2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic

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2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic

2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic By macoach.com.ng
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2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic

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2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic

2026 JAMB & WAEC Mathematics Likely questions and answers on Arithmetic

1. An amount of N300,000.00 was shared among Otobo, Ada and Adeola. Otobo received N60,000.00, Ada received \(\frac{5}{12}\) of the remainder, while the rest went to Adeola. In what ratio was the money shared? [2019/12]

  • Step 1: Calculate the remainder after Otobo's share.
    Total amount = N300,000.00
    Otobo's share = N60,000.00
    Remainder = N300,000.00 - N60,000.00 = N240,000.00
  • Step 2: Calculate Ada's share.
    Ada's share = \(\frac{5}{12} \times \text{Remainder}\)
    Ada's share = \(\frac{5}{12} \times \text{N240,000.00} = 5 \times \text{N20,000.00} = \text{N100,000.00}\)
  • Step 3: Calculate Adeola's share.
    Adeola's share = Remainder - Ada's share
    Adeola's share = N240,000.00 - N100,000.00 = N140,000.00
  • Step 4: Express the shares as a ratio.
    Otobo : Ada : Adeola
    N60,000 : N100,000 : N140,000
  • Step 5: Simplify the ratio.
    Divide all by 10,000: 6 : 10 : 14
    Divide all by 2: 3 : 5 : 7
  • Answer: The money was shared in the ratio \(\mathbf{3 : 5 : 7}\).

2. Salem, Sunday and Shaka shared a sum of N1, 100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka's share. [2013/2]

  • Step 1: Convert all amounts to the same unit (kobo or Naira).
    N1.00 = 100 kobo
    Total sum = N1,100.00 = 110,000 kobo
    Salem gets N2.00 = 200 kobo
    Sunday gets 50 kobo
    Shaka gets N2.00 = 200 kobo
  • Step 2: Determine the ratio of Salem : Sunday.
    Salem : Sunday = N2.00 : 50 kobo = 200 kobo : 50 kobo = 4 : 1
  • Step 3: Determine the ratio of Sunday : Shaka.
    Sunday : Shaka = N4.00 : N2.00 = 400 kobo : 200 kobo = 2 : 1
  • Step 4: Combine the ratios to find Salem : Sunday : Shaka.
    Salem : Sunday = 4 : 1
    Sunday : Shaka = 2 : 1
    To combine, make Sunday's ratio consistent. Multiply the first ratio by 2:
    Salem : Sunday = 8 : 2
    Sunday : Shaka = 2 : 1
    So, Salem : Sunday : Shaka = 8 : 2 : 1
  • Step 5: Calculate the total parts in the ratio.
    Total parts = 8 + 2 + 1 = 11 parts
  • Step 6: Calculate Shaka's share.
    Shaka's share = \(\left(\frac{\text{Shaka's ratio part}}{\text{Total parts}}\right) \times \text{Total sum}\)
    Shaka's share = \(\frac{1}{11} \times \text{N1,100.00} = \text{N100.00}\)
  • Answer: Shaka's share is \(\mathbf{N100.00}\).

3. The present ages of a father and his son are in the ratio 10:3. If the son is 15 years old now, in how many years will the ratio of their ages be 2:1? [2013/3]

  • Step 1: Find the father's current age.
    Let the father's current age be \(F\) and the son's current age be \(S\).
    \(F : S = 10 : 3\)
    \(S = 15\) years
    Since 3 parts = 15 years, then 1 part = \(\frac{15}{3} = 5\) years.
    Father's age (\(F\)) = \(10 \times 5 = 50\) years.
  • Step 2: Set up equations for future ages.
    Let \(x\) be the number of years from now.
    In \(x\) years, father's age = \(50 + x\)
    In \(x\) years, son's age = \(15 + x\)
    The ratio of their ages will be 2:1.
    \(\frac{50 + x}{15 + x} = \frac{2}{1}\)
  • Step 3: Solve for \(x\).
    \(50 + x = 2 \times (15 + x)\)
    \(50 + x = 30 + 2x\)
    \(50 - 30 = 2x - x\)
    \(20 = x\)
  • Answer: In \(\mathbf{20}\) years, the ratio of their ages will be 2:1.

