
The HCF and LCM of two numbers are 13 and 455, respectively. If one of the numbers lies between 75 and 125, then the number is
A. 78
B. 91
C. 104
D. 117
Answer
601.5k+ views
Hint: In this question LCM and HCF are given as two numbers, in order to find those numbers we will use the property that the product of LCM and HCF must be equal to the product of those numbers.
Complete step-by-step answer:
Let the numbers be $13a$ and $13b$ , where $a$ and $b$ are co- primes.
We know that $LCM \times HCF = {\text{ Product of two numbers}}$
Therefore we have
$
\Rightarrow 13a \times 13b = 13 \times 455 \\
\Rightarrow 169ab = 5915 \\
\Rightarrow ab = \dfrac{{5915}}{{169}} \\
\Rightarrow ab = 35 \\
$
Two co-primes with product 35 are 5 and 7.
Therefore, the required number is $(13 \times 5,13 \times 7) = (65,91)$
Hence, the number that lies between 75 and 125 is 91.
Note: In order to solve problems related to LCM and HCF, remember the concept of how to find LCM and HCF. Also remember the properties of LCM and HCF. Some of the properties of LCM and HCF are that HCF of co-prime numbers is 1. Therefore LCM of given co-prime numbers is equal to the product of the numbers. The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.
Complete step-by-step answer:
Let the numbers be $13a$ and $13b$ , where $a$ and $b$ are co- primes.
We know that $LCM \times HCF = {\text{ Product of two numbers}}$
Therefore we have
$
\Rightarrow 13a \times 13b = 13 \times 455 \\
\Rightarrow 169ab = 5915 \\
\Rightarrow ab = \dfrac{{5915}}{{169}} \\
\Rightarrow ab = 35 \\
$
Two co-primes with product 35 are 5 and 7.
Therefore, the required number is $(13 \times 5,13 \times 7) = (65,91)$
Hence, the number that lies between 75 and 125 is 91.
Note: In order to solve problems related to LCM and HCF, remember the concept of how to find LCM and HCF. Also remember the properties of LCM and HCF. Some of the properties of LCM and HCF are that HCF of co-prime numbers is 1. Therefore LCM of given co-prime numbers is equal to the product of the numbers. The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.
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