PAS MATEMATIKA BLOG: Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{ll} 66. & Suatu unit pekerjaan dapat diselesaikan oleh A \\ B, dan C bersama-sama dalam 2 jam saja. \\ Jika pekerjaan itu dapat diselesaikan oleh A dan \\ B bersama-sama dalam 2 jam 24 menit, dan oleh \\ B dan C bersama-sama dalam waktu 3 jam, \\ maka sistem persamaan berikut yang memenuhi \\ adalah . . . . \\ \begin{array}{llll} a . & {\begin{cases} A + B + C & = 2 \\ A + B & = \frac{12}{5} \\ B + C & = 3 \end{cases} \\ b . & {\begin{cases} A + B + C & = \frac{1}{2} \\ A + B & = \frac{5}{12} \\ B + C & = \frac{1}{3} \end{cases} \\ c . & {\begin{cases} \frac{1}{A} + \frac{1}{B} + \frac{1}{C} & = 2 \\ \frac{1}{A} + \frac{1}{B} & = \frac{12}{5} \\ \frac{1}{B} + \frac{1}{C} & = 3 \end{cases} \\ d . & {\begin{cases} \frac{1}{A} + \frac{1}{B} + \frac{1}{C} & = \frac{1}{2} \\ \frac{1}{A} + \frac{1}{B} & = \frac{5}{12} \\ \frac{1}{B} + \frac{1}{C} & = \frac{1}{3} \end{cases} \\ e . & {\begin{cases} \frac{1}{A} + \frac{1}{B} + \frac{1}{C} & = 2 \\ \frac{1}{A} + \frac{1}{B} - \frac{1}{C} & = \frac{12}{5} \\ - \frac{1}{A} + \frac{1}{B} + \frac{1}{C} & = 3 \end{cases} \end{array} \\ Jawab : d \\ \begin{aligned} Per & hatikan bahwa : Waktu penyelesaian \\ sua & tu pekerjaan adalah termasuk \\ per & bandingan berbalik nilai, maka \\ ∙ & A, B, dan C dalam 2 jam, artinya : \\ \frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{2}, demikian juga \\ ∙ & A dan B bersama-sama selesai dalam \\ 2 jam 24 menit atau \frac{12}{5} jam : \\ \frac{1}{A} + \frac{1}{B} = \frac{5}{12} \\ ∙ & B dan C selesai dalam 3 jam : \\ \frac{1}{B} + \frac{1}{C} = \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll} 67. & Himpunan penyelesaian dari \\ {\begin{array}{c} x + y + 4 z = 15 \\ x - y + z = 2 \\ x + 2 y - 3 z = - 4 \end{array} \\ adalah . . . . \\ \begin{array}{llll} a . & {(- 1, 1, 3)} \\ b . & {(1, 2, 3)} \\ c . & {(- 2, 1, 1)} \\ d . & {(3, 2, - 1)} \\ e . & {(1, - 2, 3)} \end{array} \\ Jawab : b \\ Semunya dikerjakan dengan metode \\ matriks (Cara Cramer) \\ \begin{aligned} x & = \frac{| \begin{array}{c} 