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100 students appeared for two testsMaths and English.
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23 Jan 2019, 18:07
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100 students appeared for two testsMaths and English. For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English. How many students passed in neither Maths nor English? (1) More than 4 students passed in both the tests. (2) Less than 6 students passed in both the tests.
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100 students appeared for two testsMaths and English.
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Updated on: 24 Jan 2019, 10:40
Gekkdude wrote: 100 students appeared for two testsMaths and English. For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English. How many students passed in neither Maths nor English?
(1) More than 4 students passed in both the tests.
(2) Less than 6 students passed in both the tests. After having a discussion with fellow gc members, i understood the question, Inline is the approach. Total students = 100 students who passed in both the tests = x Passed only in Maths = 8x Passed only in English = 9x Total students = x + 9x + 8x We have to find students passed in neither Maths nor English 8x + 9x + x = 100  N N = 100  18x Statement 1 More than 4 students passed in both the tests. x can be 5, it cant be further than this N = 100  18*5 = 100  90 = 10 Statement 2 Less than 6 students passed in both the tests. here x can be 1,2,3,4,5. When we substitute these values in N = 100  18x We get different values. OA=A
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Originally posted by KanishkM on 23 Jan 2019, 19:05.
Last edited by KanishkM on 24 Jan 2019, 10:40, edited 1 time in total.



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Re: 100 students appeared for two testsMaths and English.
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23 Jan 2019, 21:10
Official Solution  OA "A" Understand Question and Draw Inferences
Total number of students = 100 Let the number of students who passed in both tests = X Let the number of students who passed in neither Math nor English = N Therefore,
Number of students who passed only in Maths = 8X Number of students who passed only in English = 9X Representing this information in Venn Diagram we get, (sorry couldnt draw a Venn diagram)
MathCommonEngtotal students=100 (8X (X) 9X) neither Math nor English = N
8X + X + 9X + N = 100 18X + Neither = 100 N = 100 – 18X Therefore,
We need a unique value of X Also, notice here that 18X < 100,
So, the possible values of X are 1, 2, 3, 4 ,5
Statement 1: More than 4 students passed in both the tests.
X > 4 From the question stem inference, we deduced that the possible values of X are 1,2,3,4 or 5.
Since X > 4 the only possible value of X = 5 Hence, Statement 1 is sufficient.
Statement 2: Less than 6 students passed in both the tests.
X < 6 From the question stem inference, we deduced that the possible values of X are 1,2,3,4 or 5.
Therefore, with Statement 2 alone we cannot find a unique value of X .
Hence, Statement 2 is not sufficient.
Correct Answer: Option A



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100 students appeared for two testsMaths and English.
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23 Jan 2019, 21:27
Gekkdude wrote: Official Solution  OA "A" Understand Question and Draw Inferences
Total number of students = 100 Let the number of students who passed in both tests = X Let the number of students who passed in neither Math nor English = N Therefore,
Number of students who passed only in Maths = 8X Number of students who passed only in English = 9X Representing this information in Venn Diagram we get, (sorry couldnt draw a Venn diagram)
I have a quick question. Why are we multiplying thisby x shouldn't it be 8 + x and 9 + x Since the common area + Only A = A Can you please share your thoughts on this.
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Re: 100 students appeared for two testsMaths and English.
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23 Jan 2019, 21:35
simple question: For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English this the key stmt: let students who bpassed in both by x, hence passed in maths =8x and passed in Eng = 9x let who passed in neither be y therefore: x+8x+9x+y =100 18x+y =100 thus 18x<=100 now 1 stmt said x>4 let us start with 5 18*5 = 90 <100 ... satisfy 18*6 = 108>100 does not satisfy thus 5 is the only possible solution. thus A is sufficient. which gives y =10. Lets take B here 1,2,3,4,5 all satisfies. no unique solution thus B insufficient. Thus answer A
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 01:39
Gekkdude wrote: 100 students appeared for two testsMaths and English. For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English. How many students passed in neither Maths nor English?
(1) More than 4 students passed in both the tests.