4. A boy had M Dalasis (D), He spent D15 and shared the remainder equally with his sister. If the sister's share was equal to \(\frac{1}{2}\) of M, find the value of M [2012/6]

  • Step 1: Express the remainder after spending.
    Amount remaining = \(M - 15\)
  • Step 2: Express the sister's share.
    The remainder was shared equally, so the sister's share = \(\frac{M - 15}{2}\)
  • Step 3: Set up the equation based on the given information.
    Sister's share = \(\frac{1}{2}\) of \(M\)
    \(\frac{M - 15}{2} = \frac{M}{2}\)
  • Step 4: Solve for \(M\).
    Multiply both sides by 2:
    \(M - 15 = M\)
    \(-15 = M - M\)
    \(-15 = 0\)
  • Conclusion for Q4: Based on the direct interpretation, the problem leads to an inconsistent equation, suggesting it might be flawed. There is no value of M that satisfies the given conditions.

5. Ade received \(\frac{2}{3}\) of a sum of money, Nelly \(\frac{1}{2}\) of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by N3,000, how much did Ade receive? [2011/12]

  • Step 1: Assign a variable to the total sum.
    Let the total sum be \(S\).
  • Step 2: Calculate Ade's share.
    Ade's share = \(\frac{2}{3} S\)
  • Step 3: Calculate the remainder after Ade's share.
    Remainder = \(S - \frac{2}{3} S = \frac{1}{3} S\)
  • Step 4: Calculate Nelly's share.
    Nelly's share = \(\frac{1}{2} \times \text{Remainder}\)
    Nelly's share = \(\frac{1}{2} \times \frac{1}{3} S = \frac{1}{6} S\)
  • Step 5: Calculate Austin's share (the rest).
    Austin's share = Remainder - Nelly's share
    Austin's share = \(\frac{1}{3} S - \frac{1}{6} S = \frac{2}{6} S - \frac{1}{6} S = \frac{1}{6} S\)
  • Step 6: Set up the equation based on the difference between Austin's and Nelly's shares.
    Austin's share - Nelly's share = N3,000
    \(\frac{1}{6} S - \frac{1}{6} S = \text{N3,000}\)
    \(0 = \text{N3,000}\)
  • Conclusion for Q5: Based on the direct mathematical interpretation, this problem also leads to an inconsistent equation (\(0 = \text{N3,000}\)). This suggests a flaw in the problem statement, as Austin's and Nelly's shares are calculated to be equal.

6. If \(\frac{3p + 4q}{3p - 4q} = 2\), find \(p : q\) [2016/11]

  • Step 1: Cross-multiply.
    \(3p + 4q = 2 \times (3p - 4q)\)
  • Step 2: Distribute on the right side.
    \(3p + 4q = 6p - 8q\)
  • Step 3: Group terms with 'p' on one side and 'q' on the other.
    \(4q + 8q = 6p - 3p\)
    \(12q = 3p\)
  • Step 4: Express p in terms of q or q in terms of p.
    Divide by 3:
    \(4q = p\)
  • Step 5: Find the ratio \(p : q\).
    We have \(p = 4q\).
    So, \(\frac{p}{q} = \frac{4}{1}\)
    \(p : q = 4 : 1\)
  • Answer: \(\mathbf{p : q = 4 : 1}\).

