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Berikut ini tabel harga dan jumlah permintaan dan jumlah penawaran

 

 

Berdasarkan tabel di atas titik keseimbangan pasar yang tepat adalah…. space space 

  1. (50 ; 10.000) space space 

  2. (30 ; 7.500) space space 

  3. ( 30 ; 5.000) space space 

  4. (40 ; 5.000) space space 

  5. (40 ; 7.500) space space 

A. Kusumaningrum

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E. space space 

Pembahasan

Menghitung persaamaan permintaan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator P minus space P subscript d 1 end subscript over denominator P subscript d 2 end subscript minus P subscript d 1 end subscript end fraction space end cell equals cell space fraction numerator Q minus Q subscript d 1 end subscript over denominator Q subscript d 2 end subscript end fraction end cell row cell fraction numerator P space minus 5.000 over denominator 5.000 end fraction space end cell equals cell space fraction numerator Q space minus space 50 over denominator negative 20 end fraction end cell row cell negative 20 space P space plus space 100.000 space end cell equals cell space 5.000 space Q space minus space 250.000 end cell row cell negative 20 space P space end cell equals cell space 5.000 space Q space minus space 350.000 end cell row cell P subscript d space end cell equals cell space minus 250 space Q space plus space 17.500 end cell end table end style

Menghitung persamaan penawaran:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator size 14px P size 14px minus size 14px space size 14px P subscript size 14px d size 14px 1 end subscript over denominator size 14px P subscript size 14px d size 14px 2 end subscript size 14px minus size 14px P subscript size 14px d size 14px 1 end subscript end fraction size 14px space end cell size 14px equals cell size 14px space fraction numerator size 14px Q size 14px minus size 14px Q subscript size 14px d size 14px 1 end subscript over denominator size 14px Q subscript size 14px d size 14px 2 end subscript end fraction end cell row cell fraction numerator size 14px P size 14px space size 14px minus size 14px 5 size 14px. size 14px 000 over denominator size 14px 5 size 14px. size 14px 000 end fraction size 14px space end cell size 14px equals cell size 14px space fraction numerator size 14px Q size 14px space size 14px minus size 14px space size 14px 30 over denominator 20 end fraction end cell row cell size 14px 20 size 14px space size 14px P size 14px space size 14px minus size 14px space size 14px 50 size 14px. size 14px 000 size 14px space end cell size 14px equals cell size 14px space size 14px 5 size 14px. size 14px 000 size 14px space size 14px Q size 14px space size 14px minus size 14px space size 14px 150 size 14px. size 14px 000 end cell row cell size 14px 20 size 14px space size 14px P size 14px space end cell size 14px equals cell size 14px space size 14px 5 size 14px. size 14px 000 size 14px space size 14px Q size 14px space size 14px minus size 14px space size 14px 50 size 14px. size 14px 000 end cell row cell size 14px P subscript s size 14px space end cell size 14px equals cell size 14px space size 14px 250 size 14px space size 14px Q size 14px space size 14px minus size 14px space size 14px 2 size 14px. size 14px 500 end cell end table

Mengitung keseimbangan dari Pd dan Ps

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript d space end cell equals cell space P subscript s end cell row cell negative 250 space Q space plus space 17.500 space end cell equals cell space 250 space Q space minus space 2.500 end cell row cell 20.000 space end cell equals cell space 500 space Q end cell row cell Q space end cell equals cell space 40 end cell row blank blank blank row cell P subscript s space end cell equals cell space 250 space Q space minus space 2.500 end cell row blank equals cell space 250 space left parenthesis 40 right parenthesis space minus 2.500 end cell row blank equals cell space 10.000 space minus space 2.500 end cell row blank equals cell 7.500 end cell end table end style

Keseimbangan yang terjadi pada Q = 40, P= 7.500

Oleh karena itu, jawaban yang tepat adalah E. space space 

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