहले (9) वाला हर बनाएं और फिर पूरा सरल करें।
Since \(19^2\) remains, the decimal is non-terminating recurring, and the larger exponent among (2) and (5) is (4). In such questions, separate recurrence from the initial delay.
Step 2
Why this answer is correct
The correct answer is B. (4). Since \(19^2\) remains, the decimal is non-terminating recurring, and the larger exponent among (2) and (5) is (4). In such questions, separate recurrence from the initial delay.
Step 3
Exam Tip
\(19^2\) बचने से दशमलव असांत आवर्ती होगा और (2), (5) की बड़ी घात (4) आरंभिक अनावर्ती भाग देगी। ऐसे प्रश्न में आवर्तीपन और आरंभिक देरी अलग-अलग देखें।
\(448=2^6\cdot 7\), so (6) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{448}\). \(448=2^6\cdot 7\), so (6) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).
Step 3
Exam Tip
\(448=2^6\cdot 7\) है इसलिए आवर्ती भाग से पहले (6) अनावर्ती अंक आएँगे। तुलना में (2) और (5) की बड़ी घात देखें।
After cancellation, the denominator becomes \(2^4\cdot 5^2\cdot 17\). Since (17) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator becomes \(2^4\cdot 5^2\cdot 17\). Since (17) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^4\cdot 5^2\cdot 17\) बचेगा। (17) बचने से दशमलव असांत आवर्ती होगा।
B. असांत आवर्ती और (4) अनावर्ती आरंभिक अंक/Non-terminating recurring with (4) initial non-repeating digits
Step 1
Concept
Since (17) remains, the decimal is non-terminating recurring. The larger exponent in \(2^4\cdot 5^4\) gives (4) initial non-repeating digits.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती और (4) अनावर्ती आरंभिक अंक / Non-terminating recurring with (4) initial non-repeating digits. Since (17) remains, the decimal is non-terminating recurring. The larger exponent in \(2^4\cdot 5^4\) gives (4) initial non-repeating digits.
Step 3
Exam Tip
(17) बचता है इसलिए दशमलव असांत आवर्ती होगा। \(2^4\cdot 5^4\) की बड़ी घात (4) आरंभिक अनावर्ती भाग दिखाती है।
\(0.\overline{36}=\frac{36}{99}\) and \(0.\overline{63}=\frac{63}{99}\), so their sum is (1). The sum of two recurring decimals can be terminating.
Step 2
Why this answer is correct
The correct answer is A. सांत / Terminating. \(0.\overline{36}=\frac{36}{99}\) and \(0.\overline{63}=\frac{63}{99}\), so their sum is (1). The sum of two recurring decimals can be terminating.
Step 3
Exam Tip
\(0.\overline{36}=\frac{36}{99}\) और \(0.\overline{63}=\frac{63}{99}\) हैं इसलिए योग (1) है। दो आवर्ती दशमलवों का योग सांत भी हो सकता है।
Since \(320=2^6\cdot 5\), the reduced denominator is \(2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Since \(320=2^6\cdot 5\), the reduced denominator is \(2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(320=2^6\cdot 5\) कटने पर हर \(2\cdot 5^2\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
A positive power of (13) remains in the reduced denominator. Therefore the rational number has a non-terminating recurring decimal.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. A positive power of (13) remains in the reduced denominator. Therefore the rational number has a non-terminating recurring decimal.
Step 3
Exam Tip
सरलतम हर में (13) की धनात्मक घात बची है। इसलिए परिमेय संख्या का दशमलव असांत आवर्ती होगा।
The factor (41) makes the decimal recurring, and the larger exponent of (2) and (5) is (7), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.
Step 2
Why this answer is correct
The correct answer is C. (7). The factor (41) makes the decimal recurring, and the larger exponent of (2) and (5) is (7), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.
Step 3
Exam Tip
(41) के कारण दशमलव आवर्ती होगा और (2), (5) की बड़ी घात (7) अनावर्ती आरंभ देगी। मिश्रित हर में बड़ी घात से देरी मिलती है।
\(0.\overline{063}=\frac{63}{999}\), and reducing by (9) gives \(\frac{7}{111}\). An initial zero inside the repeating block is also counted as a digit.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{111}\). \(0.\overline{063}=\frac{63}{999}\), and reducing by (9) gives \(\frac{7}{111}\). An initial zero inside the repeating block is also counted as a digit.
Step 3
Exam Tip
\(0.\overline{063}=\frac{63}{999}\) और (9) से सरल करने पर \(\frac{7}{111}\) मिलता है। आवर्ती भाग में आरंभिक शून्य को भी अंक माना जाता है।
Since \(245=5\cdot 7^2\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is C. असांत आवर्ती / Non-terminating recurring. Since \(245=5\cdot 7^2\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(245=5\cdot 7^2\) कटने पर हर \(2^2\cdot 5\cdot 7\) बचता है। (7) बचने से दशमलव असांत आवर्ती होगा।
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।
The factor (17) makes the decimal recurring, and the larger exponent among (2) and (5) is (6), giving the initial non-repeating part. Understand recurrence and delay separately.
