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Multiple choice ime speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10km/hr more than the speed of going, what was the speed per hour in each direction?

  1. going speed =−30km/hr, returning=−20km/hr

  2. going speed =20km/hr, returning=30km/hr

  3. going speed =30km/hr, returning=40km/hr

  4. going speed =30km/hr, returning=20km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Letting the going speed be $v$, the returning speed is $v+10$. The time equation $150/v - 150/(v+10) = 2.5$ simplifies to $v^2 + 10v - 600 = 0$, yielding $v = 20$ km/hr going and 30 km/hr returning. Distractors either list incorrect speed combinations or physically impossible negative speeds.

Multiple choice ime speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

If a car travels 150 km in 5 hours then its speed is

  1. 150 km/hr

  2. 30 km/hr

  3. 50 km/hr

  4. 10 km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Speed=\dfrac{distance}{time}=\dfrac{150}{5}$ $=30 km/hr$