Question : Three numbers are in the ratio $\frac{1}{2}: \frac{2}{3}: \frac{3}{4}$. If the difference between the greatest number and the smallest number is 33, then the HCF of the three numbers is:
Option 1: 9
Option 2: 5
Option 3: 13
Option 4: 11
Correct Answer: 11
Solution :
Let the three numbers as $x$, $y$, and $z$. Given that the numbers are in the ratio $⇒\frac{1}{2} : \frac{2}{3} : \frac{3}{4}$.
$x : y : z = \frac{1}{2} : \frac{2}{3} : \frac{3}{4}$
Multiply all the ratios by the least common multiple (LCM) of 2, 3, and 4, which is 12.
$⇒x : y : z = 6 : 8 : 9$
So, the numbers are $6k$, $8k$, and $9k$ for some integer $k$.
Given that the difference between the greatest number and the smallest number is 33.
$⇒9k - 6k = 33$
$⇒k = 11$
Therefore, the three numbers are $6k = 66$, $8k = 88$, and $9k = 99$.
The highest common factor (HCF) of these three numbers is 11.
Hence, the correct answer is 11.
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