Step 2
Why this answer is correct
The correct answer is B. (6). The factor (17) makes the decimal recurring, and the larger exponent among (2) and (5) is (6), giving the initial non-repeating part. Understand recurrence and delay separately.
Step 3
Exam Tip
(17) के कारण दशमलव आवर्ती होगा और (2), (5) की बड़ी घात (6) आरंभिक अनावर्ती भाग देगी। आवर्तीपन और आरंभिक देरी को अलग-अलग समझें।
After cancelling \(55=5\cdot 11\), the denominator becomes \(2^2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancelling \(55=5\cdot 11\), the denominator becomes \(2^2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(55=5\cdot 11\) कटने पर हर \(2^2\cdot 5^2\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
\(0.\overline{108}=\frac{108}{999}=\frac{4}{37}\). First form the denominator with (9)'s according to the repeating digits and then reduce.
Step 2
Why this answer is correct
The correct answer is B. (37). \(0.\overline{108}=\frac{108}{999}=\frac{4}{37}\). First form the denominator with (9)'s according to the repeating digits and then reduce.
Step 3
Exam Tip
\(0.\overline{108}=\frac{108}{999}=\frac{4}{37}\) है। आवर्ती अंकों की संख्या के अनुसार पहले (9) वाला हर बनाएं फिर सरल करें।
After cancellation, the denominator is \(2^3\cdot 3\cdot 5^4\cdot 11\), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator is \(2^3\cdot 3\cdot 5^4\cdot 11\), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 3\cdot 5^4\cdot 11\) बचता है, जिसमें (3) और (11) हैं। सरलतम हर में (2) और (5) के अलावा गुणनखंड बचें तो दशमलव असांत आवर्ती होता है।
\(224=2^5\cdot 7\), so (5) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{224}\). \(224=2^5\cdot 7\), so (5) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).
Step 3
Exam Tip
\(224=2^5\cdot 7\) है इसलिए आवर्ती भाग से पहले (5) अनावर्ती अंक आएँगे। तुलना में (2) और (5) की बड़ी घात देखें।
After cancellation, the denominator becomes \(2^3\cdot 5^3\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator becomes \(2^3\cdot 5^3\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 5^3\cdot 13\) बचेगा। (13) बचने से दशमलव असांत आवर्ती होगा।
B. असांत आवर्ती और (3) अनावर्ती आरंभिक अंक/Non-terminating recurring with (3) initial non-repeating digits
Step 1
Concept
Since (11) remains, the decimal is non-terminating recurring. The larger exponent in \(2^3\cdot 5^3\) gives (3) initial non-repeating digits.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती और (3) अनावर्ती आरंभिक अंक / Non-terminating recurring with (3) initial non-repeating digits. Since (11) remains, the decimal is non-terminating recurring. The larger exponent in \(2^3\cdot 5^3\) gives (3) initial non-repeating digits.
Step 3
Exam Tip
(11) बचता है इसलिए दशमलव असांत आवर्ती होगा। \(2^3\cdot 5^3\) की बड़ी घात (3) आरंभिक अनावर्ती भाग दिखाती है।
\(0.\overline{27}=\frac{27}{99}\) and \(0.\overline{72}=\frac{72}{99}\), so their sum is (1). The sum of two recurring decimals can be terminating.
Step 2
Why this answer is correct
The correct answer is A. सांत / Terminating. \(0.\overline{27}=\frac{27}{99}\) and \(0.\overline{72}=\frac{72}{99}\), so their sum is (1). The sum of two recurring decimals can be terminating.
Step 3
Exam Tip
\(0.\overline{27}=\frac{27}{99}\) और \(0.\overline{72}=\frac{72}{99}\) हैं इसलिए योग (1) है। दो आवर्ती दशमलवों का योग सांत भी हो सकता है।
Since \(200=2^3\cdot 5^2\), the reduced denominator is \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Since \(200=2^3\cdot 5^2\), the reduced denominator is \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(200=2^3\cdot 5^2\) कटने पर हर \(5\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
A positive power of (11) remains in the reduced denominator. Therefore the rational number has a non-terminating recurring decimal.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. A positive power of (11) remains in the reduced denominator. Therefore the rational number has a non-terminating recurring decimal.
Step 3
Exam Tip
सरलतम हर में (11) की धनात्मक घात बची है। इसलिए परिमेय संख्या का दशमलव असांत आवर्ती होगा।
The factor (31) makes the decimal recurring, and the larger exponent of (2) and (5) is (6), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.
Step 2
Why this answer is correct
The correct answer is C. (6). The factor (31) makes the decimal recurring, and the larger exponent of (2) and (5) is (6), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.
Step 3
Exam Tip
(31) के कारण दशमलव आवर्ती होगा और (2), (5) की बड़ी घात (6) अनावर्ती आरंभ देगी। मिश्रित हर में बड़ी घात से देरी मिलती है।
\(0.\overline{045}=\frac{45}{999}\), 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।
