Solution Set Math II

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SOLUTION SET:

MATH 2 – INTERMEDIATE ALGEBRA

1. A (1,2) B (9,2) C (1,k) Segment AB = AC AB = X2 – X1 = 9-1 = 8 AC = Y2 – Y1 = k – 2 = 8 k – 2 = 8; k = 8 + 2 k = 10 answer: B

2. line parallel to: y = 2x – 5

Passing through pt. (1, 1) Since y = mx + b; m = 2

y = 2x + b; then substitute the coordinates of pt. (1, 1) 1 = 2(1) + b; b = 1 – 2

b = -1 y = 2x - 1 answer: B

3. Slope of the line containing pts. (2, -4) and (-5, 7)

Slope = m = 1 2 1 2 x x y y   = 2 5 ) 4 ( 7     = -7 11 answer: B

4. inspect the graph: the line passes through pts. (1, 2) and (4, -4)

slope = m = 1 4 2 4    = 3 6  = -2

y = -2x + b; then substitute any point. We use pt. (1, 2) 2 = -2(1) +b; b = 2 + 2 b = 4 y = -2x + 4 or y = 4 – 2x answer: A 5. a = b + ½ = 2 3  b b + ½ = 2 3  b

; multiply both sides by 2 2b + 1 = b + 3; 2b – b = 3 -1

b = 2

a = b + ½ = 2 + ½ = 2 ½ = 5/2 answer: C

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2x + 4y – 5(3) = 19

2x + 4y = 19 + 15; 2x + 4y = 34, divide both sides by 2 x + 2y = 17, thus, x + 2y + z = 17 + 3 = 20

answer: A 7. 32/n = n 32 = n 9

answer: B

8. 21q = 3 + q ; we raise both sides to the second power

21 + q = (3 + q)2 = 9 + 6 q + q 21 – 9 + q – q = 6 q 12 = 6 q, q = 2 q = 4 answer: B 9. 3 1  a + 2 3  a a = 2 3 1   a a , then we multiply by 2 2   a a 2 3 1   a a x 2 2   a a =

 

2 2 3 1     a a a answer: D 10. 2 2  x - 4 3 2 x = 2 2  x -

2



2

3   x x =

2



2

3 ) 2 ( 2     x x x ) 2 )( 2 ( 3 4 2     x x x = (x22x)(x72)= 4 7 2 2 x x answer: D 11. ab – 1 – b + a = ( ab – b) + ( a – 1) b ( a – 1) + ( a – 1) = ( b + 1) ( a – 1) answer: C 12. x2 – 6x + 5 = 0 ( x – 5)( x – 1) = 0 A. x2 + 1 = 0 B. x2 – x – 2 = 0, ( x -2 )( x + 1) = 0 C. 2x2 – 2 = 0, 2( x2 – 1 ) = 0, 2( x + 1)( x – 1) = 0 D. x2 – 2x – 3 = 0, ( x – 3 )( x + 1) = 0 answer: C

13. perfect square trinomial: a2x2  2abx + b2 = ( ax b )2

4x2 – 20x + 25 = ( 2x – 5 )2

answer: B

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x2 + 2xy + y2 = 20

x2 + 2(4) + y2 = 20

x2 + y2 = 20 – 8 = 12

answer: C

15. -6 is a solution to x2 + 5x + k = 7, substitute -6 to all values of x

(-6)2 + 5(-6) + k = 7, k = 7 – 36 + 30 k = 1 x2 + 5x + 1 = 7; x2 + 5x + 1 - 7 = 0 x2 + 5x – 6 = 0; factoring: ( x + 6 )( x – 1 ) = 0 x = -6 and 1 answer: A 16. ( 2x2 + 11x – p ) / ( 2x – 3 ) divisor: 2x – 3 = 0; 2x = 3; x = 3/2 using factor theorem,

2( 3/2 )2 + 11( 3/2 ) – p = 0 2(9/4) + 33/2 – p = 0 9/2 + 33/2 – p = 0 42/2 – p = 0 p = 21 answer: D 17. y = -x + 3 y = -x – 2

from the general form: y = mx + b, where the m is the slope. both equations have slopes equal to -1

since the have the same slope, these lines are parallel. answer: B

18. 2x – 3y = 12

3x + y = 7

using substitution method:

3x + y = 7, y = 7 – 3x; then substitute to the other equation 2x – 3( 7 – 3x ) = 12 2x – 21 + 9x = 12 11x = 33 x = 3 y = 7 – 3x = 7 – 3(3) = 7 – 9 y = -2 solution: ( 3,-2 ) answer: A

19. let x – smaller number

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the sum of the two numbers is 125 x + x + 17 = 125

2x = 108

x = 54 --- smaller number answer: A

20. let x – number of ducks

y – number of carabaos

there are 44 feet ( carabaos has 4 and ducks has 2 ) 2x + 4y = 44

there are 16 heads x + y = 16, x = 16 – y 2( 16 – y ) + 4y = 44 32 – 2y + 4y = 44 2y = 44 – 32 = 12 y = 6 --- carabaos answer: B 21. 4x + 2 = 3x + 9

where x is the number of students per row 4x – 3x = 9 – 2

x = 7, then substitute to above equation 4(7) + 2 = 30 --- students

answer: C

22. let x – adults

then x – 289 is the number of children there are a total of 737 persons, thus x + x – 289 = 737 2x = 1026 x = 513 --- adults children: x – 289 = 513 – 289 = 224 answer: C 23. h d

is the speed arriving 2 hrs late. Where d is the distance and h is time ( hrs.). To arrive in schedule, the train has to travel distance d in h – 2 hours.

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Rate =

time

ce

dis tan

=

h

2

d

answer: C 24. Ryan – 3 kph Jerry – 2.4 kph Ryan: ( 200m )( 1 hr / 3000m ) = 1/15 hr = 4min Jerry: ( 200m )( 1 hr / 2400m ) = 1/12 hr = 5min Difference: 1 min answer: A

25. 180 miles in a 4-hour travel

1st 3 hours: 50 mph ( 50 m / hr )( 3 hr ) = 150 miles 180 – 150 = 30 30 miles in 1 hr speed on the 4th hr: 30 mph answer: A 26. let x – son 3x – man

Now 5 yrs from now

Son x x + 5

Man 3x 3x + 5

5 yrs from now, the man’s age is 3 more than twice the age of his son. 3x + 5 = 3 + 2( x + 5 ) 3x + 5 = 2x + 10 + 3 3x – 2x = 10 + 3 – 5 x = 8 man = 3x = 3(8) = 24 answer: B

27. let x = Jill’s age

x + 14 = Jack’s age

Now 10 yrs from now

Jill x x + 10

Jack x + 14 x + 24

x + 24 = 2( x + 10 ) x + 24 = 2x + 20

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2x – x = 24 – 20 x = 4

Jack’s present age: x + 14 = 4 + 14 = 18 5 years from now: 18 + 5 = 23

answer: D 28. + = 30% 60% 50% 0.3( 10 ) + 0.6x = 0.5( 10 + x ) 3 + 0.6x = 5 + 0.5x 0.1x = 2 x = 20 L answer: C 29.

Rate Time Work

A 1/6 X x/6 B 1/4 X x/4 1 4 6   x x

; multiply both sides by 12 2x + 3x = 12

5x = 12

x = 12/5 = 2 2/5 answer: B 30.

Rate Time Work

Grace 1/45 18 18/45 Abby 1/x 18 18/x 1 18 45 18

x ; multiply both sides by 45x

18x + 18( 45 ) = 45x 45x – 18x = 810 27x = 810 x = 30 min answer: C 10 Liters x 10 + x

Figure

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