15 & 1 & 4 \\ 2 & - 1 & 1 \\ - 4 & 2 & - 3 \end{array} |}{| \begin{array}{c} 1 & 1 & 4 \\ 1 & - 1 & 1 \\ 1 & 2 & - 3 \end{array} |} \\ = \frac{15 | \begin{array}{c} - 1 & 1 \\ 2 & - 3 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ - 4 & - 3 \end{array} | + 4 | \begin{array}{c} 2 & - 1 \\ - 4 & 2 \end{array} |}{1 | \begin{array}{c} - 1 & 1 \\ 2 & - 3 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 1 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & - 1 \\ 1 & 2 \end{array} |} \\ = \frac{15 (3 - 2) - 1 (- 6 + 4) + 4 (4 - 4)}{1 (3 - 2) - 1 (- 3 - 1) + 4 (2 + 1)} \\ = \frac{15 (1) - 1 (- 2) + 4 (0)}{1 (1) - 1 (- 4) + 4 (3)} = \frac{17}{17} = 1 \\ y & = \frac{| \begin{array}{c} 1 & 15 & 4 \\ 1 & 2 & 1 \\ 1 & - 4 & - 3 \end{array} |}{| \begin{array}{c} 1 & 1 & 4 \\ 1 & - 1 & 1 \\ 1 & 2 & - 3 \end{array} |} \\ = \frac{1 | \begin{array}{c} 2 & 1 \\ - 4 & - 3 \end{array} | - 15 | \begin{array}{c} 1 & 1 \\ 1 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & 2 \\ 1 & - 4 \end{array} |}{1 | \begin{array}{c} - 1 & 1 \\ 2 & - 3 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 1 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & - 1 \\ 1 & 2 \end{array} |} \\ = \frac{1 (- 6 + 4) - 15 (- 3 - 1) + 4 (- 4 - 2)}{1 (3 - 2) - 1 (- 3 - 1) + 4 (2 + 1)} \\ = \frac{1 (- 2) - 15 (- 4) + 4 (- 6)}{1 (1) - 1 (- 4) + 4 (3)} = \frac{34}{17} = 2 \\ z & = \frac{| \begin{array}{c} 1 & 1 & 15 \\ 1 & - 1 & 2 \\ 1 & 2 & - 4 \end{array} |}{| \begin{array}{c} 1 & 1 & 4 \\ 1 & - 1 & 1 \\ 1 & 2 & - 3 \end{array} |} \\ = \frac{1 | \begin{array}{c} - 1 & 2 \\ 2 & - 4 \end{array} | - 1 | \begin{array}{c} 1 & 2 \\ 1 & - 4 \end{array} | + 15 | \begin{array}{c} 1 & - 1 \\ 1 & 2 \end{array} |}{1 | \begin{array}{c} - 1 & 1 \\ 2 & - 3 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 1 & - 3 \end{array} | + 4 | \begin{array}{c} 1 & - 1 \\ 1 & 2 \end{array} |} \\ = \frac{1 (4 - 4) - 1 (- 4 - 2) + 15 (2 + 1)}{1 (3 - 2) - 1 (- 3 - 1) + 4 (2 + 1)} \\ = \frac{1 (0) - 1 (- 6) + 15 (3)}{1 (1) - 1 (- 4) + 4 (3)} = \frac{51}{17} = 3 \end{aligned} \end{array}$

