Workout 3

5. 24,412.5

6. 24

7. 220

Answers

t . 38.5

?. 3.19

3. t19.5

4. tzgfl

Answers

l. 15 or 15.00

2. 40.80

3. 70 or 7O.OO

4. t5

Answers

1. (i) 70.8

(i i) No

2.2

3. 1.5

Answers

1. 57

2. negative

3. see solutions page

4. t5

Difficulty

(4)

(3)

(3)

(4)

Difficulty

(1)

(1)

(2)

(2)

(4)

(5)

(5)

8. t67

9. 1*

rc#

(4)

(5)

(6)

(s)

(3)

(4)

(2)

(3)

(3)

8. 25

9. 20

10.37.5

Percent Stretch

5. 64.60

6. 40

7. 0 or 0.00

* The plural form of the units will always be provided in the answer blank even if the answer appearsto reguire fhe singular form of units.

MATHCOUNTS 2009-2010

What About Math? - NSPE

4. 73,t25

5. 8047

6. 2.78

7. 1.69

What About Math? . ASA

5. 200

6. 327

7. see solutions poge

8. 900

9. (i) 4e57

(ii) t60,271

10. 51

8. 0.0005

9. 0.0003

tO. see solutions poga

Probleml0. Hal fof l000givesusthe500oddintegers.Nowwehavel ,3,5, ' l ,9, l l ,13,15, l7,19,. . .999.Thereareatotalof 1000-5:200 multiples of 5 from I to 1000. However, only half of them--rr 100 of them-are odd. Taking these out, we have 500 - 100 = 400 integersfrom I to 1000, inclusive, that are not divisible by 2 or 5.

Warm-Up 6

Probfeml. I f fourof the24crayonsareashadeofgreen,theprobabi l i tyofKiaselect ingacrayonthat isashadeofgreenis4124:116.

Problem 2. T\e 48 pieces that Evan originally grabbed was 3/7 ofthe candy. This means that ifthe candy had been evenly divided into 7portions, Evan grabbed 3 of those portions. Each portion then would be 48 = 3 : 1 6 pieces. When Evan and his six friends re-distributed thecandy evenly amongst themselves, each ofthe seven people received ll7 ofthe candy, which is 16 pieces ofcandy.

Problem3. Thevolumeoftherectangularpr ismis12x6x24:(22x3)x(2x3)x(23x31=2a x33. I f th is isalsothevolumeofourcube,thesidelengthofthecubeisthecuberootof26 x 33,whichis 22x3or 12. Thus,t}resurfaceareaofthecubeis 6 x 122= 6x 144:864.

Problem4. Theleastcommonmult ip leof3,4,5 and6is60.Since16x60=960,thereareonly16posi t iveintegerslessthan1000thatare

divisible by all four numbers.

Problem 5. Working backward, we will subtract 3 and then divide by 3 to find each previous term. The fifth term is 39, so the fourth term is (39-3)13=36/3:12.Thethirdtermis(12-3)/3=913=3.Thesecondtermis(3-3)/3:0/3:0.Final ly, thef i rst termis(0-3)/3:-313:-1.

Problem 6. In the ratio 2'.3:4, there are a total of 2 + 3 + 4 = 9 equal parts. Dividing the $45,000 trust fund by 9, we get $5000 for each part. Thethree unequal parts the fund was divided into must be $10,000, $15,000 and $20,000. Multiplying each ofthese parts by their respective interestrates, we get $10,000 x 0.04: $400, $15,000 x 0.05 = $750 and $20,000 x 0.06: $1,200. The total interest income was 400 + 750 + 1200:$23s0.

Problem 7. The 6 vertices of the octahedron will be retained and 8 new vertices will be created when a tetrahedron is elued to each of the 8 facesofthe octahedron. The resulting solid has 14 vertices.

Problem 8. Ifthe two-digit integer "AB" has the digit A in the tens place and B in the units place, then its value is l0A + B. Jane multiplies theunits digit by 5 and the tens digit by 3 and gets 58 + 3A. Since this results in the same integer, we can say that 10A + B = 58 + 34. This reducesto 7A:48, which only can mean that A = 4 and B : 7 since they each must be single digits. Jane's number must have been 47.

Problem 9. The ratio of the volumes of two similar solids is equal to the cube of the scale factor, and the ratio of the surface areas is equal to thesquare ofthe scale factor. Taking the cube root of 1/64, we find the scale factor is 1/4. Squaring this, we get the desired ratio of 1/16.

Problem 10. Lattice points are points with integer coordinates. The line y : 7 is cut by the line y : x at (7 ,7) andbythel iney:-va1 (-7,7). There are 13 lat t icepointsbetween(7, 1)and(-7,7),butwewon' tbecount ingthesesince

they are on the line y:7. Below this line, we get 1 1 lattice points inside the triangle, then 9, then 7, etc. This is atotalof l l+9+7+5+3+1:36lat t icepointsstr ict ly insidethetr iangularregion.

Workout 3

Problem 1. The statement that B is 30Vo greater than A implies that B : 1.3A. Similarly, we have C = 1.8A. Comparing C to B by division, weget 1.8A/1.3A: 1.8/1.3 i 1.3846, which means that C is 38.5% greaterthan B, to the nearest tenth.

Problem 2. Mr. Gtzzler purchased a total of 18 + 15 + 12 + l0: 55 gallons of gas. We will weight each price by the number of gallonspurchasedtocomputea"weightedaverage"asfol lows:(18x2.99+15x3.09+12x3.29+10x3.59)/55-(53.82+46.35+39.48+35.90)/55: 175.55155 = $3.19.

