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?

Swali la 10:
Juma alipewa noti mbili za shilingi elfu moja, hivyo alikuwa na jumla ya shilingi $2 \times 1000 = 2000$.
Hebu tuhesabu gharama ya kila bidhaa:

• Hatua ya 1: Hesabu gharama ya kila bidhaa.

  • Nyama: $2 \text{ kg} \times 400 \text{ sh/kg} = 800 \text{ sh}$
  • Nyanya: $2 \text{ kg} \times 300 \text{ sh/kg} = 600 \text{ sh}$
  • Vitunguu: $1\frac{1}{2} \text{ kg} \times 120 \text{ sh/kg} = 1.5 \times 120 \text{ sh} = 180 \text{ sh}$
  • Viazi: $\frac{1}{2} \text{ kg} \times 40 \text{ sh/kg} = 0.5 \times 40 \text{ sh} = 20 \text{ sh}$

• Hatua ya 2: Hesabu jumla ya gharama. Jumla ya gharama = $800 + 600 + 180 + 20 = 1600 \text{ sh}$.

• Hatua ya 3: Hesabu kiasi cha pesa alichobakiwa nacho Juma. Pesa iliyobaki = Pesa aliyokuwa nayo - Jumla ya gharama Pesa iliyobaki = $2000 \text{ sh} - 1600 \text{ sh} = 400 \text{ sh}$.

Jibu sahihi ni A. sh 400.
$\boxed{\text{sh 400}}$

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 ($h_s$) = $H - h_f = 20 \text{ m} - 10 \text{ m} = 10 \text{ m}$. Eneo la msingi mkubwa ($A_1$) = $18 \text{ m} \times 18 \text{ m} = 324 \text{ m}^2$. Eneo la msingi mdogo ($A_2$) = $9 \text{ m} \times 9 \text{ m} = 81 \text{ m}^2$.

• Hatua ya 2: Hesabu ujazo wa piramidi kubwa. Ujazo wa piramidi kubwa ($V_L$) = $\frac{1}{3} A_1 H$ $V_L = \frac{1}{3} \times 324 \text{ m}^2 \times 20 \text{ m}$ $V_L = 108 \times 20 \text{ m}^3 = 2160 \text{ m}^3$.

• Hatua ya 3: Hesabu ujazo wa piramidi ndogo iliyoondolewa. Ujazo wa piramidi ndogo ($V_S$) = $\frac{1}{3} A_2 h_s$ $V_S = \frac{1}{3} \times 81 \text{ m}^2 \times 10 \text{ m}$ $V_S = 27 \times 10 \text{ m}^3 = 270 \text{ m}^3$.

• Hatua ya 4: Hesabu ujazo wa frustum. Ujazo wa frustum ($V_f$) = Ujazo wa piramidi kubwa - Ujazo wa piramidi ndogo $V_f = 2160 \text{ m}^3 - 270 \text{ m}^3 = 1890 \text{ m}^3$.

Jibu lililohesabiwa ni $1890 \text{ m}^3$. Kati ya chaguzi zilizopo, A. $1920 \text{ m}^3$ ndiyo iliyo karibu zaidi.
$\boxed{1920 \text{ m}^3}$

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. $4s = 44 \text{ m}$ $s = \frac{44}{4} \text{ m} = 11 \text{ m}$.

• Hatua ya 2: Hesabu eneo la bustani. Eneo la mraba ($A$) = $s^2$ $A = (11 \text{ m})^2 = 121 \text{ m}^2$.

Jibu sahihi ni B. $121 \text{ m}^2$.
$\boxed{121 \text{ m}^2}$

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 $\frac{3}{4}$ 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 ($V_c$) = $L \times W \times H = 4 \text{ m} \times 3 \text{ m} \times 5 \text{ m} = 60 \text{ m}^3$.

• Hatua ya 2: Hesabu idadi ya pakiti zinazohitajika kujaza kontena zima. Ujazo wa pakiti moja ($V_p$) = $0.5 \text{ m}^3$. Idadi ya pakiti ($N$) = $\frac{V_c}{V_p} = \frac{60 \text{ m}^3}{0.5 \text{ m}^3} = 120 \text{ pakiti}$.

Jibu sahihi ni A. 120.
$\boxed{120}$

Swali la 14:

• Hatua ya 1: Hesabu ujazo wa silinda. Radius ($r$) = 7 m Urefu ($h$) = 20 m $\pi = \frac{22}{7}$ Ujazo wa silinda ($V$) = $\pi r^2 h$ $V = \frac{22}{7} \times (7 \text{ m})^2 \times 20 \text{ m}$ $V = \frac{22}{7} \times 49 \text{ m}^2 \times 20 \text{ m}$ $V = 22 \times 7 \text{ m}^2 \times 20 \text{ m}$ $V = 154 \text{ m}^2 \times 20 \text{ m} = 3080 \text{ m}^3$.

• Hatua ya 2: Badilisha ujazo kutoka mita za ujazo kwenda lita. Tunajua kuwa $1 \text{ m}^3 = 1000 \text{ lita}$. $V = 3080 \text{ m}^3 \times 1000 \text{ lita/m}^3 = 3,080,000 \text{ lita}$.

Jibu sahihi ni D. 3 080 000.
$\boxed{3\,080\,000 \text{ lita}}$

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 ($\theta$) = $85^\circ$.

• Hatua ya 2: Badilisha pembe kuwa radiani. Ili kutumia fomula ya eneo la sekta, tunahitaji pembe katika radiani au kutumia fomula inayohusisha digrii. $\theta_{\text{radiani}} = 85^\circ \times \frac{\pi}{180^\circ} = \frac{85\pi}{180} = \frac{17\pi}{36}$ radiani.

• Hatua ya 3: Hesabu eneo la sekta ya duara. Eneo la sekta ($A$) = $\frac{1}{2} r^2 \theta_{\text{radiani}}$ au $A = \frac{\theta}{360^\circ} \pi r^2$. Tutatumia fomula ya digrii. $A = \frac{85}{360} \times \frac{22}{7} \times (91 \text{ cm})^2$ $A = \frac{85}{360} \times \frac{22}{7} \times 8281 \text{ cm}^2$ $A = \frac{17}{72} \times \frac{22}{7} \times 8281 \text{ cm}^2$ $A = \frac{17 \times 11}{36 \times 7} \times 8281 \text{ cm}^2$ $A = \frac{187}{252} \times 8281 \text{ cm}^2$ $A \approx 0.74206 \times 8281 \text{ cm}^2$ $A \approx 6144.99 \text{ cm}^2$.

• Hatua ya 4: Linganisha na chaguzi. Jibu lililohesabiwa ni takriban $6145.0 \text{ cm}^2$.

Jibu sahihi ni A. $6145.03 \text{ cm}^2$.
$\boxed{6145.03 \text{ cm}^2}$