skierdude1000

Question 1: (Bluebook, page 400, #20)

The least integer of a set of consecutive integers is -25. If the sum of these integers is 26, how many integers are in this set?

a) 25
b) 26
c) 50
d) 51
e) 52

I got the right answer, but I did it out the reallllly long way, and I'm sure there is a quick way to do this... which is why I'm asking. What I did was:

An=A1+(n-1)d
An= -25 + (n-1)
An= -26 +n

Then Sum formula

Sn=26=(n/2)(A1+An)
26=(n/2)(-25-26+n)
52=n(-51+n)
0=n^2 - 51n - 52

Then I used quadratic formula to solve for n... that is not the way to do these SAT problems, I think they should take 30 seconds max.



Question 2: (Bluebook, page 424, #6) )If x^2 + y^2 = 73 and xy=24, what is the value of (x+y)^2

a) 73
b) 97
c) 100
d) 121
e) 144

I did a really long and complicated way using the quadratic formula and got that y = 8 or 3, so whichever one it equals, the answer has to be d) 121. However, that way took me way to long as well, so any suggestions on how to do this quickly would be great.

Question 3: (Bluebook, page 426, #14)
If (a+b)^.5 = (a-b)^-.5, which of the following must be true

a) b = 0
b) a+b = 1
c) a-b = 1
d) a^2 + b^2 = 1
e) a^2 - b^2 = 1

Using guess and check/random logic I decided that the answer was A, which is wrong. How the heck do I do this?

Martin

For the second question I think is something like that:
(x+y)^2=x^2+2xy+y^2, but u have that x^2 + y^2 = 73 and xy=24, so
x^2+y^2+2xy=73+48=121, so D

towerpumpkin

For the first one, you know that the sum is 26. You also know that 25+(-25)=0, 24+(-24)=0 and so on. So that means that each of the negative numbers "cancels" out with the positive numbers. The sum of the integers from -25 to 25 is 0, so your set is from -25 to 26, which would be 52 integers.

towerpumpkin

Question 3:
(a+b)^.5 = (a-b)^-.5
square both sides
(a+b)=1/(a-b)
multiply both sides by (a-b)
(a+b)(a-b)=1
a^2-b^2=1
E

jehangircama

the first question should be 52. I did it mentally by saying -25 and
+25 cancel, -24and +24 and so on. That gives 50 numbers. Then there's 0 in the middle and +26 at the end. Therefore 52.

dave123

For the 2nd one, you should just try the factors of 24, it takes no moe than 20 seconds. You guys all messed up on the first one, its 51, the question asks for the total # of INTEGERS, and 0 is not an integer.

jehangircama

No i don't think so. All whole no.s are integers so 0 should be an integer.

jehangircama

"The integers consist of the positive natural numbers (1, 2, 3, …) the negative natural numbers (-1, -2, -3, ...) and the number zero. The set of all integers is usually denoted in mathematics by in blackboard bold, which stands for Zahlen (German for "numbers")."
www.positiveintegers.org/

Hecatonchires

Yeah, before you correcting people, make sure you know what you're talking about. If you've ever taken a basic math course, you would know that integers are all whole numbers.

dave123

Oh sorry about that. I thought I remembered seeing somewhere that 0 is not an integer. I'm sorry if I sounded harsh with my phrasing.

noxiousnirvana

there's another elegant way-

notice - a^2-b^2 =(a+b)(a-b)=1
so..(a+b)=1/(a-b)

therefore, in the original identity - (a+b)^5 =1/(a-b)^5, substitute (a+b) with 1/(a-b) and voila! you get (a+b)^5 = (a-b)^-5

so..the answer is E

Ragnarok227

Question 1:

-25 + ....

the sum is 26 so for the list to be positive it must exceed the sum of

-25 +..... + -1

thus, the minimum to become equal to 0 is

-25 + ..... + 0 + ..... + 25 = 0 (50 + the zero = 51)

next integer is 26, tada! the total is 52