Question

At a game of billiards $ A $ can give $ B $ $ 15 $ points in $ 60 $ , $ A $ can give $ 20 $ point to $ C $ in $ 60 $ . How many points $ B $ give to $ C $ in a game of $ 90 $ ?
A. $ 30 $
B. $ 20 $
C. $ 10 $
D. $ 12 $

Answer 1

Hint: The problem is of ratio, where the multiplication or division of the ratios gives the required ratio. A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours. A proportion on the other hand is an equation that says that two ratios are equivalent. ... If one number in a proportion is unknown you can find that number by solving the proportion.

Complete step-by-step answer:
Game 1
 $ A $ can give $ 15 $ points to $ B $ in a game of $ 60 $ .
It means $ A $ can score $ 60 $ points while $ B $ can score points $ 15 $ less than $ A $ .
Points scored by $ B $ , $ = 60 - 15 = 45 $
The ratio between the points scored by $ A $ to $ B $ is given by,
 $ A:B = 60:45 $
In fraction it can be written as
 $ \dfrac{A}{B} = \dfrac{{60}}{{45}} $
Simplifying the ratio by dividing the numerator and denominator by $ 15 $ ,
 $ \dfrac{A}{B} = \dfrac{4}{3} \cdots \left( 1 \right) $
Game 2
 $ A $ can give $ 20 $ points to $ C $ in a game of $ 60 $ .
It means $ A $ can score $ 60 $ points while $ C $ can score $ 20 $ points less than that scored by $ A $ .
Therefore, points scored by $ C $ , $ = 60 - 20 = 40 $
The ratio between the points scored by $ A $ to $ C $ is given by,
 $ A:C = 60:40 $
In fraction it can be written as
 $ \dfrac{A}{C} = \dfrac{{60}}{{40}} $
Simplifying the ratio by dividing the numerator and denominator by 20,
 $ \dfrac{A}{C} = \dfrac{3}{2} \cdots \left( 2 \right) $
From the observation of equation (1) and (2), it is clear that to obtain the ratio $ B:C $ or \[\left( {\dfrac{B}{C}} \right)\] divide equation (2) by equation (1),
 $ \left( {\dfrac{A}{C}} \right) \div \left( {\dfrac{A}{B}} \right) = \left( {\dfrac{3}{2}} \right) \div \left( {\dfrac{4}{3}} \right) $
Solving for $ \left( {\dfrac{B}{C}} \right) $ ,
 $
  \left( {\dfrac{A}{C}} \right) \times \left( {\dfrac{B}{A}} \right) = \left( {\dfrac{3}{2}} \right) \times \left( {\dfrac{3}{4}} \right) \\
  \dfrac{B}{C} = \dfrac{9}{8} \cdots \left( 3 \right) \\
  $
On multiplying equation (3) by ,
 $ \dfrac{B}{C} = \dfrac{{90}}{{80}} \cdots \left( 4 \right) $
From equation (4) it is clear that in a game, $ B $ can score $ 90 $ points while $ B $ can give $ 90-80=10 $ points to $ C $ or $ C $ can score $ 80 $ points.
So, the correct answer is “Option C”.

Note: The important thing in the above question is to understand that it is a ratio problem. In which two ratios are given and the third ratio is to be calculated.
For instance, the ratio of $ \dfrac{x}{y} = \dfrac{1}{2} $ and $ \dfrac{y}{z}=\dfrac{2}{5} $ . The ratio for $ \dfrac{x}{z} $ is obtained by multiplying the 2 ratios.
 $
  \dfrac{x}{y} \times \dfrac{y}{z} = \dfrac{x}{z} \\
  \dfrac{x}{z} = \dfrac{1}{2} \times \dfrac{2}{5} \\
  \dfrac{x}{z} = \dfrac{1}{5} \\
  $