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How many values of x satisfy the equation x79=11? [#permalink]
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06 Feb 2018, 23:49
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[GMAT math practice question] How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5
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How many values of x satisfy the equation x79=11? [#permalink]
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07 Feb 2018, 04:07
If x7 = a, the equation will become a  9 = 11 Squaring on both sides, \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\) Solving for a, a = 20 or 2 If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2. Therefore, there are 2 values of x which satisfy the equation (Option B)
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How many values of x satisfy the equation x79=11? [#permalink]
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07 Feb 2018, 07:14
pushpitkc wrote: If x7 = a, the equation will become a  9 = 11
Squaring on both sides, \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\) Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B) Hi pushpitkcNormally valus in modulus have two solutions (one is positive and another one is negative) one cant have 5 or 4 solutions No ? Here is my solution ok so we have \(x7 9= 11\) open modulus:) step one 1.) \(x79= 11\) (rewrite as it is) \(x = 27\) step two 2.) \( (1) x7 (1)9= 11\) here i mulptiply by 1 so i get \(x+7+9=11\) \(x= 13\) is my solution/approach correct ? I wonder how did you get this and why are you squaring both sides....no square roots, or powers are mentioned....just simple modulus in question stem .. please explain \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\)



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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07 Feb 2018, 08:03
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MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 \(x79=11\) \(=> x79=11\) or \(x79=11\) \(=>x7=20\) or \(x7=2\). as Mod cannot be negative, hence this equation is not valid so we have \(x7=20\) or \(x7=20\) \(=>x=27\) or \(x=13\). We have two solutions Option B



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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07 Feb 2018, 08:07
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dave13 wrote: pushpitkc wrote: If x7 = a, the equation will become a  9 = 11
Squaring on both sides, \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\) Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B) Hi pushpitkcNormally valus in modulus have two solutions (one is positive and another one is negative) one cant have 5 or 4 solutions No ? Here is my solution ok so we have \(x7 9= 11\) open modulus:) step one 1.) \(x79= 11\) (rewrite as it is) \(x = 27\) step two 2.) \( (1) x7 (1)9= 11\) here i mulptiply by 1 so i get \(x+7+9=11\)
\(x= 13\)is my solution/approach correct ? I wonder how did you get this and why are you squaring both sides....no square roots, or powers are mentioned....just simple modulus in question stem .. please explain \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\)Hi dave13, your method is incorrect. You have two methods to solve the question in this forum, so as a learning exercise, try to find out where you made mistake and what is the correct approach. Also check your ADDITION in the highlighted portion. how can you get 13 with this approach?



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How many values of x satisfy the equation x79=11? [#permalink]
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09 Feb 2018, 00:42
=> \(x79 = 11\) \(⇔ x79 = ±11\) \(⇔ x7 = 9 ±11\) \(⇔ x7 = 20\) or \(x7 = 2\) \(⇔ x7 = 20\) since \(x7 ≥ 0\) \(⇔ x7 = ±20\) \(⇔ x = 7±20\) \(⇔ x = 27\) or \(x = 13\) Therefore, the answer is B. Answer: B
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Re: How many values of x satisfy the equation x79=11? [#permalink]
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09 Feb 2018, 13:21
niks18 wrote: MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 \(x79=11\) \(=> x79=11\) or \(x79=11\) \(=>x7=20\) or \(x7=2\). as Mod cannot be negative, hence this equation is not valid so we have \(x7=20\) or \(x7=20\) \(=>x=27\) or \(x=13\). We have two solutions Option BHello niks18 :) Why did you remove ONLY one bracket after "9" \(x79=11\) \(=> x79=11\) or \(x79=11\) Shouldnt we open modulus as whole by removing all brackets :? ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... thegmat/ do like this CASE ONE: x is positive or ZERO :) \(x79=11\) open modulus / remove all brackets \(x79=11\) \(x = 11+7+9\) > \(x = 27\):) Really ? answering mysellf :) well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID :) CASE TWO: x is negative \(x79=11\) open modulus by PUTTING minus sign in front of FIRST MODULUS \(x79=11\) so we get \(x+7+9=11\) >\(x = 1179\) > \(x = 5\) \(x = 5\) ??? but i followed this rule ? whats wrong ? how should i understand \(x = 5\):?



