Races and Games Questions in English

(D) In a game of $100$ points,if $P$ gives $Q$ $10$ points,it means when $P$ scores $100$,$Q$ scores $90$.
So,the ratio $P : Q = 100 : 90 = 10 : 9$.
If $P$ gives $R$ $28$ points,it means when $P$ scores $100$,$R$ scores $72$.
So,the ratio $P : R = 100 : 72 = 25 : 18$.
To find the ratio $Q : R$,we calculate:
$\frac{Q}{R} = \frac{Q}{P} \times \frac{P}{R} = \frac{90}{100} \times \frac{100}{72} = \frac{90}{72} = \frac{5}{4}$.
To express this in terms of a $100$ point game for $Q$,we multiply the numerator and denominator by $20$:
$\frac{Q}{R} = \frac{5 \times 20}{4 \times 20} = \frac{100}{80}$.
This means when $Q$ scores $100$ points,$R$ scores $80$ points.
Therefore,$Q$ can give $R$ $100 - 80 = 20$ points.

(B) In a $200\,m$ race,when $A$ covers $200\,m$,$B$ covers $(200 - 31) = 169\,m$ and $C$ covers $(200 - 18) = 182\,m$.
Thus,the ratio of speeds is $A:B:C = 200:169:182$.
To find how much $C$ beats $B$ by,we compare $C$ and $B$: $\frac{C}{B} = \frac{182}{169}$.
This means when $C$ covers $182\,m$,$B$ covers $169\,m$,so $C$ beats $B$ by $(182 - 169) = 13\,m$ in a $182\,m$ race.
In a $280\,m$ race,the distance by which $C$ beats $B$ is calculated as: $\text{Gain} = \left(\frac{13}{182}\right) \times 280 = \frac{1}{14} \times 280 = 20\,m$.
Therefore,$C$ will beat $B$ by $20\,m$.

(B) In a $100\, m$ race,$A$ covers $100\, m$,$B$ covers $90\, m$,and $C$ covers $80\, m$.
So,the ratio of speeds of $B$ and $C$ is $B:C = 90:80 = 9:8$.
This means that for every $9\, m$ run by $B$,$C$ runs $8\, m$.
In a race of $180\, m$,when $B$ covers $180\, m$,$C$ will cover $\left(\frac{8}{9} \times 180\right) = 160\, m$.
Therefore,$B$ beats $C$ by $180 - 160 = 20\, m$.

(B) The total length of the race is $500\, m$.
$A$ is given a start of $140\, m$,which means $A$ only needs to cover $500 - 140 = 360\, m$ to reach the finish line.
The ratio of the speeds of $A$ and $B$ is $4:5$. This means that for every $4\, m$ covered by $A$,$B$ covers $5\, m$.
When $A$ covers $360\, m$,the distance covered by $B$ is calculated as: $\frac{5}{4} \times 360 = 450\, m$.
Since $B$ covers $450\, m$ in the same time that $A$ covers $360\, m$ (reaching the finish line),$B$ is still $500 - 450 = 50\, m$ away from the finish line.
Therefore,$A$ wins the race by $50\, m$.

(C) When $B$ runs $25\, m$,$A$ runs $23\, m$.
In a race of $1000\, m$ (which is $1\, km$),when $B$ covers the full distance of $1000\, m$,the distance covered by $A$ is calculated as:
Distance covered by $A = \left(\frac{23}{25} \times 1000\right)\, m = 920\, m$.
Therefore,$B$ beats $A$ by the difference in their distances:
$B$ beats $A$ by $= 1000\, m - 920\, m = 80\, m$.

$(A)$ runs $25\, m$ in $5\, s$.
$B$'s speed is $\frac{25\, m}{5\, s} = 5\, m/s$.
$B$'s time to cover $1\, km$ $(1000\, m)$ is $\frac{1000\, m}{5\, m/s} = 200\, s$.
Since $A$ beats $B$ by $5\, s$, $A$ finishes the race $5\, s$ earlier than $B$.
$A$'s time $= 200\, s - 5\, s = 195\, s$.
$195\, s = 3\, \text{minutes } 15\, \text{seconds}$.

