GATE 2017 CS (Set 2) question paper PDF and answer key

Since $X$ is a 30-digit number starting with 47, it satisfies approximately:

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

Cubing,

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

Now,

$$4.7^3 = 103.823$$

and

$$4.8^3 = 110.592.$$

So $X^3$ lies between approximately:

$$1.03823\times 10^{89}$$

and

$$1.10592\times 10^{89}.$$

A number in this range has 90 digits.

Therefore, the correct answer is Option A.

The word seemingly shows that ideology appeared to be more urgent, but the author does not fully accept that it truly was more important.

This reveals the author's belief that literature should not be treated as inferior to ideology.

Therefore, the correct answer is Option B.

A poset is a lattice if every pair of elements has both a least upper bound and a greatest lower bound.

From the given Hasse diagram:

• $a$ acts as the least element.
• $e$ acts as the greatest element.
• For each pair among $b,c,d$, the required meet and join already exist.

Therefore, the poset is already a lattice.

No ordered pair needs to be added.

The answer is 0.

The main memory is byte-addressable with size:

$$2^{32}$$

bytes, so the address size is 32 bits.

Block size is:

$$32 = 2^5$$

bytes, so block offset bits:

$$5.$$

The cache has 512 lines:

$$512 = 2^9.$$

Since the cache is direct mapped, index bits:

$$9.$$

Therefore, tag bits:

$$32 - 9 - 5 = 18.$$

So the answer is 18.

Let the order of the B+ tree be $m$.

An internal node with $m$ pointers has:

$$m-1$$

search-key values.

Each pointer takes 2 bytes and each search-key value takes 8 bytes.

So the node size requirement is:

$$2m + 8(m-1) \le 512.$$

Simplify:

$$2m + 8m - 8 \le 512$$ $$10m \le 520$$ $$m \le 52.$$

Therefore, the maximum order of the B+ tree is 52.