kwento ni mario

KWENTO NG PAG-IBIG NI MARIO

NI: lord mv

MAYROONG ISANG BATANG LALAKI NA ANG PANGALAN AY MARIO.KAYUMANGGI,KULOT ANG BUHOK AT MATALINO.SIYA AY 13 GULANG NANG PUMASOK SIYA SA MATAAS NA PAARALANG PANG-AGHAM NG VALENZUELA.

BATA PA LAMANG SI MARIO AT MAY ANGKIN NA SIYANG TALINO.NOONG GRADE 6 AY VALEDICTORIAN SIYA NG PAARALANG  KANYANG PINASUKAN.KATULAD NG MGA BATANG LALAKING KATULAD NIYA, SIYA AY MAY HINAHANGAAN NA BABAE,SIYA SI VINNAH.

GRADE 4 PA LAMANG SI MARIO AY MAY CRUSH NA SIYA KAY VINNAH.UNA NIYA ITONG NAKITA SA MGA PALIGSAHAN AT SA MGA SEMINAR NA KANILANG DINALUHANG DALAWA.BAGAMAT HINDI NIYA KAKLASE SI VINNAH DAHIL MAGKAIBANG PAARALAN SILANG PINAPASUKAN, LAGING HINAHANGAAN NIYA SI VINNAH.

KAGAYA NG IBANG LALAKING MATATALINO, ANG KAHINAAN NI MARION AY BABAE.  IYON ANG  DAHILAN KUNG KAYA KAAGAD SIYANG NABIGHANI SA KAGANDAHAN NI VINNAH.

GAANO KAGANDA SI VINNAH?

SI VINNAH AY KATULAD NI MARIO NA MATALINO;   VALEDICTORIAN DIN SIYA SA PAARALANG PINASUKAN NIYA.HIGIT SA LAHAT AY MAGANDA. SINGKIT SIYA DAHIL MAY  HALONG INTSIK, LALONG NAGDAGDAG SA KANYANG KAGANDAHAN. MAHABA ANG KANYANG BUHOK; KASING ITIM NG GABI. AT ISA PA AY MAALINDOG.KAHIT GANYAN ANG KANYANG KATANGIAN AY PALANGITI  SIYA AT MAY KAHINHINAN ANG KILOS. LAHAT NG IYON AY TANDANGTANDA NI MARIO.

ORIENTATION DAY. DUMATING SI MARION SA KANYANG PAPASUKANG PAARALAN MGA ALAS-7 NG UMAGA. NOONG ARAW NA IYON AY IPINALIWANAG SA KANILA ANG MGA PANUNTUNAN NG PAARALANG PAPASUKAN NIYA. IBINIGAY DIN SA KANILA ANG MEMORANDUM OF AGREEMENT UPANG LAGDAAN NILA ITO.KINABUKASAN, PUMUNTA ULI SI MARIO KASAMA ANG KANYANG AMA SA PAARALAN.DOON AY KINUHA NIYA ANG MGA LIBRONG KAKAILANGANIN NIYA SA BUONG TAON. PAGKATAPOS,TININGNAN DIN NIYA KUNG ANONG SECTION NIYA, SECTION NEWTON . NUNG MAKITA NIYA IYON, NAGULAT SIYA DAHIL KAKLASE NIYA SI VINNAH. KAMAKAILAN LAMANG  AY IPINALANGIN NIYA SA PANGINOON NA SANA AY MAGING KAKLASE NIYA SI VINNAH. SALAMAT AT TINUPAD NG PANGINOON ANG  KANYANG PANALANGIN.

UNANG ARAW NG KLASE. ALAM NI MARIO  NA MAY NAGIISA LAMANG SIYANG KAKILALA SA PAARALAN NIYA, SI LAWRENCE. SECOND YEAR NA SIYA NGAYON. KATULAD NG IBANG FRESHMEN, NARANASAN DIN NI MARIO NA MANIBAGO SA BAGO NIYANG PAARALAN.

MATAPOS ANG FLAG CEREMONY, TINUNGO NA NI MARIO ANG KANYANG KWARTO. KAAGAD NA INAYOS NG GURO ANG KANILANG MGA UPUAN UPANG MAG-UMPISA NA ANG KLASE.NANG MAGHAPON NA, NARAMDAMAN NI MARIO NA SIYA AY MAGKAKASAKIT. KAYA PUMUNTA NA SIYA SA CLINIC.

NAGING MAGKAIBIGAN SI MARIO AT VINNAH NGUNIT HINDI SILA GAANONG CLOSE. SI VINNAH AY LAGING NAKIKISALAMUHA SAMGA KAPWA NIYA BABAE, SAMANTALANG SI MARIO AY ABALA SA KANILANG GRUPO KASAMA ANG KANYANG MGA KAIBIGANG LALAKI NA: SPICE GIRLS.

SA BUONG KLASE, SI MARIO ANG PINAKA KOMIKERO.SIYA AY MAY KAKAYAHANG MAPATAWA ANG KLASE KAHIT SA ISANG LINYA LAMANG.PATI ANG KANYANG MGA GALAW AY NAKAKATAWA RIN. IYON ANG DAHILAN KUNG BAKIT SIYA NAPATANYAG HINDI LANG SA KANYANG KLASE.

DAHIL SA PAGIGING ABALA SA GRUPONG SPICE GIRLS, NAPILI RIN NIYA SI JOSUE BILANG ISA SA  MGA SIR O HALIGI NG SPICE GIRLS. SIYA ANG KANILANG LORD AT FOUNDER. LINGID SA KAALAMAN NI MARIO NA SI JOSUE AY GAGAWA NG MAKAPAGPAPASAKIT NG KANYANG PUSO AT KALOOBAN.

ISANG ARAW,HABANG NAG-UUSAP SI, JOSUE, BRYAN, BYRON, AT TROY TUNGKOL SA PAG-IBIG. KAYA’T NAIINGIT SI MARIO AT SINABING “BUTI PA KAYO MAY SINASABI TUNGKOL DYAN AT AKO NANDITO LAMANG”. NAGTAWANAN SILA. TINANONG SI MARION NG ISA SA APAT KUNG SINO BA ANG HINAHANGAAN NITO.BAGO BUMABA SI MARIO SA JEEP AY SINIGAW NIYA ANG PANGALANG VINNAH!

