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A train passes a man standing on a platform in 9 sec. and pa

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Two men plan to run in opposite directions around a park [#permalink] New post 18 Aug 2012, 02:53
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Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths?

Official answer is: 250m

Many thanks!!!
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When Johns travels at 120 km/h he uses 30% more fuel [#permalink] New post 18 Aug 2012, 03:26
When Johns travels at 120 km/h he uses 30% more fuel than when he travels only at 80 km/h. If he can travel 19.5 km on one liter at 80 km/h, how far can he travel with 10 liters of fuel at 120 km/h?

The official answer: 150km

I don't get it. I always get 136,5 km

Many thanks in advance!!!
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A train passes a man standing on a platform in 9 sec. and pa [#permalink] New post 18 Aug 2012, 03:57
A train passes a man standing on a platform in 9 sec. and passes the platform in 12 sec. completely. If the platform is 50 meters long, how long is the train?

The offical answer is: 150m

Can somebody help me please?
Many thanks!!!
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Re: When Johns travels at 120 km/h he uses 30% more fuel [#permalink] New post 18 Aug 2012, 04:11
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NYC5648 wrote:
When Johns travels at 120 km/h he uses 30% more fuel than when he travels only at 80 km/h. If he can travel 19.5 km on one liter at 80 km/h, how far can he travel with 10 liters of fuel at 120 km/h?

The official answer: 150km

I don't get it. I always get 136,5 km

Many thanks in advance!!!


John travels with 1 liter of fuel at 80 km/h \, 19.5 km. For the same distance, when traveling at a speed of 120 km/h, he uses 1.3liters of fuel.
If we denote by x the distance he travels using 10 liters of fuel at the speed of 120 km/h, we can write \frac{1.3}{19.5}=\frac{10}{x}, from which we get x=\frac{10*19.5}{1.3}=150.
We used the fact that at the same speed, the fuel consumption rate is constant.
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Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink] New post 18 Aug 2012, 04:29
NYC5648 wrote:
A train passes a man standing on a platform in 9 sec. and passes the platform in 12 sec. completely. If the platform is 50 meters long, how long is the train?

The offical answer is: 150m

Can somebody help me please?
Many thanks!!!


Passing a man standing means the train travels a distance equal to its length. If we denote by L the train's length, it means the train travels the distance L in 9 seconds.
Passing the platform completely means the train travels a total distance equal to the length of the platform plus the length of itself, which is 50+L. We can deduce that the train travels 50m in 3 seconds, so in 9 seconds, it travels 3*50 =150m.
Therefore, the length of the train is L = 150m.

Try to visualize the two situations in order to understand in each case what is the distance traveled by the train.
When passing the man, timer starts when the man and the beginning of the train are aligned, and timer stops when the end of the train and the man are aligned. So, the distance traveled by the train is its length L.
When passing the platform, timer starts when the beginning of the train and one end of the platform are aligned and the timer stops when the end of the train and the other end of the platform are aligned. In this case, the train traveled 50m (the length of the platform) + its own length L.
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Re: Two men plan to run in opposite directions around a park [#permalink] New post 18 Aug 2012, 04:39
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NYC5648 wrote:
Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths?

Official answer is: 250m

Many thanks!!!


The circumference of the park is 1km = 1,000m. If T is the time passed until they met, we can write:
15*T + 25*T = 1,000 (the sum of the distances run by each is the total distance, or the circumference of the park).
We can find that T = 1,000/40 = 25 minutes.
The second runner run further by 25 * 25 - 15 * 25 = 10 * 25 = 250m.
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Re: Two men plan to run in opposite directions around a park [#permalink] New post 18 Aug 2012, 08:53
NYC5648 wrote:
Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths?

Official answer is: 250m

Many thanks!!!



Ratio of speeds 15:25 = 3:5
=> Ratio of distance covered = 3:5
=> Total Distance units = 3 + 5 = 8
=> Difference in Distance Covered = 5 -3 =2

Distance = 2*1000/8 = 150
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Re: When Johns travels at 120 km/h he uses 30% more fuel [#permalink] New post 18 Aug 2012, 08:56
NYC5648 wrote:
When Johns travels at 120 km/h he uses 30% more fuel than when he travels only at 80 km/h. If he can travel 19.5 km on one liter at 80 km/h, how far can he travel with 10 liters of fuel at 120 km/h?

The official answer: 150km

I don't get it. I always get 136,5 km

Many thanks in advance!!!


80 Kmph - 1 Lit - 19.5 Km
=> 80 Kmph - 10 Lit - 195 Km

at 120 Kmph, 30% extra fuel consumption

=> 120Kmph - 10*1.3 - 195 Km

In words, at 120 Kmph 13 Lit fuel the man can travel 195 Km
in 10 Lit the man can travel 195*10/13 = 150 Km
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Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink] New post 18 Aug 2012, 09:00
NYC5648 wrote:
A train passes a man standing on a platform in 9 sec. and passes the platform in 12 sec. completely. If the platform is 50 meters long, how long is the train?

The offical answer is: 150m

Can somebody help me please?
Many thanks!!!


If the length of the trail in L and Length of the platform is P.

