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Musa sold a bag at sh. 160,000 making a profit of sh. 20,000. Find his cost price.

Mathematics

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11. Musa sold a bag at sh. 160,000 making a profit of sh. 20,000. Find his cost price.
Musa aliuza begi kwa sh. 160,000 akapata faida ya sh. 20,000. Tafuta bei yake ya kununulia.

  1. Tambua bei ya kuuza na faida.

    Bei ya kuuza (Selling Price) = Sh. 160,000
    Faida (Profit) = Sh. 20,000

  2. Tumia fomula ya bei ya kununulia.

    Bei ya kununulia (Cost Price) = Bei ya kuuza - Faida

    Bei ya kununulia=160,000−20,000Bei\ ya\ kununulia = 160,000 - 20,000 Bei ya kununulia=140,000Bei\ ya\ kununulia = 140,000

    Bei ya kununulia ni Sh. 140,000\boxed{Sh.\ 140,000}.

    12. Solve for p. 3p≡5(mod7)3p \equiv 5 \pmod{7}
    Tafuta thamani ya p.

  3. Tafuta namba ambayo ikizidishwa na 3, inatoa salio la 5 unapogawanya kwa 7.

    Tunaweza kujaribu namba tofauti za pp kuanzia 0 hadi 6 (kwa sababu tunafanya kazi modulo 7).

    • Ikiwa p=0p=0, 3×0=0≢5(mod7)3 \times 0 = 0 \not\equiv 5 \pmod{7}
    • Ikiwa p=1p=1, 3×1=3≢5(mod7)3 \times 1 = 3 \not\equiv 5 \pmod{7}
    • Ikiwa p=2p=2, 3×2=6≢5(mod7)3 \times 2 = 6 \not\equiv 5 \pmod{7}
    • Ikiwa p=3p=3, 3×3=9≡2(mod7)3 \times 3 = 9 \equiv 2 \pmod{7}
    • Ikiwa p=4p=4, 3×4=12≡5(mod7)3 \times 4 = 12 \equiv 5 \pmod{7} (Hii ndiyo jibu sahihi)
  4. Thibitisha jibu.

    12÷7=112 \div 7 = 1 na salio la 5. Hivyo, 3×4≡5(mod7)3 \times 4 \equiv 5 \pmod{7}.
    Thamani ya pp ni 4\boxed{4}.

    13. The probability of Sandra winning a computer game is 0.6. Find the probability of her losing the game.
    Uwezekano wa Sandra kushinda mchezo wa kompyuta ni 0.6. Tafuta uwezekano wa yeye kupoteza mchezo.

  5. Tambua uwezekano wa kushinda.

    Uwezekano wa kushinda (P(W)P(W)) = 0.6

  6. Tumia fomula ya uwezekano wa kupoteza.

    Jumla ya uwezekano wote ni 1. Hivyo, uwezekano wa kupoteza (P(L)P(L)) ni 1 minus uwezekano wa kushinda.

    P(L)=1−P(W)P(L) = 1 - P(W) P(L)=1−0.6P(L) = 1 - 0.6 P(L)=0.4P(L) = 0.4

    Uwezekano wa Sandra kupoteza mchezo ni 0.4\boxed{0.4}.

    14. Using a ruler, a pencil and a pair of compasses only, construct an angle of 75∘75^\circ.
    Kwa kutumia rula, penseli na bikari pekee, chora pembe ya 75∘75^\circ.

  7. Chora mstari wa msingi.

    Chora mstari ulionyooka AB.

  8. Chora pembe ya 60∘60^\circ.

    Kutumia A kama kituo na radius yoyote, chora upinde unaokatiza AB kwenye C. Kutumia C kama kituo na radius ileile, chora upinde unaokatiza upinde wa kwanza kwenye D. Unganisha A na D. Pembe ∠DAB=60∘\angle DAB = 60^\circ.

  9. Chora pembe ya 90∘90^\circ.

    Kutumia D kama kituo na radius ileile, chora upinde unaokatiza upinde wa kwanza kwenye E. Kutumia D na E kama vituo, chora pinde mbili zinazokatizana kwenye F. Unganisha A na F. Pembe ∠FAB=90∘\angle FAB = 90^\circ.

  10. Gawanya pembe kati ya 60∘60^\circ na 90∘90^\circ.

    Pembe kati ya mstari AD (60∘60^\circ) na AF (90∘90^\circ) ni 30∘30^\circ. Gawanya pembe hii ya 30∘30^\circ mara mbili. Kutumia D na G (mahali ambapo AF inakatiza upinde wa kwanza) kama vituo, chora pinde mbili zinazokatizana kwenye H.

  11. Chora pembe ya 75∘75^\circ.

    Unganisha A na H. Mstari AH utaunda pembe ya 60∘+15∘=75∘60^\circ + 15^\circ = 75^\circ na mstari AB.
    Hivyo, ∠HAB=75∘\boxed{\angle HAB = 75^\circ}.

    15. Solve for b: 3b÷33=273^b \div 3^3 = 27
    Tafuta thamani ya b.

  12. Tumia sheria za vipeo.

    Wakati wa kugawanya namba zenye 'base' sawa, tunatoa vipeo (exponents).

    3b−3=273^{b-3} = 27
  13. Andika 27 kama kielelezo cha 3.

    Tunajua kuwa 3×3×3=273 \times 3 \times 3 = 27. Hivyo, 27=3327 = 3^3.

    3b−3=333^{b-3} = 3^3
  14. Linganisha vipeo.

    Kwa kuwa 'base' ni sawa, vipeo lazima viwe sawa.

    b−3=3b-3 = 3
  15. Tafuta thamani ya bb.

    Ongeza 3 pande zote mbili za mlinganyo.

    b=3+3b = 3 + 3 b=6b = 6

    Thamani ya bb ni 6\boxed{6}.

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