A man covered a certain distance at some speed. Had he 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. The distance (in km) is:
thankyou sir.
sir, any other simple way to solve this pblm.
distance = time * speed, so
(t-2/3)(s+3) = ts
(t+2/3)(s-2) = ts
ts - 2/3 s + 3t - 2 = ts
ts + 2/3 s - 2t - 4/3 = ts
-2s+9t = 6
2s-6t = 4
3t = 10
t = 10/3 hr
s =12 km/hr
check: the distance is 40 km; at 12 km/hr that takes 10/3 hr
40km at 15km/hr = 8/3 hr = 10/3 - 2/3
40km at 10km/hr = 4 hr = 10/3 + 2/3
To find the distance covered by the man, let's assume his original speed is x kmph, and the distance he covered is D km.
According to the problem, if the man had moved 3 kmph faster, he would have taken 40 minutes less. This can be expressed as:
D / (x + 3) = (40/60) // Converting 40 minutes to hours by dividing by 60
Simplifying this equation:
D = (2/3)(x+3) ----(Equation 1)
Similarly, if the man had moved 2 kmph slower, he would have taken 40 minutes more. This can be expressed as:
D / (x - 2) = (40/60) // Converting 40 minutes to hours by dividing by 60
Simplifying this equation:
D = (3/2)(x-2) ----(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) representing the distance covered by the man. We can solve these equations simultaneously to determine the values of x and D.
Equating the values of D from both equations, we get:
(2/3)(x+3) = (3/2)(x-2)
Now, cross-multiply and solve for x:
2(x+3) = 3(x-2)
2x + 6 = 3x - 6
6 + 6 = 3x - 2x
12 = x
Substituting the value of x into Equation 1:
D = (2/3)(12+3) = 10 km
Therefore, the distance (D) covered by the man is 10 km.
x = distance
y = speed
t = time
t = x / y
Had he moved 3 kmph faster, he would have taken 40 minutes mean:
x / y - x / ( y + 3 ) = 40 / 60 = 20 ∙ 2 / 20 ∙ 3 = 2 / 3
x / y - x / ( y + 3 ) = 2 / 3
If he had moved 2 kmph slower, he would have taken 40 minutes more mean:
x / ( y - 2 ) - x / y = 40 / 60 = 20 ∙ 2 / 20 ∙ 3 = 2 / 3
x / ( y - 2 ) - x / y = 2 / 3
First equation:
x / y - x / ( y + 3 ) = 2 / 3
x ∙ [ 1 / y - 1 / ( y + 3 ) ] = 2 / 3
x ∙ [ 1 ∙ ( y + 3 ) - 1 ∙ y ] / [ y ∙ ( y + 3 ) ] = 2 / 3
x ∙ ( y + 3 - y ) / ( y² + 3 y ) = 2 / 3
x ∙ 3 / ( y² + 3 y ) = 2 / 3
3 x / ( y² + 3 y ) = 2 / 3
Multiply both sides by y² + 3 y
3 x = 2 ∙ ( y² + 3 y ) / 3
3 x = ( 2 y² + 6 y ) / 3
Multiply both sides by 3
9 x = 2 y² + 6 y
Second equation:
x / ( y - 2 ) - x / y = 2 / 3
x ∙ [ 1 / ( y - 2 ) - 1 / y ] = 2 / 3
x ∙ [ 1 ∙ y - 1 ∙ ( y - 2 ) ] / [ y ∙ ( y - 2 ) ] = 2 / 3
x ∙ [ y - ( y - 2 ) ] / ( y² - 2 y ) = 2 / 3
x ∙ ( y - y + 2 ) / ( y² - 2 y ) = 2 / 3
x ∙ 2 / ( y² - 2 y ) = 2 / 3
2 x / ( y² - 2 y ) = 2 / 3
Divide both sides by 2
x / ( y² - 2 y ) = 1 / 3
Multiply both sides by y² - 2 y
x = ( y² - 2 y ) / 3
Multiply both sides by 3
3 x = y² - 2 y
Now you must slove system:
9 x = 2 y² + 6 y
3 x = y² - 2 y
3 x = y² - 2 y
Multiply both sides by 3
9 x = 3 y² - 6 y
9 x = 9 x
2 y² + 6 y = 3 y² - 6 y
y ∙ ( 2 y + 6 ) = y ∙ ( 3 y - 6 )
Divide both sides by y
2 y + 6 = 3 y - 6
Add 6 to both sides
2 y + 6 + 6 = 3 y - 6 + 6
2 y +12 = 3 y
Subtract 2 y to both sides
2 y +12 - 2 y = 3 y - 2 y
12 = y
y = 12 km / h
3 x = y² - 2 y
3 x = 12² - 2 ∙ 12
3 x = 144 - 24
3 x = 120
x = 120 / 3
x = 40 km