ALAM NI JOSUE ANG MGA PLANO NI MARIO UPANG LIGAWAN SI VINNAH SA TAMANG PANAHON. ALAM NIYA KUNG KAILAN GAGAWIN NI MARIO ANG KANYANG MGA HAKBANG.HANGGANG  SA HULING ARAW NG PASOK NILA BAGO MAG SEM-BREAK, SINABI NI MARIO KINA BRYAN, BYRON, JAMESH AT JOSUE KUNG ANO ANG GUSTO NIYANG REGALO SA KAARAWAN NIYA NG BIGLANG SUMAGOT SI JOSUE NA SANA IBIGAY NI MARIO SA AKIN SI VINNAH.BIGLANG NAINSULTO SI MARIO NANG BANGGITIN ITO NI JOSUE. UMALIS KAAGAD SI MARIO AT HINABOL SIYA NI JOSUE. NGUNIT HULI NA ANG LAHAT. HINDI NA NAPIGILAN NI JOSUE SI MARIO NA UMALIS.

DUMATING ANG SEMBREAK. HINDI MAALIS SA ISIPAN NI MARIO ANG SINABI NI JOSUE. KAYA WALANG ARAW AT GABI SA BUONG BAKASYON NA HINDI INISIP NI MARIO ANG MGA MANGYAYARI AT MGA HAKBANG SA DARATING NA PANAHON.

NATAPOS NA ANG BAKASYON.BALIK ESKWELA NA ULI KAYA NAKITA NA NAMAN NI MARIO ANG MAGANDANG SI VINNAH AT ANG TAKSIL NA SI JOSUE. NAGANAP NA NGA ANG MGA NAISIP NI MARIO. KINAKAIBIGAN NA NG MALALIM NI JOSUE SI VINNAH. PILIT  IPINAPALIWANAG NI BRYAN KAY MARIO ANG NAIS GAWIN NI JOSUE . NAIS MAPAGBATI NI BRYAN SINA MARIO AT JOSUE NGUNIT NAGMATIGAS SI MARIO DAHIL IBA ANG SINABI NI JOSUE KAY BRYAN TUNGKOL KAY MARIO. HINDI TUGMA ANG KANI-KANILANG MGA KWENTO. DAHIL KAY JOSUE, NABUWAG ANG GRUPO NINA MARIO, ANG SPICE GIRLS.

NABISTO NA SI JOSUE. TAMA ANG SINABI NI MARIO KAY BRYAN . SI JOSUE AY PAIBA-IBA NG SINABI. IBA ANG SINABI KAY BRYAN. IBA SA TAMA. BISTO KA RIN, JOSUE!

KAHIT NA NABISTO NA SI JOSUE, TAHIMIK PA RIN SI MARIO SA NANGYARI.HANGGANG SA ISANG ARAW, KAHIT SA IBANG TAO ITO AY PANG-AASAR, NGUNIT KAY MARIO AY IPINAPAINTINDI LAMANG NIYA KAY JOSUE ANG MGA PAGKAKAMALING NAGAWA NIYA.HANGGANG SA MGA SANDALING ITO AY GANOON PA RIN ANG GINAGAWA NI JOSUE NA WALANG SAWANG SINISIRAAN SI MARIO. BUKOD DITO, MARAMING KASALANAN DIN ANG PINAGGAGAGAWA NI JOSUE HINDI LANG SA LOOB PATI NA SA LABS NG PAARALAN.

SABI NGA NG DATING KAKLASE NI JOSUE SA ST. JOSEPH ACADEMY OF VALENZUELA AT KAKLASE NINA JOSUE AT MARIO NGAYON NA SI MACKOY,DATI DAW, SI JOSUE AY HINDI GANOON KAKULIT, AT KASAMA. NAPAKATAHIMIK DAW NG BATANG IYON.NAGULAT DIN SIYA SA MGA INAASAL NI JOSUE NGAYONG NASA SEKUNDARYA NA SILA.

DAHIL NGA SA NABISTO NA RIN SI JOSUE SA MGA KASAMAANG DULOT NIYA, INIWASAN NA RIN SIYA NG MGA KAIBIGAN NI MARIO SA NABUWAG NA SPICE GIRLS. ALAM NILA NA HINDI MAGANDA KUNG PAKIKITUNGUHAN PA SI JOSUE . HINDI NGA RIN NILA ALAM KUNG DAPAT PABANG TAWAGING KAIBIGAN SI JOSUE.HANGGANG SA MGA SANDALING ITO AY HINDI PA NAPAPATAWAD NI MARIO SI JOSUE DAHIL NANINIWALA ITO NA ANG PAGPAPATAWAD AY NARARAPAT LAMANG SA MGA TAONG DAPAT PATAWARIN AT NAKIKITANG NAGSISISI.

DAHIL NA NGA SA NANGYARI, PILIT KINAKALIMUTAN NI MARIO SI VINNAH. SA NGAYON ANG NAPILI NA NIYANG KURSO AY PAGKAPARI. PAGKATAPOS DAW NG PAG-AARAL NIYA BILANG GRADE 12 AY PAPASOK NA SIYA NG SEMINARYO.

ISA PA SA NAPILING GAWIN NI JOSUE AY TULUNGAN NA LANG NIYA ANG KANYANG KAIBIGAN NA SI CAREY KAY VINNAH. MAPUTI SI CAREY, MATANGKAD, PAYAT AT MAY KALABUAN ANG MATA. UNANG LINGGO PA LAMANG NA PASUKAN AY IPINAGTAPAT NI CAREY KAY MARIO NA MAY GUSTO NGA SIYA KAY VINNAH.

MASAKIT MAN SA LOOB NI MARIO AY TINANGGAP NIYA ANG KANYANG KAPALARAN. NAGING MAS MALAPIT SIYA SA  DIYOS AT MAS LALONG LUMAKI ANG KANYANG PAGKABANAL. DAHIL NGA SA MAGPAPARI SIYA, WALA NAMANG MALISYA ANG MAS MAIGTING NA PAKIKIPAGKAIBIGAN NIYA SA BABAE.

NAGING MAS MALAPIT NA KAIBIGAN SINA MARIO AT VINNAH. KAY MARIO MAS MABUTI ITO DAHIL NAKIKILALA NIYA SI VINNAH AT MAY MAIPAPAYO SIYA KAY  CAREY TUNGKOL SA MGA KATANGIAN NI VINNAH. LINGID SA KAALAMAN NI MARIO NA NAHUHULOG NA PALA ANG LOOB NI VINNAH KAY MARIO.

HANGGANG ISANG ARAW,PUMUNTA ANG ILANG MGA PINUNO NG ISANG SEMINARYO UPANG TANUNGIN ANG MGA ESTUDYANTE KUNG SINO ANG MAGPAPARI. BIGLANG NAGTAAS ANG KAMAY SI MARIO. AT HINDI INAASAHAN NG BUONG KLASE KUNG SINO ANG ISA PANG NAGTAAS. SIYA SI CAREY…..

