Measures of Central Tendency and DispersionMCQMTP Dec 22 Series IQuestion 2910 of 473
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In a group of persons, average weight is 60\displaystyle 60 kg. If the average of males and females taken separately is 80\displaystyle 80 kg and 50\displaystyle 50 kg respectively, find the ratio of the number of males to that of females.

Options

A2:3\displaystyle 2:3
B3:2\displaystyle 3:2
C2:1\displaystyle 2:1
D1:2\displaystyle 1:2
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Correct Answer

Option d1:2\displaystyle 1:2

All Options:

  • A2:3\displaystyle 2:3
  • B3:2\displaystyle 3:2
  • C2:1\displaystyle 2:1
  • D1:2\displaystyle 1:2

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Detailed Solution & Explanation

**Step 1: Let number of males =m\displaystyle = m and females =f\displaystyle = f.** 80m+50fm+f=60\frac{80m + 50f}{m + f} = 60 **Step 2: Solve.** 80m+50f=60m+60f80m + 50f = 60m + 60f 20m=10f20m = 10f mf=1020=12\frac{m}{f} = \frac{10}{20} = \frac{1}{2} So males : females =1:2\displaystyle = 1 : 2. **Note:** The computed ratio is **1:2** (Option D). But the `correct_option` given is A (2:3). Let us recheck: 80m+50f=60(m+f)=60m+60f20m=10fm/f=1/2\displaystyle 80m + 50f = 60(m+f) = 60m + 60f \Rightarrow 20m = 10f \Rightarrow m/f = 1/2. So males:females = 1:2. Option D says 1:2 which is the answer, but Option A says 2:3. Hence, **Option D** 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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