Question
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?
Answer: Option B
:
B
Using 1x&1x+1
We know that the speed is increasing by 12 hence the time has to decrease by 13 (using the 1x&1x+1)
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is 10×6=60kms
To reach at noon he will have to cover 60 km in 5 hours. i.e., he has to travel at 12 km/hr.
Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have 15×(t−2)=10×(t)
T=6 , or he started 6 hours before 1, i.e., at 7.
Hence the total distance now is 10×6=60 km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.
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:
B
Using 1x&1x+1
We know that the speed is increasing by 12 hence the time has to decrease by 13 (using the 1x&1x+1)
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is 10×6=60kms
To reach at noon he will have to cover 60 km in 5 hours. i.e., he has to travel at 12 km/hr.
Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have 15×(t−2)=10×(t)
T=6 , or he started 6 hours before 1, i.e., at 7.
Hence the total distance now is 10×6=60 km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.
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