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16 Dec 2019, 01:43
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[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. \(12\)
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. \(12\)
17 Dec 2019, 19:05
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MathRevolution
For the whole store, the average price per mackerel \(= \frac{2000}{1000} = $2\).
For the 50-mackerel packages, the average price per mackerel \(= \frac{97}{50} = $1.94\).
For the 30-mackerel packages, the average price per mackerel \(= \frac{67}{30} ≈ $2.23\).
For the 10-mackerel packages, the average price per mackerel \(= \frac{25}{10} = $2.50\).
The 50-mackerel average is -- by far -- the closest to the average for the whole store.
Implication:
50-mackerel packages must account for the vast majority of the packages.
Since the 3 types of packages yield a total of 1000 mackerel, we get:
50x + 30y + 10z = 1000
5x + 3y + z = 100
The answer choices indicate that the number of ten-mackerel packages is 4, 6, 8, 10 or 12.
Thus, z ≥ 4.
Case 1: x=19
Plugging x=19 into 5x + 3y + z = 100, we get:
(5*19) + 3y + z = 100
95 + 3y + z = 100
3y + z = 5
The only positive integral solution for the resulting equation is y=1 and z=2.
Since the answer choices require that z ≥ 4, Case 1 is not viable.
Case 2: x=18
Plugging x=18 into 5x + 3y + z = 100, we get:
(5*18) + 3y + z = 100
90 + 3y + z = 100
3y + z = 10
Of the five options for z, only z=4 will yield a positive value for y, with the result that y=2.
Total dollar value of 18 50-mackerel packages, 2 30-mackerel packages, and 4 10-mackerel packages = (18*97) + (2*67) + (4*25) = 1746 + 134 + 100 = 1980.
Not viable, since the total dollar value must be $2000.
Case 3: x=17
Plugging x=17 into 5x + 3y + z = 100, we get:
(5*17) + 3y + z = 100
85 + 3y + z = 100
3y + z = 15
Of the five options for z, the smallest viable option is z=6, with the result that y=3.
Total dollar value of 17 50-mackerel packages, 3 30-mackerel packages, and 6 10-mackerel packages = (17*97) + (3*67) + (6*25) = 1649 + 201 + 150 = 2000.
Success!
In Case 3, z=6.
17 Dec 2019, 21:35
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MathRevolution
Let the number of pack of fifty mackerel, pack of thirty mackerel and pack of ten mackerel be a, b and c.
so 50a+30b+10c=1000.......5a+3b+c=100....(i)
97a+67b+25c=2000....(ii)
multiply (i) by 2 and equate with (ii)....
100a+60b+20c=2000=97a+67b+25c.........3a=7b+5c.....Even if b and c are 1, a is atleast 4
multiply (i) by 25 and subtract (ii) from it....
125a+75b+25c-(97a+67b+25c)=2500-2000.......28a+8b=500.....7a+2b=125..
Now a as any odd number will give a value of b..
Least value is 5 as we have seen above and max value is 17 as b will become NEGATIVE otherwise
7*17+8b=125.....119+8b=125...b=3
If a=19, then 7*18+2b=125...133+2b=125...b=-4
Let us check when a=17 and b=3
3a=7b+5c.....3*17=7*3+5c....5c=30...c=6
see if these values fits in the equations
5a+3b+c=100...5*17+3*3+6=85+9+6=100 yes
Now let us check when a=15.......7*15+8b=125........8b=125-85.......b=5
3a=7b+5c.....3*15=7*5+5c....5c=10...c=2
see if these values fits in the equations
5a+3b+c=100...5*15+3*5+2=75+15+2=92, so NOT equal to 100 and we see the sum has started becoming less than 100.
Thus any value of a less than 17 will not SUM up to 100.
So only solution is a=17, b=3 and c=6
