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A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:

1100 m
2111 ¹/₉ m
325 m
450 m

Answer:- 1

Explanation:-

Solution:

\begin{aligned} When B runs 25m, A runs \frac{45}{2} m \\ When B runs 1000m, A runs \\ = (\frac{45}{2} \times \frac{1}{25} \times 1000) m \\ = 900 m \\ ∴ B beats A by 100m \end{aligned}

$\eqalign{ & {\text{When}}\,{\text{B}}\,\,{\text{runs}}\,{\text{25m,}}\,{\text{A}}\,{\text{runs}}\,\frac{{45}}{2}m \cr & {\text{When}}\,{\text{B}}\,\,{\text{runs}}\,{\text{1000m,}}\,{\text{A}}\,{\text{runs}} \cr & = \left( {\frac{{45}}{2} \times \frac{1}{{25}} \times 1000} \right)m \cr & = 900m \cr & \therefore {\text{B}}\,{\text{beats}}\,{\text{A}}\,{\text{by}}\,{\text{100m}} \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} When B runs 25m, A runs \frac{45}{2} m \\ When B runs 1000m, A runs \\ = (\frac{45}{2} \times \frac{1}{25} \times 1000) m \\ = 900 m \\ ∴ B beats A by 100m \end{aligned}

$\eqalign{ & {\text{When}}\,{\text{B}}\,\,{\text{runs}}\,{\text{25m,}}\,{\text{A}}\,{\text{runs}}\,\frac{{45}}{2}m \cr & {\text{When}}\,{\text{B}}\,\,{\text{runs}}\,{\text{1000m,}}\,{\text{A}}\,{\text{runs}} \cr & = \left( {\frac{{45}}{2} \times \frac{1}{{25}} \times 1000} \right)m \cr & = 900m \cr & \therefore {\text{B}}\,{\text{beats}}\,{\text{A}}\,{\text{by}}\,{\text{100m}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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