Probfem 3. Using the first pair ofweights, we can set up the proportion 50/8.3 = 7201x, where x is the unknown weight. The cross product is50x = 720 x 8.3, so .r = (720 x 8.3)150, which is 119.5 N to the nearest tenth.

Problem 4. When the 3/4-inch sides are folded up, the dimensions of the base of the box will be 12 - (2)Qg: 10 % inches by 18 (2)(314) :

16 lz inches. The volume ofthe box is (10 %)(16 %)(3/4) : 129 15116 cubic inches.

Problem5. I f thef loor is3.5metersby4.5meters, theni t is350cmby450cm.Theareaofthef loor isthen350x450:157,500sqcm. I f 1 in: 2.54 cm, then 1 sq in : 2.542 sq cm = 6.4516 sq cm. Therefore, our 157,500 sq cm is equivalent to 157 ,500 + 6.4516 x 24,412.5 square inches.

Problern6. I f60 l i ters is3%salt , then60x0.03:1.8 l i tersmustbesal t .For l .8 l i tersofsal t tobe5Toofasolut ion, thesolut ionmustbel .8-

0.05 = 1.8 x 20:36 liters. Therefore, 60 36 :24 liters ofwater must be evaporated from the orisinal solution.

Problem 7. The first card drawn determines both the suit and the rank that the next three cards have to match. The probability that the nextthree cards match the suit is ( 12151)( 1 I 150)(10/49) - (12 x 1I x 10)/(51

" 50 x 49). The probability that the next three cards match the rank is

(3151)(2150)(1/49):(3 x 2x l ) l (51 x 50 x 49). Thenumeratorof thef i rstprobabi l i ty is (12 x 11 x 10)/(3 x2x I ) :220 t imesmorethanthenumerator ofthe second probability, so k= 220.

Problem8' Threethousandtonsis3000x2000=6,000,000pounds,whichis6,000,000x4=24,000,000quarter-poundpatt ies.At l00patt ies

perminute, i twouldtake24,000,000+100:240,000minutes,or240,000+60=4000hours,or4000-24= 16Tdays,tothenearestwholenumber ofdays.

MATHCOUNTS 2OO9-2010 51

Problemg, Assumethevalueofeachor ig inalcoinis20l(s incethat 'stheiror ig inalaveragevalue).Thenweaddaquarter.Thequarter is25l ,

but the new collection ofcoins has an average value of21f. So the quarter "loses" 4l in value, and those 4l are spread to the other original coins

to boost each of their values up by 1 p. Therefore, there must have been 4 coins in the original collection. The original collection was wofth a total

of 4(201,): 801. Thus, the four original coins must have been 3 quarters and I nickel.

Problem 10. Since the cone is a right circular cone, the altitude would be perpendicular to the circular base. There is a 3-4-5 triangle with

the right-angle veftex at the center of the base of the cone, another vertex at the center ofthe sphere and the other vertex somewhere on the

circumference ofthe base ofthe cone. Since the radius ofthe sphere is 5 cm, the height ofthe cone is 3 + 5 : 8 cm. The volume ofthe cone

is (1/3) x nx 42 x 8- l28nl3 cubic cm. The volume of the sphere is (4/3) x r x 53:500n/3 cubic cm. The ratio of the two is 128/500, which

reduces to 321125.

Percent Stretch

Probleml. Twentypercent isthesameasone-f i f th,andone-f i f thof$75is(1/5X$75):$15.

Problem 2. Since the price ofthe shoes is reduced by l5o/o,the sale price will be 85% ofthe original $48, or 0.85 x $48 : $40.80.

Problem 3. Since the shoes were marked down 30%, the $49 that Janet paid for the shoes represents 70Yo of the original price. This can be

written as $49 = (0.7).rr. The original price must have been $49 + 0.7 : $70.

Problem 4, Matilda paid $86 - $73.10: $12.90 less than the original price. This amounts to a discount of 12.90186 x 100: 15%.

Problem 5. When dealing with successive discounts, we should multiply by the portion of the price that is retained. Thus, after a 20% discount

and then a 15% discount, the $95 jacket would cost $95 x 0.8 x 0.85 : $64.60.

Problem6. Adiscountof25YomeansthebuyerpaysT5To,andadiscountof2}%omeansthebuyerpays80%.Afterthetwodiscounts,thebuyer

would pay 0.75 x 0.80 : 0.60, or 60"/o of the original price. This amounts to a single discount of 40%.

Problem 7. Since multiplication is cornmutative, 0.80 x 0.75 is the same as 0.75 x 0.80. This means there is no difference in the sale price if the

coupons are used in the opposite order. Our answer is $0.

Problem 8. The discount of20% brings the price ofthe coat to 4/5 ofwhat it was. To retum to fulI value, we multiply by the reciprocal of4l5,

which is 514. The corresponding percent is 125o/o, which is a 25o/o increase.

Problemg. The$l0i temincreasedby20Yowi l lsel l for$12. I fa$l5i temisreducedto$12,theni t issel l ingfor l2 l l5or4l5ot80%ofi ts

original value. That represents a 20Vo decrease.

Problem 10. The price ofthe table is 1.6 or 8/5 times its wholesale cost. The store cannot sell it for less than 5/8 ofthe price, so the discount

cannot be greater than 318 or 37.5o/o ofthe price.

52 MATHCOUNTS 2009-2010