{"id":279692,"date":"2020-02-24T22:15:41","date_gmt":"2020-02-24T15:15:41","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279692"},"modified":"2022-03-01T18:03:34","modified_gmt":"2022-03-01T11:03:34","slug":"utbk-2020-matematika-saintek","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-matematika-saintek\/","title":{"rendered":"Prediksi Soal UTBK Matematika Saintek 2020"},"content":{"rendered":"

\"\"<\/p>\n

SBMPTN sebentar lagi, soal-soal harus segera dikuasai. Makanya, kamu harus <\/span>gercep<\/span><\/em> alias gerak cepat. Nah, Bagi kamu yang saat ini masih bingung mencari soal-soal SBMPTN, tampaknya kamu berada di <\/span>website<\/span><\/em> yang tepat.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas Prediksi UTBK SBMPTN Matematika SAINTEK 2020. Kabar baiknya, latihan soal yang disediakan Quipper Blog, lengkap dengan jawaban beserta pembahasannya, <\/span>lho<\/span><\/em>. Semakin penasaran? <\/span>Kuy<\/span><\/em>, segera kerjakan soalnya!<\/span><\/p>\n

Latihan Soal 1<\/strong><\/h2>\n

Jika\u00a0<\/span>a <\/span><\/em>log2 + <\/span>b<\/span><\/em> log3 + <\/span>c<\/span><\/em> log4 + <\/span>d<\/span><\/em> log5 + <\/span>e<\/span><\/em> log6 + <\/span>f<\/span><\/em> log7 + <\/span>g<\/span><\/em> log8 = 2.020,\u00a0maka\u00a0<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> + 2<\/span>c<\/span><\/em> + <\/span>d<\/span><\/em> + 2<\/span>e<\/span><\/em> + <\/span>f<\/span><\/em> + 3<\/span>g<\/span><\/em>\u00a0adalah…<\/span><\/p>\n

    \n
  1. 2.010<\/span><\/li>\n
  2. 3.840<\/span><\/li>\n
  3. 2.200<\/span><\/li>\n
  4. 4.040<\/span><\/li>\n
  5. 3.440<\/span><\/li>\n<\/ol>\n

    Jawaban: D<\/strong><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    Diketahui: <\/span>a <\/span><\/em>log2 + <\/span>b<\/span><\/em> log3 + <\/span>c<\/span><\/em> log4 + <\/span>d<\/span><\/em> log5 + <\/span>e<\/span><\/em> log6 + <\/span>f<\/span><\/em> log7 + <\/span>g<\/span><\/em> log8 = 2.020.<\/span><\/p>\n

    Ditanya: <\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> + 2<\/span>c<\/span><\/em> + <\/span>d<\/span><\/em> + 2<\/span>e<\/span><\/em> + <\/span>f<\/span><\/em> + 3<\/span>g<\/span><\/em>\u00a0=\u2026?<\/span><\/p>\n

    Penyelesaian:<\/span><\/p>\n

    Dengan menggunakan sifat-sifat logaritma dan sifat bilangan berpangkat, maka:<\/span><\/p>\n

    \"\"<\/p>\n

    Ingat bahwa, <\/span>x<\/span><\/em> = <\/span>y<\/span><\/em> jika <\/span>a<\/span><\/em>x<\/span><\/em><\/sup> = <\/span>a<\/span><\/em>y<\/span><\/em><\/sup> dan <\/span>a<\/span><\/em>x<\/span><\/em><\/sup> = 1 jika <\/span>x<\/span><\/em> = 0, sehingga diperoleh:<\/span><\/p>\n

    \"\"<\/p>\n

    Tentukan nilai <\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> + 2<\/span>c<\/span><\/em> + <\/span>d<\/span><\/em> + 2<\/span>e<\/span><\/em> + <\/span>f<\/span><\/em> + 3<\/span>g<\/span><\/em>.<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, nilai <\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> + 2<\/span>c<\/span><\/em> + <\/span>d<\/span><\/em> + 2<\/span>e<\/span><\/em> + <\/span>f<\/span><\/em> + 3<\/span>g<\/span><\/em> = 4.040.<\/span><\/p>\n

    Latihan Soal 2<\/strong><\/h2>\n

    Jumlah <\/span>m<\/span><\/em> suku pertama dari suatu barisan aritmetika adalah (<\/span>n<\/span><\/em> + 2). Jika jumlah (<\/span>n <\/span><\/em>+ 2) suku pertama dari barisan aritmetika adalah <\/span>m<\/span><\/em>, maka jumlah (<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> + 2) barisan tersebut adalah\u2026<\/span><\/p>\n

