Model 5 Problems with Races Practice Questions Answers Test with Solutions & More Shortcuts
time & distance PRACTICE TEST [5 - EXERCISES]
Model 1 Basic Time & Distance using formula
Model 2 Vehicles in x/y of its usual speed
Model 3 Problems on average speed
Model 4 Time & Distance with Ratios
Model 5 Problems with Races
Question : 1 [SSC CPO SI 2016]
Walking at $3/4$ of his usual speed, a man reaches his office 20 minutes late. Then his usual time for walking to his office is :
a) 30 minutes
b) 40 minutes
c) 45 minutes
d) 1 hour
Answer »Answer: (d)
Usual time = x minutes
New time = ${4x}/3$ minutes
$(Since, \text"Speed ∞ "\text"1"/\text"Time")$
According to the question,
${4x}/3 - x = 20$
$x/3 = 20$
x = 60 minutes i.e. 1 hour.
Question : 2 [SSC CPO S.I.2008]
A can give 40 metres start to B and 70 metres to C in a race of one kilometre How many metres start can B give to C in a race of one kilometre ?
a) 31$1/4$ metre
b) 32 metre
c) 31$3/4$ metre
d) 30 metre
Answer »Answer: (a)
According to the question,
When A covers 1000m, B covers
= 1000 - 40 = 960 m
and C covers =1000 - 70 = 930 m
When B covers 960m, C covers 930 m.
When B covers 1000m, C covers
= $930/960 × 1000$ = 968.75 metre
Hence, B gives C a start of
= 1000 - 968.75 = 31.25 metre
Question : 3 [SSC DEO 2009]
A jeep is chasing a car which is 5km ahead. Their respective speed are 90 km/hr and 75 km/ hr. After how many minutes will the jeep catch the car ?
a) 20 min.
b) 25 min.
c) 24 min.
d) 18 min.
Answer »Answer: (a)
Relative speed
= 95 - 75 = 15 kmph
Required time
= $\text"Distance"/ \text"Relative speed"$
= $5/15$ hours = $5/15 × 60$ minutes
= 20 minutes
Question : 4 [SSC CPO SI 2016]
Sarthak completed a marathon in 4 hours and 35 minutes. The marathon consisted of a 10 km run followed by 20 km cycle ride and the remaining distance again a run. He ran the first stage at 6 km/hr and then cycled at 16 km/ hr. How much distance did Sarthak cover in total, if his speed in the last run was just half that of his first run?
a) 35 km.
b) 45 km.
c) 40 km.
d) 5 km.
Answer »Answer: (a)
Let the total distance be x km.
Time = $\text"Distance"/ \text"Speed"$
According to the question,
$10/6 + 20/16 + {x - 30}/3$
= 4$35/60 = 4{7}/12$
$5/3 + 5/4 + x/3 - 10 = 55/12$
$x/3 + 5/3 + 5/4 - 10 = 55/12$
$x/3 + ({20 + 15 - 120}/12) = 55/12$
$x/3 - 85/12 = 55/12$
$x/3 = 85/12 + 55/12 = 140/12$
$x = 140/12 × 3$ = 35 km.
Question : 5 [SSC CGL Tier-I 2016]
Two persons ride towards each other from two places 55 km apart, one riding at 12km/hr and the other at 10 km/hr. In what time will they be 11 km apart?
a) 1 hour and 30 minutes
b) 2 hours and 45 minutes
c) 2 hours
d) 2 hours and 30 minutes
Answer »Answer: (c)
Relative speed
= 12 + 10 = 22 kmph
Distance covered
= 55 - 11 = 44 km
∴ Required time
= $(44/22)$ hours = 2 hours
IMPORTANT quantitative aptitude EXERCISES
Model 5 Problems with Races Shortcuts »
Click to Read...Model 5 Problems with Races Online Quiz
Click to Start..time & distance Shortcuts and Techniques with Examples
-
Model 1 Basic Time & Distance using formula
Defination & Shortcuts … -
Model 2 Vehicles in x/y of its usual speed
Defination & Shortcuts … -
Model 3 Problems on average speed
Defination & Shortcuts … -
Model 4 Time & Distance with Ratios
Defination & Shortcuts … -
Model 5 Problems with Races
Defination & Shortcuts …
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