question

30, 34, 43, 59, 84, 120,?

option 1) 169option 2) 148option 3) 153option 4) 176option 5) none of these
120
answer: `1/221`
solution: n/a 1, 2, 109 and 218 answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a 31, 87, 32, 21 a - ii, b - iv, c - i, d - iii Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3 finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). 1 a c d
2 a
3 c
4 b
5 c
i got a 100% you should have three choices.

 thanks for the update...i'll check it out. you can pick 1 senior in 5 ways, 1 junior in 4 ways, 1 sophomore in 3 ways and i freshman in 2 ways.
in all you can make these selections i 5 * 4 * 3 * 2 ways = 120 ways.
 
please note: selecting a committee is a "combination" problem, not a "permutation".