ScottTargetTestPrep wrote:

carcass wrote:

If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11

We can let the first odd integer = x and the next odd integer = x + 2; thus, the reciprocals are 1/x and 1/(x + 2). Thus:

1/x + 1/(x+2) = 12/35

Multiplying by 35x(x+2), we have:

35(x + 2) + 35x = 12x(x + 2)

35x + 70 + 35x = 12x^2 + 24x

12x^2 - 46x - 70 = 0

6x^2 - 23x - 35 = 0

(6x + 7)(x - 5) = 0

Thus, x = -7/6 or x = 5.

Since x is an integer, x must be 5 and the greater integer is x + 2 = 7.

Alternate solution:

We are given that the sum of the reciprocals of two consecutive odd integers is 12/35. We see that the denominator is 35. It’s not difficult to conjecture that the integers have to be 5 and 7 since 5 x 7 = 35. Finally, we can check the sum of 1/5 and 1/7 to see if they sum to 12/35:

1/5 + 1/7 = 7/35 + 5/35 = 12/35

Since they do add up to 12/35, the larger integer is 7.

Answer: C

I used the alternative route for this one but I was wondering if you could elaborate on whether there are any tips/tricks to dealing with messy quadratics, those that have a constant (for my purposes, I'm referring to a constant >1) in front of the x^2 term, since it's less straight forward and more time-consuming to figure out the roots in those situations?

I get that for quadratics with no constant in front of the x^2 term, we can use the "sum" and "product" rules to quickly figure out the form of the two roots in our head but the "sum" rule doesn't hold when we have a constant in front of the x^2 term, only the "product" rule does....I know you could technically plug values into the quadratic equation to figure out the roots but that's not a viable gmat approach so just trying to figure out what I'm missing that is keeping me from quickly solving these sorts of quadratics such as 6x^2 - 23x - 35 = 0.