{"id":279860,"date":"2020-02-24T22:04:48","date_gmt":"2020-02-24T15:04:48","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279860"},"modified":"2022-03-01T17:16:03","modified_gmt":"2022-03-01T10:16:03","slug":"utbk-2020-matematika-soshum","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-matematika-soshum\/","title":{"rendered":"Prediksi Soal UTBK Matematika Soshum 2020"},"content":{"rendered":"

\"\"<\/p>\n

SBMPTN sebentar lagi, soal-soal harus segera dikuasai. Itulah sebabnya, Quipperian 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 Soshum 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 4<\/span>m<\/span><\/em><\/sup> + 4<\/span>m<\/span><\/em> + 1<\/span><\/sup> = 15<\/span>, maka <\/span>8<\/span>m<\/span><\/em><\/sup> =\u2026<\/span><\/p>\n

    \n
  1. 2\u221a<\/span>2<\/span><\/li>\n
  2. 3\u221a<\/span>2<\/span><\/li>\n
  3. 3\u221a<\/span>3<\/span><\/li>\n
  4. 2\u221a<\/span>3<\/span><\/li>\n
  5. \u221a3<\/span><\/li>\n<\/ol>\n

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

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

    Diketahui: 4m<\/em><\/sup> + 4m<\/em> + 1<\/sup> = 15<\/span><\/p>\n

    Ditanya: <\/span>8<\/span>m<\/span><\/em><\/sup> =\u2026?<\/span><\/p>\n

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

    Mencari nilai <\/span>m<\/span><\/em>.<\/span><\/p>\n

    \"\"<\/p>\n

    Mencari nilai 8<\/span>m<\/span><\/em><\/sup>.<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, nilai 8<\/span>m<\/span><\/em><\/sup> = 3\u221a<\/span>3<\/span>.<\/span><\/p>\n

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

    Jika\u00a0<\/span>f<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em> \u2212 2<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em>\u00a0dan\u00a0<\/span>g<\/span><\/em>(<\/span>x<\/span><\/em>) = 4<\/span>bx<\/span><\/em> \u2212 1\u00a0serta\u00a08<\/span>f<\/span><\/em>(0) = 3<\/span>g<\/span><\/em>(1),\u00a0maka\u00a04<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> = …<\/span><\/p>\n

    \"\"<\/p>\n

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

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

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

    \"\"<\/p>\n

    Ditanya: 4<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> =\u2026?<\/span><\/p>\n

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

    Melihat hubungan antarfungsi, yaitu 8<\/span>f<\/span><\/em>(0) = 3<\/span>g<\/span><\/em>(1), diperoleh:<\/span><\/p>\n

    \"\"<\/p>\n

    Ini berarti, nilai 4<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em> adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

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

    Sistem persamaan\u00a02<\/span>x<\/span><\/em> + <\/span>y<\/span><\/em> = <\/span>a<\/span><\/em>, 3<\/span>x<\/span><\/em> + 2<\/span>y<\/span><\/em> = <\/span>b<\/span><\/em>,\u00a0dan\u00a07<\/span>x<\/span><\/em> + 4<\/span>y<\/span><\/em> = <\/span>c<\/span><\/em>\u00a0memiliki solusi untuk\u00a0<\/span>c<\/span><\/em> = …<\/span><\/p>\n

      \n
    1. a<\/span><\/em> + 3<\/span>b<\/span><\/em><\/li>\n
    2. a \u2013 <\/span><\/em>2<\/span>b<\/span><\/em><\/li>\n
    3. –<\/span><\/em>2a + <\/span>b<\/span><\/em><\/li>\n
    4. 2<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em><\/li>\n
    5. a <\/span><\/em>+ 2<\/span>b<\/span><\/em><\/li>\n<\/ol>\n

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

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

      Diketahui: Sistem persamaan\u00a02<\/span>x<\/span><\/em> + <\/span>y<\/span><\/em> = <\/span>a<\/span><\/em>, 3<\/span>x<\/span><\/em> + 2<\/span>y<\/span><\/em> = <\/span>b<\/span><\/em>,\u00a0dan\u00a07<\/span>x<\/span><\/em> + 4<\/span>y<\/span><\/em> = <\/span>c<\/span><\/em>.<\/span><\/p>\n

