A train covered the first 5 km of its journey at a speed of 30 km/hr , and the next 15 km at a speed of 45 km/hr . The average speed of the train was:

Option b: 40 km/hr. Solution: Step 1: Calculate total distance. Total distance = 5 + 15 = 20 km Step 2: Calculate time for each part. - Time for first 5 km at 30 km/hr: t1 = 5/30 = 1/6 hr - Time for next 15 km at 45 km/hr: t2 = 15/45 = 1/3 hr Step 3: Calculate total time. ttotal = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2 hr Step 4: Calculate average speed. v = distanceTotal time = 20/1/2 = 40 km/hr The average speed is 40 km/hr = Option B. Note: The exam key says Option D ( 42 km/hr), but the correct calculation gives 40 km/hr (Option B). Hence, Option B is the correct answer.

Measures of Central Tendency and DispersionMCQMTP Dec 22 - Series IQuestion 3019 of 473

All Questions

A train covered the first $\displaystyle 5 \text{ km}$ of its journey at a speed of $\displaystyle 30 \text{km/hr}$ , and the next $\displaystyle 15 \text{ km}$ at a speed of $\displaystyle 45 \text{km/hr}$ . The average speed of the train was:

Options

\displaystyle 38 \text{ km/hr}

\displaystyle 40 \text{ km/hr}

\displaystyle 36 \text{ km/hr}

\displaystyle 42 \text{ km/hr}

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Correct Answer

✅ Option b — $\displaystyle 40 \text{ km/hr}$

All Options:

A $\displaystyle 38 \text{ km/hr}$
B $\displaystyle 40 \text{ km/hr}$
C $\displaystyle 36 \text{ km/hr}$
D $\displaystyle 42 \text{ km/hr}$

Detailed Solution & Explanation

**Step 1: Calculate total distance.**

\text{Total distance} = 5 + 15 = 20 \text{ km}

**Step 2: Calculate time for each part.** - Time for first 5 km at 30 km/hr:

\displaystyle t_1 = \frac{5}{30} = \frac{1}{6}

hr - Time for next 15 km at 45 km/hr:

\displaystyle t_2 = \frac{15}{45} = \frac{1}{3}

hr **Step 3: Calculate total time.**

t_{\text{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \text{ hr}

**Step 4: Calculate average speed.**

\bar{v} = \frac{\text{Total distance}}{\text{Total time}} = \frac{20}{1/2} = 40 \text{ km/hr}

The average speed is

\displaystyle 40

km/hr = **Option B**. Note: The exam key says Option D (

\displaystyle 42

km/hr), but the correct calculation gives

\displaystyle 40

km/hr (Option B). 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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A train covered the first 5 km\displaystyle 5 \text{ km}5 km of its journey at a speed of 30km/hr\displaystyle 30 \text{km/hr}30km/hr, and the next 15 km\displaystyle 15 \text{ km}15 km at a speed of 45km/hr\displaystyle 45 \text{km/hr}45km/hr. The average speed of the train was: