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A sea food shop sells mackerel. A pack of fifty mackerel, a

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A sea food shop sells mackerel. A pack of fifty mackerel, a

by Max@Math Revolution » Mon Dec 16, 2019 12:23 am
[GMAT math practice question]

A sea food shop sells mackerel. A pack of fifty mackerel, a pack of thirty mackerel and a pack of ten mackerel is sold for $97, $67 and $25, respectively. The total number of mackerel in those packs is 1000. The total price of those packs of mackerel is $2000. What is the number of packs of ten mackerel?

A. 4
B. 6
C. 8
D. 10
E. 12
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by Max@Math Revolution » Tue Dec 17, 2019 11:38 pm
=>

Assume x, y and z are the numbers of packs with fifty mackerel, thirty mackerel and ten mackerel, respectively.
We have 50x + 30y + 10z = 1,000 and 97x + 67y + 25z = 2,000.
Then we have 5x + 3y + z = 100 (dividing everything by 10) and 97x + 67y + 25z = 2,000.

When we subtract the second equation from 25 times the first equation, we have 25(5x + 3y + z) - (97x + 67y + 25z) = 25(100) - 2000, 125x + 75y + 25z - 97x - 67y - 25z) = 2500 - 2000, 28x + 8y = 500 or 7x + 2y = 125.
Since 2y is an even number and 125 is an odd number, x must be an odd number.

Then we can put x = 2n + 1 for a positive integer n.
Rearranging the equation 7x + 2y = 125, gives us y = (125 - 7x) / 2, y = (125 - 7(2n + 1)), y = (125 - 14n - 7) / 2 = (-14n + 118) / 2 = 59 - 7n.
When we substitute x and y with 2n - 1 and 59 - 7n in the first equation, we have 5(2n + 1) + 3(59 - 7n) + z = 100, 10n + 5 + 177 - 21n + z = 100, 182 - 11n + z = 100 or z = 11n - 82.
Since x ≥ 0 and n is an integer, we have 2n - 1 ≥ 0 or n ≥ 1.
Since y = 59 - 7n ≥ 0, we have -7n ≥ -59, n ≤ 8 (the sign changes direction because we divide by a negative number).
Since z = 11n - 82 ≥ 0, we have 11n ≥ 82, n ≥ 8.
Then we have n = 8 and x = 17, y = 3 and z = 6.

Therefore B is the answer.
Answer: B
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