Question

question 24 1 pts how many times will the following do-while loop be executed? <img src=\test_images/f210g11g1_r.jpg\ alt=\int x = 11;do{ x += 20;} while (x <= 100);\ style=\vertical-align: -4.0px;\ />
5
4
3
1

Explanation:

Step1: Analyze the do - while loop

In a do - while loop, the body of the loop is executed at least once, and then the condition is checked. The initial value of \(x\) is \(11\). In each iteration, \(x\) is incremented by \(20\) (\(x += 20\) is equivalent to \(x=x + 20\)). The loop continues as long as \(x\leq100\).

Step2: Calculate the values of \(x\) in each iteration

- **Iteration 1**:
Initial \(x = 11\). After the loop body (we do the increment here since it's a do - while, the body is executed first), \(x=11 + 20=31\). Now we check the condition \(31\leq100\), which is true.
- **Iteration 2**:
\(x = 31+20 = 51\). Check \(51\leq100\), true.
- **Iteration 3**:
\(x=51 + 20=71\). Check \(71\leq100\), true.
- **Iteration 4**:
\(x=71+20 = 91\). Check \(91\leq100\), true.
- **Iteration 5**:
\(x=91 + 20=111\). Now we check the condition \(111\leq100\), which is false. But wait, in the do - while loop, the body is executed first, then the condition is checked. Wait, let's re - calculate:

Wait, initial \(x = 11\)
- Do - while loop: first execute the body:
- First execution: \(x=11 + 20=31\), then check \(31\leq100\) (true, loop again)
- Second execution: \(x = 31+20=51\), check \(51\leq100\) (true, loop again)
- Third execution: \(x=51 + 20=71\), check \(71\leq100\) (true, loop again)
- Fourth execution: \(x=71+20 = 91\), check \(91\leq100\) (true, loop again)
- Fifth execution: \(x=91+20 = 111\), check \(111\leq100\) (false, stop)

Wait, no, the number of times the loop body is executed: Let's list the values of \(x\) before checking the condition (since in do - while, the body is executed, then condition is checked):

- Start: \(x = 11\)
- After first execution (body): \(x = 31\), check \(31\leq100\) (true) → count = 1
- After second execution: \(x = 51\), check \(51\leq100\) (true) → count = 2
- After third execution: \(x = 71\), check \(71\leq100\) (true) → count = 3
- After fourth execution: \(x = 91\), check \(91\leq100\) (true) → count = 4
- After fifth execution: \(x = 111\), check \(111\leq100\) (false) → stop. Wait, but that would be 5 times? But wait, let's check the arithmetic sequence. The values of \(x\) after each iteration (body execution) are \(31,51,71,91,111\). The number of terms in the sequence where \(x\leq100\):

The general term of the arithmetic sequence is \(a_n=11+(n)\times20\) (since first term after execution is \(11 + 20=31\), so \(n\) starts at 1). We want \(a_n\leq100\)

\(11+20n\leq100\)

\(20n\leq89\)

\(n\leq\frac{89}{20}=4.45\)

Since \(n\) is the number of times the body is executed (because we start with \(x = 11\), execute the body, then check), so \(n = 4\) when \(x = 91\) (after 4th execution, \(x = 91\), check \(91\leq100\) → true, then 5th execution: \(x = 111\), check \(111\leq100\) → false. Wait, no, the do - while loop structure is:

do {
// body (increment x)
} while (condition);

So the steps are:

1. Enter the loop, execute the body (x becomes 31), then check condition (31 <= 100: true) → loop count 1
2. Execute body (x becomes 51), check condition (51 <= 100: true) → loop count 2
3. Execute body (x becomes 71), check condition (71 <= 100: true) → loop count 3
4. Execute body (x becomes 91), check condition (91 <= 100: true) → loop count 4
5. Execute body (x becomes 111), check condition (111 <= 100: false) → loop stops.

Wait, but that's 5 times? But the options have 5 as an option. Wait, maybe I made a mistake. Let's re - calculate:

Initial x = 11

First iteration (do): x = 11 + 20 = 31, while (31 <= 100) → true → loop again

Second iteration: x = 31+20 = 51, while (51 <= 100) → true → loop again

Third iteration: x = 51+20 = 71, while (71 <= 100) → true → loop again

Fourth iteration: x = 71+20 = 91, while (91 <= 100) → true → loop again

Fifth iteration: x = 91+20 = 111, while (111 <= 100) → false → loop stops.

So the loop body is executed 5 times? But let's check the arithmetic:

The values of x after each do - block: 31, 51, 71, 91, 111. The number of times we execute the do - block is the number of terms in this sequence where x <= 100? Wait, no, the do - block is executed, then the while condition is checked. So for x = 11, we execute the do - block (x becomes 31), then check if 31 <= 100 (yes). Then execute do - block (x becomes 51), check (yes). Then x = 71 (check yes), x = 91 (check yes), x = 111 (check no). So the number of do - block executions is 5? But let's check the difference between 11 and 100: 100 - 11 = 89. 89 divided by 20 is 4.45, so we can have 4 full increments (11+20*4 = 91) and then one more increment (91 + 20=111) which is over 100. Wait, but in do - while, the body is executed first. So when x = 11, we execute the body (x=31), then check. So the number of times the body is executed is the number of times we can do x +=20 starting from 11, such that after the increment, we check x <=100. Wait, no:

Let's list the values of x before the while check (after the do - block):

- After 1st do: x = 31, check 31 <= 100 → true (count 1)
- After 2nd do: x = 51, check 51 <= 100 → true (count 2)
- After 3rd do: x = 71, check 71 <= 100 → true (count 3)
- After 4th do: x = 91, check 91 <= 100 → true (count 4)
- After 5th do: x = 111, check 111 <= 100 → false (count 5, but loop stops)

Wait, but 11 + 20*4 = 91, which is <=100, and 11+20*5 = 111>100. So the number of times the loop runs (do - while) is 5? But let's check the options. The options are 5,4,3,1.

Wait, maybe I made a mistake in the initial value. Wait, the code is:

int x = 11;

do {

x += 20;

} while (x <= 100);

So let's track x:

- Before loop: x = 11
- Do - block 1: x = 11 + 20 = 31, then check 31 <= 100 → true → loop continues
- Do - block 2: x = 31+20 = 51, check 51 <= 100 → true → loop continues
- Do - block 3: x = 51+20 = 71, check 71 <= 100 → true → loop continues
- Do - block 4: x = 71+20 = 91, check 91 <= 100 → true → loop continues
- Do - block 5: x = 91+20 = 111, check 111 <= 100 → false → loop stops

So the do - block is executed 5 times. So the answer should be 5.

Answer:

5