{"id":280020,"date":"2020-02-25T10:53:48","date_gmt":"2020-02-25T03:53:48","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=280020"},"modified":"2020-02-25T10:53:48","modified_gmt":"2020-02-25T03:53:48","slug":"utbk-2020-tps-pemahaman-kuantitatif","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-tps-pemahaman-kuantitatif\/","title":{"rendered":"Prediksi Soal UTBK TPS – Pemahaman Kuantitatif 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 TPS (Pemahaman Kuantitatif) 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

Nilai dari\u00a00,4 + (0,8 \u00d7 0,5) + 0,7\u00a0adalah\u2026<\/span><\/p>\n

    \n
  1. 1,3<\/span><\/li>\n
  2. 1,55<\/span><\/li>\n
  3. 1,444<\/span><\/li>\n
  4. 1,44<\/span><\/li>\n
  5. 1,5<\/span><\/li>\n<\/ol>\n

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

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

    Untuk menyelesaikan soal ini, kerjakan dahulu operasi hitung di dalam tanda kurung. Kemudian, kerjakan operasi hitung penjumlahan dimulai dari sebelah kiri. Ini berarti:<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, nilai <\/span>dari\u00a00,4 + (0,8 \u00d7 0,5) + 0,7\u00a0adalah 1,5.<\/span><\/p>\n

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

    Seorang peternak membeli 5 ekor sapi dengan harga Rp7.250.000 per ekor. Kemudian, sapi tersebut dijual dengan harga Rp7.500.000 per ekor. Besar keuntungan atau kerugian yang dialami peternak tersebut adalah…<\/span><\/p>\n

      \n
    1. untung sebesar Rp1.225.000<\/span><\/li>\n
    2. untung sebesar Rp1.250.000<\/span><\/li>\n
    3. untung sebesar Rp1.500.000<\/span><\/li>\n
    4. rugi sebesar Rp1.250.000<\/span><\/li>\n
    5. rugi sebesar Rp1.260.000<\/span><\/li>\n<\/ol>\n

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

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

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

        \n
      • Harga pembelian 1 ekor sapi = Rp7.250.000<\/span><\/li>\n
      • Harga penjualan 1 ekor sapi = Rp7.500.000<\/span><\/li>\n<\/ul>\n

        Ditanya: Untung\/ rugi beserta besarannya =\u2026?<\/span><\/p>\n

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

        Oleh karena harga penjualan lebih dari harga pembelian, maka peternak mengalami untung.<\/span><\/p>\n

        \"\"<\/p>\n

        Oleh karena peternak membeli 5 ekor sapi, maka keuntungannya adalah sebagai berikut.<\/span><\/p>\n

        Total keuntungan = 5\u00a0\u00d7\u00a0Rp250.000 = Rp1.250.000<\/span><\/p>\n

        Jadi, pedagang tersebut mengalami untung sebesar Rp1.250.000.<\/span><\/p>\n

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

        15, 29, 56, 108, 208, 390, \u2026<\/span><\/p>\n

          \n
        1. 699<\/span><\/li>\n
        2. 306<\/span><\/li>\n
        3. 516<\/span><\/li>\n
        4. 406<\/span><\/li>\n
        5. 520<\/span><\/li>\n<\/ol>\n

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

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

          Diketahui pola bilangan berikut.<\/span><\/p>\n

          15,29,56,108,208,390,…<\/span><\/p>\n

          Ditanya: Bilangan yang rumpang =\u2026?<\/span><\/p>\n

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

          Untuk memperoleh bilangan yang rumpang, perhatikan pola yang digunakan antarbilangan pada tingkat\u00a01\u00a0berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Oleh karena pada tingkat\u00a01\u00a0masih belum sama, lanjutkan ke pola tingkat\u00a02.<\/span><\/p>\n

          \"\"<\/p>\n

          Pola antarbilangan pada tingkat\u00a02\u00a0tersebut juga masih belum sama. Oleh karena itu, tentukan pola pada tingkat\u00a03\u00a0seperti berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Oleh karena pola pada tingkat\u00a03\u00a0masih belum sama, lanjutkan ke pola tingkat\u00a04.<\/span><\/p>\n

          \"\"<\/p>\n

          Dari tingkat\u00a04\u00a0tersebut, diketahui bahwa pola yang digunakan adalah\u00a0+11.\u00a0Dengan demikian, pola-pola pada tingkat sebelumnya dapat dilengkapi seperti berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Dengan demikian, diperoleh\u00a0390 + 309 = 699.<\/span><\/p>\n

          \"\"<\/p>\n

          Ini berarti, bilangan yang rumpang pada barisan tersebut adalah\u00a0699.<\/span><\/p>\n

