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# 100 students appeared for two tests-Maths and English.

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Intern
Joined: 12 Dec 2018
Posts: 10
100 students appeared for two tests-Maths and English.  [#permalink]

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23 Jan 2019, 18:07
1
00:00

Difficulty:

75% (hard)

Question Stats:

43% (01:43) correct 57% (01:59) wrong based on 47 sessions

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100 students appeared for two tests-Maths 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.
VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
100 students appeared for two tests-Maths and English.  [#permalink]

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Updated on: 24 Jan 2019, 10:40
Gekkdude wrote:
100 students appeared for two tests-Maths 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
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

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.
Intern
Joined: 12 Dec 2018
Posts: 10
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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23 Jan 2019, 21:10
1
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)

|Math--Common---Eng-----total 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
VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
100 students appeared for two tests-Maths and English.  [#permalink]

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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.
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Senior Manager
Joined: 28 Jul 2016
Posts: 250
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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23 Jan 2019, 21:35
1
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
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.
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SVP
Joined: 26 Mar 2013
Posts: 2067
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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24 Jan 2019, 01:39
Gekkdude wrote:
100 students appeared for two tests-Maths 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

VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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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
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.

Hi globaldesi

Why 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.
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Senior Manager
Joined: 28 Jul 2016
Posts: 250
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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24 Jan 2019, 09:50
1
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
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.

Hi globaldesi

Why 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 re-read 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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SVP
Joined: 18 Aug 2017
Posts: 1840
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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24 Jan 2019, 10:04
Gekkdude wrote:
100 students appeared for two tests-Maths 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
Attachments

File comment: solutiom
maths english.xlsx [30.86 KiB]

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VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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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
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.

Hi globaldesi

Why 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 re-read 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 6-1 = 5
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Senior Manager
Joined: 28 Jul 2016
Posts: 250
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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24 Jan 2019, 10:25
1
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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VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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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.
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Intern
Joined: 12 Dec 2018
Posts: 10
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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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 P

---------------Math Pass---------Math Fail--------Total
---Eng Pass-----X---------------Dontknow--------9X
---Eng Fail---Dontknow-----------P--------Dontknow
Total-----------8X--------------Dontknow---------100

so per the table equation,
population of Math Pass and Eng Fail is ( 8X - X ) = 7X
population of Engl Pass and Math Fail is ( 9X - X ) = 8X

Therefore my table is now populated as,

---------------Math Pass---------Math Fail--------Total
---Eng Pass-----X---------------8X--------------9X
---Eng Fail-------7X---------------P--------7X+P
Total------------8X--------------8X+P-------100

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 IN-sufficient !

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 IN-sufficient !

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 !
VP
Joined: 09 Mar 2018
Posts: 1000
Location: India
Re: 100 students appeared for two tests-Maths and English.  [#permalink]

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24 Jan 2019, 11:40
1
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 P

---------------Math Pass---------Math Fail--------Total
---Eng Pass-----X---------------Dontknow--------9X
---Eng Fail---Dontknow-----------P--------Dontknow
Total-----------8X--------------Dontknow---------100

so per the table equation,
population of Math Pass and Eng Fail is ( 8X - X ) = 7X
population of Engl Pass and Math Fail is ( 9X - X ) = 8X

Therefore my table is now populated as,

---------------Math Pass---------Math Fail--------Total
---Eng Pass-----X---------------8X--------------9X
---Eng Fail-------7X---------------P--------7X+P
Total------------8X--------------8X+P-------100

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 IN-sufficient !

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 IN-sufficient !

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
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Re: 100 students appeared for two tests-Maths and English.   [#permalink] 24 Jan 2019, 11:40
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# 100 students appeared for two tests-Maths and English.

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