Let $x, y, z$ (all prime) be the length, breadth and height respectively of a cuboid with $x > y > z$. The volume of the cuboid is $30k^3$ cubic units, where $k$ is a natural number. What is the total surface area of the cuboid ?

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CDS - 2026

Let $x, y, z$ (all prime) be the length, breadth and height respectively of a cuboid with $x > y > z$. The volume of the cuboid is $30k^3$ cubic units, where $k$ is a natural number. What is the total surface area of the cuboid ?

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CDS 1 2026 Maths Question Paper (12-Apr-2026)

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60 square units

Cannot be determined due to insufficient data

64 square units

62 square units

The correct answer is

62 square units

Understanding the Problem Constraints

We need to find the total surface area of a cuboid. The dimensions are length ($x$), breadth ($y$), and height ($z$). We are given that $x, y, z$ are all prime numbers and satisfy the condition $x > y > z$. The volume ($V$) is given by the formula $V = 30k^3$, where $k$ is a natural number.

Relating Volume and Dimensions

The volume of a cuboid is given by the product of its dimensions: $V = x \times y \times z$. Since $x, y, z$ are prime numbers, the prime factorization of the volume $V$ must consist of exactly these three distinct primes ($x, y, z$).

We are given $V = 30k^3$. The prime factorization of 30 is $2 \times 3 \times 5$. Therefore, $V = (2 \times 3 \times 5) \times k^3$.

Analyzing the Possibilities for $k$

For the conditions to be met, the volume $V = 30k^3$ must be expressible as the product of exactly three distinct prime numbers ($x, y, z$).

Scenario 1: $k = 1$ If $k=1$, the volume is $V = 30 \times 1^3 = 30$. The prime factorization is $30 = 2 \times 3 \times 5$. This fits the requirement $V = x \times y \times z$. Given $x > y > z$, we must have $x=5$, $y=3$, $z=2$. These are indeed prime numbers. In this case, the total surface area is $A = 2(xy + yz + zx) = 2((5 \times 3) + (3 \times 2) + (2 \times 5)) = 2(15 + 6 + 10) = 2(31) = 62$ square units.
Scenario 2: $k > 1$ If $k$ is a natural number greater than 1, let its prime factorization be $k = p_1^{a_1} p_2^{a_2} \dots$. Then $k^3 = p_1^{3a_1} p_2^{3a_2} \dots$. The volume becomes $V = (2 \times 3 \times 5) \times (p_1^{3a_1} p_2^{3a_2} \dots)$. If $k > 1$, the prime factorization of $V$ will either contain more than three distinct prime factors (if $k$ has prime factors other than 2, 3, 5) or contain prime factors with exponents greater than 1 (e.g., if $k=2$, $V = 2^4 \times 3 \times 5$). In either situation, $V$ cannot be represented as the product of exactly three distinct prime numbers ($x \times y \times z$). This means that for any natural number $k > 1$, there are no prime numbers $x, y, z$ that satisfy the condition $xyz = 30k^3$.

Final Conclusion

The problem states $k$ is a natural number. The analysis reveals that the condition for $x, y, z$ being distinct primes ($x>y>z$) and their product equaling $30k^3$ is only possible if $k=1$. If $k=1$ were explicitly given or implied as the only possibility, the surface area would be 62.

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