          Jadi, jawaban yang tepat adalah\u00a0699.<\/span><\/p>\n

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

          Jika\u00a0123<\/span>cd <\/span><\/em>\u2212 4 = 344,\u00a0nilai\u00a0369<\/span>cd<\/span><\/em>\u00a0adalah\u2026<\/span><\/p>\n

            \n
          1. 1.046<\/span><\/li>\n
          2. 1.032<\/span><\/li>\n
          3. 1.044<\/span><\/li>\n
          4. 1.020<\/span><\/li>\n
          5. 1.040<\/span><\/li>\n<\/ol>\n

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

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

            Perhatikan bahwa nilai\u00a0123<\/span>cd<\/span><\/em>\u00a0dan\u00a0369<\/span>cd<\/span><\/em>\u00a0memiliki hubungan, yaitu\u00a0123<\/span>cd<\/span><\/em> \u00d7 3 = 369<\/span>cd<\/span><\/em>.\u00a0Oleh karena itu, persamaan aljabar tersebut dapat dijabarkan sebagai berikut.<\/span><\/p>\n

            \"\"<\/p>\n

            Jadi, nilai 369<\/span>cd<\/span><\/em> adalah 1.044.<\/span><\/p>\n

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

            Jika diketahui jajargenjang ABCD sebangun dengan jajargenjang PQRS, besar\u00a0<\/span>\u2220<\/span>Q\u00a0adalah…<\/span><\/p>\n

            \"\"<\/p>\n

              \n
            1. 110<\/span>o<\/span><\/sup><\/li>\n
            2. 60<\/span>o<\/span><\/sup><\/li>\n
            3. 120<\/span>o<\/span><\/sup><\/li>\n
            4. 80<\/span>o<\/span><\/sup><\/li>\n
            5. 50<\/span>o<\/span><\/sup><\/li>\n<\/ol>\n

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

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

              Perhatikan kembali jajargenjang ABCD dan PQRS berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Oleh karena kedua jajargenjang tersebut sebangun maka\u00a0<\/span>\u2220<\/span>Q = <\/span>\u2220<\/span>A = 60<\/span>\u2218<\/span>.<\/span><\/p>\n

              Selain itu, besar\u00a0<\/span>\u2220<\/span>Q\u00a0dapat ditentukan dengan menggunakan sifat sudut pada jajargenjang, yaitu jumlah sudut yang bersebelahan adalah\u00a0180<\/span>\u2218<\/span>.\u00a0Ini berarti:<\/span><\/p>\n

              \"\"<\/p>\n

              Jadi, besar <\/span>\u2220<\/span>Q adalah 60<\/span>o<\/span><\/sup>.<\/span><\/p>\n

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

              Bila ditambahkan satu garis lurus yang memotong gambar berikut ini, maksimum banyaknya bagian yang terbentuk dalam lingkaran adalah…<\/span><\/p>\n

              \"\"<\/p>\n

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

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

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

                Bagian yang terbentuk di dalam lingkaran akan maksimum jika garis yang ditambahkan melewati tiga daerah di dalam lingkaran tersebut. Perhatikan gambar berikut.<\/span><\/p>\n

                \"\"<\/p>\n

                Jadi, maksimum banyaknya bagian yang terbentuk dalam lingkaran adalah\u00a07.<\/span><\/p>\n

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

                Sebuah kereta menempuh jarak Jakarta \u2013 Surabaya dengan kecepatan\u00a060\u00a0km\/jam selama\u00a015\u00a0jam. Jika jarak tersebut ingin ditempuh dalam waktu\u00a012\u00a0jam, kecepatan kereta tersebut menjadi…<\/span><\/p>\n

                  \n
                1. 85 km\/jam<\/span><\/li>\n
                2. 95 km\/jam<\/span><\/li>\n
                3. 80 km\/jam<\/span><\/li>\n
                4. 75 km\/jam<\/span><\/li>\n
                5. 90 km\/jam<\/span><\/li>\n<\/ol>\n

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

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

                  Dari soal, diperoleh informasi berikut.<\/span><\/p>\n

                  15\u00a0jam \u2192 60\u00a0km\/jam<\/span><\/p>\n

                  12\u00a0jam \u2192 <\/span>x<\/span><\/em>\u00a0km\/jam<\/span><\/p>\n

                  Soal tersebut dapat diselesaikan dengan perbandingan berbalik nilai. Hal ini dikarenakan semakin cepat waktu yang ditempuh, kecepatan kereta akan semakin tinggi. Dengan menggunakan rumus perbandingan berbalik nilai, diperoleh:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jadi, kecepatan kereta tersebut menjadi 75 km\/jam.<\/span><\/p>\n

