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

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.

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.

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.

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

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: (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: (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