r1 r2

2012 2013 UIL Calculator Study List Problems

09A-36. The half life of Uranium 230 is 20.8 days, the time needed

for 50% to decay. How long would it take for 20% of Uranium 230

to decay? --------------------------------------------------------------------- 36= ________________ days

At = A0ekt

k = 12ln

20.8 .8 = 1 e

kt t =

12

20.8ln(.8)

ln = 6.70

09A-38. If x2 + y

2 = 152 and x

2 – y

2 = -49, what is the smallest

value of (x + y)2 ? ---------------------------------------------------------- 38= ____________________

Add equations → 2x2 = 103 x2 = 103/2 x = 103/ 2

y2 = 100.5 y = 100.5

choose x = 103/ 2 y = 100.5 (x + y)2 = 8.11

*** TI 89 solver Look at the different pairs and pick the correct pair.

09F-37. A pipe has an outside diameter of 1.25 in and an inside

diameter of 0.75 in. If Kelly hacksaws the pipe in two, what

fraction of the pipe cross sectional area is sawn when the blade

breaks through to the inside? --------------------------------------------- 37= __________________ %

cos α = .75 / 1.25 α = 53.13

θ = 2α = 106.26 → 1.85… (rad)

r1 = .75/2 , r2 = 1.25/2

22

2

2 22 1

( sin )100%

r

r r

= 22.2

09G-36. In an egg toss, a strategy is to throw the egg with the

lowest possible velocity but still cover the distance to the catcher.

If the catcher is 25 ft from the thrower, what is this velocity? ------- 36= _________________ ft/s

Lowest V occurs when θ = 45°

max

2 sin 2h

vd

g

2(1)25

32.17

v v = 28.4

25

16

1

8

25

16

9

9

Page 2

09I-36. A dog is tied to a 16 ft square shed with a 25 ft long leash.

What is percent difference in the dog’s roaming area if it is tied in

the middle of one side or on a corner? ---------------------------------- 36= __________________ %

½ (25)2 + ½ (17)

2 + ½ (1)

2 = 1437.278…

¾ (25)2 + ½ (9)

2 = 1599.856…

1437.278…, 1599.856…, % chg = 11.3

10D-37. A major league baseball pitcher throws a fast ball at

98.5 mph. If the release is 5 ft 6.7 in above the ground and

horizontal, how far would the ball travel unencumbered before

hitting the ground? --------------------------------------------------------- 37= ______________ ft(SD)

Vertical: y = y0 + v0t + ½ at2 but y and v0 = 0 so -2y0 / a = t

2

t =

06.72 52y 12

a 32.17

= .5878…

Horizontal: v = 98.5 (88/60) = 144.466.. ft/s

x = x0 + v0t + ½ at2 but x0 and a = 0 so x = (144.466…)(.5878…)

x = 84.9 {3SD}

10D-38. A red car sitting still is passed by a green car traveling at

a velocity v. After a 1.35 sec delay, the red car accelerates at

10.13 mph/sec and catches up with the green car in 13.57 sec.

How fast was the green car going?--------------------------------------- 38= ____________ mph (SD)

d = d

½ at12 = vt2

½ (10.13)(13.57)2 = v(13.57 + 1.35)

4SD 4SD 4SD

v = 62.51 {4 SD}

Page 3

10G-36. What is 349,441-902,521

? ---------------------------------------- 36= ____________________

-902,521 log 349,441 = -5003011.31907

Use the integer part less 1 for the new exponent: 10-5003012

Add +5003012 to what’s in the calculator: .68093 → 10.68093

= 4.80

= 4.80 10-5003012

10H-36. Amy can bike to the store in 7 min, and walking takes 25

min. What percent of the distance to the store was traversed by

bike if she had a flat along the way, and the total travel time was

12 minutes? ----------------------------------------------------------------- 36= __________________ %

12 = x (1 x)

1 17 25

x = .722 .722 100% = 72.2

or (1/7)t + (1/25)(12 – t) = 1 t = 91/18 (1/7)(91/18) = .722 100% = 72.2

10H-38. In Seguin, Texas one day the high was 101°F at 5 PM and

the low was 76°F at 5 AM. Assuming the temperature varied

sinusoidally, how many hours after 9 AM did the temperature

first hit 93°F? ---------------------------------------------------------------- 38= __________________ hr

T(t) = A cos (2π(f)(t)) + D

A = amplitude (H – L)/2 (101 – 76)/2 = 12.5 → -12.5

{must be negative, since we started with the high}

f = frequency (1/period) (1/24)

D = vert. displacement or centerline (76 + 101) / 2 = 88.5

93° = -12.5 cos (πt / 12) + 88.5 t = 7.4066… hours

5 AM + 7.4066797 – 9 AM = 3.41

10I-36. Two integers, each greater than 100, multiply to 46,620.

