Recommended
More Related Content
What's hot
What's hot (11)
Retroalimentación Aptitud Numérica Prueba de Aptitudes.pptx
- 3. 𝑷𝒓𝒊𝒎𝒆𝒓𝒐 𝒂𝒗𝒆𝒓𝒊𝒈𝒖𝒂𝒎𝒐𝒔, 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒒𝒖𝒆 𝒓𝒆𝒔𝒖𝒍𝒕𝒂𝒎𝒐𝒔 𝒑𝒂𝒈𝒂𝒏𝒅𝒐 𝒄𝒐𝒏 𝒆𝒔𝒕𝒆 𝒄𝒓é𝒅𝒊𝒕𝒐 𝑽𝑨𝑳𝑶𝑹 𝑫𝑬 𝑳𝑨 𝑪𝑼𝑶𝑻𝑨 ∗ 𝑪𝑨𝑵𝑻𝑰𝑫𝑨𝑫 𝑫𝑬 𝑪𝑼𝑶𝑻𝑨𝑺 𝑴𝑬𝑵𝑺𝑼𝑨𝑳𝑬𝑺 ∗ 𝑳𝑶𝑺 𝑨Ñ𝑶𝑺 𝑫𝑬 𝑷𝑳𝑨𝒁𝑶 $𝟏𝟎𝟎. 𝟎𝟎𝟎 ∗ 𝟏𝟐 ∗12 $𝟏𝟒. 𝟒𝟎𝟎. 𝟎𝟎𝟎 𝑺𝒊 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓𝒂𝒎𝒐𝒔 𝒒𝒖𝒆 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒅𝒆 𝒄𝒐𝒏𝒕𝒂𝒅𝒐, 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒆 𝒂𝒍 𝟏𝟎𝟎% 𝒅𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒎𝒐𝒔 𝒂 𝒒𝒖𝒆 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒂 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒅𝒆𝒍 𝒄𝒓é𝒅𝒊𝒕𝒐 𝒚 𝒄𝒐𝒎𝒑𝒂𝒓𝒂𝒎𝒐𝒔 $𝟏𝟎. 𝟎𝟎𝟎. 𝟎𝟎𝟎 𝟏𝟎𝟎% $𝟏𝟒. 𝟒𝟎𝟎. 𝟎𝟎𝟎 𝒙 𝒙 = $𝟏𝟒. 𝟒𝟎𝟎. 𝟎𝟎𝟎 ∗ 𝟏𝟎𝟎% $𝟏𝟎. 𝟎𝟎𝟎, 𝟎𝟎𝟎 = 𝟏𝟒𝟒% 𝟏 = 𝟏𝟒𝟒% 𝒄𝒐𝒎𝒑𝒂𝒓𝒂𝒎𝒐𝒔 𝒂𝒍 𝒓𝒆𝒔𝒕𝒂𝒓 𝟏𝟒𝟒% − 𝟏𝟎𝟎% = 𝟒𝟒%
- 6. 𝑪𝒐𝒏𝒔𝒊𝒅𝒆𝒓𝒂𝒏𝒅𝒐 𝒒𝒖𝒆 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒊𝒏𝒊𝒄𝒊𝒂𝒍, 𝒆𝒔 𝒆𝒍 𝟏𝟎𝟎%, 𝒂𝒗𝒆𝒓𝒊𝒈𝒖𝒂𝒎𝒐𝒔 𝒄𝒖𝒂𝒏𝒕𝒐 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒆 $𝟑𝟎𝟎. 𝟎𝟎𝟎 𝟏𝟎𝟎% 𝒙 𝟏𝟏𝟖% 𝒙 = $𝟑𝟎𝟎. 𝟎𝟎𝟎 ∗ 𝟏𝟏𝟖% 𝟏𝟎𝟎% 𝒙 = $𝟑. 𝟎𝟎𝟎 ∗ 𝟏𝟏𝟖 𝟏 𝒙 = $𝟑𝟓𝟒. 𝟎𝟎𝟎 𝑬𝒍 𝒗𝒂𝒍𝒐𝒓 𝒅𝒆𝒍 𝒄𝒖𝒓𝒔𝒐 𝒄𝒐𝒏 𝒍𝒂 𝒓𝒆𝒄𝒂𝒓𝒈𝒂 𝒂 𝒄𝒓é𝒅𝒊𝒕𝒐 … $𝟑𝟓𝟒. 𝟎𝟎𝟎 ÷ 𝟔 𝒂𝒍 𝟏𝟏𝟖%, 𝒚𝒂 𝒒𝒖𝒆 𝒆𝒔𝒕á 𝒉𝒂𝒃𝒍𝒂𝒏𝒅𝒐 𝒅𝒆 𝒖𝒏 𝒊𝒏𝒄𝒓𝒆𝒎𝒆𝒏𝒕𝒐 𝒅𝒆𝒍 𝟏𝟖% 𝒔𝒐𝒃𝒓𝒆 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒕𝒐𝒕𝒂𝒍. 𝑳𝒐 𝒅𝒊𝒗𝒊𝒅𝒊𝒎𝒐𝒔 𝒆𝒏𝒕𝒓𝒆 𝒆𝒍 𝒏ú𝒎𝒆𝒓𝒐 𝒅𝒆 𝒄𝒖𝒐𝒕𝒂𝒔 $𝟓𝟗. 𝟎𝟎𝟎
- 9. 𝑷𝒓𝒆𝒈𝒖𝒏𝒕𝒂 𝒄𝒐𝒏 𝒂𝒍𝒈𝒐 𝒅𝒆 𝒕𝒓𝒂𝒎𝒑𝒂, 𝒆𝒏 𝒍𝒂 𝒊𝒏𝒕𝒆𝒓𝒑𝒓𝒆𝒕𝒂𝒄𝒊ó𝒏 … 𝟏𝟒 = 𝟓 + 𝟓 + 𝟒 𝒍𝒐𝒔 𝟒 𝒔𝒊𝒏 𝒓𝒆𝒃𝒂𝒋𝒂, 𝒑𝒂𝒈𝒂𝒏 𝒆𝒍 𝒗𝒂𝒍𝒐𝒓 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒐 𝒅𝒆 $𝟓. 𝟎𝟎𝟎 $𝟓. 𝟎𝟎𝟎 − 𝟏𝟎% 𝒅𝒆 $𝟓. 𝟎𝟎𝟎 𝒂𝒓𝒎𝒂𝒎𝒐𝒔 𝒈𝒓𝒖𝒑𝒐𝒔 𝒅𝒆 𝟓, 𝒑𝒂𝒓𝒂 𝒔𝒂𝒃𝒆𝒓 𝒂 𝒄𝒖á𝒏𝒕𝒐𝒔 𝒍𝒆𝒔 𝒉𝒂𝒄𝒆𝒏 𝒓𝒆𝒃𝒂𝒋𝒂 … 𝒍𝒐𝒔 𝟏𝟎 𝒄𝒐𝒏 𝒓𝒆𝒃𝒂𝒋𝒂, 𝒑𝒂𝒈𝒂𝒏 … $𝟓. 𝟎𝟎𝟎 − $𝟓𝟎𝟎 $𝟒. 𝟓𝟎𝟎 $𝟓. 𝟎𝟎𝟎 ∗ 𝟏𝟎 𝟏𝟎𝟎 = $𝟓𝟎𝟎
