Races-and-Games

1.In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:

1. 18 m
2. 20 m
3. 27 m
4. 9 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A : B = 100 : 90 \\ A : C = 100 : 72 \\ B : C = \frac{B}{A} \times \frac{A}{C} = \frac{90}{100} \times \frac{100}{72} = \frac{90}{72} \end{aligned}

$\eqalign{ & A:B = 100:90 \cr & A:C = 100:72 \cr & B:C = {B \over A} \times {A \over C} = {{90} \over {100}} \times {{100} \over {72}} = {{90} \over {72}} \cr}$ When B runs 90 m, C runs 72 m.
When B runs 100m, C run

(\frac{72}{90} \times 100) m = 80 m

$\left( {{{72} \over {90}} \times 100} \right)m = 80m$ ∴ B can give C 20m

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2.A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

1. 5.15 kmph
2. 4.14 kmph
3. 4.25 kmph
4. 4.4 kmph

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A^{'} s s p e e d = (5 \times \frac{5}{18}) m / \sec = \frac{25}{18} m / \sec \\ T i m e t a k e n b y A t o cover 100 m = (100 \times \frac{18}{25}) \sec = 72 \sec \\ ∴ T i m e t a k e n b y B t o cover 92 m = (72 + 8) = 80 \sec \\ ∴ B^{'} s s p e e d = (\frac{92}{80} \times \frac{18}{5}) k m p h = 4.14 k m p h \end{aligned}

$\eqalign{ & A's\,speed = \left( {5 \times \frac{5}{{18}}} \right)m/\sec = \frac{{25}}{{18}}m/\sec \cr & Time\,taken\,by\,A\,to\,\operatorname{cover} \,100m = \left( {100 \times \frac{{18}}{{25}}} \right)\sec = 72\sec \cr & \therefore Time\,taken\,by\,B\,to\,\operatorname{cover} \,92m = \left( {72 + 8} \right) = 80\sec \cr & \therefore B's\,speed = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right)kmph = 4.14kmph \cr}$

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3.In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:

1. 60 m
2. 40 m
3. 20 m
4. 10 m

Answer:Option 1

Explanation:

Solution:

To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.
While A covers 3 m, B covers 4 m.
While A covers 360 m, B covers (⁴/₃ * 360) m = 480 m.
Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.

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4.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?

1. 30 points
2. 20 points
3. 10 points
4. 12 points

Answer:Option 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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5.In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:

1. 22.75 m
2. 25 m
3. 19.5 m
4. 7 ⁴/₇ m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A : B = 200 : 169 \\ A : C = 200 : 182 \\ \frac{C}{B} = (\frac{C}{A} \times \frac{A}{B}) = (\frac{182}{200} \times \frac{200}{169}) = 182 : 169 \\ W h e n C covers 182 m, B covers 169 m \\ W h e n C covers 350 m, B c o v e r s (\frac{169}{182} \times 350) m = 325 m \\ T h e r e f o r e, C b e a t s B b y (350 - 325) m \\ = 25 m \end{aligned}

$\eqalign{ & A:B = 200:169 \cr & A:C = 200:182 \cr & \frac{C}{B} = \left( {\frac{C}{A} \times \frac{A}{B}} \right) = \left( {\frac{{182}}{{200}} \times \frac{{200}}{{169}}} \right) = 182:169 \cr & When\,C\,\operatorname{covers} \,182m,\,B\,\operatorname{covers} \,169m \cr & When\,C\,\operatorname{covers} \,350m,\,B\,covers\left( {\frac{{169}}{{182}} \times 350} \right)m = 325m \cr & Therefore,\,C\,beats\,B\,by\,\left( {350 - 325} \right)m \cr & = 25m \cr}$

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6.In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:

1. 20 m
2. 25 m
3. 22.5 m
4. 9 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Distance covered by B in 9 sec \\ = (\frac{100}{45} \times 9) m = 20 m \\ ∴ A b e a t s B b y 20 m e t r e s \end{aligned}

$\eqalign{ & {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{B}}\,{\text{in}}\,{\text{9}}\,{\text{sec}} \cr & = \left( {\frac{{100}}{{45}} \times 9} \right)m = 20m \cr & \therefore A\,beats\,B\,by\,20\,metres \cr}$

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7.In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:

1. 8 points
2. 10 points
3. 14 points
4. 40 points

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A : B = 100 : 80 \\ A : C = 100 : 72 \\ ∴ \frac{B}{C} = (\frac{B}{A} \times \frac{A}{C}) = (\frac{80}{100} \times \frac{100}{72}) \\ = \frac{10}{9} = \frac{100}{90} = 100 : 90 \\ ∴ B can give C 10 points \end{aligned}

$\eqalign{ & A:B = 100:80 \cr & A:C = 100:72 \cr & \therefore \frac{B}{C} = \left( {\frac{B}{A} \times \frac{A}{C}} \right) = \left( {\frac{{80}}{{100}} \times \frac{{100}}{{72}}} \right) \cr & = \frac{{10}}{9} = \frac{{100}}{{90}} = 100:90 \cr & \therefore {\text{B}}\,{\text{can}}\,{\text{give}}\,{\text{C}}\,{\text{10}}\,{\text{points}} \cr}$

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

1. 100 m
2. 111 ¹/₉ m
3. 25 m
4. 50 m

Answer:Option 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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9.In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is:

1. 86 sec
2. 80 sec
3. 76 sec
4. None of these

Answer:Option 1

Explanation:

Solution:

\begin{aligned} B r u n s \frac{45}{2} m i n 6 \sec \\ ∴ B covers 300 m i n (6 \times \frac{2}{45} \times 300) \sec \\ = 80 \sec \end{aligned}

$\eqalign{ & B\,runs\,\frac{{45}}{2}m\,in\,6\,\sec \cr & \therefore B\,\operatorname{covers} \,300m\,in\left( {6 \times \frac{2}{{45}} \times 300} \right)\sec \cr & = 80\sec \cr}$

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10.A runs 1²/₃ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

1. 200 m
2. 300 m
3. 270 m
4. 160 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Ratio of speeds of A and B = \frac{5}{3} : 1 = 5 : 3 \\ Thus, in race of 5m, A gains 2m over B \\ 2m are gained by A in a race of 5m \\ 80m will be gained by A in a race of \\ (\frac{5}{2} \times 80) m = 200 m \\ ∴ Winning post is 200m away from the starting point \end{aligned}

$\eqalign{ & {\text{Ratio}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{speeds}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{A}}{\kern 1pt} {\text{and}}{\kern 1pt} {\text{B}} = \frac{5}{3}:1 = 5:3 \cr & {\text{Thus,}}\,{\text{in}}\,{\text{race}}\,{\text{of}}\,{\text{5m,}}\,{\text{A}}\,{\text{gains}}\,{\text{2m}}\,{\text{over}}\,{\text{B}} \cr & {\text{2m}}\,{\text{are}}\,{\text{gained}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{a}}\,{\text{race}}\,{\text{of}}\,{\text{5m}} \cr & {\text{80m}}\,{\text{will}}\,{\text{be}}\,{\text{gained}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{a}}\,{\text{race}}\,{\text{of}}\, \cr & \left( {\frac{5}{2} \times 80} \right)m = 200m \cr & \therefore {\text{Winning}}\,{\text{post}}\,{\text{is}}\,{\text{200m}}\,{\text{away}}\,{\text{from}}\,{\text{the}}\,{\text{starting}}\,{\text{point}} \cr}$

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