Races And Games — Part 1 MCQs with Explanations

$$\eqalign{ & A:B = 100:90 \cr & A:C = 100:72 \cr & B:C = {B \over A} \times {A \over C} = {{90} \over {100}} \times {{100} \over {72}} = {{90} \over {72}} \cr} $$ When B runs 90 m, C runs 72 m. When B runs 100m, C run $$\left( {{{72} \over {90}} \times 100} \right)m = 80\,m$$ ∴ B can give C 20 m

$$\eqalign{ & {\text{A's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {5 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{18}}{\text{m/sec}} \cr & {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{A}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 100 m}} \cr & = \left( {100 \times \frac{{18}}{{25}}} \right)\sec \cr & = 72\sec \cr & \therefore {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{B}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 92 m}} \cr & = \left( {72 + 8} \right) = 80\sec \cr & \therefore {\text{B's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right){\text{kmph}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.14{\text{ kmph}} \cr} $$

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 $$\left( {\frac{4}{3} \times 360} \right)$$ 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.

$$\eqalign{ & A:B = 100:90 \cr & A:C = 100:87 \cr & \frac{B}{C} = \frac{B}{A} \times \frac{A}{C} \cr & \,\,\,\,\,\,\,\, = \frac{{90}}{{100}} \times \frac{{100}}{{87}} \cr & \,\,\,\,\,\,\,\, = \frac{{30}}{{29}} \cr} $$ When B runs180m, C runs $$ \left( {\frac{{29}}{{30}} \times 180} \right){\text{m}} = 174{\text{m}}$$ ∴ B beats C by (180 - 174) m = 6 m

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

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

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

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

B runs 35m in 7sec ∴ B covers 200m in $$ {\frac{7}{{35}} \times 200} = 40\sec $$ B's time over the course = 40 sec ∴ A's time over the course (40 - 7) sec = 33 sec

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

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

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 $$\left( {\frac{5}{2} \times 80} \right){\text{m}} = 200{\text{ m}}$$ ∴ Winning post is 200 m away from the starting point

$$\eqalign{ & A:B = 100:75 \cr & B:C = 100:96 \cr & \therefore A:C = {\frac{A}{B} \times \frac{B}{C}} \cr & = {\frac{{100}}{{75}} \times \frac{{100}}{{96}}} \cr & = \frac{{100}}{{72}} \cr & 100:72 \cr & \therefore A\,\text{beats}\,C\,by \cr & = \left( {100 - 72} \right)m \cr & = 28\,m \cr} $$

B : A = 200 : 190 C : B = 200 : 180 $$\eqalign{ & = \frac{C}{A} \cr & = {\frac{C}{B} \times \frac{B}{A}} \cr & = {\frac{{200}}{{180}} \times \frac{{200}}{{190}}} \cr & = \frac{{200}}{{171}} \cr} $$ ∴ C can give to A, a start of (200 - 171) m = 29 m

In a km race, suppose A takes t sec, Then, B takes (t + 30) sec, and C takes (t + 45) sec 180 m is covered by C in 45 sec ∴ 1000 m is covered by C in = $$\left( {\frac{{45}}{{180}} \times 1000} \right)$$ sec = 250 sec ∴ A covers 1000 m in = (250 - 45) sec = 205 sec

Number of games played = 40 Number of won games = 24 Percentage of games played = $$\left( {\frac{{24}}{{40}} \times 100} \right)$$ = 60%

A : B : C = 1000 : (1000 - 50) : (1000 - 69) = 1000 : 950 : 931 In a 950 m race, B can give C a start of (950 - 931) m = 19 m In a 1000 m race, B can give C a start of = $$\left( {\frac{{19}}{{950}} \times 1000} \right)$$ = 20 m

A's speed = 5 km/hr = $$\left( {5 \times \frac{5}{{18}}} \right)$$ m/sec = $$\frac{25}{18}$$ m/sec Time taken by A to cover 100 metres = $$\left( {100 \times \frac{{18}}{{25}}} \right)$$ = 72 sec Time taken by B to cover 92 metres = (72 + 8) sec = 80 sec B's speed $$\frac{92}{80}$$ m/sec = $$\left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right)$$ km/hr = $$\frac{414}{100}$$ km/hr = 4.14 km/hr

End of Suvarna Tara Uma Vibha 4 th Game 32 32 32 32 3 rd Game 16 16 16 80 2 nd Game 8 8 72 40 1 st Game 4 68 36 20 Original Money 62 34 18 10 Survarna starts with Rs. 66

Suppose A covers 400 m in t seconds Then, B covers 385 m in (t + 5) seconds $$\eqalign{ & \therefore {\text{ B covers 400 m in}} \cr & = \left\{ {\frac{{\left( {t + 5} \right)}}{{385}} \times 400} \right\}{\text{ sec}} \cr & = \frac{{80\left( {t + 5} \right)}}{{77}}{\text{ sec}} \cr & {\text{Also, B covers 400 m in}} \cr & = \left( {t + 7\frac{1}{7}} \right){\text{ sec}} \cr & = \frac{{\left( {7t + 50} \right)}}{7}{\text{ sec}} \cr & \therefore \frac{{80\left( {t + 5} \right)}}{{77}} = \frac{{7t + 50}}{7} \cr & \Rightarrow 80\left( {t + 5} \right) = 11\left( {7t + 50} \right) \cr & \Rightarrow \left( {80t - 77t} \right) = \left( {550 - 400} \right) \cr & \Rightarrow 3t = 150 \cr & \Rightarrow t = 50 \cr & \therefore {\text{ A's speed}} \cr & = \frac{{400}}{{50}}\,m/\sec \cr & = 8\,m/\sec \cr & \therefore {\text{B's speed}} \cr & = \frac{{385}}{{55}}\,m/\sec \cr & = 7\,m/\sec \cr} $$

A : B = 1000 : 900 and B : C = 1000 : 850 $$\eqalign{ & \therefore \frac{A}{C} = \frac{A}{B} \times \frac{B}{C} \cr & \,\,\,\,\,\,\,\, = \frac{{1000}}{{900}} \times \frac{{1000}}{{850}} \cr & \,\,\,\,\,\,\,\, = \frac{{1000}}{{765}} \cr} $$ A : C = 1000 : 765 ∴ A beats C by (1000 - 765) m = 235 m

Distance covered by B in 25 sec = 200 m Distance covered by B in 22 sec =$$\left( {\frac{{200}}{{25}} \times 22} \right)$$ m = 176 m ∴ B was at a distance of (200 - 176) m = 24 m from the finishing line.

A : B = 1000 : 900, B : C = 400 : 360 = 100 : 90 = 900 : 810 ⇒ A : B : C = 1000 : 900 : 810 ⇒ A : C = 1000 : 810 ⇒ A : C = 500 : 405 ⇒ In a 500 m race, A beats C by (500 - 405) m = 95 m

B covers 35 m in 7 seconds. B covers 200 m in = $$\left( {\frac{7}{{35}} \times 200} \right)$$ sec = 40 sec Time taken by A = (40 - 7) sec = 33 sec

A's speed = 8 km/hr = $$\left( {8 \times \frac{5}{{18}}} \right)$$ m/sec = $$\frac{20}{9}$$ m/sec Time taken by A to cover 800 m = $$\left( {800 \times \frac{9}{{20}}} \right)$$ sec = 360 sec Time taken by B to cover 800 m = (360 + 15) sec = 375 sec ∴ Speed of B : = $$\frac{800}{375}$$ m/sec = $$\left( {\frac{{800}}{{375}} \times \frac{{18}}{5}} \right)$$ km/hr = $$\frac{192}{25}$$ km/hr = $$7\frac{17}{25}$$ km/hr

Karan, 91 sec → 1547 m 1 sec → 17 m 78 sec → 1326 m Arjun, 78 sec → 1547 m Karan is defeated by Arjun by = 1547 - 1326 = 221 m

76 is the correct answer to “In a linear race of 500 m, A can beat B by 50 m and in a race of 600 m, B can beat C by 60 m. By how many metres will A beat C in a race of 400 m”. This question is standard in competitive examination syllabi for Races And Games.

$$\eqalign{ & {\text{Time}} = 3\,\min 40\sec \cr & = 3\frac{{40}}{{60}} \cr & = 3\frac{2}{3} \cr & = \frac{{11}}{3}\min \cr & {\text{Circumference of circle}} = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 7 \cr & = 44{\text{ m}} \cr & {\text{Distance}} = {\text{Speed}} \times {\text{Time}} \cr & {\text{44}} \times {\text{8}} = {\text{Speed}} \times \frac{{11}}{3} \times 60 \cr & {\text{Speed}} = \frac{{32}}{{20}} \cr & {\text{Speed}} = \frac{8}{5}\,{\text{m/s}} \cr & {\text{Speed}} = \frac{8}{5} \times \frac{{18}}{5}\,{\text{km/hr}} \cr & {\text{Speed}} = \frac{{144}}{{25}}\,{\text{km/hr}} \cr} $$

Ratio of speed of A and B = 5 : 4 When A runs 500 m, B runs 400 m A will pass B every time when the distance between them is 400 m A will cover = $$\frac{{400}}{{\left( {500 - 400} \right)}}$$ = 2000 m to overtake B each time Number of time = $$\frac{{5000}}{{2000}}$$ = 2.5 ≈ 2 times (because .5 time is not possible)

