After cancelling \(385=5\cdot7\cdot11\), only \(2^3\) remains in the denominator. In exams always check the denominator in lowest form.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. After cancelling \(385=5\cdot7\cdot11\), only \(2^3\) remains in the denominator. In exams always check the denominator in lowest form.
Step 3
Exam Tip
\(385=5\cdot7\cdot11\) कटने के बाद हर में केवल \(2^3\) बचता है। परीक्षा में हमेशा सरलतम रूप के हर को देखें।
After cancelling \(231=3\cdot7\cdot11\), the denominator left is \(2\cdot5^2\). Therefore the decimal terminates.
Step 2
Why this answer is correct
The correct answer is A. समाप्त / Terminating. After cancelling \(231=3\cdot7\cdot11\), the denominator left is \(2\cdot5^2\). Therefore the decimal terminates.
Step 3
Exam Tip
\(231=3\cdot7\cdot11\) कटने के बाद हर में \(2\cdot5^2\) बचता है। इसलिए दशमलव समाप्त होगा।
\(154=2\cdot7\cdot11\), so after cancellation only \(5^2\) remains in the denominator. In exams decide from the denominator in lowest form.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. \(154=2\cdot7\cdot11\), so after cancellation only \(5^2\) remains in the denominator. In exams decide from the denominator in lowest form.
Step 3
Exam Tip
\(154=2\cdot7\cdot11\), इसलिए कटने के बाद हर में केवल \(5^2\) बचता है। परीक्षा में निर्णय हमेशा सरलतम रूप के हर से करें।
In \(\frac{63}{2^5\cdot5^2\cdot7}\), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{63}{2^5\cdot5^2\cdot7}\). In \(\frac{63}{2^5\cdot5^2\cdot7}\), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.
Step 3
Exam Tip
\(\frac{63}{2^5\cdot5^2\cdot7}\) में (63) और (7) कटने के बाद हर में केवल (2) और (5) बचते हैं। परीक्षा में पहले भिन्न को सरलतम रूप में लाएं।
A terminating decimal can be converted into \(\frac{p}{q}\) form. Hence it is rational.
Step 2
Why this answer is correct
The correct answer is A. वह परिमेय संख्या है / It is a rational number. A terminating decimal can be converted into \(\frac{p}{q}\) form. Hence it is rational.
Step 3
Exam Tip
सांत दशमलव को \(\frac{p}{q}\) रूप में बदला जा सकता है। इसलिए वह परिमेय होता है।
(0.875) is a terminating decimal so it is rational. A terminating decimal can be written as a fraction.
Step 2
Why this answer is correct
The correct answer is A. परिमेय संख्या / Rational number. (0.875) is a terminating decimal so it is rational. A terminating decimal can be written as a fraction.
Step 3
Exam Tip
(0.875) सांत दशमलव है इसलिए परिमेय है। सांत दशमलव को भिन्न में लिखा जा सकता है।
A. वह परिमेय संख्या होती है/It is a rational number
Step 1
Concept
A terminating decimal can be written in \(\frac{p}{q}\) form. So it is rational and real.
Step 2
Why this answer is correct
The correct answer is A. वह परिमेय संख्या होती है / It is a rational number. A terminating decimal can be written in \(\frac{p}{q}\) form. So it is rational and real.
Step 3
Exam Tip
सांत दशमलव को \(\frac{p}{q}\) रूप में लिखा जा सकता है। इसलिए वह परिमेय और वास्तविक होता है।
(1.25) is terminating and can be written as \(\frac{5}{4}\). Terminating decimals are rational.
Step 2
Why this answer is correct
The correct answer is A. परिमेय संख्या / Rational number. (1.25) is terminating and can be written as \(\frac{5}{4}\). Terminating decimals are rational.