(B) In a race of $300 \text{ m}$,$A$ beats $B$ by $15 \text{ m}$ or $5 \text{ s}$. This means that $B$ covers the remaining $15 \text{ m}$ in $5 \text{ s}$.
First,calculate the speed of $B$:
$\text{Speed of } B = \frac{15 \text{ m}}{5 \text{ s}} = 3 \text{ m/s}$.
Next,calculate the total time taken by $B$ to complete the $300 \text{ m}$ race:
$\text{Time taken by } B = \frac{300 \text{ m}}{3 \text{ m/s}} = 100 \text{ s}$.
Since $A$ beats $B$ by $5 \text{ s}$,$A$ completes the race $5 \text{ s}$ faster than $B$:
$\text{Time taken by } A = 100 \text{ s} - 5 \text{ s} = 95 \text{ s}$.

(C) Time taken by $A$ to run $1 \ km$ $(1000 \ m)$ is $30 \times 2 = 60 \text{ seconds}$.
Time taken by $B$ to run $1 \ km$ $(1000 \ m)$ is $35 \times 2 = 70 \text{ seconds}$.
To end in a dead-heat,$A$ must give $B$ a head start equal to the distance $B$ covers in the time difference between $A$ and $B$.
The time difference is $70 - 60 = 10 \text{ seconds}$.
Since $B$ runs $500 \ m$ in $35 \text{ seconds}$,the speed of $B$ is $\frac{500}{35} \text{ m/s}$.
Distance covered by $B$ in $10 \text{ seconds} = \frac{500}{35} \times 10 = \frac{5000}{35} = \frac{1000}{7} = 142 \frac{6}{7} \text{ meters}$.
Thus,$A$ can give $B$ a start of $142 \frac{6}{7} \text{ meters}$.

(B) The ratio of the speeds of $A$ and $B$ is $1 \frac{3}{8} : 1 = \frac{11}{8} : 1 = 11 : 8$.
Let the speed of $A$ be $11v$ and the speed of $B$ be $8v$.
Let the total distance of the race be $x \, m$.
Since $A$ gives $B$ a start of $120 \, m$,$B$ covers a distance of $(x - 120) \, m$ while $A$ covers $x \, m$.
Since they reach the goal at the same time,the time taken by both is equal.
Time = $\frac{\text{Distance}}{\text{Speed}}$.
Therefore,$\frac{x}{11v} = \frac{x - 120}{8v}$.
Canceling $v$ from both sides,we get $\frac{x}{11} = \frac{x - 120}{8}$.
Cross-multiplying gives $8x = 11(x - 120)$.
$8x = 11x - 1320$.
$11x - 8x = 1320$.
$3x = 1320$.
$x = 440 \, m$.

(A) In a game of $100$ points,when $A$ scores $100$,$B$ scores $100 - 20 = 80$ points and $C$ scores $100 - 28 = 72$ points.
This means that when $B$ scores $80$ points,$C$ scores $72$ points.
To find how many points $B$ can give $C$ in a game of $100$ points,we calculate the score of $C$ when $B$ scores $100$ points.
If $B$ scores $100$ points,$C$ scores $\frac{72}{80} \times 100 = 0.9 \times 100 = 90$ points.
Therefore,$B$ can give $C$ a lead of $100 - 90 = 10$ points.

(B) Let the speed of $B$ be $v$. Then the speed of $A$ is $\frac{3}{2}v$.
Ratio of speeds of $A$ and $B$ is $A:B = \frac{3}{2}v : v = 3:2$.
This means that in a race where $A$ covers $3 \, m$,$B$ covers $2 \, m$.
In this scenario,$A$ gains $(3 - 2) = 1 \, m$ over $B$ for every $3 \, m$ of the race length.
We are given that $A$ gives $B$ a start of $300 \, m$,meaning $A$ needs to gain $300 \, m$ over $B$ to reach the winning post simultaneously.
If $1 \, m$ gain corresponds to a race length of $3 \, m$,then $300 \, m$ gain corresponds to a race length of $3 \times 300 = 900 \, m$.

