Model 2 Vehicles in x/y of its usual speed Practice Questions Answers Test with Solutions & More Shortcuts

Question : 6 [SSC CGL Prelim 2005]

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: (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 : 7 [SSC GL Tier-I 2013]

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: (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 : 8 [SSC CGL Tier-1 2011]

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: (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

Question : 9 [SSC CGL Prelim 2005]

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: (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 : 10 [SSC CGL Tier-I 2014]

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: (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.

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