      \n
    1. \u2013<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> \u2013 2<\/span><\/li>\n
    2. m<\/span><\/em> + <\/span>n<\/span><\/em> \u2013 1<\/span><\/li>\n
    3. \u2013m<\/span><\/em> \u2013<\/span> n <\/span><\/em>\u2013 2<\/span><\/li>\n
    4. m<\/span><\/em> + <\/span>n<\/span><\/em><\/li>\n
    5. \u2013m <\/span><\/em>+ <\/span>n<\/span><\/em> \u2013 1<\/span><\/li>\n<\/ol>\n

      Jawaban: C<\/strong><\/p>\n

      Pembahasan:<\/strong><\/p>\n

      Diketahui:<\/span><\/p>\n

        \n
      • Jumlah\u00a0<\/span>m<\/span><\/em>\u00a0barisan aritmetika adalah\u00a0<\/span>n<\/span><\/em> + 2.<\/span><\/li>\n
      • Jumlah\u00a0<\/span>n<\/span><\/em> + 2\u00a0barisan aritmetika adalah\u00a0<\/span>m<\/span><\/em>.<\/span><\/li>\n<\/ul>\n

        Ditanya: Jumlah\u00a0(<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> + 2)\u00a0barisan aritmetika =\u2026?<\/span><\/p>\n

        Penyelesaian:<\/span><\/p>\n

        Oleh karena jumlah\u00a0<\/span>m<\/span><\/em>\u00a0barisan aritmetika adalah\u00a0<\/span>n<\/span><\/em> + 2,\u00a0maka:<\/span><\/p>\n

        \"\"<\/p>\n

        Oleh karena <\/span>n<\/span><\/em> + 2 barisan aritmetika adalah <\/span>m<\/span><\/em>, maka:<\/span><\/p>\n

        \"\"<\/p>\n

        Dari pengurangan persamaan (1) dan (2), didapat:<\/span><\/p>\n

        \"\"<\/p>\n

        Ini berarti jumlah (<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> + 2) barisan tersebut dinyatakan:<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, jumlah (<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> + 2 ) barisan aritmetika adalah \u2013(<\/span>m<\/span><\/em> + <\/span>n<\/span><\/em> + 2) = \u2013<\/span>m<\/span><\/em> \u2013 <\/span>n<\/span><\/em> \u2013 2.<\/span><\/p>\n

        Latihan Soal 3<\/strong><\/h2>\n

        \"\"<\/p>\n

        Jawaban: C<\/strong><\/p>\n

        Pembahasan:<\/strong><\/p>\n

        \"\"<\/p>\n

        Penyelesaian:<\/span><\/p>\n

        Dengan menggunakan sifat limit suatu fungsi, maka:<\/span><\/p>\n

        \"\"<\/p>\n

        Dengan mengeliminasi persamaan (1) dan (2) didapat:<\/span><\/p>\n

        \"\"<\/p>\n

        Ini berarti:<\/span><\/p>\n

        \"\"<\/p>\n

        \"\"<\/p>\n

        Latihan Soal 4<\/strong><\/h2>\n

        Daerah yang dibatasi oleh garis\u00a0<\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>\u00a0dan parabola\u00a0<\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>2<\/span><\/sup> \u2013 3<\/span>x<\/span><\/em>\u00a0pada interval\u00a0[0, <\/span>m<\/span><\/em>]\u00a0dengan\u00a0<\/span>m<\/span><\/em> > 4,\u00a0terdiri atas dua bagian yang mempunyai luas yang sama, maka nilai\u00a0<\/span>m\u00a0<\/span><\/em>adalah…<\/span><\/p>\n

          \n
        1. 9<\/span><\/li>\n
        2. 5<\/span><\/li>\n
        3. 7<\/span><\/li>\n
        4. 8<\/span><\/li>\n
        5. 6<\/span><\/li>\n<\/ol>\n

          Jawaban: E<\/strong><\/p>\n

          Pembahasan:<\/strong><\/p>\n

          Diketahui: Daerah yang dibatasi oleh garis\u00a0<\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>\u00a0dan parabola\u00a0<\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>2<\/span><\/sup> \u2013 3<\/span>x<\/span><\/em>\u00a0pada interval\u00a0[0, <\/span>m<\/span><\/em>]\u00a0dengan\u00a0<\/span>m<\/span><\/em> > 4,\u00a0terdiri atas dua bagian yang mempunyai luas yang sama<\/span><\/p>\n

