NABTEB MathematicsPast Questions & Explanations

Practice 52+ NABTEB 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.

   Convert $1101_2$ to base ten.

   1. A.12
   2. B.13
   3. C.14
   4. D.15

   Show explanation

   $1101_2 = (1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0) = 8+4+0+1 = 13$.
2. 2.

   Convert $25_{ten}$ to base two.

   1. A.11001
   2. B.10011
   3. C.11100
   4. D.10101

   Show explanation

   $25 = 16+8+1 = (1\times2^4)+(1\times2^3)+(0\times2^2)+(0\times2^1)+(1\times2^0) = 11001_2$.
3. 3.

   Convert $342_5$ to base ten.

   1. A.87
   2. B.92
   3. C.97
   4. D.102

   Show explanation

   $342_5 = (3\times5^2)+(4\times5^1)+(2\times5^0) = 75+20+2 = 97$.
4. 4.

   Convert $58_{ten}$ to base five.

   1. A.$203_5$
   2. B.$213_5$
   3. C.$223_5$
   4. D.$123_5$

   Show explanation

   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: $(2\times25)+(1\times5)+3 = 50+5+3 = 58$.
5. 5.

   Evaluate $101_2 + 111_2$, leaving your answer in base two.

   1. A.1011
   2. B.1100
   3. C.1101
   4. D.1110

   Show explanation

   $101_2 = 5$ and $111_2 = 7$. $5+7 = 12$. Converting $12$ back to base two gives $1100_2$.
6. 6.

   Evaluate $1011_2 - 110_2$, leaving your answer in base two.

   1. A.100
   2. B.101
   3. C.110
   4. D.111

   Show explanation

   $1011_2 = 11$ and $110_2 = 6$. $11-6 = 5$. Converting $5$ back to base two gives $101_2$.
7. 7.

   Find $11_2 \times 101_2$, leaving your answer in base two.

   1. A.1101
   2. B.1110
   3. C.1111
   4. D.10000

   Show explanation

   $11_2 = 3$ and $101_2 = 5$. $3\times5 = 15$. Converting $15$ back to base two gives $1111_2$.
8. 8.

   If $24_x = 20_{ten}$, find the value of the base $x$.

   1. A.6
   2. B.7
   3. C.8
   4. D.9

   Show explanation

   $24_x$ means $(2\times x)+4$. Setting this equal to $20$: $2x+4=20$, so $2x=16$ and $x=8$.
9. 9.

   Convert $123_4$ to base ten.

   1. A.27
   2. B.21
   3. C.24
   4. D.31

   Show explanation

   $123_4 = (1\times4^2)+(2\times4^1)+(3\times4^0) = 16+8+3 = 27$.
10. 10.

    Evaluate $1100_2 \div 10_2$, leaving your answer in base two.

    1. A.101
    2. B.110
    3. C.111
    4. D.100

    Show explanation

    $1100_2 = 12$ and $10_2 = 2$. $12 \div 2 = 6$. Converting $6$ back to base two gives $110_2$.

    Want thousands more with instant AI explanations after every answer?

    Practice free on SuperPrep
11. 11.

    If $23_r = 19_{ten}$, find the value of the base $r$.

    1. A.5
    2. B.6
    3. C.7
    4. D.8

    Show explanation

    $23_r$ means $(2\times r)+3$. Setting this equal to $19$: $2r+3=19$, so $2r=16$ and $r=8$.
12. 12.

    Evaluate $11011_2 - 1101_2$, leaving your answer in base two.

    1. A.1101
    2. B.1110
    3. C.1010
    4. D.1111

    Show explanation

    $11011_2 = 27$ and $1101_2 = 13$. $27-13 = 14$. Converting $14$ back to base two gives $1110_2$.
13. 13.

    Evaluate $\sqrt{25}$.

    1. A.3
    2. B.4
    3. C.5
    4. D.6

    Show explanation

    $\sqrt{25} = 5$ because $5^2 = 25$.
14. 14.

    Simplify $\sqrt{36} + \sqrt{49}$.

    1. A.11
    2. B.12
    3. C.13
    4. D.14

    Show explanation

    $\sqrt{36} = 6$ and $\sqrt{49} = 7$. Therefore, $6 + 7 = 13$.
15. 15.

    Find the value of $\sqrt{81} - \sqrt{16}$.

    1. A.3
    2. B.4
    3. C.5
    4. D.6

    Show explanation

    $\sqrt{81} = 9$ and $\sqrt{16} = 4$. Thus, $9 - 4 = 5$.
16. 16.

    Express $\sqrt{50}$ in its simplest form.

    1. A.$5\sqrt{2}$
    2. B.$2\sqrt{5}$
    3. C.$10\sqrt{2}$
    4. D.$25\sqrt{2}$

    Show explanation

    $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$.
17. 17.

    Simplify $3\sqrt{5} + 2\sqrt{5}$.

    1. A.$5\sqrt{5}$
    2. B.$6\sqrt{5}$
    3. C.$5$
    4. D.$\sqrt{25}$

    Show explanation

    Since the surds are alike, add the coefficients: $3 + 2 = 5$. Therefore, $3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}$.
18. 18.

    Simplify $\sqrt{12} + \sqrt{27}$.

    1. A.$5\sqrt{3}$
    2. B.$7\sqrt{3}$
    3. C.$3\sqrt{3}$
    4. D.$9\sqrt{3}$

    Show explanation

    $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$. Thus, $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$.
19. 19.

    Rationalize the denominator of $\frac{3}{\sqrt{5}}$.

    1. A.$\frac{3\sqrt{5}}{5}$
    2. B.$\frac{15}{\sqrt{5}}$
    3. C.$\frac{3}{5\sqrt{5}}$
    4. D.$\frac{\sqrt{5}}{3}$

    Show explanation

    Multiply numerator and denominator by $\sqrt{5}$: $\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$.
20. 20.

    Simplify $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})$.

    1. A.1
    2. B.2
    3. C.5
    4. D.$\sqrt{6}$

    Show explanation

    Using the difference of two squares: $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$.