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Answer :
To solve this problem, we need to find the two numbers whose Highest Common Factor (H.C.F) is 18 and Least Common Multiple (L.C.M) is 252. We also have additional conditions relating these two numbers.
Given:
- H.C.F. of the numbers is 18.
- L.C.M. of the numbers is 252.
- One number is 6 times 21. This number is computed as:
[tex]\text{Number 1} = 6 \times 21 = 126[/tex] - Six times the other number is 216.
Let's denote the two numbers as [tex]a[/tex] and [tex]b[/tex], where [tex]a = 126[/tex] and [tex]b[/tex] is the other number we need to find.
We know from the properties of H.C.F. and L.C.M. that:
[tex]a \times b = \text{H.C.F.}(a, b) \times \text{L.C.M.}(a, b)[/tex]
[tex]a \times b = 18 \times 252[/tex]
Calculate [tex]a \times b[/tex]:
[tex]a \times b = 4536[/tex]
We already know [tex]a = 126[/tex], so:
[tex]126 \times b = 4536[/tex]
Solve for [tex]b[/tex]:
[tex]b = \frac{4536}{126} = 36[/tex]
We have also been given that six times the other number is 216, so we should verify:
[tex]6b = 216[/tex]
[tex]b = \frac{216}{6} = 36[/tex]
This confirms that [tex]b = 36[/tex].
Now, we simply need to find the sum of both numbers:
[tex]\text{Sum} = a + b = 126 + 36 = 162[/tex]
Checking the options, it seems there is an inconsistency with the options provided. Should be re-checked with the correct sum expected as the result is consistent with the calculations and conditions given. The key solution steps are clear: calculating [tex]b[/tex] according to the provided requirements and ensuring the resulting operations check out.
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