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Tap A can fill a tank in 10 minutes; tap B can fill the same tank in 20 minutes. Tap C can empty the tank in 30 minutes. The three taps are left open for 5 minutes, after which tap A is closed. How long does it take to fill the remaining part of the tank?

Mathematics

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2 parts
5

2x+y≤42x + y \le 4

x−2y≥−4x - 2y \ge -4

y≥0y \ge 0

6

65\frac{6}{5}

Step-by-step solution

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Haya, twende tukatatue maswali haya.

Part 4· 5 steps

  • Kiwango cha bomba A kujaza tanki: 110\frac{1}{10} ya tanki kwa dakika.
  • Kiwango cha bomba B kujaza tanki: 120\frac{1}{20} ya tanki kwa dakika.
  • Kiwango cha bomba C kutoa maji kwenye tanki: −130-\frac{1}{30} ya tanki kwa dakika.

  1. Tafuta kiwango cha kujaza tanki wakati mabomba yote matatu yamefunguliwa.

    Kiwango cha pamoja = Kiwango A + Kiwango B + Kiwango C

    Kiwango cha pamoja=110+120−130\text{Kiwango cha pamoja} = \frac{1}{10} + \frac{1}{20} - \frac{1}{30}

    Tafuta kigawo kidogo cha shirika (LCM) cha 10, 20, na 30, ambacho ni 60.

    Kiwango cha pamoja=660+360−260=6+3−260=760 ya tanki kwa dakika\text{Kiwango cha pamoja} = \frac{6}{60} + \frac{3}{60} - \frac{2}{60} = \frac{6+3-2}{60} = \frac{7}{60} \text{ ya tanki kwa dakika}

  2. Tafuta sehemu ya tanki iliyojazwa katika dakika 5 za kwanza.

    Sehemu iliyojazwa=Kiwango cha pamoja×5 dakika=760×5=3560=712\text{Sehemu iliyojazwa} = \text{Kiwango cha pamoja} \times 5 \text{ dakika} = \frac{7}{60} \times 5 = \frac{35}{60} = \frac{7}{12}

  3. Tafuta sehemu ya tanki iliyobaki kujazwa.

    Sehemu iliyobaki=1−Sehemu iliyojazwa=1−712=512\text{Sehemu iliyobaki} = 1 - \text{Sehemu iliyojazwa} = 1 - \frac{7}{12} = \frac{5}{12}

  4. Bomba A limefungwa.

    Tafuta kiwango cha kujaza tanki wakati mabomba B na C yamefunguliwa.
    Kiwango kipya cha pamoja = Kiwango B + Kiwango C

    Kiwango kipya cha pamoja=120−130\text{Kiwango kipya cha pamoja} = \frac{1}{20} - \frac{1}{30}

    Tafuta LCM ya 20 na 30, ambayo ni 60.

    Kiwango kipya cha pamoja=360−260=160 ya tanki kwa dakika\text{Kiwango kipya cha pamoja} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \text{ ya tanki kwa dakika}

  5. Tafuta muda unaohitajika kujaza sehemu iliyobaki ya tanki.

    Muda=Sehemu iliyobakiKiwango kipya cha pamoja=512160=512×60=5×5=25 dakika\text{Muda} = \frac{\text{Sehemu iliyobaki}}{\text{Kiwango kipya cha pamoja}} = \frac{\frac{5}{12}}{\frac{1}{60}} = \frac{5}{12} \times 60 = 5 \times 5 = 25 \text{ dakika}

    Muda unaohitajika kujaza sehemu iliyobaki ya tanki ni 25 dakika.

Part 5

2x+y≤42x + y \le 4

Ili kupata usawa unaoelezea eneo lisilo na kivuli, tutapata milinganyo ya mistari mitatu na kisha kuamua usawa sahihi.

