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If the positive integer n is added to each of the integers 69, 94, and
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16 Jun 2016, 05:41
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Re: If the positive integer n is added to each of the integers 69, 94, and
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08 Dec 2016, 09:55
Bunuel wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers (2) 94 + n and 121 + n are the squares of two consecutive integers We are given that the positive integer n is added to each of the integers 69, 94, and 121, and need to determine the value of n. Statement One Alone:69 + n and 94 + n are the squares of two consecutive integers. From statement one, we can say that for some positive integer x, 69 + n = x^2 and 94 + n = (x + 1)^2. Let’s subtract the first equation from the second equation: (94 + n)  (69 + n) = (x + 1)^2  x^2 25 = x^2 + 2x + 1  x^2 25 = 2x + 1 24 = 2x 12 = x Since we know x = 12, we can substitute this into the first equation to determine the value of n: 69 + n = 12^2 69 + n = 144 n = 75 Statement one alone is sufficient to answer the question. Eliminate answer choices B, C and E. Statement Two Alone:94 + n and 121 + n are the squares of two consecutive integers. We can use the same method that we used in statement one to solve for n. Therefore, without performing the actual calculations, we can conclude that we can find a unique value for n. Statement two alone is also sufficient to answer the question. Answer: D
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If the positive integer n is added to each of the integers 69, 94, and
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Updated on: 03 Aug 2016, 09:44
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n? (1) 69 + n and 94 + n are the squares of two consecutive integers Difference between the two squares is 25 since 9469=25. This difference is unique. For example 4^2  3^2 = 7 5^2 4^2 = 9 As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT (2) 94 + n and 121 + n are the squares of two consecutive integers Difference between the squares is 27. Again this difference is unique . SUFFICIENT. (For those wondering what n is ; n=75 and the consecutive integeres are 12, 13 & 14)
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Originally posted by rishi02 on 16 Jun 2016, 06:31.
Last edited by rishi02 on 03 Aug 2016, 09:44, edited 1 time in total.




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Re: If the positive integer n is added to each of the integers 69, 94, and
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17 Jun 2016, 23:56
Let x and y be 2 consecutive squares such that y>x. Then root(y) = root (x) + 1 Now let's look at the question. 1) 94 + n and 69 + n are consecutive sqaures x = 69 + n y = 94+ n root(y) = root (x) + 1 Squaring the above equation we get: y = × + 2root (×) +1 2root (×) = y  x  1 = 94 + n  69  n  1 = 24 Root (x) = 12 x= 144 n = 144  69 = 75 y= 94 + 75 = 169 Sufficient 2) 94 + n and 121 + n are consecutive squares. Sufficient. Can be proven the same way as case 1.
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Re: If the positive integer n is added to each of the integers 69, 94, and
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18 Jun 2016, 01:12
rishi02 wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers
Difference between the two squares is 25 since 9469=25. This difference is unique. For example 4^2  3^2 = 7 5^2 4^2 = 9
As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT
(2) 94 + n and 121 + n are the squares of two consecutive integers
Difference between the squares is 27. Again this difference is unique . SUFFICIENT.
(For those wondering what n is ; n=75 and the consecutive integeres are 12, 13 & 14) Interesting application. Can you elaborate the highlighted Concept ? Thanks
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Re: If the positive integer n is added to each of the integers 69, 94, and
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18 Jun 2016, 01:57
AbdurRakib4^2  3^2 = 7 5^2 4^2 = 9 6^2 5^2 =11 100^299^2 = 199 The difference between the squares of consecutive integers always increases since a^2 b^2 = (a+b)(ab) (ab) will always be 1 since consecutive integers so as the integers increase a + b will also increase What you can also figure out from this is that a+b = 25 for this problem Therefore 2a +1 =25 and a =12 However you do not need to do this for a DS problem. Its sufficient to know that the difference is unique Hope its clear !
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Re: If the positive integer n is added to each of the integers 69, 94, and
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18 Jun 2016, 07:41
Bunuel wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers (2) 94 + n and 121 + n are the squares of two consecutive integers Statement 1. Let x and (x+1) be two consecutive integers. Then we have: 69+n=x^2 and 94+n=(x+1)^2. Substitute (69+n) into second equation to get 25+x^2=x^2 + 2x + 1 ==> 2x=24 and x=12 Hence n=75 Sufficient Statement 2. The same as Statement 1. Sufficient
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Re: If the positive integer n is added to each of the integers 69, 94, and
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19 Jun 2016, 04:06
Bunuel wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers (2) 94 + n and 121 + n are the squares of two consecutive integers Statement 1: 69 + n and 94 + n are the squares of two consecutive integers The difference between the numbers = 94  69 = 25 Let us list down some of the perfect squares. Since 69 is near to 8^2, I will start from 8^2 64, 81, 100, 121, 144, 169, 196, 225. Difference between 169 and 144 = 25 Hence 94 + n = 169, and 69 + n = 144 n = 75 SUFFICIENT Statement 2: 94 + n and 121 + n are the squares of two consecutive integers Difference between the two = 121  94 = 27 Applying the same logic and writing the perfect squares. 100, 121, 144, 169, 196, 225 Hence the numbers are 196 and 169 121 + n = 196 and 94 + n = 169 n = 75 SUFFICIENT Correct Option: D



