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An athlete walks 1 kmph faster and reaches the destination 3 hours earlier and if walks 1 kmph slower, then reaches the destination 5 hours late. If the destination is 60 km away, then the normal speed of athlete is (A) 1 kmph (C) 3 kmph (B) 2 kmph (D) 4 kmph?
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An athlete walks 1 kmph faster and reaches the destination 3 hours ear...
Given information:
- The athlete walks 1 kmph faster and reaches the destination 3 hours earlier.
- The athlete walks 1 kmph slower and reaches the destination 5 hours later.
- The distance to the destination is 60 km.

Let's assume:
- The normal speed of the athlete is x kmph.

Calculating time taken at normal speed:
- Distance = Speed × Time
- Time = Distance / Speed

Time taken at normal speed:
- Time taken at normal speed = 60 km / x kmph = 60/x hours

Time taken when walking 1 kmph faster:
- Time taken when walking 1 kmph faster = 60 km / (x + 1) kmph = 60/(x + 1) hours

Time taken when walking 1 kmph slower:
- Time taken when walking 1 kmph slower = 60 km / (x - 1) kmph = 60/(x - 1) hours

Given conditions:
- The athlete walks 1 kmph faster and reaches the destination 3 hours earlier.
- The athlete walks 1 kmph slower and reaches the destination 5 hours later.

Equations based on the given conditions:
1. 60/(x + 1) = 60/x - 3
2. 60/(x - 1) = 60/x + 5

Solving the equations:
1. 60/(x + 1) = 60/x - 3
- Multiply both sides by x(x+1) to eliminate the denominators:
60x = 60(x+1) - 3x(x+1)
60x = 60x + 60 - 3x^2 - 3x
0 = 3x^2 + 3x - 60
0 = x^2 + x - 20
0 = (x + 5)(x - 4)
x = -5 or x = 4

Since speed cannot be negative, x = 4 kmph.

2. 60/(x - 1) = 60/x + 5
- Multiply both sides by x(x-1) to eliminate the denominators:
60x = 60(x-1) + 5x(x-1)
60x = 60x - 60 + 5x^2 - 5x
0 = 5x^2 - 5x - 60
0 = x^2 - x - 12
0 = (x - 4)(x + 3)
x = 4 or x = -3

Since speed cannot be negative, x = 4 kmph.

Conclusion:
- The normal speed of the athlete is 4 kmph. Therefore, the correct answer is (D) 4 kmph.
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An athlete walks 1 kmph faster and reaches the destination 3 hours earlier and if walks 1 kmph slower, then reaches the destination 5 hours late. If the destination is 60 km away, then the normal speed of athlete is (A) 1 kmph (C) 3 kmph (B) 2 kmph (D) 4 kmph?
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An athlete walks 1 kmph faster and reaches the destination 3 hours earlier and if walks 1 kmph slower, then reaches the destination 5 hours late. If the destination is 60 km away, then the normal speed of athlete is (A) 1 kmph (C) 3 kmph (B) 2 kmph (D) 4 kmph? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about An athlete walks 1 kmph faster and reaches the destination 3 hours earlier and if walks 1 kmph slower, then reaches the destination 5 hours late. If the destination is 60 km away, then the normal speed of athlete is (A) 1 kmph (C) 3 kmph (B) 2 kmph (D) 4 kmph? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An athlete walks 1 kmph faster and reaches the destination 3 hours earlier and if walks 1 kmph slower, then reaches the destination 5 hours late. If the destination is 60 km away, then the normal speed of athlete is (A) 1 kmph (C) 3 kmph (B) 2 kmph (D) 4 kmph?.
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