The distance between Metropolis and Gotham is 450 miles. One car starts from metropolis at 3:10 PM at a speed of 20 mile/hr and another car starts from Gotham at 4:00 PM at a speed of 60 mile/hr. When the trains will meet.
Lets say the cars meet at a distance of x km from Metropolis. Also, lets call the car starting from Metropolis as A and that starting from Gotham as B. So, A with an speed of 20 mile/hr will travel a distance of x km and B with an speed of 60 mile/hr will travel 450-x km. Therefore, the time of travel of A will be x/20 and that of B will be (450-x)/60. Since they will meet at the same time, the time of travel of A will be 50 minutes more than that of B. which means that (x/20) - (50/60) = (450 - x)/60 (converting 50 minutes to hour) i.e. (3x - 50)/60 = (450 - x)/60 cancelling 60 on either side, we get, 3x - 50 = 450 - x i.e., 4x = 500 which gives us x = 125. So, the travel time of A is 125 / 20 hours i.e., 6 hours and 15 minutes. Therefore the cars will meet at 9:25 PM at a distance of 125 miles from Metropolis.
A bowl has some cookies, Jack takes one third of the cookies, Reacher and Sparrow each take one third of what is left behind. Now Sully comes and adds the same amount of cookies what is left in the bowl. The bowl has 16 cookies now. Calculate the number of cookies in the bowl at start.
Lets say the bowl had x number of cookies to start with. So, Jack took x/3 Reacher and Sparrow each took 1/3 of rest (x-x/3) cookies which comes as 2x/9 cookies each. So the bowl is left with x - (x/3 + 2x/9 + 2x/9) cookies i.e., x - 7x/9 i.e., 2x/9 cookies. Now Sully doubles the cookies in the bowl, so the bowl is now holding 4x/9 cookies. it is given that this 4x/9 =16, solving, we would get x = 36, which is the number of cookies the bowl had initially.
If 95% students in your class of 50 students are between 1m and 1.6m tall, what is the probability that height of a student 1.75m or more in the same class considering that heights of students in the class are normally distributed.
Information in the question: Heights of students normally are distributed. Class size is 50 95% of students fall between 1-1.6m Student in question is 1.75m tall. Now all we have to do is to find the z value and then by using z table for standard normal distribution, we can find the required probability. In any normal distribution, 95% of the data is within 2 standard deviations on each side of the mean. So, lets find the mean. Mean = (1+1.6)/2 = 1.3, because mean will be the middle point of 1 and 1.6. standard deviation can calculated by diving the difference of 1.6 and 1 by 4 because 95% data resides between 2 standard deviations on each side. so, SD = (1.6-1)/4 = 0.15 Now, z value = (Value in question - Mean) / Standard Deviation = (1.75-1.3)/0.15 = 3 if we take a look at the z table of standard normal distribution, we would find that area to the left of a z value of 3 is 0.99865. So, the required probability will be = 1 - the above area (because we are only seeking the probability of height being equal to or more than 1.75m.) therefore, required probability = 1 - 0.99865 =0.00135 or 0.135%