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How many values of x satisfy the equation x79=11? [#permalink]
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09 Feb 2018, 15:01
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MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 To simplify brackets within brackets, let \(u\) stand for x  7 Then u  9 = 11 Case 1: u  9 = 11 u = 20 Case 2 u  9 = 11 u = 2 We now use the values just calculated and plug in u = x  7 x  7 = 20 x  7 = 2 An absolute value is nonnegative. x  7 cannot = 2. We are left with: x  7 = 20 Case 1 x  7 = 20 x = 27Case 2 x  7 = 20 x = 13Two solutions, 27 and 13. Answer B
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Re: How many values of x satisfy the equation x79=11? [#permalink]
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09 Feb 2018, 18:22
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dave13 wrote: niks18 wrote: MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 \(x79=11\) \(=> x79=11\) or \(x79=11\) \(=>x7=20\) or \(x7=2\). as Mod cannot be negative, hence this equation is not valid so we have \(x7=20\) or \(x7=20\) \(=>x=27\) or \(x=13\). We have two solutions Option BHello niks18 :) Why did you remove ONLY one bracket after "9" \(x79=11\) \(=> x79=11\) or \(x79=11\) Shouldnt we open modulus as whole by removing all brackets :? ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... thegmat/ do like this CASE ONE: x is positive or ZERO :) \(x79=11\) open modulus / remove all brackets \(x79=11\) \(x = 11+7+9\) > \(x = 27\):) Really ? answering mysellf :) well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID :) CASE TWO: x is negative \(x79=11\) open modulus by PUTTING minus sign in front of FIRST MODULUS \(x79=11\) so we get \(x+7+9=11\) >\(x = 1179\) > \(x = 5\) \(x = 5\) ??? but i followed this rule ? whats wrong ? how should i understand \(x = 5\):? Hi dave13There are TWO mod equations here and not one. you need to solve for both mod function onebyone. Opening the bracket doesn't mean you open all the bracket at once. Even for your simple simplification questions where you use BODMAS rule you solve for different bracket taking one bracket at a time i.e you solve for small bracket first and then for big bracket.



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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09 Feb 2018, 22:17
x79=11 This gives us 2 options , the final number can be 11 or 11 before taking absolute values. Before calculating X, one thing to consider is that x7 will always a positive value. So any positive value that yields 11 or 11 when subtracting 9 from it gives us the answer. The x7 is either 20 or 2. Based on which the X is either 27 or 13. Answer: B
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Re: How many values of x satisfy the equation x79=11? [#permalink]
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10 Feb 2018, 05:27
generis wrote: MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 To simplify brackets within brackets, let \(u\) stand for x  7 Then u  9 = 11 Case 1: u  9 = 11 u = 20 Case 2 u  9 = 11 u = 2 We now use the values just calculated and plug in u = x  7 x  7 = 20 x  7 = 2 An absolute value is nonnegative. x  7 cannot = 2. We are left with: x  7 = 20 Case 1 x  7 = 20 x = 27Case 2 x  7 = 20 x = 13Two solutions, 27 and 13. Answer B Greetings generis :) you are writing solution in sui generis style :) i thought x  7 has two numbers so replacing x  7 with \(u\) made it clear :) Just one question :) when you write case Case 2 \(x  7 = 20\) ( why didnt you write like this putting minus in front of x ? > \((x  7) = 20\) ? \(x+7 =20\) but in this case i get \(x = 13\) ? why ? And one more question ibidem you say x  7 = 2 absolute cant be negative but in case TWO you got negative value \(13\) why have a great weekend



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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17 Feb 2018, 04:53
MathRevolution wrote: =>
\(x79 = 11\) \(⇔ x79 = ±11\) \(⇔ x7 = 9 ±11\) \(⇔ x7 = 20\) or \(x7 = 2\) \(⇔ x7 = 20\) since \(x7 ≥ 0\) \(⇔ x7 = ±20\) \(⇔ x = 7±20\) \(⇔ x = 27\) or \(x = 13\)
Therefore, the answer is B.
Answer: B Why mod of X7 is greater tha equal to 0 {[m]x7 ≥ 0}?



