A fruit merchant bought a certain quantity of A Grade apples at the ra

Question

A fruit merchant bought a certain nonzero quantity of A Grade apples at the rate of 3 per dollar and a certain nonzero quantity of B Grade apples at the rate of 5 per dollar, spending integer dollar amounts for each purchase. He then mixed all the apples and sold the mixture in bundles of 7 apples each at the rate of two dollars per bundle. If he made a profit of $9 from the entire transaction, what is the least number of B Grade apples he could have bought?

(A) 220
(B) 105
(C) 100
(D) 110
(E) 225

(A) 220
(B) 105
(C) 100
(D) 110
(E) 225

Advaitvarma · Accepted Answer

first convert each cost to cost per apple.
grade A: 1 apple = 1/3
grade b: 1 apple = 1/5
let x and y be the number of grade A and grade Y purchased respectively
x/3+y/5+9=(x+y)(2/7) ($9 is the profit)
upon simplification we get: y=5x/9+ 110 and when X is minimum i.e. 0 we get y=110, thus option D.

prgmatbiz · Answer

shouldn't it be y =5x/9 +  105 ?

almogsr · Answer

if you convert to price per apple it becomes an easier problem than trying to solve for groups of apples.

effatara · Answer

Edited to stipulate that the merchant bought nonzero quantities of each grade of apples.

prgmatbiz · Answer

Thanks for the clarification...now the answer would be 110... else 105...

rsrighosh · Answer

chetan2u   VeritasKarishma   Bunuel   souvik101990

Can u please share your approach.

chetan2u · Answer

rsrighosh ,

Let the dollars spent on grade A and B be a and b. 
So total apples will be 3a+5b.

Now these 3a+5b apples are sold at the rate of 7 apples per 2$.
So the amount received by selling these =  \frac{3a+5b}{\frac{7}{2}}

It is given that he earns a profit of 9$, so total amount received = $ spent + 9 = a+b+9

Thus  \frac{3a+5b}{\frac{7}{2}}=a+b+9

2(3a+5b)=7(a+b+9)

6a+10b=7a+7b+63.....3b=a+63

b=\frac{a}{3}+21

We are looking for minimum value of b, so we take minimum value of a. Minimum value of a will be 3, as a/3 has to be an integer.

b=\frac{3}{3}+21=1+21=22

If 22$ are spent on grade B apples, the number of grade B apples = 22*5=110

D

rsrighosh · Answer

chetan2u Thanks for sharing the approach Considering a$ and b$ is making the calculation easier rather than going by fractions... A lot helpful..

KarishmaB · Answer

chetan2u  's method above would be my natural method too.

Vibhatu · Answer

A grade apple: 3/dollar. B grade apple 5/dollar.

Let, merchant brought X amount of A grade apple and Y amount of B grade apple.

So, X amount of A grade apple price: X/3.
 Y amount of B grade apple price: Y/5.

Merchant mixed all the apples and sold the mixture in bundles of 7 apples for 2 dollar.

So, Mix apple= (X+Y) and Price= 2(X+Y)/7.

Now,

2(X+Y)/7- (X/3)-(Y/5) =9 (As he made 9 dollar profit)

After solving, 9Y-5X=945.

If Y=105, then 9Y=945, then he didn't buy any A grade apple, but actually he brought n onzero quantity of A Grade apples .

So, Y=110, Then 9Y= 990.   He brought some A grade apple.

So minimum value of B grade apple is 110.

ANS: D

HarshavardhanR · Answer

One way to solve this: https://gmatclub.com/forum/download/file.php?id=82556 ___ Harsha Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com GMAT-Club-Forum-qmgw165e.png

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A fruit merchant bought a certain quantity of A Grade apples at the ra

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