If you haven't taken CB SAT online Test #3 dont do this probem.
Wouldnt want to ruin it for u.
.
.
.
.
.
.
.
.
Is there a way to do this problem quickly!
Any help appreciated.
Online Test #3 sec 9, Q 16
If (s) denotes the sum of the integers from 1 to 30 inclusive, and (t) denotes the sum of the integers from 31 to 60 inclusive, what is the value of (t - s)?
(A) 30
(B) 31
(C) 180
(D) 450
(E) 900
Below is a very useful formula to remember when doing math problem solving at any level.
sum of terms from 1 to n inclusive=n(n+1)/2...Remember, this algorithm only works when you are STARTING summing from 1 and continue to a certain number n without skipping.
For, this SAT problem...
Sum from 1-30:30(30+1)/2=465(s)
Sum from 1-60:60(60+1)/2=1,830...
1,830-465=1,365-sum from 31-60(t);
(t-s)=(1,365-465)=900
I did it differently...and I think my way is easier.
Line up the numbers...start with only the first and last of each series.
1.....30
31...60
You'll notice that each number from t is 30 greater than the corresponding term from s. So for any term number x: T(x)-S(x)=30
And since there are 30 terms...the answer is 30x30=900
^Sorry about that...I just read the one were they used that formula...but I'm glad someone else can do these problems with logic like I do rather than relying on silly little formulas.
Yeah, the method I used didn't directly depend on any formula. I was just demonstrating the reasoning I used to find the answer quickly without having to remember any silly formula...Your method is more efficient though.
well...i actually thought of evil asian dictator's SAT answer first. i got that formula from calculus and i remember having to do equations using someting like that. except sigma "i" is replaced by n(n+1)/2 and sigma "i^2" is n(n+1)(2n+1)/6... so basically its not that hard to use a formula you have learned or seen before. also the pattern is also a very useful way to do this. sometimes i get mixed up with them for some reason =/.
Replies to: SAT math problem
2+29=31
3+28=31
....
15+16=31
Sum is 15X31
31+60=91
32+59=91
33+58=91
....
45+46=91
Sum is 15x91
See the pattern here??
Difference is (15x91) - (15x31)
simplifies to 15*60 = 900
pretty simple algorithm --> sum of terms = number of terms/2*(1st term +nth term)
the sum of (s) is 60+59+58+57+56....
the sum of (t) is 30+29+28+27+26....
BUT note that the difference of every couple is 30s (60-30, 59-29, 58-28, 57-27, etc)
so u have 30 digits every of which is 30 => 30x30=900
is 900 the answer ?
sum of terms from 1 to n inclusive=n(n+1)/2...Remember, this algorithm only works when you are STARTING summing from 1 and continue to a certain number n without skipping.
For, this SAT problem...
Sum from 1-30:30(30+1)/2=465(s)
Sum from 1-60:60(60+1)/2=1,830...
1,830-465=1,365-sum from 31-60(t);
(t-s)=(1,365-465)=900
Line up the numbers...start with only the first and last of each series.
1.....30
31...60
You'll notice that each number from t is 30 greater than the corresponding term from s. So for any term number x: T(x)-S(x)=30
And since there are 30 terms...the answer is 30x30=900