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A plane left half an hour than the scheduled time and in order to reach its destination $1500 \hspace{0.1cm} km$ away in time, it had to increase its speed by $33.33\%$ over its usual speed. find its increased speed?

- $250 \hspace{0.1cm} kmph$
- $500 \hspace{0.1cm} kmph$
- $750 \hspace{0.1cm} kmph$
- none

the answer given is $750 \hspace{0.1cm} kmph$ but i m getting $250$ which is correct??

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Assuming, its usual speed is $x \hspace{0.1cm} km/hr $

∴ New Speed will be $= x+ 33.33\% \hspace{0.1cm} of \hspace{0.1cm} x \\ = x+ \dfrac{1}{3} \text{ of x } \\ = \dfrac{4x}{3} \hspace{0.1cm} km/hr$

In $x \hspace{0.1cm} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{x} \hspace{0.1cm} hr$ $\qquad \left [ \because Distance = Speed \times Time \\ \qquad \qquad Or, Time = \dfrac{Distance}{Speed}\right ]$

Now In $\dfrac{4x}{3} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{\dfrac{4x}{3}} = \dfrac{1500 \times 3}{4x} \hspace{0.1cm} hr$

As the plane lates for half an hour, the difference between the usual time taken by the plane and time taken by plane in new speed will be half an hour.

$∴\dfrac{1500}{x} - \dfrac{1500 \times 3}{4x} = \dfrac{1}{2}$

Or, $\dfrac{(1500 \times 4) - (1500 \times 3)}{4x} = \dfrac{1}{2}$

Or. $\dfrac{6000 - 4500}{4x} = \dfrac{1}{2}$

Or, $\dfrac{1500}{4x} = \dfrac{1}{2}$

Or, $4x = 1500 \times 2$

Or, $x = \dfrac{3000}{4} = 750\hspace{0.1cm} km/hr$

So, the plane's usual speed is $750\hspace{0.1cm} km/hr$

∴ New speed will be $\dfrac{4x}{3}$ i.e. $\dfrac{4 \times 750}{3} = 1000 \hspace{0.1cm} km/hr$

∴ The plane's usual speed is $750 \hspace{0.1cm} km/hr$ & New speed is $1000 \hspace{0.1cm} km/hr$ which means the plane increases its speed by $250 \hspace{0.1cm} km/hr$ than its ususal speed to make up the timing .

∴ New Speed will be $= x+ 33.33\% \hspace{0.1cm} of \hspace{0.1cm} x \\ = x+ \dfrac{1}{3} \text{ of x } \\ = \dfrac{4x}{3} \hspace{0.1cm} km/hr$

In $x \hspace{0.1cm} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{x} \hspace{0.1cm} hr$ $\qquad \left [ \because Distance = Speed \times Time \\ \qquad \qquad Or, Time = \dfrac{Distance}{Speed}\right ]$

Now In $\dfrac{4x}{3} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{\dfrac{4x}{3}} = \dfrac{1500 \times 3}{4x} \hspace{0.1cm} hr$

As the plane lates for half an hour, the difference between the usual time taken by the plane and time taken by plane in new speed will be half an hour.

$∴\dfrac{1500}{x} - \dfrac{1500 \times 3}{4x} = \dfrac{1}{2}$

Or, $\dfrac{(1500 \times 4) - (1500 \times 3)}{4x} = \dfrac{1}{2}$

Or. $\dfrac{6000 - 4500}{4x} = \dfrac{1}{2}$

Or, $\dfrac{1500}{4x} = \dfrac{1}{2}$

Or, $4x = 1500 \times 2$

Or, $x = \dfrac{3000}{4} = 750\hspace{0.1cm} km/hr$

So, the plane's usual speed is $750\hspace{0.1cm} km/hr$

∴ New speed will be $\dfrac{4x}{3}$ i.e. $\dfrac{4 \times 750}{3} = 1000 \hspace{0.1cm} km/hr$

∴ The plane's usual speed is $750 \hspace{0.1cm} km/hr$ & New speed is $1000 \hspace{0.1cm} km/hr$ which means the plane increases its speed by $250 \hspace{0.1cm} km/hr$ than its ususal speed to make up the timing .

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0 votes

Answer would be **d)None**

Since speed is increased by 33.33%

so, if initial speed = v kmph ;and time = t for distance = d

then increased speed = (4/3)v kmph

so, new time = d/(4/3)v = (3/4)t

i.e., t/4 = 30 min. (as, speed is increased to compensate delay)

so, t=2 hour

given, d= 1500km

v=1500/2 =750 kmph.

thus, increased speed = (4/3)*750 = **1000 kmph**.