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NECO MathematicsPast Questions & Explanations

Practice 59+ NECO Mathematics questions from the SuperPrep question bank. Every question comes with a full explanation so you learn the method, not just the answer.

  1. 1.
    Differentiate y=3x2+5x7y = 3x^2 + 5x - 7 with respect to x.x.
    1. A.6x+56x + 5
    2. B.3x+53x + 5
    3. C.6x76x - 7
    4. D.3x2+53x^2 + 5
    Show explanation
    Apply the power rule. The derivative of 3x23x^2 is 6x6x, the derivative of 5x5x is 55, and the derivative of 7-7 is 00. Therefore, dydx=6x+5\frac{dy}{dx} = 6x + 5.
  2. 2.
    Find the determinant of the matrix [4325]\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}.
    1. A.1212
    2. B.1616
    3. C.1818
    4. D.1414
    Show explanation
    For a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is adbcad-bc. Therefore, det([4325])=(4)(5)(3)(2)=206=14\det\left(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\right)=(4)(5)-(3)(2)=20-6=14. Hence, the answer is 14\boxed{14}.
  3. 3.
    If the determinant of the matrix [x234]\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix} is equal to 1010, find the value of xx.
    1. A.22
    2. B.33
    3. C.44
    4. D.55
    Show explanation
    For the matrix [x234]\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix}, the determinant is adbcad-bc. Thus, det([x234])=4x6\det\left(\begin{bmatrix} x & 2 \\ 3 & 4 \end{bmatrix}\right)=4x-6. Given that the determinant is 1010, 4x6=104x-6=10. Adding 66 to both sides gives 4x=164x=16. Dividing by 44 gives x=4x=4. Therefore, the correct option is C\textbf{C}.
  4. 4.
    Convert 110121101_2 to base ten.
    1. A.12
    2. B.13
    3. C.14
    4. D.15
    Show explanation
    11012=(1×23)+(1×22)+(0×21)+(1×20)=8+4+0+1=131101_2 = (1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0) = 8+4+0+1 = 13.
  5. 5.
    Convert 25ten25_{ten} to base two.
    1. A.11001
    2. B.10011
    3. C.11100
    4. D.10101
    Show explanation
    25=16+8+1=(1×24)+(1×23)+(0×22)+(0×21)+(1×20)=11001225 = 16+8+1 = (1\times2^4)+(1\times2^3)+(0\times2^2)+(0\times2^1)+(1\times2^0) = 11001_2.
  6. 6.
    Convert 3425342_5 to base ten.
    1. A.87
    2. B.92
    3. C.97
    4. D.102
    Show explanation
    3425=(3×52)+(4×51)+(2×50)=75+20+2=97342_5 = (3\times5^2)+(4\times5^1)+(2\times5^0) = 75+20+2 = 97.
  7. 7.
    Convert 58ten58_{ten} to base five.
    1. A.2035203_5
    2. B.2135213_5
    3. C.2235223_5
    4. D.1235123_5
    Show explanation
    Dividing repeatedly by 5: 58÷5=1158 \div 5 = 11 remainder 33; 11÷5=211 \div 5 = 2 remainder 11; 2÷5=02 \div 5 = 0 remainder 22. Reading the remainders from last to first gives 2135213_5. Check: (2×25)+(1×5)+3=50+5+3=58(2\times25)+(1\times5)+3 = 50+5+3 = 58.
  8. 8.
    Evaluate 1012+1112101_2 + 111_2, leaving your answer in base two.
    1. A.1011
    2. B.1100
    3. C.1101
    4. D.1110
    Show explanation
    1012=5101_2 = 5 and 1112=7111_2 = 7. 5+7=125+7 = 12. Converting 1212 back to base two gives 110021100_2.
  9. 9.
    Evaluate 1011211021011_2 - 110_2, leaving your answer in base two.
    1. A.100
    2. B.101
    3. C.110
    4. D.111
    Show explanation
    10112=111011_2 = 11 and 1102=6110_2 = 6. 116=511-6 = 5. Converting 55 back to base two gives 1012101_2.
  10. 10.
    Find 112×101211_2 \times 101_2, leaving your answer in base two.
    1. A.1101
    2. B.1110
    3. C.1111
    4. D.10000
    Show explanation
    112=311_2 = 3 and 1012=5101_2 = 5. 3×5=153\times5 = 15. Converting 1515 back to base two gives 111121111_2.

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  11. 11.
    If 24x=20ten24_x = 20_{ten}, find the value of the base xx.
    1. A.6
    2. B.7
    3. C.8
    4. D.9
    Show explanation
    24x24_x means (2×x)+4(2\times x)+4. Setting this equal to 2020: 2x+4=202x+4=20, so 2x=162x=16 and x=8x=8.
  12. 12.
    Convert 1234123_4 to base ten.
    1. A.27
    2. B.21
    3. C.24
    4. D.31
    Show explanation
    1234=(1×42)+(2×41)+(3×40)=16+8+3=27123_4 = (1\times4^2)+(2\times4^1)+(3\times4^0) = 16+8+3 = 27.
  13. 13.
    Evaluate 11002÷1021100_2 \div 10_2, leaving your answer in base two.
    1. A.101
    2. B.110
    3. C.111
    4. D.100
    Show explanation
    11002=121100_2 = 12 and 102=210_2 = 2. 12÷2=612 \div 2 = 6. Converting 66 back to base two gives 1102110_2.
  14. 14.
    If 23r=19ten23_r = 19_{ten}, find the value of the base rr.
    1. A.5
    2. B.6
    3. C.7
    4. D.8
    Show explanation
    23r23_r means (2×r)+3(2\times r)+3. Setting this equal to 1919: 2r+3=192r+3=19, so 2r=162r=16 and r=8r=8.
  15. 15.
    Evaluate 1101121101211011_2 - 1101_2, leaving your answer in base two.
    1. A.1101
    2. B.1110
    3. C.1010
    4. D.1111
    Show explanation
    110112=2711011_2 = 27 and 11012=131101_2 = 13. 2713=1427-13 = 14. Converting 1414 back to base two gives 111021110_2.
  16. 16.
    Evaluate 25\sqrt{25}.
    1. A.3
    2. B.4
    3. C.5
    4. D.6
    Show explanation
    25=5\sqrt{25} = 5 because 52=255^2 = 25.
  17. 17.
    Simplify 36+49\sqrt{36} + \sqrt{49}.
    1. A.11
    2. B.12
    3. C.13
    4. D.14
    Show explanation
    36=6\sqrt{36} = 6 and 49=7\sqrt{49} = 7. Therefore, 6+7=136 + 7 = 13.
  18. 18.
    Find the value of 8116\sqrt{81} - \sqrt{16}.
    1. A.3
    2. B.4
    3. C.5
    4. D.6
    Show explanation
    81=9\sqrt{81} = 9 and 16=4\sqrt{16} = 4. Thus, 94=59 - 4 = 5.
  19. 19.
    Express 50\sqrt{50} in its simplest form.
    1. A.525\sqrt{2}
    2. B.252\sqrt{5}
    3. C.10210\sqrt{2}
    4. D.25225\sqrt{2}
    Show explanation
    50=25×2=252=52\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\sqrt{2} = 5\sqrt{2}.
  20. 20.
    Simplify 35+253\sqrt{5} + 2\sqrt{5}.
    1. A.555\sqrt{5}
    2. B.656\sqrt{5}
    3. C.55
    4. D.25\sqrt{25}
    Show explanation
    Since the surds are alike, add the coefficients: 3+2=53 + 2 = 5. Therefore, 35+25=553\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}.

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