for the

indolent wrote:

1) All of the students of Music High School are in the band, the orchestra, or both.80% of the students are in only one group.There are 119 students in the band.If 50% of the students are in the band only, how many students are in the orchestra only?

A)30 B)51 C)60 D)85 E)119

OA-(B)

2) Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females.If the ratio of females speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there?

A)192 B)195 C)200 D)205 E)208

OA- (D)

First question

let's suppose X is the no of total students

so no of students who are in one group only = 0.8X

and no of students who are only band = 0.5x

so, no of students who are orchestra only = 0.8x-0.5x=0.3x

now the no of students who are in both groups = x - 0.8x = 0.2x

so, no of students who are in band = 0.5x+0.2x

=>119 = 0.5x + 0.2x

=>119 = 0.7x

=>x = 170

no of students in orchestra only = 0.3 x = 170 * 0.3 = 51

so answer is B

For the Second question, I m getting following answer

lets suppose trouts are

r(m) are male rainbow trouts and r(f) are female rainbow trouts

s(m) are male rainbow trouts and s(f) are female rainbow trouts

since, of 645 speckled trout

s(m)+s(f) = 645,

the number of males is 45 more than twice the number of females

so s(m) = 2s(f)+45 =>3s(f)+45 = 645 => S(f) = 200

means s(f) is 200 and s(m) is 445

ratio of females speckled trout to male rainbow trout is 4:3

so, s(f)/r(m) = 4/3=> 200/r(m) = 4/3=>r(m) = 150

and the ratio of male rainbow trout to all trout is 3:20

so, r(m)/alltrouts = 3/20 => 150/alltrouts = 3/20 => all trouts = 1000

now no of all trouts = s(f) + s(m) + r(m)+r(f)

1000 = 200+445+150+r(f)=>r(f) = 1000 - 795 => 205

hence no of female rainbow trouts is 205 D

_________________

My debrief: done-and-dusted-730-q49-v40