Akshat completes his journey in 20 hours. He covers half the d... | Mathematics MCQ | Examoogle

MathematicsPrevious Year Paper 2024

Akshat completes his journey in 20 hours. He covers half the distance at 57 km/h and the rest at 95 km/h. What is the length of the journey (in Km)?

1435

1423

1416

1425

Correct Answer

📐 MA • Math Direct FormulaMathematicsPrevious Year Paper 2024

Option D

1425

Quick Summary:

Use average speed for equal half-distances: Average Speed = (2 × 57 × 95) / (57 + 95) = (2 × 57 × 95) / 152 = 71.25 km/h. Total Distance = 71.25 × 20 = 1425 km.

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📐MAMath SolutionDirect Formula

📋 Given

Total time T = 20 hours, speed for first half distance v1 = 57 km/h, speed for second half distance v2 = 95 km/h.

🔢 Formula Used

vavg=2v1v2v1+v2,D=vavg×Tv_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}, \quad D = v_{avg} \times Tvavg​=v1​+v2​2v1​v2​​,D=vavg​×T

⚡ Exam Hall Shortcut / Speed Trick

Use average speed for equal half-distances: Average Speed = (2 × 57 × 95) / (57 + 95) = (2 × 57 × 95) / 152 = 71.25 km/h. Total Distance = 71.25 × 20 = 1425 km.

⚠️ Common Student Trap / Pitfall

Taking the simple arithmetic mean of speeds ((57 + 95)/2 = 76 km/h) and multiplying by 20 gives 1520 km, which is incorrect because time spent on each half is different.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Define distance and express time for each half

Let the total length of the journey be $D$ km. The journey is divided into two equal parts, each of length $\frac{D}{2}$ km.

Distance of each half=D2\text{Distance of each half} = \frac{D}{2}Distance of each half=2D​

Set up the time equation

Time is equal to distance divided by speed. The total time for the journey is the sum of time taken for the two halves: $T = t_1 + t_2 = \frac{D/2}{57} + \frac{D/2}{95} = 20$ .

D2×57+D2×95=20\frac{D}{2 \times 57} + \frac{D}{2 \times 95} = 202×57D​+2×95D​=20

Simplify and solve for D

Factor out $\frac{D}{2}$ : $\frac{D}{2} \left( \frac{1}{57} + \frac{1}{95} \right) = 20$ . Taking LCM of $57$ and $95$ ( $19 \times 3$ and $19 \times 5$ , so LCM is $19 \times 3 \times 5 = 285$ ), we get $\frac{1}{57} + \frac{1}{95} = \frac{5 + 3}{285} = \frac{8}{285}$ .

D2×8285=20  ⟹  4D285=20\frac{D}{2} \times \frac{8}{285} = 20 \implies \frac{4D}{285} = 202D​×2858​=20⟹2854D​=20

Calculate the final distance

Multiply both sides by $285$ and divide by $4$ : $D = \frac{20 \times 285}{4} = 5 \times 285 = 1425\text{ km}$ .

D=1425 kmD = 1425\text{ km}D=1425 km

✅

D is correct because multiplying the harmonic mean average speed ( $71.25\text{ km/h}$ ) by the total time of $20\text{ hours}$ gives a total journey distance of $1425\text{ km}$ .

Core Concepts Used

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Average Speed for Equal Distances Time, Speed, and Distance Harmonic Mean of Speeds

💡 EXAM TIP

When equal distances are covered at speeds $a$ and $b$ , the overall average speed is always the harmonic mean $v_{avg} = \frac{2ab}{a+b}$ . This concept is frequently tested in both Time-Speed-Distance and Work-and-Time questions.

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