ANSWER KEY FOR MATH 10 SAMPLE EXAMINATION Instructins: If asked t label the axes please use real wrld (cntextual) labels Multiple Chice Answers: 0 questins x 1.5 = 30 Pints ttal Questin Answer Number 1 D A 3 D 4 C 5 B 6 C 7 B 8 A 9 A 10 B 11 A 1 C 13 C 14 C 15 D 16 A 17 A 18 C 19 D 0 A Cnstructed Respnse Answers 1. N wrk needs t be shwn fr this questin. a. 16 (1 pint) b. 10 (1 pint) c. Median (1 pint) LQ, UQ (1 pint) min, max (1 pint) 9 14 19 6
. a. (1, 7.00) (3, 11.00) (1 pint fr the increase f $4) (5, 15.00) b. Plt shuld be linear (slpe f and y-intercept f 5) and discrete. (1 ½ pints) Subtract ½ pint if axes are nt labelled Subtract ½ pint if data is cnnected The axes shuld be labelled and numbered if a scale ther than ne unit fr each grid mark is being used. (½ pint) Cst # Mvies c. Equatin shuld be linear. The slpe shuld be and the y-intercept 5. ( pints) N value awarded if equatin is nt linear 1 pint awarded fr crrect slpe 1 pint awarded fr crrect y-intercept Cst = 5 + (Mvies) y = x + 5 r C = n + 5 d. Methd 1 Methd Student replaces cst with 9. (1 pint) Student graphs the equatin Student slves crrectly (1 pint) and traces t determine ans. 9 = x + 5 graph shwn with pint labelled (1pt) 4 = x answer crrect (1 pt) 1 = x 1 mvies fr $9 Methd 3 Student uses a table and extends t where Cst= 9 (1 pint) Student answers questin crrectly (1 pint)
10 0 3. Students shuld set up a cutting in-equatin (1.5 pints) 10x + 0y 480 r x + y 8 60 60 Subtract ½ pint if 10x + 0y 8 Subtract ½ pint if in-equality is reversed 15 40 Students shuld set up a sewing in-equatin (1.5 pints) 15x + 40y 70 r x + 1 60 60 Subtract ½ pint if 15x + 40y 1 Subtract ½ pint if in-equality is reversed Students shuld set up the minimal prductin in-equatins (1 pint) x 0 and y 0 Subtract ½ pint if in-equality is reversed 4. Students need t use the prfit equatin P = 30x + 0y (1pint) Students need t cnsider the pints (0,70), (300, 540) and (500,300) (1pint) Students plug the pints (abve) int the prfit equatin (1pint) Students state the slutin that yields the highest prfit (1pint) Subtract ½ pint if just the crdinate is used (500, 300) withut the statement 30(300) + 0(540) = 19800 30(500) + 0(300) = 1000 30(0) + 0(70) = 14400 30(500) + 0(0) = 15000 (ptinal) Cnclusin: There shuld be 500 ftballs and 300 basketballs t prduce the maximum prfit. 5. a. Strng Psitive ( pints) (Nte: 1 pint fr strng and 1 pint fr psitive) b. Equatin is wrth pints Subtract ½ pint if nt written as equatin. Fr x = 0, 0, 40 Fr x = 1900, 190, 1940... Or y = 0.33x + 48.10 y = 0.33x 581.6 c. Students cmpare change in life expectancy t change in birth year (1 pint) Subract ½ pint if the student indicates nly why the slpe is psitive Additinal 0.33 years f life fr each extra birth year Or Additinal 3.3 years f life fr each extra decade f birth life Or The change in life expectancy cmpared t the change in birth year.
d. Student replaces the independent variable by 10 r by 00 depending n the meaning f the variable (1 pint) Slutin f 84.98 years r 87.87(if using equatin frm calculatr) (1 pint) y = 0.33(10) + 48.10 r y = 0.33(00) 581.6 y = 84.98 years 6. Methd 1: Algebraic Expressins fr height x = number f years since the blue spruce tree was planted 16 + 4(5) + 4x = height 1 pint fr describing the hemlck trees height. In 5 years 16 + 4(x +5) = height 36 + 4x = height 10 + 6x = height 1 pint fr describing the Blue 16 + 4(5) + 4x = 10 + 6x 36 + 4x = 10 + 6x 6 = x 13 = x Spruce trees height. 1 pint fr cmparing the tw heights 13 + 5 = 18 years 1 pint fr the crrect answer Nte: deduct ½ pint fr nt adding 5 t 13 (extra grwth) Methd : Using a Table Year Hemlck Height Blue Spruce Height 0 16 --- 1 0 --- 4 --- 3 8 --- 4 3 --- 5 36 10 6 40 16 7 44 8 48 8 9 5 34 10 56 40 11 60 46 1 64 5 13 68 58 14 7 64 15 76 70 16 80 76 17 84 8 18 88 88 1 pint fr year clumn (with blue spruce beginning 5 years after hemlck); 1 pint fr hemlck height clumn (increasing by 4 each year); 1 pint fr the blue spruce height clumn (increasing by 6 each year); 1 pint fr the answer (18 years)
