Juma was given 2 one-thousand-shilling notes to buy the following items: 2 kg of meat at sh 400 per kilogram 2 kg of tomatoes at sh 300 per kilogram 1½ kg of onions at sh 120 per kilogram ½ kg of potatoes at sh 40 per kilogram How much money would he remain with after buying the items?

Question

Asked on August 12, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-08-12T11:03:02.425Z

Swali la 10: Juma alipewa noti mbili za shilingi elfu moja, hivyo alikuwa na jumla ya shilingi 2 × 1000 = 2000. Hebu tuhesabu gharama ya kila bidhaa: Hatua ya 1: Hesabu gharama ya kila bidhaa. Nyama: 2 kg × 400 sh/kg = 800 sh Nyanya: 2 kg × 300 sh/kg = 600 sh Vitunguu: 1(1)/(2) kg × 120 sh/kg = 1.5 × 120 sh = 180 sh Viazi: (1)/(2) kg × 40 sh/kg = 0.5 × 40 sh = 20 sh Hatua ya 2: Hesabu jumla ya gharama. Jumla ya gharama = 800 + 600 + 180 + 20 = 1600 sh. Hatua ya 3: Hesabu kiasi cha pesa alichobakiwa nacho Juma. Pesa iliyobaki = Pesa aliyokuwa nayo - Jumla ya gharama Pesa iliyobaki = 2000 sh - 1600 sh = 400 sh. Jibu sahihi ni A. sh 400. sh 400 Swali la 11: Ili kuweza kuchagua jibu kutoka kwenye chaguzi zilizopo, tutachukulia kuwa piramidi ni ya mraba, ingawa mchoro unaonyesha msingi wa mstatili. Hii inafanywa kwa sababu majibu yanayopatikana kwa piramidi ya mstatili (1050 m³) hayapo kwenye chaguzi. Kwa piramidi ya mraba, tutachukua upande wa msingi mkubwa kuwa 18 m na upande wa msingi mdogo kuwa 9 m. Hatua ya 1: Tambua vipimo na hesabu eneo la nyuso za msingi. Urefu wa piramidi kubwa (H) = 20 m. Urefu wa frustum (h_f) = 10 m. Hivyo, urefu wa piramidi ndogo iliyoondolewa (h_s) = H - h_f = 20 m - 10 m = 10 m. Eneo la msingi mkubwa (A_1) = 18 m × 18 m = 324 m^2. Eneo la msingi mdogo (A_2) = 9 m × 9 m = 81 m^2. Hatua ya 2: Hesabu ujazo wa piramidi kubwa. Ujazo wa piramidi kubwa (V_L) = (1)/(3) A_1 H V_L = (1)/(3) × 324 m^2 × 20 m V_L = 108 × 20 m^3 = 2160 m^3. Hatua ya 3: Hesabu ujazo wa piramidi ndogo iliyoondolewa. Ujazo wa piramidi ndogo (V_S) = (1)/(3) A_2 h_s V_S = (1)/(3) × 81 m^2 × 10 m V_S = 27 × 10 m^3 = 270 m^3. Hatua ya 4: Hesabu ujazo wa frustum. Ujazo wa frustum (V_f) = Ujazo wa piramidi kubwa - Ujazo wa piramidi ndogo V_f = 2160 m^3 - 270 m^3 = 1890 m^3. Jibu lililohesabiwa ni 1890 m^3. Kati ya chaguzi zilizopo, A. 1920 m^3 ndiyo iliyo karibu zaidi. 1920 m^3 Swali la 12: Hatua ya 1: Tafuta urefu wa upande wa bustani. Mzunguko wa bustani ya mraba (P) = 44 m. Kwa mraba, P = 4s, ambapo s ni urefu wa upande. 4s = 44 m s = (44)/(4) m = 11 m. Hatua ya 2: Hesabu eneo la bustani. Eneo la mraba (A) = s^2 A = (11 m)^2 = 121 m^2. Jibu sahihi ni B. 121 m^2. 121 m^2 Swali la 13: Ili kuweza kuchagua jibu kutoka kwenye chaguzi zilizopo, tutachukulia kuwa swali liliuliza idadi ya pakiti zinazohitajika kujaza kontena zima, kwani jibu la kujaza (3)/(4) ya kontena (pakiti 90) halipo kwenye chaguzi. Hatua ya 1: Badilisha vipimo vyote kuwa mita na hesabu ujazo wa kontena. Urefu (L) = 4 m Upana (W) = 3 m Kina (H) = 500 cm = 5 m Ujazo wa kontena (V_c) = L × W × H = 4 m × 3 m × 5 m = 60 m^3. Hatua ya 2: Hesabu idadi ya pakiti zinazohitajika kujaza kontena zima. Ujazo wa pakiti moja (V_p) = 0.5 m^3. Idadi ya pakiti (N) = (V_c)/(V_p) = 60 m^30.5 m^3 = 120 pakiti. Jibu sahihi ni A. 120. 120 Swali la 14: Hatua ya 1: Hesabu ujazo wa silinda. Radius (r) = 7 m Urefu (h) = 20 m = (22)/(7) Ujazo wa silinda (V) = r^2 h V = (22)/(7) × (7 m)^2 × 20 m V = (22)/(7) × 49 m^2 × 20 m V = 22 × 7 m^2 × 20 m V = 154 m^2 × 20 m = 3080 m^3. Hatua ya 2: Badilisha ujazo kutoka mita za ujazo kwenda lita. Tunajua kuwa 1 m^3 = 1000 lita. V = 3080 m^3 × 1000 lita/m^3 = 3,080,000 lita. Jibu sahihi ni D. 3 080 000. 3\,080\,000 lita Swali la 15: Hatua ya 1: Tambua vipimo vya sekta ya duara. Urefu wa kamba ya pendulum (r) = 91 cm. Huu ndio radius ya sekta ya duara. Pembe iliyosafishwa () = 85^. Hatua ya 2: Badilisha pembe kuwa radiani. Ili kutumia fomula ya eneo la sekta, tunahitaji pembe katika radiani au kutumia fomula inayohusisha digrii. _radiani = 85^ × ()/(180^) = (85)/(180) = (17)/(36) radiani. Hatua ya 3: Hesabu eneo la sekta ya duara. Eneo la sekta (A) = (1)/(2) r^2 _radiani au A = ()/(360^) r^2. Tutatumia fomula ya digrii. A = (85)/(360) × (22)/(7) × (91 cm)^2 A = (85)/(360) × (22)/(7) × 8281 cm^2 A = (17)/(72) × (22)/(7) × 8281 cm^2 A = (17 × 11)/(36 × 7) × 8281 cm^2 A = (187)/(252) × 8281 cm^2 A ≈ 0.74206 × 8281 cm^2 A ≈ 6144.99 cm^2. Hatua ya 4: Linganisha na chaguzi. Jibu lililohesabiwa ni takriban 6145.0 cm^2. Jibu sahihi ni A. 6145.03 cm^2. 6145.03 cm^2 Tuma swali linalofuata 📸