- 12. 𝑷𝒍𝒂𝒏𝒕𝒆𝒂𝒎𝒐𝒔 … 𝑶𝒋𝒐 𝒒𝒖𝒆 𝒍𝒂 𝒓𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 𝒍𝒂 𝒑𝒊𝒅𝒆𝒏 𝒆𝒏 𝒅í𝒂𝒔 … 𝒏𝒐 𝒆𝒏 𝒉𝒐𝒓𝒂𝒔 𝒅𝒊𝒂𝒓𝒊𝒂𝒔. 𝑷𝒊𝒏𝒕𝒐𝒓𝒆𝒔 # 𝒅𝒆 𝒅í𝒂𝒔 𝑪𝒂𝒏𝒕𝒊𝒅𝒂𝒅 𝒅𝒆 𝑪𝒂𝒔𝒂𝒔 𝑹𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒄𝒐𝒎𝒑𝒖𝒆𝒔𝒕𝒂 𝒚 𝒂𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 𝒊𝒎𝒑𝒍í𝒄𝒊𝒕𝒂 𝟏 𝟏 𝟏 𝟒 𝒙 𝟏𝟐 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒊𝒓𝒆𝒄𝒕𝒂 𝒙 = 𝟏𝒑𝒊𝒏𝒕𝒐𝒓 ∗ 𝟏𝒅𝒊𝒂 ∗ 𝟏𝟐 𝒄𝒂𝒔𝒂𝒔 𝟒 𝒑𝒊𝒏𝒕𝒐𝒓𝒆𝒔 ∗ 𝟏 𝒄𝒂𝒔𝒂 = 𝟏 ∗ 𝟏𝒅𝒊𝒂 ∗ 𝟏𝟐 𝟒 ∗ 𝟏 = 𝟏𝟐𝒅í𝒂𝒔 𝟒 = 𝟑 𝒅í𝒂𝒔
- 15. 𝑨𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒖𝒏𝒂 𝒂 𝒖𝒏𝒂 𝒍𝒂𝒔 𝒐𝒑𝒄𝒊𝒐𝒏𝒆𝒔 𝒑𝒂𝒓𝒂 𝒄𝒐𝒎𝒑𝒓𝒐𝒃𝒂𝒓 𝒒𝒖𝒆 𝒔𝒆 𝒄𝒖𝒎𝒑𝒍𝒂𝒏 𝑷𝒂𝒓𝒂 𝒔𝒂𝒄𝒂𝒓 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆𝒃𝒆𝒎𝒐𝒔 𝒄𝒐𝒏𝒐𝒄𝒆𝒓 𝒍𝒐𝒔 𝒕𝒐𝒕𝒂𝒍𝒆𝒔 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒊𝒕𝒂𝒓𝒊𝒐𝒔 = 𝟐𝟎𝟎 𝑺𝒆𝒄𝒖𝒏𝒅𝒂𝒓𝒊𝒂 = 𝟏𝟔𝟎 𝑺𝒂𝒄𝒂𝒎𝒐𝒔 𝒍𝒐𝒔 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆 𝒍𝒂 𝑰𝒏𝒇. 𝑨𝒄𝒂𝒅. 𝟐𝟎𝟎 𝟏𝟎𝟎% 𝟖𝟎 𝒙 𝟏𝟔𝟎 𝟏𝟎𝟎% 𝟑𝟐 𝒙 𝒙 = 𝟖𝟎 ∗ 𝟏𝟎𝟎% 𝟐𝟎𝟎 𝒙 = 𝟑𝟐 ∗ 𝟏𝟎𝟎% 𝟏𝟔𝟎 𝒙 = 𝟒𝟎% 𝒙 = 𝟐𝟎% 𝑬𝒇𝒆𝒄𝒕𝒊𝒗𝒂𝒎𝒆𝒏𝒕𝒆 𝒆𝒔 𝒆𝒍 𝒅𝒐𝒃𝒍𝒆. 𝑶𝒑𝒄𝒊ó𝒏 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒂
- 16. 𝑨𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒖𝒏𝒂 𝒂 𝒖𝒏𝒂 𝒍𝒂𝒔 𝒐𝒑𝒄𝒊𝒐𝒏𝒆𝒔 𝒑𝒂𝒓𝒂 𝒄𝒐𝒎𝒑𝒓𝒐𝒃𝒂𝒓 𝒒𝒖𝒆 𝒔𝒆 𝒄𝒖𝒎𝒑𝒍𝒂𝒏 𝑷𝒂𝒓𝒂 𝒔𝒂𝒄𝒂𝒓 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆𝒃𝒆𝒎𝒐𝒔 𝒄𝒐𝒏𝒐𝒄𝒆𝒓 𝒍𝒐𝒔 𝒕𝒐𝒕𝒂𝒍𝒆𝒔 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒊𝒕𝒂𝒓𝒊𝒐𝒔 = 𝟐𝟎𝟎 𝑺𝒆𝒄𝒖𝒏𝒅𝒂𝒓𝒊𝒂 = 𝟏𝟔𝟎 𝑺𝒂𝒄𝒂𝒎𝒐𝒔 𝒍𝒐𝒔 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆𝒍 í𝒕𝒆𝒎 "𝒐𝒕𝒓𝒐" 𝟐𝟎𝟎 𝟏𝟎𝟎% 1𝟎 𝒙 𝟏𝟔𝟎 𝟏𝟎𝟎% 𝟖 𝒙 𝒙 = 𝟏𝟎 ∗ 𝟏𝟎𝟎% 𝟐𝟎𝟎 𝒙 = 𝟖 ∗ 𝟏𝟎𝟎% 𝟏𝟔𝟎 𝒙 = 𝟓% 𝒙 = 𝟓% 𝑬𝒇𝒆𝒄𝒕𝒊𝒗𝒂𝒎𝒆𝒏𝒕𝒆 𝒆𝒔 𝒆𝒍 𝒎𝒊𝒔𝒎𝒐. 𝑶𝒑𝒄𝒊ó𝒏 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒂
