77) How heavy is the bag

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ern5231: Master | Next Rank: 500 Posts; Posts: 226; Joined: Sun Aug 09, 2009 4:34 am

by ern5231 » Wed May 12, 2010 8:53 am

There are two bags with mixed nuts A and B. In first bag, 3/4th nuts are B and rest are A. In the second bag, the weight of B is 8 grams less than that in first bag and B constitute 7/10th of the nuts, whereas the weight of A is the same as that of the first bag. What is the weight of the first bag?

a) 35
b) 46
c) 48
d) 52
e) 54

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Source: Beat The GMAT — Problem Solving | Wed May 12, 2010 8:53 am

T11: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Jan 07, 2010 6:52 am; Thanked: 1 times; GMAT Score:700

by T11 » Wed May 12, 2010 9:26 am

Hi,

Try by plugging in numbers. Start with C.

If the weight of Bag1 is 48 then
B(3/4) = 36 and
A(1/4) = 12

Now in Bag2,
B =36-8 = 28 which is 7/10 th of the total weight
Hence A = 3/10th = (28x3)/7 = 12

The Q says the weight of A is the same as that of the first bag. Hence C is the the correct answer.

Chance favors the prepared mind.

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rajeshsources: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Sun Mar 28, 2010 2:11 am; Location: Hyderabad, India; Thanked: 9 times

by rajeshsources » Wed May 12, 2010 10:17 pm

Yes correct answer is C.

In first bag total nuts, lets say 4x (why? its easy to calculate), then
B = 3/4 of total (given) == 3x
A = rest of total(given) == x

In second bag total nuts, lets say 10x, then
B = 7/10 of total(given) == 7x
A = will be rest == 3x

Question asked for how much is the weight of nuts in first bag which we can say as they are asking for the weight of 4x nuts.

Just plug-in values and always better to start with C option. C option is 48. So, the weight of 4x nuts == 48gms
x nuts = 48/4 = 12gms. and

From this, we can calculate the weight of A&B nuts in the first bag and A&B nuts in second bag. That is,
In first bag, A= x = 12gms & B = 3x = 3*12 = 36
In second bag, B weight is 8gms less than B weight in first bag (given), then B (second bag) = B (first bag) - 8 = 36-8 = 28. Then we can say,
B nuts 7x=28gms from this, x = 4,
A nuts 3x = 3*4 = 12.

Here you can see A weight (first bag) = A weight (second bag). As it satisfies the given condition, this is answer.

HTH, GOOD LUCK,

Thanks,
Rajesh,
Loves GMAT...!!!!

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mj78ind: Master | Next Rank: 500 Posts; Posts: 265; Joined: Mon Dec 28, 2009 9:45 pmFollowed by: 2 members; Thanked: 26 times; GMAT Score:760

by mj78ind » Wed May 12, 2010 11:00 pm

Hi Was wondering how we could assume that the weight per nut of A and B is 1 gm?

rajeshsources wrote:Yes correct answer is C.

In first bag total nuts, lets say 4x (why? its easy to calculate), then
B = 3/4 of total (given) == 3x
A = rest of total(given) == x

In second bag total nuts, lets say 10x, then
B = 7/10 of total(given) == 7x
A = will be rest == 3x

Question asked for how much is the weight of nuts in first bag which we can say as they are asking for the weight of 4x nuts.

Just plug-in values and always better to start with C option. C option is 48. So, the weight of 4x nuts == 48gms
x nuts = 48/4 = 12gms. and

From this, we can calculate the weight of A&B nuts in the first bag and A&B nuts in second bag. That is,
In first bag, A= x = 12gms & B = 3x = 3*12 = 36
In second bag, B weight is 8gms less than B weight in first bag (given), then B (second bag) = B (first bag) - 8 = 36-8 = 28. Then we can say,
B nuts 7x=28gms from this, x = 4,
A nuts 3x = 3*4 = 12.

Here you can see A weight (first bag) = A weight (second bag). As it satisfies the given condition, this is answer.

HTH, GOOD LUCK,

Thanks,
Rajesh,
Loves GMAT...!!!!

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rajeshsources: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Sun Mar 28, 2010 2:11 am; Location: Hyderabad, India; Thanked: 9 times

by rajeshsources » Thu May 13, 2010 1:43 am

mj78ind ---

Actually, I haven't tried to get the weight of nut A & B. And more over, we have not assumed each nut weight as 1gm. We have calculated weight of bag of nuts having mixture of A&B by taking the fraction of A nuts & fraction of B nuts in the Mixture of A&B.

From the stems of the question "There are two bags with mixed nuts A and B." We can tell the bag has only nuts nothing else.

Generally, let's say your mother asked you to get 1 KG TOMATOES, then the store-keeper given 10 tomatoes(assume). Then, you can calculate how much is the weight of each tomato. i.e.,
1 Kg = 1000gms ---- 10 tomatoes you have got, then
each tomato == 100gms.

I have given an example only for your understanding. So, the weight of whole first bag is 48.

Hope that answers your question, GOOD LUCK,

Thanks,
Rajesh,
Loves GMAT....!!!!

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mj78ind: Master | Next Rank: 500 Posts; Posts: 265; Joined: Mon Dec 28, 2009 9:45 pmFollowed by: 2 members; Thanked: 26 times; GMAT Score:760

by mj78ind » Thu May 13, 2010 3:39 am

@Rajesh

Thanks, i get it.

So essentially we could also solve this as:
In bag 1, x - A, 3x - B; Bag 2, 3y - A, 7y - B nuts by weight

Eqtn 1 3x - 7y = 8

Eqtn 2 x = 3y

Solving above, we get x = 12 gms. Hence, Bag A = 4x = 4*12 = 48 gms

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gmatjedi: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Sat Aug 15, 2009 4:36 am; Thanked: 19 times

by gmatjedi » Thu May 13, 2010 3:40 am

Algebraic Solution:

x= total weight of nuts in bag 1
y= total weight of nuts in bag 2
A= nut A
B= nut B

Bag 1= (3/4) x B + (1/4) x A
Bag 2= (7/10) y B + (3/10) y A

Given: weight of A in Bag 1 = weight of A in Bag 2
therefore,

(1/4) x A = (3/10) y A
(12/10)y= x
so we know that the weight of bag 1 is 1.2 times the weight of bag 2 and given the above, that difference is in the weight of B nuts
so let's now change x and y as above to reflect this:
Bag 1=1.2 x
Bag 2 =x
Given: weight of B nuts in Bag 1 is 8 g greater than in Bag 2
thus:
fractional weight of B nuts in Bag 1 which is 1.2x bigger than Bag 2= fractional weight of B nuts in Bag 2 +8
(3/4)(1.2x)= (7/10)x + 8
x= 40
According to above then, Bag 1=1.2x= 48

Good Luck

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