(2) Less than 6 students passed in both the tests. Is is more of a ratio question. Both: Math : English 1x : 8x : 9x (1) More than 4 students passed in both the tests.let x = 5...total students = 90..Neither math nor English = 10 let x = 6...total students above 100........invalid. (Any x above 5 will be invalid) Sufficient (2) Less than 6 students passed in both the tests.let x = 5...total students = 90..Neither math nor English = 10 let x = 5...total students = y..Neither math nor English = another number......No need to calculate) Insufficient Answer: A



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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 09:42
globaldesi wrote: simple question: For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English this the key stmt: let students who bpassed in both by x, hence passed in maths =8x and passed in Eng = 9x let who passed in neither be y therefore: x+8x+9x+y =100 18x+y =100 thus 18x<=100 now 1 stmt said x>4 let us start with 5 18*5 = 90 <100 ... satisfy 18*6 = 108>100 does not satisfy thus 5 is the only possible solution. thus A is sufficient. which gives y =10.
Lets take B here 1,2,3,4,5 all satisfies. no unique solution thus B insufficient. Thus answer A Hi globaldesiWhy are we multiplying this by x shouldn't it be 8 + x and 9 + x isn't Only A = A  common area Can you please share your thoughts on this.
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 09:50
KanishkM wrote: globaldesi wrote: simple question: For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English this the key stmt: let students who bpassed in both by x, hence passed in maths =8x and passed in Eng = 9x let who passed in neither be y therefore: x+8x+9x+y =100 18x+y =100 thus 18x<=100 now 1 stmt said x>4 let us start with 5 18*5 = 90 <100 ... satisfy 18*6 = 108>100 does not satisfy thus 5 is the only possible solution. thus A is sufficient. which gives y =10.
Lets take B here 1,2,3,4,5 all satisfies. no unique solution thus B insufficient. Thus answer A Hi globaldesiWhy are we multiplying this by x shouldn't it be 8 + x and 9 + x isn't Only A = A  common area Can you please share your thoughts on this. The question says for every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English Let me explain this line by an example I say for every candy i give you, i take 2 candies. That would mean if you have 1 candy i will have 2 if you have 2 i will have 4 candies. If you have 3 , i will have 6 so if you have z I will have 2z. lets reread the statement The question says for every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English for every x students who passed in both, 8x passed in Maths and 9x passed in english. Compare it with candies example if 1 student passed in both , 8 passed in Maths and 9 passed in english. So let x students passed in both, thus 8x passed in Maths and 9x passed in English. total students passed = x+8x+9x Hope its clearer. I will be happy to help if still not clear
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 10:04
Gekkdude wrote: 100 students appeared for two testsMaths and English. For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English. How many students passed in neither Maths nor English?
(1) More than 4 students passed in both the tests.
(2) Less than 6 students passed in both the tests. I solved the question using 2x2 matrix ; attached .. #1 we can see that only at 5 the 2x2 matrix would stand valid ; sufficient #2 at 5 it would be valid but at 4 it wont be valid ; in sufficient IMO A
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 10:05
globaldesi wrote: KanishkM wrote: globaldesi wrote: simple question: For every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English this the key stmt: let students who bpassed in both by x, hence passed in maths =8x and passed in Eng = 9x let who passed in neither be y therefore: x+8x+9x+y =100 18x+y =100 thus 18x<=100 now 1 stmt said x>4 let us start with 5 18*5 = 90 <100 ... satisfy 18*6 = 108>100 does not satisfy thus 5 is the only possible solution. thus A is sufficient. which gives y =10.