- 17. 𝑨𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒖𝒏𝒂 𝒂 𝒖𝒏𝒂 𝒍𝒂𝒔 𝒐𝒑𝒄𝒊𝒐𝒏𝒆𝒔 𝒑𝒂𝒓𝒂 𝒄𝒐𝒎𝒑𝒓𝒐𝒃𝒂𝒓 𝒒𝒖𝒆 𝒔𝒆 𝒄𝒖𝒎𝒑𝒍𝒂𝒏 𝑷𝒂𝒓𝒂 𝒔𝒂𝒄𝒂𝒓 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆𝒃𝒆𝒎𝒐𝒔 𝒄𝒐𝒏𝒐𝒄𝒆𝒓 𝒍𝒐𝒔 𝒕𝒐𝒕𝒂𝒍𝒆𝒔 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒊𝒕𝒂𝒓𝒊𝒐𝒔 = 𝟐𝟎𝟎 𝑺𝒆𝒄𝒖𝒏𝒅𝒂𝒓𝒊𝒂 = 𝟏𝟔𝟎 𝑺𝒂𝒄𝒂𝒎𝒐𝒔 𝒍𝒐𝒔 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆𝒔, 𝒅𝒆𝒍 í𝒕𝒆𝒎 "𝒐𝒕𝒓𝒐" 𝟐𝟎𝟎 𝟏𝟎𝟎% 𝟏𝟏𝟎 𝒙 𝟏𝟔𝟎 𝟏𝟎𝟎% 𝟏𝟐𝟎 𝒙 𝒙 = 𝟏𝟏𝟎 ∗ 𝟏𝟎𝟎% 𝟐𝟎𝟎 𝒙 = 𝟏𝟐𝟎 ∗ 𝟏𝟎𝟎% 𝟏𝟔𝟎 𝒙 = 𝟓𝟓% 𝒙 = 𝟕𝟓% 𝑪𝒖𝒎𝒑𝒍𝒆 𝒖𝒏𝒂 𝒔𝒐𝒍𝒂. 𝑶𝒑𝒄𝒊ó𝒏 𝒊𝒏𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒂
- 20. 𝑯𝒂𝒍𝒍𝒂𝒎𝒐𝒔 𝒍𝒂 𝒄𝒂𝒏𝒕𝒊𝒅𝒂𝒅 𝒅𝒆 𝒄𝒂𝒅𝒂 𝒄𝒐𝒍𝒐𝒓 𝒅𝒆 𝒂𝒖𝒕𝒐𝒔 𝒅𝒆𝒑𝒆𝒏𝒅𝒊𝒆𝒏𝒅𝒐 𝒅𝒆𝒍 𝒑𝒐𝒓𝒄𝒆𝒏𝒕𝒂𝒋𝒆. 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 = 𝟑𝟎% 𝒅𝒆𝒍 𝒕𝒐𝒕𝒂𝒍 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 = 𝟖𝟎 × 𝟑𝟎 𝟏𝟎𝟎 = 𝟐𝟒 𝒂𝒖𝒕𝒐𝒔 𝑽𝒆𝒓𝒅𝒆𝒔 = 𝟕𝟓% 𝒅𝒆 𝒍𝒐𝒔 𝒒𝒖𝒆 𝒒𝒖𝒆𝒅𝒂𝒏 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 = 𝟑𝟎% 𝒅𝒆 𝟖𝟎 𝑶𝒋𝒐 … 𝒅𝒆 𝒍𝒐𝒔 𝒒𝒖𝒆 𝒒𝒖𝒆𝒅𝒂𝒏, 𝒏𝒐 𝒅𝒆𝒍 𝒕𝒐𝒕𝒂𝒍. 𝑸𝒖𝒆𝒅𝒂𝒏 … 𝑻𝒐𝒕𝒂𝒍 − 𝒃𝒍𝒂𝒏𝒄𝒐𝒔 𝑸𝒖𝒆𝒅𝒂𝒏 … 𝟖𝟎 − 𝟐𝟒 = 𝟓𝟔 𝑽𝒆𝒓𝒅𝒆𝒔 𝟕𝟓% 𝒅𝒆 𝟓𝟔 𝑽𝒆𝒓𝒅𝒆𝒔 = 𝟓𝟔 × 𝟕𝟓 𝟏𝟎𝟎 = 𝟒𝟐𝟎𝟎 𝟏𝟎𝟎 = 𝟒𝟐 𝑹𝒐𝒋𝒐𝒔 = 𝑬𝒍 𝒓𝒆𝒔𝒕𝒐 𝑹𝒐𝒋𝒐𝒔 = 𝑻𝒐𝒕𝒂𝒍 − 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 − 𝒗𝒆𝒓𝒅𝒆𝒔 𝑹𝒐𝒋𝒐𝒔 = 𝟖𝟎 − 𝟐𝟒 − 𝟒𝟐 𝑹𝒐𝒋𝒐𝒔 = 𝟏𝟒 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 + 𝑹𝒐𝒋𝒐𝒔 = 𝟐𝟒 + 𝟏𝟒 𝑩𝒍𝒂𝒏𝒄𝒐𝒔 + 𝑹𝒐𝒋𝒐𝒔 = 𝟑𝟖
