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OG 12, PS-65

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OG 12, PS-65

by Abdulla » Thu Apr 30, 2009 11:56 am
A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B
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Re: OG 12, PS-65

by kanha81 » Thu Apr 30, 2009 12:03 pm
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B
0.70 (a) + 0.50 (b) = 6.30
7a + 5b = 63
7 (4) + 5 (7) = 63

4+7 = 11 [spoiler][/spoiler]

Isn't this ambiguous?
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Re: OG 12, PS-65

by Abdulla » Thu Apr 30, 2009 12:21 pm
kanha81 wrote:
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B
0.70 (a) + 0.50 (b) = 6.30
7a + 5b = 63
7 (4) + 5 (7) = 63

4+7 = 11 [spoiler][/spoiler]

Isn't this ambiguous?


it is better to deal with integers but still you have to check what numbers are equals 63.. 7(x) + 5(y)= 63.. how to find it fast?
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Re: OG 12, PS-65

by dumb.doofus » Thu Apr 30, 2009 4:02 pm
Abdulla wrote:
kanha81 wrote:
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B


0.70 (a) + 0.50 (b) = 6.30
7a + 5b = 63
7 (4) + 5 (7) = 63

4+7 = 11 [spoiler][/spoiler]

Isn't this ambiguous?


it is better to deal with integers but still you have to check what numbers are equals 63.. 7(x) + 5(y)= 63.. how to find it fast?


ok.. so I get your point.. here is a faster way..

7x + 5y = 63 ---- (1)

x + y = T ----- (2) T = total sum that we are finding out..

Writing these two equations, will not take more than 5 secs.. I mean the info is in the question itself..

Now (1) - 7*(2) gives

2y = 7T - 63 --- (3)

We can get a even integer on the RHS only when T is odd as you can see from the equation. That means out of the given choices, you can straightaway reject a, c and e.

Now we are left with 11 and 13

quickly putting in both the values gives y = 7 or 14. It can't be 14 since the equation 1 will fail as the sum will be more than 63. So choose 11.

To do all this, I swear it didnt take more than a minute and a half.. but then its subjective.. did you find this approach quick enough?
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Re: OG 12, PS-65

by Abdulla » Thu Apr 30, 2009 4:12 pm
dumb.doofus wrote:
Abdulla wrote:
kanha81 wrote:
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B


0.70 (a) + 0.50 (b) = 6.30
7a + 5b = 63
7 (4) + 5 (7) = 63

4+7 = 11 [spoiler][/spoiler]

Isn't this ambiguous?


it is better to deal with integers but still you have to check what numbers are equals 63.. 7(x) + 5(y)= 63.. how to find it fast?


ok.. so I get your point.. here is a faster way..

7x + 5y = 63 ---- (1)

x + y = T ----- (2) T = total sum that we are finding out..

Writing these two equations, will not take more than 5 secs.. I mean the info is in the question itself..

Now (1) - 7*(2) gives

2y = 7T - 63 --- (3)

We can get a even integer on the RHS only when T is odd as you can see from the equation. That means out of the given choices, you can straightaway reject a, c and e.

Now we are left with 11 and 13

quickly putting in both the values gives y = 7 or 14. It can't be 14 since the equation 1 will fail as the sum will be more than 63. So choose 11.

To do all this, I swear it didnt take more than a minute and a half.. but then its subjective.. did you find this approach quick enough?



The odd and even role is the key here. Since we recognize that we can eliminate 3 answers. Thanks..

by the way .. the above bolded formula .. I did not understand it.. ;p but I solve the problem.
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Re: OG 12, PS-65

by dumb.doofus » Thu Apr 30, 2009 9:16 pm
Abdulla wrote:
dumb.doofus wrote:
Abdulla wrote:
kanha81 wrote:
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B


0.70 (a) + 0.50 (b) = 6.30
7a + 5b = 63
7 (4) + 5 (7) = 63

4+7 = 11 [spoiler][/spoiler]

Isn't this ambiguous?


it is better to deal with integers but still you have to check what numbers are equals 63.. 7(x) + 5(y)= 63.. how to find it fast?


ok.. so I get your point.. here is a faster way..

7x + 5y = 63 ---- (1)

x + y = T ----- (2) T = total sum that we are finding out..

Writing these two equations, will not take more than 5 secs.. I mean the info is in the question itself..

Now (1) - 7*(2) gives

2y = 7T - 63 --- (3)

We can get a even integer on the RHS only when T is odd as you can see from the equation. That means out of the given choices, you can straightaway reject a, c and e.

Now we are left with 11 and 13

quickly putting in both the values gives y = 7 or 14. It can't be 14 since the equation 1 will fail as the sum will be more than 63. So choose 11.

To do all this, I swear it didnt take more than a minute and a half.. but then its subjective.. did you find this approach quick enough?