NAGULAT ANG LAHAT. NGUNIT PARA KAY CAREY, BUO ANG KANIYANG LOOB. SUSUNDAN NIYA SI MARIO SA PAGTAHAK TUNGO SA LANDAS PARA SA DIYOS. HINDI NAKAPAGPIGIL SI VINNAH AT NAGPAALAM NA PUPUNTA SA CR. NGUNIT PAGDATING SA CR. BIGLANG BUMUHOS ANG ISANG IYAK. ANG IYAK NG PAGMAMAHAL.

PAG UWI NILA, KINAUSAP NI VINNAH SI MARIO. ISANG MATAIMTIM NA USAPAN. SINABI NI VINNAH ANG NARARAMDAMAN NIYA KAY MARIO. NAGULAT SI MARIO SA SINAMBIT NI VINNAH. HABANG NAG-UUSAP SILA, HINDI NILA NALAMAN NA MAYROON PA SILANG KASAMA AT NAGMAMASID SA KANILA. SI CAREY. SI CAREY NA NOON PA LAMANG AY MAY GUSTO KAY VINNAH. SA MGA SANDALING IYON AY HINDI MAKAPAGSALITA SI MARIO. BAGO BANGGITIN NIYA NA SASABIHIN MUNA NIYA KAY CAREY, NAGULAT ANG DALAWA NANG BIGLANG MAGSALITA SI CAREY AT NAGSABING “MARIO, SUNDIN MO ANG IYONG PUSO. ALAM NAMAN NATING DALAWA NA MAHAL NATIN SI VINNAH. NGUNIT ISA LAMANG ANG KANYANG PUSO.MAHAL KANIYA AT HANDA AKONG IBIGAY KO SIYA SAYO. ALAM KONG IBIBIGAY KO SIYA SA MABUTING KAMAY.BUO NA RIN NAMAN ANG AKING LOOB NA MAGPARI.NANGANGAKO KA BA NA IINGATAN MO SI VINNAH AT HANDANG IPAGTANGGOL HANGGANG SA MAPATID ANG IYONG HININGA?”

BIGLANG SINABI NI MARIO,“OO, KAIBIGAN, NANGANGAKO AKO NA AALAGAAN KO SI VINNAH HANGGANG SA AKING KAMATAYAN.IINGATAN KO SIYA. HINDI KO SIYA PALULUHAINO SASAKTAN MAN LANG. ASAHAN MO ITONG AKING PANGAKO.”

NAGKALIWANAGAN NA RIN SI MARIO AT CAREY. TINULOY NI CAREY ANG KANYANG  BOKASYON. NAGPARI SIYA. SINA MARIO AT VINNAH NAMAN AY NAG-ARAL NG MABUTI AT NAKAPAGTAPOS NA NG HIGHSCHOOL.ILANGTAON PA ANG LUMIPAS AT MAY DIPLOMA NA RIN SILA SA COLLEGE. NGUNIT SA MGA KABILA NG SULIRANIN,HINDI HUMINA ANG SAMAHAN NG MAGKASINTAHANG MARIO AT VINNAH BAGKUS AT LUMALIM PA AT TUMIBAY ANG KANILANG PAGSASAMA. NAPATAWAD NA RIN NILA SI JOSUE.

HANGGANG NOONG OKTUBRE 26, 2012 AY IKINASAL ANG MAG-ASAWANG SINA MARIO AT VINNAH ABOLANTE SA HARAP NI MSGR.JEREMIAH CAREY D. LEON, ANG DATI NILANG KAKLASE NA NGAYON AY MONSINYOR NA. ANG ARAW NG KANILANG KASAL AY  KAARAWAN PA NI MARIO ABOLANTE.

NAGKAROON NG APAT NA ANAK ANG MAG-ASAWA. ITO AY SINA MARVIN, MARIA VERONICA, VINCENT MARIE, AT MARIO ABOLANTE JR. SA NGAYON AY INHINYERO SI MARIO AT ISANG CEO NG KUMPANYA SI VINNAH ABOLANTE. MAYROON NA RIN SILANG MAGAGARANG BAHAY, GAMIT AT MARAMING PANG IBA. NGUNIT PINAIRAL PA RIN NG MAG-ASAWA ANG MAGING MATIPID AT MAPAGKUMBABA KUNG KAYA’T LUMAKI ANG KANILANG MGA ANAK NG MAY MABUTING ASAL. MAGANDA RIN ANG PAKIKITUNGO NG MAG-ASAWA SA KANILANG MGA TRABAHADOR AT EMPLEYADO KAYA WALA SILANG KAALITAN.

-         THE END –

PART TWO:

NAG-UUSAP ANG MAG-ANAK NA ABOLANTE.NASA EDAD NA 60 NA SI MARIO AT SI VINNAH NAMAN AY 60 NA. MAY MGA ASAWA NA RIN ANG KANILANG MGA ANAK, PAWANG MGA TAPOS NA SA KANILANG PAG-AARAL. SI  MARVIN: ENGINEERING, SI MARIA VERONICA NAMAN AY NAGING DOKTOR, SAMANTALANG SI VINCENT MARIE NAMAN AY PINAGTUUNANG PANSIN ANG PAGIGING NEGOSYANTE. AT ANG KANILANG BUNSO NA SI MARIO JR. NAMAN ANG NAMAMAHALA SA  KOMPANYA NG PAMILYA. KAHIT NA EDUKADO AT EDUKADA ANG MGA ANAK NG MAG-ASAWANG ABOLANTE, MARAMING KAPILYO AT KAPILYAHANG PINAGGAGAWA NG MGA BATA . NARIYANG GUSTONG MAGPABILI NG LOBO SA KANILANG TATAY, O LUMABAS NG BAHAY PARA BUMILI NG KWEK-KWEK AT BALUT SA MAY KABILANG KANTO.

DI LINGID SA MAG-IINA ANG SAKIT NG KANILANG TATAY. SAKIT NA MARAHIL ISA SA MGA PANGUNAHING KARAMDAMAN NG MGA TAO SA MUNDO NGAYON, ANG SAKIT SA PUSO. MEDYO MALABO NA RIN ANG MGA MATA NI MARIO. NGUNIT DAHIL HINDI KUNSUMIDO, MEDYO BATA-BATA PARIN SIYA.SI VINNAH NAMAN AY MAGANDA PARIN KAHIT MAY EDAD NA.

ISANG ARAW NAGYAYA SI MARIO JR. NA PUMUNTA SA               KANILANG BAHAY-BAKASYUNAN SA BAGUIO. BER MONTHS NA KASI AT SIGURADONG MALAMIG DOON. NAGISIP ANG MAG-ASAWA. NAPAGDESISYUNAN NILA  NA DOON NA MAGPAHINGA. SETYEMBRE NA NOON AT MALAPIT NA RIN ANG KAARAWAN NI MARIO AT ANG ANIBERSARYO NG KASAL NILA.KAYA NAPAGISIPAN NILA NA HANGGANG SA DISYEMBRE SILA DOON. ITO AY PARA NA RIN MAKAPAGPAHINGA SI MARIO DAHIL LUMALALA NA ANG KANYANG SAKIT SA PUSO.

