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।
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।
After cancelling \(22=2\cdot 11\), the denominator becomes \(2\cdot 5^4\cdot 11\). 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 \(22=2\cdot 11\), the denominator becomes \(2\cdot 5^4\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(22=2\cdot 11\) कटने पर हर \(2\cdot 5^4\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
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।
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।
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।
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।
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।
After cancelling \(14=2\cdot 7\), the denominator becomes \(2\cdot 5^3\cdot 7\). 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. After cancelling \(14=2\cdot 7\), the denominator becomes \(2\cdot 5^3\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(14=2\cdot 7\) कटने पर हर \(2\cdot 5^3\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
Since \(175=5^2\cdot 7\), 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 \(175=5^2\cdot 7\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(175=5^2\cdot 7\) कटने पर हर \(2^2\cdot 5\cdot 7\) बचता है। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancellation, the denominator becomes \(2^3\cdot 5^2\cdot 7\). 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. After cancellation, the denominator becomes \(2^3\cdot 5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 5^2\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancelling \(14=2\cdot 7\), the denominator becomes \(5^2\cdot 7\). 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. After cancelling \(14=2\cdot 7\), the denominator becomes \(5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(14=2\cdot 7\) कटने पर हर \(5^2\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
Even after \(125=5^3\) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Even after \(125=5^3\) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.
Step 3
Exam Tip
\(125=5^3\) कटने पर भी हर में (11) बचता है। सरलतम हर में (2) और (5) के अलावा कोई अभाज्य रहे तो दशमलव असांत आवर्ती होता है।
After cancellation, the denominator becomes \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
Check whether the whole power cancels or only part of it cancels. चरण 1: \(98=2\cdot 7^2\) है। चरण 2: कटौती के बाद हर \(5\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा। चरण 3: घात पूरी कटे या नहीं, यह ध्यान से देखें।
The reduced denominator becomes \(2^3\cdot 3\cdot 5^2\). Since (3) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
A prime factor may cancel only partially. चरण 1: अंश से \(2^4\cdot 3\) कटेगा। चरण 2: सरलतम हर \(2^3\cdot 3\cdot 5^2\) बचेगा। इसमें (3) बचा है, इसलिए दशमलव असांत आवर्ती होगा। चरण 3: एक ही अभाज्य गुणनखंड आंशिक रूप से कट सकता है।
The numerator (13) cancels only one factor (13) from \(13^2\).
Step 2
Why this answer is correct
The reduced denominator is \(2^2\cdot 5^2\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
Understand the difference between complete and partial cancellation. चरण 1: अंश का (13) हर के \(13^2\) में से केवल एक (13) काटेगा। चरण 2: सरलतम हर \(2^2\cdot 5^2\cdot 13\) बचेगा। (13) बचने से दशमलव असांत आवर्ती होगा। चरण 3: पूरी और आंशिक कटौती में फर्क समझें।
After cancellation, the denominator becomes \(2\cdot 5\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
After partial cancellation, always check the remaining factors. चरण 1: \(55=5\cdot 11\) है। चरण 2: कटौती के बाद हर \(2\cdot 5\cdot 11\) बचेगा। (11) बचने के कारण दशमलव असांत आवर्ती होगा। चरण 3: आंशिक कटौती के बाद बचे हुए गुणनखंडों को जरूर जाँचें।
The factor (5) and one (7) cancel, but one (7) remains. The reduced denominator is \(2^2\cdot 7\). So the decimal is non-terminating recurring.
Step 3
Exam Tip
After partial cancellation, check what factor remains. चरण 1: \(35=5\cdot 7\) है। चरण 2: हर से (5) और एक (7) कटेगा, पर एक (7) बच जाएगा। सरलतम हर \(2^2\cdot 7\) है। इसलिए दशमलव असांत आवर्ती होगा। चरण 3: आंशिक कटौती के बाद बचे गुणनखंड को जरूर देखें।
