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# A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have been 3 hours less than the scheduled time. And, if the train was slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of Journey.

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This is the basic and exam oriented question from linear equations in two variables as it was already asked in various examinations in which we have to find the distance of the journey of a train if  the train had been 5 kmph faster, it would have been 3 hours less than the scheduled time. And, if the train was slower by 4 kmph, it would have taken 3 hours more than the scheduled time.

Please give me a step by step solution for the above question

RS Aggarwal, Class 10, chapter 3E, question no 32

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1. Let the original speed be x kmph and let the time taken to complete the journey be y hours.

∴ Length of the whole journey = (xy) km

Case I: When the speed is (x + 5) kmph and the time taken is (y – 3) hrs:

Total journey = (x + 5) (y – 3) km

⇒ (x + 5) (y – 3) = xy

⇒ xy + 5y – 3x – 15 = xy

⇒ 5y – 3x = 15 ………(i)

Case II: When the speed is (x – 4) kmph and the time taken is (y + 3) hrs:

Total journey = (x – 4) (y + 3) km

⇒ (x – 4) (y + 3) = xy

⇒ xy – 4y + 3x – 12 = xy

⇒ 3x – 4y = 12 ………(ii)

On adding (i) and (ii), we get:

y = 27

On substituting y = 27 in (i), we get:  5 × 27 – 3x = 15

⇒135 – 3x = 15

⇒3x = 120

⇒x = 40

∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km

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