On question 38, it was supposed to be {n,u,m,b,e,r,s}. i forgot the s. Sorry. My bad. Thanks for the help. I agree, the approximations have been hard. I usually skip them. I didnt even work that far this time. Another question:
The product of the roots of (2x-1)(3x+2)(4x-3) = 0 is
The sum of the roots of (2x-1)(3x+2)(4x-3) = 0 is
What is the quickest way to multiply fractions like these:
Because the product of the roots is -d/a in ax^3+bx^2+cx+d=0, you get the d by multiplying all the constants in those factored ax+b forms, namely -1, 2 -3, and put them over the product of all of x's coefficients, 2, 3, 4, respectively. So the answer is (-1*2*-3)/(2*3*4)=1/4. Sum of the roots, well, it's better if you just add them, 1/2+(-2/3)+3/4=5/4-2/3, cross multiply and get 7/12. As to the multiplication of the big numbers, nah, I don't know what to do. Changing them to improper fractions is just as tedious as FOILing them.
Brad's shortcut spelt out is this: (a/b)+[b/(a+b)]=1 a^2/[b(a+b)]. Thr4, 2/3+3/5= 1 2^2/3*5=1 4/15 So take an example with bigger numbers, 7/11+11/18=1 49/198
O. Ok. I got ya. I took the District 1 numbersense test and i made a 227!!! 58/7. I guess its b cuz i have taken and looked over District 2. I thought District 1 was easier than 2. Sam, have you taken district 1 yet? If so, how did you feel about it?
Wow, congrats. 227 would be very competitive in 4A, and respectable in 5A. No, I have not yet. I'll soon tho, possibly tonight. I don't have a printer at home, so I'll just look at the screen and write down my answer, lol.
Thanks. I have been practicing really hard and i have "installed" many new tricks into my head. Hopefully i won't choke at regionals. Haha...i would never be able to look at the computer screen and write the answer. I wud only b able to work to like problem 12. lol. How is everyone preparing for the regionals numbersense test? I have taken the past two years UIL tests and Dr. Numsen tests. I also took a few TMSCA. I have a few questions off of the D1 numbersense test:
35) (2 x 3^4 + 5^6) / 7 =____ How do i do these types of questions? ans is 2
51) The product of the coefficients of (a+b)^5 is _____ I don't know what to do....obviously ans is 2500
60) 75^2 / 25^3 x 50^4 I' m not good at breaking things like this down but i am trying to improve. any suggestions on how to solve these types of questions?
25) 13 x 154
is there a trick for this? ans is 2002
This is off of Invite A 2007:
31) 13 x 385
Ans is 5005.
This is off of Invite B 2007
36) 539 x 13
ans is 7007. still don't see a trick. ehh
Thats all for now. Thanks for everyones help so far!
35 is a remainder problem, correct? What you do on these problems is take the remainders of the big numbers first before multiplying or adding everything together so that you can work with smaller but equivelant remainders.
Remainders
For the 2 --> 2/7 = 2
For the 3^4 --> do 81/7 = 4
for the 5^6 --> turn it into (5^3)^2. Now do 125/7 = 6. Square that to get 36, and then add the 8 you got from the previous numbers, and you should get 44. Take the remainder of this and you get 2.
Well. 13*77 = 1001. So... by that logic, you divide by 11, then 7, and then multiply by 1001.
So for 13*154, divide 154 by 11 to get 14, and then by 7 to get 2. Then multiply 2 by 1001 and you get 2002.
Then for 385*13, divide by 11 to get 35, then 7 to get 5. Multiply by 1001 and you get 5005.
539 * 13 becomes 49/7 = 7. Then 7*1001 = 7007.
You could memorize 1/13, 2/13, 3/13.... 12/13 of 1001 to make the process go faster like multipling by numbers that resemble fractions of 7.
and the other problem you have is matter of canceling.
75^2 *50^4 -------------- 25^3
So get rid of all your 25s, and you are left with 9*25^3*16 Which is 144*15625 = 2250000ish.
Oh and the other one you have to be able to generate pascals triangle in your head. On the fifth exponent for 2 variables, you get the following coefficients: 1,5,10,10,5,1. Multiplied together you get 2500. If you have something like (a+2b)^5 you'd do 2500*(2)^6.
