Software Development Exam  >  Software Development Tests  >  Test: Algebra - 1 - Software Development MCQ

Test: Algebra - 1 - Software Development MCQ


Test Description

15 Questions MCQ Test - Test: Algebra - 1

Test: Algebra - 1 for Software Development 2024 is part of Software Development preparation. The Test: Algebra - 1 questions and answers have been prepared according to the Software Development exam syllabus.The Test: Algebra - 1 MCQs are made for Software Development 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Algebra - 1 below.
Solutions of Test: Algebra - 1 questions in English are available as part of our course for Software Development & Test: Algebra - 1 solutions in Hindi for Software Development course. Download more important topics, notes, lectures and mock test series for Software Development Exam by signing up for free. Attempt Test: Algebra - 1 | 15 questions in 30 minutes | Mock test for Software Development preparation | Free important questions MCQ to study for Software Development Exam | Download free PDF with solutions
Test: Algebra - 1 - Question 1

Which of the following algorithms is used to efficiently compute the power of a number?

Detailed Solution for Test: Algebra - 1 - Question 1

Binary Exponentiation 1 algorithm efficiently computes the power of a number by using the binary representation of the exponent.

Test: Algebra - 1 - Question 2

What is the time complexity of the Sieve of Eratosthenes algorithm for finding prime numbers up to a given limit?

Detailed Solution for Test: Algebra - 1 - Question 2

The Sieve of Eratosthenes algorithm has a time complexity of O(n log log n), which is approximately O(n) for large values of n.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Algebra - 1 - Question 3

Which of the following algorithms is used to find the greatest common divisor (GCD) of two numbers?

Detailed Solution for Test: Algebra - 1 - Question 3

The Euclidean algorithm is used to find the greatest common divisor (GCD) of two numbers.

Test: Algebra - 1 - Question 4

In linear diophantine equations of the form ax + by = c, where a, b, and c are integers, which of the following is true?

Detailed Solution for Test: Algebra - 1 - Question 4

In linear diophantine equations, a solution exists if c is a multiple of the greatest common divisor (GCD) of a and b.

Test: Algebra - 1 - Question 5

Which of the following is an example of a linear diophantine equation?

Detailed Solution for Test: Algebra - 1 - Question 5

The equation 3x + 5y = 2 is an example of a linear diophantine equation.

Test: Algebra - 1 - Question 6

What will be the output of the code snippet?
a = 5
b = 3
c = a + b
print(c)

Detailed Solution for Test: Algebra - 1 - Question 6

The code snippet adds the values of variables a and b and assigns the result to variable c, which is then printed.

Test: Algebra - 1 - Question 7

What will be the output of the code snippet?
a = 10
b = 2
c = a / b
print(c)

Detailed Solution for Test: Algebra - 1 - Question 7

The division operation (/) in Python always returns a floating-point result, so the division of 10 by 2 will give a float value of 5.0.

Test: Algebra - 1 - Question 8

What will be the output of the code snippet?
a = 10
b = 3
c = a % b
print(c)

Detailed Solution for Test: Algebra - 1 - Question 8

The modulus operator (%) in Python returns the remainder of the division, so the remainder of 10 divided by 3 is 1.

Test: Algebra - 1 - Question 9

What will be the output of the following code snippet?
def binary_exponentiation(a, n):
    if n == 0:
        return 1
    elif n % 2 == 0:
        x = binary_exponentiation(a, n // 2)
        return x * x
    else:
        x = binary_exponentiation(a, (n - 1) // 2)
        return a * x * x

result = binary_exponentiation(2, 5)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 9

The binary_exponentiation function implements binary exponentiation algorithm. For n = 5, the function will be called recursively as follows:
binary_exponentiation(2, 5) -> binary_exponentiation(2, 2) -> binary_exponentiation(2, 1) -> binary_exponentiation(2, 0)
The function returns 2 * 2 * 2 * 2 * 2 = 32.

Test: Algebra - 1 - Question 10

What will be the output of the following code snippet?
def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a % b)

result = gcd(21, 14)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 10

The gcd function implements the Euclidean algorithm.
For a = 21 and b = 14, the function will be called recursively as follows:
gcd(21, 14) -> gcd(14, 7) -> gcd(7, 0)
The function returns 7 as the GCD of 21 and 14.

Test: Algebra - 1 - Question 11

What will be the output of the following code snippet?
def sieve_of_eratosthenes(n):
    prime = [True] * (n + 1)
    p = 2
    while p * p <= n:
        if prime[p] == True:
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1

    primes = []
    for p in range(2, n + 1):
        if prime[p]:
            primes.append(p)
    return primes

result = sieve_of_eratosthenes(20)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 11

The sieve_of_eratosthenes function implements the Sieve of Eratosthenes algorithm. It returns a list of prime numbers up to the given limit of 20.

Test: Algebra - 1 - Question 12

What will be the output of the following code snippet?
def linear_diophantine(a, b, c):
    g = gcd(a, b)
    if c % g != 0:
        return "No solution exists."
    x0, y0 = extended_euclidean(a, b)
    x = x0 * (c // g)
    y = y0 * (c // g)
    return x, y

result = linear_diophantine(5, 7, 16)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 12

The linear_diophantine function solves the linear diophantine equation 5x + 7y = 16. The solution is (3, -2).

Test: Algebra - 1 - Question 13

What will be the output of the following code snippet?

def binary_exponentiation(a, n):
    if n == 0:
        return 1
    elif n % 2 == 0:
        x = binary_exponentiation(a, n // 2)
        return x * x % 1000000007
    else:
        x = binary_exponentiation(a, (n - 1) // 2)
        return a * x * x % 1000000007

result = binary_exponentiation(2, 10)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 13

The binary_exponentiation function implements binary exponentiation algorithm.
For n = 10, the function will be called recursively as follows:
binary_exponentiation(2, 10) -> binary_exponentiation(2, 5) -> binary_exponentiation(2, 2) -> binary_exponentiation(2, 1) -> binary_exponentiation(2, 0)
The function returns 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 % 1000000007 = 1024 % 1000000007 = 1025.

Test: Algebra - 1 - Question 14

What will be the output of the following code snippet?
def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a % b)

result = gcd(24, 36)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 14

The gcd function implements the Euclidean algorithm.
For a = 24 and b = 36, the function will be called recursively as follows:
gcd(24, 36) -> gcd(36, 24) -> gcd(24, 12) -> gcd(12, 0)
The function returns 12 as the GCD of 24 and 36.

Test: Algebra - 1 - Question 15

What will be the output of the following code snippet?

def linear_diophantine(a, b, c):
    g = gcd(a, b)
    if c % g != 0:
        return "No solution exists."
    x0, y0 = extended_euclidean(a, b)
    x = x0 * (c // g)
    y = y0 * (c // g)
    return x, y

result = linear_diophantine(3, 5, 8)
print(result)

Detailed Solution for Test: Algebra - 1 - Question 15

The linear_diophantine function solves the linear diophantine equation 3x + 5y = 8. The solution is (2, -1).

Information about Test: Algebra - 1 Page
In this test you can find the Exam questions for Test: Algebra - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Algebra - 1, EduRev gives you an ample number of Online tests for practice

Top Courses for Software Development

Download as PDF

Top Courses for Software Development