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A grocery store sells apples by the pound. If the price per pound is i
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29 Jan 2018, 04:30

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[GMAT math practice question]

A grocery store sells apples by the pound. If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. What is the current price per pound of apples at the grocery store?

A grocery store sells apples by the pound. If the price per pound is i
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30 Jan 2018, 10:53

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MathRevolution wrote:

[GMAT math practice question]

A grocery store sells apples by the pound. If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. What is the current price per pound of apples at the grocery store?

A. $4 B. $4.5 C. $5 D. $5.5 E. $6

Using answer choices Start with Answer C to get a benchmark.

At $5 per pound, the number of pounds that can be purchased for $12 is

\(\frac{12}{5}= 2.4\) pounds

Price per pound increases by $1, to $6 per pound.

At $6 per pound, the number of pounds that can be purchased for $12 is

\(\frac{12}{6}= 2\) pounds

(Original number of pounds) - (more expensive number of pounds) = .4 pounds (2.4 - 2) = .4

Answer C
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Re: A grocery store sells apples by the pound. If the price per pound is i
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31 Jan 2018, 02:51

=>

Suppose p is the current price of apples at the grocery store. Then \(\frac{12}{p} – \frac{12}{(p+1)} = 0.4\) \(⇔ 12(p+1) – 12p = 0.4p(p+1)\) if we multiply both sides by p(p+1) \(⇔ 12 = 0.4(p^2+p)\) \(⇔ 30 = p^2+p\) after multiplying by \(2.5\) \(⇔ p^2+p-30 = 0\) \(⇔ (p-5)(p+6) = 0\) \(⟹ p = 5\), since prices cannot be negative.

A grocery store sells apples by the pound. If the price per pound is i
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13 Mar 2018, 05:50

Let x be the original price and p the number of pounds

Before the price increase: $12=xp.......(1) ==> p=12/x .......(2) After the price increase: $12=(x+1)(p-0.4)=xp-0.4x+p-0.4 .......(3)

By substituting (1)&(2)in (3)

\($12=12-0.4x+(12/x)-0.4\) \(0.4x-(12/x)=-0.4\) \(4x^2-120=(-0.4)(10x)\) (divided by 4) \(x^2+x-30=0\)

By Backsolving (choice c; x=5)

25+5-30=0 Choice c is the right answer
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Re: A grocery store sells apples by the pound. If the price per pound is i
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13 Mar 2018, 08:12

Top Contributor

MathRevolution wrote:

[GMAT math practice question]

A grocery store sells apples by the pound. If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. What is the current price per pound of apples at the grocery store?

A. $4 B. $4.5 C. $5 D. $5.5 E. $6

let C = CURRENT price per pound So, C - 1 = INCREASED price per pound

If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. Let's first turn this statement into a word equation We can write: pounds of apples that can be purchased with $12 at CURRENT price - 0.4 = pounds of apples that can be purchased with $12 at INCREASED price Convert to algebraic expression: 12/C - 0.4 = 12/(C - 1) Rewrite as: 12/C - 0.4C/C = 12/(C - 1) Combine terms: (12 - 0.4C)/C = 12/(C - 1) Cross multiply: (12 - 0.4C)(C - 1) = (12)(C) Expand and simplify: 0.4C² + 12.4C - 12 = 12C Rearrange: 0.4C² + 0.4C - 12 = 0 Multiply both sides by 10 to get: 4C² + 4C - 120 = 0 Divide both sides by 4 to get: C² + C - 30 = 0 Factor: (C - 5)(C + 6) = 0 So, EITHER C = 5 OR C = -6 Since C cannot have a NEGATIVE value, we can be certain that C = 5

Re: A grocery store sells apples by the pound. If the price per pound is i
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13 Mar 2018, 22:19

MathRevolution wrote:

[GMAT math practice question]

A grocery store sells apples by the pound. If the price per pound is increased by $1, $12 will buy 0.4 pounds less of apples than if the price remains at the current level. What is the current price per pound of apples at the grocery store?

A. $4 B. $4.5 C. $5 D. $5.5 E. $6

let p=current price per pound w=weight in pounds pw=$12 w=12/p substituting, p*(12/p)=(p+1)*[(12/p)-.4] ➡p^2+p-30=0 ➡(p+6)*(p-5)=0 p=$5 C

gmatclubot

Re: A grocery store sells apples by the pound. If the price per pound is i &nbs
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13 Mar 2018, 22:19