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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:

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& A:B=100:90\\ & A:C=100:72\\ & B:C=\frac{B}{A}×\frac{A}{C}=\frac{90}{100}×\frac{100}{72}=\frac{90}{72}\end{array}$
When B runs 90 m, C runs 72 m.
When B runs 100m, C run
$\left(\frac{72}{90}×100\right)m=80m$
∴ B can give C 20m

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:

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& {A}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}speed=\left(5×\frac{5}{18}\right)m/\mathrm{sec}=\frac{25}{18}m/\mathrm{sec}\\ & Time\phantom{\rule{thinmathspace}{0ex}}taken\phantom{\rule{thinmathspace}{0ex}}by\phantom{\rule{thinmathspace}{0ex}}A\phantom{\rule{thinmathspace}{0ex}}to\phantom{\rule{thinmathspace}{0ex}}\mathrm{cover}\phantom{\rule{thinmathspace}{0ex}}100m=\left(100×\frac{18}{25}\right)\mathrm{sec}=72\mathrm{sec}\\ & \therefore Time\phantom{\rule{thinmathspace}{0ex}}taken\phantom{\rule{thinmathspace}{0ex}}by\phantom{\rule{thinmathspace}{0ex}}B\phantom{\rule{thinmathspace}{0ex}}to\phantom{\rule{thinmathspace}{0ex}}\mathrm{cover}\phantom{\rule{thinmathspace}{0ex}}92m=\left(72+8\right)=80\mathrm{sec}\\ & \therefore {B}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}speed=\left(\frac{92}{80}×\frac{18}{5}\right)kmph=4.14kmph\end{array}$

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:

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.

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?

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& A:B=60:45\\ & A:C=60:40\\ & \therefore \frac{B}{C}=\left(\frac{B}{A}×\frac{A}{C}\right)=\left(\frac{45}{60}×\frac{60}{40}\right)\\ & =\frac{45}{40}=\frac{90}{80}=90:80\\ & \therefore \text{B}\phantom{\rule{thinmathspace}{0ex}}\text{can}\phantom{\rule{thinmathspace}{0ex}}\text{give}\phantom{\rule{thinmathspace}{0ex}}\text{C}\phantom{\rule{thinmathspace}{0ex}}\text{10}\phantom{\rule{thinmathspace}{0ex}}\text{points}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{game}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{90}\end{array}$

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:

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& A:B=200:169\\ & A:C=200:182\\ & \frac{C}{B}=\left(\frac{C}{A}×\frac{A}{B}\right)=\left(\frac{182}{200}×\frac{200}{169}\right)=182:169\\ & When\phantom{\rule{thinmathspace}{0ex}}C\phantom{\rule{thinmathspace}{0ex}}\mathrm{covers}\phantom{\rule{thinmathspace}{0ex}}182m,\phantom{\rule{thinmathspace}{0ex}}B\phantom{\rule{thinmathspace}{0ex}}\mathrm{covers}\phantom{\rule{thinmathspace}{0ex}}169m\\ & When\phantom{\rule{thinmathspace}{0ex}}C\phantom{\rule{thinmathspace}{0ex}}\mathrm{covers}\phantom{\rule{thinmathspace}{0ex}}350m,\phantom{\rule{thinmathspace}{0ex}}B\phantom{\rule{thinmathspace}{0ex}}covers\left(\frac{169}{182}×350\right)m=325m\\ & Therefore,\phantom{\rule{thinmathspace}{0ex}}C\phantom{\rule{thinmathspace}{0ex}}beats\phantom{\rule{thinmathspace}{0ex}}B\phantom{\rule{thinmathspace}{0ex}}by\phantom{\rule{thinmathspace}{0ex}}\left(350-325\right)m\\ & =25m\end{array}$