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Re: If the positive integer n is added to each of the integers 69, 94, and
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03 Aug 2016, 09:42
Good approach?
we know n>0 n is an integer
S1 (n69)(n94)=n*n*(n+1)*(n+1) (n69)(n94)=n^2*(n+1)*(n+1)
one variable, solved
same with S2
D



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Re: If the positive integer n is added to each of the integers 69, 94, and
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29 Oct 2016, 19:40
AbdurRakib wrote: rishi02 wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers
Difference between the two squares is 25 since 9469=25. This difference is unique. For example 4^2  3^2 = 7 5^2 4^2 = 9
As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT
(2) 94 + n and 121 + n are the squares of two consecutive integers
Difference between the squares is 27. Again this difference is unique . SUFFICIENT.
(For those wondering what n is ; n=75 and the consecutive integeres are 12, 13 & 14) Interesting application. Can you elaborate the highlighted Concept ? Thanks the BIG IDEA here: The difference between squares of two consecutive integers = Sum of the two consecutive integers eg: \(10^2  9^2 = (10+9)(109) = 19\) so on and so forth In Statement 1 we are told that (69+n) & (94+n) are the squares of two consecutive integers, So use the above idea: \((94+n)(69+n) = 25\) Since we know that the sum of the two consecutive integers is 25 & to find the individual consecutive integers: 25 = 2n+1 (since integers are consecutive) n = 12 & (n+1) = 13 Now that we have each individual integer: \(12^2 = (69+n)\) \(144 = 69 + n\) \(n = 75\) Same applies for statement 2
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Re: If the positive integer n is added to each of the integers 69, 94, and
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03 Jun 2017, 17:33
ScottTargetTestPrep wrote: Bunuel wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers (2) 94 + n and 121 + n are the squares of two consecutive integers We are given that the positive integer n is added to each of the integers 69, 94, and 121, and need to determine the value of n. Statement One Alone:69 + n and 94 + n are the squares of two consecutive integers. From statement one, we can say that for some positive integer x, 69 + n = x^2 and 94 + n = (x + 1)^2. Let’s subtract the first equation from the second equation: (94 + n)  (69 + n) = (x + 1)^2  x^2 25 = x^2 + 2x + 1  x^2 25 = 2x + 1 24 = 2x 12 = x Since we know x = 12, we can substitute this into the first equation to determine the value of n: 69 + n = 12^2 69 + n = 144 n = 75 Statement one alone is sufficient to answer the question. Eliminate answer choices B, C and E. Statement Two Alone:94 + n and 121 + n are the squares of two consecutive integers. We can use the same method that we used in statement one to solve for n. Therefore, without performing the actual calculations, we can conclude that we can find a unique value for n. Statement two alone is also sufficient to answer the question. Answer: D ScottTargetTestPrep, Could you please explain why did you subtract 69 + n = x^2 and 94 + n = (x + 1)^2 ? Thank you.



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Re: If the positive integer n is added to each of the integers 69, 94, and
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16 Aug 2018, 09:31
colorblind wrote: AbdurRakib wrote: rishi02 wrote: If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
(1) 69 + n and 94 + n are the squares of two consecutive integers
Difference between the two squares is 25 since 9469=25. This difference is unique. For example 4^2  3^2 = 7 5^2 4^2 = 9
As can be seen the difference goes on increasing and hence only one unique value is possible. SUFFICIENT
(2) 94 + n and 121 + n are the squares of two consecutive integers
Difference between the squares is 27. Again this difference is unique . SUFFICIENT.
(For those wondering what n is ; n=75 and the consecutive integeres are 12, 13 & 14) Interesting application. Can you elaborate the highlighted Concept ? Thanks the BIG IDEA here: The difference between squares of two consecutive integers = Sum of the two consecutive integers eg: \(10^2  9^2 = (10+9)(109) = 19\) so on and so forth In Statement 1 we are told that (69+n) & (94+n) are the squares of two consecutive integers, So use the above idea: \((94+n)(69+n) = 25\) Since we know that the sum of the two consecutive integers is 25 & to find the individual consecutive integers: 25 = 2n+1 (since integers are consecutive) n = 12 & (n+1) = 13 Now that we have each individual integer: \(12^2 = (69+n)\) \(144 = 69 + n\) \(n = 75\) Same applies for statement 2 Bunuel what does it mean when we write number in this way 25 = 2n+1 Does it mean a consecutive number ? but consecutive numbers are written in this form x, x+1, x+2, x+3, x+4 no ?
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Re: If the positive integer n is added to each of the integers 69, 94, and &nbs
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