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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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?



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 08:48
Refer image. Some use of number line & modulus. Ans is zero. A
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.



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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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
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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.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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).



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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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 ?
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