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

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

      \"\"<\/p>\n

      Cara 1<\/strong><\/p>\n

      Mencari nilai <\/span>x<\/span><\/em> dan <\/span>y<\/span><\/em> dalam <\/span>a<\/span><\/em> dan <\/span>b<\/span><\/em>.<\/span><\/p>\n

      \"\"<\/p>\n

      Mencari nilai <\/span>c<\/span><\/em>.<\/span><\/p>\n

      \"\"<\/p>\n

      Cara 2<\/strong><\/p>\n

      Perhatikan bentuk persamaan 1 dan persamaan 2. Jika persamaan 1 dikalikan 2 dan ditambah dengan persamaan 2, maka hasilnya:<\/span><\/p>\n

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      Jadi, nilai <\/span>c<\/span><\/em> = 2<\/span>a<\/span><\/em> + <\/span>b<\/span><\/em>.<\/span><\/p>\n

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

      Luas daerah penyelesaian sistem pertidaksamaan <\/span>x<\/span><\/em> + <\/span>y<\/span><\/em> \u2264<\/span> 6, 2<\/span>x<\/span><\/em> + 3<\/span>y<\/span><\/em> \u2265<\/span> 12, dan <\/span>x<\/span><\/em> \u2265<\/span> 0 adalah \u2026 satuan luas.<\/span><\/p>\n

        \n
      1. 6<\/span><\/li>\n
      2. 12<\/span><\/li>\n
      3. 10<\/span><\/li>\n
      4. 8<\/span><\/li>\n
      5. 14<\/span><\/li>\n<\/ol>\n

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

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

        Diketahui: Sistem pertidaksamaan <\/span>x<\/span><\/em> + <\/span>y<\/span><\/em> \u2264<\/span> 6, 2<\/span>x<\/span><\/em> + 3<\/span>y<\/span><\/em> \u2265<\/span> 12, dan <\/span>x<\/span><\/em> \u2265<\/span> 0.<\/span><\/p>\n

        Ditanya: Luas daerah penyelesaian =\u2026?<\/span><\/p>\n

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

        Mula-mula, gambar daerah penyelesaian dari sistem pertidaksamaan berikut.<\/span><\/p>\n

        1. x<\/span> + <\/span>y<\/span><\/em> \u2264 6<\/span><\/p>\n

        Titik-titik yang dilalui di antaranya (0,6) dan (6,0).<\/span><\/p>\n

        2. 2<\/span>x<\/span><\/em> + 3<\/span>y<\/span><\/em> \u2265<\/span> 12<\/span><\/p>\n

        Titik-titik yang dilalui di antaranya (0,4) dan (6,0)<\/span><\/p>\n

        3.<\/span> x<\/span><\/em> \u2265<\/span> 0\u00a0 (sumbu Y)<\/span><\/p>\n

        Ini berarti, grafiknya adalah sebagai berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        Berdasarkan grafik tersebut, luas daerah penyelesaiannya:<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, luas daerah penyelesaian pertidaksamaannya adalah 6 satuan luas.<\/span><\/p>\n

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

        \"\"<\/p>\n

          \n
        1. 42<\/span><\/li>\n
        2. 36<\/span><\/li>\n
        3. 56<\/span><\/li>\n
        4. 48<\/span><\/li>\n
        5. 32<\/span><\/li>\n<\/ol>\n

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

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

          \"\"<\/p>\n

          Ditanya: Nilai <\/span>a<\/span><\/em>2<\/span><\/sup>b<\/span><\/em> = \u2026?<\/span><\/p>\n

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

          \"\"<\/p>\n

          Ini berarti, perkalian <\/span>A<\/span><\/em>T<\/span><\/em><\/sup> dan <\/span>B<\/span><\/em>T<\/span><\/em><\/sup> adalah sebagai berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Eliminasi kedua persamaan untuk memperoleh nilai <\/span>a<\/span><\/em> dan <\/span>b<\/span><\/em>.<\/span><\/p>\n

          \"\"<\/p>\n

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

          \"\"<\/p>\n

          Jadi, nilai <\/span>a<\/span><\/em>2<\/span><\/sup>b<\/span><\/em> = 3<\/span>2<\/span><\/sup> . 4 = 9 . 4 = 36.<\/span><\/p>\n