$. Cara di atas$ full matriks-Cramer

$\begin{array}{ll} 68. & Hasil dari x y z yang memenuhi \\ {\begin{array}{c} x + y + z = 2 \\ x - y + z = - 2 \\ x - y - z = 2 \end{array} \\ adalah . . . . \\ \begin{array}{llll} a . & - 8 \\ b . & - 4 \\ c . & 2 \\ d . & 4 \\ e . & 8 \end{array} \\ Jawab : a \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{array}{c} x + y + z = 2 . . . . . (1) \\ x - y + z = - 2 . . . . . (2) \\ x - y - z = 2 . . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llc} x + y + z & = 2 \\ x - y + z & = - 2 & - \\ 2 y & = 4 \\ y & = 2 & . . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{lcc} x + y + z & = 2 \\ x - y - z & = 2 & + \\ 2 x & = 4 \\ x & = 2 & . . . . (5) \end{array} \\ Persamaan (4) & (5) ke (1) \\ \begin{aligned} x + y + z & = 2 \\ (2) + (2) + z & = 2 \\ z & = - 2 \end{aligned} \\ Jadi, x y z = (2) . (2) . (- 2) = - 8 \end{aligned} \end{array}$

$. Cara di atas$ full eliminasi-substitusi

$\begin{array}{ll} 69. & Diketahui sistem persamaan berikut \\ {\begin{array}{c} x + y + z = - 6 \\ x - 2 y + z = 3 \\ - 2 x + y + z = 9 \end{array} \\ Nilai x y z = . . . . \\ \begin{array}{llll} a . & - 30 \\ b . & - 15 \\ c . & 5 \\ d . & 30 \\ e . & 35 \end{array} \\ Jawab : d \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{array}{c} x + y + z = - 6 . . . . (1) \\ x - 2 y + z = 3 . . . . (2) \\ - 2 x + y + z = 9 . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llc} x + y + z & = - 6 \\ x - 2 y + z & = 3 & - \\ 3 y & = - 9 \\ y & = - 3 & . . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{llc} x + y + z & = - 6 \\ - 2 x + y + z & = 9 & - \\ 3 x & = - 15 \\ x & = - 5 & . . . . (5) \end{array} \\ Persamaan (4) & (5) ke (1) \\ \begin{aligned} x + y + z & = 2 \\ (- 5) + (- 3) + z & = - 6 \\ z & = 2 \end{aligned} \\ Jadi, x y z = (- 5) . (- 3) . (2) = 30 \end{aligned} \end{array}$

$\begin{array}{ll} 70. & Diketahui sistem persamaan berikut \\ {\begin{array}{c} x + 2 y + z = 4 \\ 3 x + y + 2 z = - 5 \\ x - 2 y + 2 z = - 6 \end{array} \\ Nilai x y z = . . . . \\ \begin{array}{llll} a . & - 96 \\ b . & - 24 \\ c . & 24 \\ d . & 32 \\ e . & 96 \end{array} \\ Jawab : b \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{array}{c} x + 2 y + z = 4 . . . . . . . (1) \\ 3 x + y + 2 z = - 5 . . . . . . (2) \\ x - 2 y + 2 z = - 6 . . . . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llclll} x + 2 y + z & = 4 & | \times 1 | & x + 2 y + z & = 4 \\ 3 x + y + 2 z & = - 5 & | \times 2 | & 6 x + 2 y + 4 z & = - 10 & - \\ - 5 x - 3 z & = 14 & . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{lll} x + 2 y + z & = 4 \\ x - 2 y + 2 z & = - 6 & + \\ 2 x + 3 z & = - 2 & . . . (5) \end{array} \\ Dari persamaan (4) & (5) maka, \\ \begin{array}{lll} - 5 x - 3 z & = 14 \\ 2 x + 3 z & = - 2 & + \\ - 3 x & = 12 \\ x & = - 4 & . . . . . (6) \\ didapat pula & z & = 2. . . . . . (7) \end{array} \\ Dari persamaan (6) & (7) didapatkan \\ \begin{aligned} x + 2 y + z & = 4 \\ (- 4) + 2 y + 2 & = 4 \\ y & = 3 \end{aligned} \\ Jadi, x y z = (- 4) . (3) . (2) = - 24 \end{aligned} \end{array}$ .

$\begin{array}{ll} 71. & Diketahui sistem persamaan berikut \\ {\begin{array}{c} x + y - z = 1 \\ 2 x - y + 2 z = 9 \\ x + 3 y - z = 7 \end{array} \\ Nilai \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = . . . . \\ \begin{array}{llll} a . & \frac{1}{3} \\ b . & \frac{3}{4} \\ c . & \frac{13}{12} \\ d . & \frac{5}{4} \\ e . & \frac{7}{4} \end{array} \\ Jawab : c \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{array}{c} x + y - z = 1 . . . . (1) \\ 2 x - y + 2 z = 9 . . . . (2) \\ x + 3 y - z = 7 . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llll} x + y - z & = 1 \\ 2 x - y + 2 z & = 9 & + \\ 3 x + z & = 10 & . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{llllll} x + y - z & = 1 & | \times 3 | & 3 x + 3 y - 3 z & = 3 \\ x + 3 y - z & = 7 & | \times 1 | & x + 3 y - z & = 7 & - \\ 2 x - 2 z & = - 4 \\ x - z & = 2 & . . . . (5) \end{array} \\ Dari persamaan (4) & (5) maka, \\ \begin{array}{lll} 3 x + z & = 10 \\ x - z & = - 2 & + \\ 4 x & = 8 \\ x & = 2 & . . . . . (6) \\ didapat pula & z & = 4. . . . . . (7) \end{array} \\ Dari persamaan (1) & (3) didapatkan juga \\ \begin{array}{lll} x + y - z & = 1 \\ x + 3 y - z & = - 7 & - \\ - 2 y & = - 6 \\ y & = 3 & . . . . (8) \end{array} \\ Jadi, \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12} \end{aligned} \end{array}$

