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A truck travelled to a place Q from P, the first 50 km at 10 kmph faster than the usual speed, but it returned the same distance at 10 kmph slower than usual speed. If the total time taken by the truck is 12 hours, then how many hours will travel at the faster speed?
  • a)
    8 hours
  • b)
    6 hours
  • c)
     2 hours
  • d)
    3 hours
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A truck travelled to a place Q from P, the first 50 km at 10 kmph fast...
Total time taken,
[50/(x-10)] + 50/(x +10)] = 12 hours.
By solving the equation, we get
x = 15
Time is taken by the truck at faster speed = 50/(15+10) = 2 hours.
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Most Upvoted Answer
A truck travelled to a place Q from P, the first 50 km at 10 kmph fast...
Given information:
- Distance from P to Q = Distance from Q to P = 50 km
- Time taken for the entire journey = 12 hours

Let the usual speed of the truck be x kmph.
According to the question, the truck travelled the first 50 km at a speed of (x+10) kmph and returned the same distance at a speed of (x-10) kmph.

We can use the formula:
Speed = Distance/Time

For the first 50 km:
Time taken while going from P to Q = 50/(x+10) hours
Time taken while returning from Q to P = 50/(x-10) hours

Total time taken for the first 50 km = 50/(x+10) + 50/(x-10) hours

For the entire journey:
Time taken while going from P to Q = 50/(x+10) hours
Time taken while returning from Q to P = 50/(x-10) hours

Total time taken for the entire journey = 50/(x+10) + 50/(x-10) + t hours
where t is the time taken by the truck to cover the remaining distance after the first 50 km at the usual speed x kmph.

According to the given information, the total time taken for the entire journey is 12 hours.
So we have the equation:
50/(x+10) + 50/(x-10) + t = 12

Simplifying the equation:
(5x-50)/(x^2-100) + t = 12/5
5x-50 + t(x^2-100) = 12(x^2-100)/5
5x-50 + t(x+10)(x-10) = 12(x+10)(x-10)/5
25x-250 + t(x+10)(x-10) = 12(x^2-100)
25x-250 + tx^2 - 100t = 12x^2 - 1200
12x^2 - tx^2 - 25x + 950 = 0

Solving for x using the quadratic formula:
x = [25 ± sqrt(25^2 + 4*12*950t)]/(2*12t)

Since the speed of the truck cannot be negative, we take only the positive value of x:
x = [25 + sqrt(25^2 + 4*12*950t)]/(2*12t)

We need to find the time taken by the truck to travel the first 50 km at a speed of (x+10) kmph.
Time taken = 50/(x+10)
Substituting the value of x, we get:
Time taken = 50/([25 + sqrt(25^2 + 4*12*950t)]/(2*12t) + 10)

Simplifying the expression:
Time taken = 2t[25 + sqrt(25^2 + 4*12*950t) - 20t]/[25 + sqrt(25^2 + 4*12*950t)]

We know that the total time taken for the journey is 12 hours.
So, t + 50/(x+10) + 50/(x-10) = 12

Sub
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A truck travelled to a place Q from P, the first 50 km at 10 kmph faster than the usual speed, but it returned the same distance at 10 kmph slower than usual speed. If the total time taken by the truck is 12 hours, then how many hours will travel at the faster speed?a)8hoursb)6hoursc)2 hoursd)3 hoursCorrect answer is option 'C'. Can you explain this answer?
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