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At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?

130 points
220 points
310 points
412 points

Answer:- 1

Explanation:-

Solution:

\begin{aligned} A : B = 60 : 45 \\ A : C = 60 : 40 \\ ∴ \frac{B}{C} = (\frac{B}{A} \times \frac{A}{C}) = (\frac{45}{60} \times \frac{60}{40}) \\ = \frac{45}{40} = \frac{90}{80} = 90 : 80 \\ ∴ B can give C 10 points in a game of 90 \end{aligned}

$\eqalign{ & A:B = 60:45 \cr & A:C = 60:40 \cr & \therefore \frac{B}{C} = \left( {\frac{B}{A} \times \frac{A}{C}} \right) = \left( {\frac{{45}}{{60}} \times \frac{{60}}{{40}}} \right) \cr & = \frac{{45}}{{40}} = \frac{{90}}{{80}} = 90:80 \cr & \therefore {\text{B}}\,{\text{can}}\,{\text{give}}\,{\text{C}}\,{\text{10}}\,{\text{points}}\,{\text{in}}\,{\text{a}}\,{\text{game}}\,{\text{of}}\,{\text{90}} \cr}$

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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} A : B = 60 : 45 \\ A : C = 60 : 40 \\ ∴ \frac{B}{C} = (\frac{B}{A} \times \frac{A}{C}) = (\frac{45}{60} \times \frac{60}{40}) \\ = \frac{45}{40} = \frac{90}{80} = 90 : 80 \\ ∴ B can give C 10 points in a game of 90 \end{aligned}

$\eqalign{ & A:B = 60:45 \cr & A:C = 60:40 \cr & \therefore \frac{B}{C} = \left( {\frac{B}{A} \times \frac{A}{C}} \right) = \left( {\frac{{45}}{{60}} \times \frac{{60}}{{40}}} \right) \cr & = \frac{{45}}{{40}} = \frac{{90}}{{80}} = 90:80 \cr & \therefore {\text{B}}\,{\text{can}}\,{\text{give}}\,{\text{C}}\,{\text{10}}\,{\text{points}}\,{\text{in}}\,{\text{a}}\,{\text{game}}\,{\text{of}}\,{\text{90}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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