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Odometer Log & Mileage

XAT 2012 · Quantitative Ability & Data Interpretation · Medium · Data Interpretation

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Passage / data set

Answer question based on the following information. Ramya, based in Shanpur, took her car for a 400 km trip to Rampur. She maintained a log of the odometer readings and the amount of petrol she purchased at different petrol pumps at different prices (given below). Her car already had 10 litres of petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.

StageOdometer Reading (Km)Petrol Purchased (Litre)Rate of Petrol (/litre)
Start of journey4002030
Mid-way6001535
Near Rampur6501040
End of journey800--

Question 1 of 2

What has been the mileage (in kilometers per litre) of her car over the entire trip?

  1. A.

    8.00

  2. B.

    8.50

  3. C.

    9.00

  4. D.

    9.50

  5. E.

    None of the above

Answer

A

Explanation

Total petrol available = Initial 10 + Purchased (20 + 15 + 10) = 55 litres. Remaining petrol = 5 litres. Total petrol consumed = 50 litres. Total distance = 800−400=400800 - 400 = 400 km. Mileage = 40050=8.00\frac{400}{50} = 8.00 km/litre.

Question 2 of 2

Her car's tank-capacity is 35 litres. Petrol costs 45/- litre in Rampur. What is the minimum amount of money she would need for purchasing petrol for the return trip from Rampur to Shanpur, using the same route? Assume that the mileage of the car remains unchanged throughout the route, and she did not use her car to travel around in Rampur.

  1. A.

    1714

  2. B.

    1724

  3. C.

    1734

  4. D.

    1744

  5. E.

    Data insufficient to answer.

Answer

D

Explanation

Return journey distances: Rampur (0 km) -> Pump 3 (150 km) -> Pump 2 (200 km) -> Pump 1 (400 km). Mileage = 8 km/L. Needs 50 L total for 400 km. At start (Rampur): Starts with 5 L. Can travel 5×8=405 \times 8 = 40 km. To cover 150 km to Pump 3, needs 110/8=13.75110/8 = 13.75 L at Rs. 45/L. At Pump 3 (150 km mark, 50 km to Pump 2): Needs 50/8=6.2550/8 = 6.25 L at Rs. 40/L. At Pump 2 (200 km mark, 200 km to Shanpur): Needs 200/8=25200/8 = 25 L at Rs. 35/L. Total cost = (13.75×45)+(6.25×40)+(25×35)=618.75+250+875=1743.75≈Rs. 1744(13.75 \times 45) + (6.25 \times 40) + (25 \times 35) = 618.75 + 250 + 875 = 1743.75 \approx \text{Rs. } 1744.

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