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# There are 1600 jelly beans divided between two jars, X and Y. If the

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Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 402
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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06 Feb 2015, 09:03
3
9
00:00

Difficulty:

35% (medium)

Question Stats:

73% (02:04) correct 27% (02:14) wrong based on 255 sessions

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There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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06 Feb 2015, 13:43
4
1
Hi All,

This question can certainly be solved with Algebra; with 2 variables and 2 unique equations, it's just a matter of translating the text into equations and doing "system" math.

The answers are numbers though, and there is a logical pattern in the prompt that you can use to quickly TEST THE ANSWERS.

We're told that there are a total of 1600 marbles in two jars. Jar X has 100 less than THREE TIMES the number of marbles in Jar Y. We're asked for the number of marbles in Jar X.

100 marbles is a relatively small amount, compared to the 1600 total marbles. If we ignore the "100" and do a rough estimation, Jar X would be about 1200 and Jar Y would be about 400 (which matches the information about THREE TIMES the number of marbles). Since we DO have to factor in the 100 marbles though, the number of marbles in X has to be LESS than 1200. With this deduction, the answer has to be C or D.

Let's TEST C (since it looks like the "nicer" number to deal with).

IF....
X = 1150 marbles
X + 100 = 1250
1250/3 = 416.6666 marbles
1150 + 416.66666 is NOT 1600 total marbles.
Eliminate C.

Here's the proof though:

IF....
X = 1175 marbles
X + 100 = 1275
1275/3 = 425
1175 + 425 = 1600 marbles

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Rich
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##### General Discussion
Intern
Joined: 05 Sep 2013
Posts: 9
Location: Georgia
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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06 Feb 2015, 13:28
1
1
x+y=1600 , X=1600-Y
x+100=3y, 1700=4Y
Y=425

x=1600-y=1600-425=1175
D
Director
Joined: 04 Dec 2015
Posts: 742
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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06 Nov 2018, 20:08
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350

Let Jelly beans in jar $$X = x$$ and $$Y = y$$

$$x + y = 1600$$ ----- ($$i$$)

Given jar $$X$$ has $$100$$ fewer jelly beans than three times the number of beans in jar $$Y$$.

Therefore; $$x + 100 = 3y$$

$$x = 3y - 100$$

Substituting value of $$x$$ in equation ($$i$$), we get;

$$3y - 100 + y = 1600$$

$$4y = 1600 + 100 = 1700$$

$$y = \frac{1700}{4} = 425$$

Substituting value of $$y$$ in equation ($$i$$), we get;

$$x + 425 = 1600$$

$$x = 1600$$ $$-$$ $$425 = 1175$$

Manager
Joined: 16 May 2018
Posts: 84
Location: Hungary
Schools: Queen's MBA'20
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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08 Mar 2019, 17:52
Hi,
If X has 100 fewer jelly beans than three times the number of beans in jar Y.
Then why did we write X+100 ?
Shouldnt it be X-100?
Intern
Joined: 19 Apr 2017
Posts: 21
Location: Brazil
Concentration: Finance, Entrepreneurship
GPA: 3
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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08 Mar 2019, 19:35
x+y=1600 (I)

And

3*y = x-100

Reorganizing it

3*y-100 = x (II)

Subs. (II) in (I)

3*y-100+y=1600
4*y=1700
y=425

Then

x=1175

Ans (D)
Manager
Joined: 16 May 2018
Posts: 84
Location: Hungary
Schools: Queen's MBA'20
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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09 Mar 2019, 03:13
1
3*y = x-100

Reorganizing it

3*y-100 = x (II)

But if -100 go to the other side of equation, shouldnt it become +100?
Intern
Joined: 19 Apr 2017
Posts: 21
Location: Brazil
Concentration: Finance, Entrepreneurship
GPA: 3
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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09 Mar 2019, 10:08
hsn81960 wrote:
3*y = x-100

Reorganizing it

3*y-100 = x (II)

But if -100 go to the other side of equation, shouldnt it become +100?

Sorry, it's 3*y = x+100 instead
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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06 Jun 2019, 18:08
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350

We can let y = the number of jelly beans in jar Y; thus, there are 3y - 100 jelly beans in jar X. We can now create the equation:

3y - 100 + y = 1600

4y = 1700

y = 425

Therefore, there are 3(425) - 100 = 1275 - 100 = 1175 jelly beans in jar X.

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Posts: 1
Re: There are 1600 jelly beans divided between two jars, X and Y. If the  [#permalink]

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09 Jun 2019, 23:38
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375

B. 950

C. 1150

D. 1175

E. 1350

x+y=1600 , X=1600-Y
x+100=3y, 1700=4Y
Y=425

x=1600-y=1600-425=1175
D
Re: There are 1600 jelly beans divided between two jars, X and Y. If the   [#permalink] 09 Jun 2019, 23:38
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