Measures of Central Tendency and DispersionMCQMTP June 22Question 2908 of 473
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A batsman in his 20th\displaystyle 20^{th} innings makes a score of 120\displaystyle 120 and thereby increases his average by 5\displaystyle 5. What is his average after 20th\displaystyle 20^{th} innings?

Options

A60
B55
C25
D70
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Correct Answer

Option c25

All Options:

  • A60
  • B55
  • C25
  • D70

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Detailed Solution & Explanation

**Step 1: Set up the equation.** Let the average after 19 innings =a\displaystyle = a. Then total score after 19 innings =19a\displaystyle = 19a. After 20th innings, average increases by 5: New average=a+5\text{New average} = a + 5 19a+12020=a+5\frac{19a + 120}{20} = a + 5 **Step 2: Solve.** 19a+120=20a+10019a + 120 = 20a + 100 120100=20a19a120 - 100 = 20a - 19a a=20a = 20 **Step 3: Compute average after 20th innings.** Average after 20th=a+5=20+5=25\text{Average after 20th} = a + 5 = 20 + 5 = 25 **Note:** The computed average after 20th innings is **25** (Option C), not 55 (Option B). Wait — let me reconsider. Perhaps the average increases BY 5 from 19 innings to 20, meaning each innings contributes. Let me recheck. Actually, the standard approach: Let average after 20 innings =A\displaystyle = A. Average after 19 innings =A5\displaystyle = A - 5. 19(A5)+120=20A19(A-5) + 120 = 20A 19A95+120=20A19A - 95 + 120 = 20A 25=A25 = A So average after 20th innings =25\displaystyle = 25 (Option C). Hence, **Option C** is the correct answer.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

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