Step 3
Exam Tip
(1.25) सांत दशमलव है और इसे \(\frac{5}{4}\) लिखा जा सकता है। सांत दशमलव परिमेय होते हैं।
For exactly (6) places, the larger exponent of (2) and (5) must be (6). Since \(3125=5^5\), it gives only (5) decimal places.
Step 2
Why this answer is correct
The correct answer is B. (3125). For exactly (6) places, the larger exponent of (2) and (5) must be (6). Since \(3125=5^5\), it gives only (5) decimal places.
Step 3
Exam Tip
ठीक (6) स्थानों के लिए (2) और (5) की बड़ी घात (6) होनी चाहिए। \(3125=5^5\) है, इसलिए यह केवल (5) दशमलव स्थान देगा।
Since \(750=2\cdot 3\cdot 5^3\), the reduced denominator is \(2^5\cdot 5^2\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 2
Why this answer is correct
The correct answer is C. (5) स्थान / (5) places. Since \(750=2\cdot 3\cdot 5^3\), the reduced denominator is \(2^5\cdot 5^2\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 3
Exam Tip
\(750=2\cdot 3\cdot 5^3\) कटने पर हर \(2^5\cdot 5^2\) बचता है। बड़ी घात (5) है, इसलिए दशमलव (5) स्थानों पर समाप्त होगा।
A. ठीक (5) स्थानों पर समाप्त/Terminates exactly after (5) places
Step 1
Concept
Since \(7^0=1\), the effective denominator is \(2^5\cdot 5^5=10^5\). The decimal terminates exactly after (5) places.
Step 2
Why this answer is correct
The correct answer is A. ठीक (5) स्थानों पर समाप्त / Terminates exactly after (5) places. Since \(7^0=1\), the effective denominator is \(2^5\cdot 5^5=10^5\). The decimal terminates exactly after (5) places.
Step 3
Exam Tip
\(7^0=1\) है इसलिए प्रभावी हर \(2^5\cdot 5^5=10^5\) है। दशमलव ठीक (5) स्थानों पर समाप्त होगा।
\(0.00015625=\frac{15625}{100000000}\), and reducing by (15625) gives \(\frac{1}{6400}\). Do not forget to cancel common factors in large denominators.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{6400}\). \(0.00015625=\frac{15625}{100000000}\), and reducing by (15625) gives \(\frac{1}{6400}\). Do not forget to cancel common factors in large denominators.
Step 3
Exam Tip
\(0.00015625=\frac{15625}{100000000}\) है और (15625) से सरल करने पर \(\frac{1}{6400}\) मिलता है। बड़े हर में समान गुणनखंड काटना न भूलें।
\(0.015625=\frac{15625}{1000000}=\frac{1}{64}\). 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।
A. केवल (2) और (5) हो सकते हैं/Only (2) and (5) can occur
Step 1
Concept
For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In \(q^4\), powers increase but no new prime factor appears.
Step 2
Why this answer is correct
The correct answer is A. केवल (2) और (5) हो सकते हैं / Only (2) and (5) can occur. For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In \(q^4\), powers increase but no new prime factor appears.
Step 3
Exam Tip
सांत दशमलव में सरलतम हर (q) में केवल (2) और (5) हो सकते हैं। \(q^4\) में घातें बढ़ेंगी लेकिन नया अभाज्य गुणनखंड नहीं आएगा।
\(0.046875=\frac{46875}{1000000}\), 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।
A. सांत और (5) स्थानों पर समाप्त/Terminating after (5) places
Step 1
Concept
Both \(7^0\) and \(19^0\) equal (1), so the effective denominator is \(2^5\cdot 5^2\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 2
Why this answer is correct
The correct answer is A. सांत और (5) स्थानों पर समाप्त / Terminating after (5) places. Both \(7^0\) and \(19^0\) equal (1), so the effective denominator is \(2^5\cdot 5^2\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 3
Exam Tip
\(7^0\) और \(19^0\) दोनों (1) हैं इसलिए प्रभावी हर \(2^5\cdot 5^2\) है। बड़ी घात (5) होने से दशमलव (5) स्थानों पर समाप्त होगा।
At most (9) decimal places means the fraction can be written with denominator \(10^9\). Therefore the reduced denominator must divide \(10^9\).
