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The HCF and LCM of two numbers 9 and 459 respectively. If one number is 27 .then find the other |
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Answer» second\xa0number\xa0=\xa0153Explanation:Let\xa0first\xa0number\xa0=\xa0a,second\xa0number\xa0=\xa0ba\xa0=\xa027\xa0( given )HCF(a,b)\xa0=\xa09andLCM\xa0(a,b)\xa0=\xa0459To\xa0find:Value of bsolution:We know that,$\\implies b = \\frac{9\\times 459}{27}$After cancellation, we get$\\implies b = 153$Therefore,second\xa0number\xa0=\xa0b\xa0=\xa0153 The other number is 153. Lesson 15 exercise 15.1 |
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