Praktis Kendiri 2.1a (Soalan 6 & 7) - Buku Teks Matematik Tingkatan 4 Bab 2 (Asas Nombor)

Soalan 6:
Tentukan nilai nombor berikut dalam asas sepuluh.

Penyelesaian:
(a) 23₆

\begin{matrix} Nilai nombor & = (2 \times 6^{1}) + (3 \times 6^{0}) = 12 + 3 = 15 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(2 \times 6^1\right)+\left(3 \times 6^0\right) \\ & =12+3 \\ & =15 \end{aligned}$
(b) 425₈

\begin{matrix} Nilai nombor & = (4 \times 8^{2}) + (2 \times 8^{1}) + (5 \times 8^{0}) = 256 + 16 + 5 = 277 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(4 \times 8^2\right)+\left(2 \times 8^1\right)+\left(5 \times 8^0\right) \\ & =256+16+5 \\ & =277 \end{aligned}$

(c) 110101₂

\begin{matrix} Nilai nombor = & (1 \times 2^{5}) + (1 \times 2^{4}) + (0 \times 2^{3}) + (1 \times 2^{2}) + (0 \times 2^{1}) + (1 \times 2^{0}) = & 32 + 16 + 0 + 4 + 0 + 1 = & 53 \end{matrix}

$\begin{aligned} \text { Nilai nombor }= & \left(1 \times 2^5\right)+\left(1 \times 2^4\right)+\left(0 \times 2^3\right)+\left(1 \times 2^2\right)+ \\ & \left(0 \times 2^1\right)+\left(1 \times 2^0\right) \\ = & 32+16+0+4+0+1 \\ = & 53 \end{aligned}$
(d) 338₉

\begin{matrix} Nilai nombor & = (3 \times 9^{2}) + (3 \times 9^{1}) + (8 \times 9^{0}) = 243 + 27 + 8 = 278 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(3 \times 9^2\right)+\left(3 \times 9^1\right)+\left(8 \times 9^0\right) \\ & =243+27+8 \\ & =278 \end{aligned}$

(e) 364₇

\begin{matrix} Nilai nombor & = (3 \times 7^{2}) + (6 \times 7^{1}) + (4 \times 7^{0}) = 147 + 42 + 4 = 193 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(3 \times 7^2\right)+\left(6 \times 7^1\right)+\left(4 \times 7^0\right) \\ & =147+42+4 \\ & =193 \end{aligned}$
(f) 33₄

\begin{matrix} Nilai nombor & = (3 \times 4^{1}) + (3 \times 4^{0}) = 12 + 3 = 15 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(3 \times 4^1\right)+\left(3 \times 4^0\right) \\ & =12+3 \\ & =15 \end{aligned}$

(g) 123₅

\begin{matrix} Nilai nombor & = (1 \times 5^{2}) + (2 \times 5^{1}) + (3 \times 5^{0}) = 25 + 10 + 1 = 38 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(1 \times 5^2\right)+\left(2 \times 5^1\right)+\left(3 \times 5^0\right) \\ & =25+10+1 \\ & =38 \end{aligned}$
(h) 1217₈

\begin{matrix} Nilai nombor & = (1 \times 8^{3}) + (2 \times 8^{2}) + (1 \times 8^{1}) + (7 \times 8^{0}) = 512 + 128 + 8 + 7 = 655 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(1 \times 8^3\right)+\left(2 \times 8^2\right)+\left(1 \times 8^1\right)+\left(7 \times 8^0\right) \\ & =512+128+8+7 \\ & =655 \end{aligned}$

(i) 515₆

\begin{matrix} Nilai nombor & = (5 \times 6^{2}) + (1 \times 6^{1}) + (5 \times 6^{0}) = 180 + 6 + 5 = 191 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(5 \times 6^2\right)+\left(1 \times 6^1\right)+\left(5 \times 6^0\right) \\ & =180+6+5 \\ & =191 \end{aligned}$

(j) 1121₃

\begin{matrix} Nilai nombor & = (1 \times 3^{3}) + (1 \times 3^{2}) + (2 \times 3^{1}) + (1 \times 3^{0}) = 27 + 9 + 6 + 1 = 43 \end{matrix}

$\begin{aligned} \text { Nilai nombor } & =\left(1 \times 3^3\right)+\left(1 \times 3^2\right)+\left(2 \times 3^1\right)+\left(1 \times 3^0\right) \\ & =27+9+6+1 \\ & =43 \end{aligned}$

Soalan 7:
Tentukan nilai p dan nilai q.

