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How many times is the total value of digit 3 greater than the total value of digit 2 in the number (4372) and (3742)?

Mathematics

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Answer

225 kg\text{225 kg}

Step-by-step solution

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Nimeona picha tena. Tayari tumeshughulikia maswali 1, 2, na 3a. Sasa tutaendelea na sehemu inayofuata ya swali la 3, ambayo inahusisha habari iliyobaki. Ingawa haijaandikwa wazi kama 'b)', tutatumia taarifa iliyobaki kukokotoa jumla ya mchele uliyonunuliwa.

3. A school bought some rice to be used for 6 days. On the first day, the school used 16\frac{1}{6} of the rice. On the second day, 15\frac{1}{5} of the remaining rice was used. The rest of the rice was shared equally for the remaining days. The school used 37.5 kg of rice on each of the remaining days.

b) Jumla ya mchele uliyonunuliwa na shule.

  1. Tambua idadi ya siku zilizobaki.

    Jumla ya siku ni 6. Siku 2 zimetumika (siku ya kwanza na ya pili).
    Siku zilizobaki =6−2=4= 6 - 2 = 4 siku.

  2. Kokotoa jumla ya mchele uliotumika kwa siku 4 zilizobaki.

    Shule ilitumia 37.5 kg ya mchele kwa kila moja ya siku 4 zilizobaki.
    Jumla ya mchele kwa siku 4 zilizobaki =4×37.5 kg=150 kg= 4 \times 37.5 \text{ kg} = 150 \text{ kg}.

  3. Tumia sehemu ya mchele iliyobaki (kutoka sehemu 'a') kupata jumla ya mchele uliyonunuliwa.

    Kutoka sehemu 'a', tuligundua kuwa 23\frac{2}{3} ya mchele ndiyo iliyobaki na kugawanywa kwa siku zilizobaki.
    Hivyo, 150 kg inawakilisha 23\frac{2}{3} ya jumla ya mchele uliyonunuliwa.
    Acha TT iwe jumla ya mchele uliyonunuliwa.

    23T=150 kg\frac{2}{3} T = 150 \text{ kg}
  4. Tatua kwa TT.

    Ili kupata TT, tunazidisha pande zote mbili kwa 32\frac{3}{2}:

    T=150 kg×32T = 150 \text{ kg} \times \frac{3}{2} T=(150÷2)×3 kgT = (150 \div 2) \times 3 \text{ kg} T=75×3 kgT = 75 \times 3 \text{ kg} T=225 kgT = 225 \text{ kg}

    Jumla ya mchele uliyonunuliwa na shule ni 225 kg.

    The final answer is 225 kg\text{225 kg}.

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