Explanation:

Let’s break this down point by point, walk through the logic, and tackle the options:

- 150 shooters in total, each picks at least one of the guns A, B, or C.

- 26 pick all three guns.

- 71 pick B (could include overlaps with others).

- 64 pick exactly two guns.

- 48 pick both A and C.

- 27 pick only gun A.

- 50 pick A but not C.

Let’s zone in on what they want: participants who pick both B and C.

Here’s what matters:

1. Exactly two guns = 64.

This counts people who pick:

- Only A & B

- Only B & C

- Only A & C

(not all three—those are separate).

2. All three = 26. (That’s just its own group.)

3. Both A & C = 48.

This number includes people with all three and people with only A and C.

So:

`Exactly A & C = 48 - 26 = 22`

4. Let’s use what we now know:

- Exactly two guns:

Let’s call:

- Exactly A & B = x

- Exactly B & C = y

- Exactly A & C = z = 22

So, x + y + 22 = 64 ? x + y = 42

5. We need the number who picked both B and C (anyone picking both, so including all three):

That is,

Number picking B & C (at least those two) =

= Exactly B & C + All three

= y + 26

So, finding y is the key.

6. Let’s use other info:

- Number who pick only A = 27

- Number who pick A but not C = 50

- These are people who picked A — but not C. Which is:

- only A

- A & B (not C)

So: ONLY A + ONLY A & B = 27 + x = 50 ? x = 23

Now, recall: x + y = 42 ? 23 + y = 42 ? y = 19

7. Now, just add all three:

B & C (at least) = y + 26 = 19 + 26 = 45

Here’s what all this means:

- Option 1 (45) is correct.

- Other options (66, 78, 90, 88) don’t fit the math or the logic above.

- The rest of the info (about bullets) is just extra—doesn’t change the answer for this question.

.

You walked through the right path. 45 is the right number for shooters who picked both B and C.