There are 300 seniors at Morse High School, and 40% of them have cars.

+1 for E .15X+45=120+.1X
X=1500

MAGOOSH OFFICIAL SOLUTION:

Let x = the number of students other than seniors (freshmen + sophomores + junior). We know 40% of the 300 seniors have cars. Well, 10% of 300 is 30, so 40% is 4 times this —- 4*30 = 120 seniors have cars. We know 10% of the other students have cars, so that would be 0.1*x. The total number of students with cars is 120 + 0.1x. That’s the PART.

The total number of students = 300 + x. That’s the WHOLE.

PART/ WHOLE x 100% = 15%, which means that PART/ WHOLE = 0.15, which means PART = 0.15*(WHOLE). That can be our equation.
(120 + 0.1x) = 0.15(300 + x)
120 + 0.1x = 45 + 0.15x
75 + 0.1x = 0.15x
75 = 0.15x – 0.10x = 0.05x
150 = 0.10x
1500 = x

Answer = (E)

Yes, absolutely!

Average of 10% and 40% is 15%,

wO/wS = (40 - 15)/(15 - 10) = 5/1

Ratio of number of students in other grades : Seniors = 5 : 1
Since there are 300 seniors, students in other grades = 1500

Answer (E)

We see that 300 x 0.4 = 120 seniors have cars.

If we let n = the number of students in the remaining grades, then 0.1n have cars. Since 15% of all students have cars:

(120 + 0.1n)/(n + 300) = 0.15

120 + 0.1n = 0.15n + 45

75 = 0.05n

1,500 = n

Answer: E

Hi All,

This question can be solved in a variety of ways - with ratios, as a weighted average, or even by TESTing THE ANSWERS. Most Test Takers would probably take an algebraic approach with this question (which is fine), but here's how you can use the answers to your advantage and just do Arithmetic...

We're told a few facts about a group of students:
1) There are 300 seniors and 40% of them own a car.
2) There are X non-seniors and 10% of them own a car.
3) 15% of ALL the students own a car.

We're asked for the total number of NON-seniors at the school.

From a logical standpoint, it would take a LOT more than 300 non-seniors to "bring the average down" from 40% car ownership to 15% car ownership. Given the answer choices that come with this prompt, the correct answer is likely to be one of the larger options.

Let's start with Answer C: 1200....

Seniors with a car: (.4)(300) = 120
Non-seniors w/ a car: (.1)(1200) = 120

Total with a car/All students = 240/1500
15% of 1500 = 225
This answer is NOT a match; as a percent, it is TOO BIG. We need more non-seniors to bring the percent down...

Between Answers D and E, E seems like "easier math"

Answer E: 1500....

Seniors with a car: (.4)(300) = 120
Non-seniors w/ a car: (.1)(1500) = 150

Total with a car/All students = 270/1800
15% of 1800 = 270
This IS a match, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Using a table to organise all the information we get this:
S = Sophomore and NS = Not Sophomore
Yes = Yes car and No = No car

.................Yes.................No.........All
S...............120.................280........300
NS............0.10x................................
All.........0.15(x+300)............................

So:
120 + 0.10x = 0.15(x+300)
120 + 0.10x = 0.15x + 45
75 = 0.05x
x = 1500 ANS E

Hi, can this be solved by using the weighted averages concept? VeritasKarishma

# of seniors: 300
# of Juniors (all other students other than the seniors) = x
Total # of students: x+300

Total # of seniors with a car: (40/100) * 300 = 120 seniors

Total # of juniors with a car: 0.10x

Total # of all the students with a car: (15/100) * (x+300) = 120 + 0.10x
x=1500

Hence, the total # of juniors with a car: 2500

This is how I approached this problem, please let me know if this is correct. Apologies for the awful handwriting.