A school bought 28 bags of maize and 15 bags of beans for a total of sh.263400. In the following term, the trader increased the cost of bag of maize by 20% and reduced the cost of a bag of beans by 10%,the school was then able to buy 24 bags of maize and 12 bags of beans for a total of sh.231120. Taking x to represent cost of bag of maize and y to represent cost of bag of beans; Form two simplified equations using the information above (ii)Use matrix method to find the cost of 1 bag of maize and 1 bag of beans bought from the trader initially (iii) Another school bought 30 bags of maize and 20 bags of beans after the price adjustments. Use matrix method to determine the total amount of money the school spent.

Question

Asked on August 6, 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 · 2026-08-06T19:47:24.558Z

i) Step 1: Andika mlinganyo wa kwanza kwa kutumia taarifa za awali. Shule ilinunua mifuko 28 ya mahindi kwa bei ya x na mifuko 15 ya maharagwe kwa bei ya y, kwa jumla ya Sh. 263400. 28x + 15y = 263400 Step 2: Andika mlinganyo wa pili kwa kutumia taarifa za bei zilizorekebishwa na kurahisisha. Bei ya mahindi iliongezeka kwa 20%, hivyo bei mpya ni 1.2x. Bei ya maharagwe ilipungua kwa 10%, hivyo bei mpya ni 0.9y. Shule ilinunua mifuko 24 ya mahindi na mifuko 12 ya maharagwe kwa jumla ya Sh. 231120. 24(1.2x) + 12(0.9y) = 231120 28.8x + 10.8y = 231120 Ili kuondoa desimali, zidisha mlinganyo mzima kwa 10: 288x + 108y = 2311200 Rahisisha mlinganyo kwa kugawanya kwa 36 (GCD ya 288, 108, na 2311200): (288x)/(36) + (108y)/(36) = (2311200)/(36) 8x + 3y = 64200 Milinyango iliyorahisishwa ni: 28x + 15y &= 263400 \\ 8x + 3y &= 64200 ii) Step 3: Andika milinganyo katika mfumo wa matriki na utafute kinyume cha matriki ya vigawo. Milinganyo ni: 28x + 15y = 263400 8x + 3y = 64200 Katika mfumo wa matriki: 28 & 15 \\ 8 & 3 x \\ y = 263400 \\ 64200 Wacha A = 28 & 15 \\ 8 & 3 . Kiamua cha A, det(A) = (28 × 3) - (15 × 8) = 84 - 120 = -36. Kinyume cha A, A^-1 = (1)/(-36) 3 & -15 \\ -8 & 28 . Step 4: Tumia matriki kinyume kutatua kwa x na y. x \\ y = A^-1 263400 \\ 64200 x \\ y = (1)/(-36) 3 & -15 \\ -8 & 28 263400 \\ 64200 x \\ y = (1)/(-36) (3)(263400) + (-15)(64200) \\ (-8)(263400) + (28)(64200) x \\ y = (1)/(-36) 790200 - 963000 \\ -2107200 + 1797600 x \\ y = (1)/(-36) -172800 \\ -309600 x = (-172800)/(-36) = 4800 y = (-309600)/(-36) = 8600 Gharama ya awali ya mfuko 1 wa mahindi ni Sh. 4800 na mfuko 1 wa maharagwe ni Sh. 8600. Gharama ya mfuko 1 wa mahindi = Sh. 4800 Gharama ya mfuko 1 wa maharagwe = Sh. 8600 iii) Step 5: Kokotoa bei zilizorekebishwa za mahindi na maharagwe. Bei mpya ya mahindi (x'): 1.2 × 4800 = 5760 Sh. Bei mpya ya maharagwe (y'): 0.9 × 8600 = 7740 Sh. Step 6: Tumia njia ya matriki kubainisha jumla ya pesa iliyotumiwa na shule nyingine. Shule nyingine ilinunua mifuko 30 ya mahindi na mifuko 20 ya maharagwe. Wacha Q = 30 & 20 iwe matriki ya idadi. Wacha C' = 5760 \\ 7740 iwe matriki ya bei zilizorekebishwa. Jumla ya pesa iliyotumiwa (T) ni Q · C'. T = 30 & 20 5760 \\ 7740 T = (30 × 5760) + (20 × 7740) T = 172800 + 154800 T = 327600 Jumla ya pesa iliyotumiwa na shule nyingine ni Sh. 327600. Jumla ya pesa iliyotumiwa = Sh. 327600 Tuma swali linalofuata 📸

