# Quant Boosters - Rajesh Balasubramanian - Set 4

• 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

• You may solve for D and S by writing the equations as:
D/(s−15) − D/s = 45
D/s − D/(s+10) = 20

You can solve these two equations to get D = 9750 Km which is a perfectly fine approach.

However, if you are comfortable with alligation, you may save time on such TSD question as:

Both these equations can be solved quickly to get s = 65 Km/hr, t = 150 hr. Therefore, D = 9750 Km. I want to reiterate that using the first approach is perfectly fine. That's how I would do it. It is better to spend 20 seconds extra and get the right answer instead of getting it wrong with something you are not comfortable with.

• Q23) Raj was travelling to his hometown from Mumbai. He met with a small accident 80 Km away from Mumbai and continued the remaining journey at 4/5 of his original speed and reached his hometown 1 hour and 24 minutes late. If he had met with the accident 40 Km further, he would have been an hour late.
What is Raj's normal speed?
a) 20 Km/hr
b) 15 Km/hr
c) 30 Km/hr
d) 25 Km/hr

• Scenario 1

5D1/4s − D1/s = 1 + 24/60 (always write minutes in this form if the unit is Km/hr)

D1/s = 7/5 ∗ 4 = 28/5 hours

Scenario 2

5D2/4s − D2/s = 1
=) D2/s = 4 hours

i) Therefore, in his normal speed, he can travel 40 Km in 28/5 − 4 = 8/5 hours
so, s = 40/(8/5) = 25 Km/hr. Answer d)

ii) D1/s = 28/5 =) D1 = 140 Km. Therefore, total distance D = 140 + 80 = 220 Km

• Q24) Two persons A and B start moving at each other from point P and Q respectively which are 1400 Km apart. Speed of A is 50 Km/hr and that of B is 20 Km/hr. How far is A from Q when he meets B for the 22nd time?
a) 1000 Km
b) 400 Km
c) 800 Km
d) 1400 Km

• Total distance travelled by both of them for 22nd meeting = 1400 + 21 x 2 x 1400 = 43 x 1400

Distance travelled by each will be in proportion of their speed:-

Therefore, distance travelled by A = 50/(50 + 20) x 43 x 1400 = 43000 (Note - Always do complicated calculations at last because things cancel out generally)

Now, for every odd multiple of 1400, A will be at Q and for every even multiple of 1400 A will be at P. So, at 42000 Km (1400 x 30, even multiple) A will beat P. So at their 22 meeting, A will be 1000 Km from P, therefore, 400 Km from Q

• Q25) What would happen in the previous question if both A and B had started at point P.
a) 800 Km
b) 600 Km
c) 1000 Km
d) 350 Km

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