Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 18 Feb 2017
Posts: 93

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:44
IMO A.
y+3=8+y+4−y
Let's consider the following ranges: 1. 0<y<4. All terms are positive and left is not equal to the right for any integer. 2. 8<y<0. The term 4y predominates and becomes positive. Thereby, not having any value of y which satisfies the equation. 3. y<8. Same case as above. 4. y>4. y+8 predominates and there is no integer that satisfies the equation.



Manager
Joined: 24 Jun 2019
Posts: 111

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:45
y+3=8+y+4−y
Using Trial and Error assume y = 0: 3 = 8+4..... Not Possible... so 0 is discarded
assume y=1: 4=9+3..... Not Possible.... using any positive number>1, LHS will be y+3 and RHS will be greater than y+8... so LHS will NEVER be equal to RHS....... So all positive numbers are discarded
assume y=1: 2=7+5... Not Possible assume y=2: 1=6+6.... Not Possible assume y=3: 0=5+7.... Not Possible.... till now RHS is always 12 and LHS is reducing by 1 assume y=4: 1=4+8.... Not Possible... now LHS will keep increasing by 1 assume y=5: 2=3+9... Not Possible... RHS still remains 12... it will remain 12 till y=8 assume y=8: 5=0+12... Not Possible assume y=9: 6=1+13.... Not possible... for every number after this, LHS will keep increasing by 1 and RHS will increase by 2... So LHS will never be equal to RHS....... All negative numbers are also discarded.
Ans: A  0 values will satisfy the equation
PS: Maybe there is a better way to solve this?



Manager
Joined: 28 Feb 2014
Posts: 171
Location: India
Concentration: General Management, International Business
GPA: 3.97
WE: Engineering (Education)

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:48
How many different values of y satisfy y+3=8+y+4−y ?
Lets consider different cases: Case1: y>0, as RHS has 8+y will always be greater than LHS y+3, so no value. Case 2: y=0; this is also not valid Case 3: y<0; y+3< 4y, no value exist. Hence, no value of y exists



Manager
Joined: 08 Jan 2018
Posts: 145
Location: India
Concentration: Operations, General Management
WE: Project Management (Manufacturing)

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:48
Refer image. Some use of number line & modulus. Ans is zero. A
Attachments
GOT1.jpg [ 131.52 KiB  Viewed 348 times ]



Director
Joined: 19 Oct 2018
Posts: 976
Location: India

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:51
y+3=8+y+4−y Case 1> y=< 8 y3=8y+4y y= 1 (Not possible)
Case 2 > 8<y=<3 y3=8+y+4y y=9 (not possible)
Case 3> 3<y=<4 y+3=8+y+4y y=9(not possible)
Case 4> y>4 y+3=8+y+y4 y=1 (not possible)
y can take 0 value
IMO A



Intern
Joined: 29 May 2019
Posts: 32

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:52
0 is the answer as none of the values of y satisfy the equation of absolute value when put individual values
Posted from my mobile device



Manager
Joined: 21 Jun 2018
Posts: 71

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 08:59
A Modulus can be broken with a + sign in front of it or a ve sign. so the possibilities of combinations we would have to be + + +,+ + ,+  ,+  +,  , + +, + ,  + . 8 possibilities. so the first one would be for eg y+3 = 8+y + 4y y=123 = 9. then we would enter this 9 into the equation to check if it works. 9+3 = 8 + 9 + 49 12= 17 5 12=12. Doing this for every possibility we will find only 2 values work. y=9 and y = 15
_________________
_______________________________________________________________________________________________________ Please give Kudos. Kudos encourage active discussions and help the community grow.



Intern
Joined: 10 Jan 2018
Posts: 15

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:01
Y will take 4 values only i.e 1, 1, 9, 15
Easy way is to draw this on number line and think about values of y for different range
I.e. y > 4 two mod will open with positive sign and one with negative , this will give ans Y= 1
Let's assume now y between 3 and 4 , here all mod will open with positive sign giving ans Y= 9
Now let's assume y between 8 and 3 here two mod will open with positive sign and one with negative giving ans as y= 15
Finally let's assume y less than 8 here two mod will open with negative sign giving ans as 1
Hope this explanation is right
Posted from my mobile device



Manager
Joined: 27 Mar 2018
Posts: 80
Location: India

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:10
Here the split numbers are= 8,3 and 4 For y<8, we'll have: (y3)=(8y)+(4+y) => y=9 For 8<y<3, (y3)=(8+y)+(4+y) => y=\(\frac{7}{3}\) For 3<y<4, (y+3)=(8+y)+(4+y) => y=1 And for y>4, (y+3)=(8+y)+(4y) => y=9 Clearly, there are only 3 unique solutions.
_________________
Thank you for the kudos. You are awesome!