DUMATING ANG BUWAN NG OKTUBRE. NAPAG ISIPAN NARIN NINA MARIO AT VINNAH NA SA KATEDRAL NA LANG NG BAGUIO  GAWIN ANG IKA 40 ANIBERSARYO NG KANILANG PAG-IISANG DIBDIB. DOON ULI NILA SASAMBITIN ANG “OPO PADRE”                BILANG TANDA NA TINATANGGAP NILA ANG ISA’T-ISA SA HIRAP AT GINHAWA, SA  KARAMDAMAN O KALIGAYAHAN HANGGANG SA MAPATID ANG KANILANG HININGA.

HANGGANG SA DUMATING NA ANG ARAW NA KANILANG PINAKAHIHINTAY. SA HARAP NG KANILANG KAIBIGANG PARI NA SI MSGR. CAREY, MULI SILANG NAGKAPISAN SA HARAP NG ALTAR. PAGKATAPOS NITO, NAGKAROON NG ENGGRANDENG SALU-SALO AT IMBITADO ANG LAHAT, MAHIRAP MAN O MAYAMAN.NATAPOS ANG  PROGRAMA NG MAGSAYAW ANG MAG-ASAWA SA SALIW NG KANTANG MY LOVE WILL SEE  YOU THROUGH  .

LAKING GULAT NI VINNAH ANG LAGING PAGLALAMBING NI MARIO SA KANYA. ALAM NAMAN NIYA NA ARAW-ARAW SIYANG NILALAMBING NITO NGUNIT NGAYON LANG SIYA NAKARAMDAM NG GANITO SA KANYANG ASAWA. LAGI RIN ITONG NAGTATANONG KUNG MAMATAY SIYA ANO ANG GAGAWIN NI VINNAH.SA ILANG BUWANG PANINIRAHAN NG PAMILYA SA BAGUIO AY HINDI NA MAIPINTA ANG SOBRANG KIROT NG PUSO NI MARIO KAYA KINABUKASANAY NAGPAGOT SIYA KASAMA ANG KANYANG DRIVER SA OSPITAL NG BAGUIO. NAGULAT SIYA SA SINABI NG DOKTOR NA MAY TANING NA ANG BUHAY NIYA. MASYADONG LUMALAKI RAW ANG PUSO NIYA. ISANG BUWAN NA LANG ITATAGAL NIYA SA MUNDONG IBABAW AT SIYA AY MAMAMAALAM NA……

SA ISANG BUWAN NA TANING NG BUHAY NIYA, WALA NI ISANG SEGUNDO ANG SINASAYANG NI MARIO UPANG MAKASAMA ANG KANYANG PAMILYA. HINDI ALAM NG MAG-IINA  ANG NANGYAYARI KAY MARIO. PILIT TINATAKASAN NG KANILANG PADRE DE PAMILYA ANG KATOTOHANANG BUMABALOT SA  KANYANG ISIPAN.

ISANG LINGGO NA LANG. WALA NANG SINAYANG SI MARIO NA ORAS PAGKAT KAILANGAN NIYANG PALIGAYAHIN ANG KANYANG PAMILYA. SI VINNAH NAMAN AY NAG-AALALA NA KAY MARIO. ALAM NI MARIO NA ANG PANGAKO NIYA KAY PADRE CAREY AY NAGANAP NA. INALAGAAN NIYA NG HUSTO SI VINNAH AT IPINAGTANGGOL HANGGANG SA NGAYON NA MAMAMATAY NA SIYA.WALA SIYANG SINAYANG NA PANAHON UPANG HINDI MAALAGAAN ANG KANYANG ASAWANG SI VINNAH.

PILIT INAALALA NI MARIO ANG MGA MASASAYANG ARAW NINA VINNAH. INALALA RIN ANG UNANG PAGKIKITA NINA VINNAH SA ISANG PALIGSAHAN.UNANG PAGKIKITA NILA SA VALSCI. UNANG ARAW NG PAGSASAMA NILA BILANG MAGKASINTAHAN AT UNANG ARAW NILA BILANG MAG-ASAWA.

SABADO. KAPISTAHAN NG IMACULADA CONCEPTION. PUMUNTA SINA MARIO KASAMA SINA VINNAH, MGA ANAK ASAWA NILA AT MGA APO. NAGSIMBA SILA NG MATAIMTIM.ALAM NI MARIO NA MALAPIT NA ANG KANYANG KATAPUSAN KAYA NAGPAKUMPISAL NA SIYA SA PARI.SUKAT AKALAIN NI MARIO NA SI MSGR. CAREY PALA ANG NASA KUMPISALAN.SINABI NI MARIO NA PATAWARIN SIYA SA MGA KASALANAN NIYANG NAGAWA. GINAWA NA NIYA ANG LAHAT UPANG IPAGTANNGOL, ALAGAAN AT MAHALIN ANG KANYANG PAMILYA. WALA NA RIN DAW SIYANG HIHILINGIN SA DIYOS KUNDI ANG PATAWAD. SA ISIP NI MARIO, ITO ANG PATAWAD PARA SA MAMAMATAY NA………..

ALAS-8 NG GABI. TINAWAG NI MARIO SI VINNAH AT ANG KANYANG MGA ANAK. PUMUNTA SILA SA ISANG KUBO NA KUNG SAAN SILA MAKAKAPAG-USAP BILANG ISANG PAMILYA. UNANG TININGNAN NI MARIO ANG LANGIT. SA ISANG TANAWIN MAKIKITA ANG LANGIT NA BALOT NG KADILIMAN AT MAY SINAG NNG BUWAN SA GITNA.TININGNAN DIN NIYA ANG KANYANG ASAWANG SI VINNAH. AT PANGHULI ANG KANYANG MGA ANAK.

SA KADILIMAN NG GABI AT KATAHIMIKAN NG PALIGID AY MARIRINIG ANG IYAKAN. IYAKAN DAHIL IPINAGTAPAT NA NI MARIO SA PAMILYA ANG KANYANG SAKIT. ALAM NA RIN NI MARIO NA ILANG SAGLIT NA LANG AY MAPAPATID NA ANG KANYANG HININGA.SA KANDUNGAN NI VINNAH AY NAKAHIMLAY SI MARIO. SA HULING SANDALI AY PINABAUNAN SIYA NI VINNAH NG ISANG MATAMIS NA HALIK. PAGKATAPOS NITO AY MALAYANG IPINIKIT NI MARIO ANG KANYANG MGA MATA NA PARANG NATUTULOG LANG.

