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How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 08:00
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How many different values of \(y\) satisfy \(y + 3 = 8 + y + 4  y\) ? A. 0 B. 1 C. 2 D. 3 E. 4
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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19 Jul 2019, 02:08
Bunuel wrote: How many different values of \(y\) satisfy \(y + 3 = 8 + y + 4  y\) ? A. 0 B. 1 C. 2 D. 3 E. 4
OFFICIAL EXPLANATION: FROM CRACK VERBAL: y + 3 is less than y + 8 and 4  y is never negative then the equation will be inconsistent. When y + 3 is greater than y + 8 then 4  y will be greater than y + 3 which will again make the equation inconsistent. Answer: 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, 08:38
Concept: Critical Point Approach Critical points are: y + 3 = 0; 8 + y = 0 & 4  y = 0 > y = 8,  3, 4 3 critical points > 4 cases as shown Case 1: y < 8 > l8 + yl = (8 + y); ly + 3l = (y + 3) & l4  yl = (4  y) > y + 3 = 8 + y + 4  y >  (y + 3) =  (8 + y) + (4  y) >  y  3 =  8  y + 4  y > y = 4 + 3 =  1 > y = 1; Not PossibleCase 2: 8 ≤ y < 3 > l8 + yl = (8 + y); ly + 3l = (y + 3) & l4  yl = (4  y) > y + 3 = 8 + y + 4  y >  (y + 3) = (8 + y) + (4  y) >  y  3 = 8 + y + 4  y >  y = 12 + 3 = 15 > y = 15; Not PossibleCase 3: 3 ≤ y ≤ 4 > l8 + yl = (8 + y); ly + 3l = (y + 3) & l4  yl = (4  y) > y + 3 = 8 + y + 4  y > (y + 3) = (8 + y) + (4  y) > y + 3 = 8 + y + 4  y > y = 12  3 = 9 > y = 9; Not PossibleCase 4: y > 4 > l8 + yl = (8 + y); ly + 3l = (y + 3) & l4  yl = (4  y) > y + 3 = 8 + y + 4  y > (y + 3) = (8 + y)  (4  y) > y + 3 = 8 + y  4 + y >  y = 4  3 = 1 > y = 1; Not PossibleIMO Option A Pls Hit Kudos if you like the solution
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Originally posted by Dillesh4096 on 03 Jul 2019, 08:35.
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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, 09:58
Ugh, what an ugly question.... Anyway, back to it. Let's first find intervals for each modulus. y + 3 If positive  y + 3, key point is y+3>=0 or y>=3, so we get y + 3=y+3 If negative  y + 3, key point is y+3<0 or y<3, so we get y + 3=(y+3)=y3 8 + y If positive  8 + y, key point is 8+y>=0 or y>=8, so we get 8 + y=8+y If negative  8 + y, key point is 8+y<0 or y<8, so we get 8 + y=(8+y)=y8 4  y If positive  4  y, key point is 4y>=0 or y<=4, so we get 4  y=4y If negative  4  y, key point is 4y<0 or y>4, so we get 4  y=(4y)=4+y So, we get \(1.\) \(y<8\) \(2.\) \(8\leq{y}<3\) \(3.\) \(3\leq{y}\leq4\) \(4.\) \(y>4\)
As we see no number falls between the above intervals, hence A
Originally posted by mira93 on 03 Jul 2019, 08:09.
Last edited by mira93 on 03 Jul 2019, 09:58, edited 2 times in total.



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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, 09:42
y+3 = 8+y + 4y y+3 = 8+y+4y y+3 = 12 We have four cases (1) LHS +ve ,RHS ve (2) LHS ve, RHS +ve (3) LHS +ve, RHS +ve (4) LHS ve ,RHS ve Case 3&4 and case 1&2 will give same value So take 1 &3 y+3= 12 —> y= 15 or y+3 =12 —> y=9. y=9/15
Answer C
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Originally posted by Staphyk on 03 Jul 2019, 08:09.
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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, 08:13
Start the solving by opening modulus one at a time.
1 Case : all positive
y = 9
2. first mod negative (LHS Neg) y3 = 8+y + 4y y = 15
3. rhs  one neg and one pos ; lhs pos
y+3 = 8y + 4y y = 3
4. rhs last neg
y+3 = 8+y  4+y y = 1
hence 4 different 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, 08:18
On the number line you have to check place three numbers to check conditions of these box values. numbers: 8,3,4
4 cases will be there: case 1: if x<8 all ve y3=8y4+y Solve for y . y=9;
Simirly for 8<y<3 y=7/3 3<y<4 y=9 and y>4.. y=1
therefore, three different values of y exist.