- 23. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒍𝒂 𝒓𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒄𝒐𝒎𝒑𝒖𝒆𝒔𝒕𝒂, 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒏𝒅𝒐 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 𝑵ú𝒎𝒆𝒓𝒐 𝒅𝒆 𝒄𝒖𝒂𝒅𝒓𝒊𝒍𝒍𝒂𝒔 𝒄𝒂𝒔𝒂𝒔 𝒇𝒂𝒃𝒓𝒊𝒄𝒂𝒅𝒂𝒔 𝒕𝒊𝒆𝒎𝒑𝒐 𝒆𝒏 𝒎𝒆𝒔𝒆𝒔 𝟏 𝟏 𝟒 𝟐 𝒙 𝟏 𝒂ñ𝒐 = 𝟏𝟐 𝑫𝒊𝒓𝒆𝒄𝒕𝒂 𝑫𝒊𝒓𝒆𝒄𝒕𝒂 𝒙 = 𝟐𝒄𝒖𝒂𝒅𝒓𝒊𝒍𝒍𝒂𝒔 ∗ 𝟏 𝒄𝒂𝒔𝒂 ∗ 𝟏𝟐 𝒎𝒆𝒔𝒆𝒔 𝟏 𝒄𝒖𝒂𝒅𝒓𝒊𝒍𝒍𝒂 ∗ 𝟒 𝒎𝒆𝒔𝒆𝒔 𝒙 = 𝟐𝟒 𝒄𝒂𝒔𝒂𝒔 𝟒 = 𝟔 𝒄𝒂𝒔𝒂𝒔
- 26. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒍𝒂 𝒓𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒔𝒊𝒎𝒑𝒍𝒆, 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒏𝒅𝒐 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒊𝒂 𝒓𝒆𝒄𝒐𝒓𝒓𝒊𝒅𝒂 𝑻𝒊𝒆𝒎𝒑𝒐 𝒆𝒎𝒑𝒍𝒆𝒂𝒅𝒐 𝟒𝟎𝑲𝒎 𝟐 𝒉𝒐𝒓𝒂𝒔 𝒙 𝟎, 𝟓 𝒉𝒐𝒓𝒂𝒔 𝑫𝒊𝒓𝒆𝒄𝒕𝒂 𝒙 = 𝟒𝟎𝑲𝒎 ∗ 𝟎, 𝟓 𝒉𝒐𝒓𝒂𝒔 𝟐𝒉𝒐𝒓𝒂𝒔 = 𝟐𝟎 𝑲𝒎 𝟐 = 𝟏𝟎 𝑲𝒎
- 29. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊ó𝒏 … 𝒚 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒆𝒍 𝒆𝒏𝒖𝒏𝒄𝒊𝒂𝒅𝒐, 𝒅𝒆𝒔𝒑𝒆𝒋𝒂𝒏𝒅𝒐. 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 𝒑𝒍á𝒕𝒂𝒏𝒐𝒔 = 𝟑 𝟐 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 + 𝒑𝒍á𝒕𝒂𝒏𝒐𝒔 = 𝟐𝟎𝟎 𝟐 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟑 ∗ 𝒑𝒍á𝒕𝒂𝒏𝒐𝒔 𝒑𝒍á𝒕𝒂𝒏𝒐𝒔 = 𝟐𝟎𝟎 − 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 𝟐 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟑 ∗ (𝟐𝟎𝟎 − 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔) 𝟐 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟔𝟎𝟎 − 𝟑 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔) 𝟐 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 + 𝟑 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟔𝟎𝟎 𝟓 ∗ 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟔𝟎𝟎 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟔𝟎𝟎 𝟓 𝒏𝒂𝒓𝒂𝒏𝒋𝒂𝒔 = 𝟏𝟐𝟎
