Question

How many trains do not stop at either of these stations on that day?
Oaklands and Brighton are two busy train stations on the same train line. On one particular day:
I. \[\dfrac{1}{5}th\] of the trains do not stop at Oaklands
II.45 trains do not stop at Brighton
III.60 trains stop at both Brighton and Oaklands
IV.60 trains stop at only Brighton or Oaklands (not both)

Answer 1

Hint: To find, the number of trains that do not stop at either of these stations, we need to consider all the given data and form equations such that it makes statement type questions to solve easily hence, with respect to all the given data we need to compare and find the number of trains that do not stop at either of these stations.

Complete step-by-step answer:
Let us assign letters to all the different possibilities. At this point they are unknowns, so let’s use a letter to represent them.
T= total number of trains
O= number of trains stopping only at Oaklands
B= number of trains stopping only at Brighton
N= number of trains stopping at neither
U= number of trains stopping at both
 \[\overline O \] = number of trains not stopping at Oaklands
 \[\overline B \] = number of trains not stopping at Brighton
Now, let’s list the given data:
60 trains stop at both Brighton and Oaklands
 \[ \to U = 60\]
60 trains stop at only Brighton or Oaklands (not both)
 \[ \to O + B = 60\]
 \[\dfrac{1}{5}th\] of the trains do not stop at Oaklands
 \[ \to \overline O = \dfrac{T}{5}\]
45 trains do not stop at Brighton:
 \[ \to \overline B = 45\]
Hence, total number of trains are:
 \[T = N + O + B + U\]
We are given that \[O + B = 60\] , so let’s try finding what they are in terms of other unknowns.
 \[\overline O = N + B = \dfrac{T}{5}\] (i.e., trains stopping at Neither + only B)
 \[ \Rightarrow N + B = \dfrac{1}{5}\left( {N + O + B + U} \right)\]
We, know that \[O + B = 60\] and \[U = 60\] , hence substituting it we have:
 \[ \Rightarrow N + B = \dfrac{1}{5}\left( {N + 60 + 60} \right)\]
Simplifying the terms, we get:
 \[ \Rightarrow N + B = \dfrac{1}{5}\left( {N + 120} \right)\]
 \[ \Rightarrow N + B = \dfrac{N}{5} + \dfrac{{120}}{5}\]
We need to find the number of trains stopping only at Brighton i.e., B; hence we get:
 \[ \Rightarrow B = \dfrac{N}{5} - N + \dfrac{{120}}{5}\]
 \[ \Rightarrow B = 24 - \dfrac{4}{5}N\] ………………………….. 1
We, have number of trains not stopping at Brighton i.e., \[\overline B = 45\] , with respect to number of trains stopping only at Oaklands i.e., O and number of trains stopping at neither i.e., N we get:
 \[\overline B = N + O = 45\] (trains stopping at neither + only O)
 \[ \Rightarrow O = 45 - N\] ………………………….. 2
But, \[B + O = 60\] , hence we need to find for number of trains stopping at neither i.e., N, we get:
 \[24 - \dfrac{4}{5}N + 45 - N = 60\]
Add equation 1 and equation 2 we get:
 \[ \Rightarrow - \dfrac{4}{5}N - N = 60 - 24 - 45\]
 \[ \Rightarrow - 1.8N = - 9\]
 \[ \Rightarrow N = \dfrac{{ - 9}}{{ - 1.8}}\]
 \[ \Rightarrow N = 5\] .
Therefore, 5 trains do not stop at either of these stations on that day.
So, the correct answer is “5 trains”.

Note: The key point to solve the given question, is that we must concentrate on all the given data as we need to compare all the data to find the number of trains at that particular station. As here, Oaklands and Brighton are two busy train stations on the same train line, in which we need to note the number of trains which stop and which do not stop at which place.