{"id":279151,"date":"2020-01-23T12:27:17","date_gmt":"2020-01-23T05:27:17","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279151"},"modified":"2021-11-03T18:13:15","modified_gmt":"2021-11-03T11:13:15","slug":"utbk-matematika-saintek-2017","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-matematika-saintek-2017\/","title":{"rendered":"Soal UTBK SBMPTN Matematika SAINTEK 2017"},"content":{"rendered":"

\"\"<\/p>\n

Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung mencari soal-soal SBMPTN, tampaknya kamu berada di artikel yang tepat, <\/span>nih<\/span><\/em>.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas latihan soal SBMPTN Matematika IPA TKD SAINTEK 2017. Kabar baiknya, latihan soal yang disediakan Quipper Blog, lengkap dengan jawaban beserta pembahasannya, <\/span>lho<\/span><\/em>. Semakin penasaran? Yuk, segera kerjakan soalnya!<\/span><\/p>\n

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

Segitiga ABC memiliki luas\u00a035 cm<\/span>2<\/span><\/sup>\u00a0dengan panjang alas\u00a0(3<\/span>x<\/span><\/em> + 2<\/span>y<\/span><\/em>) cm\u00a0dan tinggi\u00a05 cm. Sementara itu, segitiga DEF memiliki luas\u00a048 cm<\/span>2<\/span><\/sup>\u00a0dengan panjang alas\u00a016 cm\u00a0dan tinggi\u00a0(7<\/span>x<\/span><\/em> \u2212 2<\/span>y<\/span><\/em>) cm. Luas segitiga GHI dengan panjang alas\u00a0(\u22122<\/span>y<\/span><\/em> + 9<\/span>x<\/span><\/em>) cm\u00a0dan tinggi\u00a0(6<\/span>x<\/span><\/em> \u2212 <\/span>y<\/span><\/em>) cm\u00a0adalah\u2026<\/span><\/p>\n

    \n
  1. 60 cm<\/span>2<\/span><\/sup>\u00a0<\/span><\/li>\n
  2. 93 cm<\/span>2<\/span><\/sup>\u00a0<\/span><\/li>\n
  3. 80 cm<\/span>2<\/span><\/sup>\u00a0<\/span><\/li>\n
  4. 40 cm<\/span>2<\/span><\/sup>\u00a0<\/span><\/li>\n
  5. 30 cm<\/span>2<\/span><\/sup>\u00a0<\/span><\/li>\n<\/ol>\n

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

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

    Diketahui: Segitiga ABC memiliki luas\u00a035 cm<\/span>2<\/span><\/sup>\u00a0dengan panjang alas\u00a0(3<\/span>x<\/span><\/em> + 2<\/span>y<\/span><\/em>) cm\u00a0dan tinggi\u00a05 cm. Sementara itu, segitiga DEF memiliki luas\u00a048 cm<\/span>2<\/span>\u00a0dengan panjang alas\u00a016 cm\u00a0dan tinggi\u00a0(7<\/span>x<\/span><\/em> \u2212 2<\/span>y<\/span><\/em>) cm.<\/span><\/p>\n

    Ditanya: Luas segitiga GHI dengan panjang alas\u00a0(\u22122<\/span>y<\/span><\/em> + 9<\/span>x<\/span><\/em>) cm\u00a0dan tinggi\u00a0(6<\/span>x<\/span><\/em> \u2212 <\/span>y<\/span><\/em>) cm\u00a0=\u2026?<\/span><\/p>\n

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

    Segitiga ABC memiliki luas\u00a035 cm<\/span>2<\/span>\u00a0dengan panjang alas\u00a0(3<\/span>x<\/span><\/em> + 2<\/span>y<\/span><\/em>) cm\u00a0dan tinggi\u00a05 cm. Ini berarti:<\/span><\/p>\n

    \"\"<\/p>\n

    Sementara itu, segitiga DEF memiliki luas\u00a048 cm<\/span>2<\/span><\/sup>\u00a0dengan panjang alas\u00a016 cm\u00a0dan tinggi\u00a0(7<\/span>x<\/span><\/em> \u2212 2<\/span>y<\/span><\/em>) cm. Ini berarti:<\/span><\/p>\n

    \"\"<\/p>\n

    Dengan mengeliminasi variabel <\/span>y<\/span><\/em> pada persamaan (1) dan (2), diperoleh:<\/span><\/p>\n

    \"\"<\/p>\n

    Ini berarti, luas segitiga GHI adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, luas segitiga GHI adalah 40 cm<\/span>2<\/span><\/sup>.<\/span><\/p>\n

