Answer: Option B. -> 78 km/hr
Answer: (b)Using Rule 1,Distance = Speed × TimeSpeed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

Answer: Option B. -> 40 km./hr.
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 = 2005x = 200x = $200/5$ = 40 kmph.

Answer: Option D. -> 4$1/2$
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$ hoursUsing 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: Option C. -> 72
Answer: (c)1 m/sec = $18/5$ kmph20 m/sec = ${20 × 18}/5$= 72 kmph

Answer: Option C. -> 35 km/hr
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 = $504/300$ hrSpeed = $42/{504/300}$ = 25 kmphNow, $5/7$ usual speed = 25Usual speed = ${25 × 7}/5$ = 35 kmph

Answer: Option C. -> 20 km/hour
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 = $7/5$ hoursAverage speed= $\text"Total distance"/ \text"Total time" = (28/{7/5})$ kmph= ${28 × 5}/7$ = 20 kmph

Answer: Option B. -> 6 hours
Answer: (b)$4/3$ × usual time - usual time = 2$1/3$ usual time = 2Usual time = 2 × 3 = 6 hoursUsing 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: Option D. -> 2 hours 30 minutes
Answer: (d)Time and speed are inversely proportional.$7/6$ × Usual time - Usual time = 25 minutesUsual time $(7/6 -1)$= 25 minutesUsual time × $1/6$= 25 minutesUsual time = 25 × 6= 150 minutes= 2 hours 30 minutesUsing 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: Option C. -> 420 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)$$3y^2 + 100y = 3y^2 + 90y + 600$10y = 600 ⇒ y = 60Again from equation (i),$x/{y +10} + 1 = x/y$$x/{60 + 10} + 1 = x/60$$x/70 + 1 = x/60$${x + 70}/70 = x/60$6x + 420 = 7x7x - 6x = 420$x$ = 420 km.

Answer: Option C. -> 60 minutes
Answer: (c)$4/3$ of usual time= Usual time + 20 minutes$1/3$ of usual time = 20 minutesUsual time = 20 × 3 = 60 minutesUsing Rule 8,Here, A = 3, B= 4, t = 20 minutesUsual time taken= $\text"A"/\text"Diff of A and B"$ × time= $3/{(4 - 3)} × 20$ = 60 minutes