- 32. 𝑬𝒙𝒕𝒓𝒂𝒆𝒎𝒐𝒔 𝒍𝒐𝒔 𝒅𝒂𝒕𝒐𝒔, 𝒄𝒐𝒏𝒗𝒊𝒓𝒕𝒊𝒆𝒏𝒅𝒐 𝒕𝒐𝒅𝒐 𝒂 𝒎𝒊𝒏𝒖𝒕𝒐𝒔. 𝑨𝒕𝒓𝒂𝒔𝒐 … 𝟏 𝒉𝒐𝒓𝒂 𝒚 𝟒𝟎 𝒎𝒊𝒏𝒖𝒕𝒐𝒔 = 𝟏𝟎𝟎 𝒎𝒊𝒏𝒖𝒕𝒐𝒔 𝑬𝒍 𝒓𝒆𝒍𝒐𝒋 𝒔𝒆 𝒂𝒅𝒆𝒍𝒂𝒏𝒕𝒂 𝟑 𝒎𝒊𝒏𝒖𝒕𝒐𝒔 𝒑𝒐𝒓 𝒅í𝒂. 𝑫𝒊𝒗𝒊𝒅𝒊𝒎𝒐𝒔 𝒆𝒍 𝒂𝒕𝒓𝒂𝒔𝒐 𝒆𝒏𝒕𝒓𝒆 𝒍𝒂 𝒄𝒂𝒑𝒂𝒄𝒊𝒅𝒂𝒅 𝒅𝒆 𝒊𝒓𝒔𝒆 𝒂𝒅𝒆𝒍𝒂𝒏𝒕𝒂𝒏𝒅𝒐 … 𝟏𝟎𝟎𝒎𝒊𝒏𝒖𝒕𝒐𝒔 ÷ 𝟑𝑴𝒊𝒏𝒖𝒕𝒐𝒔/𝒅í𝒂 = 𝟑𝟑, 𝟑𝟑𝟑𝟑𝒅í𝒂𝒔 = 𝟑𝟑𝒅í𝒂𝒔 𝟏 𝟑 𝒅𝒆 𝒅í𝒂 = 𝟑𝟑𝒅í𝒂𝒔 𝟏 𝟑 𝒅𝒆 𝒅í𝒂 𝑺𝒊 𝒖𝒏 𝒅í𝒂 𝒕𝒊𝒆𝒏𝒆 𝟐𝟒 𝒉𝒐𝒓𝒂𝒔, 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝟏 𝟑 𝒅𝒆 𝒅í𝒂 … 𝟐𝟒 𝒉𝒐𝒓𝒂𝒔 ∗ 𝟏 𝟑 … = 𝟑𝟑𝒅í𝒂𝒔 𝟏 𝟑 𝒅𝒆 𝟐𝟒 𝒉𝒐𝒓𝒂𝒔 = 𝟑𝟑𝒅í𝒂𝒔 𝒚 𝟖 𝒉𝒐𝒓𝒂𝒔
- 35. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒍𝒂𝒔 𝒅𝒐𝒔 𝒓𝒂𝒛𝒐𝒏𝒆𝒔 𝑷𝑷 𝑺𝑷 = 𝟏 𝟐 𝒅𝒆𝒔𝒑𝒆𝒋𝒂𝒎𝒐𝒔 𝒍𝒂𝒔 𝒅𝒐𝒔 𝒓𝒂𝒛𝒐𝒏𝒆𝒔. 𝑷𝑷 + 𝟓𝒂ñ𝒐𝒔 𝑺𝑷 + 𝟓𝒂ñ𝒐𝒔 = 𝟐 𝟑 𝟐 ∗ 𝑷𝑷 = 𝑺𝑷 𝟑 ∗ 𝑷𝑷 + 𝟓 𝒂ñ𝒐𝒔 = 𝟐 ∗ (𝑺𝑷 + 𝟓 𝒂ñ𝒐𝒔) 𝟑 ∗ 𝑷𝑷 + 𝟏𝟓 𝒂ñ𝒐𝒔 = 𝟐 ∗ 𝑺𝑷 + 𝟏𝟎𝒂ñ𝒐𝒔 𝟏𝟓𝒂ñ𝒐𝒔 − 𝟏𝟎𝒂ñ𝒐𝒔 = 𝟐 ∗ 𝑺𝑷 − 𝟑 ∗ 𝑷𝑷 𝟓𝒂ñ𝒐𝒔 = 𝟐 ∗ 𝑺𝑷 − 𝟑 ∗ 𝑷𝑷 𝑷𝒓𝒊𝒎𝒆𝒓𝒂 𝒑𝒆𝒓𝒔𝒐𝒏𝒂 = 𝑷𝑷 𝑺𝒆𝒈𝒖𝒏𝒅𝒂𝒑𝒆𝒓𝒔𝒐𝒏𝒂 = 𝑺𝑷 𝟓𝒂ñ𝒐𝒔 = 𝟐 ∗ (𝟐 ∗ 𝑷𝑷) − 𝟑 ∗ 𝑷𝑷 𝟓𝒂ñ𝒐𝒔 = 𝟒 ∗ 𝑷𝑷 − 𝟑 ∗ 𝑷𝑷 𝟓𝒂ñ𝒐𝒔 = 𝑷𝑷 𝟐 ∗ (𝟓 𝒂ñ𝒐𝒔) = 𝑺𝑷 𝟏𝟎 𝒂ñ𝒐𝒔 = 𝑺𝑷 𝒉𝒂𝒄𝒆 𝒅𝒐𝒔 𝒂ñ𝒐𝒔 𝟑𝒂ñ𝒐𝒔 = 𝑷𝑷 8𝒂ñ𝒐𝒔 = 𝑺𝑷 𝐒𝐮𝐦𝐚 = 𝟏𝟏 𝒂ñ𝒐𝒔
- 37. 1𝟓 𝒕𝒐𝒏 𝟏𝟎𝟎% 𝟓 𝒕𝒐𝒏 𝒙 𝒙 = 𝟓𝒕𝒐𝒏 ∗ 𝟏𝟎𝟎% 𝟏𝟓𝒕𝒐𝒏 𝒙 = 𝟑𝟑, 𝟑𝟑%
- 38. 𝑻𝒐𝒕𝒂𝒍 = 𝟐𝟎 + 𝟏𝟓 + 𝟏𝟎 = 𝟒𝟓𝒕𝒐𝒏 𝟒𝟓𝒕𝒐𝒏 𝟏𝟎𝟎% 𝟏𝟓𝒕𝒐𝒏 𝒙 𝒙 = 𝟏𝟓𝒕𝒐𝒏 ∗ 𝟏𝟎𝟎% 𝟒𝟓𝒕𝒐𝒏 𝒙 = 𝟑𝟑, 𝟑𝟑%
- 43. 𝒗𝒂𝒎𝒐𝒔 𝒆𝒙𝒕𝒓𝒂𝒚𝒆𝒏𝒅𝒐 𝒅𝒂𝒕𝒐𝒔 𝒂 𝒎𝒆𝒅𝒊𝒅𝒂 𝒒𝒖𝒆 𝒗𝒂𝒎𝒐𝒔 𝒍𝒆𝒚𝒆𝒏𝒅𝒐 𝑻𝒐𝒕𝒂𝒍 = 𝟑𝟎 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 = 𝟏𝟎𝟎% 𝑩𝒖𝒆𝒏𝒐𝒔 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 = 𝟐𝟎% 𝑩𝒖𝒆𝒏𝒐𝒔 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 = 𝟐𝟎% 𝒅𝒆 𝟑𝟎 𝑹𝒆𝒔𝒕𝒐 𝒅𝒆 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 = 𝟑𝟎 − 𝟔 = 𝟐𝟒 𝒑𝒓𝒐𝒎𝒆𝒅𝒊𝒐 𝒅𝒆 𝟓 = 𝒎𝒊𝒕𝒂𝒅 𝒅𝒆 𝟐𝟒 = 𝟑𝟎 𝟐𝟎 𝟏𝟎𝟎 = 𝟔 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 = 𝟐𝟒 ∗ 𝟏 𝟐 = 𝟏𝟐 𝑸𝒖𝒆𝒅𝒂𝒏 𝟏𝟐 𝑹𝒆𝒑𝒊𝒕𝒆𝒏 𝒆𝒍 𝒆𝒙𝒂𝒎𝒆𝒏 = 𝟓𝟎% 𝒅𝒆 𝟏𝟐 = 𝟏𝟐 ∗ 𝟓𝟎 𝟏𝟎𝟎 = 𝟔 𝑷𝒊𝒆𝒓𝒅𝒆𝒏 𝒆𝒍 𝒆𝒙𝒂𝒎𝒆𝒏 = 𝟑 𝟔 𝟏𝟎𝟎% 𝟑 𝒙 𝒙 = 𝟑 ∗ 𝟏𝟎𝟎% 𝟔 𝒙 = 𝟓𝟎%
