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16 Jun 2025, 00:24
Kudos
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Dropdown 1: 15
Dropdown 2: 2900
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:18) correct
47%
(02:23)
wrong
based on 471
sessions
History
Date
Time
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Not Attempted Yet
For each of 22 drivers in a logistics company, the graph shows the driver’s monthly fuel cost (in euros, €) and the number of kilometers the driver drives per month. The graph also shows a trend line for the data.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
The trend line suggests that, for a driver who drives at least 1,000 kilometers per month, each 100-kilometer increase in distance driven corresponds to an increase of approximately euros in monthly fuel cost (rounded to the nearest €5).
For the driver with the highest monthly fuel cost, the number of kilometers driven per month is kilometers.
GMAT-Club-Forum-wk0nozm0.png [ 28.86 KiB | Viewed 2702 times ]
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
The trend line suggests that, for a driver who drives at least 1,000 kilometers per month, each 100-kilometer increase in distance driven corresponds to an increase of approximately euros in monthly fuel cost (rounded to the nearest €5).
For the driver with the highest monthly fuel cost, the number of kilometers driven per month is kilometers.
Attachment:
GMAT-Club-Forum-wk0nozm0.png [ 28.86 KiB | Viewed 2702 times ]
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Official Answer
Dropdown 1: 15
Dropdown 2: 2900
16 Jun 2025, 02:07
Kudos
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To calculate the monthly fuel cost per 1000km
We can take reference of y axis with lowest being 200 and highest being 500 and corresponding km is 1000 and 2900 respectively
So for every 1000 km there is rise of (500-200)/((2900-1000)/1000)
=300/19 = 15.78 ~ 15
Also from graph, it can be observed that the highest monthly fuel cost corresponds to 2900km
Hence answer is 15 and 2900
We can take reference of y axis with lowest being 200 and highest being 500 and corresponding km is 1000 and 2900 respectively
So for every 1000 km there is rise of (500-200)/((2900-1000)/1000)
=300/19 = 15.78 ~ 15
Also from graph, it can be observed that the highest monthly fuel cost corresponds to 2900km
Hence answer is 15 and 2900
25 Jun 2025, 08:30
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Bunuel
Solution:
Statement 1:
The trend line suggests that, for a driver who drives at least 1,000 kilometers per month, each 100-kilometer increase in distance driven corresponds to an increase of approximately ___ in monthly fuel cost.
Estimate the slope using two points on the trend line:
- At 1,000 km → approx. €200
- At 2,900 km → approx. €500
- Difference in cost: 500 − 200 = €300
- Difference in distance: 2,900 − 1,000 = 1,900 km
Code:
€300 ÷ 1,900 km × 100 km ≈ €15.8
Rounded to the nearest €5: €15Answer: €15
Statement 2:
For the driver with the highest monthly fuel cost, the number of kilometers driven per month is ___ kilometers.
From the data:
- Highest cost = €500
- Corresponding distance = 2,900 km
