MATHEMATICS

JAMB 2001 - Question 15

Mathematics 2001 JAMB Past Questions - Question 15: A man saves N100.00 in first year of work and each year saves N200.00 more than in the preceeding year.in how many years will he save N5,800.00?

A man saves N100.00 in first year of work and each year saves N200.00 more than in the preceeding year.in how many years will he save N5,800.00?
A:
B:
C:
D:
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Correct Answer

A

Explanation

Let's break down the problem step by step to find the number of years it will take for the man to save N5,800.00.In the first year, the man saves N100.00.In the second year, he saves N100.00 + N200.00 = N300.00.In the third year, he saves N300.00 + N200.00 = N500.00.In the fourth year, he saves N500.00 + N200.00 = N700.00.And so on...We can observe that the amount saved each year follows an arithmetic progression, where the first term is N100.00 and the common difference is N200.00.To find the number of years it will take for the man to save N5,800.00, we need to solve the equation:N100.00 + (N200.00) * (n - 1) = N5,800.00Simplifying the equation:N100.00 + N200.00n - N200.00 = N5,800.00Combining like terms:N200.00n - N100.00 = N5,800.00Adding N100.00 to both sides:N200.00n = N5,900.00Dividing both sides by N200.00:n = N5,900.00 / N200.00n = 29.5Since the number of years cannot be a fraction, we round up to the nearest whole number.Therefore, it will take approximately 30 years for the man to save N5,800.00.

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