Quant Boosters - Rajesh Balasubramanian - Set 4
When boat P travels downstream, it will effectively have a speed to 30kmph. Likewise, Q will have an effective speed of 10kmph. The relative speed = 40kmph. So, the two boats will meet for the first time after 300/40 hours (Distance/relative speed) = 7.5 hours (Actually, for this part we do not need the speed of the stream)
The second part is more interesting, because the speed of the boats change when they change direction. Boat P is quicker, so it will reach the destination sooner. Boat P will reach City B in 10 hours 300/30. When boat P reaches city B, boat Q will be at a point 100kms from city B.
After 10 hours, both P and Q will be travelling upstream,
P's speed = 20 km/hr
Q's speed = 10 km/hr
Relative speed = 10km/hr
Q is ahead of P by 100 kms
P will catch up with Q after 10 more hours Distance/RelativeSpeed − 100/10.
So, P and Q will meet after 20 hours at a point 200 kms from city B.
Correct Answer: 7.5 hours and 20 hours
Q11) Cities M and N are 600km apart. Bus A starts from city M towards N at 9AM and bus B starts from city N towards M at the same time. Bus A travels the first one-third of the distance at a speed of 40kmph, the second one-third at 50kmph and the third one-third at 60kmpr. Bus B travels the first one-third of the total time taken at a speed of 40kmph, the second one-third at 50kmph and the third one-third at 60kmph. When and where will the two buses cross each other?
a) 300 kms from A
b) 280 kms from A
c) 305 kms from A
d) 295 kms from A
Travels 200km @ 40kmph
the next 200km @ 50kmph and
the final 200km @ 60kmph
So, Bus A will be at a distance of 200km from city M after 5 hours, and at a distance of 400km after 9 hours, and reach N after 12 hours and 20 mins
Travels at an overall average speed of 50kmph, so will take 12 hours for the entire trip.
So, Bus B will travel
160 kms in the first 4 hours
200 kms in the next 4 hours
and 240 in the final 4 hours
So, both buses cross each other when they are in their middle legs.
After 5 hours, bus A will be at a position 200kms from city M. At the same time, bus B will be at a distance 210kms from city N (4 * 40 + 50).
The distance between them will be 190kms (600 - 200 - 210). Relative speed = Sum of the the two speeds = 50 + 50 = 100 kmph.
Time taken = 190/100 = 1.9 hours. = 1 hour and 54 minutes. So, the two buses will meet after 6 hours and 54 minutes. Bus B will have travelled 210 + 95 = 305 kms. So, the two buses will meet at a point that is 305 kms from City N and 295 kms from city A.
Q12) A car of length 4m wants to overtake a trailer truck of length 20m travelling at 36km/hr within 10 seconds. At what speed should the car travel?
a) 12 m/s
b) 14.8 m/s
c) 12.4 m/s
d) 7.6 m/s
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