Both are divisible by 6, and the smaller number is divisible by 5

as well. What is the sum of the two integers? ------------------------ 36= ______________ integer

46620 = 215.916…

215 is divisible by 5.

However, it does not divide evenly into 46620 and is not divisible by 6..

Try next lower number divisible by 5 → 210. This is also divisible by 6.

This does divide evenly into 46620. → 222. This is also divisible by 6.

210 + 222 = 432

8d

12

18

12

3960

3960

x

Page 4

10I-37. What is the minimum elevation above the earth’s surface

necessary to have a direct line of sight to both Fiji and New Zealand

if they are 1314 mi apart? Neglect refraction. -------------------------- 37= _________________ mi

arc length = r θ 1314 = 3960(θ) θ = 19.011…°

cos (19.011…° / 2) = 3960 / (3960 + x)

x = 55.1

11B-38. What is the electrical resistance of an immersion hot water

heater necessary to bring 8-oz of water from room temperature, 25°C, to

its boiling point, 100°C, in 60 sec? The specific heat of water is 4.19

J/gK. Power generated by the heater in watts (= J/s = Vamp = V2/ohm)

is the product of the voltage, 110 V, and the current in amps. The

current is the voltage divided by the resistance in ohms. ------------------ 38= _____________ ohm

Q = energy P = power V = voltage I = current

R = resistance T = temperature t = time m = mass c = specific heat

Q = mcT P = VI P = Q / t Q / t = VI Q = VIt I = V/R

mcT = VIt = V(V/R)t

R = (V2

t) / (mcT) and m = 8 oz.(liq) = 236.588 ml = 236.588 g

R = (1102(60)) / (236.588)(4.19)(100 – 25) = 9.76

11D-38. Mike and Mary stand unmoving 24 ft apart. Ned stands 18 ft from

Mike and 8 ft from Mary, forming a scalene triangle. What is the

shortest distance Ned can move to create a right triangle? ---------------- 38= _______________ ft

If let “d’ be the median to the 24 foot side and use the formula

for finding median lengths in triangles:

2 2 22

2c

a b cm

; where a + b > c.

d = ½ 2 2 22(18 8 ) 24 d = 7.071… 12 – d = 4.93

167

S95

A

C

D

Page 5

11E-26. Harvey in Aspermont, TX wants to arrange a meeting with Mike

in Childress which is 95 mi north, and also with Harry who lives in

Decatur, 17 mi due east. If the meeting site selected was Seymour

which is equidistant from all three towns, how far does each person

travel? ---------------------------------------------------------------------------- 26= ______________ mi

S is located at the circumcenter of a right triangle.

2 295 167

2

= 96.1

11E-38. The distance x a carbon atom moves in solid iron equals

Q0 RT2D t exp where D0 = 0.2 cm2/s, t = elapsed time, Q = 32,000 cal/mole, R = 1.987 cal/(mole-K) and T = absolute temperature (K).

Carbon is placed on the surface of iron powder 55m in diameter.

Calculate the minimum temperature necessary to diffuse the carbon to the

center of the iron particles in 3 hr.? ------------------------------------------- 38= ______________ °C

(55 10-4

cm) / 2 = .00275

.00275 = 32,000

1.9872(.2)(3 3600)

Te

Solve: T = 798.7117…K 798.7117… – 273.15 = 526

11F-37. A kid used a clothes pin to attach a playing card to a spoke

on the front wheel of his bike. Once every wheel rotation, the card

“clicked” when it struck the bike frame yoke. If the front wheel has an

18-in diameter, what is the bike velocity when clicks are ¼ s apart? ---- 37= _____________ mph

(18π in/rev) (4 rev/s) (1 ft/12 in) (1 mi/5280 ft) (3600 s/hr) = 12.9

11G-26. A surveyor estimated a rectangular field to be 2.93 acres. The

subsequent survey produced side dimensions of 221.52 ft and 589.91 ft.

