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I KVS-JMO(S) 1997

Q1 (a). Find the difference between the largest and the smallest numbers that can be formed with six digits. (b) The average of nine consecutive natural numbers is 81. Find the largest of these numbers. (c) What will be 77% of a number whose 55% is 240? (d) Flowers are dropped in a basket which become double after every minute. The basket became full in 10 minutes. After how many minutes the basket was half full? Ans1.(a) There are two possibilities- (i) If digits can be repeated- Largest number of six digits = 999999 Smallest number of six digits=100000 Difference=899999 (ii) If digits are not repeated- Largest number of six digits = 987654 Smallest number of six digits=102345 Difference=885305 (b) Let numbers be x. x+1, x+2,x+3,x+4,x+5,x+6,x+7,x+8 Average=81 =>9x+36/9=81 => x+4=81 =>x=77 Largest number =85 (c) 55% of x=240 => x=240*100/55 77%of x=(77*240*100)/55*100=336 (d) Flowers in basket become double after every minute. Therefore, Basket was half full 1 minute before it becomes full i.e. in 9 minutes. Q2. A number consists of 3 digits whose sum is 7. The digit at the units place is twice the digit at the ten’s place. If 297 is added to the number, the digits of the number are reversed. Find the number. Ans2. Let, Digit at Hundred’s place=x Digit at Ten’s place=y Then Digit at one’s place=2y Also sum of digits=7 => x+y+2y=7

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=> x=7-3y Number=100(7-3y)+10y+2y=700-288y On reversing digits new number=100(2y)+10y+(7-3y)=207y+7 Since on adding 297 digits are reversed Therefore, 700-288y+297=207y+7 => 495y=990 y=2 Hence, Number=124 Q3 (a) When an integer ‘n’ is divided by 1995.The remainder is 75. What is the remainder when ‘n’ is divided by 57? (b) In a cube with each edge of length 1 metre, denote one vertex by the letter O. Find the sum of distances from O to each of the other vertices of the cube. Ans3(a). Let, The Number be n n when divided by 1995 leaves remainder 75. => n=1995xq+75 => n=57x35q+57+18 => n=57(35q+1) +18 Hence, remainder will be 18 when the number is divided by 57. (b) From O Sum of distances of three vertices along edges=3x1metre=3metre Sum of distances of other three vertices along diagonals of faces in which O lies

=3x metre=3 metre

Distances of opposite vertex= metre

Sum of distances of all other vertices= (3+3 + ) metre =8.974 metre app.

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Q4. ABC is an equilateral triangle. Points D,E, F are taken on sides AB, BC, CA respectively so that AD=BE=CF. Show that AE,BF,CD enclose an equilateral triangle. Ans4.

Triangle ABC is equilateral => AB=BC=CA Also, AD=BE=CF Therefore, BD=AF=CE In tr. BAF and tr.CBD BA=BC AF=BD <BAF=<CBD=600 By SAS congruence axiom => <BCD=<ABF => <BCQ=<DBQ (i) But <PQR =<QBC+<BCQ {Exterior angle} =<QBC+<DBQ {From i } = <ABC =600 Similarly <QPR and <QRP can be shown as 600 Therefore, is equilateral. i.e. AE,BF and CD enclose an equilateral triangle. Q5(a). Find the degree of the polynomial given by expression.

(b)

Ans5 (a) Expression = 2{x3+3x(x5-1)} {(a+b)3+(a-b)3=2{a3+3ab2} = 2{3x6+x3-3x}

E

D

C B

A

F

Q P

R

E

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Therefore, Degree of expression =6

(b)

=>

=>

=> 6x-12-2x-4=0 since x-1 0

Therefore, x=4 Q6. An ant crawls around the outside of a square of side 1 metre, at all times keeping exactly 1 metre from the boundary of the square. Find the area enclosed, in square metres, in one complete circuit by the ant. Ans6.

Ant crawls such that its distance from square is always 1m. Along the sides it will move in a line

parallel to the sides while at the four corners it will move in circular path.

Hence Area enclosed by the path =

=8.14

Q7. (a) The length of a rectangular sheet of paper is twice its breadth. Show how to cut the paper into three pieces which can be rearranged to form a square. (b) Through two given points A and B in two parallel straight lines, LM and XY respectively, draw straight lines which will form a rhombus with the given parallel lines. Give reasons for your construction. What happens when AB is perpendicular to LM

1 m

1 m

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Ans7(a).

