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01 Jan 2020, 22:51
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Everyday I cover the distance from my home and office at a usual speed and take a certain time at the usual speed. When I increase my usual speed by 5 kmph, I take 10 minutes less than usual. If I reduce my usual speed by 5 kmph, I take 15 minutes more than usual. Find the distance from home to office.
A. 25 km
B. 30 km
C. 40 km
D. 45 km
E. 50 km
A. 25 km
B. 30 km
C. 40 km
D. 45 km
E. 50 km
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02 Jan 2020, 11:36
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Usual - Distance (D) = Speed(S) *time(t) -----I
1st case when speed is increased by 5 kmph
D = (S+5) *(t-10/60) --- II
2nd case when speed is reduced by 5 kmph
D = (S-5)*(t+15/60) ----- III
Equating 1 and II
st = st -s/6 +5t- 5/6
s-30t+5 =0 -----A
equating I and III
st = st +s/4 -5t -5/4
s-20t-5=0 ---B
Subtrating A from B
(30t-20t) = 10
t = 1 hour
So S= 25kmph
Hence Distance D = 25*1 = 25 km
Answer is A
1st case when speed is increased by 5 kmph
D = (S+5) *(t-10/60) --- II
2nd case when speed is reduced by 5 kmph
D = (S-5)*(t+15/60) ----- III
Equating 1 and II
st = st -s/6 +5t- 5/6
s-30t+5 =0 -----A
equating I and III
st = st +s/4 -5t -5/4
s-20t-5=0 ---B
Subtrating A from B
(30t-20t) = 10
t = 1 hour
So S= 25kmph
Hence Distance D = 25*1 = 25 km
Answer is A
04 Jan 2020, 20:19
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Bunuel
Everyday I cover the distance from my home and office at a usual speed and take a certain time at the usual speed. When I increase my usual speed by 5 kmph, I take 10 minutes less than usual. If I reduce my usual speed by 5 kmph, I take 15 minutes more than usual. Find the distance from home to office.
A. 25 km
B. 30 km
C. 40 km
D. 45 km
E. 50 km
A. 25 km
B. 30 km
C. 40 km
D. 45 km
E. 50 km
Noting that 10 minutes = ⅙ hour and 15 minutes = ¼ hour, we can create the equations:
rt = d
(r + 5)(t - 1/6) = d
and
(r - 5)(t + 1/4) = d
Equating the first and second equations, we have:
(r + 5)(t - 1/6) = rt
rt - r/6 + 5t - 5/6 = rt
-r/6 + 5t - 5/6 = 0
Multiplying the above equation by -6, we have:
r - 30t + 5 = 0 → Eq. I
Similarly, equating the first and third equations, we have:
(r - 5)(t + 1/4) = rt
rt + r/4 - 5t - 5/4 = rt
r/4 - 5t - 5/4 = 0
Multiplying the above equation by 4, we have:
r - 20t - 5 = 0 → Eq. II
Subtracting Eq. I from Eq. II, we have:
10t - 10 = 0
10t = 10
t = 1
Substituting 1 for t in Eq. II, we have:
r - 20(1) - 5 = 0
r = 25
So d = rt = 25(1) = 25.
Answer: A
