SNBT Zone : Aljabar
Table of Contents
Tipe:
No.
25% dari K adalah 40 dan \(\dfrac27\) dari Y adalah 28. Nilai- 254
- 256
- 258
- 268
- 276
ALTERNATIF PENYELESAIAN
\(\eqalign{
25\%K&=40\\
\dfrac14K&=40\\
K&=160
}\)
\(\eqalign{ \dfrac27Y&=28\\ Y&=28\cdot\dfrac72\\ &=98 }\)
\(\eqalign{ \dfrac27Y&=28\\ Y&=28\cdot\dfrac72\\ &=98 }\)
\(\eqalign{
K+Y&=160+98\\
&=\boxed{\boxed{258}}
}\)
Jadi, nilai K + Y adalah 258.
JAWAB: C
JAWAB: C
No.
Diberikan persamaan berikut: \[\dfrac{x}{a+b-c}=\dfrac{y}{b+c-a}=\dfrac{z}{c+a-b}\] Berapakah nilai dariALTERNATIF PENYELESAIAN
Misal \(\dfrac{x}{a+b-c}=\dfrac{y}{b+c-a}=\dfrac{z}{c+a-b}=p\)
\(\begin{aligned} (a-b)x+(b-c)y+(c-a)z&=(a-b)(a+b-c)p+(b-c)(b+c-a)p+(c-a)(c+a-b)p\\ &=p\left(a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab\right)\\ &=p\left(0\right)\\ &=\boxed{\boxed{0}} \end{aligned}\)
\(\begin{aligned} (a-b)x+(b-c)y+(c-a)z&=(a-b)(a+b-c)p+(b-c)(b+c-a)p+(c-a)(c+a-b)p\\ &=p\left(a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab\right)\\ &=p\left(0\right)\\ &=\boxed{\boxed{0}} \end{aligned}\)
Jadi, (a − b)x + (b − c)y + (c − a)z = 0 .
No.
Jika $\dfrac{a}b+\dfrac{c}d=16$ dan $\dfrac{a}c+\dfrac{b}d=36$ dengan b, c, dan d ≠ 0, maka $\dfrac{c}b=$ ....- $\dfrac34$
- $\dfrac23$
- $\dfrac29$
- $\dfrac43$
- $\dfrac49$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{a}b+\dfrac{c}d&=16\\
\dfrac{ad+bc}{bd}&=16\\
ad+bc&=16bd
\end{aligned}\)
\(\begin{aligned} \dfrac{a}c+\dfrac{b}d&=36\\ \dfrac{ad+bc}{cd}&=36\\ ad+bc&=36cd \end{aligned}\)
\(\begin{aligned} 16bd&=36cd\\ 4bd&=9cd\\ \dfrac{c}b& \color{blue}\boxed{\boxed{\color{black}=\dfrac49}} \end{aligned}\)
\(\begin{aligned} \dfrac{a}c+\dfrac{b}d&=36\\ \dfrac{ad+bc}{cd}&=36\\ ad+bc&=36cd \end{aligned}\)
\(\begin{aligned} 16bd&=36cd\\ 4bd&=9cd\\ \dfrac{c}b& \color{blue}\boxed{\boxed{\color{black}=\dfrac49}} \end{aligned}\)
Jadi, $\dfrac{c}b=\dfrac49$.
JAWAB: E
JAWAB: E
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