 PART 3

INUWI NINA VINNAH ANG BANGKAY NG KANYANG ASAWA SA KAILANG TIAHAN SA VALENZUELA. DOON INAYUSAN SI MARIO.

MARAMING DUMATING SA BUROL NI MARIO. KAHIT NA MAYAMAN NA, MARAMI PARING MGA MARALITANG DUMALO. ITO AY KANILANG PASASALAMAT SA KABAITANG IPINAPAKITA SA KANILA NI MARIO. NANDYAN DIN ANG MGA KAKLASE NINA MARIO AT VINNAH. ANG GRUPONG SPICE GIRLS NA SINA BYRON, BRYAN, EJ, DAN, JAMESH, VINCE AT ANG DATING TRAYDOR NGUNIT PINATAWAD NANINA MARIO NA SI JOSUE. PILIT ISINASANTABI NI VINNAH ANG PAGDADALAMHATI SA PAGPANAW NG KANYANG KABIYAK. INAALIW SIYA NG KANYANG MGA ANAK. PATI NAANG DATING MAY GUSTO SA KANYA AT PAI NA SI MSGR. CAREY. ALAM NI CAREY NA NAGING MASAYA SI VINNAH SA PILING NI MARIO. TALAGANG TINUPAD NI MARIO ANG KANYANG PANGAKO. WALANG ARAW AT GABI NA BINANTAYAN NI VINNAH ANG NAKAHIMLAY NA KATAWAN NI MARIO.NAPAGDESISYUNAN NI VINNAH NA HINDI I-CREMATE ANG BANGKAY NG KANYANG ASAWA BAGKUS AY ILIBING SIYA SA KANILANG MUSULEO.

LINGGO. ITO ANG ARAW NA AALIS NA SI MARIO AT PUPUNTA NA SA KANYANG HULING HANTUNGAN. MAAGANG NAGHANDA SI VINNAH AT ANG KANYANG MGA ANAK. MEDYO MAY KALAYUAN ANG SIMBAHAN PARA SA HULING PAGBABASBAS NG KATAWAN NI MARIO.

IKA -1O NG UMAGA. SA SALIW NG KANTANG AVE MARIA AY TUMUNGO ANG PRUSISYON NG KARO SA SIMBAHAN NG SAN DIEGO DE ALCALA SA VALENZUELA. PAGKATAPOS NG MISA AY PUMUNTA SILA SA PAARALANG PINASUKAN NI MARIO NOONG SIYA AY NASA ELEMENTARYA PA.SUMUSUD AY ANG MATAAAS NA PAARALANG PANG-AGHAM NG VALENUELA NA KUNG SAAN NAGING KAKLASE NI MARIO SI VINNAH. AT ANG HULING DESTINASYON AY SA HULING HANTUNGAN NI MARIO SA VALENZUELA MEMORIAL PARK. SA SALIW NG KANTANG MY LOVE WILL SEE YOU THROUGH NA KANILANG THEME SONG, IBINABA ANG LABI NI MARIO MULA SA KARO. SA HULING SANDALI AY NAKITA NI VINNAH ANG  MAALIWALAS NA MUKHA NI MARIO. PINABAUNAN NIYA ITO NG FLYING KISS NA PARANG DATI NUNG SILA AY MAGKASINTAHAN PA LAMANG. PAGKATAPOS NITO AY IPINASOK NA SI MARIO SA NITSO AT SINARHAN. TODO ANG IYAK NG MGA KAMAG-ANAK NI MARIO.

MAHAL NA MAHAL NI VINNAH SI MARIO. KAHIT NAMAYAPA NA ANG KANYANG ASAWA. NI ISANG BESES AY HINDI PA SIYA NAGHANAP NG PAPALIT SA POSISYON NI MARIO.ALAM NIYA KUNG GAANO NIYA KAMAHAL SI MARIO. KUNG HINDI SINABI NI CAREY NOONG UNA PA LAMANG NA PINAPAYAGAN NIYA SINA MARIO AY HINDI NIYA MAPAPANGASAWA SI MARIO AT MAPUPUNTA SIYA SA IBANG LALAKI NA HINDI  NIYA NAMAN MAHAL. HINDI RIN SIYA MAGKAKAANAK NG APAT NA MABABAIT AT MABUBUTING BATA KUNG HINDI DAHIL KAY MARIO. ANG MGA TAONG INILAGI NI MARIO SA MUNDO AY PATUNGKOL SA PAGMAMAHAL NI MARIO KAY VINNAH.

NASA EDAD 70 NA SI VINNAH NGUNIT MALAKAS PA SIYA. KAYA NAABUTAN PA NIYA ANG MGA APONG NASA KOLEHIYO NA. HANGGANG SA ISANG ARAW AY NAMAYAPA NA SIYA. DAHIL KAHIT MALAKS PA SIYA AY LAGI NIYANG IPINAPANALANGIN SA DIYOS NA SANA AY KUHANIN NA SIYA UPANG MAGKASAMA NA  SILA NINA MARIO. TANGGAP ITO NG KANYANG MGA ANAK. ALAM NG KANYANG MGA ANAK KUNG GAANO NAGMAMAHALAN ANG DALAWA.

NGAYONG NASA LANGIT NA ANG DALAWA, ANG KANILANG PAGMAMAHALAN AY HINDI NAWAWALA BAGKUS AY LALO PANG TUMITIBAY. NANINIWALA ANG DALAWA NA MAKIKITA RIN NILA ANG KANILANG MGA MAHAL SA BUHAY SA DARATING NA MGA ARAW.

ITO ANG WALANG KAMATAYANG PAGMAMAHALAN NINA MARIO AT VINNAH

THE END

solution problems

A chemist has on hand a supply of a 50% methyl alchohol solution and a 20% methyl alcohol solution.  How much of each kind should he mix to produce 300 grams of a 30% methyl alcohol solution?

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The micro-world of Solutions

Solution problems are "chemistry-like" problems.  They are frequently stated using the terms solution and concentration (or strength), and may refer toconcentration as a percent (%) solution (of the principal ingredient).

To make these concepts concrete, consider the following simple procedure: we mix 8 oz. of water with 2 oz. of salt to obtain a solution of salt in water (commonly called "saltwater" or "brine"):

In this picture (as in all to follow), we show the constituents of the mixture as being "separated" so we can see their numerical relationshops.  Of course, in a real brine solution, the salt and water would be completely mixed and indistinguishable one from the other.

The salt is referred to as the solute, the water is the solvent, and the resulting mixture as the solution. We need a way of quantifying the concentration of salt (that is, a number that expresses how salty the salt water really is).  "Concentration" is defined to be

the ratio of the amount of solute (salt), to the total amount of solution (water + salt)

In this case, the total amount of solution is 8 + 2 = 10 oz.  Since the amount of salt is 2 oz., the concentration of salt in this solution will be 2/10 = .20 (or 20%).

