Math Problem Statement
Solution
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Extracted Problem:
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Problem 79: The expenditure of a man for the first 3 months is ₹200 and the expenses increase uniformly by ₹40. His expenditure for the next 3 months will be:
(A) ₹320 (B) ₹360 (C) ₹400 (D) ₹480
Solution:
The problem states that the man’s expenditure increases uniformly by ₹40 each month after the first 3 months. Let's analyze it step by step:
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Initial 3-month Expenditure:
- For the first 3 months, his expenditure is ₹200.
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Monthly Increase:
- From the 4th month, his monthly expenditure increases by ₹40.
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Expenditure for Next 3 Months:
- 4th month: ₹200 + ₹40 = ₹240
- 5th month: ₹240 + ₹40 = ₹280
- 6th month: ₹280 + ₹40 = ₹320
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Total Expenditure for the Next 3 Months:
- Sum = ₹240 + ₹280 + ₹320 = ₹840
However, since the problem seems to be asking for the monthly expenditure at the end of the 3rd month, the expenditure on the 6th month would be ₹320.
So, the correct answer for the monthly expenditure at the end of 6 months would be ₹320 (Option A).
Additional Questions:
- If the expenditure increases by ₹50 each month instead of ₹40, what would be the expenditure for the next 3 months?
- What would be the total expenditure for the first 6 months?
- How would the expenditure change if the initial expenditure was ₹300 instead of ₹200?
- If the expenditure was calculated for 12 months instead of 6, how would you approach the problem?
- How can the concept of arithmetic progression be applied to solve this problem more generally?
Tip: Always double-check the terms and structure of arithmetic progressions when dealing with uniformly increasing or decreasing sequences.
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Math Problem Analysis
Mathematical Concepts
Arithmetic Progression
Linear Sequences
Formulas
Sum of first n terms of an arithmetic progression: S_n = n/2 * (2a + (n-1)d)
nth term of an arithmetic progression: a_n = a + (n-1)d
Theorems
Arithmetic Progression Formula
Suitable Grade Level
Grades 7-9
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