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07 Apr 2021, 02:15
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B
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Difficulty:
55%
(hard)
Question Stats:
63% (02:06) correct
37%
(02:10)
wrong
based on 103
sessions
History
Date
Time
Result
Not Attempted Yet
In a certain year, Carrier X traveled 101,098 kilometers and consumed 9,890 lites of diesel fuel, while in the same year, Carrier Y traveled 203,000 kilometers and consumed 24,896 lites of diesel fuel. The fuel mileage is defined as kilometers per liter of fuel. The mileage of Carrier X is approximately what percent greater or lesser than that of Carrier Y?
(A) 20%
(B) 25%
(C) 33.33%
(D) 37.50%
(E) 40%
(A) 20%
(B) 25%
(C) 33.33%
(D) 37.50%
(E) 40%
09 Apr 2021, 00:02
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Bunuel
Since we are looking for approximate values, we can round off the digits in a way that it eases our calculation.
But as the options are close, we cannot approximate beyond a point. Here luckily, all the numbers are close to the number that we would round off to.
Carrier X traveled 101,098 kilometers and consumed 9,890 liters of diesel fuel => \(\frac{101098}{9890}=\frac{100000}{10000}=10km/hr\)
Carrier Y traveled 203,000 kilometers and consumed 24,896 liters of diesel fuel => \(\frac{203000}{24896}=\frac{200000}{25000}=8km/hr\)
\(100*\frac{10-8}{8}=100*\frac{2}{8}\) = 25%
B
16 Apr 2021, 09:27
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Bunuel
Solution:
Carrier X’s fuel mileage is 101,098/9890 ≈ 100,000/10,000 = 10 km/L. Similarly, Carrier Y’s fuel mileage is 203,000/24,896 ≈ 200,000/25,000 = 8 km/L. Therefore, Carrier X’s fuel mileage is approximately 2 km/L more than Carrier Y’s fuel mileage, or 2/8 = 25% more than Carrier’s Y’s fuel mileage.
Answer: B
