Rd1_solutions_2011.2012
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SML Round 1 2011-2012
Short Answers
1. E 2 3^ 2 3 6^5 7776
2. B Let s be the amount of the stock. Monday 1.2 s , Tuesday 1.2 .1(1.2 ) 1.08 s s s ,
Wednesday 11.08 (1.08 ) 1.266
s s s , a rise of 26%
3. D 2 3 8 and 2 3 20 the solution to this system is 7, 2. So 9a b a b a b a b
4. D By trial and error,6 2 2
3 21 29 2011 . So 3 21 29 53a b c
5. C 0 red and any amount of white gives 1 mix; 1 red and 0,1,2,3,4, or 5 white gives 6 more;
2 red and 1,3, or 5 white gives 3 more; 3 red and 1,2,4, or 5 white gives 4 more; and
4 red and 1,3, or 5 white gives 3 more. So there are 1+6+3+4+3 = 17 shades.
6. A 2 2 must equal 2 or 6. So 2, 2 x x
7. B 3 3 2
1 2 1 2When 3, and .t t t P Pe P Pe P P P
8. C ( 8)( 1) 9. ( 4)( 2) 6. x x b x x c So 3b c
9. B Let x = distance traveled in 30 minutes.2
( 2.5) ( 5) ( 7.5) ( 10) ( 12.5) ( 15) 197.5 36.753
x x x x x x x x
In 2 hours, 4 15 4(36.75 15) 132 x miles.
10. D Amount of antifreeze: 2 .2 2.5 0.625 x x x
11. E (7,1) (7,2) (7,3) (7,4) (7,5) (7,6) 8659 P P P P P P
12. D The given lines intersect at the point (-6,12). Construct a line through the origin that is
parallel to each:2 3
and3 2
y x y x . The first one goes through (3,-2) and the other
goes through (-2,3). These two lines are symmetric with respect to the line y = x. The given
lines are then symmetric with respect to the line that goes through (-6,12) and is parallel tothe line y = x. That line has equation 12 6 or 18 y x y x so m + b = 19.
13. B If a is the length of a side of the square inscribed in C, then
is the diameter of the circle C. Let b be the length
of a side of a square inscribed in a semicircle of C. Then 2 2 2 2(2 ) 90 18b b d b
14. C 1572 157 651 2
a a a
d b
b
a 2
a
a
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15. A Total # of boxes = 6 5n . # empty = 5 5 5 4n n n .
# full = (6 5 ) (5 4 ) 1n n n . 1 18n #empty = 5 4 17 73
16. D Al – chemistry, Bo – physics, Cy – biology, Di – math
17. C 3 9! 71.3 So the smallest possible value of P is 72.
18. D # red in box 10 9 8 7 6 5 4 3 2 1 0
# red in bag 0 0-1 0-2 0-3 0-4 0-5 0-6 0-6 0-6 0-6 1-6
# ways 1 2 3 4 5 6 7 7 7 7 6
Total # ways = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 7 + 7 + 7 + 6 = 55
19. D
BD lies along the line y x and EC lies along the line1
362
y x . Set these equal
and solve for x. 24 x 20. E
5 2 5 2
4 9 5 1 4 9 14 9 14 19
4 45 4 9 14 5
5 5
45
5
4 4
5 2 2
2
79 7 9 0 1
9
2 ( ) ( 2 3 )
( 2 3 )1 1
So,1
81Thus,
(1 ) 497
9
k
x x x x
y r r kr x x x x x x
x x x x x x x y
x x
x y x y
x
x x y
x x
81 + 49 = 130
G
B
D A
E
C