Download Let the line be represented by equation for a real

ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
1. Let the line L1 be represented by equation 3x  ay  6 for a real constant a .
Let the line L2 be represented by equation bx  y  6 for a real constant b .
Let the line L3 be represented by the equation 2 x  8 y  c for a real constant c .
A = the value of a so that L1 is perpendicular to L3 .
B = the value of b so that L2 has an x-intercept that is half of its y-intercept.
C = the value of c so that L3 has an x-intercept that is the reciprocal of the x-intercept of L1 .
D = the value of b so that the line L2 passes through the point (2, 10) .
-----------------------------------------------------------------------------------------------------------------------------2. E1 is the equation 2 x  3 y  10 ;
E2 is the equation 4 x  y  1 ;
E3 is the equation 5 x  6 y  63 ; E4 is the equation 7 x  y  16 .
A= the x-coordinate of the solution to the system of equations E1 and E2 .
B= the y-coordinate of the solution to the system of equations E2 and E3 .
C= the x-coordinate of the solution to the system of equations E1 and E4 .
D= the y-coordinate of the solution to the system of equations E1 and E4 .
----------------------------------------------------------------------------------------------------------------------------3. A  the least real solution to the equation 4  6 x  10 .
B  the greatest real solution to the equation 8  3x  1  3 .
1
C  the number of distinct real solutions to the equation 4  x  1  0 .
2
1
D  the greatest integer solution to the inequality 6  x  2 .
3
---------------------------------------------------------------------------------------------------------------------------4. (Note: A, B, C and D are algebraic expressions.)
Let A be the common factor of 2 x 2  x  15 and 2 x 2  13x  20 , given that all factors
have positive integer leading coefficients.
Let B be the factor of 6 x 2  23x  20 that is NOT a factor of 2 x 2  x  15 given that all
factors have positive integer leading coefficients.
Let C be the common factor of 6 x 2  23x  20 and 3x 2  23x  36 given that all
factors have positive integer leading coefficients.
Let D be the product of the following: the factor of 3x 2  23x  36 with leading coefficient 3,
and the binomial ( x  1) . Write your answer in standard form, with decreasing powers on
terms.
-----------------------------------------------------------------------------------------------------------------------------
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
5. Container J1 contains 20 ounces of pure water, and 4 ounces of pure salt, well mixed.
These are the only ingredients in J1 .
Container J 2 contains 50 ounces of a well-mixed solution, 5% of which is pure salt.
A = the number of ounces of pure salt that must be added to J1 so that J1 contains exactly
30% salt. A should be written as a simplified fraction.
B = the number of ounces of pure water that must be added to J1 so that J1 contains
exactly 10% salt.
C = the number of ounces of pure salt that must be added to J 2 so that J 2 contains
exactly 12% salt. C should be written as a simplified fraction.
D = the fraction of pure salt in the total mixture if half of container J1 , and all of J 2 are
combined. D should be written as a simplified fraction.
------------------------------------------------------------------------------------------------------------------------------6. Maritha has 50 coins, quarters and nickels only. The total value of the coins is $12.10.
A = the number of quarters that Maritha has.
Jackie has 21 coins, dimes and nickels only. She has twice as many nickels as dimes.
B = the value of all of Jackie's coins, in dollars.
Samuel has $50, with only one-dollar bills and half-dollar coins. He has ten more coins
than bills. C = the number of half-dollar coins that Samuel has.
Nikita has the same number of dimes as nickels as half-dollar coins. She has only
these three types of coins. She has $7.80 total. D = the number of coins that Nikita has.
-------------------------------------------------------------------------------------------------------------------------------
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
7.
Sickles Invitational
L1 has equation y  4 x  4 . L2 has equation y  11  x .
L1 and L2 are shown in the figures, but not labeled.
y
figure 1
A = the area of the triangular region in Quadrant I so that
the sides of the triangle are on L1 , L2 and the x-axis.
See figure 1 which may not be drawn to scale.
x
B = the sum of the coordinates of intersection point of the
graphs of L1 and L2 .
y
C : the horizontal line with equation y  4 intersects
the triangle described previously (part A) at points
E and F, shown in figure 2. C is the distance from E to F.
D = the area above the line with equation y  4 and
in the shaded region shown in figure 2.
figure 2
E
F
x
-------------------------------------------------------------------------------------------------------------------------1
1
1
E3 :  3  ( x  1)  ;
8. E1 : 6  2(4 x  5)  2 x  28 ; E2 : x  x  x  x  ;
2
4
2
3
5
1 
1 
E4 :