Note that concentration is usually expressed as a percent (obtained from the decimal concentration by multiplying by 100).  In general, in the statement of problems and their answers, concentrations will be expressed using %. But in any calculations involving concentrations, we will always use the decimal form of the concentration.

In the next example, we are presented with 100 grams of a 20% solution of salt in water:

It is reasonable to ask the question "How much salt is there in this solution?".  Mathematically, we are asking: "What is 20% of 100?".  With the reminder that "of" usually is translated into the mathematical operation of "times" (multiplication), we then refine the question to "What is 20% times 100?".   Finally, by converting the 20% to decimal form, we can answer the original question:

amount of salt = (.20)(100) = 20 grams

We can compute the amount of salt (solute) in the solution by multiplying the concentration (converted to decimal form by dividing by 100) by the total amount of solution.

Note that the solute in any solution problem can be identified as that component of the solution referred to using the % sign; for example, if a problem refers to a 7% solution of pure hydrochloric acid in water, the hydrochloric acid is the solute, and the water is the solvent.

---

What numerical values represent valid concentrations?

Concentration will always be a number on the range from 0% - 100% (0.0 - 1.0 in decimal form).

A concentration of 0% (0.0) represents the absence of any solute.  We can, if we wish, refer to pure water as a 0% concentration of salt in water!

On the other hand, a 100% solution represents pure solute (no solvent).  A 100% solution of ethyl alcohol is pure (undiluted) ethyl alcohol.

Our calculational formulas still hold in these extreme cases:  if we have 13 oz. of a 100% solution of anti-freeze, the amount of anti-freeze is given by (1.0)(13) = 13 oz. of anti-freeze (naturally!).  On the other hand, 13 oz. of a 0% solution of anti-freeze contains (0.0)(13) = 0 oz. of anti-freeze (naturally!).

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The vocabulary of solutions
 

Word	Definition
solution	a mixture of two chemicals or substances
solute	that component of a solution that is referred to by its concentration (strength), usually as a %
solvent	the non-solute component of the solution, sometimes water
concentration	a measure of the strength of the solute in the solution, given by the formula: (concentration) =(amount of solute)                          -------------------                      (total amount of solution) The resulting number is usually expressed as a % by multiplying by 100.  The mathematical model for solutions According to the above discussion, we measure concentration by the following formula: (concentration) =(amount of solute)                                                     -------------------                                                 (total amount of solution) By rearranging this equation, we have (amount of solute) = (concentration)(total amount of solution) It was this form of the math model that was used in the opening discussion to compute the amount of salt in 100 grams of a 20% sulution of salt in water: amount of salt = (.20)(100) = 20 grams This is the math model that will be used extensively in what follows. Concentration as used in this formula (and in any equation) will be the decimal form of concentration (a number on the range 0.0 - 1.0).   Combining two or more solutions So far, we have discussed situations involving only one solution at a time.  Most solution word problems create a little more interest by referring to two (or more) solutions of different concentrations that are combined to form a new solution of yet another concentration. For example, consider the following scenario: 10 oz. of a 20% solution of salt in water (solution A) is combined with 12 oz of a 50% solution of salt in water (solution B) to create a combined solution (solution C). It is not too hard to figure out the amount of salt in solution C: (amount of salt in solution C) = (amount of salt in solution A) + (amount of salt in solution B) = (.20)(10) + (.50)(12) = 2 + 6 = 8 oz. of salt In fact, we can also compute the concentration of salt in solution C. Since the total amount of solution C is 10 + 12 = 22 oz., of which 8 oz. is salt (we just computed that), we have: (concentration of salt in solution C) = 8/22 = .36   (36%) Summing up, we can state that the combined solution (solution C) consists of 22 oz. of a 36% solution of salt in water.      The Rough Equation for solution problems Consider our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? There is a fundamental property possesed by this problem situation (and all others like it).  We have already implicitly used this property in the above discussion "Combining Solutions".  Note well: the first component of the solution has a certain amount of alcohol, the second component has a certain amount of alcohol, and we are combining these two solutions.  It stands to reason that, since this process neither creates or destroys alcohol, the amount of alcohol in the combined solution will be the sum of the amounts of alcohol in the constituent solutions This observation is basic to the solution of all solution problems, and forms the basis for creating the Solution Problem Rough Equation. We call it the Solute Equation (amount of solute in solution 1) + (amount of solute in solution 2) = (amount of solute in combined solution) The corresponding rough equation for our opening problem will be: (amt. alcohol in the 50% sol.) + (amt. alcohol in the 20% sol.) = (amt. alcohol in the 30% sol.) From English to Algebra In the above exercises, you were able to use the math model to compute concentration (or, in its other form, to compute the amount of solute) in situations where only numerical quantities were involved.  In the solution of algebra problems, the various elements of the problem are not always expressible numerically, because of the presence of an "unknown".  Consider again our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? The final solution of this problem will start with a statement that identifies one of the unknowns of the problem, for example: let x = number of oz. of 50% alcohol solution to be used Accordingly, in the course of solving the problem, x can be used, wherever it is needed, to represent the amount of 50% alcohol to be used. But how much pure alcohol is in this x ounces of 50% alcohol?  Of course, if we knew that x = 100, we could readily compute the amount of pure alcohol: (.50)(100) = 50 oz. of alcohol.  But we don't know what x is (after all, it is the unknown!). The best we can do is multiply .50 times the x and express the result as an algebraic expression, not a number.  So what we get is (.50)(x) = .50x    (oz. of pure alcohol in x ounces of 50% alcohol solution) This may not seem very informative, but it is: (1) correct (by virtue of the math model) (2) the best we can do for now (3) very handy in finding out later exactly what x is! More English to Algebra Unknowns can be combined in other ways to create "correct" algebraic expressions to be used in our "calculations". Suppose we mix 10 oz. of solution A with a certain amount of solution B to obtain 30 oz. of the combined solution. How much solution B was used? I think we would all agree that the answer is 20 oz. How did you get that answer?  You probably did it very quickly, almost without thinking about it, as follows: (amt. of solution B) = (combined amt.) - (amt. of solution A) = 30 - 10 = 20 oz. On the other hand, suppose we don't know the actual amount of solution A -- only that it is an unknown which we have called x. How do we now answer the above question? By the very same process, except that now we cannot actually do the subtraction, but only indicate it symbolically. So the answer is: 30 - x     (oz.) This is correct, and the best we can do. Returning again to our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? As before, we start with: let x = amount of 50% alcohol to be used Now we can write an expression for the amount of 20% alcohol to be used: 300 - x     (grams) Solving solution problems With the above preparation, we are ready to attack our opening problem, and see it through to completion: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?  Step 1: Identify what it is we are looking for. According to the problem statement, we need to identify two things: (1) the amount of 50% alcohol, and (2) the amount of 20% alcohol to be combined to form the 30% solution. We will be using a method of solution that involves only one unknown.   Since one unknown cannot stand for two things at once, we have to choose (the choice is arbitrary).  I made the choice: #1: let x = the amount of 50% alcohol to be used  Step 2: Draw a picture for problems involving lengths or distances. This problem does not, so: #2: N/A  Step 3: (1)  Write the rough equation, based on the Solute Equation, as stated above: #3: rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution) = (amount of alcohol in 30% solution) (2)  Refine the rough equation into a finished equation. We do so by making repeated use of the math model: (amount of solute) = (concentration)(total amount of solution) This will turn our rough equation into the following: (conc. of 50% solution)(amt. of 50% solution) + (conc. of 20% solution)(amt. of 20% solution) = (conc. of 30% solution)(amt. of 30% solution) We are now ready to refine further, by making use of things we know: (1) The various concentrations are easy: .50, .20, and .30 respectively. (2) The amount of 50% solution is also easy: it is simply x (see the "let statement"!). (3) But what about the amount of 20% solution?  Since the total amount is to be 300 ozs.,      that leaves 300 - x for the amount of 20% solution. Assembling all this information, we arrive at the finished equation: .50x + .20(300 - x) = .30(300) Step 4: Solve and check the finished equation: #4: .50x + (.20)(300) - .20x = 90 .30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90   (it checks) Step 5: State the answer. Since the question was "how much of each solution must we mix", we must provide two "answers" . The answer to the amount of 50% solution is given by the solution to our equation: x = 100 oz..  What about the amount of 20% solution?  Since there is to be 300 oz. altogether, this leaves 200 oz. for the 20% solution.  Our answer looks like this: #5: 100 oz. of 50% methyl alcohol, 200 oz. of 30% methyl alcohol. Note that we can check our answer by doing the following calculations:   amt. alcohol in 50% solution: 50% of 100 oz. = 50 oz of alcohol amt. alcohol in 20% solution: 20% of 200 oz. = 40 oz. of alcohol for a total of 90 oz of alcohol in the resulting mixture. What will be the concentration of alcohol in the resulting mixture? It will be 90/300 = .30  (30%) which satisfies the conditions of the problem, that we are to create 300 oz. of a 30% solution. Recapitulation of the finished solution We repeat the whole solution as it should be written out by the solver: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?   #1: let x = the amount of 50% alcohol to be used #2:  N/A #3: rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution) = (amount of alcohol in 30% solution) finished equation: .50x + .20(300 - x) = .30(300) #4: .50x + (.20)(300) - .20x = 90 .30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90    (it checks) #5:  100 oz. of 50% methyl alcohol, 200 oz. of 20% methyl alcohol

