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Bunuel
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Distance between Andy's office and home is 100 Km. One day, he was lat 25 Apr 2023, 01:30
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55% (hard)

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67% (02:34) correct 33% (02:49) wrong based on 136 sessions

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Distance between Andy's office and home is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 km/hr and reached the office at the normal time. What was Andy's speed at that day?

A. 50 Km/hr
B. 30 Km/hr
C. 25 Km/hr
D. 20 Km/hr
E. 16 Km/hr


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C
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$!vakumar.m
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Distance between Andy's office and home is 100 Km. One day, he was lat Updated on: 25 Apr 2023, 04:54
Originally posted by $!vakumar.m on 25 Apr 2023, 03:12.
Last edited by $!vakumar.m on 25 Apr 2023, 04:54, edited 1 time in total.
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Distance between Andy's office and home is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 km/hr and reached the office at the normal time. What was Andy's speed at that day?

let s be the speed and t be the time taken
then, 100 = st......1
and 100 = (s+5)(t-1).......2
Solving the equation we get t = 5 and s = 20km/h
On that day, Andy's speed is s+5 = 25 km/hr.
Answer C

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Distance between Andy's office and home is 100 Km. One day, he was lat 25 Apr 2023, 04:26
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sivakumarm786
Distance between Andy's office and home is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 km/hr and reached the office at the normal time. What was Andy's speed at that day?

let s be the speed and t be the time taken
then, 100 = st......1
and 100 = (s+5)(t-1).......2
Solving the equation we get t = 5 and s = 20km/h
Answer D
Show more



We are asked what was Andy's speed at that day, that means we need s+5 i.e. 25km/h
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Distance between Andy's office and home is 100 Km. One day, he was lat 28 Apr 2023, 14:56
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sivakumarm how did you solve the following quickly?

Quote:
Solving the equation we get t = 5 and s = 20km/h
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I got to this but seems like testing numbers takes too long:
100 = st
100 = (s + 5)(t-1) --> 105 = st - s + 5t
substitute st --> 105 = 100 - s + 5t --> 5 = 5t - s

now with 5 = 5t - s it's unclear how to quickly get to t = 5 and s = 20
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Distance between Andy's office and home is 100 Km. One day, he was lat 28 Apr 2023, 23:54
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Bunuel
Distance between Andy's office and home is 100 Km. One day, he was late by an hour than the normal time to leave for the office, so he increased his speed by 5 km/hr and reached the office at the normal time. What was Andy's speed at that day?

A. 50 Km/hr
B. 30 Km/hr
C. 25 Km/hr
D. 20 Km/hr
E. 16 Km/hr
Show more
Solution:

  • Ideal scenario:
    • \(Speed = x\) km/hr (let) and \(Distance=100\) km
    • \(Time=\frac{Distance}{Speed}=\frac{100}{x}\)
  • One day:
    • \(Speed = x+5\) km/hr (let) and \(Distance=100\) km
    • \(Time=\frac{Distance}{Speed}=\frac{100}{x+5}\)

  • According to the question, \(\frac{100}{x}-\frac{100}{x+5}=1\)
    \(⇒\frac{100(x+5)-100x}{x(x+5)}=1\)
    \(⇒100x+500-100x=x(x+5)\)
    \(⇒500=x(x+5)\)
  • From this we can say \(x=20\)
  • So, speed at that day \(=x+5=20+5=25\) km/hr


Hence the right answer is Option C
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