What is the percent error of the estimate? ----------------------------------- 26= ___________ %(SD)

A = (221.52)(589.91) = 130,676.86…

A (640 acres / 52802) = 2.99992… = B

[(2.93 / B) {3SD} – 1] {2SD} 100% = -2.3

or

B, 2.93, % chg = -2.3

y = b

r

ds

Page 6

11G-37. The lines y = 12x, y = 4x+8 and y = b intersect to form a

triangle. Solve for b if the length of the line segment associated with

y = b equals 15 and b is negative. --------------------------------------------- 37= _________________

x2 – x1 = 15

y / 12 – (y – 8) / 4 = 15

y = -78.0

or

12x = 4(x – 15) + 8

x = -6.5

y = 12(-6.5) = -78.0

11H-37. It takes Abe 30 min to row his boat 2 mi up river. He rows back

to where he started in 10 min. What is the river velocity? ---------------- 37= _____________ mph

(b – c) (30/60) = 2 (b + c)(10/60) = 2

Solve: b = 8 c = 4.00

11I-38. Marin runs a mile in 5 min 30 sec. She and Mary, who runs a

mile in 7 min 26 sec, start at the south end of a ¼ mi circular track.

Marin takes off running on the track, and at the same time, Mary takes

off in a straight line off the track, meeting up with Marin before she

finishes one lap. Considering due east to be 0°, at what positive angle

does Mary need to run? --------------------------------------------------------- 38= ______________ rad

rt = d

t/446 = d t/330 = s t = 330s

sin (θ/2) = d/2r s = θ r θ = s/r

d/s = 330 / 446 d = 330 s / 446

nsolve sin (x/2) = 330x / [446(2)] x = 2.60706487 rad

(π – x)/2 = .267263892 π /2 - .2672 = 1.30

12B-37. Bulk potatoes are sliced into 0.02 in thick slices to make potato chips. A potato loses 75% of its mass during frying. Potatoes on average are 6 in long, weigh 0.8 lb and cost $0.88/lb. An 11-oz bag of chips costs $2.58 and has 128 chips. What is the ratio of the cost of chips to the cost of the starting bulk potatoes needed to make those chips (on a mass basis)? ------------------------- 37= ___________________ 6 / .02 = 300 slices 128/300 × (.8 lb) × ($.88/lb) = $0.30037… 2.58 / .30037… = 8.59

12

3

6

9

Page 7

12D-36. How long after 5:30 do the minute and hour hands of a clock first align? ------------------------------------------------------------------------------------- 36= ___________________ min 11/12 T = 30 + 27.5 T = 62.7 Alternate Solution:

3011 30 3060

5.5 / min

2D-38.One thousand synchronized lights are place in a line spaced 1 ft apart. The first light flashes on/off, then after a time delay τ, the second light flashes similarly, then the third, etc., such that the light blip “moves” along the row with constant velocity. What is τ if the apparent blip velocity is 1.5 times the speed of light, 186,000 mi/s? ------------------------------------------------------------------- 38= __________________ μsec 1.5(186,000) t = 1 / 5280 (10

6) t = .000679

12E-37. It takes 1.5 J (= 1 Nm) of energy to knock over a bowling pin and move it out of the way. What is the minimum velocity of a 15-lb bowling ball to bowl a strike, knocking over all 10 pins? The bowling ball energy is 0.5 mv

2 where m is

the ball mass and v is its velocity. 1 N = 1 kgm/s2. ------------------------------ 37= ___________________ fps

15 lb → 6.80388… kg (HP conversion) 10(1.5) = ½ (6.80388..)v

2

v = 2.09982… m/s × 100 cm/m × 1 in/ 2.54 cm × 1 ft/12 in. = 6.89

12F-37. In Olympic archery, the archer shoots an arrow with an average speed of 320 ft/s at a target 70 meters away. Assuming the arrow is released at an elevation equivalent to the target’s bulls eye, what should the archer’s release angle be (positive, less than 45° with 0° parallel to the ground)? ------------ 37= ___________________ deg 70 m × 100 cm/ 1 m × 1 in/ 2.54 cm × 1 ft/ 12 in. = 229.658… ft {A} dh max = (v

2 sin 2θ) / g {A} = 320

2 sin 2θ / 32.17 θ = 2.07

12G-36. Sam begs for bread from a local baker. The first day, Sam received a full loaf of bread. On the second and subsequent days, Sam got 40% of the preceding day’s allocation. How much total bread will Sam ultimately receive? 36= _____________ loaves S = 1 / (1 -.4) S = 1.67

12H-37. A chair was caught up in a tornado, rotating around a 1500 ft circumference with a speed of 110 mph. After picking up the chair, the tornado traveled 5 mi at 35 mph before dissipating. How far did the chair travel? - 37= ____________________ mi 5/35 = 1/7 hr 110(1/7) + 5 = 20.7