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How many values of x satisfy the equation x79=11? [#permalink]
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17 Feb 2018, 07:36
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Salamander911 wrote: MathRevolution wrote: =>
\(x79 = 11\) \(⇔ x79 = ±11\) \(⇔ x7 = 9 ±11\) \(⇔ x7 = 20\) or \(x7 = 2\) \(⇔ x7 = 20\) since \(x7 ≥ 0\) \(⇔ x7 = ±20\) \(⇔ x = 7±20\) \(⇔ x = 27\) or \(x = 13\)
Therefore, the answer is B.
Answer: B Why mod of X7 is greater tha equal to 0 {\(x7 ≥ 0}? Hi Salamander911Mod means magnitude or distance from 0 on a number line. Distance is never negative, it is either positive or 0. Hence mod is never negative it is either positive or 0, that's why [m]x7≥0\)



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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18 Feb 2018, 00:50
Ix7I9=11 or Ix7I9=11 Ix7I=20 or Ix7I=2 x7=20 or x7=20 , Ix7I cannot take a negative value because modulus of a number is non negative x=27 or x=13
So x can have two values
So answer is B



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Re: How many values of x satisfy the equation x79=11? [#permalink]
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18 Feb 2018, 04:30
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pushpitkc wrote: If x7 = a, the equation will become a  9 = 11
Squaring on both sides, \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\) Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B) H pushpitkc its me again i have one question regarding this \(a^2  18a  40 = 0\) you say now we need to solve for a. did you solve it using discriminant ? or through expansion ? like this \(a(a18)40= 0\) i remember for example if equation were of this form \(5a^2  18a  40 = 0\) then i wouldnt be able to take \(a\) from the brackets  right ? and would need to solve it through discriminant thank you



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How many values of x satisfy the equation x79=11? [#permalink]
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18 Feb 2018, 04:43
dave13 wrote: pushpitkc wrote: If x7 = a, the equation will become a  9 = 11
Squaring on both sides, \(a^2  18a + 81 = 121\) \(a^2  18a  40 = 0\) Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B) H pushpitkc :) its me again i have one question regarding this \(a^2  18a  40 = 0\) you say now we need to solve for a. did you solve it using discriminant ? or through expansion ? like this \(a(a18)40= 0\) i remember for example if equation were of this form \(5a^2  18a  40 = 0\) then i wouldnt be able to take \(a\) from the brackets  right ? and would need to solve it through discriminant thank you :) Hey dave13In order to solve the equation a quadratic equation \(a^2  18a  40 = 0\) we need to understand that the product of the roots are 40 and sum is 18 A combination of numbers which gives us this is 20 and +2 Hence, the equation can be rewritten as \(a^2  20a + 2a  40 = 0\) \(a(a  20) + 2(a  20) = 0\) \((a + 2)(a  20) = 0\) > a = 2 or 20 You might want to read these links which I got from the Quantitative MegaThread http://www.purplemath.com/modules/factquad.htmhttp://www.purplemath.com/modules/solvquad.htmHope this helps you!
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Re: How many values of x satisfy the equation x79=11? [#permalink]
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18 Feb 2018, 21:16
dave13 wrote: niks18 wrote: MathRevolution wrote: [GMAT math practice question]
How many values of x satisfy the equation x79=11? A. 1 B. 2 C. 3 D. 4 E. 5 \(x79=11\) \(=> x79=11\) or \(x79=11\) \(=>x7=20\) or \(x7=2\). as Mod cannot be negative, hence this equation is not valid so we have \(x7=20\) or \(x7=20\) \(=>x=27\) or \(x=13\). We have two solutions Option BHello niks18 Why did you remove ONLY one bracket after "9" \(x79=11\) \(=> x79=11\) or \(x79=11\) Shouldnt we open modulus as whole by removing all brackets ? like here in the first sample question ? https://www.veritasprep.com/blog/2014/0 ... thegmat/ do like this CASE ONE: x is positive or ZERO \(x79=11\) open modulus / remove all brackets \(x79=11\) \(x = 11+7+9\) > \(x = 27\):) Really ? answering mysellf well yes, its valid because our initial condition was x is positive HENCE we get x= 27 VALID CASE TWO: x is negative \(x79=11\) open modulus by PUTTING minus sign in front of FIRST MODULUS \(x79=11\) so we get \(x+7+9=11\) >\(x = 1179\) > \(x = 5\) \(x = 5\) ??? but i followed this rule ? whats wrong ? how should i understand \(x = 5\):? We can't remove all brackets at once. We can remove brackets one by one with carefulness. If \(b ≥0\), \(xa = b ⇔ xa = ±b\) If \(c ≥0\), \(xa  b = c ⇔ xa  b = ±c ⇔ xa = b ±c ⇔ xa = ±(b ±c) ⇔ x = a ±(b ±c)\)
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