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

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& \text{Distance}\phantom{\rule{thinmathspace}{0ex}}\text{covered}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{B}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{9}\phantom{\rule{thinmathspace}{0ex}}\text{sec}\\ & =\left(\frac{100}{45}×9\right)m=20m\\ & \therefore A\phantom{\rule{thinmathspace}{0ex}}beats\phantom{\rule{thinmathspace}{0ex}}B\phantom{\rule{thinmathspace}{0ex}}by\phantom{\rule{thinmathspace}{0ex}}20\phantom{\rule{thinmathspace}{0ex}}metres\end{array}$

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

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& A:B=100:80\\ & A:C=100:72\\ & \therefore \frac{B}{C}=\left(\frac{B}{A}×\frac{A}{C}\right)=\left(\frac{80}{100}×\frac{100}{72}\right)\\ & =\frac{10}{9}=\frac{100}{90}=100:90\\ & \therefore \text{B}\phantom{\rule{thinmathspace}{0ex}}\text{can}\phantom{\rule{thinmathspace}{0ex}}\text{give}\phantom{\rule{thinmathspace}{0ex}}\text{C}\phantom{\rule{thinmathspace}{0ex}}\text{10}\phantom{\rule{thinmathspace}{0ex}}\text{points}\end{array}$

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

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& \text{When}\phantom{\rule{thinmathspace}{0ex}}\text{B}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{runs}\phantom{\rule{thinmathspace}{0ex}}\text{25m,}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{runs}\phantom{\rule{thinmathspace}{0ex}}\frac{45}{2}m\\ & \text{When}\phantom{\rule{thinmathspace}{0ex}}\text{B}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{runs}\phantom{\rule{thinmathspace}{0ex}}\text{1000m,}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{runs}\\ & =\left(\frac{45}{2}×\frac{1}{25}×1000\right)m\\ & =900m\\ & \therefore \text{B}\phantom{\rule{thinmathspace}{0ex}}\text{beats}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{100m}\end{array}$

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

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& B\phantom{\rule{thinmathspace}{0ex}}runs\phantom{\rule{thinmathspace}{0ex}}\frac{45}{2}m\phantom{\rule{thinmathspace}{0ex}}in\phantom{\rule{thinmathspace}{0ex}}6\phantom{\rule{thinmathspace}{0ex}}\mathrm{sec}\\ & \therefore B\phantom{\rule{thinmathspace}{0ex}}\mathrm{covers}\phantom{\rule{thinmathspace}{0ex}}300m\phantom{\rule{thinmathspace}{0ex}}in\left(6×\frac{2}{45}×300\right)\mathrm{sec}\\ & =80\mathrm{sec}\end{array}$

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?

Answer:Option 1

Explanation:

Solution:
$\begin{array}{rl}& \text{Ratio}\phantom{\rule{1pt}{0ex}}\text{of}\phantom{\rule{1pt}{0ex}}\text{speeds}\phantom{\rule{1pt}{0ex}}\text{of}\phantom{\rule{1pt}{0ex}}\text{A}\phantom{\rule{1pt}{0ex}}\text{and}\phantom{\rule{1pt}{0ex}}\text{B}=\frac{5}{3}:1=5:3\\ & \text{Thus,}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{race}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{5m,}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{gains}\phantom{\rule{thinmathspace}{0ex}}\text{2m}\phantom{\rule{thinmathspace}{0ex}}\text{over}\phantom{\rule{thinmathspace}{0ex}}\text{B}\\ & \text{2m}\phantom{\rule{thinmathspace}{0ex}}\text{are}\phantom{\rule{thinmathspace}{0ex}}\text{gained}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{race}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{5m}\\ & \text{80m}\phantom{\rule{thinmathspace}{0ex}}\text{will}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\text{gained}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{A}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{race}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\\ & \left(\frac{5}{2}×80\right)m=200m\\ & \therefore \text{Winning}\phantom{\rule{thinmathspace}{0ex}}\text{post}\phantom{\rule{thinmathspace}{0ex}}\text{is}\phantom{\rule{thinmathspace}{0ex}}\text{200m}\phantom{\rule{thinmathspace}{0ex}}\text{away}\phantom{\rule{thinmathspace}{0ex}}\text{from}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{starting}\phantom{\rule{thinmathspace}{0ex}}\text{point}\end{array}$

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