E5 :   6  x   10 ; E6 : 4  x  2 x  1  3
;
x 1 x 1
2 
2 
A = the number of equations above ( E1 through E6 ) with an integer solution.
B = the number of equations above ( E1 through E6 ) with a positive solution.
C = the sum of the solutions of the equations above ( E1 through E6 ). Write C
in decimal form.
D = the product of the solutions to equations E1 and E2 .
----------------------------------------------------------------------------------------------------------------------------
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
9.
Sickles Invitational
A: the value of the exponent of the simplified expression:
That is, find A if
B=
23 23 210
, with base 2.
26
23 23 210
is simplified to 2 A .
26
1
 1 
 3  evaluated at x  . B should be written as a simplified fraction
2
 2x 
or an integer, whichever applies.
 2x 
3
3
C= The value of C for which
3.11020
 5.0  10C .
6.2 102
x10 x15
 x 2 for all positive values of x .
12
k
x x
-------------------------------------------------------------------------------------------------------------------------P2 : 2 x 4  6 x3  x 2  10 x  2 ;
P3 : 5 x 4  8 x 2  x  4
10. P1 : 3x 4  5 x3  x 2  10 x  1 ;
D = the value of k for which the expression
A = the coefficient of the x 2 term of the polynomial, written in standard form, when
P1 and P2 are multiplied.
B = the coefficient of the x 3 term of the polynomial, written in standard form, when
P1 and P3 are multiplied.
C = the degree of the polynomial, written in standard form, when polynomial P3 is
squared, and then multiplied by polynomial P1 .
D = the sum of the coefficients of the difference  P3  P2  , when written in standard form.
------------------------------------------------------------------------------------------------------------------------6
 5 . A should be written in decimal form.
11. A = the value of x so that
6
4x 
6
4x 
4 x  ...
1 3

