<p>There are two numbers a and b whose product is 192 and the quotient of AM by HM of their greatest common divisor and least common multiple is \(\frac{169}{48}\). The smaller of a and b is</p>
Step-by-Step Solution
Key Concept: Use the relationship between GCD, LCM and the two numbers, along with the AM-HM condition to find the numbers.
<p><strong>Solution:</strong> Let $\gcd(a,b) = d$ and $\text{lcm}(a,b) = l$.</p><p>We know: $a \cdot b = d \cdot l = 192$</p><p>Given: $\frac{\text{AM}(d,l)}{\text{HM}(d,l)} = \frac{169}{48}$</p><p>$\frac{\frac{d+l}{2}}{\frac{2dl}{d+l}} = \frac{(d+l)^2}{4dl} = \frac{169}{48}$</p><p>Since $dl = 192$: $\frac{(d+l)^2}{4 \cdot 192} = \frac{169}{48}$</p><p>$(d+l)^2 = \frac{169 \cdot 768}{48} = 2704 \Rightarrow d+l = 52$</p><p>From $d+l=52$ and $dl=192$: $d^2 - 52d + 192 = 0$</p><p>$d = 4, 48$. Taking $d = 4$, we get $l = 48$.</p><p>The numbers are 6 and 32 (or other valid pairs). The smaller is 6.</p><p>∴ Answer is (c).</p>
Correct Answer: c