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Solved Examples(Set 1) - Races and Games

1. A is $2\dfrac{1}{3}$ times as fast as B. If A gives B a start of $80$ metres, how long should the race course be so that both of them reach at the same time?
A. $170$ metresB. $140$ metres
C. $160$ metresD. $150$ metres
Discuss
answer with explanation

Answer: Option B

Explanation:

Solution 1

speed of A : speed of B = $\dfrac{7}{3}:1=7:3$

It means, in a race of $7$ metres, A gains $(7-3)=4$ metres.

If A needs to gain $80$ metres, race should be of $\dfrac{7}{4}×80$ $=140$ metres.

Solution 2

If A is $n$ times as fast as B and A gives B a start of $x$ metres, then the length of the race course required so that A and B reaches the winning post at the same time is $x\left(\dfrac{n}{n-1}\right)$ metres.

read more ...

Here, $n=2\dfrac{1}{3}=\dfrac{7}{3},\quad x=80$

Length of the race course $=x\left(\dfrac{n}{n-1}\right)$
$=80\left(\dfrac{\dfrac{7}{3}}{\dfrac{7}{3}-1}\right)=80\left(\dfrac{7}{7-3}\right)=140$ metres

2. A can run $224$ metres in $28$ seconds and B in $32$ seconds. By what distance can A beat B?
A. $32$ metreB. $24$ metre
C. $36$ metreD. $28$ metre
Discuss
answer with explanation

Answer: Option D

Explanation:

Solution 1

Clearly, A can beat B by $4$ seconds. Let's find out how much B will run in these $4$ seconds.

Speed of B $=\dfrac{224}{32}=7$ m/s
Distance covered by B in $4$ seconds $=7×4=28$ metres.

That is, A can beat B by $28$ metres.

Note: Distance covered by B in $4$ seconds can be directly obtained as $\dfrac{224}{8}=28$ metres without calculating speed of B. Here we divided $224$ by $8$ because $32$ seconds divided by $8$ gives $4$ seconds.

Solution 2

If A can run $x$ metre race in $t_1$ seconds and B in $t_2$ seconds, where $t_1 \lt t_2$, then A beats B by a distance of $\dfrac{x}{t_2}(t_2-t_1)$ metres.

read more ...

$x=224$ metres
$t_1=28$ seconds
$t_2=32$ seconds

Required distance $=\dfrac{224}{32}(32-28)=28$ metres.

3. In a $100$ metre race, A can give B $10$ metres and C $28$ metres. In the same race B can give C :
A. $15$ metresB. $10$ metres
C. $20$ metresD. $25$ metres
Discuss
answer with explanation

Answer: Option C

Explanation:

Solution 1

A : B $=100:90$
A : C $=100:72$

=> B : C $=90:72=100:\dfrac{72×100}{90}$ $=100:80$

That is, B can give C $20$ metres.

Solution 2

"A can give B $10$ metres".
=> In a $100$ metre race, if A starts from the starting point and B starts $10$ metres ahead of the starting point at the same time, both reach the end point at the same time.

"A can give C $28$ metres"
=> In a $100$ metre race, if A starts from the starting point and C starts $28$ metres ahead of the starting point at the same time, both reach the end point at the same time.

That is, in a $(100-10)=90$ metre race, B can give C $(28-10)=18$ metres

=> In a $100$ metre race, B can give C $\left(\dfrac{18}{90}×100\right)=20$ metres.

Solution 3

"A can give B $10$ metres".
=> In a $100$ metre race, if A starts from the starting point and B starts $10$ metres ahead of the starting point at the same time, both reach the end point at the same time.

"A can give C $28$ metres"
=> In a $100$ metre race, if A starts from the starting point and C starts $28$ metres ahead of the starting point at the same time, both reach the end point at the same time.

That is, while A covers $100$ metre, B covers $(100-10)=90$ metres and C covers $(100-28)=72$ metres.

=> When B covers $90$ metres, C covers $72$ metres.

=> When B covers $100$ metres, C covers $\left(\dfrac{72×100}{90}\right)$ $=80$ metres.

That is, in a $100$ metre race, B can give C $(100-80)$ $=20$ metres.

4. At a game of billiards, A can give B $15$ points in $60$ and A can give C $20$ points in $60.$ How many points can B give C in a game of $90?$
A. $20$ pointsB. $12$ points
C. $22$ pointsD. $10$ points
Discuss
answer with explanation

Answer: Option D

Explanation:

Solution 1

A : B $=60:45$
A : C $=60:40$

=> B : C $=45:40=90:80$

That is, B can give C $10$ points.

Solution 2

"A can give B $15$ points in $60$"
=> In a game of $60,$ if A starts with $0$ points and B starts with $15$ points, they both score $60$ points at the same time.

"A can give C $20$ points in $60$"
=> In a game of $60,$ if A starts with $0$ points and C starts with $20$ points, they both score $60$ points at the same time.

That is, in a game of $(60-15)=45,$ B can give C $(20-15)$ $=5$ points.

=> That is, in a game of $90,$ B can give C $(5×2)$ $=10$ points

Solution 3

"A can give B $15$ points in $60$"
=> In a game of $60,$ if A starts with $0$ points and B starts with $15$ points, they both score $60$ points at the same time.

"A can give C $20$ points in $60$"
=> In a game of $60,$ if A starts with $0$ points and C starts with $20$ points, they both score $60$ points at the same time.

That is, while A scores $60$ points, B scores $(60-15)=45$ points and C scores $(60-20)=40$ points.

=> When B scores $45$ points, C scores $40$ points.

=> When B scores $90$ points, C scores $80$ points.

That is, in a game of $90,$ B can give C $(90-80)=10$ points.

5. In a $500$ metre race, the ratio of the speeds of two contestants A and B is $3:4.$ A has a start of $140$ metres. Then, A wins by:
A. $20$ metresB. $60$ metres
C. $40$ metresD. $10$ metres
Discuss
answer with explanation

Answer: Option A

Explanation:

Solution 1

A speed : B speed $=3:4$

=> A distance : B distance $=3:4=360:480$

That is, A wins by $20$ metres.

Solution 2

To reach the winning point, A will have to cover a distance of $(500-140)=360$ metres.

speed of A : speed of B $=3:4$
That is, when A covers $3$ metres, B covers $4$ metres.

=> When A covers $360$ metres, B covers $\left(\dfrac{4}{3}×360\right)$ $=480$ metres.

Remaining distance B has to cover $=(500-480)$ $=20$ metres.
Therefore, A wins by $20$ metres.

Comments(7)

profileTeesta
2014-03-13 08:20:48 
south-west
like 0 dislike 0 reply
profilesrk
2013-12-08 08:52:18 
3. south west
like 0 dislike 0 reply
profileVISHAL
2013-11-22 09:04:37 
I am Standing at the centre of a circular field. I go down south to the edge of the field and then turning left I walk along the boundary of the field equal to three-eights of its length. Then I turn west and go right across to the opposite point to the boundary. In
which direction am I from the starting point?

(1) North-west     (2)North                       
(3) South-west     (4)West
(5) None of these
like 0 dislike 0 reply
profilesaurabh tripathi
2015-01-23 20:07:48 
north west is the correct answer
like 0 dislike 0
profilep
2014-09-01 17:31:27 
north west

like 0 dislike 0
profileSam
2013-11-24 16:41:23 
                      |
                      |
    |                 |
    |                 |
    |                 |
    ___________

Hence, answer is north
like 0 dislike 0
profileSam
2014-03-14 14:25:55 
Ok, South West seems to be the answer
like 0 dislike 0

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