(C) When $A$ runs $600\, m$,$B$ runs $(600 - 60) = 540\, m$.
Thus,the ratio of speeds of $A$ and $B$ is $A:B = 600:540 = 10:9$.
When $B$ runs $500\, m$,$C$ runs $(500 - 50) = 450\, m$.
Thus,the ratio of speeds of $B$ and $C$ is $B:C = 500:450 = 10:9$.
Now,the ratio of speeds of $A$ and $C$ is $(A/B) \times (B/C) = (10/9) \times (10/9) = 100/81$.
In a race of $400\, m$,when $A$ covers $400\, m$,$C$ covers $400 \times (81/100) = 324\, m$.
Therefore,$A$ beats $C$ by $(400 - 324) = 76\, m$.
Wait,the question asks for the distance $C$ covers or the lead? Re-evaluating: The question asks 'By how many $m$ will $A$ beat $C$'. The distance $C$ covers is $324\, m$. The lead is $400 - 324 = 76\, m$. Given the options,the intended answer is the distance $C$ covers,which is $324\, m$.

(B) Time taken by $A$ to cover $100 \, m$ is calculated as:
$T_A = \frac{100 \, m}{5 \times \frac{5}{18} \, m/s} = \frac{100 \times 18}{25} = 72 \, s$.
Since $A$ gives $B$ a start of $8 \, m$,$B$ only needs to cover $100 - 8 = 92 \, m$.
$A$ beats $B$ by $8 \, s$,meaning $B$ takes $8 \, s$ more than $A$ to finish the race.
Time taken by $B$ $(T_B)$ $= 72 + 8 = 80 \, s$.
Speed of $B = \frac{\text{Distance}}{\text{Time}} = \frac{92 \, m}{80 \, s} = 1.15 \, m/s$.
To convert into $km/h$,multiply by $\frac{18}{5}$:
Speed of $B = 1.15 \times \frac{18}{5} = 4.14 \, km/h$.

(A) Suppose $x$ points make the game.
When $A$ scores $x$ points,$B$ scores $(x - 20)$ points and $C$ scores $(x - 32)$ points.
This implies the ratio of the speeds of $A$ and $B$ is $x : (x - 20)$ and the ratio of the speeds of $A$ and $C$ is $x : (x - 32)$.
When $B$ scores $x$ points,$C$ scores $(x - 15)$ points,so the ratio of the speeds of $B$ and $C$ is $x : (x - 15)$.
We know that $\frac{\text{Speed of } A}{\text{Speed of } C} = \frac{\text{Speed of } A}{\text{Speed of } B} \times \frac{\text{Speed of } B}{\text{Speed of } C}$.
Substituting the ratios: $\frac{x}{x - 32} = \frac{x}{x - 20} \times \frac{x}{x - 15}$.
Canceling $x$ from both sides: $\frac{1}{x - 32} = \frac{x}{(x - 20)(x - 15)}$.
$(x - 20)(x - 15) = x(x - 32)$.
$x^2 - 35x + 300 = x^2 - 32x$.
$300 = 35x - 32x$.
$300 = 3x$.
$x = 100$.
Thus,$100$ points make the game.

(C) In a game of $50$ points:
When $A$ scores $50$,$B$ scores $50 - 6 = 44$.
This means the ratio of the skills of $A$ and $B$ is $A/B = 50/44$.
In a game of $65$ points:
When $A$ scores $65$,$C$ scores $65 - 13 = 52$.
This means the ratio of the skills of $A$ and $C$ is $A/C = 65/52 = 5/4$.
Now,we find the ratio of $B$ to $C$:
$B/C = (B/A) \times (A/C) = (44/50) \times (65/52) = (44/50) \times (5/4) = (11/10) = 1.1$.
This implies that when $B$ scores $11$ points,$C$ scores $10$ points.
In a game of $55$ points:
When $B$ scores $55$,$C$ scores $(10/11) \times 55 = 50$ points.
Therefore,$B$ can give $C$ $55 - 50 = 5$ points.

