Math Case Study Questions Class 10

Math Case Study Questions Class 10

Understanding Math Case Study Questions Class 10 is crucial for exams. Students often practice with math case study questions class 10 to build confidence. These problems involve sequences where each term follows a rule. Additionally, solving Case Study math questions for class 10 and math case study questions class 10 prepares students for higher-level concepts. Therefore, focusing on this topic strengthens both analytical and problem-solving skills effectively.

Math Case Study Questions Class 10

In Case Study Class 10 , questions often highlight real-life situations. For example, calculating savings, distances, or patterns in daily life. Practicing math case study questions helps students connect classroom learning with practical use. Moreover, working on Case Study Questions Class 10 ensures better exam preparation. As a result, students gain confidence, accuracy, and speed. Importantly, these case studies encourage logical thinking along with systematic application of formulas.

Case Study 2

Meena decides to start saving money for her higher education. She deposits money in her piggy bank according to a specific pattern. On the first day, she saves 15 rupees, on the second day 20 rupees, on the third day 25 rupees, and so on. She wishes to know how much she will save after a certain number of days and on which day she will achieve her target of saving 3000 rupees. Her daily savings thus form an Arithmetic Progression (AP). Let us solve the following questions using the properties of AP.

The general form of an Arithmetic Progression is:

\[a, a+d, a+2d, a+3d, \ldots\]

where \(a\) is the first term and \(d\) is the common difference.

The \(n\)th term is given by:

\[a_n = a + (n-1)d\]

The sum of the first \(n\) terms is given by:

\[S_n = \frac{n}{2} \big(2a + (n-1)d\big)\]

1. What is the common difference \(d\) in Meena’s saving pattern?

  • A) 4
  • B) 5
  • C) 6
  • D) 7
Answer: B) 5
Solution: First three terms: 15, 20, 25. \(d = 20 – 15 = 5\).

2. How much will Meena save on the 12th day?

  • A) 65
  • B) 70
  • C) 75
  • D) 80
Answer: B) 70
Solution: \(a=15\), \(d=5\), \(n=12\). \(a_{12} = 15 + (12-1)(5) = 15 + 55 = 70\).

3. How much will she save in the first 30 days?

  • A) 2500
  • B) 2550
  • C) 2600
  • D) 2650
Answer: 2625 rupees (Note: None of the options match the correct calculation)
Solution: \(a=15\), \(d=5\), \(n=30\). \(S_{30} = \dfrac{30}{2}[2(15) + (30-1)(5)] = 15[30 + 145] = 15 \times 175 = 2625\).

4. On which day will Meena save exactly 115 rupees?

  • A) 19th day
  • B) 20th day
  • C) 21st day
  • D) 22nd day
Answer: C) 21st day
Solution: \(a_n = 15 + (n-1)(5) = 115\). \((n-1)(5) = 100 \implies n-1 = 20 \implies n=21\).

5. After how many days will Meena save exactly 3000 rupees in total?

  • A) 30 days
  • B) 31 days
  • C) 32 days
  • D) 33 days
Answer: C) 32 days
Solution: We need \(S_n = 3000\). \(S_n = \dfrac{n}{2}[2(15) + (n-1)(5)] = \dfrac{n}{2}(5n+25) = \dfrac{5n^2 + 25n}{2} = 3000\). Solving the equation \(n^2 + 5n – 1200 = 0\), we get \(n=32\).

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