(a) 1101_{2} = (1 \times 2^{p}) + (1 \times q) + (1 \times 2^{0})

$\text { (a) } 1101_2=\left(1 \times 2^p\right)+(1 \times q)+\left(1 \times 2^0\right)$

(b) 375_{8} = (3 \times 8^{p}) + (q \times 8^{1}) + (5 \times 8^{0})

$\text { (b) } 375_8=\left(3 \times 8^p\right)+\left(q \times 8^1\right)+\left(5 \times 8^0\right)$

(c) 1321_{4} = (1 \times p^{q}) + (3 \times 4^{2}) + (2 \times 4^{1}) + (1 \times 4^{0})

$\text { (c) } 1321_4=\left(1 \times p^q\right)+\left(3 \times 4^2\right)+\left(2 \times 4^1\right)+\left(1 \times 4^0\right)$

Penyelesaian:
(a)

1101_{2} = (1 \times 2^{p}) + (1 \times q) + (1 \times 2^{0})

$1101_2=\left(1 \times 2^p\right)+(1 \times q)+\left(1 \times 2^0\right)$

(1 \times 2^{3}) + (1 \times 2^{2}) + (1 \times 2^{0}) = (1 \times 2^{p}) + (1 \times q) + (1 \times 2^{0})

$\left(1 \times 2^3\right)+\left(1 \times 2^2\right)+\left(1 \times 2^0\right)=\left(1 \times 2^p\right)+(1 \times q)+\left(1 \times 2^0\right)$

\begin{matrix} Secara bandingan, \begin{matrix} 2^{3} & = 2^{p} & 2^{2} & = q p & = 3 & q & = 2^{2} \end{matrix} \end{matrix}

$\begin{aligned} &\text { Secara bandingan, }\\ &\begin{array}{rlrl} 2^3 & =2^p & 2^2 & =q \\ p & =3 & q & =2^2 \end{array} \end{aligned}$

(b)

375_{8} = (3 \times 8^{p}) + (q \times 8^{1}) + (5 \times 8^{0})

$375_8=\left(3 \times 8^p\right)+\left(q \times 8^1\right)+\left(5 \times 8^0\right)$

(3 \times 8^{2}) + (7 \times 8^{1}) + (5 \times 8^{0}) = (3 \times 8^{p}) + (q \times 8^{1}) + (5 \times 8^{0})

$\left(3 \times 8^2\right)+\left(7 \times 8^1\right)+\left(5 \times 8^0\right)=\left(3 \times 8^p\right)+\left(q \times 8^1\right)+\left(5 \times 8^0\right)$

\begin{matrix} Secara bandingan, \begin{matrix} 8^{2} & = 8^{p} & q = 7 p & = 2 \end{matrix} \end{matrix}

$\begin{aligned} &\text { Secara bandingan, }\\ &\begin{array}{rlr} 8^2 & =8^p & q=7 \\ p & =2 & \end{array} \end{aligned}$

(c)

1321_{4} = (1 \times p^{q}) + (3 \times 4^{2}) + (2 \times 4^{1}) + (1 \times 4^{0})

$1321_4=\left(1 \times p^q\right)+\left(3 \times 4^2\right)+\left(2 \times 4^1\right)+\left(1 \times 4^0\right)$

\begin{matrix} (1 \times 4^{3}) + (3 \times 4^{2}) + (2 \times 4^{1}) + (1 \times 4^{0}) = (1 \times p^{q}) + (3 \times 4^{2}) + (2 \times 4^{1}) + (1 \times 4^{0}) \end{matrix}

$\begin{aligned} & \left(1 \times 4^3\right)+\left(3 \times 4^2\right)+\left(2 \times 4^1\right)+\left(1 \times 4^0\right)= \\ & \left(1 \times p^q\right)+\left(3 \times 4^2\right)+\left(2 \times 4^1\right)+\left(1 \times 4^0\right) \end{aligned}$

\begin{matrix} Secara bandingan, \begin{matrix} 4^{3} & = p^{q} p & = 4 q & = 3 \end{matrix} \end{matrix}

$\begin{aligned} &\text { Secara bandingan, }\\ &\begin{aligned} 4^3 & =p^q \\ p & =4 \\ q & =3 \end{aligned} \end{aligned}$

Praktis Kendiri 2.1a (Soalan 6 & 7) – Buku Teks Matematik Tingkatan 4 Bab 2 (Asas Nombor)

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