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# m11 06

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m11 06 [#permalink]  08 Feb 2009, 16:30
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Two sacks together contained 43.25 kg of sugar. If after 4.75 kg of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 73% the weight of the sugar in the second, what was the original difference in the weights of the sacks?

(A) 1.25 kg
(B) 2.00 kg
(C) 2.75 kg
(D) 3.25 kg
(E) 3.50 kg

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

REVISED VERSION OF THIS QUESTION IS HERE: m11-75557.html#p1174923
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Re: m11 06 [#permalink]  15 Feb 2009, 09:16
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ConkergMat wrote:
Two sacks together contained 43.25 kg of sugar. After 4.75(4) kg of sugar was taken from the first sack and poured into the second, the weight of the sugar in the first sack became 73% the weight of the sugar in the second. What was the original difference in weights of the sacks?

I can't check right now but am quite positive. Please double check.

1: x + y = 43.25.............................................I

2: x - 4.75 = 73% (y + 4.75)
x - 0.73y = 4.75 + (0.73 x 4.75)
x - 0.73y = 4.75 (1.73) ..............................II

Deduct II from I:
1.73y = 43.25 - (4.75x1.73)
y = (43.25 - (4.75x1.73))/1.73
y = 20.25
x = 23
x - y = 23 - 20.25 = 2.75
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Re: m11 06 [#permalink]  20 Jan 2010, 05:42
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given info
1.x+y=43.25
2.x-4.75=.73(y+4.75) ==> x-4.75 =.73y+3.4675 ==> x-.73y=8.2175

now we got 2 equations

x+y=43.25
x-.73y=8.2175

solving this we get
y = 20.25
so then x= 23

so difference b/w th wt.of the sacks x-y =23-20.25=2.75

so ans is C
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Re: m11 06 [#permalink]  20 Jan 2010, 05:43
has anyone got better shortcuts or any other diff approach..
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Re: m11 06 [#permalink]  20 Jan 2010, 05:51
Is there a faster way to solve this without doing the calculations?
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Re: m11 06 [#permalink]  20 Jan 2010, 06:44
Question like this are not difficult but calculation can kill the time in exam....

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Re: m11 06 [#permalink]  20 Jan 2010, 06:58
really time consuming calculation ...i hope some one has a better way !!! would really like to know that !
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Re: m11 06 [#permalink]  20 Jan 2010, 08:27
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Seems like the decimal calculations are a little extreme for a GMAT question.
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Re: m11 06 [#permalink]  20 Jan 2010, 11:48
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Let x and y be the new wt. of sacks. Then y+0.73y = 43.25, so y = 43.35/1.73 = 25 and x = 43.25 - 25 = 18.25. The original wt. are 18.25 + 4.75 = 23
and 43.25 - 23 = 20.25. The difference is 23- 20.75 = 2.75.
A bit simpler, i guess?

Last edited by nvgroshar on 24 Jan 2010, 09:48, edited 1 time in total.
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Re: m11 06 [#permalink]  23 Jan 2010, 22:31
The only short cut i can think of is to use 45 instead of 43.25 and 75% instead of 73%. However, I wouldn't risk it on the test unless I was super tight on time:

x-4.75 = 3/4 (45-x+4.75)

x-4.75 = 3/4 (49.75-x)

7/4x= 3/4 * 49 (3/4) - 4(3/4) ( don't do the multiplication yet but time 4 for both sides first)
7x =3*199/4 -19/4 = 588/4 = 147 =>x=21 y=24 so the difference is about 3 but it's hard to tell if the answer is C 2.75 or D 3.25.

The more educated answer would be C because the total is really 43.25 so y=22.25 thus 1.25 however, x is estimated in this case so the real answer should be between 1.25 and 3, thus C

I am not sure if i would think this crystal clear on the real test and if it is even going to save you any time - I would totally put in the time to do it step by step without seeking for any shortcut.
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Re: m11 06 [#permalink]  25 Jan 2011, 10:38
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Here is the solution from one smart fellow residing on Russian math guru forum:
After the sugar was moved, in the first sack the amount of sugar accounted for 73% of the second, therefore in the second it was 100%. Total of them equals to 173%. Let's find out what amount of sugar falls at 1%, it's 43.25/173% = 0.25 kg.
Hence, after shifting in second sack it was 100*0.25 = 25, and in the first 43.25-25 = 18.25.
Let's calculate the difference before the changes: 18.25+4.75 - (25-4.75) = 2.75.
Using this technique it seems possible to complete this problem within a minute for almost no time consuming calculation involved.

I've just noticed, that nvgroshar offered the same but using variables. It's good too.
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Re: m11 06 [#permalink]  27 Jan 2012, 02:06
Question is easy but calculation takes time , is there some other to do it?
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Re: m11 06 [#permalink]  27 Jan 2012, 03:42
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Expert's post
ashut27 wrote:
Question is easy but calculation takes time , is there some other to do it?

Generally if on the real test you are faced with some hard calculations, something like 1.79*4.89, then it might indicate two cases: either there is a shortcut solution for this problem, which won't involve such tedious math or answer choices are handy to deal with it.

