UN Matematika SMP 2018

Soal 1.
Hasil dari $6\times (-7+4) : (-6-3)= ....$
A. $-6$
B. $-2$
C. $2$
D. $6$
Pembahasan: C
$6\times (-7+4) : (-6-3)=6\times (-3) : (-9)= (-18) : (-9) =2$

Soal 2.
Hasil dari $\frac{\frac{3}{5} + \frac{1}{2}}{\frac{3}{5} - \frac{1}{2}}$ adalah ....
A. $\frac{1}{10}$
B. $\frac{1}{11}$
C. $10$
D. $11$
Pembahasan: D

$\frac{\frac{3}{5} + \frac{1}{2}}{\frac{3}{5} - \frac{1}{2}}=\frac{\frac{6+5}{10}}{\frac{6-5}{10}}=\frac{\frac{11}{10}}{\frac{1}{10}}=\frac{11}{1}=11$

Soal 3.
Suhu udara di wilayah Jakarta $25^{\circ}C$. Di saat yang sama suhu udara di wilayah Tokyo Jepang $-6^{\circ}C$. Berapakah perbedaan suhu di kedua wilayah tersebut?
A. $31^{\circ}C$
B. $19^{\circ}C$
C. $-19^{\circ}C$
D. $-31^{\circ}C$
Pembahasan:
$25^{\circ}C - (-6^{\circ}C)=25^{\circ}C +6^{\circ}C=31^{\circ}C$

Soal 4.
Panitia kegiatan sosial menerima sumbangan gula pasir beratnya $22\frac{3}{4}$ kg, $25\frac{1}{2}$ kg, dan $24\frac{1}{4}$ kg untuk dibagikan kepada sekelompok warga. Setiap warga menerima $2\frac{1}{2}$ kg. Berapa banyak warga yang menerima gula pasir tersebut?
A. $36$
B. $35$
C. $29$
D. $27$
Pembahasan:
Banyak gula pasir yang diterima:
$22\frac{3}{4}+25\frac{1}{2}+24\frac{1}{4} = (22+25+24) + (\frac{3}{4}+\frac{1}{2}+\frac{1}{4})=71 + (\frac{3}{4}+\frac{1}{4}+\frac{1}{2})=71 + (1+\frac{1}{2})=72\frac{1}{2}$
Banyak warga:
$72\frac{1}{2}: 2\frac{1}{2} = \frac{145}{2} : \frac {5}{2}=\frac{145}{2} \times \frac {2}{5}=\frac {145}{5}=29$

Soal 5.
Hasil dari $(-5)^3 +(-5)^2 +(-5)^1 +(-5)^0$ adalah ....
A. $156$
B. $30$
C. $-104$
D. $-105$
Pembahasan: C
$(-5)^3 +(-5)^2 +(-5)^1 +(-5)^0= -125 + 25 +(-5) +1=-104$

Soal 6.
Hasil dari $\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}$ adalah ....
A. $-24 + 14 \sqrt{3}$
B. $24 + 14 \sqrt{3}$
C. $24 - 14 \sqrt{3}$
D. $-24 - 14 \sqrt{3}$
Pembahasan: A
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{4\cdot 3\sqrt{2} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}} \times \frac{3\sqrt{2} - 2 \sqrt{6}}{3\sqrt{2} - 2 \sqrt{6}}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{(12\sqrt{2} - 6 \sqrt{6})(3\sqrt{2} - 2 \sqrt{6})}{(3\sqrt{2})^2 - (2\sqrt{6})^2}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{36 \cdot 2 - 24 \cdot \sqrt{12} - 18\sqrt{12} + 12 \cdot 6}{18-24}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 24 \cdot 2\sqrt{3} - 18\cdot 2\sqrt{3} + 72 }{18-24}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 24 \cdot 2\sqrt{3} - 18\cdot 2\sqrt{3} + 72 }{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 48\sqrt{3} - 36\sqrt{3} + 72}{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{144 - 84\sqrt{3}}{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=-24 + 14 \sqrt{3}$

Soal 7.
Tiga suku berikutnya dari barisan 2, 3, 7, 16, ... adalah ....
A. $-24 + 14 \sqrt{3}$
B. $24 + 14 \sqrt{3}$
C. $24 - 14 \sqrt{3}$
D. $-24 - 14 \sqrt{3}$
Pembahasan: A
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{4\cdot 3\sqrt{2} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}} \times \frac{3\sqrt{2} - 2 \sqrt{6}}{3\sqrt{2} - 2 \sqrt{6}}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{(12\sqrt{2} - 6 \sqrt{6})(3\sqrt{2} - 2 \sqrt{6})}{(3\sqrt{2})^2 - (2\sqrt{6})^2}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{36 \cdot 2 - 24 \cdot \sqrt{12} - 18\sqrt{12} + 12 \cdot 6}{18-24}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 24 \cdot 2\sqrt{3} - 18\cdot 2\sqrt{3} + 72 }{18-24}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 24 \cdot 2\sqrt{3} - 18\cdot 2\sqrt{3} + 72 }{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{72 - 48\sqrt{3} - 36\sqrt{3} + 72}{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=\frac{144 - 84\sqrt{3}}{-6}$
$\frac{4\sqrt{18} - 6 \sqrt{6}}{3\sqrt{2} + 2 \sqrt{6}}=-24 + 14 \sqrt{3}$