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

Question : 11 [SSC CGL Tier-I 2016]

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: (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 : 12 [SSC CGL Prelim 2002]

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: (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 is
Time = $\text"A"/\text"(Difference of A and B)"$ × 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 : 13 [SSC CHSL 2015]

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: (d)

1 kmph = $5/18$ m/sec

54 kmph = $5/18 × 54$

= 15 m/sec.

Question : 14 [SSC CGL Tier-1 2011]

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: (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

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