Download 2010 March Regional Algebra 1 Individual Test

2010 March Regional Algebra 1 Individual Test Solutions:
7
1.
13 + 3
7
13 −3
4
–
−
2
7– 3

2( 7+ 3)
4
7
13−3
13 + 3
13−3
–
2( 7+ 3)
7 – 3 ( 7+ 3)

 D.
2. 5x + 8y =3, m = -5/8. Perpendicular line will have opposite reciprocal of the slope 
new slope = 8/5. Using m = 8/5 and (1, 4), we find y = (8/5)(x) + (12/5). Writing the
equation in standard form, we obtain
5y – 8x = 12  C.
3. (1/20) + (1/15) + (1/x) = 1/5  15 + 20 + (300/x) = 60  25 x = 300  x = 12 minutes
 D.
4. By definition, it is the Commutative Property  B.
5. 1569978 is divisible by 1, 2, 3, 6, and 9. The sum of these divisors is 21  B.
6. 119 is divisible by 7 and by 17  C.
7. (x3 + 15x + 2)2 + (–x6 – 20x5 –4x3 - 225x2 – 60x4 – 4) 
x6 + 30x4 + 225x2 + 4x3 + 60x + 4 – 4 – x6 – 20x5 – 4x3 - 225x2 – 60x4 
-30x4 – 20x5 + 60x  10x(-3x3 – 2x4 + 6)  E.
8. (x + y)2 = 225  x + y = 15, x + y = -15 (second possibility is ruled out, though, because
both integers must be positive).
(x – y)2 = 49  x – y = 7, x - y = -7
(x + y) + (x – y) = 2x = 22  x = 11.
11 + y = 15, y  4.
OR
(x+y) + (x – y) = 15 – 7 = 8 = 2x  x = 4
x + y = 15, 4 + y = 15  y = 11.
In either case, the difference between x and y is 7 units. However, the magnitude refers
to the value of the number, and not to its sign  answer is E.
9. 36e2g2 – 36e2 – 4g2 + 4  36e2(g2 – 1) – 4(g2 – 1)  (36e2 – 4) (g2 – 1) 
4(9e2 – 1)(g+1)(g-1)  4(3e +1)(3e – 1)(g+1)(g-1)  C.
10.
243𝑦 −
147𝑦 +
75𝑦  9 3𝑦 − 7 3𝑦 + 5 3𝑦 = 7 3𝑦
C = 7, D = 3. C2 - D2 = 49 – 9 = 40  A.
11. C = (7/10)(W), 2(C – 11) = W – 11 + 9. Substituting, and simplifying:
2C – 22 = W – 2  2C = W + 20  (7/5)(W) = W + 20
(2/5)(W) = 20  W = 50. If W = 50, then C = (7/10)(50) = 35.
Wenxuan is currently 50 years old, while Clara is 35  C.
12. y = k/x2. y = 4, x = 100.  k =(y)(x2) = 4(100)2 = 40,000.
y = 40,000/x2; x = 160, y = (40,000)/(160)2 = 1.5625  C.
13. I – False. There are 8 primes between 1 and 20 (2,3,5,7,11,13,17,19)
II - False. There are 5 triangular numbers between 1 and 20 (1,3,6,10,15).
III - False – 3/2 is both rational and an integer.
IV – True. Answer – C.
14. x – 5 = 3x + 3  x2 – 10x +25 = 3x + 3  x2 – 13x + 22 = 0
(x – 2)(x – 11) = 0  x = 2, 11. However, x =2 is an extraneous solution since it yields an
impossible solution ( 3x + 3 = −3). Hence the answer is B.
15. f(x) = 5x2 – 23x + 17. The discriminant of a function is equal to B2 – 4AC  (-23)2 –
4(5)(17) = 529 – 340 = 189  C.
16. f(x) = 2x2 – 21x – 65. To find the x-intercepts, we set the function equal to
zero. 2x2 – 21x – 65 = 0  2x2 – 26x + 5x – 65 = 0 
2x(x -13) + 5(x – 13) = 0  x = -2.5, x = 13.
X-intercepts are then (-2.5, 0) & (13,0)  E.
17. If ΔABC is a right triangle, then it is true that a2 + b2 = c2. Therefore,
(2)2 + (x)2 = (2x2 – 9x + 12)  4 + x2 = 2x2 – 9x + 12  x2 – 9x + 8 = 0 
x =1, 8. However, we cannot evaluate the value of M(x) for a prime solution to the triangle since
neither 1 nor 8 is prime  answer is E.
18.
1
2187
= 35y-13  3-7 = 35y-13  5y – 13 = –7  5y = 6  y = 6/5 or B) 1.2.
19. 0= h +vt – 16t2. In this case, –16t2 + 32t + 40 = 0. This
parabola achieves its maximum value at t = -b/2a or -32/-32 = 1.
At t = 1, the book is at a height of -16(1)2 + 32 + 40 = 56 feet above the
ground. However, Bo is 40 feet above the ground. Therefore the distance
between him and the book is only 16 feet  B.
20. 3*y = (9 – 3y +y)(3y) = 30  y(9 – 2y) = 10  2y2 – 9y + 10 = 0 
(2y – 5)(y-2) = 0  y = 2.5, 2. Since 2.5 is not an integer, the only solution
is y = 2. 2*4 = 0  C.
21) f(x) = 2x2 + 5x + 3, then f(x+2) = 2(x+2)2 + 5(x+2) + 3 = 2x2 + 13x + 21= 0
 (2x + 7)(x +3) = 0  x = –3.5, –3. A2 + B2 = (-3.5)2 + (-3)2 =
(49/4) + (36/4) = 85/4 = 21.25  C.
22) g(x) = 8719x2 + 1085x – 18. Sum of the roots = -b/a  -1085/8719  B.
23) y = 4x – 5, 3x = 2y – 4 
-3y = -12x +15
8y – 16 = 12x
5y – 16 = 15  5y = 31, y = 31/5. If y = 31/5, then x = 14/5  -D/C = -31/14  B.
24) Distance between (a,b) and (c,d) = [(c-a)2 + (d-b)2]
Plugging in for (1,14) and (-4,2), we get a distance of 13 units  C.
25) C – (a/r) = L +(o/s)  L + (o/s) – C = - a/r  r = -a/[L + (o/s) – C] 
r = -as/(Ls + o – Cs)  as/(Cs – Ls – o)  C.
26) By definition, the domain is {1,2,3}  B
27) 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32….the pattern for the last digit repeats
every fourth power. 2010/4 = 502 with remainder 2. The remainder of
two indicates that the last digit will be equal to 4  B.
28) Answer is E. 8[2(62 – 52) + 11(–2)] simplifies out to zero, and
multiplication by zero always implies a product of zero.
29) (a + 3)(a – 1)(a + 4)(a + 5) = (a + 5)(a + 7)(a – 1)
a + 5 = a2 + 7a + 12  a2 + 6a + 5 = 0  a = -1, -5. However, if a = -5, the
denominator of the original expression is equal to zero  a = -1 is the
only solution  C.
30) The term with the highest sum of exponents determines the degree of the
polynomial. In this case, 8xy5z9 has the greatest sum of 15  C.