B = the reciprocal of the value of 2 4 .
3
6
2
C = 3402  3382  5002  4992 .
3x
6.
3x
6
3x
6
3x
6
...
------------------------------------------------------------------------------------------------------------------------------
D = the value of x so that
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
12. A is the value of k if the average (arithmetic mean) of the slopes of the lines with
equations 3x  2 y  8, kx  3 y  12 and x  2 y  5 is 6 .
B is the average (arithmetic mean) of the y-intercepts of the lines with equations
3x  2 y  8, kx  3 y  12 and x  2 y  5 . B must be written in reduced fraction form.
C is the value of k if the average (arithmetic mean) of the x-intercepts of the lines with
equations 3x  2 y  8, kx  3 y  12 and x  2 y  5 is 3 .
D is the average of the ten numbers in set R S (sets R and S combined into one
set of ten elements) if the average of the 3 numbers in set R is 10 and the average
of the 7 numbers in set S is 40.
----------------------------------------------------------------------------------------------------------------------------13. A rectangle has length 3x  4
and width 3x 1 . A second and
smaller rectangle has length x  4
and width x .
3x 1
x
x4
3x  4
A = the value of x if the positive difference of the
perimeters of the two rectangles is 78, and x  0 .
B = the area of the shaded region between the two rectangles, if x=3.
C = the value of x if the area of the shaded region between the two rectangles
is 9, and x  0 .
D = the least integer value for x so that x  2 , and the sum of the perimeters of the two
rectangles is an integer evenly divisible by 10.
------------------------------------------------------------------------------------------------------------------------14.
S1 : 2  (3  x)  (2  3)  x
S 2 : 2  (3  x)  2  ( x  3)
S 3 : 2  (3  x)  (3  x)  2
S4 :
2(3  x)  2(3)  2 x
S6 :
(3  x)2  2(3  x)
S5 :
2(3  x)  2( x  3)
A = the number of statements ( S1 through S 6 ) above that represent the
Commutative Property in any form (over addition or multiplication).
B = the number of statements ( S1 through S 6 ) above that represent the
Associative Property in any form (over addition or multiplication).
C = the number of statements above ( S1 through S 6 ) above that represent the
Symmetric Property of Equality.
D = the number of statements above ( S1 through S 6 ) above that represent the
Distributive Property of Multiplication over Addition.
--------------------------------------------------------------------------------------------------------------------
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
15. Pipe 1, when open, will fill my initially empty pool in 3 hours.
Pipe 2, when open, will fill the same pool in 4 hours.
Drain 1, when open, will allow the same pool to completely drain in 5 hours, given
that it was completely full initially.
A = the number of minutes it would take the empty pool to be completely filled if pipe 1
and pipe 2 are both open. The drain is closed. Write A as a reduced fraction.
B = the number of minutes it would take the empty pool to be completely filled if pipe 2
and drain 1 are both open. (Pipe 1 is closed.)
C = the number of minutes it would take the empty pool to be completely filled if pipe 1
and drain 1 are both open. (Pipe 2 is closed.)
D= the number of minutes it would take the empty pool to be completely filled, if both
pipe 1 and pipe 2 are open, and drain 1 is also open. Write D as a reduced fraction.
ANSWERS!!
Part A
1
Part B
2
Part C
1
Part D
8
13
2
2
10
or
3
equivalent
3x  4
0
23
3x  4
3x 2  x  4
9
124
36
10
3
3
or 0.75
4
1
or 0.5
2
1
4
2x  5
5
6
20
7
48
1.40 or 1.4
175
44
40
7
8
40
5
11
3
5
47.75
9
10
17
10
11
12
103
1.55
57
12
2355
9
25
24
31
13
14
15
10
4
720
7
1
16
59
18
5
6
83
1
1200
1
or 0.5
2
11
1
0
450
6
1
3600
23
2
20
notes
part D must be in standard
form.
parts A, C and D must be
in fraction form
B is the value in dollars,
but does not need $ sign
C must be in decimal form
B must be fraction form.
A must be in decimal form
B must be in fraction form
A and D must be in
fraction form
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
SOLUTIONS:
3
1 3
3
. Slope of L3 is  .  4 at a  . A=3/4.
a
4 a
4
6
6
B: x-intercept of L2 is ; the y-intercept is -6.  3 . b=B= -2.
b
b
c
c 1
C: the x-intercept of L3 is ; the x-intercept of L1 is 2.  . c=1.
2
2 2
D: bx  y  6  -2x=10+6. x= -8.
2. A: 2 x  3 y  10  2 x  3 y  10
14x=7.
3( 4 x  y  1 )  12 x  3 y  3
x=0.5
B: 5( 4 x  y  1 )  20x+5y= -5
-19y=247
-4( 5 x  6 y  63 ) -20x-24y= 252
y= -13
C: 2 x  3 y  10  2 x  3 y  10
-19x= -38
-3( 7 x  y  16 ) 21x  3 y  48
x=2.
D: 2 x  3 y  10 using part C, x=2, gives 4-3y=10. y= -2.
3. A: 4  6 x  10 or  4  6 x  10 . x= -1 or 7/3. The least real solution is -1.
1. A: Slope of L1 is
B: 3x  1  11 so 3x  1  11 or  3x  1  11 . x=10/3 or -4. The greatest is 10/3.
1
x  1 is not possible, and so there are 0 distinct solutions. C=0.
2
1
1
1
D: 6  x  2  6  x  2 and  6  x  2 , solves to x  12 and x  24 . The greatest
3
3
3
integer in that set is 23.
4. A: 2 x 2  x  15 = (2 x  5)( x  3) . 2 x 2  13x  20 = (2 x  5)( x  4) . The common factor is 2 x  5 .
B: 6 x 2  23x  20 = (2 x  5)(3x  4) . 2 x 2  x  15 = (2 x  5)( x  3) . The uncommon factor of
6 x 2  23x  20 is 3x+4.
C: 6 x 2  23x  20 = (2 x  5)(3x  4) . 3x 2  23x  36 = (3x  4)( x  9) . The common factor is 3x+4
C: 4 
D: (3x  4)( x  1)  3x2  x  4 .
x4
30

5. A:
. 100x  400  30x  600 . 70x=200. x=20/7.
x  20 100
4
1
 . x  20  40. x  20 .
B:
x  20 10
2.5  x 12

C: J 2 has 5% salt which is 50(0.05)=2.5 oz.
. 250  100x  12x  600 .
x  50 100
88x=350. x=350/88 = 175/44.
D: Half of J1 is 2 oz of salt out of 12 oz total. Add J 2 to get 4.5 out of 62.
That is 45/620 = 9/124
6. A: Q  N  50, 0.25Q  0.05 N  12.10 25Q  5(50  Q)  1210 . 20Q=1210-250=960. Q=48.
B: D  2 D  21 gives 7 dimes and 14 nickels. That is 0.70+0.70=1.40 in dollars.
C: (H-10)+0.5(H)=50. 1.5H=60. 600/15=H=40.
D: 10D+5D+50D=780. 65D=780. D=12. So the total coins is 3(12)=36.
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
7. A: 4x-4=11-x, so 5x=15. The point of intersection is (3, 8). X-intercepts are 1 and 11
so the base of the triangle is 10. 1/2(10)(8)=40.
B: Since the second equation gives x+y=11, the sum of the coordinates is 11.
C: 4x-4=4 gives x=2 for E. 11-x=4 gives x=7 for F. EF=5.
D: Height of this triangle is 8-4=4 (see part A for the 8 number). Part C gives base 5.
Area is 1/2(4)(5)=10.
1
1
8. E1 : 6  2(4 x  5)  2 x  28 has solution x= -2; E2 : x  x  x  x  solves to x=  ;
2
4
1
1
3
5
E3 :  3  ( x  1)  has solution 15; E4 :

has solution x=4;
4
2
x 1 x 1
1 
1 
E5 :   6  x   10 has solution 31; E6 : 4  x  2 x  1  3 has solution 0.
2 
2 
A = 5 integer solutions. B=3 positive solutions. C=sum 47.75. D= E1 ( E2 )=0.5.
9. A:
23 23 210
26
B:  2 x
3