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

    \"\"<\/p>\n

    \"\"<\/p>\n

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

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

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

    \"\"<\/p>\n

    Ditanya: Besarnya anuitas =\u2026?<\/span><\/p>\n

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

    Berdasarkan rumus anuitas, diperoleh:<\/span><\/p>\n

    \"\"<\/p>\n

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

    Diketahui\u00a0(<\/span>x<\/span><\/em> + 3)\u00a0merupakan salah satu faktor dari suku banyak\u00a0<\/span>P<\/span><\/em>(<\/span>x<\/span><\/em>)=<\/span>x<\/span><\/em>4 <\/span><\/sup>\u2212 4<\/span>x<\/span><\/em>3<\/span><\/sup> \u2212 7<\/span>x<\/span><\/em>2<\/span><\/sup> + 34<\/span>x<\/span><\/em> + <\/span>n<\/span><\/em>. Peluang selisih pasangan akarnya bernilai lebih dari 3 adalah…<\/span><\/p>\n

    \"\"<\/p>\n

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

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

    Diketahui:<\/span> (<\/span>x<\/span><\/em> + 3)\u00a0merupakan salah satu faktor dari suku banyak P<\/em>(x<\/em>)=x<\/em>4 <\/sup>\u2212 4x<\/em>3<\/sup> \u2212 7x<\/em>2<\/sup> + 34x<\/em> + n<\/em><\/span>.<\/span><\/p>\n

    Ditanya: Peluang selisih pasangan akarnya bernilai lebih dari 3 =\u2026?<\/span><\/p>\n

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

    Peluang sampelnya (<\/span>S<\/span><\/em>) adalah hasil selisih dari tiap pasangan akar P<\/em>(x<\/em>)=x<\/em>4 <\/sup>\u2212 4x<\/em>3<\/sup> \u2212 7x<\/em>2<\/sup> + 34x<\/em> + n<\/em><\/span>.<\/span><\/p>\n

    Oleh karena (<\/span>x<\/span><\/em> + 3) merupakan salah satu faktor (akar) dari <\/span>P<\/span><\/em>(<\/span>x<\/span><\/em>), maka sisa pembagian <\/span>P<\/span><\/em>(<\/span>x<\/span><\/em>) oleh <\/span>x<\/span><\/em> = -3 adalah 0.<\/span><\/p>\n

    Dengan cara Horner, diperoleh:<\/span><\/p>\n

    \"\"<\/p>\n

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

    \"\"<\/p>\n

    Dengan demikian, akar-akar <\/span>P<\/span><\/em>(<\/span>x<\/span><\/em>) adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Selisih dari pasangan akarnya adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Dengan demikian, <\/span>S<\/span><\/em> = {1, 2, 3, 4, 5, 7} dan <\/span>n<\/span><\/em>(<\/span>S<\/span><\/em>) = 6.<\/span><\/p>\n

    Misalkan A adalah kejadian selisih pasangan akarnya bernilai lebih dari 3, sehingga:<\/span><\/p>\n

    A<\/span><\/em> = {4, 5, 7} dan <\/span>n<\/span><\/em>(<\/span>A<\/span><\/em>) = 3.<\/span><\/p>\n

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

    \"\"<\/p>\n

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

    Diketahui fungsi\u00a0<\/span>f<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>k<\/span><\/em><\/sup>\u00a0<\/span>dan\u00a0<\/span>g<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>. Misalkan\u00a0<\/span>D<\/span><\/em>\u00a0adalah daerah yang dibatasi oleh kurva\u00a0<\/span>g<\/span><\/em>, sumbu X, dan\u00a0<\/span>x<\/span><\/em> = 1. Kurva\u00a0<\/span>f<\/span><\/em>\u00a0membagi daerah\u00a0<\/span>D<\/span><\/em>\u00a0menjadi daerah\u00a0<\/span>D<\/span><\/em>1<\/span><\/em><\/sub>\u00a0dan\u00a0<\/span>D<\/span><\/em>2<\/span><\/sub>, dengan perbandingan\u00a01 : 2. Jika\u00a0<\/span>D<\/span><\/em>1<\/span>\u00a0adalah daerah yang dibatasi oleh kurva\u00a0<\/span>f<\/span><\/em>\u00a0dan\u00a0<\/span>g<\/span><\/em>, nilai\u00a0<\/span>k<\/span><\/em> =\u00a0…<\/span><\/p>\n

    \"\"<\/p>\n

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

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

    Diketahui: <\/span>f<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>k<\/span><\/em><\/sup>\u00a0<\/span>dan\u00a0<\/span>g<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>.<\/span><\/p>\n

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

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

    D<\/span><\/em>\u00a0adalah daerah yang dibatasi oleh kurva\u00a0<\/span>g<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>, sumbu X, dan\u00a0<\/span>x <\/span><\/em>= 1. Dengan demikian, diperoleh gambar berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Kurva\u00a0<\/span>f<\/span><\/em>(<\/span>x<\/span><\/em>) = <\/span>x<\/span><\/em>k<\/span><\/em><\/sup>\u00a0membagi daerah\u00a0<\/span>D<\/span><\/em>\u00a0menjadi daerah\u00a0<\/span>D<\/span><\/em>1<\/sub>\u00a0<\/span>dan\u00a0<\/span>D<\/span><\/em>2<\/span><\/sub>, sehingga diperoleh gambar berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Oleh karena perbandingan luas daerah <\/span>D<\/span><\/em>1<\/sub>\u00a0<\/span>:\u00a0<\/span>D<\/span><\/em>2<\/span><\/sub> = 1 : 2, maka:<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, nilai <\/span>k<\/span><\/em> = 2.<\/span><\/p>\n

