Question

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

The initial value of \(x\) is \(11\). In a do - while loop, the loop body is executed at least once, and then the condition is checked. The loop body is \(x += 20\) (which is equivalent to \(x=x + 20\)). The condition for the loop to continue is \(x\leq100\).

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

• Iteration 1:

After the first execution of the loop body, \(x=11 + 20=31\). Since \(31\leq100\), the loop will continue.

• Iteration 2:

After the second execution of the loop body, \(x = 31+20 = 51\). Since \(51\leq100\), the loop will continue.

• Iteration 3:

After the third execution of the loop body, \(x=51 + 20=71\). Since \(71\leq100\), the loop will continue.

• Iteration 4:

After the fourth execution of the loop body, \(x=71+20 = 91\). Since \(91\leq100\), the loop will continue.

• Iteration 5:

After the fifth execution of the loop body, \(x=91 + 20=111\). Now, \(111>100\), so the loop will stop.

Wait, let's re - check. Wait, the initial value is \(x = 11\). Let's list the values step by step:

First iteration (do part): \(x=11 + 20=31\), then check while condition (\(31\leq100\)): true.

Second iteration: \(x = 31+20=51\), check \(51\leq100\): true.

Third iteration: \(x=51 + 20=71\), check \(71\leq100\): true.

Fourth iteration: \(x=71+20 = 91\), check \(91\leq100\): true.

Fifth iteration: \(x=91+20 = 111\), check \(111\leq100\): false.

Wait, but the do - while loop executes the body first and then checks the condition. So the number of times the loop body is executed is 5? Wait, no, wait the options are 5,4,3,1. Wait, maybe I made a mistake.

Wait, let's re - calculate:

Start with \(x = 11\).

Do - while loop:

1st execution: \(x=11 + 20=31\), condition \(31\leq100\) → true.

2nd execution: \(x = 31+20=51\), condition \(51\leq100\) → true.

3rd execution: \(x=51 + 20=71\), condition \(71\leq100\) → true.

4th execution: \(x=71+20 = 91\), condition \(91\leq100\) → true.

5th execution: \(x=91+20 = 111\), condition \(111\leq100\) → false.

So the loop body is executed 5 times? But let's check the options. Wait, maybe the initial value is different? Wait the code is \(int\ x = 11;do\{x += 20;\}while(x <= 100);\)

Wait, let's solve the inequality \(11+20n\leq100\), where \(n\) is the number of times we add 20 (the number of loop iterations).

\(20n\leq100 - 11=89\)

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

Since \(n\) is the number of times we execute the loop (and in do - while, we execute at least once), and \(n\) must be an integer. Wait, but when \(n = 4\), \(x=11+20\times4=11 + 80 = 91\), then we execute the loop again (n = 5) to get \(x=91 + 20=111\), which is greater than 100. Wait, the loop body is executed 5 times? But the options have 5 as an option. Wait, maybe I made a mistake in the inequality.

Wait, the loop is do - while, so the steps are:

• Execute the body (x +=20)
• Check the condition.

So the first execution: x becomes 31, condition true.

Second: 51, true.

Third:71, true.

Fourth:91, true.

Fifth:111, false.

So the loop body is executed 5 times. So the answer should be 5.

Answer:

5