A → 250 m → 25 second 1000 m → 100 second B → 2500 m → 30 second 1000 m → 120 second B should run 20 ahead, i.e. 30 30 second → 250 20 second → $$\frac{{250 \times 20}}{{30}}$$ = 166.66 = 166.67 m

Beenu, 50 sec → 400 m 1 sec → 8 m 44 sec → 44 × 8 Amit,

$$\eqalign{ & {\text{Speed of B}} = \frac{{50}}{5} = 10{\text{ m/sec}} \cr & {\text{Time same}} \cr & {\text{Then,}} \cr} $$ $${\text{Difference in speed}} = \frac{{200}}{1} - \frac{{10}}{1} = \frac{{10}}{{19}}$$

$$\eqalign{ & {\text{Babita's speed}} = 1 \cr & {\text{Geeta's speed}} = \frac{5}{2} \cr} $$ $$\eqalign{ & \Rightarrow {\text{Relative speed in same direction}} \cr & = \frac{5}{2} - 1 = \frac{3}{2} \cr & \Rightarrow {\text{Time}} = \frac{{40}}{{\frac{3}{2}}} = \frac{{80}}{3} = 26.66 \cr & {\text{Distance travel by Babita}} \cr & = 1 \times 26.66\, - - - - - \,26.66 \cr & {\text{Total distance}} = 40 + 26.66 \cr & = 66.66 \cr & = 66.67 \cr} $$

$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{500 + 500}}{{625 + 375}} \cr & = \frac{{1000}}{{1000}} \cr & = 1{\text{ m/s}} \cr} $$

$$\eqalign{ & \frac{{1300}}{{\text{Q}}} - \frac{{1500}}{{\text{P}}} = 8{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{and, }}\frac{{1100}}{{\text{Q}}} = \frac{{1500}}{{\text{P}}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{{\text{P}}}{{\text{Q}}} = \frac{{15}}{{11}} = \frac{{15x}}{{11x}} \cr & {\text{From equation }}\left( {\text{i}} \right) \cr & \frac{{1300}}{{11x}} - \frac{{1500}}{{15x}} = 8 \cr & 19500 - 16500 = 8 \times 11x \times 15x \cr & 3000 = 8 \times 11 \times 15x \cr & x = \frac{{25}}{{11}} \cr & {\text{Time of P to travel 1500m}} = \frac{{1500}}{{15x}} \cr & = \frac{{1500 \times 11}}{{15 \times 25}} \cr & = 44\sec \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr} $$ $$\eqalign{ & 11\mu \to 25 \cr & 1\mu \to \frac{{25}}{{11}} \cr & 15\mu \to \frac{{15 \times 25}}{{11}} \cr & {\text{Speed of Q}} = \frac{{200}}{8} = 25 \cr & {\text{Time of P}} = \frac{{1500 \times 11}}{{15 \times 25}} = 44\sec \cr} $$

$${\text{A beat C by}} = \frac{{100}}{{25}} \times 9 = 36\,{\text{m}}$$

B's speed = 6 m/s B's distance in 10 second = 10 × 6 = 60 m \[\begin{array}{*{20}{c}} {}&{\text{A}}&{\text{B}} \\ {{\text{Distance}} \to }&{1500}&{1440} \\ {{\text{Speed}} \to }&x&6 \\ {{\text{Time}} \to }&{240{\text{ second}}}&{4\min } \end{array}\] ∵ Time same Time of B = Time of A Time of A = 4 min

\[\begin{array}{*{20}{c}} {\text{B}}&{\text{C}} \\ {600}&{500} \\ {600\,{\text{m}}}&{{\text{Race}}} \end{array}\] $${\text{Speed}} = \frac{{170}}{{30}} = \frac{{17}}{3}{\text{m/s}}$$

Let the time taken by A to cover 1 km = $$x$$ seconds Time taken by B and C = $$x$$ + 30 seconds and $$x$$ + 45 seconds $$\eqalign{ & \frac{x}{{x + 45}} = \frac{{41}}{{50}} \cr & 50x = 41x + 1845 \cr & x = 205{\text{ seconds}} \cr} $$

\[\begin{array}{*{20}{c}} {}&{\text{A}}&{\text{B}} \\ {{\text{Distance}} \to }&{400}&{385} \end{array}\] $$\eqalign{ & \frac{{400}}{{\text{A}}} = \frac{{385}}{{\text{B}}} - 10 \cr & \frac{{400}}{{16}} = \frac{{385}}{{\text{B}}} - 10 \cr & 25 + 10 = \frac{{385}}{{\text{B}}} \cr & {\text{B}} = 11 \cr & {\text{Speed of B}} = 11{\text{ m/s}} \cr} $$