Model 2 Vehicles in x/y of its usual speed 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) 2 hours 10 minutes
(b) 2 hours 25 minutes
(c) 2 hours 15 minutes
(d) 2 hours 30 minutes
The correct answers to the above question in:
Answer: (d)
Time and speed are inversely proportional.
$7/6$ × Usual time - Usual time = 25 minutes
Usual time $(7/6 -1)$
= 25 minutes
Usual time × $1/6$
= 25 minutes
Usual time = 25 × 6
= 150 minutes
= 2 hours 30 minutes
Using Rule 8,
Here, A = 6, B = 7, t = $25/60 = 5/12$ hrs.
Usual time = $\text"A"/\text"Diff of A and B"$ × time
= $6/{(7 - 6)} × 5/12 = 5/2$ hrs.
= 2 hours 30 minutes
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Read more usual vs actual speed Based Quantitative Aptitude Questions and Answers
Question : 1
A train running at $7/11$ of its own speed reached a place in 22 hours. How much time could be saved if the train would run at its own speed?
a) 16 hours
b) 8 hours
c) 7 hours
d) 14 hours
Answer »Answer: (b)
Since the train runs at $7/11$ of its own speed,
the time it takes is $11/7$ of its usual speed.
Let the usual time taken be t hours.
Then we can write, $11/7$ t = 22
t = ${22 × 7}/11$ = 14 hours
Hence, time saved
= 22 - 14 = 8 hours
Question : 2
A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is
a) 620 km
b) 600 km
c) 420 km
d) 540 km
Answer »Answer: (c)
Fixed distance = x km and certain speed = y kmph (let).
Case I,
$x/{y +10} = x/y$ - 1
$x/{y +10} + 1 = x/y$ --- (i)
Case II,
$x/{y + 20} = x/y - 1 - 3/4$
= $x/y - {4 + 3}/4$
$x/{y + 20} + 7/4 = x/y$ --- (ii)
From equations (i) and (ii),
$x/{y +10} + 1 = x/{y + 20} + 7/4$
$x/{y +10} - x/{y + 20} = 7/4$ - 1
$x({y + 20 - y - 10}/{(y + 10)(y + 20)})$
= ${7 - 4}/4 = 3/4$
${x × 10}/{(y + 10)(y + 20)} = 3/4$
3 (y + 10) (y + 20) = 40 x
${3(y + 10)(y + 20)}/40$ = x --(iii)
From equation (i),
${3(y + 10)(y + 20)}/{40(y + 10)}$ + 1
= ${3(y + 10)(y + 20)}/{40y}$
3 (y +20) + 40 = ${3(y + 10)(y + 20)}/y$
$3y^2 + 60y + 40y$
= $3(y^2 + 30y + 200)$
$3y^2 + 100y = 3y^2 + 90y + 600$
10y = 600 ⇒ y = 60
Again from equation (i),
$x/{y +10} + 1 = x/y$
$x/{60 + 10} + 1 = x/60$
$x/70 + 1 = x/60$
${x + 70}/70 = x/60$
6x + 420 = 7x
7x - 6x = 420
$x$ = 420 km.
Question : 3
A car moving in the morning fog passes a man walking at 4 km/ h. in the same direction. The man can see the car for 3 minutes and visibility is upto a distance of 130 m. The speed of the car is :
a) 5 km. per hour
b) 7 km. per hour
c) 6$3/5$ km. per hour
d) 7$3/5$ km. per hour
Answer »Answer: (c)
Speed of car = x kmph.
Relative speed = (x - 4) kmph.
Time = 3 minutes
= $3/60$ hour = $1/20$ hour
Distance = 130 metre
= $130/1000$ km. = $13/100$ km.
Relative speed = $\text"Distance"/ \text"Time"$
$x - 4 = 13/100$ × 20
5x - 20 = 13
5x = 20 + 13 = 33
$x = 33/5 = 6{3}/5$ kmph.
Question : 4
By walking at $3/4$ of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is
a) 30 minutes
b) 40 minutes
c) 60 minutes
d) 75 minutes
Answer »Answer: (c)
$4/3$ of usual time
= Usual time + 20 minutes
$1/3$ of usual time = 20 minutes
Usual time = 20 × 3 = 60 minutes
Using Rule 8,
Here, A = 3, B= 4, t = 20 minutes
Usual time taken
= $\text"A"/\text"Diff of A and B"$ × time
= $3/{(4 - 3)} × 20$ = 60 minutes
Question : 5
Walking at three-fourth of his usual speed, a man covers a certain distance in 2 hours more than the time he takes to cover the distance at his usual speed. The time taken by him to cover the distance with his usual speed is
a) 5 hours
b) 6 hours
c) 5.5 hours
d) 4.5 hours
Answer »Answer: (b)
$4/3$ × usual time - usual time = 2
$1/3$ usual time = 2
Usual time = 2 × 3 = 6 hours
Using Rule 8,
Here, $\text"A"/ \text"B"= 3/4$, time= 2 hrs.
Usual Speed
= $\text"A"/\text"Diff of A and B"$ × time
= $3/{(4 - 3)} × 2$ = 6 hours
Question : 6
A car travelling with $5/7$ of its usual speed covers 42 km in 1 hour 40 min 48 sec. What is the usual speed of the car?
a) 30 km/hr
b) 25 km/hr
c) 35 km/hr
d) 17$6/7$ km/hr
Answer »Answer: (c)
1 hr 40 min 48 sec
= 1 hr $(40 + 48/60)$ min
= 1 hr $(40 + 4/5)$ min
= 1 hr $204/5$ min
= $(1 + 204/300)$ hr = $504/300$ hr
Speed = $42/{504/300}$ = 25 kmph
Now, $5/7$ usual speed = 25
Usual speed = ${25 × 7}/5$ = 35 kmph
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
-
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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