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ChandlerBong
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 07 Nov 2023, 08:54
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C
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75% (hard)

Question Stats:

67% (02:28) correct 33% (02:36) wrong based on 1143 sessions

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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. \(24 - \frac{L}{500} - \frac{S}{2}\)

B. \(24 - \frac{L}{500}-\frac{S}{20}\)

C. \(26 - \frac{L}{500}-\frac{S}{20}\)

D. \(44 - \frac{L}{500}-\frac{S}{2}\)

E. \(44 - \frac{L}{500}-\frac{S}{20}\)
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C
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gmatophobia
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 07 Nov 2023, 10:27
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ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. 24 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

B. 24 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

C. 26 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

D. 44 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

E. 44 - \(\frac{L}{500}\) - \(\frac{S}{20}\)
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Let's assume that Solomon's truck had the following speed and weight

  • Speed = 50mph
  • Weight = 500 pounds

Expected Mileage = \(24 - 1 - \frac{1}{2} = 22.50\)
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  • We have substracted 1 because the load of the truck is 500 pounds
  • We have substracted 1/2 because the truck is moving 10 miles above 40
Answer choice elimination

A. 24 - \(\frac{L}{500}\) - \(\frac{S}{2}\) → \(24 - \frac{500}{500} - \frac{50}{2} = 24 - 1 - 25 = -2\) ⇒ Eliminate

B. 24 - \(\frac{L}{500}\) - \(\frac{S}{20}\) → \(24 - \frac{500}{500} - \frac{50}{20} = 24 - 1 - 2.5 = 20.50\) ⇒ Eliminate

C. 26 - \(\frac{L}{500}\) - \(\frac{S}{20}\) → \(26 - \frac{500}{500} - \frac{50}{20} = 26 - 1 - 2.5 = 22.50\) ⇒ Answer

D. 44 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

E. 44 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

Option C
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 08 Nov 2023, 08:35
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ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. 24 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

B. 24 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

C. 26 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

D. 44 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

E. 44 - \(\frac{L}{500}\) - \(\frac{S}{20}\)
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Solution:

  • Decrease because of L pounds weight \(= -\frac{L}{500}\)
  • Decrease because of S speed \(=-\frac{1}{2}\times \frac{(S-40)}{10}=-\frac{S-40}{20}\)

  • Final mileage \(=24-\frac{L}{500}-\frac{S-40}{20}\)
    \(⇒24-\frac{L}{500}-\frac{S}{20}+\frac{40}{20}\)
    \(⇒24-\frac{L}{500}-\frac{S}{20}+2\)
    \(⇒26-\frac{L}{500}-\frac{S}{20}\)

Hence the right answer is Option C
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 09 Nov 2024, 01:46
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Detailed video solution:


IMO C

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ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. \(24 - \frac{L}{500} - \frac{S}{2}\)

B. \(24 - \frac{L}{500}-\frac{S}{20}\)

C. \(26 - \frac{L}{500}-\frac{S}{20}\)

D. \(44 - \frac{L}{500}-\frac{S}{2}\)

E. \(44 - \frac{L}{500}-\frac{S}{20}\)
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 11 Nov 2024, 05:36
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ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. \(24 - \frac{L}{500} - \frac{S}{2}\)

B. \(24 - \frac{L}{500}-\frac{S}{20}\)

C. \(26 - \frac{L}{500}-\frac{S}{20}\)

D. \(44 - \frac{L}{500}-\frac{S}{2}\)

E. \(44 - \frac{L}{500}-\frac{S}{20}\)
Show more

When the options have variables in them, plugging in values works.
Plug in the easiest values. Say L = 0 and S = 40. We are given that speed will be 24 mpg in that case.
Only options (C) and (D) are possible since they give 24.

Put L = 0 and S = 50 now. An increase of 10 in speed will reduce mileage by 1/2 and make it 23.5.
Only (C) works.

Answer (C)

Check another question with variables in options and how to attempt it: https://anaprep.com/arithmetic-work-rat ... s-and-all/
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 06 Aug 2025, 14:05
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Let's Assume the L = 500 and S = 60..so 20 mph extra .
Total reduction should be ....2 miles..... so 22 miles....
Input values..C ...26 - 1 (500/500) - 60/20(S/20).....Gives 22... Answer! C

ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. \(24 - \frac{L}{500} - \frac{S}{2}\)

B. \(24 - \frac{L}{500}-\frac{S}{20}\)

C. \(26 - \frac{L}{500}-\frac{S}{20}\)

D. \(44 - \frac{L}{500}-\frac{S}{2}\)

E. \(44 - \frac{L}{500}-\frac{S}{20}\)
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 12 Aug 2025, 11:27
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This is exactly same way i solved this in the mock and this was correct answer, but I was wondering, according to this formula, if his speed were to decrease, his mileage would increase by 0.5 miles per galon.

The question doesn't mention or eny anything about decreasing speed so i think this would be okay.

I was curious:
If it were said it decreases both way, we could use mod, but what would be the formula if reducing speed didn't affect the mileage

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ChandlerBong
Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed of 40 mph (miles per hour) with no load. His truck's gas mileage decreases 1 mpg for each 500 pounds of added load and it decreases \(\frac{1}{2}\) mpg for each 10 mph of increased speed over 40 mph. Which of the following is a formula for the gas mileage of Solomon's truck when its speed is S mph (S >= 40) and it has a load of L pounds?

A. 24 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

B. 24 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

C. 26 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

D. 44 - \(\frac{L}{500}\) - \(\frac{S}{2}\)

E. 44 - \(\frac{L}{500}\) - \(\frac{S}{20}\)

Solution:

  • Decrease because of L pounds weight \(= -\frac{L}{500}\)
  • Decrease because of S speed \(=-\frac{1}{2}\times \frac{(S-40)}{10}=-\frac{S-40}{20}\)

  • Final mileage \(=24-\frac{L}{500}-\frac{S-40}{20}\)
    \(⇒24-\frac{L}{500}-\frac{S}{20}+\frac{40}{20}\)
    \(⇒24-\frac{L}{500}-\frac{S}{20}+2\)
    \(⇒26-\frac{L}{500}-\frac{S}{20}\)

Hence the right answer is Option C
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 17 Sep 2025, 18:44
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Solomon's truck gets 24 mpg (miles per gallon) when driven at a speed 12 Jul 2026, 12:55
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Here's my video solution:
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