          Ditanya: <\/span>m<\/span><\/em> =\u2026?<\/span><\/p>\n

          Penyelesaian:<\/span><\/p>\n

          Mula-mula. Gambarkan grafik garis <\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>\u00a0dan parabola\u00a0<\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>2<\/span><\/sup> \u2013 3<\/span>x<\/span><\/em>.<\/span><\/p>\n

          Untuk parabola <\/span>y<\/span><\/em> = <\/span>x<\/span><\/em>2<\/span><\/sup> \u2013 3<\/span>x<\/span><\/em>:<\/span><\/p>\n

          \"\"<\/p>\n

          Dari persamaan di atas, diperoleh:<\/span><\/p>\n

            \n
          • titik potong dengan sumbu X adalah (0, 0) dan (3,0);<\/span><\/li>\n
          • titik puncaknya adalah sebagai berikut.<\/span><\/li>\n<\/ul>\n

            \"\"<\/p>\n

            Sketsa grafiknya adalah sebagai berikut.<\/span><\/p>\n

            \"\"<\/p>\n

            Berdasarkan informasi pada soal, diketahui bahwa:<\/span><\/p>\n

            \"\"<\/p>\n

            Ini berarti:<\/span><\/p>\n

            \"\"<\/p>\n

            \"\"<\/p>\n

            Jadi, nilai <\/span>m<\/span><\/em> adalah 6.<\/span><\/p>\n

            Latihan Soal 5<\/strong><\/h2>\n

            \"\"<\/p>\n

              \n
            1. 55o<\/sup><\/span><\/li>\n
            2. 90o<\/sup><\/span><\/li>\n
            3. 115o<\/sup><\/span><\/li>\n
            4. 100o<\/sup><\/span><\/li>\n
            5. 45o<\/sup><\/span><\/li>\n<\/ol>\n

              Jawaban: B<\/strong><\/p>\n

              Pembahasan:<\/strong><\/p>\n

              \"\"<\/p>\n

              Ditanya: Sudut A =\u2026?<\/span><\/p>\n

              Penyelesaian:<\/span><\/p>\n

              Sketsa segitiganya adalah sebagai berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Berdasarkan aturan trigonometri dalam segitiga, maka:<\/span><\/p>\n

              \"\"<\/p>\n

              Jadi, besar sudut A adalah 90<\/span>o<\/span><\/sup>.<\/span><\/p>\n

              Latihan Soal 6<\/strong><\/h2>\n

              \"\"<\/p>\n

                \n
              1. 5<\/span><\/li>\n
              2. -5<\/span><\/li>\n
              3. 7<\/span><\/li>\n
              4. -7<\/span><\/li>\n
              5. 3<\/span><\/li>\n<\/ol>\n

                Jawaban: D<\/strong><\/p>\n

                Pembahasan:<\/strong><\/p>\n

                Diketahui:<\/span><\/p>\n

                \"\"<\/p>\n

                Ditanya: Nilai <\/span>y<\/span><\/em> \u2013 <\/span>z<\/span><\/em> =\u2026?<\/span><\/p>\n

                Penyelesaian:<\/span><\/p>\n

                \"\"<\/p>\n

                \"\"<\/p>\n

                Ini berarti:<\/span><\/p>\n

                \"\"<\/p>\n

                Lakukan eliminasi pada persamaan (1) dsn (2).<\/span><\/p>\n

                \"\"<\/p>\n

                Jadi, nilai <\/span>y<\/span><\/em> \u2013 <\/span>z<\/span><\/em>\u00a0 = -7.<\/span><\/p>\n

                Latihan Soal 7<\/strong><\/h2>\n

                \"\"<\/p>\n

                Jawaban: A<\/strong><\/p>\n

                Pembahasan:<\/strong><\/p>\n

                \"\"<\/p>\n

                Ditanya: Persamaan kurva =\u2026?<\/span><\/p>\n

                Penyelesaian:<\/span><\/p>\n

                \"\"<\/p>\n

                Ini berarti:<\/span><\/p>\n

                \"\"<\/p>\n

                Kurva semula:<\/span><\/p>\n

                \"\"<\/p>\n

                Jadi, kurva semulanya adalah <\/span>y<\/span><\/em> = 2<\/span>x<\/span><\/em>2<\/span><\/sup> \u2013 <\/span>x<\/span><\/em> \u2013 1.<\/span><\/p>\n

