Measures of Central Tendency and DispersionMCQPYQ June 24 Series IIQuestion 2932 of 473
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The mean weight of 15 students is 110 kg. The mean weight of 5 of them is 100 kg, and that of another five students is 125 kg., then the mean weight of the remaining students is:

Options

A120
B105
C115
DNone of these
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Correct Answer

Option b105

All Options:

  • A120
  • B105
  • C115
  • DNone of these

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Detailed Solution & Explanation

**Step 1: Total weight of 15 students.** Total=15×110=1650 kg\text{Total} = 15 \times 110 = 1650 \text{ kg} **Step 2: Sum of two known groups.** - Group 1 (5 students): 5×100=500\displaystyle 5 \times 100 = 500 kg - Group 2 (5 students): 5×125=625\displaystyle 5 \times 125 = 625 kg **Step 3: Remaining 5 students.** Sum of remaining=1650500625=525 kg\text{Sum of remaining} = 1650 - 500 - 625 = 525 \text{ kg} Mean=5255=105 kg\text{Mean} = \frac{525}{5} = 105 \text{ kg} Hence, **Option B** 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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