Rocket City Math League
2011-2012
Round 2
Mercury Test Solutions
1. Let the mass of each rough planet be x and smooth be y. Hence the equations 5y+3x=86 (equation 1) and
2y+7x=104 (equation 2) can be obtained. 2*(equation 1) gives 10y+6x=172. 5*(equation 2) gives
35x+10y=5(104)=520. After solving using elimination (5*eq2-2*eq1), 29x=348 is left. Thus we get x=12. Using
the first equation 5y+3x=86, we substitute 12 for x, we get 5y+3(12)=86. 5y+36=86 5y=50 y=10. The question
asks for 3y+2x, hence 3(10)+2(12) 30+24 54 megagrams.
2. 60 rolls/15 cents=1 roll/n cents 60n=15 n=1/4. Therefore each roll cost ¼ cent, or 1/400 dollar during the
sale. (1/400)*(100/85)=1/340 dollar each
54
1/340
3. Let what John-john has at the beginning be x. We multiply his money by the fraction he has left after paying for
each item. Hence 18= (3/4){[1/3][(1/2)(x-24)]} 24 = (1/3)(1/2)(x-24) 72=(1/2)(x-24) 144=x-24 x=168
**We multiply ¾ when he used ¼ of his remaining money, 1/3 when he used 2/3, etc.
168
4. The journey requires 2+(7/12)*24 or 16 pounds of yogurt per day. Given that the spaceship can only hold 400
pounds of yogurt, Mikey can hence travel 400/16 or 25 days, which is 600 hours without running out of yogurt. The
spaceship can travel 1.5 km per hour, hence travelling for 600 hours gives 900 kilometers’ progress.
900
5. (x-5)(3x+2)+(18x+11)+(2x+1)(x-9)=120
3 x 2  13 x  10  18 x  11  2 x 2  17 x  9  120
5 x 2  12 x  8  120
5 x 2  12 x  128  0
-4, 32/5
( x  4)(5 x  32)
Therefore x = -4 or 32/5.
2
2 2
) = 12  a = 75. The sum of
5
6. Let the first term be a. If the third term is 12 and the common ratio is , then 𝑎(
5
an infinite geometric sequence with first term a and common ratio r is
𝑎
1−𝑟
, which in this case equals
75
1−
2
5
= 125.
x 2  7 x  18
x  5 x3  12 x 2  17 x  90
125
𝒙𝟐 − 𝟕𝒙 − 𝟏𝟖
7.
8. Since distance doesn’t matter, let it be 560 kilometers for each trip. The former trip averages 800km/h, thus it
take 0.7 hour to reach the destination. The latter trip is much slower. Since he travelled only 280km/h, it takes 2
hours to reach the destination. The average speed=(distanced travelled)/(time used), which equals to
(560*2)/(2+0.7)=1120/2.7=11200/27
𝟏𝟏𝟐𝟎𝟎
𝟐𝟕
9. 515312126=515312262636=515212318=101253318. Thus there are 12 consecutive zeroes from the units digit, so A=0;
there are 10 letters in antecedent, with e appearing 3 times, n appearing 2 times, and t appearing 2 times. Hence
B=10!/(3!2!2!)=10*9*8*7*6*5*4*3*2/(3*2*2*2)=10*9*8*7*6*5=151200; there are only 15 possible values (note
that 5 is missing on purpose) for each digit, thus there are 15*14*13*12 combinations that meet the criteria.
15*14*13*12=32760 total possible combinations. A+B-C=0+151200-32760=118440
3𝑥 + 2𝑦 + 7𝑧 = 9
10. 4𝑥 − 5𝑦 + 3𝑧 = 53
2𝑥 + 𝑦 − 6𝑧 = −13
Let the first row = A, the second row = B, and the third row = C.
A-2C  -x+19z=35. Call this equation D.
B+5C  14x-27z=-12 Call this equation E.
14D+E  239z =478  z = 2.
By D, -x+38=35  x= 3.
By C, 6+y-12=-13  y = -7.
(3, -7, 2)
118400
(3, -7, 2)
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