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On a Monday, John left from his home for his office at 09:05 AM
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13 May 2018, 10:49
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63% (01:55) correct 38% (02:15) wrong based on 56 sessions
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On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home? A. 50 B. 55 C. 60 D. 65 E. 75
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Re: On a Monday, John left from his home for his office at 09:05 AM
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13 May 2018, 11:46
Let speed be V
Monday V*55min = Distance
Tuesday (V+15)*44min = Distance
Equating V=60
Distance would be 55



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Re: On a Monday, John left from his home for his office at 09:05 AM
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13 May 2018, 13:16
saswata4s wrote: On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?
A. 50 B. 55 C. 60 D. 65 E. 75 On Monday John left his home @9:05 AM and reached office @10AM, so in total he took 55 minutes to cover the distance On tuesday he left house 15 minutes later @9:20 and reached office @ 10:04 taking 44 minutes to cover the distance to his office Let's assume his speed on monday was x km/hr, so his speed on tuesday = x + 15 km/hr On both the days he covered same distance so, 55*x = 44(x + 15) that is x = 60 km/hr so the distance in km = 55 / 60 * 60 = 55km Hence B is the correct Answer



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Re: On a Monday, John left from his home for his office at 09:05 AM
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13 May 2018, 17:12
zishu912 wrote: saswata4s wrote: On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?
A. 50 B. 55 C. 60 D. 65 E. 75 On Monday John left his home @9:05 AM and reached office @10AM, so in total he took 55 minutes to cover the distance On tuesday he left house 15 minutes later @9:20 and reached office @ 10:04 taking 44 minutes to cover the distance to his office Let's assume his speed on monday was x km/hr, so his speed on tuesday = x + 15 km/hr On both the days he covered same distance so, 55*x = 44(x + 15) that is x = 60 km/hr so the distance in km = 55 / 60 * 60 = 55km Hence B is the correct Answer Thanks for the explanation!



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On a Monday, John left from his home for his office at 09:05 AM
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13 May 2018, 22:36
saswata4s wrote: On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?
A. 50 B. 55 C. 60 D. 65 E. 75 Total time taken for Monday  9:0510:00 i.e. 55 mins Total time taken for Tuesday  9:2010:04 i.e. 44 mins Let's assume speed for Monday is V km/h Then Speed for Tuesday will be (V+15) km/h However, distance travelled is same for both days and distance = speed*time taken This means  V*55 = (V+15)*44 On Solving, V = 60 km/h Distance (Converting minutes to hours) = V*(55/60) = 55 Km



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Re: On a Monday, John left from his home for his office at 09:05 AM
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14 Jul 2018, 19:17
saswata4s wrote: On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?
A. 50 B. 55 C. 60 D. 65 E. 75 We can let 55/60 = 11/12 = John’s time, in hours, on Monday and r = John’s rate on Monday On Tuesday his time was 44/60 = 11/15 hours, and his rate was (r + 15). Since the distance from home to office was the same for the two days, we can create the equation: r(11/12) = (r + 15)(11/15) r/12 = (r + 15)/15 12(r + 15) = 15r 12r + 180 = 15r 180 = 3r 60 = r So, John’s rate on Monday was 60 km/h. Since it took John 55 minutes = 55/60 hours to reach his office, the distance between his home and his office is 60 x 55/60 = 55 km. Answer: B
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Re: On a Monday, John left from his home for his office at 09:05 AM &nbs
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