Find the value of the expression : \(\frac{(0.12×0.12×0.12)+(0.012×0.012×0.012)…
2024
Find the value of the expression :
\(\frac{(0.12×0.12×0.12)+(0.012×0.012×0.012)}{(0.48×0.48×0.48)+(0.048×0.048×0.048)}\)
Answer: D. \(\frac{1}{64}\) — ConceptIf every term of a sum is multiplied by the same factor k, the whole sum is multiplied by k. For cube terms, (ka)3 = k3a3. Thus a scale factor k in…
- A.
\(\frac{1}{16}\)
- B.
\(\frac{1}{4}\)
- C.
4
- D.
\(\frac{1}{64}\)
Attempted by 1 students.
Show answer & explanation
Correct answer: D
Concept
If every term of a sum is multiplied by the same factor k, the whole sum is multiplied by k.
For cube terms, (ka)3 = k3a3. Thus a scale factor k in each base becomes k3 in the sum of cubes.
Application
Let N = (0.12)3 + (0.012)3 denote the numerator.
The corresponding denominator bases are four times as large: 0.48 = 4 × 0.12 and 0.048 = 4 × 0.012.
Therefore D = (4 × 0.12)3 + (4 × 0.012)3 = 43[(0.12)3 + (0.012)3] = 64N.
Hence \(\frac{N}{D}=\frac{N}{64N}=\frac{1}{64}\).
Cross-check
Directly, N = 0.001728 + 0.000001728 = 0.001729728, while D = 0.110592 + 0.000110592 = 0.110702592. Since \(\frac{D}{N}=64\), the required value is \(\frac{1}{64}\).