Let, Sides of rectangle be x and 2x . Area of rectangle = 2x.x=2x2 If rectangle is to be converted into square then areas will be equal

Side of square = x which can be obtained as hypotenuse of an isosceles right triangle with equal sides x each Hence, Taking E Mid point of side BC ,cut along AE and DE making them as sides of square from other side. AD will be diagonal of such square. (b) L A P M

X Q B Y

Join AB Draw perpendicular bisector of AB, to intersect LM at P and XY at Q. Join AQ and BP Quad. AQBP will be rhombus. Justification: If diagonals of a quadrilateral are perpendicular bisector of each other then the quadrilateral is a rhombus. If AB is perpendicular to LM then its perpendicular bisector will be parallel to LM and hence no rhombus will form.

E

D

C B

A 2x

x

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Q8. In a toy a cylinder is mounted on flat surface of a hemisphere of diameter 6cm. A cone of slant height 5cm is further mounted on the cylinder. Total height of the toy from the bottom of the hemisphere to the vertex of the cone is 9cm. Find ( i ) The total surface area of toy. (ii) The total volume of the material of toy. Ans8.

Slant height of cone=5cm Radius of cone =Radius of cylinder=3cm

Therefore, Height of cone= =4cm Also, Radius and height of hemisphere=Radius of cylinder=3cm So, Height of cylinder= 9-(4+3) cm=2cm Hence, Total surface area of toy= =141.3

Total volume of toy= = 152.72

Q9.How many different triangles can be formed by joining the points A,B,C,D,E,F,G,H as shown in the figure? Justify your answer. A B C D E F

H

Ans9. Triangles can be formed in two ways (i ) By taking two points from line segment AF and one from G or H Number of such triangles= (6x5/2)x2 =30

G

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(Any of the 6 points can be taken as first vertex and out of other 5 we can have second but in this way each side will be counted twice. Therefore the product is divided by 2 and third vertex can be taken either G or H) (ii) By taking two vertices from F,G,H and one from A,B,C,D,E Number of such triangles= 1x5=5 Therefore, Total number of possible triangles=30+5=35 Q10.In a certain town there are ten restaurants and ‘n’ theatres. A group of tourists spent a few days in the town and visited the theatres and the restaurants during their stay. At the end of their stay it was found that restaurant was visited exactly by 4 tourists and that each theatre was visited exactly by 6 tourists. Given that each tourist visited exactly 5 restaurants and 3 theatres, find ’n’, the number of theatres. Ans10. Let, Number of tourists =x There are 10 restaurants and each restaurant was visited by 4 tourists and each tourist visited 5 restaurants. Therefore, Total visits to restaurants= 5x=10.4 => x=8 Also, Each tourist visited 3 theatres and each theatre was visited 6 times Therefore, Total visits to theatres=6n=3.8 Hence, Number of theatres n=4

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II KVS-JMO(S) 1998 Q1.Fill in the blank: The multiple of 11 nearest to 1000 is …………. The ratio of areas of shaded rectangles A and B is ………

( c) The continued product of (y-2)(y+2)(y2+4)(y4+16) is equal to ………………… (d) If x+1/x =3 ,then x3 +1/x3 is equal to ……………. Ans1 (a) 1001 (b) 1:1 ( c) y8-256 ( d) x3 +1/x3 =18 {Solve and justify answers yourself} Q2. Find the missing digits in the following multiplication sum:

3 5 9 7

* * * ------------------------------ * * * * * * * * * * * * * * * -------------------------------- * * * * 5 4 1 --------------------------------- Ans2. 3 5 9 7

7 5 3

------------------------------ 1 0 7 9 1 1 7 9 8 5 2 5 1 7 9 -------------------------------- 2 7 0 8 5 4 1 --------------------------------- {Solve and justify answers yourself, Explanation required in Examination}

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Q3. ( a) Solve the equation: .

(b) Find the largest prime factor of 314+313-12

Ans3(a). .

