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# Of the 645 speckled trout in a certain fishery that contains

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Director
Joined: 12 Jun 2006
Posts: 516
Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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Updated on: 22 Sep 2013, 22:58
2
20
00:00

Difficulty:

55% (hard)

Question Stats:

74% (02:34) correct 26% (03:14) wrong based on 559 sessions

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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

Originally posted by ggarr on 20 May 2007, 13:59.
Last edited by Bunuel on 22 Sep 2013, 22:58, edited 1 time in total.
Edited the question, added the OA and the options.
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Joined: 02 Sep 2009
Posts: 51122
Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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22 Sep 2013, 23:15
7
5
ggarr 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

{All} = {speckled} + {rainbow}

The ratio of male rainbow to all is 3x:20x:
20x = {speckled} + {female rainbow} + 3x

Since given that there are 645 speckled, then:
20x = 645 + {female rainbow} + 3x
{female rainbow} + 645 = 17x = {multiple of 17}

So, we have that {correct answer} + 645 = {multiple of 17}. Only option D works.

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Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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07 Nov 2014, 18:25
9
2
Great question. You need to set it out in a table - it just makes it so much simpler to digest the information.

We are given 45 + 2F + F=645. We can then use the ratios to fill the rest of the table out from 2) onwards.

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##### General Discussion
Director
Joined: 13 Mar 2007
Posts: 532
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20 May 2007, 22:18

"ratio of male rainbow trout to all trout is 3:20"

Does that mean Rm/645 = 3/20 ? [Rm = male raibow trout]

In that case, by approxmation, Rm = 3 x 645 / 20 = 99

Now, Sf/Rm = 4/3 => Sf = 4 x 99 / 3 = 132 !
Manager
Joined: 14 Mar 2007
Posts: 94

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21 May 2007, 03:11
9
1
205.

It will be easier if you draw a table for this one. Let T be the total trout.

total speckled trout (ST) = 645
If female speckled trout = fST
then male speckled trout = 45 + 2fST
Therefore 45+ 2fST + fST = 645
We solve and get fST = 200
and mST= 445

Ratio of fST to mRT = 4:3
We have fST = 200
Solve ratio to get mRT = 150

Now ratio of mRT to T = 3:20
We have mRT = 150
Solve ratio to get T = 1000

Now unless youve drawn the table it might get a little complex.
From above we have mST and mRT
mST + mRT = 445 + 150 = 595
So T - 595 = total female trouts
1000 - 595 = 405 female trouts
We have from above that fST = 200

So (whew!) fRT = 405 - 200 = 205
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Joined: 11 Jul 2013
Posts: 33
Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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16 Oct 2013, 03:08
Bunuel wrote:
ggarr 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

{All} = {speckled} + {rainbow}

The ratio of male rainbow to all is 3x:20x:
20x = {speckled} + {female rainbow} + 3x

Since given that there are 645 speckled, then:
20x = 645 + {female rainbow} + 3x
{female rainbow} + 645 = 17x = {multiple of 17}

So, we have that {correct answer} + 645 = {multiple of 17}. Only option D works.

question says total speclked = 645 and female speck to male speck = 4: 3
so 7x = 645 = 92.14

so males speck = 3*92.14=276.42

what am i doing wrong ????
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Posts: 51122
Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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16 Oct 2013, 03:15
tyagigar wrote:
Bunuel wrote:
ggarr 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

{All} = {speckled} + {rainbow}

The ratio of male rainbow to all is 3x:20x:
20x = {speckled} + {female rainbow} + 3x

Since given that there are 645 speckled, then:
20x = 645 + {female rainbow} + 3x
{female rainbow} + 645 = 17x = {multiple of 17}

So, we have that {correct answer} + 645 = {multiple of 17}. Only option D works.

question says total speclked = 645 and female speck to male speck = 4: 3
so 7x = 645 = 92.14

so males speck = 3*92.14=276.42

what am i doing wrong ????

Stem says: the ratio of female speckled trout to male rainbow trout is 4:3.
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Posts: 46
Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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19 Apr 2015, 14:01
1
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?

