Aptitude Overflow

+1 vote

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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Best answer

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 .

0

what i am feeling that i am not understanding the meaning of last line of question becz on reading it with answer 250 it seems correct and with 1000 it also seems correct so what is tha language of question actually??

+1

If the question is "How much speed the train is increased than its usual speed" $\rightarrow$ Answer would be $250 km/hr$

and if the question asks us to find the increased speed then answer would be $1000 km/hr$

and if the question asks us to find the increased speed then answer would be $1000 km/hr$

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**.

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