Lets take B here 1,2,3,4,5 all satisfies. no unique solution thus B insufficient. Thus answer A Hi globaldesiWhy are we multiplying this by x shouldn't it be 8 + x and 9 + x isn't Only A = A  common area Can you please share your thoughts on this. The question says for every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English Let me explain this line by an example I say for every candy i give you, i take 2 candies. That would mean if you have 1 candy i will have 2 if you have 2 i will have 4 candies. If you have 3 , i will have 6 so if you have z I will have 2z. lets reread the statement The question says for every student who passed in both the tests, 8 students passed only in Maths and 9 students passed only in English for every x students who passed in both, 8x passed in Maths and 9x passed in english. Compare it with candies example if 1 student passed in both , 8 passed in Maths and 9 passed in english. So let x students passed in both, thus 8x passed in Maths and 9x passed in English. total students passed = x+8x+9x Hope its clearer. I will be happy to help if still not clear Thank you for the quick response. As per the candy example, we were discussing in terms of the ratio, right ?? This was not a typical sets question then i believe...., was there any trigger word which makes them apart from the inline So how different the question would have been, if the inline logic had to be used Only A = Total A  Both Example been, There are 10 students, 6 Study English and 5 Study Mathematics, 1 study Both here only English would be 61 = 5
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 10:25
Well this is a nice question that involves both ratio and sets. However such questions are not uncommon in GMAT. You will get a lot of such questions. The inline statement will be straight such as one student passed in both and eight passed in maths and 9 passed in english. You realize the only difference is word "for"
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 10:28
globaldesi wrote: Well this is a nice question that involves both ratio and sets. However such questions are not uncommon in GMAT. You will get a lot of such questions. The inline statement will be straight such as one student passed in both and eight passed in maths and 9 passed in english. You realize the only difference is word "for" Yes, now i understood it. Thank you, for your help at this.
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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 11:03
Can someone help me on my thought process please? My test is in 2 day and I've spent over 2 hours trying to figure out where I went wrong..... So my thought process is as follows, Let the one's who failed in both Math and English be PMath PassMath FailTotal Eng PassXDontknow9X Eng FailDontknow PDontknow Total8XDontknow100 so per the table equation, population of Math Pass and Eng Fail is ( 8X  X ) = 7Xpopulation of Engl Pass and Math Fail is ( 9X  X ) = 8XTherefore my table is now populated as, Math PassMath FailTotal Eng PassX 8X9X Eng Fail 7X P 7X+ PTotal8X 8X+ P100 And therefore going by the equations, I get 16X + P = 100 Now lets go to (1) (1) More than 4 students passed in both the tests. X>4, so by substituting X=5, P = 20 and life is good........ Substituting X=6, P = 4 and life is good again...... So (A) is INsufficient ! Now lets go to (2) (2) Less than 6 students passed in both the tests. X<6, so by substituting X = 5, 4, 3, 2, 1 it all gives an OK result; so (B) is INsufficient ! Taking (1) and (2) together gives us X=5; Therefore answer should be (C) as both are needed to give a unique solution  doesn't match OA !



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Re: 100 students appeared for two testsMaths and English.
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24 Jan 2019, 11:40
Gekkdude wrote: Can someone help me on my thought process please? My test is in 2 day and I've spent over 2 hours trying to figure out where I went wrong..... So my thought process is as follows, Let the one's who failed in both Math and English be PMath PassMath FailTotal Eng PassXDontknow9X Eng FailDontknow PDontknow Total8XDontknow100 so per the table equation, population of Math Pass and Eng Fail is ( 8X  X ) = 7Xpopulation of Engl Pass and Math Fail is ( 9X  X ) = 8XTherefore my table is now populated as, Math PassMath FailTotal Eng PassX 8X9X Eng Fail 7X P 7X+ PTotal8X 8X+ P100 And therefore going by the equations, I get 16X + P = 100 Now lets go to (1) (1) More than 4 students passed in both the tests. X>4, so by substituting X=5, P = 20 and life is good........ Substituting X=6, P = 4 and life is good again...... So (A) is INsufficient ! Now lets go to (2) (2) Less than 6 students passed in both the tests. X<6, so by substituting X = 5, 4, 3, 2, 1 it all gives an OK result; so (B) is INsufficient ! Taking (1) and (2) together gives us X=5; Therefore answer should be (C) as both are needed to give a unique solution  doesn't match OA ! The pass equation shouldnt be 16x + N =100 I just followed the table, made by you. Pass in English + Pass in Maths + Pass in Both =100  N 8x + 9x +x =100  N 18x + N =100 Your approach was good Posted from my mobile device
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Re: 100 students appeared for two testsMaths and English.
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