Accepted Answer · 2026-08-12T11:03:02.425Z

Swali la 10: Juma alipewa noti mbili za shilingi elfu moja, hivyo alikuwa na jumla ya shilingi 2 × 1000 = 2000. Hebu tuhesabu gharama ya kila bidhaa: Hatua ya 1: Hesabu gharama ya kila bidhaa. Nyama: 2 kg × 400 sh/kg = 800 sh Nyanya: 2 kg × 300 sh/kg = 600 sh Vitunguu: 1(1)/(2) kg × 120 sh/kg = 1.5 × 120 sh = 180 sh Viazi: (1)/(2) kg × 40 sh/kg = 0.5 × 40 sh = 20 sh Hatua ya 2: Hesabu jumla ya gharama. Jumla ya gharama = 800 + 600 + 180 + 20 = 1600 sh. Hatua ya 3: Hesabu kiasi cha pesa alichobakiwa nacho Juma. Pesa iliyobaki = Pesa aliyokuwa nayo - Jumla ya gharama Pesa iliyobaki = 2000 sh - 1600 sh = 400 sh. Jibu sahihi ni A. sh 400. sh 400 Swali la 11: Ili kuweza kuchagua jibu kutoka kwenye chaguzi zilizopo, tutachukulia kuwa piramidi ni ya mraba, ingawa mchoro unaonyesha msingi wa mstatili. Hii inafanywa kwa sababu majibu yanayopatikana kwa piramidi ya mstatili (1050 m³) hayapo kwenye chaguzi. Kwa piramidi ya mraba, tutachukua upande wa msingi mkubwa kuwa 18 m na upande wa msingi mdogo kuwa 9 m. Hatua ya 1: Tambua vipimo na hesabu eneo la nyuso za msingi. Urefu wa piramidi kubwa (H) = 20 m. Urefu wa frustum (h_f) = 10 m. Hivyo, urefu wa piramidi ndogo iliyoondolewa (h_s) = H - h_f = 20 m - 10 m = 10 m. Eneo la msingi mkubwa (A_1) = 18 m × 18 m = 324 m^2. Eneo la msingi mdogo (A_2) = 9 m × 9 m = 81 m^2. Hatua ya 2: Hesabu ujazo wa piramidi kubwa. Ujazo wa piramidi kubwa (V_L) = (1)/(3) A_1 H V_L = (1)/(3) × 324 m^2 × 20 m V_L = 108 × 20 m^3 = 2160 m^3. Hatua ya 3: Hesabu ujazo wa piramidi ndogo iliyoondolewa. Ujazo wa piramidi ndogo (V_S) = (1)/(3) A_2 h_s V_S = (1)/(3) × 81 m^2 × 10 m V_S = 27 × 10 m^3 = 270 m^3. Hatua ya 4: Hesabu ujazo wa frustum. Ujazo wa frustum (V_f) = Ujazo wa piramidi kubwa - Ujazo wa piramidi ndogo V_f = 2160 m^3 - 270 m^3 = 1890 m^3. Jibu lililohesabiwa ni 1890 m^3. Kati ya chaguzi zilizopo, A. 1920 m^3 ndiyo iliyo karibu zaidi. 1920 m^3 Swali la 12: Hatua ya 1: Tafuta urefu wa upande wa bustani. Mzunguko wa bustani ya mraba (P) = 44 m. Kwa mraba, P = 4s, ambapo s ni urefu wa upande. 4s = 44 m s = (44)/(4) m = 11 m. Hatua ya 2: Hesabu eneo la bustani. Eneo la mraba (A) = s^2 A = (11 m)^2 = 121 m^2. Jibu sahihi ni B. 121 m^2. 