Model 2 Vehicles in x/y of its usual speed Section-Wise Topic Notes With Detailed Explanation And Example Questions

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

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

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

Answer: (d)

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.

Answer: (c)

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

Answer: (b)

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.