                Latihan Soal 8<\/strong><\/h2>\n

                Lima orang karyawan mempunyai pendapatan sebagai berikut.<\/span><\/p>\n

                  \n
                • Pendapatan karyawan pertama satu setengah kali karyawan kelima.<\/span><\/li>\n
                • Pendapatan karyawan kedua lebih\u00a0Rp500.000\u00a0dari karyawan pertama.<\/span><\/li>\n
                • Pendapatan karyawan ketiga lebih\u00a0Rp650.000\u00a0dari karyawan pertama.<\/span><\/li>\n
                • Pendapatan karyawan keempat kurang\u00a0Rp450.000\u00a0dari karyawan pertama.<\/span><\/li>\n<\/ul>\n

                  Bila rata-rata kelima karyawan tersebut\u00a0Rp10.500.000,\u00a0maka pendapatan karyawan kelima adalah…<\/span><\/p>\n

                    \n
                  1. Rp7.400.000<\/span><\/li>\n
                  2. Rp8.500.000<\/span><\/li>\n
                  3. Rp6.500.000<\/span><\/li>\n
                  4. Rp8.000.000<\/span><\/li>\n
                  5. Rp6.000.000<\/span><\/li>\n<\/ol>\n

                    Jawaban: A<\/strong><\/p>\n

                    Pembahasan:<\/strong><\/p>\n

                    Diketahui:<\/span><\/p>\n

                      \n
                    • Pendapatan karyawan pertama satu setengah kali karyawan kelima.<\/span><\/li>\n
                    • Pendapatan karyawan kedua lebih\u00a0Rp500.000\u00a0dari karyawan pertama.<\/span><\/li>\n
                    • Pendapatan karyawan ketiga lebih\u00a0Rp650.000\u00a0dari karyawan pertama.<\/span><\/li>\n
                    • Pendapatan karyawan keempat kurang\u00a0Rp450.000\u00a0dari karyawan pertama.<\/span><\/li>\n
                    • Pendapatan rata-rata kelima karyawan tersebut\u00a0Rp10.500.000.<\/span><\/li>\n<\/ul>\n

                      Ditanya: Pendapatan karyawan kelima =\u2026?<\/span><\/p>\n

                      Penyelesaian:<\/span><\/p>\n

                      Misalkan: <\/span>kn<\/span><\/em> = pendapatan karyawan ke-<\/span>n<\/span><\/em> dengan <\/span>n<\/span><\/em> = 1, 2, 3, 4, 5.<\/span><\/p>\n

                      Ini berarti:<\/span><\/p>\n

                      \"\"<\/p>\n

                      Oleh karena rata-rata pendapatan kelima karyawan\u00a0Rp10.500.000,\u00a0maka:<\/span><\/p>\n

                      \"\"<\/p>\n

                      Jadi, pendapatan karyawan kelima adalah Rp7.400.000.<\/span><\/p>\n

                      Latihan Soal 9<\/strong><\/h2>\n

                      Dari\u00a012\u00a0orang karyawan yang terdiri atas\u00a04\u00a0pria dan\u00a08\u00a0wanita, akan dibentuk tim kerja yang beranggotakan\u00a06\u00a0orang. Tim yang akan dibentuk harus terdiri atas paling sedikit\u00a04\u00a0orang wanita. Banyaknya cara membentuk tim kerja tersebut adalah…<\/span><\/p>\n

                        \n
                      1. 990<\/span><\/li>\n
                      2. 672<\/span><\/li>\n
                      3. 436<\/span><\/li>\n
                      4. 774<\/span><\/li>\n
                      5. 821<\/span><\/li>\n<\/ol>\n

                        Jawaban: B<\/strong><\/p>\n

                        Pembahasan:<\/strong><\/p>\n

                        Diketahui:<\/span><\/p>\n

                          \n
                        • 12\u00a0orang karyawan yang terdiri atas\u00a04\u00a0pria dan\u00a08\u00a0wanita.<\/span><\/li>\n
                        • Akan dibentuk tim kerja yang beranggotakan\u00a06\u00a0orang.<\/span><\/li>\n
                        • Tim harus terdiri atas paling sedikit\u00a04\u00a0orang.<\/span><\/li>\n<\/ul>\n

                          Ditanya: Banyaknya cara membentuk tim kerja =\u2026?<\/span><\/p>\n

                          Pembahasan:<\/span>
                          \n<\/span>Tim kerja dapat beranggotakan\u00a04\u00a0wanita dan\u00a02\u00a0pria atau\u00a05\u00a0wanita dan\u00a01\u00a0pria atau\u00a06\u00a0wanita. Ini berarti:<\/span><\/p>\n

