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Last semester, Professor K taught two classes, A and B. Each student
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24 Jun 2018, 21:24
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Question Stats:
86% (01:07) correct 14% (01:15) wrong based on 665 sessions
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Last semester, Professor K taught two classes, A and B. Each student in class A handed in 7 assignments, and each student in class B handed in 5 assignments. How many students were in class A ?
(1) The students in both classes combined handed in a total of 85 assignments. (2) There were 10 students in class B.
Re: Last semester, Professor K taught two classes, A and B. Each student
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24 Jun 2018, 21:51
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Bunuel wrote:
Last semester, Professor K taught two classes, A and B. Each student in class A handed in 7 assignments, and each student in class B handed in 5 assignments. How many students were in class A ?
(1) The students in both classes combined handed in a total of 85 assignments. (2) There were 10 students in class B.
To find the number of students in class A, we need to know the total number of assignments they (class A) handed in. We'll look for a statement that gives us this information, a Logical approach.
Neither (1) nor (2) gives us this information so they are both insufficient. Combined: We can now calculate our missing information (as 85 - 10*5 = total number of assignments handed in by class A) Sufficient
Re: Last semester, Professor K taught two classes, A and B. Each student
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25 Jun 2018, 00:19
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Say class A has 'a' students and class B has 'b' students. We need to find a = ?
(1) The students in both classes combined handed in a total of 85 assignments. 7a + 5b = 85 Note that a, b are pos. integers.
There are only two possible solutions (which I got by putting numbers in the equation; tedious, I know) a = 5, b = 10 a = 10, b = 3 Since we have two possible values of a, Statement 1 is not sufficient.
(2) There were 10 students in class B. b = 10 Does not help with a. Insufficient.
Re: Last semester, Professor K taught two classes, A and B. Each student
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25 Jun 2018, 01:20
sandman13 wrote:
There are only two possible solutions (which I got by putting numbers in the equation; tedious, I know) a = 5, b = 10 a = 10, b = 3 Since we have two possible values of a, Statement 1 is not sufficient.
Hi,
In many cases, there is built in pattern that allow you solve without any tedious work. I solved this equation in less than 30 seconds.
Re: Last semester, Professor K taught two classes, A and B. Each student
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25 Jun 2018, 01:43
Bunuel wrote:
Last semester, Professor K taught two classes, A and B. Each student in class A handed in 7 assignments, and each student in class B handed in 5 assignments. How many students were in class A ?
(1) The students in both classes combined handed in a total of 85 assignments. (2) There were 10 students in class B.
From the information given, let A = the number of student in class A and B = the number of student in class B (1) The students in both classes combined handed in a total of 85 assignments. So we got the equation 7A + 5B = 85 let consider some possible combination between A and B which are 1. A = 5 and B = 10 2. A = 10 and B = 3 Since there are 2 possible combinations. (1) is insufficient
(2) There were 10 students in class B. This statement alone just tell us that the total assignments from class B is 10*5=50 which is insufficient to answer what is the value of A
But when we combine both statements. It is sufficient to answer the question. So the answer is "C"
Re: Last semester, Professor K taught two classes, A and B. Each student
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25 Jun 2018, 01:47
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1
Mo2men wrote:
sandman13 wrote:
There are only two possible solutions (which I got by putting numbers in the equation; tedious, I know) a = 5, b = 10 a = 10, b = 3 Since we have two possible values of a, Statement 1 is not sufficient.
Hi,
In many cases, there is built in pattern that allow you solve without any tedious work. I solved this equation in less than 30 seconds.
We should consider any possible cases from any statements in data sufficiency problem especially these type of question which sometime can be really tricky. Let say if the statement (1) is changed a little into "The students in both classes combined handed in a total of 40 assignments." Considered from the given info that 7A + 5B = 40 ---> the only possible combination is A=5 and B=1 which this statement alone is sufficient to answer the question.
Last semester, Professor K taught two classes, A and B. Each student
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25 Jun 2018, 01:57
1
Bunuel wrote:
Last semester, Professor K taught two classes, A and B. Each student in class A handed in 7 assignments, and each student in class B handed in 5 assignments. How many students were in class A ?
(1) The students in both classes combined handed in a total of 85 assignments. (2) There were 10 students in class B.
Let A = # of student in class A & B = # of student in class B
Total Assignment of class A = 7A
Total Assignment of class B = 5B
(1) The students in both classes combined handed in a total of 85 assignments.
Equation: 7A + 5B = 85 Let's look how we can solve in less with minimum effort. Look to number properties (in same cases unit digit plays great role too but not here)
'85' is multiple of 5 & '5B' is multiple of 5......Hence 7A MUST be multiple of 5.
Therefore 'A' can take values 5, 10, 15...etc
Let A=5.......Then B=10
Let A=10.......Then B=3
We can't proceed more as it will exceed 85
Insufficient
(2) There were 10 students in class B.
No info about A.
Insufficient
Combine 1 & 2 We have only one clear answer when B=10 so A = 5
Re: Last semester, Professor K taught two classes, A and B. Each student
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28 Jun 2018, 09:20
1
there are 2 kinds of 2 variable equation on gmat. if the numbers are big, 2 variables need 2 equation to solve. if numbers are small, one equation is enough to find 2 variable . this is the lession. gmat is a little unfair when playing this trick
Last semester, Professor K taught two classes, A and B. Each student
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03 Nov 2019, 00:37
I would say its a tricky question.
We need, How many students were in class A?
Statement 1: Total assignment combined = 85 Because each student of Class A handed in 7 assignments, and class B handed in 5 assignments. Class A student's number would be multiple of 7 and the student number of class B would be multiple of 5.
Here possible combinations are A(X7) B 7 78 14 71 21 64 28 57 35 50 42 43 49 36 56 29 63 22 70 15 77 8 84 1 Here two combinations are possible So, Stat 1 Insufficient. (If there was one combination Stat 1 would have been sufficient)
Statement 2: B=10 clearly Insufficient
Combined, Sufficient
Answer : C
gmatclubot
Last semester, Professor K taught two classes, A and B. Each student
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03 Nov 2019, 00:37
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86% (01:07) correct 
