Of the 645 speckled trout in a certain fishery that contains

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pm; Followed by:23 members

Of the 645 speckled trout in a certain fishery that contains

by BTGmoderatorDC » Sun Oct 07, 2018 2:08 am

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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

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Source: Beat The GMAT — Problem Solving | Sun Oct 07, 2018 2:08 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pm; Thanked: 59 times; Followed by:33 members

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)!

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

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

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

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

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