ABC26GN0734 · Decimal Fractions
Subject: General Aptitude · Chapter: Decimal Fractions · Exam: · Marks: · Difficulty:
If $1^{3}+2^{3}+\ldots . .+9^{3}=2025$, then the value of $(0.11)^{3}$ $+(0.22)^{3}+\ldots . .+(0.99)^{3}$ is close to:
(a)0.2695
(b)0.3695
(c)2.695
(d)3.695
Answer
Explanation
$$\begin{aligned} & (0.11)^{3}+(0.22)^{3}+\ldots . .+(0.99)^{3}=(0.11)^{3}\left(1^{3}+2^{3}+\ldots .+9^{3}\right) \\ & \quad=0.001331 \times 2025=2.695275 \approx 2.695 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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