Quant Boosters - Rajesh Balasubramanian - Set 4
car of length = 4 m
length of truck = 20 m
Speed of truck = 36 km/hr
Total distance to be covered by the car 'd' = 20 + 4 = 24 m
Let speed of car be S1 m/s
Speed of truck S2 = 36 km/hr = 10 m/s
Both the car and the truck are travelling in the same direction and we know that to overtake the truck, the speed of the ca should be more than that of the truck.
Hence, relative speed = |S1 - S2|
Time to overtake = 10s
S = d/t
|S1 - 10| = 24/10
S1 - 10 = 2.4
S1 = 10 + 2.4 = 12.4 m/s
Q13) Train A travelling at 63 kmph takes 27 to sec to cross Train B when travelling in opposite direction whereas it takes 162 seconds to overtake it when travelling in the same direction. If the length of train B is 500 meters, find the length of Train A.
a) 400 m
b) 810 m
c) 500 m
d) 310 m
Let the length of Train A be x meters
Let speed of Train B be y kmph
Relative distance = Relative speed * time taken to cross/overtake
Relative speed of 2 trains = 63 + y
Time taken to cross = 27 sec or 27/3600 hrs
Relative distance between 2 trains = Length of Train A + length of train B = (x + 0.5) km
Therefore, x + 0.5 = (63 + y) * 27 / 3600 ----- (1)
Relative speed of 2 trains = 63 – y
Time taken to overtake = 162 sec or 162/3600 hrs
Relative distance between 2 trains = x + 0.5
Therefore, x + 0.5 = (63 – y) * 162/3600 --- (2)
From (1) and (2), solve for y.
(63 + y) * 27 = (63 – y) * 162
27y + 162 y = 63*162 – 63 *27
189y = 63 * 135 or y = 45 kmph
Substitute in (2) to get x.
x + 0.5 = (63 – 45) * 162/3600
Or x = 0.31 km or 310 meters
Q14) P cycles at a speed of 4 m/s for the first 8 seconds, 5 m/s for the next 8 seconds, 6 m/s for the next 8 and so on. Q cycles at a constant speed of 6.5 m/s throughout. If P and Q had to cycle for a 400 m race, how much lead in terms of distance, can P give Q and still finish at the same time as Q?
a) 43.4 m
b) 56.6 m
c) 32.1 m
d) P cannot give a lead as Q is always ahead of P
Distance covered by P in 8 sec = 4 * 8 = 32 m
Distance covered by P in 16 sec = 32 + 5 *8 = 72 m
Distance covered by P in 24 sec = 72 + 6 * 8 = 120 m
Distance covered by P in 32 sec = 120 + 7 * 8 = 176 m
Distance covered by P in 40 sec = 176 + 8 * 8 = 240 m
Distance covered by P in 48 sec = 240 + 9 * 8 = 312 m
Total distance in 48 sec = 312m
To cover balance 86 m with speed = 10 m/sec, time taken = 86/10 = 8.6 sec
So P would finish 400 m in 48 + 8.6 = 56.6 seconds.
In 56.6 seconds, Q cycles 56.6 * 6.5 = 367.9 m or B should have a 32.1 m lead to result in a dead heat.
Q15) A bus starts from a bus stop P and goes to another bus stop Q. In between P and Q, there is a bridge AB of certain length. A man is standing at a point C on the bridge such that AC:CB = 1:3. When the bus starts at P and if the man starts running towards A, he will meet the bus at A. But if he runs towards B, the bus will overtake him at B. Which of the following is true?
a) Bus travels 3x times faster than the man
b) Bus travels 2x times faster than the man
c) The bus and the man travel at the same speed
d) 4x the speed of the man is equal to 3x the speed of the bus
Let the speed of the bus be ‘b’ and man be ‘m’.
Let the distance between P and A be y.
AC:CB = 1:3 implies AC = x and CB = 3x
When bus goes from P to A and the man from C to A, time taken by both are equal and they meet at A.
y/b = x/m
or b/m = y/x --- (1)
When bus goes from P to B and the man from C to B, again time taken by both are equal and they meet at B.
(y+x+3x)/b = 3x/m
b/m = (4x+y)/3x ---(2)
Equating (1) and (2), we get y/x = (4x+y)/3x
3y = 4x + y or y = 2x
Therefore ratio of speeds b/m = 2x/x or 2 : 1
Q16) Ramesh takes 6.5 hours to go from city A to city B at 3 different speeds 30 kmph, 45 kmph, and 60 kmph covering the same distance with each speed. The respective mileages per liter of fuel are 11 km, 14 km and 18 km for the above speeds. Ramesh's friend Arun is an efficient driver and wants to minimise his friend's car's fuel consumption. So he decides to drive Ramesh's car one day from city A to city B. How much fuel will he be able to save?
a) 4.2 liters
b) 4.5 liters
c) 0.7 liters
d) 0.3 liters
Total time = 6.5 hrs
Let the total distance be x
Ramesh's speeds = 30, 45, 60 kmph
6.5 = (x/3)/30 + (x/3)/45 + (x/3)/60
6.5 = (2x + 4x/3 + x)/180
6.5 = (6x+4x+3x)/540
or x = 270 km
With each speed distance covered = 270/3 = 90 km
Fuel consumed when driving at 30 = 90/11 = 8.1 l
Fuel consumed when driving at 45 = 90/14 = 6.4 l
Fuel consumed when driving at 60 = 90/18 = 5
Total fuel = 8.1 + 6.4 + 5 = 19.5 liters
When Arun drives the car, he should drive at 60 kmph throughout so that he gets the maximum mileage!
