A train travels at certain speed a distance of 63 km and then Travels a distance of 72 km at an average speed it taken 3 hours to complete total Jannat is original average
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Your question is incomplete .
complete question is ---->a train travels at a certain average speed of distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed.if it take 3 hours to com[let the total journey , what is the original average speed ]
Let the original speed of train is x km/h
time taken to cover 63 km with speed x km/h, T₁ = distance/time = 63/x hours
Again, question said , speed of train now (x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h , T₂= 72/(x + 6) km/h
A/C to question ,
T₁ + T₂ = 3 hours
⇒63/x + 72/(x + 6) = 3
⇒ 21/x + 24/(x + 6) = 1
⇒ (21x + 126 + 24x) = x(x + 6)
⇒45x + 126 = x² + 6x
⇒ x² - 39x - 126 = 0
⇒x² - 42x + 3x - 126 = 0
⇒ x(x - 42) + 3(x - 42) = 0
⇒(x + 3)(x - 42) = 0
∴x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative
Hence, original speed of train = 42 km/h
complete question is ---->a train travels at a certain average speed of distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed.if it take 3 hours to com[let the total journey , what is the original average speed ]
Let the original speed of train is x km/h
time taken to cover 63 km with speed x km/h, T₁ = distance/time = 63/x hours
Again, question said , speed of train now (x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h , T₂= 72/(x + 6) km/h
A/C to question ,
T₁ + T₂ = 3 hours
⇒63/x + 72/(x + 6) = 3
⇒ 21/x + 24/(x + 6) = 1
⇒ (21x + 126 + 24x) = x(x + 6)
⇒45x + 126 = x² + 6x
⇒ x² - 39x - 126 = 0
⇒x² - 42x + 3x - 126 = 0
⇒ x(x - 42) + 3(x - 42) = 0
⇒(x + 3)(x - 42) = 0
∴x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative
Hence, original speed of train = 42 km/h
Answered by
303
Please try to post the complete question , so that we an answer appropriately.
Complete Question is "
A train travels at a certain speed of a distance of 63 km and then Travels a distance of 72 km at an average speed 6 km per hour more than its original speed its takes 3 hours to complete total journey what is the original Average speed"
Let the speed of the train is Xkm/h
Time taken to cover 63km with X km/h speed is :
T1= Distance/SpeedT1=63/X hours---------(1)
It's given that train Travels a distance of 72 km at an average speed 6 km per hour more than its original speed.
Time taken to cover 72km with speed (x+6)Km/H
T2=72/x+6 hr------------(2)
But given that total time is 3hrs.T1+T2=3hrs
63/x+72(/x+6)=3taking 3 common,
21/x+24/(x+6)=121
(x+6)+24x=x(x+6)
21x+126+24x=x²+6x
x²+6x-45x-126=0
x²-39x-126=0
x²-42x+3x-126=0
x(x-42)+3(x-42)=0
(x-42)(x+3)=0
x=42, -3
since speed cannot be negative:x=42Km/h
∴Original speed of train=42km/h
Complete Question is "
A train travels at a certain speed of a distance of 63 km and then Travels a distance of 72 km at an average speed 6 km per hour more than its original speed its takes 3 hours to complete total journey what is the original Average speed"
Let the speed of the train is Xkm/h
Time taken to cover 63km with X km/h speed is :
T1= Distance/SpeedT1=63/X hours---------(1)
It's given that train Travels a distance of 72 km at an average speed 6 km per hour more than its original speed.
Time taken to cover 72km with speed (x+6)Km/H
T2=72/x+6 hr------------(2)
But given that total time is 3hrs.T1+T2=3hrs
63/x+72(/x+6)=3taking 3 common,
21/x+24/(x+6)=121
(x+6)+24x=x(x+6)
21x+126+24x=x²+6x
x²+6x-45x-126=0
x²-39x-126=0
x²-42x+3x-126=0
x(x-42)+3(x-42)=0
(x-42)(x+3)=0
x=42, -3
since speed cannot be negative:x=42Km/h
∴Original speed of train=42km/h
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