$\begin{array}{ll} 72. & Diketahui sistem persamaan berikut \\ {\begin{array}{c} x + y + z = 5 \\ x + y - 4 z = 10 \\ - 2 x + y + z = 0 \end{array} \\ Nilai dari \frac{x z}{y} adalah . . . . \\ \begin{array}{llll} a . & - \frac{6}{13} \\ b . & - \frac{5}{13} \\ c . & - \frac{1}{13} \\ d . & \frac{1}{13} \\ e . & \frac{7}{13} \end{array} \\ Jawab : b \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{array}{c} x + y + z = 5 . . . . . (1) \\ x + y - 4 z = 10 . . . . . (2) \\ - 2 x + y + z = 0 . . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llll} x + y + z & = 5 \\ x + y - 4 z & = 10 & - \\ 5 z & = - 5 \\ z & = - 1 & . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{llllll} x + y + z & = 5 \\ - 2 x + y + z & = 0 & - \\ 3 x & = 5 \\ x & = \frac{5}{3} & . . . . (5) \end{array} \\ Dari persamaan (4) & (5) maka, \\ \begin{aligned} x + y + z & = 5 \\ \frac{5}{3} + y - 1 & = 5 \\ y & = 5 + 1 - \frac{5}{3} = \frac{13}{3} \end{aligned} \\ Jadi, \frac{x z}{y} = \frac{(\frac{5}{3}) . (- 1)}{\frac{13}{3}} = - \frac{5}{13} \end{aligned} \end{array}$ .

$\begin{array}{ll} 73. & Himpunan penyelesaian dari \\ {\begin{cases} \frac{1}{x} + \frac{2}{y} + \frac{3}{z} & = 8 \\ \frac{2}{x} + \frac{2}{y} + \frac{4}{z} & = 10 \\ \frac{2}{x} + \frac{4}{y} + \frac{2}{z} & = 4 \end{cases} \\ adalah {(x, y, z)}, maka x + 3 z = . . . . \\ \begin{array}{llll} a . & 0 \\ b . & \frac{1}{3} \\ c . & 1 \\ d . & 3 \\ e . & 5 \end{array} \\ Jawab : d \\ \begin{aligned} Diketahui sistem persamaan \\ {\begin{cases} \frac{1}{x} + \frac{2}{y} + \frac{3}{z} & = 8 . . . . (1) \\ \frac{2}{x} + \frac{2}{y} + \frac{4}{z} & = 10 . . . . . (2) \\ \frac{2}{x} + \frac{4}{y} + \frac{2}{z} & = 4 . . . . . . . . . . . (3) \end{cases} \\ Saat (1) + (2), maka \\ \begin{array}{llll} \frac{1}{x} + \frac{2}{y} + \frac{3}{z} & = 8 \\ \frac{2}{x} + \frac{2}{y} + \frac{4}{z} & = 10 & - \\ - \frac{1}{x} - \frac{1}{z} & = - 2 \\ \frac{1}{x} + \frac{1}{z} & = 2 & . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{lllllll} \frac{2}{x} + \frac{2}{y} + \frac{4}{z} & = 8 & | \times 2 | & \frac{4}{x} + \frac{4}{y} + \frac{8}{z} & = 16 \\ \frac{2}{x} + \frac{4}{y} + \frac{2}{z} & = 4 & | \times 1 | & \frac{2}{x} + \frac{4}{y} + \frac{2}{z} & = 4 & - \\ \frac{2}{x} + \frac{6}{z} & = 12 \\ \Leftrightarrow & \frac{1}{x} + \frac{3}{z} & = 6 & . . . (5) \end{array} \\ Dari persamaan (4) & (5) maka, \\ \begin{array}{llll} \frac{1}{x} + \frac{3}{z} & = 6 \\ \frac{1}{x} + \frac{1}{z} & = 2 - \\ \frac{2}{z} & = 4 \\ z & = \frac{1}{2} . . . . . . (6) \\ x & = 2 - \frac{1}{2} = \frac{3}{2} \end{array} \\ Jadi, x + 3 z = \frac{3}{2} + 3. \frac{1}{2} = 3 \end{aligned} \end{array}$ .