solution problems

Solution Problems

A chemist has on hand a supply of a 50% methyl alchohol solution and a 20% methyl alcohol solution.  How much of each kind should he mix to produce 300 grams of a 30% methyl alcohol solution?

---

The micro-world of Solutions

Solution problems are "chemistry-like" problems.  They are frequently stated using the terms solution and concentration (or strength), and may refer toconcentration as a percent (%) solution (of the principal ingredient).

To make these concepts concrete, consider the following simple procedure: we mix 8 oz. of water with 2 oz. of salt to obtain a solution of salt in water (commonly called "saltwater" or "brine"):

In this picture (as in all to follow), we show the constituents of the mixture as being "separated" so we can see their numerical relationshops.  Of course, in a real brine solution, the salt and water would be completely mixed and indistinguishable one from the other.

The salt is referred to as the solute, the water is the solvent, and the resulting mixture as the solution. We need a way of quantifying the concentration of salt (that is, a number that expresses how salty the salt water really is).  "Concentration" is defined to be

the ratio of the amount of solute (salt), to the total amount of solution (water + salt)

In this case, the total amount of solution is 8 + 2 = 10 oz.  Since the amount of salt is 2 oz., the concentration of salt in this solution will be 2/10 = .20 (or 20%).

Note that concentration is usually expressed as a percent (obtained from the decimal concentration by multiplying by 100).  In general, in the statement of problems and their answers, concentrations will be expressed using %. But in any calculations involving concentrations, we will always use the decimal form of the concentration.

In the next example, we are presented with 100 grams of a 20% solution of salt in water:

It is reasonable to ask the question "How much salt is there in this solution?".  Mathematically, we are asking: "What is 20% of 100?".  With the reminder that "of" usually is translated into the mathematical operation of "times" (multiplication), we then refine the question to "What is 20% times 100?".   Finally, by converting the 20% to decimal form, we can answer the original question:

amount of salt = (.20)(100) = 20 grams

We can compute the amount of salt (solute) in the solution by multiplying the concentration (converted to decimal form by dividing by 100) by the total amount of solution.

Note that the solute in any solution problem can be identified as that component of the solution referred to using the % sign; for example, if a problem refers to a 7% solution of pure hydrochloric acid in water, the hydrochloric acid is the solute, and the water is the solvent.

---

What numerical values represent valid concentrations?

Concentration will always be a number on the range from 0% - 100% (0.0 - 1.0 in decimal form).

A concentration of 0% (0.0) represents the absence of any solute.  We can, if we wish, refer to pure water as a 0% concentration of salt in water!

On the other hand, a 100% solution represents pure solute (no solvent).  A 100% solution of ethyl alcohol is pure (undiluted) ethyl alcohol.

Our calculational formulas still hold in these extreme cases:  if we have 13 oz. of a 100% solution of anti-freeze, the amount of anti-freeze is given by (1.0)(13) = 13 oz. of anti-freeze (naturally!).  On the other hand, 13 oz. of a 0% solution of anti-freeze contains (0.0)(13) = 0 oz. of anti-freeze (naturally!).

