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

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!!!

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!!!

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 \(1.3\)liters 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.

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.

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!

What is the approximate area of a circle that has an equilateral triangle with an area of 4√3 inscribed in it?

Image attached
O is the center of the circle and triangle ABC is equilateral.
AD is the angle bisector and is perpendicular bisector of the other side...

So, AO:OD = 2:1
(Radius = 2x)
In Triangle ADB, applying Pythagorean theorem we have
a^2 = (3x)^2 + (a/2)^2

=> x = a/(2*(sqrt 3))

given that area of triangle = 4 sqrt3 = (sqrt 3)*a^2/4

so, a = 4

so, x = a/(2*(sqrt 3)) = 2/(sqrt 3)
radius = 2x = 4/ (sqrt3)

Area of circle = pie r^2 = (22/7) * 16/3 = 352/21

Hope it helps!

If it takes Mike M days to finish the job alone and Tina T days to do the same job, then we can write:

\(\frac{1}{M}+\frac{1}{T}=\frac{1}{5}\) and
\(\frac{2}{M}+\frac{1}{3T}=\frac{1}{3}.\)

Multiply the first equation by 2, subtract from it the second equation, and solve for T.
You will get \(T=25.\)

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.

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

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

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

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!!

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, \(R=88/22=4m/sec.\)

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?




Thanks guys!!! Appreciate your help and effort!

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

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.