Manager
Status: Victory is never a one time thing.
Joined: 14 Jan 2018
Posts: 57
Location: Oman
Concentration: Finance, Marketing
GMAT 1: 590 Q49 V21 GMAT 2: 650 Q47 V33
GPA: 3.8

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:10
Though I am sure there has to be a quicker way to approach this problem, I took the traditional (and sure) way of solving this problem.
Since there are are 3 modulus, 6 cases are possible. Solving the 6 cases, you will realise that y can take four values which are 9, 1, 7/3 and 15. However upon substituting each of these values one by one in the question, you will find that none of these values satisfy the modulus equation.
Hence option A is the correct answer.



Intern
Joined: 17 Aug 2016
Posts: 39

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:13
y + 3 = 8 + y + 4  y
(Y+3)= (8+y) + (4y)
The answer is zero, nothing will satisfy



Senior Manager
Joined: 18 Jan 2018
Posts: 307
Location: India
Concentration: General Management, Healthcare
GPA: 3.87
WE: Design (Manufacturing)

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:14
How many different values of y satisfy y+3=8+y+4−y
mod value of any number is always positive so value of y+3 ,8+y and 4y always positive >=0
Case 1 : y is positive if y is positive y+3 < 8+y  this is always true So no Y is possible Case 2 : y is negative when y is negative 4y will become greater than y+3
So no value of y satisfies the above the equation Option A is correct



Intern
Joined: 30 Aug 2018
Posts: 15

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:16
Okey What i have learned is that when you see a question just as above where there is one variable (y in above case) and more than one modulas then follow the critical point approach. Critical Point Approach: Step 1: equate each term of modulas with 0. In above case we will have three terms bcoz we have 3 modulas. a) y+3=0> i.e. y=3 b) 8+y=0> i.e. y=8 c) 4y=0> i.e. y=4
Step 2) Put the equated value in number line and it becomes _______8______3_____4______
Step 3) As u see we have 8,3,4 as determiners, now i) try putting value beyond 8 (say 9) on our y+3> 8+3=5 (remember u can put any negative value beyond 8 and when you add that value to y+3 you get negative value. Try putting the same picked value in 8+y> 8+(9)> 1 and you will again get negative value. Now putting same value (9) in 4y> u get 4(9)=13 which is positive in this case.
Now what you have to do is Put (1) just before the terms on which u got negative while doing above step and (1) on terms where you got positive value.
it becomes (1)(y+3)= (1)(8+y) +(1)(4y) solving this you will get y= 1 In the number line, you can see that we took 8 as limit and took 9 for the equation and while solving we got value as 1. The thing to consider here is that we had a limit of infinitive to 8 but we got ans as 1 and 1 does not lies in that infinitive to 8. So rule out this ans. Lets say if we had got value beyond 8 then we would have got the left side limiting value of our equation.
ii) in our i) we did put value beyond 8. Now in our second case we will put value which is between 8 and 3 bcoz we have case of beyond 8, between 8 and 3, between 3 and 4, and beyond 4.
Lets take (5) and we will repeat same steps and we will get y+3> 5+3=2 (ve) 8+y> 8+(5)=3 (+ve) 4y> 4(5)=9 (+ve) then our equation becomes as (1)(y+3)= (1)(8+y)+(1)(4y) y=15 which do not fall in between 8 and 3.
when you repeat the same steps for 3 to 4 and beyond 4. You will come to see that neither of them will have values which lie withing the limiting number line range.
So, when none of the equation has satisfied the limiting range, you have eliminated all. This way ans comes to 0.
(Note): If any two y= something have lied in the limiting number line lets say our beyond 8 term have given y=12 and lets say our 3 to 4 would have given y= 2,then you would have got the range of the equation and that range would have been 12<=y<=2. This means we would have (12,11,10,9,8,7,6,5,4,3,2,1,0,1,2)= 15 values of Y satisfying the modulas equation of our question. Hope this is perfectly understandable. And any queries are welcome even on PM.
Posted from my mobile device



Senior Manager
Joined: 13 Feb 2018
Posts: 450

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:17
We can observe that if y=0 y+3=8+y+4−y Does not hold true
y>0 y+3 will always be less than 8+y
y<0 y+3 will always be less than 4−y
SO we cannot find any value neither in negatives nor in positives to hold the equation true and y=0 is not a valid option as well
IMO Ans: A