7. A man left N5, 720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive? [2004/6]

  • Step 1: Assign variables for shares.
    Let the son's share be \(S\).
    Let each daughter's share be \(D\).
  • Step 2: Relate daughter's share to son's share.
    \(D = \frac{3}{4} S\)
  • Step 3: Write the total amount in terms of \(S\).
    Total amount = Son's share + (3 \(\times\) Daughter's share)
    N5,720 = \(S + 3 \times \left(\frac{3}{4} S\right)\)
    N5,720 = \(S + \frac{9}{4} S\)
  • Step 4: Combine the terms with \(S\).
    N5,720 = \(\frac{4}{4} S + \frac{9}{4} S\)
    N5,720 = \(\frac{13}{4} S\)
  • Step 5: Solve for \(S\) (son's share).
    \(S = \text{N5,720} \times \frac{4}{13}\)
    \(S = \left(\frac{\text{N5,720}}{13}\right) \times 4\)
    \(\frac{\text{N5,720}}{13} = \text{N440}\)
    \(S = \text{N440} \times 4\)
    \(S = \text{N1,760}\)
  • Answer: The son received \(\mathbf{N1,760}\).

8. A map is drawn to a scale of 1:20,000. Use it to calculate the: (i) distance, in kilometers, represented by 4.5 cm on the map; (ii) distance, in metres, on the map, between two towns 16 km apart; (iii) area on the map, of a forest, which covers 85 km². [2006/6]

Scale: \(1 : 20,000\) means 1 unit on the map represents 20,000 units on the ground.

(i) Distance, in kilometers, represented by 4.5 cm on the map.

  • Step 1: Convert map distance to ground distance in cm.
    Map distance = 4.5 cm
    Ground distance = \(4.5 \text{ cm} \times 20,000 = 90,000 \text{ cm}\)
  • Step 2: Convert ground distance from cm to km.
    1 m = 100 cm
    1 km = 1,000 m = \(1,000 \times 100 \text{ cm} = 100,000 \text{ cm}\)
    Ground distance in km = \(\frac{90,000 \text{ cm}}{100,000 \text{ cm/km}} = 0.9 \text{ km}\)
  • Answer (i): \(\mathbf{0.9 \text{ km}}\)

(ii) Distance, in metres, on the map, between two towns 16 km apart.

  • Step 1: Convert actual ground distance to cm.
    Actual ground distance = 16 km
    \(16 \text{ km} = 16 \times 100,000 \text{ cm} = 1,600,000 \text{ cm}\)
  • Step 2: Calculate map distance in cm.
    Map distance = \(\frac{\text{Actual ground distance}}{\text{Scale factor}}\)
    Map distance = \(\frac{1,600,000 \text{ cm}}{20,000} = 80 \text{ cm}\)
  • Step 3: Convert map distance from cm to metres.
    Map distance in metres = \(\frac{80 \text{ cm}}{100 \text{ cm/m}} = 0.8 \text{ m}\)
  • Answer (ii): \(\mathbf{0.8 \text{ m}}\)

(iii) Area on the map, of a forest, which covers 85 km².

  • Step 1: Determine the area scale.
    Linear scale = \(1 : 20,000\)
    Area scale = \((1)^2 : (20,000)^2 = 1 : 400,000,000\)
    This means 1 unit\(^2\) on the map represents 400,000,000 units\(^2\) on the ground.
  • Step 2: Convert ground area from km² to cm².
    \(1 \text{ km}^2 = (1,000 \text{ m})^2 = (1,000 \times 100 \text{ cm})^2 = (100,000 \text{ cm})^2 = 10,000,000,000 \text{ cm}^2 \text{ (10}^{10} \text{ cm}^2)\)
    Actual ground area = \(85 \text{ km}^2 = 85 \times 10,000,000,000 \text{ cm}^2 = 850,000,000,000 \text{ cm}^2\)
  • Step 3: Calculate map area in cm².
    Map area = \(\frac{\text{Actual ground area}}{\text{Area scale factor}}\)
    Map area = \(\frac{850,000,000,000 \text{ cm}^2}{400,000,000}\)
    Map area = \(\frac{8500}{4} \text{ cm}^2 = 21.25 \text{ cm}^2\)
  • Answer (iii): \(\mathbf{21.25 \text{ cm}^2}\)

