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March Statewide Invitational
Answers and Solutions
1.
2.
3.
4.
5.
B
D
B
C
A
6. B
7. B
8. A
9. D
10. C
11.
12.
13.
14.
15.
D
D
E (17)
C
D
Algebra 1 Individual
16.
17.
18.
19.
20.
B
C
C
C
E (48)
21.
22.
23.
24.
25.
B
B
B
C
B
26.
27.
28.
29.
30.
C
B
C
C
B
æ -3x 2 y 5 ö (−3)3 𝑥 6 𝑦 15 -27x 6 y15
1. B – ç
=
÷=
23
8
è 2 ø
3
2. D – Distance=rate * time; upstream 3=rate * (1.5), so rate = 2mph and downstream 3= rate * (1) so rate = 3mph
rower-current= 2mph; using elimination
r-c =2
r+c=3
2r = 5
r=2.5
Using substitution r + c = 3, 2.5 +c =3, Solution: D. 0.5mph
3. B 3𝑥 2 − 7𝑥 − 40 where factored form: (3𝑥 + 8)(𝑥 − 5) where A=3, B=8, C=1, D= -5, 3+8+1+9-5) = 7
3
4. C. slope (𝑦2 − 𝑦1 )/(𝑥2 − 𝑥1 )= (25 − 22)/(2 − 1)= 1 perpendicular lines have opposite, reciprocal slopes, so the
slope must be
−1
3
solution is C. y = - 13 x + 2
x2 - x - 6
(𝑥−3)(𝑥+2)
factors to:
where the domain excluded values are when denominator is = 0, if x+5=0, x
2
(𝑥+5)(𝑥+2)
x + 7x +10
x-3
cannot equal -5 and if x+2=0, x cannot equal -2, therefore the solution is (A)
x ¹ -5,-2
x +5
6. B
Collin’s remainder is -30 the square of which (−30)2 = 900
5. A –
7. B – Let x = #mL of the 2% acid solution and let y= #mL of the 8% solution. Therefore x= 75-y and 0.02x +
0.08y=0.05(75) so 0.02x +0.08y=3.75. Using substitution, 0.02(75-y) + 0.08y = 3.75; 1.5-0.02y +0.08y = 3.75;
0.06y=2.25 so y= 37.5 mL x= 75-37.5= 37.5 mL of the 2% solution.
8. A –
8b2 - 4b 2b -1 4𝑏(2𝑏 −1)
=
×
¸
3𝑏2
9b
3b2
9𝑏
36𝑏2 (2𝑏−1)
=
=
(2𝑏−1) 3𝑏2 (2𝑏−1)
12
9. D – 7x-5y=28 x-intercept is when y=0 so 7x=28, x=4 and y-intercept is when x=0 so -5y=28, y= −
the intercepts would be 4 +
28
(− 5 )=
D.
8
−5
28
5
where the sum of
10. C – y = 2x 2 + 6x -8 so 2𝑥 2 + 6𝑥 − 8 = 0 factored 2(𝑥 2 + 3𝑥 − 4) = 0 2(𝑥 + 4)(𝑥 − 1) =0, x= -4, x= 1 sum of
which (-4) +1= -3
11. D 9 - 4 - x = 2 =(( √9 − √4 − 𝑥))2 = (2)2 = 9 − √4 − 𝑥 = 4 ; −√4 − 𝑥 = −5; √4 − 𝑥 = 5; 4 − 𝑥 = 25; 𝑥 =
−21 and 2𝑥 − 3; 2(−21) − 3 = −42 − 3 = −45
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March Statewide Invitational
12. D. If slope is
𝚫𝒚
𝚫𝒙
=
𝒚𝟐 −𝒚𝟏
=
𝒙𝟐 −𝒙𝟏
Algebra 1 Individual
choice D , (11,13), (8,18)=
18−13
8−11
=
5
−3
=
5
−3
13. E. (17) Perpendicular lines have opposite reciprocal slopes; First, calculating the slope expression of the line through
6−3
9−8
3
−1
the points (-4,3) and (𝑘, 6) would be k−(−4) = − (𝑘−7)−3 and simplified − k+4 = 𝑘−10 which through cross multiplication:
-1(k+4) = -3(k-10); -k-4= -3k+30; solving to 2k= 34; k=17 None of the above listed solutions.
14. C. if -3 9x + 4
= -21; then |9𝑥 + 4| = 7 which would simplify to 9𝑥 + 4 = 7 or 9𝑥 + 4 = −7, solving each
1
equation you get 9𝑥 = 3, which x = 3 and 9x = −11, which x = −
15. D. If y =
11
9
7
7
7 ; then
= 𝑥(𝑥−3)(𝑥+3) therefore the excluded values are those which would make the
x (x2 −9)
x - 9x
3
denominator =0, x=0, x-3=0 and x+3= 0 which will result in the excluded values from the domain of 0, 3, and -3
16. B. Let x= # oz of the 10% solution where the following table can be used to find the solution:
Total amount of solution % saline
Total amount saline
Starting
Solution added
Resulting Solution
50
x
50 +x
25%
10%
15%
12.5
0.1x
0.15(50+x)
If 12.5 + 0.10x = 0.15(50 +x) then 12.5 +0.10 x = 7.5 + 0.15x and solving the equation for x; 12.5= 7.5 + 0.05x followed
by 5.0= 0.05x and x= 100 ounces of the 10% solution
17. C. 4𝑥 3 + 12𝑥 2 − 16𝑥 = 0 factored form being 4𝑥(𝑥 2 + 3𝑥 − 4) = 0 factoring the trinomial would lead to:
4𝑥(𝑥 − 1)(𝑥 + 4) = 0 after which, 4x=0 so x= 0; x-1= 0 so x=1, and x+4=0 so x= -4. Sum of which 0+1+(-4)= -3
18. C. let t = traveler’s walking speed and let w= the walkway speed so that 𝑡 − 𝑤 = 2 and 𝑡 + 𝑤 = 8 Using elimination,
𝑡−𝑤 =2
𝑡+𝑤 =8
2𝑡 = 10 where t=5 ft/sec (traveler’s speed)