=>

=>

=> 3x2-9x-4x+12=3x2-9x+6 Therefore, x= -3/2 (b) 314+313-12 =313(3+1) -12 =3.4(312-1) =3.4(36-1)(36+1) =3.4.(32-1)(34+32+1)(32+1)(34-32+1) =3.4.8.91.10.73 =26.3.5.7.13.73 Largest prime factor of 314+313-12 =73 Q4. What is the surface area of a cube which just fits inside a sphere of radius 1 cm? Ans4. Cube which just fits inside a sphere will be such that its diagonal is equal to diameter of sphere. Diagonal of cube = 2cm

=2 cm

Side of cube = cm

Therefore,

Surface area of cube=6a2 =6x cm2= 8 cm2

Q5 (a). An equilateral triangle and a regular hexagon have equal perimeters. What is the ratio of the area of the triangle to the area of the hexagon? (b) A 2 by 2 square has semicircles drawn on each edge inside the square. The overlaps create four shaded ‘petals’. Find the shaded area. Ans5a. Let, Side of triangle=x Side of hexagon=y Since perimeters of the two figures are equal Therefore, 3x=6y => x=2y

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Hence , Ratio of areas =

= =

(b) Draw figure yourself All four semicircles will intersect at centre of circle. Required area can be taken as -a triangle formed by edge and vertices of this edge joining the centre is excluded from semicircle on the edge. Total four such portion on each edge will give the required area. Hence

Required area =

=

Q6. Price of one kg tea and three kg of sugar is Rs.128. If the rate of sugar increases by 50% and that of tea by 10%,their price becomes Rs160.Find separately the rates of tea and sugar per kg. Ans6. Let, Price of tea = Rs x /kg Price of sugar=Rs y /kg As per statements of problem; x+3y=128 ( i) x(1+1/10) +3y(1+1/2)=160 (ii) Solving we get, y=16 and x=80 Price of one kg tea = Rs80 Price of one kg sugar = Rs16 Q7 (a).In the trapezium PQRS angle QRS is twice the angle QPS.QR has length 8cm and RS is 12cm. What is the length of PS? Explain. (b) Determine the angle x in the diagram.

1400

x 1400

1400

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Ans7(a)

Draw RT bisecting <QRS Since <QRS= 2<QPS <QRT= <SRT= <QPT Also QR is parallel to PS <QRT = <STR Hence , <TRS=<RTS From these we conclude, Quad PTRQ is a parallelogram and triangle RTS is isosceles with TS=RS=12cm PT=Qr=8cm Hence PS=(8+12)cm =20cm (b)

Two exterior angles of triangle are 1400. Therefore, <FBC=3600 –(1400+1400) =800 <BFE =1800-1400=400 Therefore x=1800-(800+400) =600

P

1400

1400

1400

C

A

B E

F

1400

8cm

8cm R Q

T

S

12cm

x

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Q8. Given a triangle PQR, construct a triangle ABC with P,Q and R as mid points of its sides. Give

the steps of construction and justify.

Ans8. { Do construction your self} Steps of construction: I. Draw line parallel to PQ ,QR and RP through R,P and Q respectively II. Name the points where these lines intersect in pairs as A,B and C . Triangle ABC will be the required triangle. Justification: PR is parallel to QC, QR parallel to PC Therefore, quad. QPRA is a parallelogram Similarly, quad. QPRA is a parallelogram Hence PR=QC PR=QA => QC= QA similarly P and R can be shown mid points of BC and AB respectively. Q9(a). Calculate the area represented by the shaded part of the given rectangle. Dimensions

are marked in meters.

(b) What is the sum of the four angles a, b, c, d in the diagram?

d

30

6 6

c

b

a

6

6

18

20 8

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Ans9a. From the figure it is clear that the shaded region consists of one rectangle of dimensions 18cmX12cm and four squares of side 6cm each. Therefore, area of shaded region=(18x8+4x6x6) cm2 = 288 cm2 (b) Join initial points from where two parallel rays originate. Now, In the figure a, b, c, d form a quadrilateral and two consecutive interior angles between parallel lines Hence, a+b+c+d =3600+1800 = 5400 Q10. In how many ways can the word ‘MATHS’ be traced out in this diagram,if you are only allowed to move one step at a time horizontally or vertically, up or down, backwards or forwards?

S S H S

S H T H S S H T A T H S

S H T A M A T H S S H T A T H S

S H T H S S H S

S Ans10. There is M at the center and around it there are four A’s and ther is symmetry. Therefore ,we can count MATHS in any one direction and total words will be its four times. Process can be seen by this tree diagram:

M

A A A A

T T T H H H H H H H S S S S S S S S S S S S S S S There are 15 words for each A Hence, Total number of such words=15x4=60

17 LPS, KVRKT

KVS-JMO 1999

Q1.Fill in the blanks:

(a) (0.4)2-(0.1)2 equals ………..

(b) The angles of a triangle are in the ratio 2:3:4.The greatest angle in degrees is ……………..

(c) If the radius of a circle is increased by 100%, the area is increased by……%.

(d) If log102=a and log103=b, then log1012 equals………

(e) Two circles of equal size are contained in a rectangle as shown below. If the radius of

each circle is 1cm, then the area of the shaded portion in cm2 is ………..

Ans 1(a). (0.4)2-(0.1)2 =(0.4-0.1)(0.4+0.1) =0.15 (b) Greatest angle=4/9x1800=800 ( c) Increase in area=300% (d) log1012= log10(22x3)=2log102+log3=2a+b (e) Since circles touch sides of rectangle and each other. Therefore dimensions of rectangle will be 4cmx2cm. Area of shaded portion=(4x2-2πx12)cm2 =0.72 cm2

Q2. Find the greatest number of four digits which when divided by 2,3,4,5,6,7 leaves a remainder 1 in each case. Ans2. Required number will be 1 more than greatest four digit multiple of 2,3,4,5,6,7 . LCM of 2,3,4,5,6,7 =420 10000=420x23+340 Greatest four digit multiple of 2,3,4,5,6,7 =420x23=9660 Required number=9660+1=9661 Q3 ( a)How many prime numbers between 10 and 99 remain prime when the order of their digits is reversed? (b) Exactly one of the numbers 234,2345,23456,234567,2345678,23456789 is a prime. Which one must it be? Ans 3(a). There are 9 numbers between 10 and 99 which remain prime when the order of their digits is reversed. These are- 11,13,17,31,37,71,73, 79 and 97. (b) 234,23456,2345678 are even 2345 is divisible by 5 234567 is divisible by 3 Hence, 23456789 is prime.

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Q4 (a) A two-digit number is such that if a decimal point is placed between its two digits, the resulting number is one-quarter of the sum of two digits. What is the original number?

(b) Solve:

Ans4(a). Let, Number=10x+y If decimal is placed- x+y/10=1/4(x+y) y=5x Only possible value for x is 1 Therfore, Number=15

(b)

=>

=>

=>

=> =>4x=-6 Therefore, X=-3/2

Q5.In the figure shown below, AB=AD= cm and BEDC is a square.

Also,the area of tri. AEB=area of square BEDC, Find the area of BEDC.

Ans 5.Let, side of square=acm

ar( ADB)=1/2(AO)xBD

=

E

A

D

B

C

O

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=

=

ar( AOB)=

Now, ar( AEB)= ar( AOB)- ar( EOB)= ar(sq BEDC)

26a2=260 a2=10 Hence, area of sq BEDC=10cm2

Q6. A sphere just fits inside a cube, and the cube just fits inside a cylinder (touching the sides

and both the top and bottom faces).What fraction of the cylinder is occupied by the sphere?

Ans. 6. Let, side of cube=2r Then, Radius of sphere =r

Radius of cylinder= { Diameter of cylinder will be equal to diagonal of cube} Height of cylinder=2r

Volume of cylinder= =

Volume of sphere=

Fraction of cylinder occupied by sphere=

Note: Draw figure yourself Q7(a). Find the remainder when x3-19x+38 is divided by x+5. (b). Factorize x3-19x+30 Ans.7(a ).Remainder when x3-19x+38 is divided by x+5 =(-5)3-19x(-5)+38=8 {Using remainder theorem} (b) x3-19x+30 = (x-2)(x-3)(x+5) {Write explanation using Factor Theorem} Q8. Show how to construct a square having given the sum of a diagonal and a side. Give the steps of construction and justify the same.

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Ans8.

Steps of Construction:

Let, Side of square + Diagonal=k =a+ a ,where a is length of side ( I) Draw XY= k (II) Draw a ray making 900 to XY at X

(III) Draw (22 )0 with XY at Y intersecting ray from X at A.

