"the placement test for a college has scores that are normally distributed with a mean of 600 and a standard deviation of 60.if the college accepts only the top 1% of examinees, what is the cutoff score on the test for admission? "

Given that only the top 1% are admitted, the cutoff will be as follows:
the z-score is given  by:
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
P(x>X)=1-P(z<Z)=0.01
⇒P(z<Z)=1-0.01=0.99
thus:
the value of z that corresponds to 0.99 is 2.33
thus:
2.33=(x-600)/60
solving for x we get:
139.8=x-600
x=739.8
thus the cutoff was 739.8