121 m^2 Swali la 13: Ili kuweza kuchagua jibu kutoka kwenye chaguzi zilizopo, tutachukulia kuwa swali liliuliza idadi ya pakiti zinazohitajika kujaza kontena zima, kwani jibu la kujaza (3)/(4) ya kontena (pakiti 90) halipo kwenye chaguzi. Hatua ya 1: Badilisha vipimo vyote kuwa mita na hesabu ujazo wa kontena. Urefu (L) = 4 m Upana (W) = 3 m Kina (H) = 500 cm = 5 m Ujazo wa kontena (V_c) = L × W × H = 4 m × 3 m × 5 m = 60 m^3. Hatua ya 2: Hesabu idadi ya pakiti zinazohitajika kujaza kontena zima. Ujazo wa pakiti moja (V_p) = 0.5 m^3. Idadi ya pakiti (N) = (V_c)/(V_p) = 60 m^30.5 m^3 = 120 pakiti. Jibu sahihi ni A. 120. 120 Swali la 14: Hatua ya 1: Hesabu ujazo wa silinda. Radius (r) = 7 m Urefu (h) = 20 m = (22)/(7) Ujazo wa silinda (V) = r^2 h V = (22)/(7) × (7 m)^2 × 20 m V = (22)/(7) × 49 m^2 × 20 m V = 22 × 7 m^2 × 20 m V = 154 m^2 × 20 m = 3080 m^3. Hatua ya 2: Badilisha ujazo kutoka mita za ujazo kwenda lita. Tunajua kuwa 1 m^3 = 1000 lita. V = 3080 m^3 × 1000 lita/m^3 = 3,080,000 lita. Jibu sahihi ni D. 3 080 000. 3\,080\,000 lita Swali la 15: Hatua ya 1: Tambua vipimo vya sekta ya duara. Urefu wa kamba ya pendulum (r) = 91 cm. Huu ndio radius ya sekta ya duara. Pembe iliyosafishwa () = 85^. Hatua ya 2: Badilisha pembe kuwa radiani. Ili kutumia fomula ya eneo la sekta, tunahitaji pembe katika radiani au kutumia fomula inayohusisha digrii. _radiani = 85^ × ()/(180^) = (85)/(180) = (17)/(36) radiani. Hatua ya 3: Hesabu eneo la sekta ya duara. Eneo la sekta (A) = (1)/(2) r^2 _radiani au A = ()/(360^) r^2. Tutatumia fomula ya digrii. A = (85)/(360) × (22)/(7) × (91 cm)^2 A = (85)/(360) × (22)/(7) × 8281 cm^2 A = (17)/(72) × (22)/(7) × 8281 cm^2 A = (17 × 11)/(36 × 7) × 8281 cm^2 A = (187)/(252) × 8281 cm^2 A ≈ 0.74206 × 8281 cm^2 A ≈ 6144.99 cm^2. Hatua ya 4: Linganisha na chaguzi. Jibu lililohesabiwa ni takriban 6145.0 cm^2. Jibu sahihi ni A. 6145.03 cm^2. 6145.03 cm^2 Tuma swali linalofuata 📸

Juma was given 2 one-thousand-shilling notes to buy the following items: 2 kg of meat at sh 400 per kilogram 2 kg of tomatoes at sh 300 per kilogram 1½ kg of onions at sh 120 per kilogram ½ kg of potatoes at sh 40 per kilogram How much money would he remain with after buying the items?

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Juma was given 2 one-thousand-shilling notes to buy the following items: 2 kg of meat at sh 400 per kilogram 2 kg of tomatoes at sh 300 per kilogram 1½ kg of onions at sh 120 per kilogram ½ kg of potatoes at sh 40 per kilogram How much money would he remain with after buying the items?

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