9. The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7:2. Find their present ages. [2001/2]

  • Step 1: Set up equations for present ages.
    Let the woman's current age be \(W\) and the daughter's current age be \(D\).
    \(W + D = 46\) (Equation 1)
  • Step 2: Set up equations for ages in 4 years.
    In 4 years:
    Woman's age = \(W + 4\)
    Daughter's age = \(D + 4\)
    Ratio: \(\frac{W + 4}{D + 4} = \frac{7}{2}\)
  • Step 3: Cross-multiply the ratio equation.
    \(2 \times (W + 4) = 7 \times (D + 4)\)
    \(2W + 8 = 7D + 28\)
    \(2W - 7D = 28 - 8\)
    \(2W - 7D = 20\) (Equation 2)
  • Step 4: Solve the system of linear equations.
    From Equation 1, \(W = 46 - D\). Substitute this into Equation 2.
    \(2 \times (46 - D) - 7D = 20\)
    \(92 - 2D - 7D = 20\)
    \(92 - 9D = 20\)
    \(92 - 20 = 9D\)
    \(72 = 9D\)
    \(D = \frac{72}{9}\)
    \(D = 8\) years (daughter's present age)
  • Step 5: Find the woman's present age.
    \(W = 46 - D\)
    \(W = 46 - 8\)
    \(W = 38\) years (woman's present age)
  • Answer: The woman's present age is \(\mathbf{38}\) years and her daughter's present age is \(\mathbf{8}\) years.

10. Ali, Musah and Yusif shared N420,000.00 in the ratio 3:5:8 respectively. Find the sum of Ali and Yusif's shares. [2019/3]

  • Step 1: Calculate the total parts in the ratio.
    Ratio Ali : Musah : Yusif = 3 : 5 : 8
    Total parts = \(3 + 5 + 8 = 16\) parts
  • Step 2: Calculate the value of one part.
    Value of one part = \(\frac{\text{Total amount}}{\text{Total parts}}\)
    Value of one part = \(\frac{\text{N420,000.00}}{16} = \text{N26,250.00}\)
  • Step 3: Calculate Ali's share.
    Ali's share = \(3 \times \text{N26,250.00} = \text{N78,750.00}\)
  • Step 4: Calculate Yusif's share.
    Yusif's share = \(8 \times \text{N26,250.00} = \text{N210,000.00}\)
  • Step 5: Calculate the sum of Ali's and Yusif's shares.
    Sum = Ali's share + Yusif's share
    Sum = N78,750.00 + N210,000.00 = N288,750.00
  • Answer: The sum of Ali and Yusif's shares is \(\mathbf{N288,750.00}\).

11. A man left town M at 10:00 a.m. and travelled by car to town N at an average speed of 72 km/h. He spent 2 hours for a meeting and returned to town M by bus at an average speed of 40 km/h. If the distance covered by the bus was 2 km longer than that of the car and he arrived at town M at 1:55 p.m., calculate the distance from M to N. [2021/2]

  • Step 1: Define variables.
    Let \(D\) be the distance from M to N (by car).
    Distance by bus = \(D + 2\) km.
    Speed by car (\(V_c\)) = 72 km/h
    Speed by bus (\(V_b\)) = 40 km/h
  • Step 2: Calculate time for each leg of the journey.
    Time by car (\(T_c\)) = \(\frac{\text{Distance}}{\text{Speed}} = \frac{D}{72}\) hours
    Time by bus (\(T_b\)) = \(\frac{\text{Distance}}{\text{Speed}} = \frac{D + 2}{40}\) hours
  • Step 3: Calculate total elapsed time.
    Start time = 10:00 a.m.
    Arrival time = 1:55 p.m.
    Total time elapsed = (13 hours 55 minutes) - (10 hours 00 minutes) = 3 hours 55 minutes
    Convert total time to hours: \(3 \text{ hours} + \frac{55}{60} \text{ hours} = 3 + \frac{11}{12} \text{ hours} = \frac{47}{12} \text{ hours}\)
  • Step 4: Formulate the total time equation.
    Total time elapsed = Time by car + Meeting time + Time by bus
    \(\frac{47}{12} = \frac{D}{72} + 2 + \frac{D + 2}{40}\)
  • Step 5: Solve the equation for \(D\).
    Subtract 2 from both sides:
    \(\frac{47}{12} - 2 = \frac{D}{72} + \frac{D + 2}{40}\)
    \(\frac{47}{12} - \frac{24}{12} = \frac{D}{72} + \frac{D + 2}{40}\)
    \(\frac{23}{12} = \frac{D}{72} + \frac{D + 2}{40}\)
    Find a common denominator for 12, 72, and 40. LCM(12, 72, 40) = 360.
    Multiply the entire equation by 360:
    \(360 \times \frac{23}{12} = 360 \times \frac{D}{72} + 360 \times \frac{D + 2}{40}\)
    \(30 \times 23 = 5D + 9 \times (D + 2)\)
    \(690 = 5D + 9D + 18\)
    \(690 = 14D + 18\)
    \(690 - 18 = 14D\)
    \(672 = 14D\)
    \(D = \frac{672}{14}\)
    \(D = 48 \text{ km}\)
  • Answer: The distance from M to N is \(\mathbf{48 \text{ km}}\).