- 46. 𝟑𝟔 𝒍𝒊𝒕𝒓𝒐𝒔 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏 𝒂 𝒍𝒐𝒔 𝟑 𝟕 𝒇𝒂𝒍𝒕𝒂𝒏 𝒑𝒐𝒓 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒂𝒓 𝒆𝒍 𝒃𝒂𝒓𝒓𝒊𝒍 𝟒 𝟕 𝒚𝒂 𝒒𝒖𝒆 𝒆𝒍 𝒕𝒐𝒕𝒂𝒍 𝒆𝒔 𝟕 𝟕 𝟑𝟔 𝒍𝒊𝒕𝒓𝒐𝒔 𝟑 𝟕 𝒅𝒆𝒍 𝒃𝒂𝒓𝒓𝒊𝒍 𝒙 𝟒 𝟕 𝒅𝒆𝒍 𝒃𝒂𝒓𝒓𝒊𝒍 𝒙 = 𝟒 𝟕 𝒅𝒆𝒍 𝒃𝒂𝒓𝒓𝒊𝒍 ∗ 𝟑𝟔 𝒍𝒊𝒕𝒓𝒐𝒔 𝟑 𝟕 𝒅𝒆𝒍 𝒃𝒂𝒓𝒓𝒊𝒍 𝒙 = 𝟒 ∗ 𝟑𝟔 𝒍𝒊𝒕𝒓𝒐𝒔 𝟑 = 𝟒𝟖𝒍𝒊𝒕𝒓𝒐𝒔
- 49. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒂𝒄𝒐𝒓𝒅𝒆 𝒂𝒍 𝒆𝒏𝒖𝒏𝒄𝒊𝒂𝒅𝒐 … 𝑪𝒍𝒂𝒖𝒅𝒊𝒂 = 𝑪 𝑨𝒏𝒂 = 𝑨 4 ∗ 𝑪 − 𝟐𝟎 = 𝑨 𝑨 𝟑 + 𝟐 = 𝑪 4 ∗ 𝑪 − 𝟐𝟎 𝟑 + 𝟐 = 𝑪 4 ∗ 𝑪 − 𝟐𝟎 𝟑 = 𝑪 − 𝟐 4 ∗ 𝑪 − 𝟐𝟎 = 𝟑 ∗ (𝑪 − 𝟐) 4 ∗ 𝑪 − 𝟐𝟎 = 𝟑 ∗ 𝑪 − 𝟔 4 ∗ 𝑪 − 𝟑 ∗ 𝑪 = 𝟐𝟎 − 𝟔 𝑪 = 𝟏𝟒𝐚ñ𝐨𝐬
- 52. 𝑺𝒆 𝒅𝒆𝒃𝒆 𝒄𝒖𝒎𝒑𝒍𝒊𝒓 𝒄𝒐𝒏 𝒖𝒏𝒂 𝒂𝒔𝒊𝒔𝒕𝒆𝒏𝒄𝒊𝒂 𝒊𝒈𝒖𝒂𝒍 𝒐 𝒔𝒖𝒑𝒆𝒓𝒊𝒐𝒓 𝒂𝒍 𝟖𝟓%, 𝟏𝟓% 𝒅𝒆 𝟒𝟖 𝒄𝒍𝒂𝒔𝒆𝒔 𝑺𝒆 𝒑𝒖𝒆𝒅𝒆 𝒇𝒂𝒍𝒕𝒂𝒓 𝒉𝒂𝒔𝒕𝒂 𝒆𝒍 𝟏𝟓% 𝟒𝟖 𝒄𝒍𝒂𝒔𝒆𝒔 ∗ 𝟏𝟓 𝟏𝟎𝟎 𝟒𝟖 𝒄𝒍𝒂𝒔𝒆𝒔 ∗ 𝟏𝟓 𝟏𝟎𝟎 = 𝟕, 𝟐 𝒄𝒍𝒂𝒔𝒆𝒔
- 55. 𝑬𝒏 𝒐𝒇𝒆𝒓𝒕𝒂 𝒔𝒆 𝒑𝒂𝒈𝒂 𝒆𝒍 𝟗𝟕% 𝒑𝒐𝒓 𝒄𝒂𝒅𝒂 𝒂𝒓𝒕í𝒄𝒖𝒍𝒐 𝟗𝟕% 𝒅𝒆 $𝟏𝟓𝟎𝟎 $𝟏𝟓𝟎𝟎 ∗ 𝟗𝟕 𝟏𝟎𝟎 = $𝟏. 𝟒𝟓𝟓 𝑷𝒐𝒓 𝟕 𝒂𝒓𝒕í𝒄𝒖𝒍𝒐𝒔 𝒑𝒂𝒈𝒂 … = $𝟏. 𝟒𝟓𝟓 ∗ 𝟕 = $𝟏𝟎. 𝟏𝟖𝟓
- 58. 𝑰𝒏𝒊𝒄𝒊𝒂𝒍𝒎𝒆𝒏𝒕𝒆 𝒕𝒊𝒆𝒏𝒆 𝒗𝒂𝒍𝒐𝒓 𝒅𝒆𝒍 𝟏𝟎𝟎% 𝑺𝒊 𝒂𝒖𝒎𝒆𝒏𝒕𝒂 𝒖𝒏 𝟓𝟎% … 𝑺𝒊 𝒍𝒖𝒆𝒈𝒐 𝒅𝒊𝒔𝒎𝒊𝒏𝒖𝒚𝒆 𝒖𝒏 𝟓𝟎% … 𝑸𝒖𝒆𝒅𝒂 𝒄𝒐𝒏 𝒖𝒏 𝒗𝒂𝒍𝒐𝒓 𝒅𝒆𝒍 𝟏𝟓𝟎% … 𝒄𝒐𝒏 𝒓𝒆𝒔𝒑𝒆𝒄𝒕𝒐 𝒂𝒍 𝒗𝒂𝒍𝒐𝒓 𝒊𝒏𝒊𝒄𝒊𝒂𝒍 𝑬𝒍 𝟓𝟎% 𝒅𝒆 𝒂𝒍𝒈𝒐, 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒆 𝒂 𝒍𝒂 𝒎𝒊𝒕𝒂𝒅 … 𝑳𝒂 𝒎𝒊𝒕𝒂𝒅 𝒅𝒆 𝟏𝟓𝟎% 𝒆𝒔 𝟕𝟓% 𝑪𝒐𝒎𝒑𝒂𝒓𝒂𝒅𝒐 𝒄𝒐𝒏 𝒆𝒍 𝒑𝒓𝒆𝒄𝒊𝒐 𝒊𝒏𝒊𝒄𝒊𝒂𝒍, 𝒃𝒂𝒋ó 𝒖𝒏 𝟐𝟓%
- 60. chocolate?
- 61. chocolate? 𝑫𝒆 𝟖𝟎𝟎, 𝒆𝒍 𝟑𝟎% 𝒔𝒐𝒏 𝒏𝒊ñ𝒐𝒔 … 𝟑𝟎% 𝒅𝒆 𝟖𝟎𝟎 𝟖𝟎𝟎 ∗ 𝟑𝟎 𝟏𝟎𝟎 = 𝟐𝟒𝟎 𝑫𝒆 𝒆𝒔𝒐𝒔 𝟐𝟒𝟎 … 𝒆𝒍 𝟖𝟎% 𝒑𝒓𝒆𝒇𝒊𝒆𝒓𝒊𝒆𝒓𝒐𝒏 𝒔𝒂𝒃𝒐𝒓 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒕𝒆 𝒂 𝒄𝒉𝒐𝒄𝒐𝒍𝒂𝒕𝒆 𝑫𝒆 𝒆𝒔𝒐𝒔 𝟐𝟒𝟎 … 𝟐𝟎% 𝒑𝒓𝒆𝒇𝒊𝒆𝒓𝒊𝒆𝒓𝒐𝒏 𝒔𝒂𝒃𝒐𝒓 𝒂 𝒄𝒉𝒐𝒄𝒐𝒍𝒂𝒕𝒆 𝟐𝟎% 𝒅𝒆 𝟐𝟒𝟎 𝟐𝟒𝟎 ∗ 𝟐𝟎 𝟏𝟎𝟎 = 𝟒𝟖
- 62. chocolate?
