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

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.

$7/6$ × 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

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

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: (c)

1 m/sec = $18/5$ kmph

20 m/sec = ${20 × 18}/5$

= 72 kmph

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.