Math Problem Statement

Three athletes, Samrat, Sarath, and Sridhar, run a race. Sarath finished 24 meters ahead of Sridhar and 36 meters ahead of Samrat, while Sridhar finished 16 meters ahead of Samrat. If each athlete runs the entire distance at their respective constant speeds, what is the length of the race?

Solution

From the uploaded image, the problem states:

Three athletes, Samrat, Sarath, and Sridhar, run a race. Sarath finishes 24 meters ahead of Sridhar and 36 meters ahead of Samrat, while Sridhar finishes 16 meters ahead of Samrat. If each athlete runs the entire distance at their respective constant speeds, what is the length of the race?


Solution:

Let the length of the race be LL.

  • Assume their speeds:
    • Speed of Samrat = SSamratS_{\text{Samrat}},
    • Speed of Sarath = SSarathS_{\text{Sarath}},
    • Speed of Sridhar = SSridharS_{\text{Sridhar}}.

Each athlete completes the race in their own time. The relationship between speed, distance, and time is Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}.

Key relationships given:

  1. Sarath finishes 24 meters ahead of Sridhar:

    • When Sarath finishes the race (i.e., runs LL meters), Sridhar has run L24L - 24 meters.

    Hence, their times are equal: LSSarath=L24SSridhar.\frac{L}{S_{\text{Sarath}}} = \frac{L - 24}{S_{\text{Sridhar}}}.

  2. Sarath finishes 36 meters ahead of Samrat:

    • When Sarath finishes the race, Samrat has run L36L - 36 meters.

    Hence: LSSarath=L36SSamrat.\frac{L}{S_{\text{Sarath}}} = \frac{L - 36}{S_{\text{Samrat}}}.

  3. Sridhar finishes 16 meters ahead of Samrat:

    • When Sridhar finishes the race, Samrat has run L16L - 16 meters.

    Hence: LSSridhar=L16SSamrat.\frac{L}{S_{\text{Sridhar}}} = \frac{L - 16}{S_{\text{Samrat}}}.


Eliminating speeds:

From these equations, solve step-by-step:

(i) From the first equation: LSSarath=L24SSridhar    SSridharSSarath=L24L.\frac{L}{S_{\text{Sarath}}} = \frac{L - 24}{S_{\text{Sridhar}}} \implies \frac{S_{\text{Sridhar}}}{S_{\text{Sarath}}} = \frac{L - 24}{L}.

(ii) From the second equation: LSSarath=L36SSamrat    SSamratSSarath=L36L.\frac{L}{S_{\text{Sarath}}} = \frac{L - 36}{S_{\text{Samrat}}} \implies \frac{S_{\text{Samrat}}}{S_{\text{Sarath}}} = \frac{L - 36}{L}.

(iii) From the third equation: LSSridhar=L16SSamrat    SSamratSSridhar=L16L.\frac{L}{S_{\text{Sridhar}}} = \frac{L - 16}{S_{\text{Samrat}}} \implies \frac{S_{\text{Samrat}}}{S_{\text{Sridhar}}} = \frac{L - 16}{L}.


Combining ratios:

Using the above relationships, solve for LL:

  1. Combine: SSridharSSarathSSarathSSamratSSamratSSridhar=1.\frac{S_{\text{Sridhar}}}{S_{\text{Sarath}}} \cdot \frac{S_{\text{Sarath}}}{S_{\text{Samrat}}} \cdot \frac{S_{\text{Samrat}}}{S_{\text{Sridhar}}} = 1.

  2. Substitute the ratios: L24LL36LLL16=1.\frac{L - 24}{L} \cdot \frac{L - 36}{L} \cdot \frac{L}{L - 16} = 1.

  3. Simplify: (L24)(L36)L(L16)=1.\frac{(L - 24)(L - 36)}{L(L - 16)} = 1.

  4. Expand and solve: (L24)(L36)=L(L16).(L - 24)(L - 36) = L(L - 16). Expand both sides: L260L+864=L216L.L^2 - 60L + 864 = L^2 - 16L. Simplify: 60L+864=16L.-60L + 864 = -16L. 864=44L.864 = 44L. L=86444=19.636meters.L = \frac{864}{44} = 19.636 \, \text{meters}.


Thus, the length of the race is approximately 19.64 meters.


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Math Problem Analysis

Mathematical Concepts

Algebra
Ratios and Proportions
Linear Equations

Formulas

Time = Distance / Speed
Ratio comparison of speeds
Expansion of binomials for solving quadratic equations

Theorems

Proportionate speeds in uniform motion
Quadratic equation solutions

Suitable Grade Level

Grades 9-11