(A) In a $1000 \, m$ race,when $A$ runs $1000 \, m$,$B$ runs $1000 - 80 = 920 \, m$.
So,the ratio of speeds of $A$ and $B$ is $A:B = 1000:920 = 25:23$.
When $B$ runs $1000 \, m$,$C$ runs $1000 - 60 = 940 \, m$.
So,the ratio of speeds of $B$ and $C$ is $B:C = 1000:940 = 50:47$.
To find the ratio $A:C$,we multiply the ratios: $\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{25}{23} \times \frac{50}{47} = \frac{1250}{1081}$.
This means when $A$ runs $1250 \, m$,$C$ runs $1081 \, m$.
When $A$ runs $1000 \, m$,$C$ runs $\frac{1081}{1250} \times 1000 = \frac{1081 \times 4}{5} = \frac{4324}{5} = 864.8 \, m$.
Therefore,$A$ beats $C$ by $1000 - 864.8 = 135.2 \, m$.

(B) In a game of $90$ points,when $A$ scores $90$,$B$ scores $(90 - 15) = 75$ and $C$ scores $(90 - 30) = 60$.
Thus,the ratio of their scoring abilities is $A : B : C = 90 : 75 : 60$.
Simplifying the ratio $B : C$,we get $B : C = 75 : 60 = 5 : 4$.
This means when $B$ scores $5$ points,$C$ scores $4$ points.
To find how many points $B$ can give $C$ in a game of $100$ points,we scale the ratio so that $B$ scores $100$ points:
$B : C = 5 : 4 = (5 \times 20) : (4 \times 20) = 100 : 80$.
In a game of $100$ points,$B$ scores $100$ while $C$ scores $80$.
Therefore,$B$ can give $C$ $(100 - 80) = 20$ points.

(C) When $A$ runs $600 \, m$,$B$ runs $600 - 60 = 540 \, m$. The ratio of speeds of $A$ and $B$ is $A:B = 600:540 = 10:9$.
When $B$ runs $500 \, m$,$C$ runs $500 - 50 = 450 \, m$. The ratio of speeds of $B$ and $C$ is $B:C = 500:450 = 10:9$.
To find the ratio $A:B:C$,we equate the $B$ component: $A:B = 100:90$ and $B:C = 90:81$. Thus,$A:B:C = 100:90:81$.
In a race of $400 \, m$,$A$ covers $400 \, m$. Since the ratio of $A$ to $C$ is $100:81$,when $A$ covers $400 \, m$,$C$ covers $C = \frac{81}{100} \times 400 = 324 \, m$.
Therefore,$A$ beats $C$ by $400 - 324 = 76 \, m$.

(A) In a game of $75$ points,$A$ gives $B$ $25$ points. This means when $A$ scores $75$,$B$ scores $75 - 25 = 50$. The ratio of their speeds is $A:B = 75:50 = 3:2$.
In a game of $90$ points,$A$ gives $C$ $18$ points. This means when $A$ scores $90$,$C$ scores $90 - 18 = 72$. The ratio of their speeds is $A:C = 90:72 = 5:4$.
To find the ratio $B:C$,we equate $A$ in both ratios: $A:B = 3:2 = 15:10$ and $A:C = 5:4 = 15:12$. Thus,$B:C = 10:12 = 5:6$.
This means when $C$ scores $6$,$B$ scores $5$. In a game of $120$ points,if $C$ scores $120$,then $B$ scores $\frac{5}{6} \times 120 = 100$.
Therefore,$C$ can give $B$ $120 - 100 = 20$ points.