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

    \"\"<\/p>\n

      \n
    1. y = -5<\/em><\/li>\n
    2. y = 2<\/em><\/li>\n
    3. y = 6<\/em><\/li>\n
    4. y = 3<\/em><\/li>\n
    5. y = -3<\/em><\/li>\n<\/ol>\n

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

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

      \"\"<\/p>\n

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

      Dengan teknik faktorisasi, diperoleh:<\/span><\/p>\n

      \"\"<\/p>\n

      Fungsi\u00a0<\/span>y<\/span><\/em>\u00a0memiliki asimtot tegak jika penyebutnya bernilai sama dengan nol. Bentuk\u00a0(<\/span>x<\/span><\/em>2 <\/span><\/sup>+ 2)\u00a0selalu bernilai positif dan tidak mungkin bernilai nol. Ini berarti, bentuk\u00a0(<\/span>x<\/span><\/em> + <\/span>a<\/span><\/em>)\u00a0haruslah dapat disederhanakan (dicoret) dengan pembilang agar\u00a0<\/span>y<\/span><\/em>\u00a0tidak memiliki asimtot tegak. Ini berarti:<\/span><\/p>\n

      \"\"<\/p>\n

      Dengan mensubstitusikan nilai <\/span>a<\/span><\/em> = <\/span>b<\/span><\/em> ke persamaan:<\/span><\/p>\n

      \"\"<\/p>\n

      Dengan demikian, persamaan asimtot datarnya adalah sebagai berikut.<\/span><\/p>\n

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

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

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

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

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

      \"\"<\/p>\n

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

      \"\"<\/p>\n

      \"\"<\/p>\n

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

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

      \"\"<\/p>\n

      Diketahui lingkaran besar dan lingkaran kecil dengan jari-jari masing-masing 8 cm dan\u00a04\u221a<\/span>2<\/span>\u00a0cm. Lingkaran kecil melalui pusat lingkaran besar. Jika ruas garis yang menghubungkan titik potong kedua lingkaran merupakan diameter lingkaran kecil, perbandingan luas daerah irisan kedua lingkaran dengan luas daerah lingkaran kecil yang tidak beririsan adalah…<\/span><\/p>\n

      \"\"<\/p>\n

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

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

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

        \n
      • Jari-jari lingkaran besar = <\/span>R<\/span><\/em> = 8 cm<\/span><\/li>\n
      • Jari-jari lingkaran kecil = <\/span>r<\/span><\/em> = 4\u221a<\/span>2<\/span>\u00a0cm<\/span><\/li>\n<\/ul>\n

        Ditanya: Perbandingan luas daerah irisan kedua lingkaran dengan luas daerah lingkaran kecil yang tidak beririsan adalah =\u2026?<\/span><\/p>\n

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

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

        \"\"<\/p>\n

        Misalkan\u00a0<\/span>L<\/span><\/em>\u00a0adalah luas irisan kedua lingkaran.<\/span><\/p>\n

        Luas irisan kedua lingkaran merupakan jumlah luas tembereng lingkaran kecil dan tembereng lingkaran besar.<\/span><\/p>\n

        Untuk tembereng lingkaran kecil:<\/span><\/p>\n

        \"\"<\/p>\n

        \"\"<\/p>\n

        Untuk tembereng lingkaran besar:<\/span><\/p>\n

        \"\"<\/p>\n

        Luas tembereng lingkaran besar (daerah berwarna biru) diperoleh dari luas juring AOB dikurangi luas segitiga AOB.<\/span><\/p>\n

        AOB adalah sudut keliling lingkaran kecil yang menghadap ke diameternya (AB), sehingga besar sudut AOB adalah\u00a090<\/span>\u2218<\/span>.\u00a0Dengan demikian, diperoleh:<\/span><\/p>\n

        \"\"<\/p>\n

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

        \"\"<\/p>\n

        Misalkan luas daerah lingkaran kecil yang tidak beririsan adalah\u00a0L3.<\/span><\/p>\n

        \"\"<\/p>\n

        Dengan demikian, perbandingan luas daerah irisan kedua lingkaran dengan luas daerah lingkaran kecil yang tidak beririsan adalah sebagai berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan latihan soal SBMPTN Matematika IPA TKD SAINTEK 2017 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":"

        Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung…<\/p>\n","protected":false},"author":99400369,"featured_media":279197,"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":[679385508],"ppma_author":[679386827],"class_list":["post-279151","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn","tag-soal-utbk-saintek-2017"],"acf":[],"yoast_head":"\nSoal UTBK SBMPTN Matematika SAINTEK 2017 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Buat kamu yang sedang mencari kumpulan soal Matematika SAINTEK SBMPTN 2017, cek di sini saja! 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