If mean of 5 observations x+1, x+3, x+5, x+7 and x+9 is given 15, then the value of x will be:

Option a: 10. Solution: Step 1: Compute the mean of the 5 observations. Mean = (x+1)+(x+3)+(x+5)+(x+7)+(x+9)/5 = 5x + (1+3+5+7+9)/5 = 5x + 25/5 = x + 5 Step 2: Set equal to 15 and solve. x + 5 = 15 x = 10 Hence, Option A is the correct answer.

Measures of Central Tendency and DispersionMCQPYQ Dec 23Question 2875 of 473

All Questions

If mean of 5 observations $\displaystyle x+1, x+3, x+5, x+7$ and $\displaystyle x+9$ is given 15, then the value of $\displaystyle x$ will be:

Options

\displaystyle 10

\displaystyle 12

\displaystyle 8

\displaystyle 11

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle 10$

All Options:

A $\displaystyle 10$
B $\displaystyle 12$
C $\displaystyle 8$
D $\displaystyle 11$

Detailed Solution & Explanation

**Step 1: Compute the mean of the 5 observations.**

\text{Mean} = \frac{(x+1)+(x+3)+(x+5)+(x+7)+(x+9)}{5}

= \frac{5x + (1+3+5+7+9)}{5} = \frac{5x + 25}{5} = x + 5

**Step 2: Set equal to 15 and solve.**

x + 5 = 15 \implies x = 10

Hence, **Option A** 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.

Related Comparison Tables

Mean vs Median vs Mode: Measures of Central Tendency Fiscal Policy vs Monetary Policy: CA Foundation Economics

Ready to Master Measures of Central Tendency and Dispersion?

Practice all 473 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

If mean of 5 observations x+1,x+3,x+5,x+7\displaystyle x+1, x+3, x+5, x+7x+1,x+3,x+5,x+7 and x+9\displaystyle x+9x+9 is given 15, then the value of x\displaystyle xx will be: