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2010PascalSolution滑铁卢竞赛题答案

1.In cents,the?ve given choices are50,90,95,101,and115cents.

The di?erences between each of these and$1.00(or100cents),in cents,are

100?50=50100?90=10100?95=5101?100=1115?100=15 The di?erence between$1.01and$1.00is the smallest(1cent),so$1.01is closest to$1.00.

Answer:(D) https://www.sodocs.net/doc/475526523.html,ing the correct order of operations,

(20?16)×(12+8)

4×20

=20

Answer:(C)

3.We divide the750mL of?our into portions of250mL.We do this by calculating750÷250=3.

Therefore,750mL is three portions of250mL.

Since50mL of milk is required for each250mL of?our,then3×50=150mL of milk is required in total.

Answer:(C) 4.There are8?gures in total.Of these,3are triangles.

Therefore,the probability is3.

Answer:(A) 5.We simplify the left side and express it as a fraction with numerator1:

1 9+

Therefore,the number that replaces the is6.

Answer:(C) 6.There are16horizontal segments on the perimeter.Each has length1,so the horizontal

segments contribute16to the perimeter.

There are10vertical segments on the perimeter.Each has length1,so the vertical segments contribute10to the perimeter.

Therefore,the perimeter is10+16=26.

(We could arrive at this total instead by starting at a?xed point and travelling around the outside of the?gure counting the number of segments.)

Answer:(E) 7.Since33=3×3×3=3×9=27,then

√

33+33+33=√

27+27+27=

√

81=9

Answer:(B)

8.The di?erence between the two given numbers is7.62?7.46=0.16.

This length of the number line is divided into8equal segments.

The length of each of these segments is thus0.16÷8=0.02.

Point P is three of these segments to the right of7.46.

Thus,the number represented is7.46+3(0.02)=7.46+0.06=7.52.

Answer:(E)

9.A 12by 12grid of squares will have 11interior vertical lines and 11interior horizontal lines.(In the given 4by 4example,there are 3interior vertical lines and 3interior horizontal lines.)Each of the 11interior vertical lines intersects each of the 11interior horizontal lines and creates an interior intersection point.

Thus,each interior vertical line accounts for 11intersection points.

Therefore,the number of interior intersection points is 11×11=121.

Answer:(B)

10.Because the central angle for the interior sector “Less than 1hour”is 90?,then the fraction of the students who do less than 1hour of homework per day is 90?360?=14.In other words,25%of the students do less than 1hour of homework per day.

Therefore,100%?25%=75%of the students do at least 1hour of homework per day.

Answer:(E)

11.Solution 1

Since there is more than 1four-legged table,then there are at least 2four-legged tables.

Since there are 23legs in total,then there must be fewer than 6four-legged tables,since 6four-legged tables would have 6×4=24legs.

Thus,there are between 2and 5four-legged tables.

If there are 2four-legged tables,then these tables account for 2×4=8legs,leaving 23?8=15legs for the three-legged tables.

Since 15is divisible by 3,then this must be the solution,so there are 15÷3=5three-legged tables.

(We can check that if there are 3or 4four-legged tables,then the number of remaining legs is not divisible by 3,and if there are 5four-legged tables,then there is only 1three-legged table,which is not allowed.)

Solution 2

Since there is more than 1table of each type,then there are at least 2three-legged tables and 2four-legged tables.

These tables account for 2(3)+2(4)=14legs.

There are 23?14=9more legs that need to be accounted for.These must come from a combination of three-legged and four-legged tables.

The only way to make 9from 3s and 4s is to use three 3s.

Therefore,there are 2+3=5three-legged tables and 2four-legged tables.

Answer:(E)

12.Solution 1

The total area of the rectangle is 3×4=12.

The total area of the shaded regions equals the total area of the rectangle (12)minus the area of the unshaded region.

The unshaded region is a triangle with base of length 1and height 4;the area of this region is 12(1)(4)=2.Therefore,the total area of the shaded regions is 12?2=10.

Solution 2

The shaded triangle on the left has base of length 2and height of length 4,so has an area of 12

(2)(4)=4.

The shaded triangle on the right has base of length3(at the top)and height of length4,so

has an area of1

2(3)(4)=6.

Therefore,the total area of the shaded regions is4+6=10.

Answer:(C)

13.Since the ratio of boys to girls at Cayley H.S.is3:2,then3

3+2=3

of the students at Cayley

H.S.are boys.

Thus,there are3

5(400)=1200

=240boys at Cayley H.S.

Since the ratio of boys to girls at Fermat C.I.is2:3,then2

2+3=2

of the students at Fermat

C.I.are boys.

Thus,there are2

5(600)=1200

=240boys at Fermat C.I.

There are400+600=1000students in total at the two schools.

Of these,240+240=480are boys,and so the remaining1000?480=520students are girls.

Therefore,the overall ratio of boys to girls is480:520=48:52=12:13.

Answer:(B) 14.When the given net is folded,the face numbered5will be opposite the face numbered1.

Therefore,the remaining four faces share an edge with the face numbered1,so the product of the numbers is2×3×4×6=144.

Answer:(B)

15.The percentage10%is equivalent to the fraction1

10.

Therefore,t=1

10s,or s=10t.

Answer:(D)

16.Since the base of the folded box measures5cm by4cm,then the area of the base of the box

is5(4)=20cm2.

Since the volume of the box is60cm3and the area of the base is20cm2,then the height of the box is60=3cm.

Therefore,each of the four identical squares has side length3cm,because the edges of these squares form the vertical edges of the

box.

Therefore,the rectangular sheet measures3+5+3=11cm by3+4+3=10cm,and so has area11(10)=110cm2.

Answer:(B) 17.Solution1

Since SUR is a straight line,then∠RUV=180??∠SUV=180??120?=60?.

Since P W and QX are parallel,then∠RV W=∠V T X=112?.

Since UV W is a straight line,then∠RV U=180??∠RV W=180??112?=68?.

Since the measures of the angles in a triangle add to180?,then

∠URV=180??∠RUV?∠RV U=180??60??68?=52?

Solution2

Since SUR is a straight line,then∠RUV=180??∠SUV=180??120?=60?.

Since P W and QX are parallel,then∠RST=∠RUV=60?.

Since ST X is a straight line,then∠RT S=180??∠V T X=180??112?=68?.

Since the measures of the angles in a triangle add to180?,then

∠URV=∠SRT=180??∠RST?∠RT S=180??60??68?=52?

Answer:(A) 18.Solution1

When Catherine adds30litres of gasoline,the tank goes from1

8full to3

full.

Since3

4?1

,then5

of the capacity of the tank is30litres.

Thus,1

8of the capacity of the tank is30÷5=6litres.Also,the full capacity of the tank is

8×6=48litres.

To?ll the remaining1

4of the tank,Catherine must add an additional1

×48=12litres of gas.

Because each litre costs$1.38,it will cost12×$1.38=$16.56to?ll the rest of the tank. Solution2

Suppose that the capacity of the gas tank is x litres.

Starting with1of a tank,30litres of gas makes the tank3full,so1x+30=3x or5x=30or x=48.

The remaining capacity of the tank is1

4x=1

(48)=12litres.

At$1.38per litre,it will cost Catherine12×$1.38=$16.56to?ll the rest of the tank.

Answer:(C) 19.The area of a semi-circle with radius r is1πr2so the area of a semi-circle with diameter d is

1 2π(1

d)2=1

πd2.

The semicircles with diameters UV,V W,W X,XY,and Y Z each have equal diameter and

thus equal area.The area of each of these semicircles is1

8π(52)=25

π.

The large semicircle has diameter UZ=5(5)=25,so has area1

8π(252)=625

π.

The shaded area equals the area of the large semicircle,minus the area of two small semicircles, plus the area of three small semicircles,which equals the area of the large semicircle plus the area of one small semicircle.

Therefore,the shaded area equals625

8π+25

π=650

π=325

π.

Answer:(A)

20.The sum of the odd numbers from5to21is

5+7+9+11+13+15+17+19+21=117

Therefore,the sum of the numbers in any row is one-third of this total,or39.

This means as well that the sum of the numbers in any column or diagonal is also39.

Since the numbers in the middle row add to39,then the number in the centre square is 39?9?17=13.

Since the numbers in the middle column add to39,then the number in the middle square in the bottom row is39?5?13=21.

91317

x21

Since the numbers in the bottom row add to39,then the number in the bottom right square is39?21?x=18?x.

Since the numbers in the bottom left to top right diagonal add to39,then the number in the top right square is39?13?x=26?x.

Since the numbers in the rightmost column add to39,then(26?x)+17+(18?x)=39or 61?2x=39or2x=22,and so x=11.

We can complete the magic square as follows:

19515

91317

11217

Answer:(B) 21.We label the numbers in the empty boxes as y and z,so the numbers in the boxes are thus

8,y,z,26,x.

Since the average of z and x is26,then x+z=2(26)=52or z=52?x.

We rewrite the list as8,y,52?x,26,x.

Since the average of26and y is52?x,then26+y=2(52?x)or y=104?26?2x=78?2x.

We rewrite the list as8,78?2x,52?x,26,x.