  1. Tafuta mlinganyo wa mstari wa kwanza (kutoka juu kushoto kwenda chini kulia).

    Mstari huu unapita kwenye pointi (0,4)(0, 4) na (2,0)(2, 0).
    Mteremko m=0−42−0=−42=−2m = \frac{0-4}{2-0} = \frac{-4}{2} = -2.
    Y-intercept c=4c = 4.
    Mlinganyo wa mstari ni y=−2x+4y = -2x + 4, au 2x+y=42x + y = 4.
    Eneo lisilo na kivuli liko chini ya mstari huu, kwa hivyo usawa ni 2x+y≤42x + y \le 4.

  2. Tafuta mlinganyo wa mstari wa pili (kutoka chini kushoto kwenda juu kulia).

    Mstari huu unapita kwenye pointi (−4,0)(-4, 0) na (0,2)(0, 2).
    Mteremko m=2−00−(−4)=24=12m = \frac{2-0}{0-(-4)} = \frac{2}{4} = \frac{1}{2}.
    Y-intercept c=2c = 2.
    Mlinganyo wa mstari ni y=12x+2y = \frac{1}{2}x + 2. Kuzidisha kwa 2 kunatoa 2y=x+42y = x + 4, au x−2y=−4x - 2y = -4.
    Eneo lisilo na kivuli liko juu ya mstari huu, kwa hivyo usawa ni x−2y≥−4x - 2y \ge -4.

  3. Tafuta mlinganyo wa mstari wa tatu (mstari wa x-axis).

    Mstari huu ni y=0y = 0.
    Eneo lisilo na kivuli liko juu ya mstari huu, kwa hivyo usawa ni y≥0y \ge 0.

    Usawa unaoelezea eneo lisilo na kivuli ni:

    2x+y≤4\boxed{2x + y \le 4} x−2y≥−4\boxed{x - 2y \ge -4} y≥0\boxed{y \ge 0}

Part 6

65\frac{6}{5}

Tutatathmini usemi bila kutumia kikokotoo.

12 of 312+112(212−23)34 of 212÷12\frac{\frac{1}{2} \text{ of } 3\frac{1}{2} + 1\frac{1}{2} (2\frac{1}{2} - \frac{2}{3})}{\frac{3}{4} \text{ of } 2\frac{1}{2} \div \frac{1}{2}}

  1. Badilisha namba mchanganyiko kuwa sehemu zisizofaa.

    312=723\frac{1}{2} = \frac{7}{2}
    112=321\frac{1}{2} = \frac{3}{2}
    212=522\frac{1}{2} = \frac{5}{2}

  2. Tathmini mabano kwenye nambari.

    212−23=52−23=156−46=1162\frac{1}{2} - \frac{2}{3} = \frac{5}{2} - \frac{2}{3} = \frac{15}{6} - \frac{4}{6} = \frac{11}{6}

  3. Tathmini "of" (kuzidisha) na kuzidisha kwenye nambari.

    12 of 312=12×72=74\frac{1}{2} \text{ of } 3\frac{1}{2} = \frac{1}{2} \times \frac{7}{2} = \frac{7}{4} 112(212−23)=32×116=3312=1141\frac{1}{2} (2\frac{1}{2} - \frac{2}{3}) = \frac{3}{2} \times \frac{11}{6} = \frac{33}{12} = \frac{11}{4}

  4. Jumlisha nambari.

    Nambari=74+114=184=92\text{Nambari} = \frac{7}{4} + \frac{11}{4} = \frac{18}{4} = \frac{9}{2}

  5. Tathmini "of" (kuzidisha) kwenye denominata.

    34 of 212=34×52=158\frac{3}{4} \text{ of } 2\frac{1}{2} = \frac{3}{4} \times \frac{5}{2} = \frac{15}{8}

  6. Tathmini kugawa kwenye denominata.

    Denominata=158÷12=158×21=308=154\text{Denominata} = \frac{15}{8} \div \frac{1}{2} = \frac{15}{8} \times \frac{2}{1} = \frac{30}{8} = \frac{15}{4}

  7. Gawanya nambari kwa denominata.

    Jibu=92154=92×415=3630=65\text{Jibu} = \frac{\frac{9}{2}}{\frac{15}{4}} = \frac{9}{2} \times \frac{4}{15} = \frac{36}{30} = \frac{6}{5}

    Jibu la usemi ni 65\boxed{\frac{6}{5}}.

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