                          \"\"<\/p>\n

                          Jadi, banyaknya cara membentuk tim kerja adalah 672.<\/span><\/p>\n

                          Latihan Soal 10<\/strong><\/h2>\n

                          Dilakukan survei terhadap\u00a02.000\u00a0orang lulusan SMA dan diperoleh data\u00a0900\u00a0orang mengikuti SNMPTN IPA,\u00a01.300\u00a0siswa mengikuti SNMPTN IPS, dan\u00a0350\u00a0siswa mengambil kedua tes tersebut. Berdasarkan data tersebut dapat disimpulkan bahwa persentase peluang siswa lulusan SMA yang tidak mengikuti tes SNMPTN adalah…<\/span><\/p>\n

                            \n
                          1. 7,5%<\/span><\/li>\n
                          2. 10%<\/span><\/li>\n
                          3. 15%<\/span><\/li>\n
                          4. 12,5%<\/span><\/li>\n
                          5. 17,5%<\/span><\/li>\n<\/ol>\n

                            Jawaban: A<\/strong><\/p>\n

                            Pembahasan:<\/strong><\/p>\n

                            Diketahui:<\/span><\/p>\n

                              \n
                            • 900\u00a0orang mengikuti SNMPTN IPA.<\/span><\/li>\n
                            • 1.300\u00a0siswa mengikuti SNMPTN IPS.<\/span><\/li>\n
                            • 350\u00a0siswa mengambil kedua tes tersebut.<\/span><\/li>\n
                            • Survei dilakukan terhadap\u00a02.000\u00a0lulusan SMA.<\/span><\/li>\n<\/ul>\n

                              Ditanya: Peluang siswa lulusan SMA yang tidak mengikuti tes SNMPTN =\u2026?<\/span><\/p>\n

                              Pembahasan:<\/span>
                              \n<\/span>Misalkan:<\/span>
                              \n<\/span>n<\/span><\/em>(<\/span>S<\/span><\/em>)=\u00a0banyaknya orang yang disurvei<\/span>
                              \n<\/span>n<\/span><\/em>(<\/span>A<\/span><\/em>)=\u00a0banyaknya orang yang mengikuti SNMPTN IPA<\/span>
                              \n<\/span>n<\/span><\/em>(<\/span>B<\/span><\/em>)=\u00a0banyaknya orang yang mengikuti SNMPTN IPS<\/span><\/p>\n

                              Ini berarti:<\/span><\/p>\n

                              \"\"<\/p>\n

                              Persentase peluang siswa lulusan SMA yang tidak mengikuti SNMPTN adalah sebagai berikut.<\/span><\/p>\n

                              \"\"<\/p>\n

                              Jadi, persentase peluang siswa SMA yang tidak mengikuti SNMPTN adalah\u00a07,5%.<\/span><\/p>\n

                              Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan Prediksi UTBK SBMPTN Matematika SAINTEK 2020 di atas? Agar pemahamanmu semakin terasah, sering-seringlah mengerjakan latihan soal. Ingat, selangkah lagi kamu akan memasuki PTN impian.\u00a0<\/span><\/p>\n

                              Untuk mendukung kesuksesanmu masuk PTN, Quipper Video hadir dengan Paket Intensif UTBK SBMPTN 2020. Semua kebutuhan sudah tersedia di dalamnya, mulai dari materi persiapan, materi pemantapan, latihan soal, bank soal, hingga pembahasan lengkapnya. Jadi, tunggu apalagi. <\/span>Kuy<\/span><\/em>, gabung Quipper Video sekarang!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

                              SBMPTN sebentar lagi, soal-soal harus segera dikuasai. Makanya, kamu harus gercep alias gerak cepat. Nah, Bagi kamu yang saat ini masih bingung mencari soal-soal…<\/p>\n","protected":false},"author":99400369,"featured_media":279744,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[679385273,679385209],"tags":[],"ppma_author":[679386827],"class_list":["post-279692","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn"],"acf":[],"yoast_head":"\nPrediksi Soal UTBK Matematika Saintek 2020 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Buat kamu yang sedang kebingungan mencari soal UTBK 2020 matematika saintek, langsung saja nih cek di sini. Ada jawaban dan pembahasan lengkapnya juga!\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.quipper.com\/id\/blog\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-matematika-saintek\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prediksi Soal UTBK Matematika Saintek 2020 - Quipper Blog\" \/>\n<meta property=\"og:description\" content=\"Buat kamu yang sedang kebingungan mencari soal UTBK 2020 matematika saintek, langsung saja nih cek di sini. 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