- 64. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒓𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒄𝒐𝒎𝒑𝒖𝒆𝒔𝒕𝒂 𝒚 𝒂𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 𝑵ú𝒎𝒆𝒓𝒐 𝒅𝒆 𝒔𝒆𝒄𝒓𝒆𝒕𝒂𝒓𝒊𝒂𝒔 𝒄𝒂𝒏𝒕𝒊𝒅𝒂𝒅 𝒅𝒆 𝒄𝒆𝒓𝒕𝒊𝒇𝒊𝒄𝒂𝒅𝒐𝒔 𝒕𝒊𝒆𝒎𝒑𝒐 𝒆𝒎𝒑𝒍𝒆𝒂𝒅𝒐 𝟏 𝒔𝒆𝒄𝒓𝒆𝒕𝒂𝒓𝒊𝒂 𝟏𝟓 𝒄𝒆𝒓𝒕𝒊𝒇𝒊𝒄𝒂𝒅𝒐𝒔 𝟒 𝒉𝒐𝒓𝒂𝒔 𝟔 𝒔𝒆𝒄𝒓𝒆𝒕𝒂𝒓𝒊𝒂𝒔 𝟗𝟎 𝒄𝒆𝒓𝒕𝒊𝒇𝒊𝒄𝒂𝒅𝒐𝒔 𝒙 𝒅𝒊𝒓𝒆𝒄𝒕𝒂 𝒅𝒊𝒓𝒆𝒄𝒕𝒂 𝒙 = 𝟏 𝒔𝒆𝒄𝒓𝒆𝒕𝒂𝒓𝒊𝒂∗𝟗𝟎 𝒄𝒆𝒓𝒕𝒊𝒇𝒊𝒄𝒂𝒅𝒐𝒔∗𝟒 𝒉𝒐𝒓𝒂𝒔 𝟔 𝒔𝒆𝒄𝒓𝒆𝒕𝒂𝒓𝒊𝒂𝒔 ∗𝟏𝟓 𝒄𝒆𝒓𝒕𝒊𝒇𝒊𝒄𝒂𝒅𝒐𝒔 𝒙 = 𝟒 𝒉𝒐𝒓𝒂𝒔
- 67. 𝑬𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒍𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆 𝒂𝒄𝒖𝒆𝒓𝒅𝒐 𝒂𝒍 𝒆𝒏𝒖𝒏𝒄𝒊𝒂𝒅𝒐 … 𝟐𝟎% 𝒅𝒆 𝑩 + 𝟐𝟓% 𝒅𝒆 𝑨 = 𝟕 𝟐 ∗ 𝟓𝟎% 𝒅𝒆𝒍 𝟐𝟎%𝒅𝒆 𝑩 = 𝟓 𝟏𝟎𝟎% 𝒅𝒆𝒍 𝟐𝟎%𝒅𝒆 𝑩 = 𝟓 𝑩 ∗ 𝟏𝟎𝟎 𝟏𝟎𝟎 ∗ 𝟐𝟎 𝟏𝟎𝟎 = 𝟓 𝑩 = 𝟓 ∗ 𝟏𝟎𝟎 𝟐𝟎 𝑩 = 𝟐𝟓 𝟐𝟎% 𝒅𝒆 𝟐𝟓 + 𝟐𝟓% 𝒅𝒆 𝑨 = 𝟕 𝟐𝟓 ∗ 𝟐𝟎 𝟏𝟎𝟎 + 𝑨 ∗ 𝟐𝟓 𝟏𝟎𝟎 = 𝟕 𝟓 + 𝑨 ∗ 𝟐𝟓 𝟏𝟎𝟎 = 𝟕 𝑨 ∗ 𝟐𝟓 𝟏𝟎𝟎 = 𝟕 − 𝟓 𝑨 ∗ 𝟐𝟓 𝟏𝟎𝟎 = 𝟐 𝑨 = 𝟐 ∗ 𝟏𝟎𝟎 𝟐𝟓 𝑨 = 𝟖 𝒎𝒊𝒕𝒂𝒅 𝒅𝒆 𝑨 = 𝟒
- 70. 𝑺𝒊 𝒔𝒆 𝒉𝒂 𝒑𝒂𝒈𝒂𝒅𝒐 𝒍𝒐𝒔 𝟑 𝟓 𝒅𝒆 𝒖𝒏𝒂 𝒅𝒆𝒖𝒅𝒂, 𝒇𝒂𝒍𝒕𝒂𝒏 𝒑𝒐𝒓 𝒑𝒂𝒈𝒂𝒓 𝟐 𝟓 . $𝟏. 𝟓𝟎𝟎 𝟑 𝟓 𝒅𝒆 𝒍𝒂 𝒅𝒆𝒖𝒅𝒂 𝒙 𝟐 𝟓 𝒅𝒆 𝒍𝒂 𝒅𝒆𝒖𝒅𝒂 𝒙 = 𝟐 𝟓 𝒅𝒆 𝒍𝒂 𝒅𝒆𝒖𝒅𝒂 ∗ $𝟏. 𝟓𝟎𝟎 𝟑 𝟓 𝒅𝒆 𝒍𝒂 𝒅𝒆𝒖𝒅𝒂 𝒙 = 𝟐 ∗ $𝟏. 𝟓𝟎𝟎 𝟑 = $𝟏. 𝟎𝟎𝟎
- 73. 𝒇𝒂𝒍𝒕𝒂𝒃𝒂𝒏 𝟏𝟖% 𝒑𝒐𝒓 𝒆𝒏𝒇𝒆𝒓𝒎𝒆𝒅𝒂𝒅 𝒚 𝟖% 𝒑𝒐𝒓 𝒓𝒆𝒖𝒏𝒊𝒐𝒏𝒆𝒔, 𝒕𝒐𝒕𝒂𝒍 𝟐𝟔% 𝟑𝟕 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 𝟕𝟒% 𝒙 𝟐𝟔% 𝒙 = 𝟑𝟕 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 ∗ 𝟐𝟔% 𝟕𝟒% 𝒙 = 𝟏𝟑 𝒂𝒍𝒖𝒎𝒏𝒐𝒔 𝑨𝒔𝒊𝒔𝒕í𝒂𝒏 𝟕𝟒% 𝒒𝒖𝒆 𝒄𝒐𝒓𝒓𝒆𝒔𝒑𝒐𝒏𝒅𝒆𝒏 𝒂 𝟑𝟕 𝒂𝒍𝒖𝒎𝒏𝒐𝒔.
