Model 5 Problems with Races Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 5 EXERCISES
The following question based on time & distance topic of quantitative aptitude
(a) 30 minutes
(b) 40 minutes
(c) 45 minutes
(d) 1 hour
The correct answers to the above question in:
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.
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Read more problems on races Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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.
GET time & distance PRACTICE TEST 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
time & distance Shortcuts and Techniques with Examples
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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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