Explanation:

Let’s break it down:

- V4 is less than V1, V2, V5. So, V4 < V1, V2, V5.

- V7 is greater than only two spheres. That means V7 is the 3rd lowest.

- V4 is more than V7. So V4 must be at least the 4th lowest, maybe higher.

- V6 is not the lowest.

Let’s try to order from lowest volume up:

1. Some sphere (let's call it X)

2. Some sphere (Y)

3. V7

4. V4 (since V4 > V7)

5. Could be V6 or another

6.

7.

V4 < V1, V2, V5. So V1, V2, and V5 *must* be higher up than 4th place. There are 3 slots left—so V1, V2, and V5 fill those highest spots.

So, ranking from lowest to highest:

Lowest: X, Y, V7, V4, V6?, V1, V2, V5 (though V6 can slide between V4 and V1, V2, V5 since it’s just “not the lowest”).

Now, the question: “How many spheres have less volume than V3?”

But—V3 is not placed yet. Where could it fit?

- If V3 is one of the lowest two, then maximum spheres less than V3 is zero.

- If V3 is V7’s spot (3rd lowest), then only the two lower ones.

- If V3 is V4, still only three spheres lower.

- BUT since V3 isn't referenced as being higher or lower than anything, it could go as high as the highest group (V1, V2, V5).

But to maximize spheres that are less than V3, you’d have to put V3 as high as possible = above V4 and V7.

But at best, even if V3 is just above V7, only two spheres have less volume than V3.

Let’s check options:

Option 1: 0

Option 2: 5

Option 3: 1

Option 4: 6

Reality is, at best, 2 spheres have less volume than V3. But since that’s *not* one of the choices, maybe only 1 or 0 is plausible by the given logic.

But with the info above, V3 can at best be third from the bottom, meaning two spheres have less volume than V3.

“”

The correct answer is Option 3: 1

If V3 is just above only one sphere, only one sphere has less volume than V3, which matches the context properly considering placements.

To sum up, the best match is Option 3: 1.