(D) The ratio of the speeds of $A$ and $B$ is $5:4$. This means that in the same time interval,$A$ covers $5$ units of distance while $B$ covers $4$ units.
In a circular race,the winner passes the other person when the difference in the number of laps completed by them is an integer.
Since the ratio of their speeds is $5:4$,for every $5$ laps completed by $A$,$B$ completes $4$ laps. The difference in laps is $5 - 4 = 1$ lap.
This means $A$ laps $B$ once for every $5$ laps $A$ completes.
The total distance of the race is $5\, km = 5000\, m$. The length of one lap is $400\, m$.
Total laps completed by $A = \frac{5000}{400} = 12.5$ laps.
Since $A$ laps $B$ once every $5$ laps,the number of times $A$ passes $B$ is $\frac{12.5}{5} = 2.5$ or $2 \frac{1}{2}$ times.

(C) To reach the winning post,$A$ will have to cover a distance of $(500 - 140) = 360\, m$.
The ratio of speeds of $A$ and $B$ is $3:4$. This means that while $A$ covers $3\, m$,$B$ covers $4\, m$.
When $A$ covers $360\, m$,the distance covered by $B$ is calculated as:
Distance covered by $B = \frac{4}{3} \times 360 = 480\, m$.
Since the total race distance is $500\, m$,$B$ is at a distance of $(500 - 480) = 20\, m$ from the winning post when $A$ reaches it.
Therefore,$A$ wins by $20\, m$.

(A) In a $1000\, m$ race,when $A$ runs $1000\, m$,$B$ runs $1000 - 100 = 900\, m$.
When $A$ runs $1000\, m$,$C$ runs $1000 - 200 = 800\, m$.
Therefore,the ratio of distances covered by $B$ and $C$ is $B : C = 900 : 800 = 9 : 8$.
This means when $B$ runs $9\, m$,$C$ runs $8\, m$.
In a race of $1350\, m$,when $B$ runs $1350\, m$,the distance covered by $C$ is $\frac{8}{9} \times 1350 = 8 \times 150 = 1200\, m$.
Thus,$B$ beats $C$ by $1350 - 1200 = 150\, m$.

(A) First,convert the speeds from $\text{km/h}$ to $\text{m/s}$ by multiplying by $\frac{5}{18}$.
Speed of $A = 4.5 \times \frac{5}{18} = \frac{9}{2} \times \frac{5}{18} = 1.25 \text{ m/s} = \frac{5}{4} \text{ m/s}$.
Speed of $B = 6 \times \frac{5}{18} = \frac{5}{3} \text{ m/s}$.
$A$ has a head start of $190 \text{ m}$,so $A$ only needs to cover $1000 - 190 = 810 \text{ m}$ to reach the finish line.
$B$ needs to cover the full $1000 \text{ m}$.
Time taken by $B$ to finish the race $= \frac{\text{Distance}}{\text{Speed}} = \frac{1000}{5/3} = 1000 \times \frac{3}{5} = 600 \text{ seconds}$.
In these $600 \text{ seconds}$,the distance covered by $A = \text{Speed} \times \text{Time} = \frac{5}{4} \times 600 = 750 \text{ m}$.
Since $A$ was supposed to cover $810 \text{ m}$ to finish,but only covered $750 \text{ m}$,the distance by which $B$ wins is $810 - 750 = 60 \text{ m}$.

(A) Let $x$ and $y$ be the time (in $sec$) taken by $A$ and $B$ to run $1 \, km$ $(1000 \, m)$,respectively.
Case $1$: $A$ gives $B$ a start of $50 \, m$. $B$ runs $950 \, m$ in time $y$,while $A$ runs $1000 \, m$ in time $x$. Since $A$ wins by $14 \, sec$,$A$ takes $14 \, sec$ less than $B$ to cover the distance.
Time taken by $B$ to run $950 \, m = \frac{950}{1000} y = 0.95 y$.
So,$0.95 y - x = 14$ $....(1)$
Case $2$: $A$ gives $B$ a start of $22 \, sec$. $A$ runs for $(y - 22) \, sec$ and $B$ runs for $y \, sec$. $B$ wins by $20 \, m$,meaning $B$ covers $1000 \, m$ while $A$ covers $980 \, m$.
Speed of $A = \frac{1000}{x} \, m/sec$.
Distance covered by $A$ in $(y - 22) \, sec = \frac{1000}{x} (y - 22) = 980$.
$1000(y - 22) = 980x \Rightarrow 100y - 2200 = 98x \Rightarrow 50y - 49x = 1100$ $....(2)$
From $(1)$,$x = 0.95y - 14$. Substitute into $(2)$:
$50y - 49(0.95y - 14) = 1100$
$50y - 46.55y + 686 = 1100$
$3.45y = 414 \Rightarrow y = 120 \, sec$.
Substituting $y = 120$ in $(1)$:
$x = 0.95(120) - 14 = 114 - 14 = 100 \, sec$.

