Speed , Time and Distance

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

Speed , Time and Distance

by harsh.champ » Mon Feb 08, 2010 2:55 pm

If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/ hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

(A)11 Km/hr
(B)12 Km/hr
(C)13 Km/hr
(D)14 Km/hr
(E)15 Km/hr

The OA is B.

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Source: Beat The GMAT — Problem Solving | Mon Feb 08, 2010 2:55 pm

akahuja143: Master | Next Rank: 500 Posts; Posts: 182; Joined: Mon Apr 20, 2009 7:09 pm; Thanked: 1 times; Followed by:1 members

by akahuja143 » Mon Feb 08, 2010 7:10 pm

Distance = Speed X time

first case when speed is 10 distance =10X (say x hr is the time)
Second case distance =15(x -2)1Pm and 11 am

both of should be equal and as we solve x we will get x =6 and thereby distance will be 60 Km and 12 noon would 5 hours of time 60/5 =12

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pm; Location: Arabian Sea; Thanked: 125 times; Followed by:2 members

by ajith » Mon Feb 08, 2010 9:42 pm

harsh.champ wrote:If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/ hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

(A)11 Km/hr
(B)12 Km/hr
(C)13 Km/hr
(D)14 Km/hr
(E)15 Km/hr

The OA is B.

say the man was riding for x hours to reach at 11
15x is the distance
distance = 10 (x+2)
15x = 10x +20

20 = 5x
x=4

60 is the distance to reach in 5 hours he must travel at 60/5 = 12 Kmph

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Tue Feb 09, 2010 9:17 am

another approach using percentages in product constancy .

Concept :

Let product C(constant) = A * B

if A increases by 25% => B has to decrease by 20% ( in order to keep the product constant)

for. e.g. A*B = C

if A becomes 1.25*A ( 25% increase)

=> B should become .8*B (20% decrease) to keep product equal to C (A*B).

i.e 1/4th increase in A => 1/5th decrease in B

similarly 1/5th increase in A => 1/6th decrease in B

and 1/6th increase in A => 1/7th decrease in B

See a pattern here?

Now coming to our problem : (Distance (constant) = Speed * Time)

It is given that increase in speed from 10Kmph to 15Kmph (50% or 1/2 increase) resulted in time to decrease by 2hrs.

But, from Product constancy => we know that a a 50 % or 1/2 increase means 1/3 decrease in time.

OR 2 hours reduction is given an 1/3, hence 1 hour reduction = 1/6 decrease

=> 1/5 increase in other variable (product constancy)

=> 1/5 * 10 Kmph increase = 2 Kmph increase

or the speed should be increased to 10 + 2 = 12 Kmph.

Once this concept is internalized, problems like these can be done orally.

Regards,
Harsha

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Tue Feb 09, 2010 10:01 am

harsh.champ wrote:If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/ hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

(A)11 Km/hr
(B)12 Km/hr
(C)13 Km/hr
(D)14 Km/hr
(E)15 Km/hr

The OA is B.

Let the distance be X km and time taken with 10km/hr be t
so X/10=t, X/15=t-2 --> t=6
now if he reaches at noon time =t-1= 5 hrs. so speed = X/5 =10t/5 = 10x6/5 =[spoiler]12 Km/hr. (b)[/spoiler]

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by Scott@TargetTestPrep » Fri Jan 10, 2020 11:50 am

harsh.champ wrote:If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/ hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

(A)11 Km/hr
(B)12 Km/hr
(C)13 Km/hr
(D)14 Km/hr
(E)15 Km/hr

The OA is B.

We can let t = the number of hours it takes to arrive at noon. We can create the equation:

10(t + 1) = 15(t - 1)

10t + 10 = 15t - 15

25 = 5t

5 = t

Thus, the distance is 10(5 + 1) = 60. To arrive in 5 hours, he must cycle at a rate of 60/5 = 12 km/hr

Answer: B

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