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

          Jika perbandingan suku pertama dengan suku keempat suatu barisan aritmetika adalah\u00a02:5,\u00a0maka perbandingan suku kedua dan kelima barisan tersebut adalah…<\/span><\/p>\n

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

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

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

            Diketahui: Barisan aritmetika dengan <\/span>u<\/span><\/em>1<\/span><\/em><\/sub> : <\/span>u<\/span><\/em>4<\/span><\/em><\/sub> = 2 : 5.<\/span><\/p>\n

            Ditanya: <\/span>u<\/span><\/em>2<\/span><\/em><\/sub> : <\/span>u<\/span><\/em>5<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

            Mula-mula, carilah hubungan antara <\/span>a<\/span><\/em> (suku awal) dan <\/span>b<\/span><\/em> (beda suku).<\/span><\/p>\n

            \"\"<\/p>\n

            Oleh karenanya, perbandingan suku kedua dengan kelimanya adalah sebagai berikut.<\/span><\/p>\n

            \"\"<\/p>\n

            Jadi, perbandingan suku kedua dan kelima barisan tersebut adalah 4 : 7.<\/span><\/p>\n

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

            \"\"<\/p>\n

              \n
            1. 16<\/span><\/li>\n
            2. 14<\/span><\/li>\n
            3. 12<\/span><\/li>\n
            4. 10<\/span><\/li>\n
            5. 8<\/span><\/li>\n<\/ol>\n

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

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

              Diketahui: Bentuk integral.<\/span><\/p>\n

              \"\"<\/p>\n

              Pengintegralan (untuk memudahkan, kita hilangkan terlebih dahulu batasnya).<\/span><\/p>\n

              Diperoleh hasil berikut.<\/span><\/p>\n

              \"\"<\/p>\n

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

              Diketahui persegi panjang ABCD dengan\u00a0AB = 4\u221a<\/span>3<\/span>\u00a0cm dan\u00a0BC = 4\u00a0cm. Jika E adalah titik potong AC dan BD, maka besar\u00a0<\/span>\u2220<\/span>BAE\u00a0adalah…<\/span><\/p>\n

                \n
              1. 90<\/span>o<\/span><\/sup><\/li>\n
              2. 75<\/span>o<\/span><\/sup><\/li>\n
              3. 60<\/span>o<\/span><\/sup><\/li>\n
              4. 45<\/span>o<\/span><\/sup><\/li>\n
              5. 30<\/span>o<\/span><\/sup><\/li>\n<\/ol>\n

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

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

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

                \"\"<\/p>\n

                Ditanya: Besar <\/span>\u2220<\/span>BAE\u00a0=\u2026?<\/span><\/p>\n

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

                Perhatikan gambar berikut.<\/span><\/p>\n

                \"\"<\/p>\n

                Oleh karena segitiga ABC adalah segitiga siku-siku, maka dapat langsung menggunakan perbandingan trigonometri berikut.<\/span><\/p>\n

                \"\"<\/p>\n

                Jadi, besar <\/span>\u2220<\/span>BAE\u00a0= 30<\/span>o<\/span><\/sup>.<\/span><\/p>\n

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

                Nilai rata-rata ulangan harian matematika suatu kelas yang terdiri atas\u00a032\u00a0siswa adalah\u00a072\u00a0dengan jangkauan\u00a080.\u00a0Jika nilai siswa terendah dan tertinggi tidak diikutsertakan, maka nilai rata-ratanya menjadi\u00a072,8.\u00a0Nilai tertinggi tes tersebut adalah…<\/span><\/p>\n

                  \n
                1. 97<\/span><\/li>\n
                2. 96<\/span><\/li>\n
                3. 98<\/span><\/li>\n
                4. 95<\/span><\/li>\n
                5. 100<\/span><\/li>\n<\/ol>\n