$\begin{array}{ll} 74. & Diketahui tiga buah bilangan berturut-turut \\ a, b, dan c . Rata-rata dari ke tiga bilangan \\ itu adalah 12. Bilangan kedua sama dengan \\ jumlah bilangan yang lain dikurangi 12 . \\ Jika bilangan ke tiga sama dengan jumlah \\ bilangan yang lain, maka nilai 2 a + b - c = . . . . \\ \begin{array}{llll} a . & - 42 \\ b . & - 36 \\ c . & - 18 \\ d . & - 12 \\ e . & - 6 \end{array} \\ Jawab : e \\ \begin{aligned} Model matematika dari persamaan di atas \\ {\begin{array}{c} a + b + c = 36 . . . . (1) \\ - a + b - x = 12 . . . . (2) \\ a + b - c = 0 . . . . (3) \end{array} \\ Saat (1) + (2), maka \\ \begin{array}{llll} a + b + c & = 36 \\ - a + b - c & = 12 & + \\ 2 b & = 48 \\ b & = 24 & . . . (4) \end{array} \\ Saat (1) + (3), maka \\ \begin{array}{llllll} a + b + c & = 36 \\ a + b - c & = 0 & - \\ 2 c & = 36 \\ c & = 18 & . . . . (5) \end{array} \\ Dari persamaan (4) & (5) maka, \\ \begin{aligned} a + b + c & = 36 \\ a + 24 + 18 & = 36 \\ a & = 36 - 42 \\ = - 6 \end{aligned} \\ Jadi, 2 a + b - c = 2 (- 6) + 24 - 18 = - 6 \end{aligned} \end{array}$

$\begin{array}{ll} 75. & Jumlah uang terdiri atas koin pecahan R p 500, 00 \\ R p 200, 00 d a n R p 100, 00 dengan nilai total \\ R p 100.000, 00. Jika nilai uang pecahan 500-an \\ setengah dari nilai uang pecahan 200-an, tetapi \\ tiga kali uang pecahan 100-an, maka banyak koin \\ adalah . . . . \\ \begin{array}{llll} a . & 460 \\ b . & 440 \\ c . & 420 \\ d . & 380 \\ e . & 350 \end{array} \\ Jawab : a \\ \begin{aligned} Model matematika dari kasus di atas \\ {\begin{array}{c} A (500) + B (200) + C (100) = 100.000 . . . . (1) \\ A (500) = \frac{1}{2} B (200) . . . . (2) \\ A (500) = 3 C (100) . . . . (3) \end{array} \\ Dari persamaan (2) didapatkan \\ 2 A (500) = B (200) \\ Dari persamaan (3) akan didapatkan \\ \frac{1}{3} A (500) = C (100) \\ Dari persamaan (1) maka, \\ A (500) + B (200) + C (100) = 100.000 \\ A (500) + 2 A (500) + \frac{1}{3} A (500) = 100.000 \\ \frac{10}{3} A (500) = 100.000 \Leftrightarrow A (500) = 30.000 \\ maka akan didapatkan \\ B (200) = 2 (30.000) = 60.000 \\ C (100) = \frac{1}{3} (30.000) = 10.000 \\ {\begin{cases} A (500) & = 30.000 \Rightarrow A = \frac{30.000}{500} = 60 \\ B (200) & = 60.000 \Rightarrow B = \frac{60.000}{200} = 300 \\ C (100) & = 10.000 \Rightarrow C = \frac{10.000}{100} = 100 \end{cases} \\ Jadi, A + B + C = 60 + 300 + 100 = 460 \end{aligned} \end{array}$ .

PAS MATEMATIKA BLOG

Latihan Soal 8 Persiapan PAS Gasal Matematika Wajib Kelas X

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