Accepted Answer · 2026-08-06T19:47:24.558Z

i) Step 1: Andika mlinganyo wa kwanza kwa kutumia taarifa za awali. Shule ilinunua mifuko 28 ya mahindi kwa bei ya x na mifuko 15 ya maharagwe kwa bei ya y, kwa jumla ya Sh. 263400. 28x + 15y = 263400 Step 2: Andika mlinganyo wa pili kwa kutumia taarifa za bei zilizorekebishwa na kurahisisha. Bei ya mahindi iliongezeka kwa 20%, hivyo bei mpya ni 1.2x. Bei ya maharagwe ilipungua kwa 10%, hivyo bei mpya ni 0.9y. Shule ilinunua mifuko 24 ya mahindi na mifuko 12 ya maharagwe kwa jumla ya Sh. 231120. 24(1.2x) + 12(0.9y) = 231120 28.8x + 10.8y = 231120 Ili kuondoa desimali, zidisha mlinganyo mzima kwa 10: 288x + 108y = 2311200 Rahisisha mlinganyo kwa kugawanya kwa 36 (GCD ya 288, 108, na 2311200): (288x)/(36) + (108y)/(36) = (2311200)/(36) 8x + 3y = 64200 Milinyango iliyorahisishwa ni: 28x + 15y &= 263400 \\ 8x + 3y &= 64200 ii) Step 3: Andika milinganyo katika mfumo wa matriki na utafute kinyume cha matriki ya vigawo. Milinganyo ni: 28x + 15y = 263400 8x + 3y = 64200 Katika mfumo wa matriki: 28 & 15 \\ 8 & 3 x \\ y = 263400 \\ 64200 Wacha A = 28 & 15 \\ 8 & 3 . Kiamua cha A, det(A) = (28 × 3) - (15 × 8) = 84 - 120 = -36. Kinyume cha A, A^-1 = (1)/(-36) 3 & -15 \\ -8 & 28 . Step 4: Tumia matriki kinyume kutatua kwa x na y. x \\ y = A^-1 263400 \\ 64200 x \\ y = (1)/(-36) 3 & -15 \\ -8 & 28 263400 \\ 64200 x \\ y = (1)/(-36) (3)(263400) + (-15)(64200) \\ (-8)(263400) + (28)(64200) x \\ y = (1)/(-36) 790200 - 963000 \\ -2107200 + 1797600 x \\ y = (1)/(-36) -172800 \\ -309600 x = (-172800)/(-36) = 4800 y = (-309600)/(-36) = 8600 Gharama ya awali ya mfuko 1 wa mahindi ni Sh. 4800 na mfuko 1 wa maharagwe ni Sh. 8600. Gharama ya mfuko 1 wa mahindi = Sh. 4800 Gharama ya mfuko 1 wa maharagwe = Sh. 8600 iii) Step 5: Kokotoa bei zilizorekebishwa za mahindi na maharagwe. Bei mpya ya mahindi (x'): 1.2 × 4800 = 5760 Sh. Bei mpya ya maharagwe (y'): 0.9 × 8600 = 7740 Sh. Step 6: Tumia njia ya matriki kubainisha jumla ya pesa iliyotumiwa na shule nyingine. Shule nyingine ilinunua mifuko 30 ya mahindi na mifuko 20 ya maharagwe. Wacha Q = 30 & 20 iwe matriki ya idadi. Wacha C' = 5760 \\ 7740 iwe matriki ya bei zilizorekebishwa. Jumla ya pesa iliyotumiwa (T) ni Q · C'. T = 30 & 20 5760 \\ 7740 T = (30 × 5760) + (20 × 7740) T = 172800 + 154800 T = 327600 Jumla ya pesa iliyotumiwa na shule nyingine ni Sh. 327600. Jumla ya pesa iliyotumiwa = Sh. 327600 Tuma swali linalofuata 📸

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