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

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

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

                  \"\"<\/p>\n

                  Ditanya: Nilai tertinggi tes tersebut =\u2026?<\/span><\/p>\n

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

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

                  x<\/span><\/em> = nilai tertinggi<\/span><\/p>\n

                  y<\/span><\/em> = nilai tertinggi<\/span><\/p>\n

                  n<\/span><\/em> = total nilai untuk 30 anak lainnya<\/span><\/p>\n

                  Ini berarti, rata-rata kelas dengan 32 siswa dapat ditulis dengan:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jangkauan kelas:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jika\u00a0<\/span>x\u00a0<\/span><\/em>dan\u00a0<\/span>y<\/span><\/em>\u00a0tidak diikutsertakan, maka rata-rata kelas dapat ditulis dengan:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Substitusikan nilai <\/span>n<\/span><\/em> ke dalam persamaan (1), maka:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Eliminasi persamaan (2) dengan (3) untuk mencari nilai\u00a0<\/span>x<\/span><\/em>,\u00a0sehingga diperoleh:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jadi, nilai tertinggi di kelas tersebut adalah 100.<\/span><\/p>\n

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

                  Sebuah toko ingin membuat nomor undian yang disusun dari angka\u00a01,2,4,4,7.\u00a0Jika kupon tersebut disusun berdasarkan kodenya mulai dari yang terkecil sampai yang terbesar, maka banyaknya kupon dengan kode yang lebih besar dari\u00a024.000\u00a0adalah…<\/span><\/p>\n

                    \n
                  1. 22 kupon<\/span><\/li>\n
                  2. 32 kupon<\/span><\/li>\n
                  3. 56 kupon<\/span><\/li>\n
                  4. 67<\/span> kupon<\/span><\/li>\n
                  5. 45 kupon<\/span><\/li>\n<\/ol>\n

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

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

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

                      \n
                    • Pilihan angka 1, 2, 4, 4, dan 7.<\/span><\/li>\n
                    • Kupon disusun dari yang terkecil sampai terbesar.<\/span><\/li>\n<\/ul>\n

                      Ditanya: Banyak kupon dengan kode lebih besar dari 24.000 =\u2026?<\/span><\/p>\n

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

                      X adalah banyaknya angka yang tersedia.<\/span><\/p>\n

                      Ada empat kemungkinan, yaitu sebagai berikut.<\/span><\/p>\n

                      \"\"<\/p>\n

                      Jadi, banyak kupon dengan kode lebih besar dari 24.000 sebanyak 6 + 3 + 24 + 12 = 45 kupon.<\/span><\/p>\n

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

                      Lima orang akan berfoto bersama-sama, yaitu Ade, Bimo, Cici, Dedy, dan Eko. Peluang Ade dan Bimo duduk berdampingan adalah…<\/span><\/p>\n

                      \"\"<\/p>\n

                      Jawaban: E<\/strong> \u00a0<\/strong><\/p>\n

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

                      Diketahui: <\/span>Ada lima orang (Ade, Bimo, Cici, Dedy, dan Eko) ingin berfoto bersama-sama.<\/span><\/p>\n

                      Ditanya: Peluang Ade dan Bimo duduk berdampingan = …?<\/span><\/p>\n

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

                        \n
                      • Menentukan ruang sampel\u00a0<\/span>n<\/span><\/em>(<\/span>S<\/span><\/em>)<\/span>
                        \n<\/span>Banyaknya cara\u00a05\u00a0orang duduk\u00a0= 5!<\/span><\/li>\n
                      • Menentukan banyak kejadian\u00a0<\/span>n<\/span><\/em>(<\/span>A<\/span><\/em>)<\/span>
                        \n<\/span>Banyaknya\u00a02\u00a0orang duduk selalu berdampingan\u00a0= 2!<\/span>
                        \n<\/span>Oleh karena Ade dan Bimo selalu duduk berdampingan, maka dihitung satu AB. Ini berarti, ada empat (AB, C, D, E) yang duduk berdampingan\u00a0= 4!<\/span><\/li>\n
                      • Peluang 2 orang duduk selalu berdampingan dirumuskan sebagai berikut.<\/span><\/li>\n<\/ul>\n

                        \"\"<\/p>\n

                        Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan Prediksi UTBK SBMPTN Matematika SOSHUM 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. Itulah sebabnya, Quipperian harus gercep alias gerak cepat. Nah, Bagi kamu yang saat ini masih bingung mencari…<\/p>\n","protected":false},"author":99400369,"featured_media":279902,"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-279860","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn"],"acf":[],"yoast_head":"\nPrediksi Soal UTBK Matematika Soshum 2020 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Quipperian, daripada bingung-bingung cari soal UTBK 2020 Matematika Soshum, mampir ke artikel ini saja! 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