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) 5 km. per hour
(b) 7 km. per hour
(c) 6$3/5$ km. per hour
(d) 7$3/5$ km. per hour
The correct answers to the above question in:
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.
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Read more usual vs actual speed Based Quantitative Aptitude Questions and Answers
Question : 1
A man with $3/5$ of his usual speed reaches the destination 2$1/2$ hours late. Find his usual time to reach the destination.
a) 4$1/2$ hours
b) 3$3/4$ hours
c) 3 hours
d) 4 hours
Answer »Answer: (b)
$3/5$ of usual speed will take $5/3$ of usual time.
[Since, time & speed are inversely proportional]
$5/3$ of usual time = usual time + $5/2$
$2/3$ of usual time = $5/2$
usual time = $5/2 × 3/2$
= $15/4 = 3{3}/4$ hours.
Using Rule 8,If an object travels certain distance with the speed of $A/B$ of its original speed and reaches its destination 't' hours before or after, then the taken time by object travelling at original speed isTime = $\text"A"/\text"(Difference of A and B)"$ × time (in hour)
Here, A = 3, B = 5, t = 2$1/2$
Usual time = $\text"A"/\text"Diff of A and B"$ × time
= $3/{5 - 3} × 2{1}/2$
= $3/2 × 5/2$
= $15/4 = 3{3}/4$ hours
Question : 2
The speed of a car is 54 km/hr. What is its speed in m/sec?
a) 150 m/sec
b) 194.4 m/sec
c) 19-44 m/sec
d) 15 m/sec
Answer »Answer: (d)
1 kmph = $5/18$ m/sec
54 kmph = $5/18 × 54$
= 15 m/sec.
Question : 3
Walking $6/7$th of his usual speed, a man is 12 minutes late. The usual time taken by him to cover that distance is
a) 1 hour 20 minutes
b) 1 hour 15 minutes
c) 1 hour 12 minutes
d) 1 hour
Answer »Answer: (c)
Time and speed are inversely proportional.
Usual time × $7/6$ - usual time
= 12 minutes
Usual time × $1/6$ = 12 minutes
Usual time = 72 minutes
= 1 hour 12 minutes
Using Rule 8,
Here, A = 6, B = 7, t = $12/60 = 1/5$ hrs.
Usual time= $\text"A"/\text"Diff of A and B"$ × time
= $6/{(7 - 6)} × 1/5 = 1{1}/5$ hrs.
= 1 hrs. 12 minutes
Question : 4
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 : 5
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 : 6
Walking at $6/7$th of his usual speed a man is 25 minutes late. His usual time to cover this distance is
a) 2 hours 10 minutes
b) 2 hours 25 minutes
c) 2 hours 15 minutes
d) 2 hours 30 minutes
Answer »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
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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