CAT Previous Year Questions: Time & Distance Take (July 2) - CAT MCQ

The distance travelled before the trip (x in km.) is a palindrome.
Let the average speed of Brishti be s kmph.
The total distance covered by Brishti after the trip is 26862 km.
The distance covered by Brishti during the trip is 8×s = 8s km.
x = 26862 – 8s
For x to be a palindrome, of the given options, s can only be 100.
Therefore, the average speed of Brishti during the trip is 100 kmph.

Let the distance between city A and the meeting point M be D1 and the distance between city B and the meeting point M be D2.
Arvind covers D1 in t hours and D2 in 6 hours.
Let the distance between city A and the meeting point M be D1 and the distance between city B and the meeting point M be D2.
Arvind covers D1 in t hours and D2 in 6 hours.

t² = 6 × 24 = 12²
t = 12
Total time taken by Arvind to travel from city A to city B = (t + 6) = 18 hours.
Speed of Arvind = 54 kmph
Distance between city A to city B = 54 × 18 = 972 km. 

The speeds of Ravi and Ashok converted to meters per second are
 respectively.
Let M be the meeting point between Ravi and Ashok. 

We are given the definition of the function, f(x) = ax² + bx + c , and been told that b²<4ac . This means that f(x) has imaginary roots and therefore no real roots at all.
If a function has no real roots, the graph of the function can never touch the x-axis, because touching the x-axis means, for some real value x, the value of f(x) is 0. In other words, it means that the function has roots for some real value x.

When we graph this function quadratic function f(x), we get a parabola, that should never touch the x-axis. Such a parabola should be completely above the x-axis or completely below the x-axis.

If it is completely above the x-axis:
It means that the value of f(x) is always positive for any value of x.
The set of values of x that satisfies the condition that f(x) < 0 is an empty set.
If it is completely below the x-axis:
It means that the value of f(x) is always negative for any value of x.
The set of values of x that satisfies the condition that f(x) < 0 is the set of all real numbers.
Since set S contains all the integers ‘m’ that satisfy the above conditions, Set S is either an empty set or the set of all integers.

Speed of the faster train = 160/12 = 40/3 m/s
Speed of the slower train = 
Sum of speeds (when the trains travel towards each other) = 
Let the slower train be xx metres long; then: 
On solving, x = 190 m

Considering the length of train A = La, length of train B = Lb.
The speed of train A be 5x, speed of train B be 3x.
From the information provided :
The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.
In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5x + 3x = 8x because both the trains are traveling in the opposite direction.
Hence (8x) x (46) = Lb.
In the information provided :
It took another 69 seconds for the rear ends of the trains to cross each other.
In the next 69 seconds
The train B traveled a distance equivalent to the length of train A in this 69 seconds.
Hence (8x) x (69) = La.
La/Lb = 69/46 = 3/2 = 3 : 2

Considering the distance travelled by Mira in one minute = M,
The distance traveled by Amal in one minute = A.
Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3 x (A+M) = C. (1)
C is the circumference of the circle.
Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :
Hence 45 x (A-M) = 3C. (2)
Multiplying (1)*15 we have :
45A + 45M = 15C.
45A - 45M = 3C.
Adding the two we have A = 18C/90
Subtracting the two M = 12C/90
Since Mira travels 12C/90 in one minute, in one hour she travels 12C/90 . 60 = 8C
Hence a total of 8 rounds.

Let the length of the train be l and its speed be s. Given

⇒ 
∴ Length of the train =500 m.
Hence the required time to cross a lamp post = 
i.e. 81.81 (or) 82 sec.

In this particular case, we know Ratio of speeds = 
C₁ : C₂ = √20 : 45 i.e., 2: 3
As the speed of Car C₁ is 60 kmph, the speed of Car C₂ is 90 kmph

Let the usual speed of the train be s and time taken at that speed be 't'.
Given by travelling at s/3, it reached 30 min late. Hence the usual time:
Distance travelled = s x t
Distance travelled in the first 5min = s  x  t / 3. D
Distance to be travelled in the last 6min = 2 st / 3
Required speed to cover that distance on time = 
Hence the percentage increase in its speed = (2 / 3) x 100
i.e.,