Page 8

12I-36. Sam makes coffee by adding one teaspoon of coffee concentrate to 6 oz of water. How much coffee concentrate is needed to make 25 gallons of coffee? 36= ______________ qt

(1 / 6)

6 (1 / 6) 25(128)

x

x = 86.486… oz x / 32 oz = 2.70

12I-38. Sonya drove 30% of the distance to her destination at 56 mph. She then sped up so her total average trip velocity was 63 mph. What was her velocity on the second leg of the trip? ----------------------------------------------- 38= __________________ mph

Let distance = 63 mi. .3(63) .7(63) 63

56 63V V = 66.6

sin 82 / 389 = sin A / 182

A = 27.6

180 - (82 + 27.6) = B = 70.4

sin 70.4 / 518 = sin x / 389

x = 45.0

tanA = 1

2 3 2 3

x

x 16.102

tanB = 30

tanC = 180 - (A + B) = 133.89…

½ bh - 2 sin sin

2sin

c A B

C = shaded area

½ (x)( 2 3x ) - 2( 3) (sin16.102)(sin30

2(sin133.89...)

x = .904

2

2 23 (sin16.102)(sin30)3 (1.44...) .9042(sin133.89...)

xx x

x = .7914… 2x = 1.58

04A-69.

04A-69=_______________________________

SCALENE TRIANGLES

82

182

389 518

deg?

REGULAR HEXAGON

04B-70=_______________________________

04B-70.

Shaded Area = 0.904

?

midpoint

center

Page 9

Use the small similar triangle to find the angle:

Tan θ = (5.69 – 4.67) / 1.44 = 1.02 / 1.44

θ = .616

w2 = (2.14)

2 + (4.05)

2 – 2(2.14)(4.05) cos .87

w = 3.131242

sin .87 / w = sin θ / 4.05 = 1.4196 rad

But 1.4196 rads is too small ( < /2).

This is an example of the ambiguous case for a SSA

triangle.

So: 1.4196 = 1.7219658 rads =

θ = 1.7219658 rad

β = π – (.65 + θ)

β = .7696268 rad

α = π – (1.41 + β)

α = .9619658 rad

sin α / z = sin 1.41 / w

z = 2.60218

x2 = 3.02

2 + z

2 – 2(3.02) z cos (.65)

x = 1.84

05C-60.

05C-60 = _______________________

CIRCULAR ARC, RIGHT TRIANGLES

rad? 1.44

5.694.67

rad? 1.44

5.694.67

5.69 - 4.67

06E-60.

06E-60. = _______________________

SCALENE TRIANGLES

?

4.051.41 rad

0.65 rad 0.87 rad

3.02 2.14

x

4.051.41

.65 .87

3.02 2.14

zw

A

1- A

1

Page 10

1.72 = 2.18

2 + 1.18

2 – 2(2.18)(1.18) cos A

A = 50.7549…

B = A + 21.5 = 72.2549

x2 = 2.18

2 + 1.82

2 – 2(2.18)(1.82) cos B

x = 2.38

Let side = 1

A = ½ bh

A = ½ (1)(1 - A ) Use TI solver.

A = .25 → A = .5

tan θ = (1 - .5)/ 1 = ½

θ = 26.6

07E-60.

SCALENE TRIANGLES

07E-60 = _______________________

2.18

?

1.70

1.82

1.18

21.5o

08C-60.

08C-60 = _______________________

AREA(TRIANGLE) = A

SQUARE AND RIGHT TRIANGLE

deg?

A

?

747 747

r

60

747 747

w

508

830

992

h

508

992

xw

x

Page 11

360 (60 )

360

π(747)

2 = 1.33 10

6

θ = 26.87...

cos θ = w / 747

w = 666.326…

2w = 1332.65…

2w – 747 = 586

5082 = 508

2 + 992

2 – 2(508)(992) cos x

x = 12.478…

tan x = w/ 830 w = 183.67…

h = 2w = 367.35…

tan θ = 367.35… / 992 θ = .355

08E-60.

08E-60 = _______________________

?

AREA(SECTOR) = 1.33X106

SECTOR, EQUILATERAL AND ISOSCELES TRIANGLES

747

09F-60.

09F-60 = _______________________

RECTANGLE AND ISOSCELES TRIANGLES

AB is a straight line segment

508 B

A

830 992

rad?

x

14o

r

w

2

Page 12

πr2/2 = ½ bh b = r

πr2

= rh

h = πr

let r = 1 h = π

tan θ = π/2

θ = 57.5

w = r tan 14°

height = r

rect = wr = r2 tan 14°

570 = ¾ πr2 + r

2 tan 14°

r = 14.8

57,200 = πR

2 – π(R – 91.5)

2

R = 145.2…

r = 145.2… - 91.5 = 53.7

10D-60.