Manager
Joined: 06 Jun 2019
Posts: 133

How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
Updated on: 03 Jul 2019, 10:13
\(y+3=8+y+4−y\) Let's first identify the intervals as Bunuel does: \(1.\) \(y<8\) \(2.\) \(8\leq{y}<3\) \(3.\) \(3\leq{y}\leq4\) \(4.\) \(y>4\) Let's solve the equality for each interval: 1. \(y<8\) for this interval we open each modulus with proper sign: \((y+3) = (8+y) + 4  y\) \(y = 1\) If we combine the intervals y = 1 and \(y<8\) then there is NO solution. 2. \(8\leq{y}<3\) for this interval we open each modulus with proper sign: \((y+3) = 8 + y + 4  y\) \(y = 15\) If we combine the intervals y = 15 and \(8\leq{y}<3\) then there is NO solution. 3. \(3\leq{y}\leq4\) for this interval we open each modulus with proper sign: \(y+3 = 8 + y + 4  y\) \(y = 9\) If we combine the intervals y = 9 and \(3\leq{y}\leq4\) then there is NO solution. 4. \(y>4\) for this interval we open each modulus with proper sign: \(y+3 = 8 + y  (4  y)\) \(y = 1\) If we combine the intervals y = 1 and \(y>4\) then there is NO solution. Hence A
_________________
Bruce Lee: “I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times.”GMAC: “I fear not the aspirant who has practiced 10,000 questions, but I fear the aspirant who has learnt the most out of every single question.”
Originally posted by JonShukhrat on 03 Jul 2019, 09:22.
Last edited by JonShukhrat on 03 Jul 2019, 10:13, edited 1 time in total.



Manager
Joined: 03 Oct 2012
Posts: 159
Location: India
Concentration: Entrepreneurship, Strategy
WE: Brand Management (Pharmaceuticals and Biotech)

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:36
When y>0: y+3=8+y+4−y y+3 = 8+y+4y y+3 = 12 y = 9 ................................ (I) When y<0: y+3 = 8+y+4y y3 = 8+y+4y y3 = 12 y = 15 y = 15 ..............................(II) Putting these 2 values of y in the equation and checking reveals that only y=9 holds true. y = 15 yield an equation which is not possible. The question asks How many different values of y satisfy the given equation? Answer: only 1 equation satisfies the equation. Therefore choice B is the correct answer.



Manager
Joined: 15 Jun 2019
Posts: 205

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 09:43
y+3=8+y +4−y. There are 3 key points here: 8, 3, 4. So we have 4 conditions: a) y<−8 gives −(y+3)=−(8+y)+(4−y)> y=−1. We reject the solution because our condition is not satisfied (1 is not less than 8) b) −8≤y<−3. −(y+3)=(8+y)+(4−y) > y=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.) c) for −3≤y<4 gives (y+3)=(8+y)+(4−y) > y=9. We reject the solution because our condition is not satisfied (9 is not within (3,4) interval.) d) for y≥4. gives (y+3)=(8+y)(4−y) > y=−1. We reject the solution because our condition is not satisfied (1 is not greater than 4) hence ans is 0 ie A
_________________
please do correct my mistakes that itself a big kudo for me,
thanks



Intern
Joined: 08 Nov 2016
Posts: 19
Location: India
GPA: 3.99
WE: Web Development (Computer Software)

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 10:03
How many different values of y satisfy y + 3 = 8 + y + 4  y ?
There can be 6 different combinations tried out to find out value of y.
For example  y+3 can be (y+3) or (y3) depending on whether y> 3 OR y<3 Exactly same way all other component can have 2 values each, and if you try it all possible combinations, there no value which satisfy this equation. So answer is choice A (0).



Intern
Joined: 06 Jan 2019
Posts: 2
Location: Thailand
GPA: 3.31

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 10:04
According to the absolute property,  y+3 can be answered in 2 ways y+3= y+3 = y3 so we can prove this equation by that value and the one and only value of y is 9
so B is the correct answer.



Manager
Joined: 13 Nov 2018
Posts: 118
Location: India

Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
Show Tags
03 Jul 2019, 10:04
Solving above equation we get
y=1, 9,15,7/3
Plugin the values in eqn only for y=9
it satisfies the eqn
SO only 1 values satisfies the eqn




Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
[#permalink]
03 Jul 2019, 10:04



Go to page
Previous
1 2 3 4
Next
[ 64 posts ]



