SAT math problem

Test Preparation
1

<p>If you haven't taken CB SAT online Test #3 dont do this probem.
Wouldn’t want to ruin it for u.
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<p>Is there a way to do this problem quickly!
Any help appreciated.</p>

<p>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</p>

2

<p>1+30 =31
2+29=31
3+28=31
....
15+16=31</p>

<p>Sum is 15X31</p>

<p>31+60=91
32+59=91
33+58=91
....
45+46=91</p>

<p>Sum is 15x91</p>

<p>See the pattern here??</p>

<p>Difference is (15x91) - (15x31)
simplifies to 15*60 = 900</p>

<p>pretty simple algorithm --> sum of terms = number of terms/2*(1st term +nth term)</p>

3

<p>well,</p>

<p>the sum of (s) is 60+59+58+57+56....
the sum of (t) is 30+29+28+27+26....</p>

<p>BUT note that the difference of every couple is 30s (60-30, 59-29, 58-28, 57-27, etc)</p>

<p>so u have 30 digits every of which is 30 => 30x30=900</p>

<p>is 900 the answer ?</p>

4

<p>Below is a very useful formula to remember when doing math problem solving at any level.</p>

<p>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.</p>

<p>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</p>

5

<p>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</p>

<p>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</p>

6

<p>Brilliant!</p>

7

<p>s snack, my way is the same :)</p>

8

<p>^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.</p>

9

<p>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.</p>

10

<p>^ Sorry for being so vague, I was only referring to the evil asian dictator's post.</p>

11

<p>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 =/.</p>