12. Two positive numbers are in the ratio 3: 4. The sum of thrice the first number and twice the second is 68. Find the smaller number. [2022/1]

  • Step 1: Represent the numbers using the ratio.
    Let the first number be \(3x\).
    Let the second number be \(4x\).
  • Step 2: Set up the equation based on the given sum.
    Thrice the first number = \(3 \times (3x) = 9x\)
    Twice the second number = \(2 \times (4x) = 8x\)
    Sum = \(9x + 8x = 68\)
  • Step 3: Solve for \(x\).
    \(17x = 68\)
    \(x = \frac{68}{17}\)
    \(x = 4\)
  • Step 4: Find the smaller number.
    The smaller number is \(3x\).
    Smaller number = \(3 \times 4 = 12\)
  • Answer: The smaller number is \(\mathbf{12}\).

13. A car travels a distance of 112 km at an average speed of 70 km/h. It then travels further for 60 km at an average speed of 50 km/h. Calculate, for the entire journey, the total time taken. [2023/1]

  • Step 1: Calculate time for the first part of the journey.
    Distance 1 (\(D_1\)) = 112 km
    Speed 1 (\(S_1\)) = 70 km/h
    Time 1 (\(T_1\)) = \(\frac{D_1}{S_1} = \frac{112 \text{ km}}{70 \text{ km/h}} = 1.6 \text{ hours}\)
  • Step 2: Calculate time for the second part of the journey.
    Distance 2 (\(D_2\)) = 60 km
    Speed 2 (\(S_2\)) = 50 km/h
    Time 2 (\(T_2\)) = \(\frac{D_2}{S_2} = \frac{60 \text{ km}}{50 \text{ km/h}} = 1.2 \text{ hours}\)
  • Step 3: Calculate the total time taken.
    Total Time = \(T_1 + T_2\)
    Total Time = \(1.6 \text{ hours} + 1.2 \text{ hours} = 2.8 \text{ hours}\)
  • Step 4: Convert total time to hours and minutes (optional but often good practice).
    \(2.8 \text{ hours} = 2 \text{ hours} + (0.8 \times 60) \text{ minutes} = 2 \text{ hours and } 48 \text{ minutes}\).
  • Answer: The total time taken for the entire journey is \(\mathbf{2.8 \text{ hours (or 2 hours and 48 minutes)}}\).

14. A shop owner gave an end-of-year bonus to two of his attendants, Kontor and Gapson in the ratio of their ages. Kontor's age is one and half times that of Gapson who is 20 years old. If Kontor received Le 200,000.00, find: (i) the total amount shared; (ii) Gapson's share. [2023/6]

  • Step 1: Calculate Kontor's age.
    Gapson's age = 20 years
    Kontor's age = \(1.5 \times \text{Gapson's age} = 1.5 \times 20 = 30\) years
  • Step 2: Determine the ratio of their ages.
    Kontor : Gapson = 30 : 20
    Simplify the ratio: 3 : 2

(i) Find the total amount shared.

