IMO D...both the stmts can be used to get the answer..

stmt1 : 210 = 1000((1+r/100)^2 - 1)

--> (1+r/100)^2 = 121/100

--> 1+r/100 = 11/10

--> r = 10 > 8 --suff

stmt 2

(1+r/100)^2 > 1.15

1.08^2 = 1.1464

here 1+r/100 > (1.15)^1/2

1.15^1/2 will be greater than 1.08

hence r will be greater than 8

suff

ans d

I thought D as well, but why is he saying A is the answer

i will stand by my answer...because i see no mistake in my approach and calculations {thank god!! i am improoving else i am doing 30* 3 = 30 these days }

2010gmat wrote:IMO D...both the stmts can be used to get the answer..

stmt1 : 210 = 1000((1+r/100)^2 - 1)

--> (1+r/100)^2 = 121/100

--> 1+r/100 = 11/10

--> r = 10 > 8 --suff

stmt 2

(1+r/100)^2 > 1.15

1.08^2 = 1.1464

here 1+r/100 > (1.15)^1/2

1.15^1/2 will be greater than 1.08

hence r will be greater than 8

suff

ans d

The statement highlighted in bold is not true . (1.08 )^ 2 = 1.166 whcih is greater than 1.15 . So rate could be < 8 or > 8 .. statement 2 is not sufficient..

Might be a little longer to solve, but to check, I used r = 7.5%, so 1.075^2 = 1.155, which is greater than 1.15, so it disproves statement 2.

rohan_vus wrote:

2010gmat wrote:IMO D...both the stmts can be used to get the answer..

stmt1 : 210 = 1000((1+r/100)^2 - 1)

--> (1+r/100)^2 = 121/100

--> 1+r/100 = 11/10

--> r = 10 > 8 --suff

stmt 2

(1+r/100)^2 > 1.15

1.08^2 = 1.1464
here 1+r/100 > (1.15)^1/2

1.15^1/2 will be greater than 1.08

hence r will be greater than 8

suff

ans d

The statement highlighted in bold is not true . (1.08 )^ 2 = 1.166 whcih is greater than 1.15 . So rate could be < 8 or > 8 .. statement 2 is not sufficient..

2010gmat wrote:i will stand by my answer...because i see no mistake in my approach and calculations {thank god!! i am improoving else i am doing 30* 3 = 30 these days }

god help me...

I finally got it. Statement 2 is too difficult to solve for r without a calculator, so just plug in 8.1 and 7.9 for r. (Pain in the butt, but solveable) They both make the statement true, but give different answers to the question, making statement 2 insufficient.

2010gmat: just use the calculator and try the 2 values I suggested.

no need to try 2 values...trying 8 will solve the purpose...

i had used calc only to calculate 1.08*1.08

and then wrote 1.1664 as 1.464 dont know why i am screwing up when it comes to calculations

The (single) equation in the question involves three unknown values: I, n, and r. Therefore, if a statment provides us with values for two of these unknowns, we will be able to compute the value of the third.

Because statement one provides us with two values (I=210, and n=2), we can compute r, and so would be able to determine whether or not it is greater than 8. Statement 1 is sufficient.

Remember, in data sufficiency, we always want to refrain from actually doing the math whenever possible. One of the best tactics in data sufficiency involves simply counting the number of unknowns, and then counting the number of distinct linear equations. Say you have two unknowns; then you require two different equations to solve for the value of either of the two unknowns. There is, of course, a caveat to the tactic. If you need to solve for a relationship between n unknowns (x/y or x-y, etc), then you don't need n equations (typically, however, you will need n-1 equations). A special equation (x-y=5) that relates the two unknowns properly will be sufficient.

For statement 2:
(1 +r/100)^2 > 1.15
1+r/100 > sqrt1.15
1+ r/100 > sqrt 115/100

Because 10^2 =100 and b/c 11^2 = 121, sqrt 115 is approximately 10.7. (And our approximation should be 10.7, not 10.8 b/c we are squaring). Therefore:

1 + r/100 > 10.7/10
r/100 > 10.7/10 - 10/10
r/100 > 0.7/10
r/100 > 0.07
r > 7 (approximately, would actually be just a touch bigger than 7)

Knowing that r is greater than approximately 7 is insufficient information for the purpose of concluding whether r is greater than 8.

Statement 1 is sufficient; statement 2 is not sufficient. The answer should be A.

Note: 10^2 =100, and 11^2 = 121. On the number line, 115 is about three times as far away from 100 than it is from 121. So approximating sqrt 115 as 10.75 (the number that is three times as far away from 10 as it is from 11) would also have been a great strategy that would lead to a correct judgment about the second statement's sufficicency.