BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Of the 645 speckled trout in a certain fishery that contains

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members
by BTGmoderatorDC » Sun Oct 07, 2018 2:08 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

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 female 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

Source: Manhattan Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

Of the 645 speckled trout in a certain fishery that contains

by fskilnik@GMATH » Sun Oct 07, 2018 7:43 am
BTGmoderatorDC wrote: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 female 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
Source: Manhattan Prep
Excellent opportunity for the 'grid' (a.k.a double-matrix, table, you-name-it) and the 'k technique' (working together)!

Image

$$? = \left( {20k - 645} \right) - 3k$$
$$\left( {45 + 8k} \right) + 4k = 645\,\,\,\,\, \Rightarrow \,\,\,\,\,k = 50$$
$$? = \underleftrightarrow {\left( {1000 - 645} \right) - 150} = 205$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8151
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Thu Apr 18, 2019 5:43 pm
BTGmoderatorDC wrote: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 female 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

Source: Manhattan Prep
We can let m = male speckled trout, n = male rainbow trout, f = female speckled trout and g = female rainbow trout. Therefore, we have:

m + f = 645

m = 2f + 45

f/n = 4/3

And

n/(645 + n + g) = 3/20

Substitute 2f + 45 for m in the first equation, we have:

2f + 45 + f = 645

3f = 600

f = 200

Substitute 200 for f in the third equation, we have:

200/n = 4/3

4n = 600

n = 150

Finally, substitute 150 for n in the fourth equation, we have:

150/(645 + 150 + g) = 3/20

150/(795 + g) = 3/20

3000 = 2385 + 3g

615 = 3g

205 = g

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1