A. To cross the platform, the train has to travel L + P distance. 12 Sec
B. To cross the man, the train has to travel on L distance. In 9 Secs

A-B = [(L+P) - L] Distance in 12-9 Secs = 3 secs
=> P distance in 3 Secs
=> 50 m in 3 Secs
=> 150 m in 9 Secs (multiply 3)

And as in B, the length of the train is the distance traveled in 9 Secs. i.e. 150
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If Brad gives Tom a head start of 10 meters to tie and Tom [#permalink] New post 18 Aug 2012, 11:45
If Brad gives Tom a head start of 10 meters to tie and Tom gives John a head start of 10 meters to tie in a 100 meter race, how many meters must Brad give John in a 100 meter race to tie?


Official answer: 19

Can somebody help me with this one please?

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In a 100 meter race Greg gives Ralph a 12 meter head start a [#permalink] New post 18 Aug 2012, 11:51
In a 100 meter race Greg gives Ralph a 12 meter head start and still wins the race by 2 seconds. If Greg can run the race at an average speed of 5 m/sec., what is Ralph’s average speed for the race?

Official answer is: 4m/sec

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Re: If Brad gives Tom a head start of 10 meters to tie and Tom [#permalink] New post 18 Aug 2012, 12:44
NYC5648 wrote:
If Brad gives Tom a head start of 10 meters to tie and Tom gives John a head start of 10 meters to tie in a 100 meter race, how many meters must Brad give John in a 100 meter race to tie?


Official answer: 19

Can somebody help me with this one please?

Thanks!!


Brad runs 100m while Tom runs just 90m.
While Tom runs 100m, John runs only 90m. Therefore, if Tom runs 90m, John runs only 90*(90/100)=81m. The distances run by John and Tom are proportional, the ratio being 9:10.
So, while Brad runs 100m, John runs only 81m.
Brad can give John a head start of 100 - 81 = 19m.
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Re: In a 100 meter race Greg gives Ralph a 12 meter head start a [#permalink] New post 18 Aug 2012, 12:56
NYC5648 wrote:
In a 100 meter race Greg gives Ralph a 12 meter head start and still wins the race by 2 seconds. If Greg can run the race at an average speed of 5 m/sec., what is Ralph’s average speed for the race?

Official answer is: 4m/sec

Thanks guys! Appreciate your time and effort.


If we denote by t the time Ralph is running, by G Greg's speed, and by R Ralph's speed, then we can write the following equations:
G*(t-2)=100 and R*t=88. Greg is running 100m in 2 seconds less than Ralph is running 100 - 12 = 88m.

Since G=5m/sec, we find t=22sec. Hence, R=88/22=4m/sec.
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At summer camp 33 teens play frisbee, 26 play golf, 20 swim, [#permalink] New post 21 Aug 2012, 11:06
At summer camp 33 teens play frisbee, 26 play golf, 20 swim, 12 play frisbee and golf, 7 play golf and swim, 8 play frisbee and swim and 3 don’t do anything. What is the maximum and minimum number of teens at camp?

[Reveal] Spoiler:
OA: Min 55; Max 62



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Re: At summer camp 33 teens play frisbee, 26 play golf, 20 swim, [#permalink] New post 21 Aug 2012, 12:10
I hope it is a set problem where maximum number of students are U-(AUB)' =33-3 = 30
and minimum number is = AUBUC
Please verify
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Re: At summer camp 33 teens play frisbee, 26 play golf, 20 swim, [#permalink] New post 21 Aug 2012, 13:03
Hi soumyaranjandash,

I am sorry buddy but it is wrong. The official answer is:

OA: Min 55; Max 62

Unfortunately I dont have a detailed solution.
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Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink] New post 22 Aug 2012, 01:11
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Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths?

Suppose they cross paths in time 't' mins

Then, distance travelled by first man + distance travelled by Second man = 1km = 1000m

=> 15 * t + 25 * t = 1000
=> t = 25

Distance travelled by Second runner = 25meters/min * t = 25*25m = 625 meters.
Distacne travelled by First runner = 15meters/min * t = 15*25m = 375meters

How much farther has Second runner traveleld = Distance travelled by Second runner - that by first = 625m - 375m = 250 meters

Hope it helps!
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Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink] New post 22 Aug 2012, 01:25
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At summer camp 33 teens play frisbee, 26 play golf, 20 swim, 12 play frisbee and golf, 7 play golf and swim, 8 play frisbee and swim and 3 don’t do anything. What is the maximum and minimum number of teens at camp?


Total Number of students in the camp (according to the file attached with this post are) = 33 + 26 + 20 - (12-x) - (7-x) - (8-x) - 2x + 3
= 55 + x

For minimum value x should be 0 so Minimum value is 55
For maximum value x should be maximum , now we have to calculate the maximum value of x

since, 12-x, 7-x and 8-x have to be non negative so maximum value of x can be 7

So, Maximum number of students = 55 + 7 = 62

Hope it helps!
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What is the approximate area of a circle that has an equilat [#permalink] New post 22 Aug 2012, 11:26
What is the approximate area of a circle that has an equilateral triangle with an area of 4√3 inscribed in it?

[Reveal] Spoiler:
OA: 16


Can anyone please explain a step-by-step solution. I guess I am missing a concept here.

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If one of the angles of a parallelogram is 120 degrees, [#permalink] New post 22 Aug 2012, 12:12
If one of the angles of a parallelogram is 120 degrees, what is the ratio of the lengths of the diagonals inside the parallelogram?


[Reveal] Spoiler:
OA: 1: sqr3


Can anybody help me with this please?

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If one of the angles of a parallelogram is 120 degrees,   [#permalink] 22 Aug 2012, 12:12
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