3
233106  210 . A=10.
6
1
22 1
1
1
 1 
9 1 
6
 3  = 8 x  3   4 x at x  . 4    6  4 
2
2
2 16
2
 2x 
 2x 
3.11020
 0.5 1018  5.0 1017 . C=17.
2
6.2 10
x10 x15
D: 12 k  x 2 so 10 15 12  k  2 . k=11.
x x
10. A: x 2 term of (3x 4  5x3  x 2  10 x  1)(2 x 4  6 x3  x 2  10 x  2) : x 2 (2)  10 x(10 x)  1( x 2 )
gives 2+100+1= 103.
B: x 3 term of (3x4  5x3  x 2  10 x  1)(5x 4  8x 2  x  4) . -5(-4)+1(1)+10(-8)=20+1-80= -59
C: When P3 is squared, we get degree 8, multiplied by P1 gives degree 12.
C:
D:
 P3  P2  , sum of coefficients, gives sum of
P3 minus sum of P2 , -6 – 19= -25
(5x 4  8x 2  x  4)  (2 x 4  6 x3  x 2  10 x  2) has coefficients 3-6-7-9-6= -25
6
 5 . 20x-25=6. 20x=31. x= 31/20 (multiply numerator and denominator by 5
11. A:
4x  5
to get A=1.55.
1 3
4  
1
2 4 23
 . The reciprocal is B = 18.
B: 
=
3
 6  24 18
4  6
2

2
C = 340  3382  5002  4992 = (340  338)(340  338)  (500  499)(500  499) =
678(2)+999=1356+999=1355+1000=2355.
3x
 6 . 3x=72. x=24.
D:
66
3 k 1

 18 . Times 6: 9  2k  3  108 . k=57.
12. A: 
2 3
2
ALGEBRA ONE TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
B: -4+4+2.5=2.5. Divided by 3 is
Sickles Invitational
5 1 5

2 3 6
8 12
  5  9 . Times 3k gives 8k+36+15k=27k. k=9
3 k
D: If the average of the first set is 10, with 3 numbers, the sum is 30. Similarly, the
second set has a sum of 280. Total is 310, divided by 10 gives 31.
13. A: Perimeters are 4x+8 and 12x+6. The positive difference for x  0 is 8x-2, so
8x-2=78 gives x=10.
B: If x=3 then the areas of the rectangles are 3(7)=21 and 8(13)=104. Area=104-21=83.
C: (3x  1)(3x  4)  x( x  4)  9 . 9 x 2  9 x  4  x 2  4 x  9 . 8 x 2  5 x  13  0 .
(8 x  13)( x  1)  0 . If x  0 then x=1.
D: Sum of the perimeters is 16x+14. If x>2 and the sum is divisible by 10 then use trial
and error. x=3 has units digit 2, x=4…8, x=5…4, x=6….0 so it is divisible by 10. D=6.
14.
S1 : 2  (3  x)  (2  3)  x Assoc. of Add.
S 2 : 2  (3  x)  2  ( x  3) . Commut. of Add
C:
S 3 : 2  (3  x)  (3  x)  2 Comm. of Add
S 4 : 2(3  x)  2(3)  2 x Distributive…
S 5 : 2(3  x)  2( x  3) Comm. of Add
S 6 : (3  x)2  2(3  x) Comm. of Mult.
A =Commutative Property in any form= 4 statements
B = Associative Property in any form = 1 statement
C = Symmetric Property of Equality=0 statements
D =Distributive Property of Multiplication over Addition=1 statement.
1 1
7
12
720
t  1. t 
15. A: t  t  1 .
hours, which is
minutes.
3 4
12
7
7
1 1
B: t  t  1 . t  20 hours, which is 1200 minutes.
4 5
1 1
15
t  t  1. t 
C:
hours, which is 450 minutes.
3 5
2
1 1 1
t  t  t  1 . 20t 15t 12t  60 . 23t=60. t=60/23 hours, which is 3600/23 min.
D:
3 4 5