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C

lets say the current cost of the apple be "A".

no of pounds that can be bought with when the cost is A for $12 = 12/A.

now new cost is A+1.

no of pounds that can be bought with when the cost is 'A+1' for $12 = 12/(A+1)

per question stem we have Current pound when cost is A - pound when cost is "A+1" = 0.4

12/A-(12/(A+1)) = 0.4

Solving, gives a quadratic equation with A=-6,5. As the cost is positive A=5.
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=>

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.

Therefore, the answer is C.

Answer: C
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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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MathRevolution
[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
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