Why conditional logic is needed in Go - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print "Adult" if age is 18 or more.

Go

if age [1] 18 {
    fmt.Println("Adult")
}

Drag options to blanks, or click blank then click option'

A!=

B<

C==

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>=' will check the wrong condition.

Using '==' only checks if age is exactly 18, not more.

The condition age >= 18 checks if age is 18 or more, which means the person is an adult.

2fill in blank

medium

Complete the code to print "Even" if the number is divisible by 2.

Go

if num [1] 2 == 0 {
    fmt.Println("Even")
}

Drag options to blanks, or click blank then click option'

A*

B/

C%

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' divides the number but does not check remainder.

Using '*' or '-' does not help check evenness.

The modulus operator % gives the remainder. If num % 2 == 0, the number is even.

3fill in blank

hard

Fix the error in the condition to check if score is less than 50.

Go

if score [1] 50 {
    fmt.Println("Fail")
}

Drag options to blanks, or click blank then click option'

A<

B==

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' checks if score is greater than 50, which is opposite.

Using '==' checks equality, not less than.

The condition score < 50 checks if the score is less than 50, meaning fail.

4fill in blank

hard

Fill both blanks to print "Teenager" if age is between 13 and 19 inclusive.

Go

if age [1] 13 && age [2] 19 {
    fmt.Println("Teenager")
}

Drag options to blanks, or click blank then click option'

A>=

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '<=' excludes 19.

Using '>' instead of '>=' excludes 13.

The condition age >= 13 && age <= 19 checks if age is between 13 and 19 inclusive.

5fill in blank

hard

Fill all three blanks to create a map of words to their lengths for words longer than 3 characters.

Go

lengths := make(map[string]int)
for _, [3] := range words {
    if len([3]) > 3 {
        lengths[[1]] = [2]
    }
}

Drag options to blanks, or click blank then click option'

Aword

Blen(word)

Dw

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Using the wrong expression for the length.

This code creates a map where each word is a key and its length len(word) is the value, but only for words longer than 3 characters.