12th Grade Math: Hardest Quiz!

Reviewed by Janaisa Harris
Janaisa Harris, BA (Mathematics) |
High School Math Teacher
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Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a bachelor's degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher.
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By Max
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Max
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Quizzes Created: 37 | Total Attempts: 60,550
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  • 1/10 Questions

     What is |-26|?

    • -26
    • 26
    • 0
    • 1
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About This Quiz

Do you like solving math problems? Take this 12th Grade's hardest Math quiz and see how well you understand the calculation and formulas. Mathematics is a pure and simple idea, but it can get complicated. Math is also exciting but can sometimes get frustrating. Are you ready to take on a set of highly complex mat questions? Fasten your seat belts, and let's go on the math-filled roller coaster! All the best for a perfect score on this math quiz.

12th Grade Math: Hardest Quiz! - Quiz

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  • 2. 

     Which is the smallest?

    •  -1

    • -1/2

    •  0

    • 3

    Correct Answer
    A.  -1
    Explanation
    The given options are -1, -1/2, 0, and 3. Among these options, -1 is the smallest number.

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  • 3. 

     Simplify: (4 – 5) – (13 – 18 + 2).

    •  -1

    • –2

    • 1

    • 2

    Correct Answer
    A. 2
    Explanation
    The given expression simplifies to (-1) - (13 - 18 + 2). First, we simplify the inner parentheses by performing the operations inside it, which gives us (-1) - (13 - 16). Next, we simplify the remaining parentheses by performing the operation inside it, which gives us (-1) - (-3). Finally, we subtract -3 from -1, which gives us 2.

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  • 4. 

     Factor: 3y(x – 3) -2(x – 3)

    •  (x – 3)(x – 3)

    • (x – 3)2

    • (x – 3)(3y – 2)

    • 3y(x – 3)

    Correct Answer
    A. (x – 3)(3y – 2)
    Explanation
    The given expression can be factored as (x - 3)(3y - 2). This can be determined by distributing the terms in the expression and combining like terms. The first term, 3y(x - 3), can be expanded as 3xy - 9y. The second term, -2(x - 3), can be expanded as -2x + 6. Combining these terms gives 3xy - 9y - 2x + 6, which can be rearranged as (x - 3)(3y - 2).

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  • 5. 

    Factor: 5x2 – 15x – 20

    • 5(x-4)(x+1)

    • -2(x-4)(x+5)

    • -5(x+4)(x-1)

    • 5(x+4)(x+1)

    Correct Answer
    A. 5(x-4)(x+1)
    Explanation
    The given expression is a quadratic trinomial that can be factored using the method of factoring by grouping. By looking at the coefficients of the terms, we can see that the common factor is 5. By factoring out 5, we get 5(x^2 - 3x - 4). Now, we need to factor the quadratic trinomial (x^2 - 3x - 4). This can be factored as (x - 4)(x + 1). Therefore, the correct answer is 5(x - 4)(x + 1).

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  • 6. 

     Multiply: (x – 4)(x + 5)

    •  x2 + 5x - 20

    • X2 - 4x - 20

    •  x2 - x - 20

    • X2 + x - 20

    Correct Answer
    A. X2 + x - 20
    Explanation
    The given expression is a product of two binomials, (x - 4) and (x + 5). To find the product, we can use the FOIL method, which stands for First, Outer, Inner, Last. Multiplying the first terms of each binomial gives us x * x = x^2. Multiplying the outer terms gives us x * 5 = 5x. Multiplying the inner terms gives us -4 * x = -4x. Multiplying the last terms gives us -4 * 5 = -20. Combining like terms, we get x^2 + 5x - 4x - 20, which simplifies to x^2 + x - 20. Therefore, the correct answer is x^2 + x - 20.

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  • 7. 

     Solve for x: 2x – y = (3/4)x + 6

    • (y + 6)/5

    • 4(y + 6)/5

    •  (y + 6)

    • 4(y - 6)/5

    Correct Answer
    A. 4(y + 6)/5
    Explanation
    The correct answer is 4(y + 6)/5. To solve for x, we need to isolate the variable. First, we can simplify the equation by multiplying both sides by 4 to get rid of the fraction. This gives us 8x - 4y = 3x + 24. Next, we can combine like terms by subtracting 3x from both sides, resulting in 5x - 4y = 24. Finally, we can divide both sides by 5 to solve for x, which gives us x = (4y + 24)/5. This matches the given answer of 4(y + 6)/5.

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  • 8. 

    What is the radius of a circle that has a circumference of 3.14 meters?

    • 0.2 meter

    • 0.3 meter

    • 0.4 meter

    • 0.5 meter

    Correct Answer
    A. 0.5 meter
    Explanation
    The circumference of a circle is calculated using the formula C = 2πr, where C is the circumference and r is the radius. In this case, the given circumference is 3.14 meters. By rearranging the formula, we can solve for the radius: r = C / 2π. Plugging in the given circumference, we get r = 3.14 / (2 * 3.14) = 0.5 meters. Therefore, the radius of the circle is 0.5 meters.

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  • 9. 

     Simplify:(4x2 - 2x) - (-5x2 - 8x)

    • 9x(3x + 2)

    • 8x(3x + 2)

    • 4x(3x + 2)

    • 3x(3x + 2)

    Correct Answer
    A. 3x(3x + 2)
    Explanation
    The given expression is a subtraction of two polynomials. To simplify it, we need to distribute the negative sign to each term inside the parentheses of the second polynomial. This gives us: (4x^2 - 2x) + (5x^2 + 8x). Now, we can combine like terms by adding the coefficients of the same degree terms. In this case, the x^2 terms have coefficients of 4 and 5, so we add them to get 9x^2. The x terms have coefficients of -2 and 8, so we add them to get 6x. Therefore, the simplified expression is 9x^2 + 6x, which can be factored as 3x(3x + 2).

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  • 10. 

     Factor: 3x4y3 – 48y3

    • 3y3(x2 + 4)(x + 2)(x -2)

    • 4y3(x2 + 4)(x + 2)(x -2)

    • 3y3(x2 + 3)(x + 2)(x -2)

    • 3y3(x2 + 3)(x + 1)(x -1)

    Correct Answer
    A. 3y3(x2 + 4)(x + 2)(x -2)
    Explanation
    The given expression can be factored by finding the greatest common factor (GCF) of the terms. The GCF of 3x^4y^3 and 48y^3 is 3y^3. Dividing both terms by 3y^3 gives us x^4 - 16. This can be further factored as the difference of squares: (x^2 + 4)(x + 2)(x - 2). Therefore, the correct answer is 3y^3(x^2 + 4)(x + 2)(x - 2).

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Janaisa Harris |BA (Mathematics) |
High School Math Teacher
Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a bachelor's degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher.

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