Since the average of8and52?x is78?2x,then

8+(52?x)=2(78?2x)

60?x=156?4x

3x=96

x=32

Therefore,x=32.

Answer:(D) 22.Since JKLM is a rectangle,then the angles at J and K are each90?,so each of SJP and

QKP is right-angled.

By the Pythagorean Theorem in SJP,we have

SP2=JS2+JP2=522+392=2704+1521=4225

Since SP>0,then SP=√

4225=65.

Since P QRS is a rhombus,then P Q=P S=65.

By the Pythagorean Theorem in QKP,we have

KP2=P Q2?KQ2=652?252=4225?625=3600

Since KP>0,then KP=√

3600=60.

(Instead of using the Pythagorean Theorem,we could note instead that SJP is a scaled-up version of a3-4-5right-angled triangle and that QKP is a scaled-up version of a5-12-13 right-angled triangle.This would allow us to use the known ratios of side lengths to calculate the missing side length.)

Since KQ and P Z are parallel and P K and W Q are parallel,then P KQW is a rectangle,and

so P W=KQ=25.

Similarly,JP ZS is a rectangle and so P Z=JS=52.

Thus,W Z=P Z?P W=52?25=27.

Also,SY RM is a rectangle.Since JM and KL are parallel(JKLM is a rectangle),JK and ML are parallel,and P Q and SR are parallel(P QRS is a rhombus),then∠MSR=∠KQP and∠SRM=∠QP K.

Since SMR and QKP have two equal angles,then their third angles must be equal too.

Thus,the triangles have the same proportions.Since the hypotenuses of the triangles are equal, then the triangles must in fact be exactly the same size;that is,the lengths of the corresponding sides must be equal.(We say that SMR is congruent to QKP by“angle-side-angle”.) In particular,MR=KP=60.

Thus,ZY=SY?SZ=MR?JP=60?39=21.

Therefore,the perimeter of rectangle W XY Z is2(21)+2(27)=96.

Answer:(D) 23.First,we note that2010=10(201)=2(5)(3)(67)and so20102=223252672.

Consider N consecutive four-digit positive integers.

For the product of these N integers to be divisible by20102,it must be the case that two di?erent integers are divisible by67(which would mean that there are at least68integers in the list)or one of the integers is divisible by672.

Since we want to minimize N(and indeed because none of the answer choices is at least68), we look for a list of integers in which one is divisible by672=4489.

Since the integers must all be four-digit integers,then the only multiples of4489the we must consider are4489and8978.

First,we consider a list of N consecutive integers including4489.

Since the product of these integers must have2factors of5and no single integer within10 of4489has a factor of25,then the list must include two integers that are multiples of5.To minimize the number of integers in the list,we try to include4485and4490.

Thus our candidate list is4485,4486,4487,4488,4489,4490.

The product of these integers includes2factors of67(in4489),2factors of5(in4485and 4490),2factors of2(in4486and4488),and2factors of3(since each of4485and4488is divisible by3).Thus,the product of these6integers is divisible by20102.

Therefore,the shortest possible list including4489has length6.

Next,we consider a list of N consecutive integers including8978.

Here,there is a nearby integer containing2factors of5,namely8975.

So we start with the list8975,8976,8977,8978and check to see if it has the required property.

The product of these integers includes2factors of67(in8978),2factors of5(in8975),and2 factors of2(in8976).However,the only integer in this list divisible by3is8976,which has only1factor of3.

To include a second factor of3,we must include a second multiple of3in the list.Thus,we extend the list by one number to8979.

Therefore,the product of the numbers in the list8975,8976,8977,8978,8979is a multiple of 20102.The length of this list is5.

Thus,the smallest possible value of N is5.

(Note that a quick way to test if an integer is divisible by3is to add its digit and see if this total is divisible by3.For example,the sum of the digits of8979is33;since33is a multiple of3,then8979is a multiple of3.)

Answer:(A)

24.We label the terms x1,x2,x3,...,x2009,x2010.

Suppose that S is the sum of the odd-numbered terms in the sequence;that is,

S=x1+x3+x5+···+x2007+x2009

We know that the sum of all of the terms is5307;that is,

x1+x2+x3+···+x2009+x2010=5307

Next,we pair up the terms:each odd-numbered term with the following even-numbered term.

That is,we pair the?rst term with the second,the third term with the fourth,and so on,until we pair the2009th term with the2010th term.There are1005such pairs.

In each pair,the even-numbered term is one bigger than the odd-numbered term.That is, x2?x1=1,x4?x3=1,and so on.

Therefore,the sum of the even-numbered terms is1005greater than the sum of the odd-numbered terms.Thus,the sum of the even-numbered terms is S+1005.

