There are two numbers *A* and *B*, where *A* is 33.33\% less than *B*. The product of the LCM and HCF of *A* and *B* is 294.

Compare the following quantities:

**Quantity I:** 450

**Quantity II:** LCM of $(6A - 20)$ and $(3B - 7)$

Question

There are two numbers *A* and *B*, where *A* is 33.33\% less than *B*. The product of the LCM and HCF of *A* and *B* is 294.

Compare the following quantities:

**Quantity I:** 450

**Quantity II:** LCM of $(6A - 20)$ and $(3B - 7)$

Accepted Answer

**Solution:**

1. Given *A* is 33.33\% less than *B*.
   - 33.33\% = $\frac{1}{3}$
   - So, $A = B - \frac{1}{3}B = \frac{2}{3}B$

2. Let $B = 3x$, then $A = 2x$.

3. The product of LCM and HCF of *A* and *B* is given as 294.
   - We know for any two numbers $a$ and $b$, $\text{LCM}(a,b) \times \text{HCF}(a,b) = a \times b$.
   - So, $A \times B = 294$
   - Substitute $A = 2x$ and $B = 3x$:
     $$ 2x \times 3x = 6x^2 = 294 $$
   - $x^2 = \frac{294}{6} = 49$
   - $x = 7$

4. Therefore,
   - $A = 2x = 2 \times 7 = 14$
   - $B = 3x = 3 \times 7 = 21$

5. Calculate $6A - 20$ and $3B - 7$:
   - $6A - 20 = 6 \times 14 - 20 = 84 - 20 = 64$
   - $3B - 7 = 3 \times 21 - 7 = 63 - 7 = 56$

6. Find LCM of 64 and 56:
   - Prime factorization:
     - 64 = $2^6$
     - 56 = $2^3 \times 7$
   - LCM takes highest powers:
     - LCM = $2^6 \times 7 = 64 \times 7 = 448$

7. Compare Quantity I and Quantity II:
   - Quantity I = 450
   - Quantity II = 448

8. Since 450 > 448,

**Answer:** (a) Quantity I > Quantity II

**Note:** The correct answer is option (a) Quantity I > Quantity II.

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