A car travels $\\frac{3}{4}$th of the distance at a speed of $60 \\text{ km\/hr}$ and the remaining $\\frac{1}{4}$th of the distance at a speed of $v \\text{ km\/hr}$. If the average speed for the full journey is $50 \\text{ km\/hr}$, then the value of $v$ is<\/p>\n
[amp_mcq option1=”40″ option2=”30″ option3=”100\/3″ option4=”35″ correct=”option3″]<\/p>\n
Part 2: Distance $D_2 = D – D_1 = D – \\frac{3}{4}D = \\frac{1}{4}D$. Speed $S_2 = v \\text{ km\/hr}$.
\nTime taken for Part 2: $T_2 = \\frac{D_2}{S_2} = \\frac{\\frac{1}{4}D}{v} = \\frac{D}{4v}$ hours.<\/p>\n
Total distance for the journey is $D$.
\nTotal time for the journey is $T_{total} = T_1 + T_2 = \\frac{D}{80} + \\frac{D}{4v}$.<\/p>\n
The average speed for the full journey is given as $50 \\text{ km\/hr}$.
\nAverage Speed = $\\frac{\\text{Total Distance}}{\\text{Total Time}}$
\n$50 = \\frac{D}{\\frac{D}{80} + \\frac{D}{4v}}$<\/p>\n
We can factor out $D$ from the denominator:
\n$50 = \\frac{D}{D\\left(\\frac{1}{80} + \\frac{1}{4v}\\right)}$
\n$50 = \\frac{1}{\\frac{1}{80} + \\frac{1}{4v}}$<\/p>\n
Taking the reciprocal of both sides:
\n$\\frac{1}{50} = \\frac{1}{80} + \\frac{1}{4v}$<\/p>\n
Now, solve for $v$:
\n$\\frac{1}{4v} = \\frac{1}{50} – \\frac{1}{80}$<\/p>\n
Find a common denominator for the right side, which is 400:
\n$\\frac{1}{4v} = \\frac{8}{400} – \\frac{5}{400}$
\n$\\frac{1}{4v} = \\frac{8 – 5}{400}$
\n$\\frac{1}{4v} = \\frac{3}{400}$<\/p>\n
Cross-multiply:
\n$4v \\times 3 = 1 \\times 400$
\n$12v = 400$<\/p>\n
Divide by 12:
\n$v = \\frac{400}{12}$<\/p>\n
Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 4:
\n$v = \\frac{400 \\div 4}{12 \\div 4} = \\frac{100}{3}$<\/p>\n