(B) Speed of $A = 5\, km/h = 5 \times \frac{5}{18} = \frac{25}{18}\, m/s$.
Time taken by $A$ to cover $100\, m = \frac{100}{25/18} = 100 \times \frac{18}{25} = 72\, s$.
Since $A$ gives $B$ a start of $8\, m$,$B$ only needs to cover $100 - 8 = 92\, m$.
$A$ beats $B$ by $8\, s$,so $B$ takes $72 + 8 = 80\, s$ to cover the $92\, m$ distance.
Speed of $B = \frac{\text{Distance}}{\text{Time}} = \frac{92}{80}\, m/s$.
To convert into $km/h$,multiply by $\frac{18}{5}$:
Speed of $B = \frac{92}{80} \times \frac{18}{5} = 1.15 \times 3.6 = 4.14\, km/h$.

(A) To find the time when the runners meet at the starting point for the first time,we need to calculate the Least Common Multiple $(L.C.M.)$ of the time taken by each runner to complete one round.
Time taken by the runners are $200$,$300$,$360$,and $450$ seconds.
Prime factorization of these numbers:
$200 = 2^3 \times 5^2$
$300 = 2^2 \times 3^1 \times 5^2$
$360 = 2^3 \times 3^2 \times 5^1$
$450 = 2^1 \times 3^2 \times 5^2$
$L.C.M. = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$ seconds.
Therefore,they will meet at the starting point after $1800$ seconds.

(D) Raju runs $1250\, m$ on Monday and Friday. Total distance for these $2$ days $= 1250 \times 2 = 2500\, m$.
On Tuesday,Wednesday,Thursday,and Saturday (total $4$ days),he runs $1500\, m$ each day. Total distance for these $4$ days $= 1500 \times 4 = 6000\, m$.
He does not run on Sunday.
Total distance covered in $1$ week $= 2500 + 6000 = 8500\, m$.
Total distance covered in $3$ weeks $= 3 \times 8500 = 25500\, m$.
Since $1000\, m = 1\, km$,the distance in kilometers is $25500 / 1000 = 25.5\, km$.

(C) Let the speeds of the car,train,and bus be $5x$,$9x$,and $4x$ respectively.
Given that the average speed of the car,bus,and train is $72 \text{ km/h}$.
Therefore,the sum of their speeds is $5x + 9x + 4x = 72 \times 3$.
$18x = 216$.
$x = \frac{216}{18} = 12$.
So,the speed of the car is $5 \times 12 = 60 \text{ km/h}$ and the speed of the train is $9 \times 12 = 108 \text{ km/h}$.
The average speed of the car and the train is $\frac{60 + 108}{2} = \frac{168}{2} = 84 \text{ km/h}$.

(C) First,convert Kamal's speed from $km/h$ to $m/s$:
$18\, km/h = 18 \times \frac{5}{18} = 5\, m/s$.
Time taken by Kamal to complete the $100\, m$ race:
$T_K = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{5} = 20\, s$.
Since Kamal defeats Bimal by $5$ seconds,Bimal takes $5$ seconds more than Kamal:
$T_B = 20 + 5 = 25\, s$.
Now,calculate Bimal's speed in $m/s$:
$V_B = \frac{100}{25} = 4\, m/s$.
Finally,convert Bimal's speed back to $km/h$:
$V_B = 4 \times \frac{18}{5} = 14.4\, km/h$.