Step 2
Why this answer is correct
The correct answer is C. \(10^9\). At most (9) decimal places means the fraction can be written with denominator \(10^9\). Therefore the reduced denominator must divide \(10^9\).
Step 3
Exam Tip
अधिकतम (9) दशमलव स्थानों का अर्थ है भिन्न को \(10^9\) हर के साथ लिखा जा सकता है। इसलिए सरलतम हर \(10^9\) का भाजक होगा।
A. ठीक (7) स्थानों पर समाप्त/Terminates exactly after (7) places
Step 1
Concept
The reduced denominator is \(10^7\), so the decimal terminates exactly after (7) places. If the denominator is reduced, do not assume further cancellation.
Step 2
Why this answer is correct
The correct answer is A. ठीक (7) स्थानों पर समाप्त / Terminates exactly after (7) places. The reduced denominator is \(10^7\), so the decimal terminates exactly after (7) places. If the denominator is reduced, do not assume further cancellation.
Step 3
Exam Tip
सरलतम हर \(10^7\) है इसलिए दशमलव ठीक (7) स्थानों पर समाप्त होगा। सरलतम हर दिया हो तो अंश से और कटौती नहीं माननी चाहिए।
After cancellation, the denominator becomes \(2^5\cdot 5^4\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 2
Why this answer is correct
The correct answer is B. (5). After cancellation, the denominator becomes \(2^5\cdot 5^4\). The larger exponent is (5), so the decimal terminates after (5) places.
Step 3
Exam Tip
कटौती के बाद हर \(2^5\cdot 5^4\) बचेगा। बड़ी घात (5) है इसलिए दशमलव (5) स्थानों पर समाप्त होगा।
A. कथन और कारण दोनों सही हैं तथा कारण सही व्याख्या है/Both are true and the reason explains it
Step 1
Concept
Since \(169=13^2\), the reduced denominator is \(2^3\cdot 5^4\). Therefore the reason correctly explains the terminating decimal rule.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं तथा कारण सही व्याख्या है / Both are true and the reason explains it. Since \(169=13^2\), the reduced denominator is \(2^3\cdot 5^4\). Therefore the reason correctly explains the terminating decimal rule.
Step 3
Exam Tip
\(169=13^2\) कटने पर हर \(2^3\cdot 5^4\) बचता है। इसलिए कारण सांत दशमलव के नियम को सही तरह समझाता है।
\(0.00084=\frac{84}{100000}\), 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।
For exactly (8) places, the larger exponent of (2) and (5) must be (8). Only \(2^8\cdot 5^3\) satisfies this.
Step 2
Why this answer is correct
The correct answer is B. \(2^8\cdot 5^3\). For exactly (8) places, the larger exponent of (2) and (5) must be (8). Only \(2^8\cdot 5^3\) satisfies this.
Step 3
Exam Tip
ठीक (8) स्थानों के लिए (2) और (5) की बड़ी घात (8) होनी चाहिए। दिए विकल्पों में केवल \(2^8\cdot 5^3\) यह शर्त पूरी करता है।
The denominator has only (2) and (5), so the decimal terminates with the larger exponent (9). In exams, use the larger exponent instead of adding exponents.
Step 2
Why this answer is correct
The correct answer is C. (9) स्थान / (9) places. The denominator has only (2) and (5), so the decimal terminates with the larger exponent (9). In exams, use the larger exponent instead of adding exponents.
Step 3
Exam Tip
हर में केवल (2) और (5) हैं इसलिए दशमलव सांत होगा और स्थान बड़ी घात (9) के बराबर होंगे। परीक्षा में घातों को जोड़ने की जगह बड़ी घात देखें।