Fuel consumption when driving at 60 for entire journey = 270/18 = 15 liters
Difference = 19.5 – 15 = 4.5
Q17) Amar, Akbar and Antony decide to have a 'x' m race. Antony completes the race 14 m ahead of Amar. Akbar finishes 20 m ahead of Antony and 32 m ahead of Amar. What is Amar’s speed?
a) 9/10th of Antony's speed
b) 5/8th of Akbar's speed
c) 14/15th of Antony's speed
d) 10/7th of Akbar's speed
When Antony completes, Amar would have run (x – 14) m; Time taken is the same
Ratio of Amar’s speed : Antony’s speed = (x–14)/t1 : x/t1
Or (x–14)/x ---- (1)
When Akbar finishes, Amar would have run (x – 32) m; Antony would have run (x – 20) m; Time taken is the same
At this point, the ratio of Amar’s speed : Antony’s speed = (x–32)/t2 : (x−20)/t2
Or (x–32)/(x−20) ---- (2)
Equating (1) and (2), we get
(x–14)/x = (x–32)/(x−20)
x^2 – 34x + 280 = x^2 – 32x
2x = 280
Or x = 140 m
Ratio of Amar’s speed : Antony’s speed = (140–14)/140 = 126/140 = 9/10
Q18) Tom, Jerry and Bill start from point A at the same time in their cars to go to B. Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B. When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B. If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed, what could be the distance between the points A and B
a) 40 miles
b) 24 miles
c) 31 miles
d) 63 miles
Let the distance between A and B be ‘x’ miles.
Let their speeds be T, J and B in miles/hour
When Tom meets Jerry:
Distance travelled by Tom = x + 9
Distance travelled by Jerry = x – 9
Time taken will be the same => (x+9)/T= (x–9)/J
Or T/J = (x+9)/(x−9) -- (1)
When Jerry meets Bill:
Distance travelled by Jerry = x + 7
Distance travelled by Bill = x – 7
Time taken will be the same => (x+7)/J= (x–7)/B
Or J/B = (x+7)/(x−7) -- (2)
Given 3T = 5B
or T/B = 5/3
From (1) and (2)
T/B = T/J * J/B
5/3 = (x+9)/(x−9) * (x+7)/(x−7)
5(x - 9)(x - 7) = 3 (x + 9) (x + 7)
5x^2 – 80x + 315 = 3x^2 + 48x + 189
2x^2 – 128x + 126 = 0
x^2 – 64x + 63 = 0
x = 63 or x = 1
x has to be 63
Q19) Kumar started from Chennai at x hrs y minutes and travelled to Vellore. He reached Vellore at y hrs z minutes. If the total travel time was z hrs and x minutes, his starting time in Chennai could have been ______ (Assume clock format to be 0 to 24 hrs).
a) 02:08 hrs
b) 13:03 hrs
c) 00:02 hrs
d) 12:01 hrs
x, y or z cannot be greater than 24
Also x + y + z cannot be greater than 24
Given x hrs + z hrs = y hrs
Also y min + x min = z min
The only value of x that will satisfy the above 2 equations is 0.
Hence, from among the choices given, the time could be 00:02 hrs
Q20) When Sourav increases his speed from 20 Km/hr to 25 Km/hr, he takes one hour less than the usual time to cover a certain distance. What is the distance usually covered by him?
a) 125 Km
b) 100 Km
c) 80 Km
d) 120 Km
Here, the distance travelled by Sourav is constant in both the cases. So,
20 * t = 25 * (t-1) = D
20t = 25t - 25
5t = 25
t = 5 hrs
So, Distance travelled = 20 * 5 = 100 Km.
Q21) Distance between the office and the home of Alok is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 Km/hr and reached office at the normal time. What is the changed speed of Alok?
a) 25 Km/hr
b) 20 Km/hr
c) 16 Km/hr
d) 50 Km/hr
Here, again the distance is constant. However if we write the equation as we did in previous question, we get:
s * t = (s + 5) * (t - 1) = 100
In this case, we can identify other constant as the time which Alok took less to cover the distance i.e. 1 hour. So, we can write:
100/s−100/(s+5) = 1 (Difference of time take in both the cases is 1 hour)
=) 100s + 500 - 100s = s(s+5)
=) s(s+5) = 500 (One should practice to solve these kind of equations directly by factorizing 500 into 20 X 25, instead of solving the entire quadratic. However, if you are more comfortable with quadratic, keep using it. More important thing is to avoid silly mistakes once you have got the concept right)
Solving we get, s = 20 Km/hr
Therefore, Increased speed = s + 5 = 25 km/hr
We could have alternatively identified increase in speed by 5 Km/hr as a constant. Therefore, we could write the equations as:
100/(t−1) − 100/t = 5
Solving, we get t = 5 hrs, t-1 = 4hrs. Therefore, increased speed = 100/(t−1) = 25 Km/hr.
Q22) Akash when going slower by by 15 Km/hr, reaches late by 45 hours. If he goes faster by 10 Km/hr from his original speed, he reaches early by by 20 hours than the original time. Find the distance he covers.
a) 8750 Km
b) 9750 Km
c) 1000 Km
d) 3750 Km