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M20-27

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M20-27 [#permalink]

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New post 16 Sep 2014, 00:09
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If 8 apples and 10 oranges cost as much as 22 apples and 4 oranges, how much does a combination of 7 oranges and 3 apples cost?


(1) An orange costs $0.01 more than an apple

(2) The ratio of the price of an orange to the price of an apple is 7:3
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 00:09
Official Solution:


Let \(o\) denote the price of an orange and \(a\) the price of an apple. We know from the stem that \(8a + 10o = 22a + 4o\) or\(14a = 6o\) or \(o:a = \frac{7}{3}\).

Statement (1) by itself is sufficient. S1 provides another equation: \(o = a + 0.01\). After solving the system of two linear equations for \(a\) and \(o\) we will be able to answer the question.

Statement (2) by itself is insufficient. S2 adds no new information.


Answer: A
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M20-27 [#permalink]

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New post 04 Aug 2015, 19:54
Bunuel wrote:
If 8 apples and 10 oranges cost as much as 22 apples and 4 oranges, how much does a combination of 7 oranges and 3 apples cost?


(1) An orange costs $0.01 more than an apple

(2) The ratio of the price of an orange to the price of an apple is 7:3



Hello Bunuel,

In the math book, it states, under the subtitle, following cases, order is important: "If a problems says ‘the ratio of x and y’, it means ‘x divided by y’ NOT 'y divided by x'" <http://gmatclub.com/forum/word-problems-made-easy-87346.html>

So, by the definition the second statement backwards. I mean, the ratio of the price of an orange to the price of an apple is \(3:7\), not \(7:3\)
\(a =\) apples
\(o =\) oranges
\(8a + 10o = 22a + 4o\)
\(6o = 14a\)
If \(3o = 7a\), then the ratio is \(3:7\)

I'm sure most people understand what the statement intends to mean, but I thought the problem could be less confusing for others who are working on ratio problems (like myself :) ). After all, it's not the first time I've seen two values be set up to equal \(0\), especially after the section on moduli. Let me know your thoughts. Thanks

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New post 20 Aug 2015, 08:07
mejia401 wrote:
Bunuel wrote:
If 8 apples and 10 oranges cost as much as 22 apples and 4 oranges, how much does a combination of 7 oranges and 3 apples cost?


(1) An orange costs $0.01 more than an apple

(2) The ratio of the price of an orange to the price of an apple is 7:3



Hello Bunuel,

In the math book, it states, under the subtitle, following cases, order is important: "If a problems says ‘the ratio of x and y’, it means ‘x divided by y’ NOT 'y divided by x'" <http://gmatclub.com/forum/word-problems-made-easy-87346.html>

So, by the definition the second statement backwards. I mean, the ratio of the price of an orange to the price of an apple is \(3:7\), not \(7:3\)
\(a =\) apples
\(o =\) oranges
\(8a + 10o = 22a + 4o\)
\(6o = 14a\)
If \(3o = 7a\), then the ratio is 3:7

I'm sure most people understand what the statement intends to mean, but I thought the problem could be less confusing for others who are working on ratio problems (like myself :) ). After all, it's not the first time I've seen two values be set up to equal \(0\), especially after the section on moduli. Let me know your thoughts. Thanks


\(3o = 7a\)

\(o:a=7:3\).

No?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M20-27 [#permalink]

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New post 20 May 2016, 16:04
Bunuel wrote:
Official Solution:


Let \(o\) denote the price of an orange and \(a\) the price of an apple. We know from the stem that \(8a + 10o = 22a + 4o\) or\(14a = 6o\) or \(o:a = \frac{7}{3}\).


I'm sorry but how do we get to \(o:a = \frac{7}{3}\) from 14a = 6o? Should it not be a = \(\frac{3}{7}\)o after dividing 14 from both sides?
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Re: M20-27 [#permalink]

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New post 20 May 2016, 19:04
redfield wrote:
Bunuel wrote:
Official Solution:


Let \(o\) denote the price of an orange and \(a\) the price of an apple. We know from the stem that \(8a + 10o = 22a + 4o\) or\(14a = 6o\) or \(o:a = \frac{7}{3}\).


I'm sorry but how do we get to \(o:a = \frac{7}{3}\) from 14a = 6o? Should it not be a = \(\frac{3}{7}\)o after dividing 14 from both sides?


Hi,
yes a = \(\frac{3}{7}\)o ....
If I divide both sides by o... \(\frac{a}{o}=\frac{3}{7}\)...... so a:o = 3/7, which is same as the opposite \(o:a = \frac{7}{3}\) ..
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: M20-27 [#permalink]

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New post 20 May 2016, 19:31
chetan2u wrote:
redfield wrote:
Bunuel wrote:
Official Solution:


Let \(o\) denote the price of an orange and \(a\) the price of an apple. We know from the stem that \(8a + 10o = 22a + 4o\) or\(14a = 6o\) or \(o:a = \frac{7}{3}\).


I'm sorry but how do we get to \(o:a = \frac{7}{3}\) from 14a = 6o? Should it not be a = \(\frac{3}{7}\)o after dividing 14 from both sides?


Hi,
yes a = \(\frac{3}{7}\)o ....
If I divide both sides by o... \(\frac{a}{o}=\frac{3}{7}\)...... so a:o = 3/7, which is same as the opposite \(o:a = \frac{7}{3}\) ..


Woops I think I was overlooking the ratio that prefaced the fraction (o:a), thanks for breaking it down.
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Re: M20-27 [#permalink]

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New post 19 Oct 2017, 22:48
The ratio of price of apples to price of oranges is 3:7. To find 7O+3A , we need to find A and O. Option B does not give us any new input. Option A does help us find A and O. Hence the answer is option A.
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Re: M20-27   [#permalink] 19 Oct 2017, 22:48
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