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Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin

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Bunuel
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Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   25 Mar 2020, 03:27
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Walking at \(\frac{4}{7}\)th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. How many minutes does he need to cover that distance at his usual speed?

A. 20 min
B. 24 min
C. 25 min
D. 27 min
E. 30 min


M37-55

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Bunuel
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   26 Nov 2021, 08:18
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Official Solution:

Walking at \(\frac{4}{7}\)th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. How many minutes does he need to cover that distance at his usual speed?

A. \(20\)
B. \(24\)
C. \(25\)
D. \(27\)
E. \(30\)


Say the usual speed and the usual time are \(s\) and \(t\), respectively. Then the distance is \(d=st\).

The distance can also be expressed as \(d=(\frac{4}{7}*s)(t+15)\)

Equate: \(st=(\frac{4}{7}*s)(t+15)\);

Reduce by \(s\) and solve to get \(t=20\) minutes.


Answer: A
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   25 Mar 2020, 03:30
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Bunuel wrote:
Walking at 4/7th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. What is the time he needs to cover that distance at his usual speed?

A. 20 min
B. 24 min
C. 25 min
D. 27 min
E. 30 min


Given: Walking at 4/7th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work.

Asked: What is the time he needs to cover that distance at his usual speed?

Let the time he needs to cover that distance at his usual speed = t min

v1/v2 = t2/t1
4/7 = t/( t+15)
7t = 4t + 60
3t = 60
t = 20 mins

IMO A
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   25 Mar 2020, 03:54
OA is A=20 minutes
Distance in km.(D)
Speed in km/h.(S)
Time in hour.(T)
D=ST-------(1)
D=4S/7*T+1/4-----(2)
ST=4S/7*(4T+1)/4
28ST=16ST+4S
12ST=4S
T=1/3
1/3 hour= 20 minutes.
OA is A

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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   25 Mar 2020, 04:49
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Bunuel wrote:
Walking at 4/7th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. What is the time he needs to cover that distance at his usual speed?

A. 20 min
B. 24 min
C. 25 min
D. 27 min
E. 30 min


Solution:
Let 7x be the usual speed of Randy.
    • \(\frac{4}{7}\) of the usual speed = \(\frac{4}{7}*7x = 4x\)
Let D be the total distance between home and office.
    • Time taken to reach office at usual speed = \(\frac{D}{7x}\) …(i)
    • Time taken to reach office at new speed = \(\frac{D}{4x}\)
According to the condition given,
    • \(\frac{D}{4x} – \frac{D}{7x} = 15\)
    o \(D*3 = 15*28x\)
    o \(D = 5*28x\)
On substituting D = 5*28x in equation (i), we get
    • Time taken to reach office at usual speed = \(5*\frac{28x}{7x} =5*4 = 20\) minute.
Hence, the correct answer is Option A.
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   25 Mar 2020, 05:02
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Bunuel wrote:
Walking at 4/7th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. What is the time he needs to cover that distance at his usual speed?

A. 20 min
B. 24 min
C. 25 min
D. 27 min
E. 30 min

Solution:
Let 7x be the usual speed of Randy.
    • \(\frac{4}{7}\) of the usual speed =\( \frac{4}{7}*7x = 4x \)
Let t1 and t2 be the time taken to reach office at new and usual speed respectively.
We know that speed and time are inversely proportional to each other.
    • \(t1: t2 = 7x: 4x\)
Now,
    • Extra time taken = 15min
    • \(7x -4x = 15\)
    • \(3x = 15\)
    • \(x = 5\)
    •Time taken to reach office at usual speed = \(t2 = 4*x = 4*5 =20\) min
Hence, the correct answer is Option A.
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   20 May 2021, 23:23
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Bunuel wrote:
Walking at 4/7th of his usual speed, Randy takes 15 minutes longer to cover the distance from home to work. What is the time he needs to cover that distance at his usual speed?

A. 20 min
B. 24 min
C. 25 min
D. 27 min
E. 30 min


We know, \(Speed ∝\frac{1}{time}\)

Let, his usual speed be x. If his speed is x, time taken is 1/x, if his speed is 4x/7 then the time taken is 7x/4.

Here, late time~usual time = 15 minute
So, 7x/4~x=15
=>x=20.

Answer is A
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink] New post   02 Jun 2021, 01:45
The distance is same; the difference is in time and rate so,
R*T=D
Let rate be x, x*T=D
As per the question, 4x/7*(T+15)=D
So, x*T=4x/7*(T+15)
xt=20x = t=20 (A)
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Re: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Fin [#permalink]
02 Jun 2021, 01:45
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