If I have an equation and an inequality - normally I would insert the equation into the inequality. Like this for example:
x>y
y=10
hence x>10
However, take a look at this problem here:
Some sequence is such that Sn = 1/n - 1/(n+1) for all integers n=>1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
(1) k>10
(2) k<10
Here is how I solved this:
1) I took some primes and played around with Sn = 1/n - 1/(n+1) to figure out that the sum of any consecutive integers will be 1/first less 1/last.
2) I took a look at the first statement, k>10. Ok let's try k=10 this gave me 9/10. The question asks if the sum is greater than 9/10, so I must figure out which way the inequality points. Because I didn't know how to do it, I simply took the next number, i.e. 11 and got the answer 10/11, which is bigger, so YES. Also this means SUFFICIENT.
3) Statement 2 obviously was NOT suff, since it could go all the way to 0 and from my above experimentation I already figured out which way the inequality points.
My question:
I have seen MGMAT write down fractions in this form: something over greater than something. So for example 2/>4. So I was wondering - is there anyway I could have build a similar equation with 10 when I was trying to figure out which way the sign of the inequality was pointing? In other words, how could I have shortened / eliminated the need to do both 10 and then 11 to figure out where the inequality points?
Am I completely offtrack here? Is there a much quicker solution I am missing?
Took me 2:44![]()
x>y
y=10
hence x>10
However, take a look at this problem here:
Some sequence is such that Sn = 1/n - 1/(n+1) for all integers n=>1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
(1) k>10
(2) k<10
Here is how I solved this:
1) I took some primes and played around with Sn = 1/n - 1/(n+1) to figure out that the sum of any consecutive integers will be 1/first less 1/last.
2) I took a look at the first statement, k>10. Ok let's try k=10 this gave me 9/10. The question asks if the sum is greater than 9/10, so I must figure out which way the inequality points. Because I didn't know how to do it, I simply took the next number, i.e. 11 and got the answer 10/11, which is bigger, so YES. Also this means SUFFICIENT.
3) Statement 2 obviously was NOT suff, since it could go all the way to 0 and from my above experimentation I already figured out which way the inequality points.
My question:
I have seen MGMAT write down fractions in this form: something over greater than something. So for example 2/>4. So I was wondering - is there anyway I could have build a similar equation with 10 when I was trying to figure out which way the sign of the inequality was pointing? In other words, how could I have shortened / eliminated the need to do both 10 and then 11 to figure out where the inequality points?
Am I completely offtrack here? Is there a much quicker solution I am missing?
Took me 2:44