THREE-QUARTER CIRCLE AND RECTANGLE

Total Area = 570

10D-60 = _______________________

R = ?

14.0o

10F-60.

10F-60 = _______________________

Hatched Area = 57,200

CONCENTRIC CIRCLES

R = ?

91.5

10A-60.

RIGHT TRIANGLE AND SEMICIRCLE

10A-60 = _______________________

Semicircle Area = Right Triangle Area

Radius

deg?

3.89

2.05

2.05

d

2.05d

Page 13

Vc - Vf = Volume

V = π(r)2h – 1/3 π (r1

2 + r2

2 + r1r2)h

V = π(90.5/2)2(88.8) – 1/3 π [(90.5/2)

2 + (40.5/2)

2 +

(90.5/2)(40.5/2)]88.8

V = 257,000

d2 + 2.05

2 = 3.89

2

d = 3.3059…

tan θ = 2.05/(2.05 + d) θ = 20.9°

11B-49.

CYLINDER WITH FRUSTUM CAVITY

11B-49 = _______________________

Volume = ?

88.8

40.5

90.5

11D-60.

SQUARE AND RIGHT TRIANGLES

11D-60 = _______________________

3.89

2.05

deg?

L 86.4

44.8

w

419 - x

Page 14

176 – 89.6 = 86.4

44.82 + 86.4

2 = L

2

L = 97.32420…

A = ½ (89.6)L = 4360.12…

TA = 4A + 5(89.6)2

TA = 57,600

V = (1/3)BH B = ½ bh

V = (1/3)(1/2)(419 – x)(w)(h)

V = [(419 – x)(w)(h)] / 6

(419 x)(w)(h)

6(419 x)(w)(h)

419(w)(h)6

= .1

419 x

(419 x)6 419

6

= .1 Solve x = 190

11E-49.CUBE AND PYRAMID

Total Surface Area = ?

11E-49 = _______________________

89.6

176

11G-50.

TRUNCATED RECTANGULAR SOLID

?

11G-49 = _______________________

419

Missing Corner VolumeTruncated Solid Volume

= 0.1

Page 15

40.92 = 44.3

2 + 18.1

2 – 2(44.3)(18.1)cos θ

θ = 67.360…

sin θ / 40.9 = sin / 18.1

= 24.107…

½ (18.1)(x)sin θ = ½ (40.9)(44.3 – x)sin

x = 22.15

AB2 = x

2 + (18.1)

2 – 2(x)(18.1)cos θ

AB = 22.6

Let radius of smaller sphere = 1. (4/3)πr

3 = 9.56 (2/3)π(1)

3

2r

3 = 9.56 r = 1.6845…

Sin θ = r/(r+1) θ = 38.9

11I-60.

11I-60 = _______________________

SCALENE TRIANGLES WITH EQUAL AREA

AB = ?

40.9

A

18.1

B

44.3

12A-49.

SPHERE AND HEMISPHERE

deg?

12A-49 = _______________________

Volume (Sphere) =9.56 Volume (Hemisphere)

r

r + 1

θ

12A-60.RECTANGLE

AB = BC = 0.539 BD = ?

12A-60 = _______________________

0.194

0.468

A

B

CD

Page 16

AC

2 = .194

2 + .468

2 AC = .5066….

BC2 = AC

2 + AB

2 – 2(AC)(AB) cos BAC

BAC = 61.9683…

tan CAD = CD / AD

CAD = 67.4845…

BAD = BAC +CAD = 129.4528…

BD2 = AD

2 + AB

2 – 2(AD)(AB) cosBAD

BD = .679

V = πr

2h + (1/3)π r

2 h

V = (4/3)π r

2 h

.0634 = (4/3)π (.331/2)

2 h

h = .553

12E-49.SLANT CIRCULAR CYLINDER AND CONE

Combined Total Volume = 0.0634

12E-49 = _______________________

?

0.331 0.331

560b

45o

60o

75o

560 2560 2

Page 17

cos 75° = b / 560 2 b = 204.974… A = ½ ab sin C

A = ½ (560 2 )(204.974…) sin 75°

A = 78,400

12I-60.SQUARE, EQUILATERAL AND RIGHT TRIANGLE

12I-60 = _______________________

Hatched Area = ?

560