Methd 3: Trial and Errr Methd Students cmpare 36 + 4(#)t 10 + 6(#) ( pints) Students use sequential chice f numbers (1 pint) Answer (18 years) (1 pint) Or Students cmpare 6 + 4(# + 5) t 10 + 6(#) ( pints) Students use sequential chice f numbers (1 pint) Answer (18 years) (1 pint) 7. a. Students are able t state the equatin f the quadratic ( pints) 1 pint subtracted fr n reflectin ½ pint subtracted if sign errr in translatin N pints awarded if the student des nt have a quadratic equatin. y = (x )(x + ) r (y 4) = (x + 0) r y = - x + 4 r - y + 4 = x (y 4) = x b. Students state the factrs (1 pint) Students set each factr t zer (1 pint) Students state the slutins ( pints) ½ pint subtracted if the factred quadratic (x + 6)(x 1) is nt equated t zer (x + 6)(x 1) = 0 x + 6 = 0 x 1 = 0 x = 6 x = ½ 8. a and b. Students use tw different methds frm the fllwing. Pythagrean Therem (1 pint fr setting up crrectly) Use Pythagrean therem t slve crrectly (1 pint) Trignmetric Ratis (1 pint fr setting up crrectly) Use Trignmetric rati t slve crrectly (1 pint) Pythagrean Therem 6 = 11.4 + p 546.04 = p 3.37 m = p Trignmetric Ratis P x sin 64 = tan 64 = 6 11.4 P = 6(sin 64 ) x = 11.4(tan 64 ) P = 3. 37 m x = 3. 37 m
c. Students set up a crrect prprtin ( 1 pint) Students use the prprtin t determine height crrectly (1 pint) 159.9 P = 1 pint is subtracted if the prprtin 78 106 is used yielding P = 177. 1 cm Zer pints given if methds ther than similar triangles is used. 159.9 P = 78 1140 159.9 P = 1140 78 P = 337 cm P = 3. 37 m 9. a. N wrk needs t be shwn. 1 pint fr each crrect measure. 180 5 15 = DBA = 140 140 180 140 = 40 180 40 90 = BAC = 50 50 b. Students use a crrect Trignmetric Ratin (1 pint) Students use the trignmetric rati crrectly t btain crrect answer (1 pint) sin 5 = 1 x 1 x = sin 5 x = 8.39
c. Methd 1: (fllw thrugh errrs frm part a) Students use trig rati (½ pint) and Pythagrean therem (½ pint) Students btain the length BC (½ pint) Students add t btain the height requested (½ pint) cs 50 = 1 x = cs 50 x = 18.67 1 x y + 1 = 18.67 y = 04.57 y = 14.30 14.30 + 11.4 + 4.9 = 30.6 Methd : (fllw thrugh errrs frm part a) Students use trig rati (1 pint) Students btain the length BC (½ pint) Students add t btain the height requested (½ pint) tan 50 X 50º 1 tan 50 14.3 1 14.30 + 11.4 + 4.9 = 30.6 30. a (1 pint fr the diagram; distances d nt have t be labelled) A 150 A 450 450 B 550 B 1000 300
b Students set up triangle f interest (½ pint) Students set up and trig rati t determine bearing (½ pint) Students slve fr angle in triangle and determine bearing (1 pint) Students set up either Pythagrean therem r trig rati crrectly t slve fr AB (1 pint) Students slve fr the distance AB (1 pint) Tan A 150 = 450 A = 18.43 Bearing is 360 18.43 = 341.57 150 + 450 = C r cs 18.43 C = 474.34. r sin 18.43. x = 474.34 x = 474.34
31. a Students state angles and sides fr a triangular regin (1 pint) Students determine crrectly height f the triangle (1.5 pint) Students determine the area the triangle (1.5 pint) Students determine the area f the prism s base (1 pint) Students determine the vlume f the prism (1 pint) 60 360 6 = 60 30 4 cm 60 h tan 60 = h = 3.46 cm 1 A = bh 1 A = (4)(3.46) A = 6.9 cm 1 A = bh 1 A = ()(3.46) A = 3.46 cm 6A = 41.57 cm r 6A = 41.5 cm 1A = 41.57 cm r 1A = 41.5 cm V V V = = = bh 41.57 6 r 41.5 6 49.4 cm 3 r 49.1 cm 3 b. Students replace vlume by 300 and height by 6 ( 1 pint) Students slve fr r (1 pint) Student determines the crrect slutin (1 pint) V cylinder = π r h 300 = π (r) (6) 50 = r π ; r = 3.99 cm r cm
3. a. Students let the sum f the three sides ttal 71 ( 1 pint) Students simplify crrectly the sum perimeter expressin crrectly (1.5 pints) Students slve fr x crrectly (1/ pint) 3( x + 14) + ( + 3x) + 3 + x = 71 3x + 4 4 + 6x + 3 + x = 71 5x + 41 = 71 5x = 30 x = 6 b. ½ (x + 1)(4x + 8) = (x - 1)(x + 7) Expressin fr area f triangle (1 pint) (1 pint) (1 pint) Expressin fr area f rectangle (1 pint) Setting tw areas equal t each ther (1/ pint) Slving fr x (1.5 pints) (x + 1)(x + 4) = (x - 1)(x + 7 4x + x + 8x + 4 = 4x x + 14x 7 10x + 4 = 1x 7 11 = x 11 = x x = 5.5