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g106
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8(t+5) = 10(t-5)
8t+40 = 10t-50
=>t = 45 minutes

distance = 8*((45+5)/60)= 8*(5/6) = 6.66 km

Answer is B.
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Using RT=D

Distance= 8 km/h * (t+5)m
Distance= 10 km/h * (t-5)m
Therefore, 8 km/h * (t+5)m = 10 km/h * (t-5)m
solving for t is = 45 min

Distance= 8 km/h * (45+5) min
= 8 km/h * 50/60 hr
= 6.66 km
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Assume distance is D.
d/8-8/10 = 10/60
10d-8d/80 = 1/6
2d/80 = 1/6
d= 1/6*40
d = 6.66

answer is B
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Yes ratioes is quick.
Speed 8:10 => 4:5
So, Time is 5:4.
Bt given diff is 5-(-5) = 10 min
=> Time is 50:40

Hence distance = 8 * 50/60 = 6.67 Km
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distance = rate * time

(1) d = 8km/h * (t+5)
(2) d = 10 km/h * (t-5)

equalize equations: 8 * (t+5) = 10 * (t-5) --> 8t + 40 = 10t - 50 --> 90 = 2t --> 45 = t (in minutes)

now insert into equation (1) or (2): d = 8 km/h * (45+5) --> 8km/h * 5/6 h --> 40/6 km = 6.67 km. B.
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gopu106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?

A)9KM
B)6.66KM
C)7.25KM
D)12.40KM
E)13KM

Alternatively, don't assume any variables.

Ratio of speed in 2 cases = 8:10 = 4:5
Ratio of time in the 2 cases = 5:4
The difference between the time is total 10 mins so time taken in the two cases must be 50 min and 40 mins respectively.
Distance = 8 km/hr * (50/60) hrs = 6.66 km

You will be surprised how quickly you can solve questions using ratios (and without writing anything down once you get used to it - that probably accounts for a lot of time saving)

To learn more about ratios, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... of-ratios/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... os-in-tsd/
Karishma,
Can you please explain how you got that the times must be 40 and 50 minutes?
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VeritasPrepKarishma
gopu106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?

A)9KM
B)6.66KM
C)7.25KM
D)12.40KM
E)13KM

Alternatively, don't assume any variables.

Ratio of speed in 2 cases = 8:10 = 4:5
Ratio of time in the 2 cases = 5:4
The difference between the time is total 10 mins so time taken in the two cases must be 50 min and 40 mins respectively.
Distance = 8 km/hr * (50/60) hrs = 6.66 km

You will be surprised how quickly you can solve questions using ratios (and without writing anything down once you get used to it - that probably accounts for a lot of time saving)

To learn more about ratios, check out:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... of-ratios/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... os-in-tsd/
Karishma,
Can you please explain how you got that the times must be 40 and 50 minutes?

Ratio of time taken is 5:4. This is a difference of 1.
Actually the difference between time taken is 10 mins. This means, actual values have a multiplier of 10.
So time taken must be 5*10 = 50 mins and 4*10 = 40 mins

Read up on ratio scale and actual values on the links given in the post above.
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[quote="VeritasPrepKarishma"][quote="gopu106"]Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?


please explain me how the time ratio is 5:4 ?
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rahulnagarajan
VeritasPrepKarishma
gopu106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?


please explain me how the time ratio is 5:4 ?

The two speeds are 8 and 10. This is a ratio of 4:5.
The ratio of time taken is inverse of the ratio of speeds if distance covered is the same in two cases.
So given that the distance covered is same, if ratio of speeds is 4:5, the ratio of time taken will be 5:4.