The odd and even role is the key here. Since we recognize that we can eliminate 3 answers. Thanks..

by the way .. the above bolded formula .. I did not understand it.. ;p but I solve the problem.


oh k... no worries .. lemme show u..

I have actually written the same thing above too..

Now (1) - 7*(2) gives

2y = 7T - 63 --- (3)

That essentially means multiply second equation by 7 and then subtract it from the first.. and that would give you

2y = 7T - 63..

It's quite simple.. isnt it?
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by maihuna » Fri May 01, 2009 11:48 am
A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?
===================================
0.7x + 0.5y = 6.30
7x + 5y = 63


x y total
========
9 0 9
4 7 11 => ans
========

tric: We have an upper bound on number of apples, as it doesnt match we need to reduce its number in a muliple to give unit remainder o 5 for the total.
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Re: OG 12, PS-65

by Ian Stewart » Fri May 01, 2009 11:58 am
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B
Alternatively, you could do:

0.7*a + 0.5*b = 6.3
7a + 5b = 63
5b = 63 - 7a
5b = 7(9 - a)

Since the right hand side of this last equation is a multiple of 7, the left hand side must be as well, so b must be divisible by 7. Since b can't be 14 (then the bananas alone would cost more than $6.30), then b must be 7, and a must then be 4, as you can quickly see by plugging into the last equation above. So a+b = 11.
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by maihuna » Fri May 01, 2009 12:01 pm
Great Ian, so its an HCF question combined with brutal prime. I think you have indicated the best way here. Can you please suggest if all such puzzles are bound to work this way?

Regards,
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by Abdulla » Fri May 01, 2009 1:08 pm
Ian.. This is the best approach I have seen for these types of questions.

What if we have another situation where we cannot factor 63 and 7a ??
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by awesomeusername » Fri May 01, 2009 1:42 pm
Here's a no-brainer way of solving it. (I can't formulate such a complex solution as Ian in a timed situation)

Map out multiples of .70

(1) .70
(2) 1.40
(3) 2.10
(4) 2.80
(5) 3.50
(6) 4.20
(7) 4.90
(8) 5.60
(9) 6.30

Since 9 + 0 is no a solution, cross off (9). We know that a multiple of .70 + X = 6.30.

So 5.60 + X = 6.30. Since X is NOT a multiple of .50, this can't be a solution.
4.90 + x = 6.30 No Solution, x not mutliple of .50
4.20 + x = 6.30 No Solution, x not mutliple of .50
3.50 + x = 6.30 No Solution, x not mutliple of .50
2.80 + x = 6.30 Since we know that 2.80 + blah.50 = 6.30, we've found a solution.

so

(4) apples and (6.30-2.80)/.50 bananas, so (7) bananas...

(11)

Maybe not so simple? lol
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by Ian Stewart » Fri May 01, 2009 1:56 pm
Abdulla wrote:Ian.. This is the best approach I have seen for these types of questions.

What if we have another situation where we cannot factor 63 and 7a ??
You can use similar logic to what I did above in this question, which is in the new OG:

www.beatthegmat.com/0-15-stamps-and-0-2 ... 16955.html

If you couldn't apply some kind of logic to quickly cut down the number of cases to consider, problems like these could potentially be pretty time consuming. That doesn't mean you'll never see one where you need to test several possible values, but on the GMAT it's unlikely - they normally build shortcuts into their questions.
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by typhoonguywlblwu » Fri May 01, 2009 2:10 pm
What a brilliant solution ian.I am becoming a fan of your innovative approach.

Also,awesome...good effort.Made me think of a solution for these kind of problems and your method could be quite useful.
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by alivapriyada » Thu Sep 30, 2010 11:51 am
Ian Stewart wrote:
Abdulla wrote:A certain fruit stand sold apples for $0.70 each and bananas for $0.50. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchased?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

It took too long to check each possibility, can we do it in a faster way ?

OA is B
Alternatively, you could do:

0.7*a + 0.5*b = 6.3
7a + 5b = 63
5b = 63 - 7a
5b = 7(9 - a)

Since the right hand side of this last equation is a multiple of 7, the left hand side must be as well, so b must be divisible by 7. Since b can't be 14 (then the bananas alone would cost more than $6.30), then b must be 7, and a must then be 4, as you can quickly see by plugging into the last equation above. So a+b = 11.
simply awesome explnation!!!
many thanks :)
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by artistocrat » Wed May 25, 2011 10:57 pm
I think the key to this question is to keep subtracting $0.50 from $6.30 until you get a number divisible by $0.70, or in other words, keep subtracting 5 from 63 until you get a number divisible by 7. I think the magic number that is divisible by 7 is 28, or $2.80. It's that simple. You can do it in your head.
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