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A cooperative society received 18,357, 248 cartons of avocados from farmers in one season. How many times is the total value of digit 3 greater than the total value of digit 2 in the number? (4 mks)

Mathematics

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Huu hapa ni ufumbuzi wa maswali uliyotuma:

1. A Cooperative Society received 18,357,248 cartons of avocados from farmers in one season. How many times is the total value of digit 3 greater than the total value of digit 2 in the number?

Ili kujibu swali hili, tunahitaji kwanza kupata thamani ya kila tarakimu katika nambari 18,357,248.

  1. Pata thamani ya tarakimu 3.

    Tarakimu 3 iko katika nafasi ya mamia elfu.
    Thamani ya tarakimu 3 ni 3×100,000=300,0003 \times 100,000 = 300,000.

  2. Pata thamani ya tarakimu 2.

    Tarakimu 2 iko katika nafasi ya mamia.
    Thamani ya tarakimu 2 ni 2×100=2002 \times 100 = 200.

  3. Gawanya thamani ya tarakimu 3 kwa thamani ya tarakimu 2.

    Ili kujua ni mara ngapi thamani ya tarakimu 3 ni kubwa kuliko thamani ya tarakimu 2, tunagawanya:

    300,000200=1,500\frac{300,000}{200} = 1,500

    Thamani ya tarakimu 3 ni mara 1,500 zaidi ya thamani ya tarakimu 2.

    The final answer is 1,500 times\boxed{\text{1,500 times}}.

    2. A fisherman on a boat saw a fish at a depth of 5m. He noticed a bird vertically above the fish at a height of 12m from the water surface. Use a number line to determine the distance between the fish and the bird.

  4. Tambua nafasi ya samaki.

    Samaki yuko chini ya usawa wa maji kwa mita 5. Hii inaweza kuwakilishwa kama −5-5 mita kwenye mstari wa nambari.

  5. Tambua nafasi ya ndege.

    Ndege yuko juu ya usawa wa maji kwa mita 12. Hii inaweza kuwakilishwa kama +12+12 mita kwenye mstari wa nambari.

  6. Kokotoa umbali kati ya samaki na ndege.

    Umbali kati ya vitu viwili kwenye mstari wa nambari hupatikana kwa kutoa nafasi ndogo kutoka kwa nafasi kubwa, au kwa kuongeza umbali kutoka sehemu ya kuanzia (usawa wa maji) hadi kila kitu.
    Umbali = (urefu wa ndege kutoka usawa wa maji) + (kina cha samaki kutoka usawa wa maji)
    Umbali =12 m−(−5 m)= 12 \text{ m} - (-5 \text{ m})
    Umbali =12 m+5 m= 12 \text{ m} + 5 \text{ m}
    Umbali =17 m= 17 \text{ m}

    The final answer is 17 m\boxed{\text{17 m}}.

    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.

    a) The fraction of rice that was shared equally for the remaining days.

  7. Pata sehemu ya mchele iliyobaki baada ya siku ya kwanza.

    Mchele wa awali ni 1 (au 66\frac{6}{6}).
    Siku ya kwanza ilitumika 16\frac{1}{6}.
    Mchele uliobaki baada ya siku ya kwanza =1−16=56= 1 - \frac{1}{6} = \frac{5}{6}.

  8. Pata sehemu ya mchele iliyotumika siku ya pili.

    Siku ya pili ilitumika 15\frac{1}{5} ya mchele uliobaki.
    Mchele uliotumika siku ya pili =15×56=16= \frac{1}{5} \times \frac{5}{6} = \frac{1}{6}.

  9. Pata sehemu ya mchele iliyobaki baada ya siku ya pili.

    Mchele uliobaki baada ya siku ya pili =56−16=46=23= \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}.
    Hii ndiyo sehemu ya mchele iliyobaki ambayo iligawanywa kwa usawa kwa siku zilizobaki.

    The final answer is 23\boxed{\frac{2}{3}}.

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