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If x is an integer, What is the value of x+5+ x3?
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10 Jul 2019, 22:53
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If x is an integer, What is the value of x+5+ x3? \(A. x^2<16\) \(B. (x+1)^2<25\)
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"Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources:  Efficient LearningTele: +911140396815 Mobile : +919910661622 Email : kinshook.chaturvedi@gmail.com



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If x is an integer, What is the value of x+5+ x3?
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Updated on: 07 Aug 2019, 08:03
The range 5≤x≤3 will give a unique value for x+5+ x3
If x is outside this range, we can get multiple values
(1) \(x^2<16\)
x<4 or 4<x<4
Since x is an integer, the minimum value x can take is 3 and the maximum is 3
This lies within our favorable region 5≤x≤3
Sufficient
(2) \((x+1)^2<25\)
x+1<5 or 5<x+1<5 or 6<x<4
This means the minimum value x can take is 5 and the maximum is 3
Once again, this is within our favorable region 5≤x≤3
Sufficient
Answer is (D)
Originally posted by firas92 on 11 Jul 2019, 01:17.
Last edited by firas92 on 07 Aug 2019, 08:03, edited 1 time in total.



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Re: If x is an integer, What is the value of x+5+ x3?
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11 Jul 2019, 23:05
Bunuel I do not really follow the solution above. Some concepts seem to have been used without proper explanation. Could you please provide a solution to this that a beginner would understand? Thanks in advance.
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Re: If x is an integer, What is the value of x+5+ x3?
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13 Jul 2019, 16:45
Before we dive into this question, let's try and understand the properties of the modulus function 1. x+a has a domain (infinity, infinity) and range [0, infinity). 2. x+a can be rewritten as x(a) and is nothing but the distance of x from (a). 2. x+a divides the x axis into 3 regions: a. x < a b. x = a c. x > a Therefore, the range of the function x+5+ x3 can be divided into 3 segments: 1. x<5 2. 5≤x≤3 3. x>3 When x<5: 1. One approach would be taking 2 random values of x (x1 and x2) such that x<5 and checking the values of f(x) = x+5+ x3. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range. 2. Second approach would be to check out the graphs of the two functions, x+5 and x3 in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x. When x>3: 1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = x+5+ x3. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range. 2. Second approach would be to check out the graphs of the two functions, x+5 and x3 in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x. When 5≤x≤3: The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at 5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8. The same can be verified by trying random values of x. Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not. Statement 1: x^2<16 => x<4 => 4<x<4. Since x is an integer, this translates to 3≤x≤3. This lies in the favorable range. Therefore this is sufficient. Statement 2: (x+1)^2 <25 => x+1<25 => 5<x+1<5 => 6<x<4. Since x is an integer, this translates to 5≤x≤3. This lies in the favorable range. Therefore this is sufficient. Therefore, the answer is (D). Posted from my mobile device
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If x is an integer, What is the value of x+5+ x3?
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26 Jul 2019, 11:52
Kinshook wrote: If x is an integer, What is the value of x+5+ x3?
\(A. x^2<16\) \(B. (x+1)^2<25\) given: x=integer, x+5+x3=? (1) \(x^2<16\): then x<4,…4>x>4={3,2,1,0,1,2,3}, when you replace any of these values in the equation, you get the same result, 8; sufficient. (2) \((x+1)^2<25\): then, x+1<5,…6<x<4={5,4…}, when you replace any of these values in the equation, you get the same result, 8; sufficient. Answer (D). Hey Bunuel or chetan2u, instead of testing all the values in the range, isn't there some way better?



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Re: If x is an integer, What is the value of x+5+ x3?
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07 Aug 2019, 08:08
Can you explain why this happened wizardofoddzSince x is an integer, this translates to 3≤x≤3 Why suddenly range changed from 4 to 4 to 3 to 3? wizardofoddz wrote: Before we dive into this question, let's try and understand the properties of the modulus function
1. x+a has a domain (infinity, infinity) and range [0, infinity). 2. x+a can be rewritten as x(a) and is nothing but the distance of x from (a). 2. x+a divides the x axis into 3 regions: a. x < a b. x = a c. x > a
Therefore, the range of the function x+5+ x3 can be divided into 3 segments: 1. x<5 2. 5≤x≤3 3. x>3
When x<5:
1. One approach would be taking 2 random values of x (x1 and x2) such that x<5 and checking the values of f(x) = x+5+ x3. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range. 2. Second approach would be to check out the graphs of the two functions, x+5 and x3 in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x.
When x>3:
1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = x+5+ x3. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range. 2. Second approach would be to check out the graphs of the two functions, x+5 and x3 in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x.
When 5≤x≤3:
The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at 5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8. The same can be verified by trying random values of x.
Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not.
Statement 1: x^2<16
=> x<4 => 4<x<4. Since x is an integer, this translates to 3≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Statement 2: (x+1)^2 <25
=> x+1<25 => 5<x+1<5 => 6<x<4. Since x is an integer, this translates to 5≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Therefore, the answer is (D).
Posted from my mobile device



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If x is an integer, What is the value of x+5+ x3?
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07 Aug 2019, 11:08
x+5+ x3 = (x+5)  (x3) = 8 > if both x+5>0 and x3<0 are satisfied. That gives us a combined range 5≤x≤3. If x is outside that range, we can get many x value.
(1) x^2<16 4<x<4 where x is n integer. x is within the abovementioned range 5≤x≤3. Thus, the equation give a singular result, that is a constant 8. SUFFICIENT
(2) (x+1)^2<25 5<x+1<5 or 6<x<4 where x is an integer. In other word, x is within the abovementioned range 5≤x≤3. Thus, the equation gives a singular solution, that is a constant 8. SUFFICIENT
Answer is (D)




If x is an integer, What is the value of x+5+ x3?
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07 Aug 2019, 11:08