Since the sum of all of the terms equals the sum of the odd-numbered terms plus the sum of the even-numbered terms,then S+(S+1005)=5307or2S=4302or S=2151.

Thus,the required sum is2151.

Answer:(C) 25.Before we answer the given question,we determine the number of ways of choosing3objects

from5objects and the number of ways of choosing2objects from5objects.

Consider5objects labelled B,C,D,E,F.

The possible pairs are:BC,BD,BE,BF,CD,CE,CF,DE,DF,EF.There are10such pairs.

The possible triples are:DEF,CEF,CDF,CDE,BEF,BDF,BDE,BCF,BCE,BCD.There are10such triples.

(Can you see why there are the same number of pairs and triples?)

Label the six teams A,B,C,D,E,F.

We start by considering team A.

Team A plays3games,so we must choose3of the remaining5teams for A to play.As we saw above,there are10ways to do this.

Without loss of generality,we pick one of these sets of3teams for A to play,say A plays B,C and D.

We keep track of everything by drawing diagrams,joining the teams that play each other with

a line.

Thus far,we have

B C D

There are two possible cases now–either none of B,C and D play each other,or at least one pair of B,C,D plays each other.

Case1:None of the teams that play A play each other

In the con?guration above,each of B,C and D play two more games.They already play A and cannot play each other,so they must each play E and F.

This gives

E F

B C D

No further choices are possible.

There are10possible schedules in this type of con?guration.These10combinations come from choosing the3teams that play A.

Case2:Some of the teams that play A play each other

Here,at least one pair of the teams that play A play each other.

Given the teams B,C and D playing A,there are3possible pairs(BC,BD,CD).

We pick one of these pairs,say BC.(This gives10×3=30con?gurations so far.)

B C D

It is now not possible for B or C to also play D.If it was the case that C,say,played D,then we would have the con?guration

B C D

E F

Here,A and C have each played3games and B and D have each played2games.Teams E and F are unaccounted for thus far.They cannot both play3games in this con?guration as the possible opponents for E are B,D and F,and the possible opponents for F are B,D and E,with the“B”and“D”possibilities only to be used once.

A similar argument shows that

B cannot play D.

Thus,B or C cannot also play D.So we have the con?guration

B C D

Here,A has played3games,B and C have each played2games,and D has played1game.

B and

C must play1more game and cannot play

D or A.

They must play E and F in some order.There are2possible ways to assign these games(BE and CF,or BF and CE.)This gives30×2=60con?gurations so far.

Suppose that B plays E and C plays F.

B C D

E F

So far,A,B and C each play3games and E,F and D each play1game.

The only way to complete the con?guration is to join D,E and F.

B C D

E F

Therefore,there are60possible schedules in this case.

In total,there are10+60=70possible schedules.

Answer:(E)

Hypatia滑铁卢数学竞赛(Grade 11)-数学Mathematics-2008-试题 exam

2008Hypatia Contest(Grade11) Wednesday,April16,2008 1.For numbers a and b,the notation a?b means2a+b2+ab. For example,1?2=2(1)+22+(1)(2)=8. (a)Determine the value of3?2. (b)If x?(?1)=8,determine the value of x. (c)If4?y=20,determine the two possible values of y. (d)If(w?2)?w=14,determine all possible values of w. 2.(a)Determine the equation of the line through the points A(7,8)and B(9,0). (b)Determine the coordinates of P,the point of intersection of the line y=2x?10and the line through A and B. (c)Is P closer to A or to B?Explain how you obtained your answer. 3.In the diagram,ABCD is a trapezoid with AD parallel to BC and BC perpendicular to AB.Also,AD=6,AB=20,and BC=30. (a)Determine the area of trapezoid ABCD. (b)There is a point K on AB such that the area of KBC equals the area of quadrilateral KADC.Determine the length of BK. (c)There is a point M on DC such that the area of MBC equals the area of quadrilateral MBAD.Determine the length of MC. C 4.The peizi-sum of a sequence a1,a2,a3,...,a n is formed by adding the products of all of the pairs of distinct terms in the sequence.For example,the peizi-sum of the sequence a1,a2,a3,a4 is a1a2+a1a3+a1a4+a2a3+a2a4+a3a4. (a)The peizi-sum of the sequence2,3,x,2x is?7.Determine the possible values of x. (b)A sequence has100terms.Of these terms,m are equal to1and n are equal to?1.The rest of the terms are equal to2.Determine,in terms of m and n,the number of pairs of distinct terms that have a product of1. (c)A sequence has100terms,with each term equal to either2or?1.Determine,with justi?cation, the minimum possible peizi-sum of the sequence.

初中物理《物态变化》测试题(含答案)

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物态变化竞赛题

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