  • Step 3: Relate Kontor's share to the ratio.
    Kontor's share = Le 200,000.00
    From the ratio, Kontor has 3 parts.
    So, 3 parts = Le 200,000.00
  • Step 4: Calculate the value of one part.
    1 part = \(\frac{\text{Le }200,000.00}{3} \approx \text{Le }66,666.67\)
  • Step 5: Calculate the total parts and total amount.
    Total parts = 3 (Kontor) + 2 (Gapson) = 5 parts
    Total amount = \(5 \times \left(\frac{\text{Le }200,000.00}{3}\right) = \frac{\text{Le }1,000,000.00}{3} \approx \text{Le }333,333.33\)
  • Answer (i): The total amount shared is \(\mathbf{\text{Le }333,333.33}\).

(ii) Find Gapson's share.

  • Step 6: Calculate Gapson's share.
    Gapson has 2 parts.
    Gapson's share = \(2 \times \left(\frac{\text{Le }200,000.00}{3}\right) = \frac{\text{Le }400,000.00}{3} \approx \text{Le }133,333.33\)
  • Answer (ii): Gapson's share is \(\mathbf{\text{Le }133,333.33}\).

15. (a). In a man's will, he gave \(\frac{1}{3}\) of the total acres of his cocoa farm to the wife and \(\frac{2}{5}\) of what is left to the family. The rest of the farm was to be shared amongst his three children in the ratio 3:5: 2. Given that, the child who had the least share received 8 acres, calculate the: (i) total acres the man left; (ii) number of acres the wife received.(b) The list price of a Television set is $ 1,600.00. It can be purchased by a deposit of $ 400.00 and the rest of the amount paid by 12 monthly installment at 25 % per annum simple interest. If the Television set is purchased by installment, find the total cost. [2024/10]

(a) Cocoa Farm Problem

  • Step 1: Define variables.
    Let \(T\) be the total acres of the farm.
    Wife's share = \(\frac{1}{3} T\)
  • Step 2: Calculate the remainder after the wife's share.
    Remainder 1 = \(T - \frac{1}{3} T = \frac{2}{3} T\)
  • Step 3: Calculate the family's share (excluding children's specific shares yet).
    Family's share = \(\frac{2}{5} \text{ of Remainder 1} = \frac{2}{5} \times \frac{2}{3} T = \frac{4}{15} T\)
  • Step 4: Calculate the rest of the farm (for the children).
    Remainder 2 (for children) = Remainder 1 - Family's share
    Remainder 2 = \(\frac{2}{3} T - \frac{4}{15} T = \frac{10}{15} T - \frac{4}{15} T = \frac{6}{15} T = \frac{2}{5} T\)
  • Step 5: Allocate children's shares based on their ratio.
    Children's ratio = 3 : 5 : 2
    Total parts for children = \(3 + 5 + 2 = 10\) parts
    The child with the least share has 2 parts.
    2 parts = 8 acres
  • Step 6: Calculate the value of one part for the children.
    1 part = \(\frac{8 \text{ acres}}{2} = 4 \text{ acres}\)
  • Step 7: Calculate the total acres for the children (Remainder 2).
    Total acres for children = \(10 \times 4 \text{ acres} = 40 \text{ acres}\)
    So, Remainder 2 = 40 acres.
  • (i) Calculate the total acres the man left (T).
    From Step 4, Remainder 2 = \(\frac{2}{5} T\)
    \(40 \text{ acres} = \frac{2}{5} T\)
    \(T = 40 \times \frac{5}{2} = 20 \times 5 = 100 \text{ acres}\)
  • Answer (a)(i): The total acres the man left is \(\mathbf{100 \text{ acres}}\).
  • (ii) Calculate the number of acres the wife received.
    Wife's share = \(\frac{1}{3} T\)
    Wife's share = \(\frac{1}{3} \times 100 \text{ acres} = \frac{100}{3} \text{ acres} \approx 33.33 \text{ acres}\)
  • Answer (a)(ii): The wife received \(\mathbf{33 \frac{1}{3} \text{ acres (or approximately 33.33 acres)}}\).