The first sentence contanes some amiguity in my eyes. I spent over and hour to figure out teh answer but went to totlay wrong direction taking the " the number of males is 45 more than twice the number of females" as a value for all the males in teh fishery and not for the Spackeled torut, having that done, I simply could not figure out cuz one variable was mising in my equations. but than when I looked at teh posts i see what I did wrong. otehrwise it is solvabale in 2 mints not hard at all, but I got stucked

Did anyone else experianced this kind of ambiguity issue
Whats is teh source of the question.
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GPA: 4
Re: Of the 645 speckled trout in a certain fishery that contains only spec  [#permalink]

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20 Jul 2015, 10:36
Use a table set up as:

Gender Speckled Rainbow Total
Male 445 150 495

Female 200 205 405

Total 645 355 1000

1. To find female speckled trout:
45+2ST+ST = 645
3ST=600
ST = 200

2. Ratio of female speckled trout to male rainbow trout is 4:3
4/3 = 200/x
600=4x
x=150

3. Ratio of male rainbow trout to all trout is 3:20
3/20 = 150/x
3000=3x
x=1000

4. Fill in the blanks for total rainbow trout. 1000-645 = 355

5. Total rainbow trout - male rainbow trout. 355-150=205

Hope this helps
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Of the 645 speckled trout in a certain fishery that contains only spec  [#permalink]

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20 Jul 2015, 11:02
1
Turkish 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

Text in red : calculated values from the question.

Text in black: given information

Now per the question, total speckled trouts = 645 and out of these, the males are 45 more than twice the female speckled trouts

----> If female speckled trouts = x ---> male speckled trouts = 2x+45

Additionally, ratio of female speckled trouts (x) to male rainbow trouts = 4:3 ---> male rainbow trouts = 3x/4

Assume, number of female rainbow trouts = y ----> total rainbow trouts = male rainbow trouts + female rainbow trouts = 3x/4 + y

Finally, you have been given that the ratio of male rainbow trouts to total trouts = 3/20 ----> (3x/4) / total = 3/20

Thus, total = 5x

---> 2x+45+x= 645 ---> x = 200

and as 645+3x/4+y = 5x (susbtitute x = 200) to get y = 205. Thus D is the correct answer.
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Intern
Joined: 17 Oct 2016
Posts: 1
Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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06 Nov 2016, 05:13
kzivrev 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?

The first sentence contanes some amiguity in my eyes. I spent over and hour to figure out teh answer but went to totlay wrong direction taking the " the number of males is 45 more than twice the number of females" as a value for all the males in teh fishery and not for the Spackeled torut, having that done, I simply could not figure out cuz one variable was mising in my equations. but than when I looked at teh posts i see what I did wrong. otehrwise it is solvabale in 2 mints not hard at all, but I got stucked

Did anyone else experianced this kind of ambiguity issue
Whats is teh source of the question.

I faced the same issue. I read the question to mean the number of males (across Speckled and Rainbow) to be 2x + 45 than females (across Speckled and Rainbow). When we reduce it to a data matrix, it starts getting tricky. Why do people automatically assume that they are referring to Speckled trout. What am I missing here ?!!!
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Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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17 Nov 2016, 00:37
1
ggarr 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

M spec = 445 and Female spec = 200

200/ Male rain = 4/3 , thus Male rainbow = 600/4 = 150 ( total male are thus 150+445 = 595)

150/t = 3/20 thus T = 1000 , thus total female = 1000-595 = 405

Total female = 200+ Female rainbow = 405 , thus Female rainbow = 205
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Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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21 May 2017, 16:52
ggarr 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

This problem is best done via an overlapping sets table.

We can set ST, RT, and Total as the first row column headers, and M/F/Total as the first column rows headers. Then we know that 2x+ 45 + x = 645, where x is the number of female speckled trout. Once we solve for that value, we can quickly plug it into the ratio 4:3 of female speckled trout to male rainbow trout to determine the number of male rainbow trout. However, this stage attempts to trick you, because 150 is an answer choice, but the question asks for the number of female speckled trout, so you have to then subtract 105 from 355, the total number of rainbow trout, to arrive at 205, the number of female rainbow trout.
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Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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19 Jun 2018, 07:52
a very confusing yet great question. lets go step by step.
We know that total nu of sp is 645, out of which 2x+45 is male, while x is female. we get easy equation 2x+x+45=645 or x=200, so females of sp is 200, male of sp is 445.
Then the question tells us that nu of f sp:m.r is 4:3. Since we have f.sp 200, we get a ration of 4/3=200/x or 150, so we have 150 of m.r
last ratio from stem tells us that m.r to total is 3/20. we know that m.r is 150, easy ratio we get is 150/x=3/20 or x is 1000, so in total of everything we have 1000.
lets sum up:
We have f.s 200,, m.s 445 total 645
We have f.r x, m.r. 150 total x
We have total of females 200+x m 595 and total 1000,
from this table it is easy to identify number of f.r. which is 205
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Re: Of the 645 speckled trout in a certain fishery that contains  [#permalink]

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07 Oct 2018, 07:53
ggarr 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

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.
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Re: Of the 645 speckled trout in a certain fishery that contains &nbs [#permalink] 07 Oct 2018, 07:53
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