Jack drove from Point A to Point B at a rate of 1.2 minutes per kilo

At 1.2 minute per km, Jack would be traveling \(\frac{60}{1.2}\) or 50 km in an hour. This is the
speed from Point A to B. At 1 minute per km, Jack would be traveling \(\frac{60}{1}\) or 60 kmph,
which is the speed from Point B to A.

As he took a minimum of 3 hours to travel the distance and a maximum of 5 hours, the distance
between Points A and B is between 150 and 250 km respectively. Also, he would travel between
the 120 and 180 km respectively on his way back(since the time taken is between 2 and 3 hours)

The only answer option which lies in both the ranges is 160 km(Option B) and is our answer!

Hey dave13

Yes, your solution is correct here. But, whenever you have a similar
problem with a solution which has 2 values from the answer options,
in those cases, you will need to check the other range as well

For instance, if you had only tried the information from A to B - the
distance could have between 150 and 250(and three of the answer
options - Option B, C, and D could be the answer! So, in order to
eliminate and get one answer, we use the other condition as well.

Hope this helps you!

Let’s first determine Jack’s rate, in km per hour, from Point A to Point B. Since 60/1.2 = 50, Jack drove from Point A to Point B at 50 km per hour. Similarly, his return trip rate is 60/1. Thus, he drove back from Point B to Point A at 60 km per hour.

Let’s use the formula: rate x time = distance. Since it took him between 3 hours to 5 hours to travel from Point A to Point B, the distance between Point A and Point B could be anywhere from 50 x 3 = 150 km to 50 x 5 = 250 km.

Since it took him between 2 hours to 3 hours to travel from Point B to Point A, the distance between Point A and Point B could be anywhere from 60 x 2 = 120 km to 60 x 3 = 180 km.

The only distance in the given choices that fits into both ranges is 160 km.

Answer: B
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Two possible traps here:
1) min/max distance; and
2) distance is ONE way ("between ... A and B")

Leg 1: from A to B

Rate: \(\frac{1km}{1.2mins}*\frac{50}{50}=\frac{50km}{60mins}=\frac{50km}{1hr}\)

Distance? He drove between 3 and 5 hours
\(\frac{50km}{hr}*3hrs=150\) kms
\(\frac{50km}{hr}*5hrs=250\) kms

\(150 \leq D \leq 250\) kms

Leg 2: from B to A
Rate: \(\frac{1km}{1min}=\frac{60km}{60mins}=\frac{60km}{1hr}\)

Distance? He drove between 2 and 3 hours
\(\frac{60km}{hr}*2hrs=120\) kms
\(\frac{60km}{hr}*3hrs=180\) kms

\(120 \leq D \leq 180\) kms

For Leg 1 the MINIMUM distance is 150 kms, because the minimum time is 3 hours.
3 hours' distance of 150 kms "wins" over Leg 2 minimum of 120 kms. (Distance depends on time traveled, and time is defined strictly for each leg.)*

The distance between A and B cannot be less than 150 km.

For Leg 2, from B to A, the MAXIMUM distance is 180 kms. The distance from B to A (=A to B) cannot be greater than 180 km

He traveled at least 150 km and at most 180 km: \(150< D<180\) kms

Only 160 kms fits in that range

Answer B

*Track on time to decide minimum and maximum
Minimum
-- the least time for Leg 1 is 3 hours
-- The least time Leg 2 is 2 hours
-- He cannot travel 2 hours for Leg 1
3 hours (+ corresponding distance at rate in Leg 1) "wins."
Minimum distance is 150

Maximum
-- The most time in Leg 2 is 3 hours
-- The most time in Leg 1 is 5 hours
-- He cannot travel 5 hours during Leg 2
3 hours (+ corresponding distance at rate in Leg 2) "wins."
Maximum distance is 180 km
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generis pushpitkc is my solution correct ?

IMO word problems are like CRs,
it was a hot summer day and Jonh was looking for a watering hole He embarked on a fantastic adventure through stunning mountain ridges, he was heading like snail (at a speed of 1.2 min per klm ) to turtle lake; en route he took selfies, pics, videos and posted on Instagram

So, when John finally reached the lake and wanted desperately to dive in, he realized that he forgot to take his swimsuit so he drove back home straight away between 2 hours to 3 hours so 60 * 3 = 180, since it says BETWEEN 2 and 3 hrs, distance must be a little less than 180. Hence I picked 160 the distance from A to B cant contradict the distance from B to A cause, he spent less time on his way back home. hence I distance from B to A carries more weight than from A to B.

So basically I used information from B to A, and ignored info from A to B. these word problems love "feeding" us with words

YAY !

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