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At 3 different speeds of 30 kmph, 45 kmph, and 60 kmph, it takes Adam
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12 Mar 2022, 22:44
12 Mar 2022, 22:44
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At 3 different speeds of 30 kmph, 45 kmph, and 60 kmph, it takes Adam 6.5 hours to go from city A to city B while covering the same distance with each speed. For the above speeds, mileages per litre of fuel are 11km, 14km and 18km respectively. Eden, who is Adam's firend, wants to minimise his friend's car's fuel consumption. How much fuel can he save if he decides to drive Adam's car one day from city A to city B?
A. 4.2 liters
B. 4.5 liters
C. 3.7 litres
D. 0.7 liters
E. 0.3 liters
OA edited on 25 Mar 2022
A. 4.2 liters
B. 4.5 liters
C. 3.7 litres
D. 0.7 liters
E. 0.3 liters
OA edited on 25 Mar 2022
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: At 3 different speeds of 30 kmph, 45 kmph, and 60 kmph, it takes Adam
[#permalink]
25 Mar 2022, 01:05
25 Mar 2022, 01:05
1
OA Explanation:
....while covering the same distance with each speed, means each portion is \(D\) Km long
\(6.5 = \frac{D}{30} + \frac{D}{45} + \frac{D}{60}\)
\(6.5 = \frac{D}{15}[\frac{1}{2} + \frac{1}{3} + \frac{1}{4}]\)
\((6.5)(15) = D(\frac{6 + 4 + 3)}{12})\)
\(D = 90\) Km
Total distance = \(3(90) = 270\) Km
Mileage for the first portion is 11 km/lt i.e. \(\frac{90}{11} = 8.18\) lt of fuel is being consumed
Mileage for the second portion is 14 km/lt i.e. \(\frac{90}{14} = 6.42\) lt of fuel is being consumed
Mileage for the third portion is 18 km/lt i.e. \(\frac{90}{18} = 5.00\) lt of fuel is being consumed
Total fuel consumption = \((8.18 + 6.42 + 5.0) = 19.6\) lt
Since the fuel consumption need to be minimised, Eden must drive his car at 60 Kmph through the entire distance of 270 Km
This means - Mileage for the entire distance will be 18 km/lt
i.e. \(\frac{270}{18} = 15.00\) lt of fuel will be consumed
Fuel saved = 19.6 - 15 = 4.6 lt
Hence, option B
....while covering the same distance with each speed, means each portion is \(D\) Km long
\(6.5 = \frac{D}{30} + \frac{D}{45} + \frac{D}{60}\)
\(6.5 = \frac{D}{15}[\frac{1}{2} + \frac{1}{3} + \frac{1}{4}]\)
\((6.5)(15) = D(\frac{6 + 4 + 3)}{12})\)
\(D = 90\) Km
Total distance = \(3(90) = 270\) Km
Mileage for the first portion is 11 km/lt i.e. \(\frac{90}{11} = 8.18\) lt of fuel is being consumed
Mileage for the second portion is 14 km/lt i.e. \(\frac{90}{14} = 6.42\) lt of fuel is being consumed
Mileage for the third portion is 18 km/lt i.e. \(\frac{90}{18} = 5.00\) lt of fuel is being consumed
Total fuel consumption = \((8.18 + 6.42 + 5.0) = 19.6\) lt
Since the fuel consumption need to be minimised, Eden must drive his car at 60 Kmph through the entire distance of 270 Km
This means - Mileage for the entire distance will be 18 km/lt
i.e. \(\frac{270}{18} = 15.00\) lt of fuel will be consumed
Fuel saved = 19.6 - 15 = 4.6 lt
Hence, option B
Re: At 3 different speeds of 30 kmph, 45 kmph, and 60 kmph, it takes Adam
[#permalink]
17 Apr 2025, 20:04
17 Apr 2025, 20:04
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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