If the positive integer n is added to each of the integers 69, 94, and

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 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 94-69=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)

AbdurRakib

4^2 - 3^2 = 7
5^2 -4^2 = 9
6^2 -5^2 =11
100^2-99^2 = 199

The difference between the squares of consecutive integers always increases since a^2 -b^2 = (a+b)(a-b)

(a-b) 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 2b +1 =25 and b =12, a=13

However you do not need to do this for a DS problem. Its sufficient to know that the difference is unique

Hope it is clear!

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.

Interesting application.

Can you elaborate the highlighted Concept ?

Thanks

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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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

the BIG IDEA here:
The difference between squares of two consecutive integers = Sum of the two consecutive integers
eg: \(10^2 - 9^2 = (10+9)(10-9) = 19\) 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

This is typical DS !

Hope this note helps !

Answer: Option D

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ScottTargetTestPrep please could you help me understand what prompted you to subtract the first and second equations? I want to understand your thought process while attempting this question

We are trying to solve a system of equations consisting of 69 + n = x^2 and 94 + n = (x + 1)^2. We have two equations and two unknowns, so the natural thing to is to eliminate one of the variables. Subtracting the equations is one way of doing it, you could also write n = x^2 - 69 using the first equation and substitute this for n in the second equation:

94 + n = (x + 1)^2

94 + (x^2 - 69) = x^2 + 2x + 1

25 = 2x + 1

24 = 2x

12 = x

As you can see, we obtain the same result. So pretty much the only reason I subtracted the equations is so that one of the variables will be eliminated.
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n = I+ | what is n?

1. (94+n) - (69+n) = (x+1)^2 - x^2. The equation can be solved to get the ans. (Sufficient)
2. Same as 1. (Sufficient)
Ans D