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# M20-27

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Math Expert
Joined: 02 Sep 2009
Posts: 51159

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16 Sep 2014, 00:09
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Difficulty:

45% (medium)

Question Stats:

67% (01:15) correct 33% (01:11) wrong based on 189 sessions

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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 _________________ Math Expert Joined: 02 Sep 2009 Posts: 51159 Re M20-27 [#permalink] ### Show Tags 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 _________________ Senior Manager Joined: 15 Sep 2011 Posts: 326 Location: United States WE: Corporate Finance (Manufacturing) M20-27 [#permalink] ### Show Tags 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
Math Expert
Joined: 02 Sep 2009
Posts: 51159

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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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Senior Manager
Joined: 18 Aug 2014
Posts: 324

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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?
_________________

Math Expert
Joined: 02 Aug 2009
Posts: 7108

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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 $$o:a = \frac{7}{3}$$ ..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Senior Manager
Joined: 18 Aug 2014
Posts: 324

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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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Senior Manager
Joined: 08 Jun 2015
Posts: 436
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

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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 &nbs [#permalink] 19 Oct 2017, 22:48
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# M20-27

Moderators: chetan2u, Bunuel

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