Practice Usual vs actual speed - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   Walking at $3/4$ of his usual speed, a man is 1$1/2$ hours late. His usual time to cover the same distance, (in hours) is

(a)

(b)

(c)

(d)

Explanation:

Time and speed are inversely proportional.

$4/3$ of usual time –usual time = $3/2$

$1/3$ usual time = $3/2$

Usual time = ${3 × 3}/2$

$9/2 = 4{1}/2$ hours

Using Rule 8,

Here, A = 3, B = 4, t= $3/2$

Usual time = $\text"A"/\text"Diff of A and B"$ × time

= $3/{(4 - 3)} × 3/2 = 4{1}/2$ hrs.


Q-2)   A car goes 20 metres in a second. Find its speed in km/hr.

(a)

(b)

(c)

(d)

Explanation:

1 m/sec = $18/5$ kmph

20 m/sec = ${20 × 18}/5$

= 72 kmph


Q-3)   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)

(b)

(c)

(d)

Explanation:

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.


Q-4)   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)

(b)

(c)

(d)

Explanation:

$4/3$ × 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


Q-5)   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)

(b)

(c)

(d)

Explanation:

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


Q-6)   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)

(b)

(c)

(d)

Explanation:

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


Q-7)   The speed of a car is 54 km/hr. What is its speed in m/sec?

(a)

(b)

(c)

(d)

Explanation:

1 kmph = $5/18$ m/sec

54 kmph = $5/18 × 54$

= 15 m/sec.


Q-8)   A car covers a certain distance in 25 hours. If it reduces the speed by $1/5$th, the car covers 200 km. less in that time. The speed of car is

(a)

(b)

(c)

(d)

Explanation:

Speed of car = x kmph.

Distance = Speed × Time = 25x km.

Case II,

Speed of car = ${4x}/5$ kmph.

Distance covered

= ${4x}/5 × 25$ = 20x km.

According to the question,

25x - 20x = 200

5x = 200

x = $200/5$ = 40 kmph.


Q-9)   A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/ hour and 60 km/hour respectively. Its average speed over this distance is

(a)

(b)

(c)

(d)

Explanation:

Total distance

= 7 × 4 = 28 km.

Total time

= $(7/10 + 7/20 + 7/30 + 7/60)$ hours

= $({42 + 21 + 14 + 7}/60)$ hours

= $84/60$ hours = $7/5$ hours

Average speed

= $\text"Total distance"/ \text"Total time" = (28/{7/5})$ kmph

= ${28 × 5}/7$ = 20 kmph


Q-10)   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)

(b)

(c)

(d)

Explanation:

$4/3$ 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