19. C.
6 3
9 6
1
1
- = -10 and + = 27 let 𝑚 = 𝑥 and 𝑛 = 𝑦 so that using substitution,
x y
x y
6𝑚 − 3𝑛 = −10
→
2(6𝑚 − 3𝑛 = −10) → 12𝑚 − 6𝑛 = −20
9𝑚 + 6𝑛 = 27 and using elimination,
9𝑚 + 6𝑛 = 27
1
21𝑚 = 7 and 𝑚 = therefore,
3
1
3
1
𝑥
= , 𝑤ℎ𝑒𝑟𝑒 𝑥 = 3
20. E(48) Let x= # kg of 30% copper alloy and let y= # kg of 75% copper alloy
0.30x + 0.75y = 0.66(60) and x+y = 60; s= 60-y so that using substitution, 0.30(60 − 𝑦) + 0.75𝑦 = 39.6 and
18 − 0.3𝑦 + 0.75𝑦 = 39.6, 0.45𝑦 = 39.6 − 18; 0.45𝑦 = 21.6; 𝑦 = 48 kg of the 75% copper alloy E None of the
above
21. B 212 x 2 y 3 - 3x5 y8 + xy 4 -5x 4 y 4 by definition, the degree of a polynomial is the highest degree of the terms, (the
highest sum of the exponents of each term of the polynomial) −3𝑥 5 𝑦 8 has the highest term degree, 5+8= 13
22. B 1700 ∗ 0.90 = $1530 after the 1st price drop and then increasing it by 10% would make it 110% of the new price,
1530 ∗ 1.10 = $1683
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March Statewide Invitational
Algebra 1 Individual
23. B. If lines are collinear, it means they have equal slopes. Therefore, (-5,y) must have the same slope with one of the
two given points as the two given points have with one another. Slope of the line through (5, -4) and (4, 23) would be:
23−(−4) 27
27 y−(−4)
= −1 = −27 therefore using either of the 2 given points, if (5,-4) and (-5, y) are collinear, then − 1 = −5−5
4−5
which by using the rules of proportional equations, if you simplify and cross multiply 270 = 𝑦 + 4, therefore y = 266
24. C
y2 −5y+6
y2 +4𝑦−21
factors to:
(y−3)(y−2)
(y−3)(y+7)
where the excluded values from the domain are if denominator = 0;
y − 3 = 0; y = 3 and y + 7 = 0, y = −7 sum of excluded values is 3 + (-7)= -4
25. B. If
-3 2x -8 +15 = -3(5- x) then −3|2𝑥 − 8| + 15
= −15 + 3𝑥 and isolating the absolute value expression,
3(𝑥−10)
−3|2𝑥 − 8| = −15 + 3𝑥 − 15 , −3|2𝑥 − 8| = 3𝑥 − 30 dividing both sides by -3, yields: |2𝑥 − 8| = −3 so that:
|2𝑥 − 8| = −(𝑥 − 10) being that the absolute value is = variable expression (not a negative integer), you can solve this
by continuing :|2𝑥 − 8| = 10 − 𝑥 therefore 𝑎) 2𝑥 − 8 = 10 − 𝑥 and 𝑏) 2𝑥 − 8 = −(10 − 𝑥) solving the first equation,
𝑎) 2𝑥 − 8 = 10 − 𝑥 therefore 3𝑥 = 18 and 𝑥 = 6 and 𝑏) 2𝑥 − 8 = −(10 − 𝑥) so that 2𝑥 − 8 = 𝑥 − 10 and 𝑥 = −2
the absolute value of the difference of the solutions would be |6-(-2)| = |8| = 8
26. C. let c= price the store pays (the cost of the book) and let s= the book priced to sell. s= 1.8c therefore 6.30= 1.8c and
c= $3.50
27. B.
Connor’s remainder is 5
Sally’s remainder is 10 ; Sum of the remainders is 5 +10= 15
28. C In general, with n people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n-1)/2, we need to solve the equation n(n-1)/2 = 153 where if simplified and set the quadratic equal to
zero you get, 𝑛2 − 𝑛 − 306 = 0 the factored form of which is (𝑛 − 18)(𝑛 + 17) = 0 since you cannot have a negative
number of people, n= 18. There are 18 people at the party
29. C Distance = rate * time, where the first rate of 35mph let t = time; An hour later, the time traveled would be t-1 and
the rate was 40mph therefore, The 1st train distance will equal the 2nd train distance ; 35𝑡 = 40(𝑡 − 1) So that
35𝑡 = 40𝑡 − 40 solving for t, 5𝑡 = 40 where 𝑡 = 8 Distance traveled will be rate (35) multiplied by the time (8)= 280
miles.
30. B. Let x represent the part of the entire time cleaning such that :
1
1
1
1
𝑥 + 𝑥= 1 solving the equation by multiplying by the LCD of 90, you get 90( 𝑥 + 𝑥=1)90 and you get
30
45
30
45
3𝑥 + 2𝑥=90 where 5x= 90 and x= 18 total minutes to clean the kitchen.
3