Ternary operator in Javascript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to assign 'adult' if age is 18 or more, otherwise 'minor'.

Javascript

const status = age [1] 18 ? 'adult' : 'minor';

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

A>=

B==

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '>=' causes wrong classification.

Using '<' reverses the logic.

The ternary operator checks if age is greater than or equal to 18 to assign 'adult'.

2fill in blank

medium

Complete the code to assign 'even' if number is divisible by 2, else 'odd'.

Javascript

const type = (num % 2) [1] 0 ? 'even' : 'odd';

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

A>

B<

C===

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' causes odd numbers to be classified as even.

Using '>' or '<' is incorrect for equality check.

The ternary checks if remainder is exactly 0 to determine even numbers.

3fill in blank

hard

Fix the error in the ternary condition to check if score is passing (>= 60).

Javascript

const result = score [1] 60 ? 'pass' : 'fail';

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

A!=

B<

C==

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' reverses the pass/fail logic.

Using '==' only passes if score is exactly 60.

The condition must check if score is greater than or equal to 60 to pass.

4fill in blank

hard

Fill both blanks to create a ternary that returns 'positive' if num > 0, else 'non-positive'.

Javascript

const sign = num [1] 0 ? 'positive' : [2];

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

A>

B'non-positive'

C<=

D'negative'

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' in first blank reverses the logic.

Using 'negative' instead of 'non-positive' changes meaning.

The ternary checks if num is greater than 0; if not, it returns 'non-positive'.

5fill in blank

hard

Fill all three blanks to create a ternary that returns uppercase key if value > 10, else lowercase key if value > 5, else 'low'.

Javascript

const label = value [1] 10 ? key.[2]() : (value [3] 5 ? key.toLowerCase() : 'low');

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

A>

BtoUpperCase

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>' changes the logic.

Using 'toLowerCase' in second blank returns lowercase instead of uppercase.

The first condition checks if value is greater than 10, then returns uppercase key. The second checks if value is greater than 5, then returns lowercase key. Otherwise, returns 'low'.