Unfortunately neither case is applicable for this particular problem: you have to do some ugly math on one stage or another and answer choices are too close to each other to use some kind of approximation method. There are some shortcuts suggested above though they still involve division of 43.25 by 1.73.
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Re: m11 06 [#permalink]  27 Jan 2012, 04:20
Bunuel wrote:
ashut27 wrote:
Question is easy but calculation takes time , is there some other to do it?

Generally if on the real test you are faced with some hard calculations, something like 1.79*4.89, then it might indicate two cases: either there is a shortcut solution for this problem, which won't involve such tedious math or answer choices are handy to deal with it.

Unfortunately neither case is applicable for this particular problem: you have to do some ugly math on one stage or another and answer choices are too close to each other to use some kind of approximation method. There are some shortcuts suggested above though they still involve division of 43.25 by 1.73.

Very true, I tried a lot to solve this question with approximation but it led to me wrong choice 3.25 (I got 3.1 as value after solving with approximation), I believe it is better to move on rather than spending too much time on this question.
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Re: m11 06 [#permalink]  27 Jan 2012, 19:48
I used approximations.

let a be the weight of the first bag
let b be the weight of the second bag

100a-73b = 821.75
a+b = 43.25
100a+100b = 4325

b = 3503.25/173 ~ 3500/175= 20 (approx)
a = 43.25-20 = 23.25

a-b = 3.25 hence D is the answer.

Although you can approximate in this question, it's tricky because all the answers are only 0.25 apart.
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Re: m11 06 [#permalink]  28 Jan 2012, 02:54
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mourinhogmat1 wrote:
I used approximations.

let a be the weight of the first bag
let b be the weight of the second bag

100a-73b = 821.75
a+b = 43.25
100a+100b = 4325

b = 3503.25/173 ~ 3500/175= 20 (approx)
a = 43.25-20 = 23.25

a-b = 3.25 hence D is the answer.

Although you can approximate in this question, it's tricky because all the answers are only 0.25 apart.

That's the problem with approximation for this question: OA is C not D (see my post above about this kind of questions).
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Re: m11 06 [#permalink]  28 Jan 2012, 03:49
Bunuel wrote:
mourinhogmat1 wrote:
I used approximations.

let a be the weight of the first bag
let b be the weight of the second bag

100a-73b = 821.75
a+b = 43.25
100a+100b = 4325

b = 3503.25/173 ~ 3500/175= 20 (approx)
a = 43.25-20 = 23.25

a-b = 3.25 hence D is the answer.

Although you can approximate in this question, it's tricky because all the answers are only 0.25 apart.

That's the problem with approximation for this question: OA is C not D (see my post above about this kind of questions).

I was going to say the same thing that here approximation led to wrong result (It was same for me). At the same time approximation also took time as calculation is not very easy and options are very close.
Approximation is a fantastic approach whenever we have options those are not very close.
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Re: m11 06 [#permalink]  29 Jan 2013, 05:31
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ConkergMat wrote:
Two sacks together contained 43.25 kg of sugar. If after 4.75 kg of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 73% the weight of the sugar in the second, what was the original difference in the weights of the sacks?

(A) 1.25 kg
(B) 2.00 kg
(C) 2.75 kg
(D) 3.25 kg
(E) 3.50 kg

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

BELOW IS REVISED VERSION OF THIS QUESTION:
Two sacks together contained 40 kilograms of sugar. If after 1 kilogram of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 60% the weight of the sugar in the second, what was the original difference in the weights of the sacks?

A. 4
B. 6
C. 8
D. 10
E. 12

Say the weight of the second sack after change is $$x$$ kilograms, then the weight of the first sack after change would be $$0.6x$$. Since the total weight of sugar in both sacks remained the same then $$x+0.6x=40$$ --> $$x=25$$.

Now, if the weight of the second sack after change is 25 kilograms then initially it was 25-1=24 kilograms and the initial weight of the first sack was 40-24=16 kilograms, so the difference was 24-16=8 kilograms.

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Re: m11 06 [#permalink]  29 Jan 2013, 08:16
I guess the easiest and quickest approximation is to just round up 1.73 to 1.75

Sack2 = 4325/175

we can easily identify that both are divisible by 25. hence Sack2 = 173/7 ~ 24.7

since we know we rounded UP, hence result will also be a little UP, ie 25

Rest i guess i no prob.
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Re: m11 06 [#permalink]  29 Jan 2013, 20:41
Another way which I found to be simpler is using mixed fractions rather than decimals and the componendo - dividendo rule . Here is my approach.

A+B=43.25 or A+B= 173/4 ...(I)

We know that (A - 4.75)/(B+4.75) = 73/100

Now we know that if X/Y = A/B then (X+Y)/(X-Y) = (A+B)/(A-B). This is the componendo - dividendo rule. Using this we have:

(A+B)/(A-B-9.5) = 173 / -27...(II)

Using (I) and (II), we substitute 173/4 for A+B
= (173/4)/(A-B-19/2) = 173 / -27 ........... (173 gets cancelled from the numerator)
= -27/4 = A-B -19/2
= 19/2 - 27/4 = A - B (Note that we require the value of A-B or B-A)
= (38-27)/4
= 11/4
= 2.75

Re: m11 06   [#permalink] 29 Jan 2013, 20:41
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