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Of the final grades received by the students in a certain [#permalink]
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I guess the more staightforward way to do this is giving a variable for the total number of students, say T:

Then:
(1/5)T + (1/4)T + (1/2)T + 10 = T
T/5 + T/4 + T/2 + 10 = T
(4T + 5T + 10T) / 20 = T - 10 --> we multiplied with the LCM and moved 10 to the other side.
19T = 20 (T - 10)
19T = 20T - 200
T = 200, ANS D
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Of the final grades received by the students in a certain [#permalink]
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HI All,

If you plan to take a "math approach" to this question, you might find that converting the fractions to decimals saves you some work:

1/5 = .2 are As
1/4 = .25 are Bs
1/2 = .5 are Cs
X = Total grades given

Sub-Total = .95(X) of the grades are A/B/C, so the other .05(X) are Ds

.05(X) = 10
5X = 1,000
X = 200

Final Answer:

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Re: Of the final grades received by the students in a certain [#permalink]
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I would rather use decimals for this question.
Grades A:0,2, B:0,25, C:0,5 and the remaining(0,2+0,25+0,5=0,95) 0,05*X=10 --> So, X=200 (D)
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Re: Of the final grades received by the students in a certain [#permalink]
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Hi All,

This question can also be solved by TESTing THE ANSWERS.

We're given data on the "breakdown" of grades in a course: 1/5 are A's, 1/4 are B's, 1/2 are C's, and the remaining 10 grades are D's. We're asked for the total number of students in the course.

Let's TEST Answer C: 160 students

Total = 160
1/5 are A's = 32
1/4 are B's = 40
1/2 are C's = 80
Sub-total = 152, which leaves 8 students with D's. This is TOO LOW (there are supposed to be 10 students with D's). Thus we need MORE total students.
Eliminate Answers A, B and C.

Since 8 is relatively close to 10, we probably don't need that many more students, so let's TEST D next....

Answer D: 200 students

Total = 200
1/5 are A's = 40
1/4 are B's = 50
1/2 are C's = 100
Sub-total = 190, leaving 10 for D's. This is a MATCH for what we were told, so this MUST be the answer.

Final Answer:

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Re: Of the final grades received by the students in a certain [#permalink]
Walkabout wrote:
Of the final grades received by the students in a certain math course, 1/5 are A's, 1/4 are B's, 1/2 are C's, and the remaining 10 grades are D's. What is the number of students in the course?

(A) 80
(B) 110
(C) 160
(D) 200
(E) 400





1/5 + 1/4 + 1/2 + x =1=T somehow

x= 1/20

1/20 T=10

T=200
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Walkabout wrote:
Of the final grades received by the students in a certain math course, 1/5 are A's, 1/4 are B's, 1/2 are C's, and the remaining 10 grades are D's. What is the number of students in the course?

(A) 80
(B) 110
(C) 160
(D) 200
(E) 400


We start by creating a variable for the total number of students in the math course. We can say:

T = Total number of students in the math course

Next, we can use variable T in an equation that we translate from the given information. We are given that, of the final grades received by the students in a certain math course, 1/5 are A's, 1/4 are B's, 1/2 are C's, and the remaining 10 grades are D's. Since this represents all the grades in the class, it represents all the students in the class. Thus we know:

#A’s + #B’s + #C’s + #D’s = Total number of students in the class

1/5(T) + ¼(T) + ½(T) + 10 = T

We can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have:

4T + 5T + 10T + 200 = 20T

19T + 200 = 20T

200 = T

There are a total of 200 students in the math class.

Answer is D.
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