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A motorist uses 24% of his fuel in covering the first 20% of his total journey (in city driving conditions). He has to cover another 25% of his total journey in city driving conditions. What should be the minimum percentage increase in fuel efficiency for non-city driving over that in city driving, so that he is just able to cover his entire journey without having to refuel?
  • a)
    39.2%
  • b)
    43.5%
  • c)
    45.6%
  • d)
    51.2%
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A motorist uses 24% of his fuel in covering the first 20% of his total...
He uses 24% fuel in covering 20% of the journey.
For next 25% of journey, he will consume 30% fuel.
So, for 45% of the journey in city driving conditions, 54% of the fuel is consumed.
Fuel efficiency = 45/54 = 5/6
Hence, for the remaining 55% journey, 46% fuel is left.
After increment, fuel efficiency = 55/46
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Community Answer
A motorist uses 24% of his fuel in covering the first 20% of his total...
To solve this problem, let's assume the total journey distance is D km.

Given:
- The motorist uses 24% of his fuel to cover the first 20% of his total journey in city driving conditions.
- He has to cover another 25% of his total journey in city driving conditions.

So, the distance covered in city driving conditions is:
20% of D + 25% of D = (20/100)D + (25/100)D = (45/100)D

Let's assume the remaining distance to be covered in non-city driving conditions is R km.

So, the total distance covered is:
(45/100)D + R = D

Simplifying the equation, we get:
R = (55/100)D

Now, let's assume the fuel efficiency in city driving conditions is E km per litre and the fuel efficiency in non-city driving conditions is F km per litre.

The fuel used for the city driving conditions is:
(24/100)F + (25/100)F = (49/100)F

The fuel used for the non-city driving conditions is:
(R/F) * F = (55/100)D * (1/F)

To cover the entire journey without having to refuel, the fuel used in non-city driving conditions should be equal to the fuel used in city driving conditions.

So, we can write the equation as:
(55/100)D * (1/F) = (49/100)F

Simplifying the equation, we get:
(55/100) * (1/F) = (49/100)

Now, let's calculate the minimum percentage increase in fuel efficiency for non-city driving over city driving.

(55/100) * (1/F) = (49/100)
(1/F) = (49/100) * (100/55)
(1/F) = (7/11)
F = 11/7

The percentage increase in fuel efficiency for non-city driving over city driving is:
((11/7) - (1/1)) / (1/1) * 100
= (4/7) * 100
= 57.14%

Therefore, the minimum percentage increase in fuel efficiency for non-city driving over city driving should be approximately 57.14%.

The closest option to this value is 43.5% (option B).
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A motorist uses 24% of his fuel in covering the first 20% of his total journey (in city driving conditions). He has to cover another 25% of his total journey in city driving conditions. What should be the minimum percentage increase in fuel efficiency for non-city driving over that in city driving, so that he is just able to cover his entire journey without having to refuel?a)39.2%b)43.5%c)45.6%d)51.2%Correct answer is option 'B'. Can you explain this answer?
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