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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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Bunuel
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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Bunuel
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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Bunuel
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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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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