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The distance travelled by vehicle F on Day 2 was approximately what percent of the distance travelled by it on Day 1?
Answer & Solution
Correct Answer:
Option
E
$$\eqalign{
& \text{Speeds of vehicles on Day 1:} \cr
& \text{A → } \frac{832}{16}\text{ km/hr}=\text{52 km/hr} \cr
& \text{B → } \frac{516}{12}\text{ km/hr}=\text{43 km/hr} \cr
& \text{C → } \frac{693}{11}\text{ km/hr}=\text{63 km/hr} \cr
& \text{D → } \frac{552}{12}\text{ km/hr}=\text{46 km/hr} \cr
& \text{E → } \frac{935}{17}\text{ km/hr}=\text{55 km/hr} \cr
& \text{F → } \frac{703}{19}\text{ km/hr}=\text{37 km/hr} \cr
& \cr
& \text{Speed of vehicles on Day 2:} \cr
& \text{A → } \frac{864}{16}\text{ km/hr}=\text{54 km/hr} \cr
& \text{B → } \frac{774}{18}\text{ km/hr}=\text{43 km/hr} \cr
& \text{C → } \frac{810}{18}\text{ km/hr}=\text{45 km/hr} \cr
& \text{D → } \frac{765}{15}\text{ km/hr}=\text{51 km/hr} \cr
& \text{E → } \frac{546}{14}\text{ km/hr}=\text{39 km/hr} \cr
& \text{F → } \frac{636}{12}\text{ km/hr}=\text{53 km/hr} \cr} $$
$$\eqalign{ & \text{Required %} \cr & = \left(\frac{636}{703}\times100\right)\% \cr & = \frac{63600}{703}\% \cr & = 90.46\% \cr & \approx 90\% \cr} $$
$$\eqalign{ & \text{Required %} \cr & = \left(\frac{636}{703}\times100\right)\% \cr & = \frac{63600}{703}\% \cr & = 90.46\% \cr & \approx 90\% \cr} $$
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LoginQuestion is saying to find distance and solution is about finding speed..... please clarify if I’m wrong.