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

                  Jika setiap sel memiliki sisi 1 cm dan garis PC membagi gambar menjadi dua daerah, perbandingan luas kedua daerah tersebut adalah…<\/span><\/p>\n

                  \"\"<\/p>\n

                    \n
                  1. 3 : 4<\/span><\/li>\n
                  2. 4 : 5<\/span><\/li>\n
                  3. 2 : 3<\/span><\/li>\n
                  4. 5 : 6<\/span><\/li>\n
                  5. 1 : 1<\/span><\/li>\n<\/ol>\n

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

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

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

                    \"\"<\/p>\n

                    Luas daerah I merupakan jumlah dari luas \u2206PAC dan persegi panjang ABED. Ini berarti:<\/span><\/p>\n

                    \"\"<\/p>\n

                    Luas daerah II merupakan jumlah dari luas \u2206PCL, persegi panjang LKGF, dan persegi KJIH. Ini berarti:<\/span><\/p>\n

                    \"\"<\/p>\n

                    Jadi, perbandingan kedua luas daerah tersebut adalah 5 : 6.<\/span><\/p>\n

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

                    Sebuah bangun ruang dibentuk dari\u00a03\u00a0kubus kosong yang masing-masing mempunyai rusuk\u00a02\u00a0cm. Jika terdapat bangun prisma B pejal menempati ruang tersebut, volume ruang yang tersisa adalah…<\/span><\/p>\n

                    \"\"<\/p>\n

                      \n
                    1. 12 cm<\/span>3<\/span><\/sup><\/li>\n
                    2. 18 cm<\/span>3<\/span><\/sup><\/li>\n
                    3. 14 cm<\/span>3<\/span><\/sup><\/li>\n
                    4. 16 cm<\/span>3<\/span><\/sup><\/li>\n
                    5. 20 cm<\/span>3<\/span><\/sup><\/li>\n<\/ol>\n

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

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

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

                      \"\"<\/p>\n

                      Mula-mula, tentukan volume prisma B pejal.<\/span><\/p>\n

                      \"\"<\/p>\n

                      Kemudian, tentukan volume tiga buah kubus.<\/span><\/p>\n

                      \"\"<\/p>\n

                      Dengan demikian, volume ruang yang tersisa adalah sebagai berikut.<\/span><\/p>\n

                      \"\"<\/p>\n

                      Jadi, volume ruang yang tersisa adalah\u00a020 cm<\/span>3<\/span><\/sup>.<\/span><\/p>\n

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

                      Umur Dina dua kali umur Deni, sedangkan umur Dika\u00a010\u00a0tahun lebih muda dari umur Dina. Jika rata-rata umur ketiganya sama dengan umur Dika, perbandingan umur Dina, Deni, dan Dika adalah\u2026<\/span><\/p>\n

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

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

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

                        Misalkan umur Dina adalah\u00a0<\/span>x<\/span><\/em>, umur Deni adalah\u00a0<\/span>y<\/span><\/em>, dan umur Dika adalah\u00a0<\/span>z<\/span><\/em>.<\/span><\/p>\n

                        Oleh karena umur Dina dua kali umur Deni dan umur Dika\u00a010\u00a0tahun lebih muda dari umur Dina, maka:<\/span><\/p>\n

                        \"\"<\/p>\n

                        Oleh karena rata-rata umur Dina, Deni, dan Dika adalah sama dengan umur Dika, maka:<\/span><\/p>\n

                        \"\"<\/p>\n

                        Dengan mensubstitusikan persamaan\u00a0(1)\u00a0dan\u00a0(2)\u00a0ke persamaan\u00a0(3), diperoleh:<\/span><\/p>\n

                        \"\"<\/p>\n

                        Ini berarti, umur Dina, Deni, dan Dika berturut-turut adalah\u00a040, 20,\u00a0dan\u00a030.<\/span><\/p>\n

                        Jadi, perbandingan umur Dina, Deni, dan Dika adalah\u00a040 : 20 : 30 = 4 : 2 : 3.<\/span><\/p>\n

                        Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan Prediksi UTBK SBMPTN TPS (Pemahaman Kuantitatif) 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. Yuk, 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":280198,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_wpcom_ai_launchpad_first_post":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-280020","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn"],"acf":[],"yoast_head":"\nPrediksi Soal UTBK TPS - Pemahaman Kuantitatif 2020 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Quipperian, bagi kamu yang sedang mencari prediksi soal UTBK 2020 TPS Pemahaman Kuantitatif, cek di sini ya! 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