Question

If average speed of car is increased by 10km/hr,the time required to travel a certain distance is reduced by 36 min. If the average speed is decreased by 15km/hr, then 1hr 12 min more are required to travel the same distance. Find the average speed of the car and the distance travelled.​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

Let, the Average speed is =$v_{avg} \:km.h{}^{-1}$

Now , let the distance is= x km

now ,the average speed is increased by 10km/hr

so the new average speed is now

$\therefore v_{avg(increased)}=(v_{avg}+10)\:km.h{}{-1}$

now ,the average speed is decreased by 15km/hr

so the new average speed is now

$\therefore v_{avg(decreased)}=(v_{avg}-15)\:km.h{}{-1}$

now , therefore the required time to travel with the average speed is

$\rightarrow \frac{x}{v_{avg}} \: hr$

now , therefore the required time to travel with the new increased average speed is

$\rightarrow \frac{x}{v_{avg(increased)}} \: hr$

therefore the required to travel with the new dcreased speed is

$\rightarrow \frac{x}{v_{avg(decreased)}} \: hr$

Now, According to the Question

•

If average speed of car is increased by 10km/hr,the time required to travel a certain distance is reduced by 36 min.

$\therefore \implies \frac{x}{v_{avg(increased)}}+\frac{32}{60}=\frac{x}{v_{avg}} \: hr\:\:\:\\(\because \:32min=\frac{32}{60}\:hr)\\ \implies \frac{x}{(v_{avg}+10)}+\frac{32}{60}=\frac{x}{v_{avg}}\\ \implies \frac{x}{v_{avg}}- \frac{x}{(v_{avg}+10)}+\frac{32}{60}=\frac{32}{60}\\ \implies \frac{x(v_{avg}+10)-xv_{avg}}{v_{avg}(v_{avg}(10)}=\frac{32}{60}\\ \implies \frac{\cancel{xv_{avg}}+10x-\cancel{xv_{avg}}}{v_{avg}(v_{avg}+10)}=\frac{32}{60}\\ \implies \boxed{\bf\red{150x=8v{}^{2}_{avg}+80v_{avg}}}..........…(i)$

•

If the average speed is decreased by 15km/hr, then 1hr 12 min more are required to travel the same distance.

$\therefore \implies \frac{x}{v_{avg(decreased)}}-\frac{72}{60}=\frac{x}{v_{avg}} \: hr\:\:\:\\(\because \:1hr 12min=1\frac{12}{60}\:hr=\frac{72}{60} hr\\ \implies \implies \frac{x}{(v_{avg}-15)}-\frac{x}{v_{avg}}=\frac{6}{5}\\ \implies \frac{xv_{avg}-x(v_{avg}-15)}{v_{avg}(v_{avg}-15)}=\frac{6}{5}\\ \implies \frac{xv_{avg}-xv_{avg}+15x)}{v_{avg}(v_{avg}-15)}=\frac{6}{5}\\ \implies 75x=6v{}^{2}_{avg}-90v_{avg}\\ \implies \boxed{\bf\red{150x= 12v{}^{2}_{avg}-180v_{avg}}}........(ii)$

comparing equations (i) and (ii)....

$\implies 8v{}^{2}_{avg}+80v_{avg} =12v{}^{2}_{avg}-180v_{avg} \\ \implies =12v{}^{2}_{avg}- 8v{}^{2}_{avg}=80v_{avg}+180v_{avg}\\ \implies 4v{}^{2}_{avg}=260v_{avg}\\ \implies v{}^{2}_{avg}- 65v_{avg}=0\\ \implies v_{avg}(v_{avg}-65)=0\\ \rightarrow either \\v_{avg}=0(not \:taken\: into\: consideration)\\ \rightarrow or\\ v_{avg}-65=0\implies \boxed{\bf\red{v_{avg}=65 \:km.h{}^{-1}}}$

from equation (i).....

$150x=8v{}^{2}_{avg}+80v_{avg}.\\ \implies 150x=8(65){}^{2}+80(65)\\ \implies x=\frac{39000}{150}\\ \implies \boxed{\bf\red{x=260\:km}}$

$\therefore$ average speed is =65 km/h

$\therefore$ distance travelled=260 km

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \#\mathcal{answer with quality }\: \: \& \: \: \#BAL }$

Answer 2

Given :-----

• If average Speed is increased by 10km/h , Time is reduced by 36 min or = 36/60 = 3/5 hours .
• if average speed is decreased by 15km/h , Time is Increased by 1hr12 min = 72 min = 72/60 = 6/5 hours .

To Find :-----

• Average Speed and Distance Travelled by Car .. ?

we can solve this by 2 methods, First we will Try by basic Method, than i will Tell You Direct Formula for this type of Problem....

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Let speed original is S , Distance is D , Time is T ,

Than , By first case we get,

S. T. (D = S × T)

(S+10)(T-3/5) = ST

→ ST -3/5S + 10T - 6 = ST

→ 10T - 3/5S = 6

→ 50T - 3S = 30 ---------------------Equation (1)

Now, By case (2) we get,

→ (S-15)(T+6/5) = ST ( As DisTance is same in Both case)

→ -15T -18 + 6S/5 = 0

→ 6S - 75T = 90 ---------------------------- Equation (2)

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Now, Multiply Equation (1) by 2 and Adding Both We get,

→ 2(50T-3S) + ( 6S - 75T ) = 2(30) + 90

→ 100T - 6S + 6S - 75T = 150

→ 25T = 150

→ T = 6 Hours ..

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Putting in Any Equation Now we get,

→ 50×6 -3S = 30

→ 300 - 30 = 3S

→ 3S = 270

→ S = 90 km/h .

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Hence,

→ D = S × T = 90 × 6 = 540km/h ....

So, original Average Speed was 90km/h , and original Distance of The journey was 540km ....

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( By Direct Formula, we will use Cross - Multiplication , for Solving the Equation )

Try Yourself the calculation ..

I m Telling you Formula ,..

original S = S1*S2*(T1+T2)/(S2*T1-S1*T2)

Original T = T1*T2*(S1+S2)/(S2*T1-S1*T2)

put S1 = 10, S2 = 15 , T1 = 3/5 , T2= 6/5 ,

I solved , you will get, same Original S = 90km/h , and original T = 6 hours ...

(Try now and if any doubt ask ..)

(Hope it Helps you)

#BAL ..