time and distance Model Questions & Answers, Practice Test for ssc cgl tier 1 2024
ssc cgl tier 1 2024 SYLLABUS WISE SUBJECTS MCQs
Number System
Ratio & Proportion
Percentage
Time & Work
Time & Distance
Trigonometric Ratios & Identities
The distance between two points (A and B) is 110 km. X starts running from point A at a speed of 60 km/h and Y starts running from point B at a speed of 40 km/h at the same time. They meet at a point C, somewhere on the line AB. What is the ratio of AC to BC?
Answer: (c)
Distance between two points = 110 km
Relative speed = 60 + 40 = 100 km/h
Time after which they meet
= ${\text"Total distance"}/{\text"Relative speed"} = {110}/{100}$ = 1.10 h
Distance corvered by A in 1.10 h = AC = 60 × 1.10
= 66 km
Remaining distance = BC = 110 – 66 = 44 km
Required ratio = AC : BC = 66 : 44 = 3 : 2
Nilesh goes to school from his village & returns at the speed of 4 km/hr. If he takes 6 hours in all, then what is the distance between the village and the school?
Answer: (e)
Let the distance between the village and the school be x km.
According to the question,
$x/4+x/2=6$
or, ${x+2x}/4=6$
or, 3x = 6 × 4
∴$x={6×4}/3=8$ km
A man starts from B to K and another from K to B at the same time. After passing each other they complete their journeys in $3{1/3}$ and $4{4/5}$ hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.
Answer: (a)
Ratio of speed = $\text"1st man 's speed"/\text"2nd man 's speed"$ = $√b/√a=√b/√a=√{{4{4/5}}/{3{1/3}}}$
= $ √{24/5×3/10}=√{36/25}=6/5$
∴$12/{2nd man 's speed} = 6/5$
∴$2^{nd}$ man's speed = $60/6$ = 10 km/hr.
A 200 meter long train crosses a platform double its length in 36 seconds. What is the speed of the train in km/hr ?
Answer: (c)
Speed of train
= ${(200+400)}/36×18/5$
= 60 km/hr.
Two trains each 200 m long move towards each other on parallel lines with velocities 20 km/h and 30 km/h, respectively. What is the time that elapses when they first meet until they have cleared each other?
Answer: (a)
Relative speed of trains = (20 + 30) km/h = 50 km/h
= 50 km/h = 50 × $5/{18}$ m/s
Total relative distance = 200 + 200 = 400 m
∴ Required time = ${400 × 18}/{50 × 5}$ = 28.8 s
ssc cgl tier 1 2024 IMPORTANT QUESTION AND ANSWERS
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