---

The vocabulary of solutions
 

Word	Definition
solution	a mixture of two chemicals or substances
solute	that component of a solution that is referred to by its concentration (strength), usually as a %
solvent	the non-solute component of the solution, sometimes water
concentration	a measure of the strength of the solute in the solution, given by the formula: (concentration) =(amount of solute)                          -------------------                      (total amount of solution) The resulting number is usually expressed as a % by multiplying by 100.  The mathematical model for solutions According to the above discussion, we measure concentration by the following formula: (concentration) =(amount of solute)                                                     -------------------                                                 (total amount of solution) By rearranging this equation, we have (amount of solute) = (concentration)(total amount of solution) It was this form of the math model that was used in the opening discussion to compute the amount of salt in 100 grams of a 20% sulution of salt in water: amount of salt = (.20)(100) = 20 grams This is the math model that will be used extensively in what follows. Concentration as used in this formula (and in any equation) will be the decimal form of concentration (a number on the range 0.0 - 1.0).   Combining two or more solutions So far, we have discussed situations involving only one solution at a time.  Most solution word problems create a little more interest by referring to two (or more) solutions of different concentrations that are combined to form a new solution of yet another concentration. For example, consider the following scenario: 10 oz. of a 20% solution of salt in water (solution A) is combined with 12 oz of a 50% solution of salt in water (solution B) to create a combined solution (solution C). It is not too hard to figure out the amount of salt in solution C: (amount of salt in solution C) = (amount of salt in solution A) + (amount of salt in solution B) = (.20)(10) + (.50)(12) = 2 + 6 = 8 oz. of salt In fact, we can also compute the concentration of salt in solution C. Since the total amount of solution C is 10 + 12 = 22 oz., of which 8 oz. is salt (we just computed that), we have: (concentration of salt in solution C) = 8/22 = .36   (36%) Summing up, we can state that the combined solution (solution C) consists of 22 oz. of a 36% solution of salt in water.      The Rough Equation for solution problems Consider our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? There is a fundamental property possesed by this problem situation (and all others like it).  We have already implicitly used this property in the above discussion "Combining Solutions".  Note well: the first component of the solution has a certain amount of alcohol, the second component has a certain amount of alcohol, and we are combining these two solutions.  It stands to reason that, since this process neither creates or destroys alcohol, the amount of alcohol in the combined solution will be the sum of the amounts of alcohol in the constituent solutions This observation is basic to the solution of all solution problems, and forms the basis for creating the Solution Problem Rough Equation. We call it the Solute Equation (amount of solute in solution 1) + (amount of solute in solution 2) = (amount of solute in combined solution) The corresponding rough equation for our opening problem will be: (amt. alcohol in the 50% sol.) + (amt. alcohol in the 20% sol.) = (amt. alcohol in the 30% sol.) From English to Algebra In the above exercises, you were able to use the math model to compute concentration (or, in its other form, to compute the amount of solute) in situations where only numerical quantities were involved.  In the solution of algebra problems, the various elements of the problem are not always expressible numerically, because of the presence of an "unknown".  Consider again our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? The final solution of this problem will start with a statement that identifies one of the unknowns of the problem, for example: let x = number of oz. of 50% alcohol solution to be used Accordingly, in the course of solving the problem, x can be used, wherever it is needed, to represent the amount of 50% alcohol to be used. But how much pure alcohol is in this x ounces of 50% alcohol?  Of course, if we knew that x = 100, we could readily compute the amount of pure alcohol: (.50)(100) = 50 oz. of alcohol.  But we don't know what x is (after all, it is the unknown!). The best we can do is multiply .50 times the x and express the result as an algebraic expression, not a number.  So what we get is (.50)(x) = .50x    (oz. of pure alcohol in x ounces of 50% alcohol solution) This may not seem very informative, but it is: (1) correct (by virtue of the math model) (2) the best we can do for now (3) very handy in finding out later exactly what x is! More English to Algebra Unknowns can be combined in other ways to create "correct" algebraic expressions to be used in our "calculations". Suppose we mix 10 oz. of solution A with a certain amount of solution B to obtain 30 oz. of the combined solution. How much solution B was used? I think we would all agree that the answer is 20 oz. How did you get that answer?  You probably did it very quickly, almost without thinking about it, as follows: (amt. of solution B) = (combined amt.) - (amt. of solution A) = 30 - 10 = 20 oz. On the other hand, suppose we don't know the actual amount of solution A -- only that it is an unknown which we have called x. How do we now answer the above question? By the very same process, except that now we cannot actually do the subtraction, but only indicate it symbolically. So the answer is: 30 - x     (oz.) This is correct, and the best we can do. Returning again to our opening problem: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton? As before, we start with: let x = amount of 50% alcohol to be used Now we can write an expression for the amount of 20% alcohol to be used: 300 - x     (grams) Solving solution problems With the above preparation, we are ready to attack our opening problem, and see it through to completion: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?  Step 1: Identify what it is we are looking for. According to the problem statement, we need to identify two things: (1) the amount of 50% alcohol, and (2) the amount of 20% alcohol to be combined to form the 30% solution. We will be using a method of solution that involves only one unknown.   Since one unknown cannot stand for two things at once, we have to choose (the choice is arbitrary).  I made the choice: #1: let x = the amount of 50% alcohol to be used  Step 2: Draw a picture for problems involving lengths or distances. This problem does not, so: #2: N/A  Step 3: (1)  Write the rough equation, based on the Solute Equation, as stated above: #3: rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution) = (amount of alcohol in 30% solution) (2)  Refine the rough equation into a finished equation. We do so by making repeated use of the math model: (amount of solute) = (concentration)(total amount of solution) This will turn our rough equation into the following: (conc. of 50% solution)(amt. of 50% solution) + (conc. of 20% solution)(amt. of 20% solution) = (conc. of 30% solution)(amt. of 30% solution) We are now ready to refine further, by making use of things we know: (1) The various concentrations are easy: .50, .20, and .30 respectively. (2) The amount of 50% solution is also easy: it is simply x (see the "let statement"!). (3) But what about the amount of 20% solution?  Since the total amount is to be 300 ozs.,      that leaves 300 - x for the amount of 20% solution. Assembling all this information, we arrive at the finished equation: .50x + .20(300 - x) = .30(300) Step 4: Solve and check the finished equation: #4: .50x + (.20)(300) - .20x = 90 .30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90   (it checks) Step 5: State the answer. Since the question was "how much of each solution must we mix", we must provide two "answers" . The answer to the amount of 50% solution is given by the solution to our equation: x = 100 oz..  What about the amount of 20% solution?  Since there is to be 300 oz. altogether, this leaves 200 oz. for the 20% solution.  Our answer looks like this: #5: 100 oz. of 50% methyl alcohol, 200 oz. of 30% methyl alcohol. Note that we can check our answer by doing the following calculations:   amt. alcohol in 50% solution: 50% of 100 oz. = 50 oz of alcohol amt. alcohol in 20% solution: 20% of 200 oz. = 40 oz. of alcohol for a total of 90 oz of alcohol in the resulting mixture. What will be the concentration of alcohol in the resulting mixture? It will be 90/300 = .30  (30%) which satisfies the conditions of the problem, that we are to create 300 oz. of a 30% solution. Recapitulation of the finished solution We repeat the whole solution as it should be written out by the solver: A chemist has on hand a supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?   #1: let x = the amount of 50% alcohol to be used #2:  N/A #3: rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution) = (amount of alcohol in 30% solution) finished equation: .50x + .20(300 - x) = .30(300) #4: .50x + (.20)(300) - .20x = 90 .30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90    (it checks) #5:  100 oz. of 50% methyl alcohol, 200 oz. of 20% methyl alcohol