use the answer provided to back into the answer. I.e. Start with the middle value and see whether you should go higher or lower.

I started with 24km and with 24km/hr it will take 60 minutes to arrive. Since this is 5 minutes late, I subtract 5 minutes from 60 to get a desired time of 55 minutes. The desired time is the same for both trials.

Since I know my desired time is 55 minutes, and that with a 25% increase in speed, which is 30km/hour, then I need to be 4 minutes quicker than my desired time, or actual time of 51 minutes at 30 km/hour. Divide 30 into 24 and you get .8hours or 48 minutes. The difference between my desired time of 55 minutes and 48 minutes is 7 minutes which is too great.

I know that Distance= Rate x Time and rearranged, Time = Distance/Rate. I know that my Time needs to get smaller and the only way to do that is to decrease the numerator or increase the Denominator. Since our rate is given (30 km/hour) I can't increase my denominator and therefore must decrease my numerator, Distance. Therefore move onto to a smaller distance.

The answer is 18. Work out the same way as desrcibed above.




***** You can solve this using alegbra too

24km/hour = 24km/60 minutes = 2/5 km/min=.4km/min
30 km/hour= 30km/60 minutes = 1/2 km/min=.5km/min

Distance is unknown X
Desired time is unknown y

1) X/2/5= y +5

2) X/1/2= y -4

Solve for y using either equation

(using no. 2) y= X/1/2 + 4

Substitute for y, X/2/5= y + 5, X/2/5= X/1/2 + 4 + 5
Convert Fractions 5x/2 (or 2.5) = 2X + 9
Subtract 2x from both sides = 2.5X-2X = 9, .5x=9
Multiply both sides by 2,
x=18km
_________________

There are three things in life: Price, Quality, and Speed. You only get two out of the three.

A (18 km) will be the answer.

Let t be the time

25% faster than 24 km/h = 30 km/h

(t+(5/60))24 = (t-(4/60))30
(t+ (1/12))24 = ( t - (1/15))30
(12t+1)2 = (15t-1)2
t = 2/3 hours

Distance can be calculated by substituting the time in any one of the equations
(t+(5/60))24 OR (t-(4/60))30

Distance = 18 km Answer A

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Very Easy And Alternate Way


let Speed S1 =24 km/Hr then time taken T1 & Distance =D
where T1 =T+5 (says reaches 5 min late)

Now if 25% faster
Then S2= 125%*24= 30 Km/hr then time taken T2 & Distance=D (distance is same)
where T2 =T-4 (says reaches 4 min early)

by speed formula:

D=S1*T1 ....i)

D=S2*T2......ii)

Taking taking equation i) &ii)

S1*T1=S2*T2

24*(T+5)=30(T-4)
24T +120=30T -120
6T=240
T=240/6=40 (remember this is in min)

so time taken when speed is 24 Km is 45 min...

45 min = 3/4 hrs
Plugging it in any eq say i)
24=D/(3/4)
24*3=4D
D=18 Km

Hope this helps.... kindly give ur feedback about the method...

Karan

This is the best and shortest explaination

A

key is find T and then to find D and not stop at T.
_________________

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my way:
V1=24km/h
V2=V1*1.25=30km/h
Distance1=Distance2
T1=t+5 (5 minutes late = more time to cover the distance)
T2=t-4 (4 minute early = less time to cover the distance)
V1T1=V2T2
24(t+5)=30(t-4)
24t+120=30t-120
240=6t
t=40 (in minutes)
Distance = 30*[(40-4)/60] or 24*[(40+5)/60)] = 18km

Another fast approach to solve this problem:

At 25% faster, speed is 30 Km/hr.
At 30km/hr, lets say time taken is T.

Distance still to be covered @ 24 km/hr speed (this is after time T) =
24 * 9/60 = 3.6 km

At 6 km/hr (which is the speed difference) in time T, Distance covered =
3.6 km (calculated above)

So at 30 km/hr distance covered in time T =
3.6 * 5 = 18kms.

Great explanation.

I didn't check view all threads . however I'll provide a method with which you can solve practically any problem whether speed or time is given.

24km/hr --- 5 mins late
30km/hr--- 4 mins early
Diff in time = 9mins.

(had he reached at 8:00am ideal time; 24km/hr --- reaches at 8:05
30km/hr --- reaches at 7:56.)

lcm ( 24 and 30) =120..
hence if the distance was 120km..
24km/hr would take 5hrs
30km/hr would take 4hrs.
Diff in time = 1 hr..

apply Unitary method

dist............time diff.
120...........1hr(60min)
? ............. 9mins.

(120*9)/60= 18km


** Note-- if only the time taken is given , you can use the same method to solve . Only replace speed with time for the lcm..

Cheers