X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number $X^3$ will have

Select one answerSingle correct

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Solution

Step-by-step verified solution

Official answer: A) 90 digits

Quantitative Aptitude

Given that $X$ is a 30-digit number beginning with $47\dots$ , we can write:

4.7 \times 10^{29} \le X < 4.8 \times 10^{29}

Cubing the boundaries:

(4.7)^3 \times 10^{87} \le X^3 < (4.8)^3 \times 10^{87}

Calculating the powers:

$(4.7)^3 = 103.823$
$(4.8)^3 = 110.592$

Therefore:

103.823 \times 10^{87} \le X^3 < 110.592 \times 10^{87}

1.03823 \times 10^{89} \le X^3 < 1.10592 \times 10^{89}

A number in the range $[10^{89}, 10^{90})$ has exactly $89 + 1 = 90$ digits.

Hence, $X^3$ has 90 digits, and option (A) is correct.

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This Quantitative Aptitude previous year question is Q63 of the GATE CSE 2017 Set 2 paper. Find more GATE CSE PYQs on Quantitative Aptitude in General Aptitude, or browse all GATE CSE PYQs.

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Quantitative Aptitude in GATE CSE

Quantitative Aptitude has 53 GATE CSE PYQs across 13 papers from 2011 to 2026. 27 of them are 2-mark MCQs like this one, and GATE CSE 2017 Set 2 has 3.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.