For more details on this concept, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... os-in-tsd/
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Guys

Perhaps I am overlooking the issue- I am making some silly mistake please correct me
I am getting confused with units (km/hr & mins)

I converted 5 mins to 'hr' since the rate given is km/hr

so i got

8(t-1/12) = 10(t+1/12)

I am getting incorrect answer , please let me know how can you directly consider t+5 & t-5

Thanks in advance!!
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seemachandran
Guys

Perhaps I am overlooking the issue- I am making some silly mistake please correct me
I am getting confused with units (km/hr & mins)

I converted 5 mins to 'hr' since the rate given is km/hr

so i got

8(t-1/12) = 10(t+1/12)

I am getting incorrect answer , please let me know how can you directly consider t+5 & t-5

Thanks in advance!!

1/12 is fine since you converted 5 mins to 5/60 = 1/12 hrs. You assumed that t is in hrs and hence took the speed as 8 and 10 - perfect.

The issue was in the equation you made.
At speed 8, he takes 5 mins extra so time taken is t + 1/12. At speed 10, time taken is t - 1/12

So 8(t+1/12) = 10(t-1/12)
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g106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?

A)9KM
B)6.66KM
C)7.25KM
D)12.40KM
E)13KM

Since time differences given are in minutes so let's convert speed in Km per minute

8 kmph = 8/60 km per minute = (2/15) km per minute
10 kmph = 10/60 km per minute = (1/6) km per minute


Let, Time of train to depart from station = t minutes

Distance between Point A and Railway Station = Speed*Time = (2/15)*(t+5)

Distance between Point A and Railway Station = Speed*Time = (1/6)*(t-5)

i.e. (2/15)*(t+5) = (1/6)*(t-5)
i.e. 12t+60 = 15t - 75

i.e. 3t = 135
i.e. t = 45 minutes

i.e. Distance = (1/6)*(t-5) = (1/6)*(45-5) = 40/6 = 20/3 Km = 6.66 Km

Answer: Option B
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seemachandran
Guys

Perhaps I am overlooking the issue- I am making some silly mistake please correct me
I am getting confused with units (km/hr & mins)

I converted 5 mins to 'hr' since the rate given is km/hr

so i got

8(t-1/12) = 10(t+1/12)

I am getting incorrect answer , please let me know how can you directly consider t+5 & t-5

Thanks in advance!!

Don't always make mathematical equation but also look at relevance of the equation formed

you formed an equation 8(t-1/12) = 10(t+1/12)

when you know speed is less (i.e.8 kmph) then time should be more (i.e. (t+1/12) ) whereas you have taken less speed and less time on one sie and more speed and more time on other side and have equated them which is not logically relevant.

Less speed will take more time and More speed will take less time to travel the same distance.

I hope this helps!!!

P.S. (Similar logical aspect of an equation/inequation are greatly helpful in topic like Inequality)
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Thanks Karishma & GMATinsight.. understood the issue
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g106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?

A)9KM
B)6.66KM
C)7.25KM
D)12.40KM
E)13KM


Another approach is testing the answers.

We need to find the distance that makes the difference between times 10 min.

convert speeds to km/min

S1= 2/15 & S2=1/6. To be strategic I will start by choice 'B' or 'D'

When scanning the answer all fraction is easy to recognize

Chioice B= 6.66= 6 2/3=20/3.................T1=20/3 *15/2=50 min & T2= 20/3* 6= 40 min...... Then difference is 10 min

Answer: B
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g106
Walking at 8KMPH From Point A to Point railway station, mike will miss the train by 5 minutes. If he walks at 10KMPH he will reach the station by 5 minutes early.What is the distance between Point A to Railway station?

A)9KM
B)6.66KM
C)7.25KM
D)12.40KM
E)13KM

We can let t = the time Mike would arrive at the railway station if he were on time.

Since when he walks at 8 KMPH, he arrives 5 minutes late, when he walks at 10 KMPH, he arrives 5 minutes early, and the distances in each scenario are the same, we have:

(remember, 5 minutes = 1/12 hour)

8(t + 1/12) = 10(t - 1/12)

8t + 2/3 = 10t - 5/6

Multiplying the equation by 6, we have:

48t + 4 = 60t - 5

9 = 12t

3/4 = t

Thus, the distance is:

8(3/4 + 1/12) = 6 + ⅔ = 6.66 KM

Answer: B
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