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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, 08:25
to find the possible values of y in \(y+3=8+y+4−y\), there are 4 intervals to examine:\(x≥4\) > y+3 = 8+y + y4 > \(y= 1\) > invalid because not within stated range \(4>x ≥3\) > y+3 = 8+y + 4y > \(y= 9\) > invalid because not within stated range \(3>x ≥8\) > y3 = 8+y + 4y > \(y= 15\) > invalid because not within stated range \(8 ≥x\) > y3 = 8y + 4y > \(y= 1\) > invalid because not within stated range so there is acceptable value upon validating, so 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, 08:26
Only one solution when y=9 (y+3=8+y+4y gives y=123=9) satisfies the equation. When all modulus is negative still y=9 (y3=8y4+y gives y=9; y=9). IMO B
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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, 08:27
How many different values of y satisfy y+3=8+y+4−y ?
A. 0 B. 1 C. 2 D. 3 E. 4
Answer: A
y+3 = 8+y+4y => y+8 y+3 + y4 = 0
range of y: y+8 y+3 y4 y >= 4 : + + + => y+8 y3+y4 = 0 => y = 1, but y can't be 1 because y has to be >=4 > no solution 3 <=y<4 : + +  => y+8 y 3 y +4 =0 => y = 9 , but y can't be 9 because y has to be < 4 > no solution 8 <=y<3: +   => y+8 +y +3 y +4 =0 => y = 15 , but y can't be 15 because y has to be >= 8 > no solution y < 8 :    => y+8 y3+y4 = 0 => y = 1, but y can't be 1 because y has to be < 8 > no solution



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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, 08:29
The intervals should be divided on the number line for 8, 3, 4. In none of the intervals, the equation holds for a particular value of y. So answer should be A Posted from my mobile device
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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, 08:33
solving the expression does not given any value to satisfy the given relation IMO A; 0
How many different values of yy satisfy y+3=8+y+4−yy+3=8+y+4−y ?
A. 0 B. 1 C. 2 D. 3 E. 4



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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, 08:34
How many different values of yy satisfy y+3=8+y+4−y A. 0 B. 1 C. 2 D. 3 E. 4 if y is positive then y+3 will always be less than 8+y i.e. y+3 can never be equal to 8+y+4−y If y is Negative then y+3 will always be less than 4−y i.e. y+3 can never be equal to 8+y+4−y i.e. NO SOLUTION Answer: Option 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, 08:35
PS Question (Algebra):
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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, 08:35
Zeros of the equation: y+3=0 8+y=0 4y=0
y=8; 3; 4
1st case: y<=8
y3=8y+4y >> y=1 (doesn't meet the criteria y<=8)
2nd case: 8<=y<=3 y3=8+y+4y y3=12 y=15 y=15 (doesn't meet the criteria 8<=y<=3)
3rd case: 3<=y<=4 y+3=8+y+4y y+3=12 y=9(doesn't meet the criteria 8<=y<=3)
4th case: y>=4 y+3=8+y4+y y=1 (doesn't meet the criteria y>=4)
No solution



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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, 08:35
Quote: y+3=8+y+4−yy+3=8+y+4−y
By substituting values in the equation, we realise in all cases LHS < RHS. for ve values 4y will always be more than y+3 for +ve value 8+y will always be more than y+3 Hence 0, Option 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, 08:36
y + 3 = 8 + y + 4  y
We will check for mod points at 3,8,4 if y>4 Then: \(y+3=8+y4+y\) y=1. Not possible since Y>4 3<y<4 Then: \(y+3=8+y+4y\) y=11. Not possible since 3<y<4 8<y<3 Then: \(y3=8+y+4y\), y=15.Not possible since 8<y<3 8>y \(y3=8y+4y\) y=1. Not possible since 8>y
zero solution. 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, 08:41
How many different values of y satisfy y+3=8+y+4−y? A. 0 B. 1 C. 2 D. 3 E. 4 Let us divide the number line into 4 different regions:  Region 1 where y>4 Region 2 where 3<y<=4 Region 3 where 8<y<=3 Region 4 where y<=8 Region 1 where y>4 expression becomes y+3 = y+8 + y4 => y = 1 Not possible in the region since y>4 Region 2 where 3<y<=4 y+3= y+8+4y => y=9 Not possible in the region since 3<y<=4 Region 3 where 8<y<=3 y3 = y+8 +4y => y = 15 Not possible in the region since 8<y<=3 Region 4 where y<=8 y3 = y8 +4y => y = 1 Not possible in the region since y<=8 Therefore, it may be concluded that no value of y satisfies the above equation. => Zero solutions (0) IMO 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, 08:44
Let us draw a number line: 8________3_________0_________4 Clearly the distance of any possible value >4 is farthest from 8, So it's not possible that distance between that number and 3 = distance between that number & 4 + distance between that number & 8 Same concept applies for all the other zones, <8, 8<y<4. So zero 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, 08:44
Step1 Identify the the positive and negative intervals against each absolute expression. Step2 Use the property of absolute function, x=x when x>0 and x=x when x<0 Now the intervals are : 1)y<8 2) 8<=y<3 3) 3<=y<=4 4) y>4 Now testing the expression at the intervals , we have No solution for y. Ans. (A)
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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