(b) Television Set Installment Problem

  • Step 1: Identify given values.
    List price = $1,600.00
    Deposit = $400.00
    Installment period = 12 months (1 year)
    Interest rate = 25% per annum simple interest
  • Step 2: Calculate the amount to be paid by installment.
    Amount for installment = List price - Deposit
    Amount for installment = $1,600.00 - $400.00 = $1,200.00
  • Step 3: Calculate the simple interest.
    Interest (\(I\)) = Principal (\(P\)) \(\times\) Rate (\(R\)) \(\times\) Time (\(T\))
    \(P = \$1,200.00\)
    \(R = 25\% = 0.25\)
    \(T = 1\) year (12 months)
    \(I = \$1,200.00 \times 0.25 \times 1 = \$300.00\)
  • Step 4: Calculate the total amount paid in installments.
    Total installment amount = Principal for installment + Interest
    Total installment amount = $1,200.00 + $300.00 = $1,500.00
  • Step 5: Calculate the total cost of the Television set.
    Total Cost = Deposit + Total installment amount
    Total Cost = $400.00 + $1,500.00 = $1,900.00
  • Answer (b): The total cost of the Television set is \(\mathbf{\$1,900.00}\).

16. In a certain year, the consumption pattern of electricity charges in a town was as follows: the cost of the first 30 units was $1.00 per unit, the cost of the next 30 units was $7.00 per unit, the cost of each additional unit was $5 00. (a) If Amaka used 420 units of electricity in January that year, calculate the amount paid. (b) If Amaka paid $2,740.00 in the month of February, calculate the number of units of electricity consumed. (c) Find, correct to two decimal places, the percentage change in units of electricity consumed by Amaka in January and February. [2025/6]

(a) Amaka's bill for January (420 units)

  • Step 1: Calculate cost for the first 30 units.
    Cost = \(30 \text{ units} \times \$1.00/\text{unit} = \$30.00\)
  • Step 2: Calculate cost for the next 30 units.
    Cost = \(30 \text{ units} \times \$7.00/\text{unit} = \$210.00\)
  • Step 3: Calculate remaining units.
    Total units used = 420
    Units accounted for = \(30 + 30 = 60\) units
    Remaining units (additional units) = \(420 - 60 = 360\) units
  • Step 4: Calculate cost for the additional units.
    Cost = \(360 \text{ units} \times \$5.00/\text{unit} = \$1,800.00\)
  • Step 5: Calculate the total amount paid.
    Total bill = \$30.00 + \$210.00 + \$1,800.00 = \$2,040.00
  • Answer (a): Amaka paid \(\mathbf{\$2,040.00}\) in January.

(b) Units consumed in February (Bill = $2,740.00)

  • Step 1: Account for the fixed tiers.
    Cost of first 30 units = \$30.00
    Cost of next 30 units = \$210.00
    Total cost for first 60 units = \$30.00 + \$210.00 = \$240.00
  • Step 2: Calculate the cost attributed to additional units.
    Remaining bill = Total bill - Cost for first 60 units
    Remaining bill = \$2,740.00 - \$240.00 = \$2,500.00
  • Step 3: Calculate the number of additional units.
    Cost per additional unit = \$5.00
    Number of additional units = \(\frac{\text{Remaining bill}}{\text{Cost per unit}}\)
    Number of additional units = \(\frac{\$2,500.00}{\$5.00/\text{unit}} = 500 \text{ units}\)
  • Step 4: Calculate the total units consumed.
    Total units = First 60 units + Additional units
    Total units = \(60 + 500 = 560 \text{ units}\)
  • Answer (b): Amaka consumed \(\mathbf{560 \text{ units}}\) in February.

(c) Percentage change in units consumed by Amaka in January and February.

