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# Time and Distance #6

 6. A man covered a certain distance at some speed. If he had moved $3$ kmph faster, he would have taken $40$ minutes less. If he had moved $2$ kmph slower, he would have taken $40$ minutes more. What is the distance in km? A. $36$ B. $40$ C. $38$ D. $42$

Explanation:

# Solution 1

reference: formula 5.4 - special case

$\text{speed}=\dfrac{2v_1v_2}{v_1-v_2}=\dfrac{2×3×2}{3-2}=12\text{ km/hr}$

$\text{distance }=vt_1\left(1+\dfrac{v}{v_1}\right)\\=12×\dfrac{40}{60}\left(1+\dfrac{12}{3}\right)=40\text{ km}$

# Solution 2

Let distance $=x$ km,
his speed $=v$ kmph

Time taken when moving at normal speed - Time taken when moving $3$ kmph faster $=40$ minutes
$\Rightarrow \dfrac{x}{v}-\dfrac{x}{v+3}=\dfrac{40}{60}\\\Rightarrow x\left[\dfrac{1}{v}-\dfrac{1}{v+3}\right]=\dfrac{2}{3}\\\Rightarrow x\left[\dfrac{v+3-v}{v(v+3)}\right]=\dfrac{2}{3}\\\Rightarrow 2v(v+3)=9x~\cdots(1)$

Time taken when moving $2$ kmph slower - Time taken when moving at normal speed $=40$ minutes
$\Rightarrow \dfrac{x}{v-2}-\dfrac{x}{v}=\dfrac{40}{60}\\\Rightarrow x\left[\dfrac{1}{v-2}-\dfrac{1}{v}\right] = \dfrac{2}{3}\\\Rightarrow x\left[\dfrac{v-v+2}{v(v-2)}\right]=\dfrac{2}{3}\\\Rightarrow x\left[\dfrac{2}{v(v-2)}\right] = \dfrac{2}{3}\\\Rightarrow x\left[\dfrac{1}{v(v-2)}\right] = \dfrac{1}{3}\\\Rightarrow v(v-2)=3x~\cdots(2)$

$\dfrac{(1)}{(2)}\Rightarrow \dfrac{2(v+3)}{(v-2)}=3\\\Rightarrow 2v+6=3v-6\\ \Rightarrow v=12$

Substituting this value of $v$ in $(1)$
$\Rightarrow 2×12×15=9x\\\Rightarrow x =\dfrac{2×12×15}{9}=\dfrac{2×4×15}{3}\\=2×4×5=40$

Hence distance $=40$ km

arunkumar
2016-08-21 13:50:06
3 kmph Increase: 40 minutes less
2 kmph Decrease: 40 minutes More

Firstly calculate the original speed by mentioned below trick

=2*Net Increment * Net decrement / Difference of Net increment & decrement
=2*3*2/3-2
=12 kmph

Now Equation became mentioned below
3 kmph Increase= 12+3 => 15 Kmph: 40 minutes less
2 kmph Decrease =12-2 => 10 Kmph: 40 minutes More

Now calculating the distance
=  total time difference * products of speed / difference of speed
=  80/60 ( converting into hour )* 15*10 /(15-10 )
= 40 Km
jiju
2016-08-21 14:24:39
@Arunkumar, this is useful one. But this formula can be used only when the increase in time is equal to decrease in time (40 minutes here). Let's see how the formula used can be derived.

x/v - x/(v+3)= 40/60 -------(1)
x/(v-2) - x/v = 40/60 -------(2)

(1)=(2)
x/v - x/(v+3) = x/(v-2) - x/v
=> 1/v - 1/(v+3) = 1/(v-2) - 1/v
=> 3/v(v+3) = 2/v(v-2)
=> 3/(v+3) = 2/(v-2)
=> 3(v-2) = 2(v+3)
=> 3v - 3×2 = 2v + 2×3
=> v(3-2) =  2×3 + 3×2
=> v(3-2) =  2×2×3
=> v =  2×2×3/(3-2)
=> v =  2 × net increment × net decrement/(difference of net increment & decrement)

This is why your formula for original speed working fine. Note that this can be used only when the increment in time and decrement in time are same (here 40 minutes) because (1)=(2) is used to derive this formula.

Similarly,  let's take a look on the second formula used by you for calculating distance.

(1)+(2) gives
x/(v-2) - x/(v+3)= 40/60+40/60
x[(v+3)-(v-2)]/(v-2)(v+3)= 40/60+40/60
(x × difference of speed)/products of speed= total time difference
x = total time difference × products of speed /difference of speed
0 0
sanju
2015-07-06 07:11:46
3*40=120
2*40=80
120-80=40
Amrik
2016-08-20 08:46:03
How did you do this?
0 0
Anju
2013-11-07 16:15:52
please... can we have any logic without doing all these process?