- 76. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒓𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒚 𝒂𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 … 𝟕 𝒄𝒂𝒃𝒂𝒍𝒍𝒐𝒔 𝟕𝟖 𝒅𝒊𝒂𝒔 𝟐𝟏 𝒄𝒂𝒃𝒂𝒍𝒍𝒐𝒔 𝒙 𝒙 = 𝟕 𝒄𝒂𝒃𝒂𝒍𝒍𝒐𝒔 ∗ 𝟕𝟖 𝒅í𝒂𝒔 𝟐𝟏 𝒄𝒂𝒃𝒂𝒍𝒍𝒐𝒔 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 … 𝒙 = 𝟐𝟔 𝒅í𝒂𝒔
- 79. 𝑬𝒔𝒕𝒂𝒃𝒍𝒆𝒄𝒆𝒎𝒐𝒔 𝒓𝒆𝒈𝒍𝒂 𝒅𝒆 𝒕𝒓𝒆𝒔 𝒄𝒐𝒎𝒑𝒖𝒆𝒔𝒕𝒂 𝒚 𝒂𝒏𝒂𝒍𝒊𝒛𝒂𝒎𝒐𝒔 𝒍𝒂 𝒑𝒓𝒐𝒑𝒐𝒓𝒄𝒊𝒐𝒏𝒂𝒍𝒊𝒅𝒂𝒅 … 20 𝒄𝒂𝒋𝒂𝒔 𝟏𝟎 𝒎𝒆𝒔𝒆𝒔 𝟏𝟎𝟎% 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝟏𝟔 𝒄𝒂𝒋𝒂𝒔 𝟓 𝒎𝒆𝒔𝒆𝒔 𝒙 𝒅𝒊𝒓𝒆𝒄𝒕𝒂 𝒙 = 𝟏𝟔 𝒄𝒂𝒋𝒂𝒔 ∗ 𝟏𝟎 𝒎𝒆𝒔𝒆𝒔 ∗ 𝟏𝟎𝟎% 𝟐𝟎 𝒄𝒂𝒋𝒂𝒔 ∗ 𝟓 𝒎𝒆𝒔𝒆𝒔 𝒙 = 𝟏𝟔𝟎% 𝒙 = 𝟏𝟔𝟎% − 𝟏𝟎𝟎% = 𝟔𝟎%
- 82. 𝑬𝒙𝒕𝒓𝒂𝒆𝒎𝒐𝒔 𝒍𝒐𝒔 𝒅𝒂𝒕𝒐𝒔 𝒂𝒍 𝒍𝒆𝒆𝒓 𝒃𝒊𝒆𝒏 𝒆𝒍 𝒆𝒏𝒖𝒏𝒄𝒊𝒂𝒅𝒐 … 𝑵 𝒕𝒐𝒎𝒂𝒕𝒆𝒔 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒏 𝟏𝟎 𝒕𝒂𝒓𝒓𝒐𝒔 𝒅𝒆 𝒔𝒂𝒍𝒔𝒂 𝟏 𝒑𝒆𝒅𝒊𝒅𝒐, 𝟓𝟎𝟎 𝒄𝒂𝒋𝒂𝒔 𝒅𝒆 𝟑𝟎 𝒕𝒂𝒓𝒓𝒐𝒔 𝒄𝒂𝒅𝒂 𝒖𝒏𝒂 … 𝟏 𝒑𝒆𝒅𝒊𝒅𝒐, 𝟓𝟎𝟎 ∗ 𝟑𝟎 = 𝟏𝟓. 𝟎𝟎𝟎 𝒕𝒂𝒓𝒓𝒐𝒔 𝟐 𝒑𝒆𝒅𝒊𝒅𝒐𝒔 𝒑𝒐𝒓 𝒔𝒆𝒎𝒂𝒏𝒂, 𝒆𝒏 𝒖𝒏 𝒎𝒆𝒔 𝒒𝒖𝒆 𝒕𝒊𝒆𝒏𝒆 𝟓 𝒔𝒆𝒎𝒂𝒏𝒂𝒔 … 𝒔𝒐𝒏 𝟏𝟎 𝒑𝒆𝒅𝒊𝒅𝒐𝒔 𝟏 𝒑𝒆𝒅𝒊𝒅𝒐 𝒕𝒊𝒆𝒏𝒆 𝟏𝟓. 𝟎𝟎𝟎 𝒕𝒂𝒓𝒓𝒐𝒔, 𝒑𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝟏𝟎 𝒑𝒆𝒅𝒊𝒅𝒐𝒔 𝒕𝒊𝒆𝒏𝒆 𝟏𝟓𝟎. 𝟎𝟎𝟎 𝒕𝒂𝒓𝒓𝒐𝒔 𝑺𝒊 𝑵 𝒕𝒐𝒎𝒂𝒕𝒆𝒔, 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒏 𝟏𝟎 𝒕𝒂𝒓𝒓𝒐𝒔, 𝒑𝒂𝒓𝒂 𝒑𝒓𝒐𝒅𝒖𝒄𝒊𝒓 𝟏𝟓𝟎. 𝟎𝟎𝟎 𝒕𝒂𝒓𝒓𝒐𝒔 … 𝑵 𝒕𝒐𝒎𝒂𝒕𝒆𝒔 𝟏𝟎 𝒕𝒂𝒓𝒓𝒐𝒔, 𝒅𝒆 𝒔𝒂𝒍𝒔𝒂 𝒙 𝟏𝟓𝟎. 𝟎𝟎𝟎 𝒕𝒂𝒓𝒓𝒐𝒔, 𝒅𝒆 𝒔𝒂𝒍𝒔𝒂 𝒙 = 𝟏𝟓𝟎. 𝟎𝟎𝟎 ∗ 𝑵 𝒕𝒐𝒎𝒂𝒕𝒆𝒔 𝟏𝟎 𝒙 = 𝟏𝟓. 𝟎𝟎𝟎 ∗ 𝑵 𝒕𝒐𝒎𝒂𝒕𝒆𝒔
- 85. 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒎𝒐𝒔 𝒆𝒍 𝒂𝒛ú𝒄𝒂𝒓 𝒚 𝒍𝒂 𝒇𝒓𝒖𝒕𝒂 𝒑𝒐𝒓 𝟓𝟎 𝟏 𝒌𝒍 𝒎𝒆𝒓𝒎𝒆𝒍𝒂𝒅𝒂 𝟎, 𝟓 𝒌𝒍 𝒂𝒛ú𝒄𝒂𝒓 𝒚 𝟎, 𝟔 𝒌𝒍 𝒅𝒆 𝒇𝒓𝒖𝒕𝒂 𝟓𝟎 𝒌𝒍 𝒎𝒆𝒓𝒎𝒆𝒍𝒂𝒅𝒂 𝒙 𝒙 = 𝟓𝟎𝒌𝒍 ∗ (𝟎, 𝟓𝒌𝒍 𝒂𝒛ú𝒄𝒂𝒓 𝒚 𝟎, 𝟔 𝒌𝒍 𝒇𝒓𝒖𝒕𝒂 ) 𝟏 𝒌𝒍 𝒙 = 𝟐𝟓 𝒌𝒍 𝒂𝒛ú𝒄𝒂𝒓 𝒚 𝟑𝟎 𝒌𝒍 𝒅𝒆 𝒇𝒓𝒖𝒕𝒂
- 88. 𝑷𝒂𝒓𝒂 𝒔𝒐𝒍𝒖𝒄𝒊𝒐𝒏𝒂𝒓𝒍𝒐, 𝒍𝒐𝒔 𝒅𝒂𝒕𝒐𝒔 𝒆𝒔𝒕á𝒏 𝒊𝒏𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒐𝒔 … 𝑷𝒆𝒓𝒐 𝒑𝒐𝒅𝒆𝒎𝒐𝒔 𝒂𝒏𝒂𝒍𝒊𝒛𝒂𝒓𝒍𝒂𝒔 𝒓𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂𝒔 𝒆𝒏 𝒃𝒂𝒔𝒆 𝒂 𝒍𝒐𝒔 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒐𝒔 𝒒𝒖𝒆 𝒎𝒆𝒏𝒄𝒊𝒐𝒏𝒂 𝒆𝒍 𝒆𝒏𝒖𝒏𝒄𝒊𝒂𝒅𝒐 …