  • Step 1: Identify units for each month.
    Units in January = 420 units
    Units in February = 560 units
  • Step 2: Calculate the change in units.
    Change = February units - January units
    Change = \(560 - 420 = 140\) units (increase)
  • Step 3: Calculate the percentage change.
    Percentage Change = \(\left(\frac{\text{Change}}{\text{Original units (January)}}\right) \times 100\%\)
    Percentage Change = \(\left(\frac{140}{420}\right) \times 100\%\)
    Percentage Change = \(\left(\frac{1}{3}\right) \times 100\% \approx 33.33\%\) (correct to two decimal places)
  • Answer (c): The percentage change in units of electricity consumed is \(\mathbf{33.33\%\) increase.

17. Yaro drove from a town Gaja to Banga. After 2 hours in the journey, he observed that he had covered 80 km and realized that if he continued driving at same average speed, he would end up being late for 15 minutes. If he decided to increase the average speed by 10 km/h, he would arrive at Banga 36 minutes earlier. Find the distance between Gaja and Banga. [2025/7]

  • Step 1: Determine the initial speed.
    After 2 hours, 80 km covered.
    Initial speed (\(S\)) = \(\frac{\text{Distance}}{\text{Time}} = \frac{80 \text{ km}}{2 \text{ h}} = 40 \text{ km/h}\)
  • Step 2: Define variables for the remaining journey.
    Let \(D_{\text{rem}}\) be the remaining distance to Banga.
    Let \(T_{\text{expected}}\) be the expected time for the remaining journey (if on time).
  • Step 3: Formulate equations based on the "late" scenario.
    If he continues at \(S = 40 \text{ km/h}\), he would be 15 minutes late.
    Actual time to cover \(D_{\text{rem}}\) at \(S = 40 \text{ km/h} = \frac{D_{\text{rem}}}{40}\) hours
    This actual time is 15 minutes (\(\frac{15}{60} = 0.25\) hours) longer than the expected time.
    So, \(\frac{D_{\text{rem}}}{40} = T_{\text{expected}} + 0.25\) (Equation 1)
  • Step 4: Formulate equations based on the "earlier" scenario.
    If he increases speed by 10 km/h, new speed (\(S'\)) = \(40 + 10 = 50 \text{ km/h}\).
    He would arrive 36 minutes earlier.
    Actual time to cover \(D_{\text{rem}}\) at \(S' = 50 \text{ km/h} = \frac{D_{\text{rem}}}{50}\) hours
    This actual time is 36 minutes (\(\frac{36}{60} = 0.6\) hours) shorter than the expected time.
    So, \(\frac{D_{\text{rem}}}{50} = T_{\text{expected}} - 0.6\) (Equation 2)
  • Step 5: Solve the system of equations for \(D_{\text{rem}}\).
    From Equation 1: \(T_{\text{expected}} = \frac{D_{\text{rem}}}{40} - 0.25\)
    Substitute \(T_{\text{expected}}\) into Equation 2:
    \(\frac{D_{\text{rem}}}{50} = \left(\frac{D_{\text{rem}}}{40} - 0.25\right) - 0.6\)
    \(\frac{D_{\text{rem}}}{50} = \frac{D_{\text{rem}}}{40} - 0.85\)
    Group terms with \(D_{\text{rem}}\):
    \(0.85 = \frac{D_{\text{rem}}}{40} - \frac{D_{\text{rem}}}{50}\)
    \(0.85 = \frac{5D_{\text{rem}} - 4D_{\text{rem}}}{200}\) (Common denominator for 40 and 50 is 200)
    \(0.85 = \frac{D_{\text{rem}}}{200}\)
    Solve for \(D_{\text{rem}}\):
    \(D_{\text{rem}} = 0.85 \times 200\)
    \(D_{\text{rem}} = 170 \text{ km}\)
  • Step 6: Calculate the total distance between Gaja and Banga.
    Total distance = Distance already covered + Remaining distance
    Total distance = \(80 \text{ km} + 170 \text{ km} = 250 \text{ km}\)
  • Answer: The distance between Gaja and Banga is \(\mathbf{250 \text{ km}}\).

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