Everyday Ashwin starts at 3:00 pm from his house to pick up his son from school. They reach their house at 5:00 pm. one day school was over at 3:00 pm. Ashwin, not aware of this, started from home as usual. He met his son on the way and they reached home 20 minutes earlier than usual. If the speed of his car is 55 kmph, find his son speed (kmph).
(A) 10
(B) 11
(C) 12
(D) 15
(E) 16
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Everyday Ashwin starts at 3:00 pm from his house to pick...
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If Ashwin drives at 55 km/hour and his roundtrip drive takes 2 hours, then each leg of the trip takes one hour, and is exactly 55km.LUANDATO wrote:Everyday Ashwin starts at 3:00 pm from his house to pick up his son from school. They reach their house at 5:00 pm. one day school was over at 3:00 pm. Ashwin, not aware of this, started from home as usual. He met his son on the way and they reached home 20 minutes earlier than usual. If the speed of his car is 55 kmph, find his son speed (kmph).
(A) 10
(B) 11
(C) 12
(D) 15
(E) 16
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
If they arrive home 20 minutes, or 1/3 hour early, we know he must have saved 10 minutes or 1/6 hour on each leg of the trip. Put another way, to pick up his son, Ashwin traveled for 5/6 of an hour, instead of a full hour. In that remaining 1/6 hour, Ashwin would have traveled an additional 55 * (1/6) = 55/6 km, so this must be the distance that his son covered. (Alternatively, we know that Ashwin covered 55 * (5/6) miles. If the total distance is 55 km, then his son must cover the remainder. 55 - 55 * (5/6) = 55 * (1/6) or 55/6.
To summarize. In 5/6 of an hour, the son would cover 55/6 km. If his speed is r, then r * (5/6) = 55/6); r = 11. The answer is B
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We see that the distance between Ashwin's home and his son's school is 55 km since it normally takes him two hours for the round trip (and thus one hour for a one-way trip). Furthermore, since on that particular day when his son walked part of the way from school to home and they reached home 20 minutes earlier than usual, it must be true that Ashwin only drove 1 hour 40 minutes for the round trip and 50 minutes for a one-way trip (here, the one-way trip is the distance from home to the place where he met his son). Thus, if we let r = his son's speed in kmph, we can create an equation based on the 50 minutes he drove from home when he met his son (notice that his son had walked 50 minutes from school when he met his father and his 50 minutes = 5/6 hour):BTGmoderatorLU wrote:Everyday Ashwin starts at 3:00 pm from his house to pick up his son from school. They reach their house at 5:00 pm. one day school was over at 3:00 pm. Ashwin, not aware of this, started from home as usual. He met his son on the way and they reached home 20 minutes earlier than usual. If the speed of his car is 55 kmph, find his son speed (kmph).
(A) 10
(B) 11
(C) 12
(D) 15
(E) 16
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
5/6 * 55 + 5/6 * r = 55
Multiplying the equation by 6, we have:
275 + 5r = 330
5r = 55
r = 11
Answer: B
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