I wouldn't apply backwards solution or number plugging for most of the DS questions. This may only confuse you, as majority of the DS questions would require you to define sufficiency and not to find value. However, here you need to find value and may go with plugging numbers too ...

Question implies 500*n + 500*(n*n)
st(1) 500*n=1/16 *(500*n +500*n^2)
16*500*n=500*n+500*n^2, cancel 500*n on both sides 15=n Sufficient
st(2) 500*n + 500*(n*n)=120,000
n+n^2=240 and (n+16)(n-15)=0 by using FOIL, n=15 Sufficient

d

I have not seen here a big room for backwards solution, rather than in st(2) after I got n=15 in st(1). I immediately found coefficients 15 and 16 to make 240 in FOIL and reduced time otherwise needed to be spent on Discriminant calc.

kishokbabu wrote:Mary persuaded n friends to donate $500 each to her
election campaign, and then each of these n friends
persuaded n more people to donate $500 each to
Mary's campaign. If no one donated more than once
and if there were no other donations, what was the
value of n ?

(1) The first n people donated 1/16 of the total
amount donated.
(2) The total amount donated was $120,000.

can anyone explain this problem with number plugging?

You don't need any math here..

You have one unknown and each option sets up an equation for 'n' here using which 'n' can be found.

Stop right there and chose D

also agree with Shankar, in st(1) you have reduced to linear eq. and done
in st(2) you have 500*n + 500*(n*n)=120,000 which is quadratics of type a^2+a-C=0

there's a small trick with quadratics when you know that you get one root +ve and the other is -ve. This trick is more understandable with FOIL application: coefficient for a is (root1+root2) and value C is (root1*root2). Coefficient of a is +ve and coefficient of C is -ve. Hence you may get only one root +ve and one root -ve to comply with the signs of a and C. Decide in st(2) right away - one +ve root (your only doubt that quadratics might have two values for n is gone) and mark D.