PDA

View Full Version : One more ETS question

neokeynesian
11-14-2004, 08:18 AM
3)

Col A

The sum of all the integers from 19 to 59, inclusive

Col B

The sum of all the integers from 22 to 60, inclusive

What is the best way to approach this type of question?

thebullfighter
11-14-2004, 10:56 AM
3)

Col A

The sum of all the integers from 19 to 59, inclusive

Col B

The sum of all the integers from 22 to 60, inclusive

What is the best way to approach this type of question?
either subtract 18 from 19 & 59 and then use sum of frist n natural numbers formulae & add 41*19 to it. (there are 41 numbers in all.59-19=40, 40+1 as inclusive.)
similarly for cal. B. subtract 21 frm 22 & 60. find sum, add 39*21

or better, use sum of A.P (arth. progression), with d=1 & sum= n/2(a+l), where a-1st term ie 19 & l-last term, ie 59

so, Col. A is 41/2 (59+19) = 41*39
Col. B: 39/2 (22+60) = 39*40

so, A=B

HTH

krakow
11-14-2004, 10:57 AM
just a formula..?

a_1+..+a_n = ((a_1 + a_n) *n ) /2

19+..+59; n=59-19 + 1 = 41
19+..+59=(19+59)*41/2 = 41*39 = 1599

22+..+60; n=39
22+..+60=(22+60)*39/2 = 41*39

so, A=B

sebman
11-14-2004, 02:43 PM
There still remains this method
sum of numbers)/(total numbers)=average
implies sum of numbers = total numbers*average.

Average=(19+59)/2 = 39
total numbers=59-19+1=41
sum of numbers=(41)(39)

Average=(22+62)/2=41
Total numbers=62-22+1=39
Sum of numbers=(41)(39)
And as u can c by comparism the two columns are the ame.

I hope this helps and it serves as a short cut when u come across similar question in the future.[dance]

thebullfighter
11-14-2004, 07:33 PM
There still remains this method
sum of numbers)/(total numbers)=average
implies sum of numbers = total numbers*average.

Average=(19+59)/2 = 39
total numbers=59-19+1=41
sum of numbers=(41)(39)

Average=(22+62)/2=41
Total numbers=62-22+1=39
Sum of numbers=(41)(39)
And as u can c by comparism the two columns are the ame.

I hope this helps and it serves as a short cut when u come across similar question in the future.[dance]
u did just th same thing. infact even krakow did th same thing."((a_1 + a_n) *n ) /2"

either subtract 18 from 19 & 59 and then use sum of frist n natural numbers formulae & add 41*19 to it. (there are 41 numbers in all.59-19=40, 40+1 as inclusive.)
similarly for cal. B. subtract 21 frm 22 & 60. find sum, add 39*21

or better, use sum of A.P (arth. progression), with d=1 & sum= n/2(a+l), where a-1st term ie 19 & l-last term, ie 59

so, Col. A is 41/2 (59+19) = 41*39
Col. B: 39/2 (22+60) = 39*40

so, A=B

HTH
all are just th same, it's th formulae for sum of AP.

u used [(a+l)/2]*n, kwakow used "((a_1 + a_n) *n ) /2", & i hd used (a+l)*n/2.

all are same. :)

HTH[dance]

777
11-14-2004, 07:36 PM
This is an interesting question. I solved it this way--

First compare the two numbers.

19 - 59 = 19, 20, 21...59

22 - 60 = 22, 23, 24....60

Columns A & B contain duplicate numbers that you should just negate. They are 22-59. Now compare

Col. A -- 19 + 20 + 21 = 60

Col. B -- 60

So they're both equal

krakow
11-14-2004, 08:35 PM
u did just th same thing. infact even krakow did th same thing."((a_1 + a_n) *n ) /2"
HTH[dance]
look at my and your posts' time, I sent it one minute later; I didn't see your reply before.. ;)

hinanshaikh
12-06-2004, 07:17 PM
find the sum of the first 19 digits and then the sum of the first 59 and then subtract to find the sum of the numbers in between and then add 19 because both are included.

n*(n+1) /2 is the sum of the first n digits.

do the same with the second part

for both u get 1599, thus A = B

vineetgsvm
12-06-2004, 11:37 PM
i think we shouldn't take these formulae personally. well till n till u can get the correct answer by ne means it's ok. afterall ets won't crosscheck the scratch sheet for the rough work.

hinanshaikh
12-07-2004, 06:12 PM
yeaa i know
but its always handy to know a few rather than depend on ure intuition and brainstorm over a qs!!

777
12-07-2004, 07:54 PM
I only spent 15 secs on the problems bc the solution just jumped out at me. My goal is to score high on the GRE not write a math thesis. If I had a choice of plugging in the formula and doing the math (1 min.?) I'd rather solve it my way bc its the quickest way possible. I gain time to spare on the harder more time consuming problems even when I know the formula for the question. What do you think is the point of multiple choice?

hinanshaikh
12-07-2004, 08:24 PM
i know...but not eveyrne is good at the same kind of problems...so everyone to suit themeselves ... u do what ure comfortable with and the way u can save time..

byeee

777
12-07-2004, 08:26 PM
BTW, I've tried both ways, and frankly I've found that the only way to complete the exam on time is to skip these time-consuming steps.