time and distance Model Questions & Answers, Practice Test for ssc chsl tier 1 2023
ssc chsl tier 1 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
Simplification
Powers Surds & Indices
Ratio Proportion & Partnership
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
A bus started its journey from Ramgarh and reached Devgarh in 44 min with its average speed of 50 km/hr. If the average speed of the bus is increased by 5 km/hr, how much time will it take to cover the same distance ?
Answer: (c)
Distance between Ramgarh and Devgarh = ${50 × 44}/{60} = {110}/3$
Average speed of the bus is increased by 5 km/hr then the speed of the bus = 55 km/hr
Required time = ${\text"Distance"}/{\text"Speed"} = {110}/3 × {60}/{55}$ = 40 min
Starting from his house, one day a student walks at a speed of $2{1/2}$km / hr and reaches his school 6 minutes late. Next day he increases his speed by 1 km/hr and reaches the school 6 minutes early. How far is the school from his house?
Answer: (a)
Distance from his house
= $\text"Product of speed" /\text"Difference of speed"$ × total time
= ${5/2×7/2}/1×12/60$
=$35/4×1/5=7/4=1.75$km
A man travelled from the village to post office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hr 48 min, then the distance of post office from the village is
Answer: (c)
Let time taken from village to post office (one side) = t hrs
Time taken for whole journey = 5 hrs 48 min
= 5hr$48/60$hr=$5{4/5}$ hrs.
Now, 25 × t = $4{(5{4/5-t)}$
⇒25t=${29×4}/5-4t⇒29t={29×4}/5$
⇒t=$4/5$hrs
∴ Distance =$25×4/5=20$ km
An increase in the speed of car by 10 km per hour saves 1 hour in a journey of 200 km, find the initial speed of the car.
Answer: (d)
Let initial speed = x km/h
∴ $200/x-200/{x+10}=1$
x( x+ 10) = 2000
$x^2$+10x-2000=0
x = 40 km/h
∴ Initial speed of car = 40 km/h
A man travels 600 km by train at 80 km/hr, 800 km by ship at 40 km/hr, 500 km by aeroplane at 400 km/hr and 100 km by car at 50 km/hr. The average speed for entire distance is
Answer: (b)
Average speed = $\text"Total distance"/\text"Total time"$
${600+800+500+100}/{600/80+800/40+500/400+100/50}=65{5/123}$km/h
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