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NEW QUESTION # 112
Given the following code:
document.body.addEventListener(' click ', (event) => {
if (/* CODE REPLACEMENT HERE */) {
console.log('button clicked!');
)
});
Which replacement for the conditional statement on line 02 allows a developer to correctly determine that a button on page is clicked?
Answer: D
NEW QUESTION # 113
A developer has a fizzbuzz function that, when passed in a number, returns the following:
'fizz' if the number is divisible by 3.
'buzz' if the number is divisible by 5.
'fizzbuzz' if the number is divisible by both 3 and 5.
Empty string '' if the number is divisible by neither 3 nor 5.
Which two test cases properly test scenarios for the fizzbuzz function?
Answer: B,C
Explanation:
:
First, recall the fizzbuzz rules:
If n divisible by 3 and not by 5 → return 'fizz'.
If n divisible by 5 and not by 3 → return 'buzz'.
If n divisible by both 3 and 5 → return 'fizzbuzz'.
Otherwise → return '' (empty string).
Evaluate each test:
Option A:
let res = fizzbuzz(true);
console.assert(res === '');
In JavaScript, when using arithmetic operations with true, it is coerced to the number 1. A typical implementation of fizzbuzz would treat the argument as a number and compute:
true % 3 → 1 % 3 → 1
true % 5 → 1 % 5 → 1
So true is effectively treated as 1, which is divisible by neither 3 nor 5. The function should return ''. The assertion res === '' will pass.
This test covers the scenario "neither divisible by 3 nor 5".
Option B:
let res = fizzbuzz(3);
console.assert(res === '');
3 is divisible by 3 but not by 5.
According to fizzbuzz rules, fizzbuzz(3) should return 'fizz'.
This test expects '', so it is incorrect; the assertion would fail in a correct implementation.
Option C:
let res = fizzbuzz(5);
console.assert(res === 'fizz');
5 is divisible by 5 but not by 3.
According to fizzbuzz rules, fizzbuzz(5) should return 'buzz'.
This test expects 'fizz', so it is incorrect.
Option D:
let res = fizzbuzz(15);
console.assert(res === 'fizzbuzz');
15 is divisible by both 3 and 5.
According to fizzbuzz rules, fizzbuzz(15) should return 'fizzbuzz'.
This assertion correctly matches the expected result, so it is a proper test case.
Thus, the two test cases that correctly test valid fizzbuzz behavior are:
Answe r: A, D
Study Guide / Concept Reference (no links):
Modulo operator (%) for divisibility checks
Truthy/number coercion of boolean true in arithmetic (Number(true) === 1) Designing unit tests with console.assert Typical fizzbuzz logic structure and expected outputs
NEW QUESTION # 114
A team at Universal Containers works on a big project and uses Yarn to deal with the project's dependencies. A developer added a dependency to manipulate dates and pushed the updates to the remote repository. The rest of the team complains that the dependency does not get downloaded when they execute yarn.
What could be the reason for this?
Answer: B
Explanation:
If the dependency was added as a dev dependency (yarn add --dev), it goes into devDependencies.
When NODE_ENV=production, Yarn (and npm) typically skip devDependencies on install.
Thus the package doesn't get installed for other team members in a production environment.
YARN_ENV is not a standard variable, --save is npm syntax (and Yarn saves by default), and --add is not how Yarn works (the command itself is yarn add).
NEW QUESTION # 115
A developer uses a parsed JSON string to work with user information as in the block below:
01 const userInformation = {
02 "id": "user-01",
03 "email": "user01@universalcontainers.demo",
04 "age": 25
05 };
Which two options access the email attribute in the object?
Answer: B,D
Explanation:
Comprehensive and Detailed Explanation From JavaScript Knowledge:
userInformation is a plain JavaScript object:
{
id: "user-01",
email: "user01@universalcontainers.demo",
age: 25
}
You can access object properties in two standard ways:
Dot notation: object.propertyName
Bracket notation: object['propertyName']
Apply to each option:
A . userInformation.email
Correct dot notation; accesses "user01@universalcontainers.demo".
B . userInformation.get("email")
.get() is a method on Map instances, not on plain objects.
A regular object does not have a .get method; this would be TypeError.
C . userInformation["email"]
Correct bracket notation; string "email" is used as the property key.
Returns "user01@universalcontainers.demo".
D . userInformation[email]
Here email is treated as a variable, not a string literal.
Unless there is a variable named email defined with a value matching a property key, this will either:
Throw a ReferenceError (if email is not defined), or
Use the value of email as a computed key (not what is intended here).
It is not the correct way to access the "email" property literal.
Therefore, the correct ways are:
Answe r: A, C
Relevant concepts: object property access (. vs []), plain objects vs Map, literal property names vs computed property names.
NEW QUESTION # 116
Refer to the following code:
What is the output of line 11?
Answer: D
NEW QUESTION # 117
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