NECO Mathematics Practice Questions & Explanations

Try 20 of 120+ NECO Mathematics questions as a free quiz — select your answers, submit, and see your score with a full explanation for every one.

1.
Differentiate $y = 3x^2 + 5x - 7$ with respect to $x.$
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Apply the power rule. The derivative of $3x^2$ is $6x$ , the derivative of $5x$ is $5$ , and the derivative of $-7$ is $0$ . Therefore, $\frac{dy}{dx} = 6x + 5$ .
2.
Find the determinant of the matrix $\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}$ .
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For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $ad-bc$ . Therefore, $\det\left(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\right)=(4)(5)-(3)(2)=20-6=14$ . Hence, the answer is $\boxed{14}$ .
3.
If the determinant of the matrix $\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix}$ is equal to $10$ , find the value of $x$ .
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For the matrix $\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix}$ , the determinant is $ad-bc$ . Thus, $\det\left(\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix}\right)=4x-6$ . Given that the determinant is $10$ , $4x-6=10$ . Adding $6$ to both sides gives $4x=16$ . Dividing by $4$ gives $x=4$ . Therefore, the correct option is $\textbf{C}$ .
4.
Convert $1101_2$ to base ten.
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$1101_2 = (1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0) = 8+4+0+1 = 13$ .
5.
Convert $25_{ten}$ to base two.
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$25 = 16+8+1 = (1\times2^4)+(1\times2^3)+(0\times2^2)+(0\times2^1)+(1\times2^0) = 11001_2$ .
6.
Convert $342_5$ to base ten.
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$342_5 = (3\times5^2)+(4\times5^1)+(2\times5^0) = 75+20+2 = 97$ .
7.
Convert $58_{ten}$ to base five.
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Dividing repeatedly by 5: $58 \div 5 = 11$ remainder $3$ ; $11 \div 5 = 2$ remainder $1$ ; $2 \div 5 = 0$ remainder $2$ . Reading the remainders from last to first gives $213_5$ . Check: $(2\times25)+(1\times5)+3 = 50+5+3 = 58$ .
8.
Evaluate $101_2 + 111_2$ , leaving your answer in base two.
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$101_2 = 5$ and $111_2 = 7$ . $5+7 = 12$ . Converting $12$ back to base two gives $1100_2$ .
9.
Evaluate $1011_2 - 110_2$ , leaving your answer in base two.
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$1011_2 = 11$ and $110_2 = 6$ . $11-6 = 5$ . Converting $5$ back to base two gives $101_2$ .
10.
Find $11_2 \times 101_2$ , leaving your answer in base two.
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$11_2 = 3$ and $101_2 = 5$ . $3\times5 = 15$ . Converting $15$ back to base two gives $1111_2$ .
11.
If $24_x = 20_{ten}$ , find the value of the base $x$ .
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$24_x$ means $(2\times x)+4$ . Setting this equal to $20$ : $2x+4=20$ , so $2x=16$ and $x=8$ .
12.
Convert $123_4$ to base ten.
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$123_4 = (1\times4^2)+(2\times4^1)+(3\times4^0) = 16+8+3 = 27$ .
13.
Evaluate $1100_2 \div 10_2$ , leaving your answer in base two.
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$1100_2 = 12$ and $10_2 = 2$ . $12 \div 2 = 6$ . Converting $6$ back to base two gives $110_2$ .
14.
If $23_r = 19_{ten}$ , find the value of the base $r$ .
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$23_r$ means $(2\times r)+3$ . Setting this equal to $19$ : $2r+3=19$ , so $2r=16$ and $r=8$ .
15.
Evaluate $11011_2 - 1101_2$ , leaving your answer in base two.
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$11011_2 = 27$ and $1101_2 = 13$ . $27-13 = 14$ . Converting $14$ back to base two gives $1110_2$ .
16.
Evaluate $\sqrt{25}$ .
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$\sqrt{25} = 5$ because $5^2 = 25$ .
17.
Simplify $\sqrt{36} + \sqrt{49}$ .
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$\sqrt{36} = 6$ and $\sqrt{49} = 7$ . Therefore, $6 + 7 = 13$ .
18.
Find the value of $\sqrt{81} - \sqrt{16}$ .
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$\sqrt{81} = 9$ and $\sqrt{16} = 4$ . Thus, $9 - 4 = 5$ .
19.
Express $\sqrt{50}$ in its simplest form.
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$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$ .
20.
Simplify $3\sqrt{5} + 2\sqrt{5}$ .
Show explanation
Since the surds are alike, add the coefficients: $3 + 2 = 5$ . Therefore, $3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}$ .

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