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(x+1)^2-x^2=91
2x+1=91
X+X+1=91
Sum=91

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Bunuel
The difference of the squares of two consecutive numbers is 91. What is the sum of these numbers?

(A) 0
(B) 46
(C) 54
(D) 91
(E) None of these


This is not a good question. The sum can also be -91. What if the consecutive numbers are -45 and -46?
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Bunuel
The difference of the squares of two consecutive numbers is 91. What is the sum of these numbers?

(A) 0
(B) 46
(C) 54
(D) 91
(E) None of these


This is not a good question. The sum can also be -91. What if the consecutive numbers are -45 and -46?


Yes, experts...what if they are -45 & -46...then sum becomes -91 and hence (E) NOTA...
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Bunuel
The difference of the squares of two consecutive numbers is 91. What is the sum of these numbers?

(A) 0
(B) 46
(C) 54
(D) 91
(E) None of these

Bunuel the question is wrong due to insufficient information. You said two consequences number. In this case it can be negative or positive both. If we take negative number as -45 and -46 then the answer is different. Please provide full information as negative or positive number.

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The difference of the squares of two consecutive numbers is 91. What is the sum of these numbers?

Let's call the two consecutive numbers "x" and "x + 1" because consecutive numbers differ by 1.

The square of the first number (x) is x^2, and the square of the second number (x + 1) is (x + 1)^2. According to the problem, the difference of the squares of these two consecutive numbers is 91. So, we can write the equation:

(x + 1)^2 - x^2 = 91

Now, let's simplify and solve for x:

(x + 1)^2 - x^2 = 91

Expand (x + 1)^2:

(x^2 + 2x + 1) - x^2 = 91

Now, simplify and combine like terms:

2x + 1 = 91

Subtract 1 from both sides:

2x = 91 - 1
2x = 90

Divide by 2:

x = 45

So, one of the consecutive numbers is 45, and the other is 45 + 1 = 46.

Now, you can find the sum of these two numbers:

Sum = 45 + 46 = 91

The sum of these two consecutive numbers is 91.

Hence D
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