(IV) Draw perpendicular bisector of AY intersecting XY at B, Join AB Draw perpendiculars from B and A to intersect at C. Square AXBC is the required square. Justification: Since perpendicular bisector of AY intersects at B Therefore BA=BY <ABX= <BAY+<BYA=450 AX=BX , <AXB=900

AB= XY=XB+BY

=XB+AB= a+ a Q9. A bus is going from P to Q at a constant speed. After covering a distance of 50km,it develops a fault( at R) and covers the remaining distance at 4/5 of its former speed. It reaches Q 45 minutes late. If the fault had occurred 20km further on ( at S),it would have reached Q only 33 minutes late. Find the speed of the bus and distance from P to Q. P Q Ans9.Let, Distance= D Speed=x km/hr

(i)

and,

(ii)

R S

B

A

Y X

C

50 km 20km

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Subtracting I from ii x=25 and D= 125 Speed=25km/hr, Distance=125km Q10.Find the numberof perfect cubes between 1 and 8000001 which are exactly divisible by 45. Ans10.Perfect cubes between 1 and 8000001 which are divisible by 45 must be cubes of numbers divisible by 3 and 5 8000000= (200)3 Number of such perfect cubes will be number of numbers divisible by 15 from 1 to 200

= =13

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KVS –JMO(S) 1999

Q1. Find the greatest number of five digits which is divisible by 56, 72, 84 and 96 leaves

remainders 48, 64, 76 and 88 respectively.

Let, number be x

8896768464724856 dcbax

8)1(968)1(8481728156 dcba Number must be 8 less than a multiple of 56, 72, 84, 96

L.C.M of 56,72,84 1nd 96 = 2016 Greatest number of 5 digits

121549201699999

Largest multiple of 5 digits = 121599999

98784 Required number = 98776

Q2 (a) If 0182 xx , find the value of 3

3 1

xx .

0182 xx

81

xx

3

3 1

xx =

xx

xx

13

13

= 8383

= 512 + 24 = 536

Q2 ( b) Which is greater: 1411 1731 or ?

1411 1731

3

11

11 1717

3117 4913

17

3117

11

11

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= 217

311

11

11

217

31

204817

3111

0491317

3111

01731 1411 1114 3117

Q3 (a) Show that 9999999999 54321 is exactly divisible by 5.

9999999999

9999999999

53241

54321

each is divisible by 5 {x n + y n is divisible by x + y when n is odd}

Q 3 ( b)Solve: 2

3

3

3

3

3

x

x

x

x

2

3

3

3

3

3

x

x

x

x

2

33392

22

x

xx

2

3

9

3222x

x

xx 892

0982 xx

019 xx

9x or 1x

Q 4. All the measurements in the adjoining figure are in centimeters. What is the total

area of the shaded region? 4 3 6 3 2

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On observation we conclude that the shaded portion is forming a rectangle of

dimensions 186 i.e.18=( 2+3+6+3+4)

Total area of shaded region 108186 cm2

Q 5(a) Find the number of perfect cubes between 1 and 1000001 which are exactly

divisible by 7.

Number of perfect cubes between 1 and 1000001 which are exactly divisible by 7 must be

cubes of numbers between 1 and 100 that are exactly divisible by 7. Therefore, requied

number of such cubes = 14

Q5 (b) An unknown polynomial leaves a remainder 53x when it is divided by

)7)(6( xx . What remainder is obtained if this polynomial is divided by 6x ?

53)(76)( xxqxxxp = (x-6 ){(x-7)q(x) + 3} + 13

Hence Required remainder when p(x) is divided by 6x = 13

Q 6. In a rectangle ABCD, AB = BC2 . Triangle ABE overlaps the rectangle and is

equilateral, and M is the mid – point of BE. Find (i) CMB (ii) the fraction of the

2

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rectangle ABCD that is covered by triangle ABE.

BM = BC

BCMBMC

also 30MBC

75CMB

Area of two triangles not included =3

.3

2BCBC

BC

Fraction of triangle in rectangle = 2

2

2

2

32

BC

BCBC

=32

132

Q7. How many numbers from 1 to 50 are divisible by neither 5 nor 7, and have neither 5 nor 7

as a digit.

Number of numbers divisible neither by by 5 nor by 7 = 50 -10 – 7 +1=34 Numbers having 5 or/and 7 as digit in above numbers are 17 , 27, 37 and 47 Hence, Required number of numbers = 34 - 4=30

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Q8 Three circles ,each of unit radius, are so drawn that the centre of each is an intersection of

other two as shown in the adjoining figure. Find the area of the shaded

portion.

Here, AB=BC=CA=1 AB=AD=DB=1 Therefore, < BAD =600

Area of smaller segment of circle III cut off by AB = Area of segment of II cut off by AD Area of shaded portion ABD= area of sector BAD By symmetry, Area of shaded portion are equal Area=3 xarea(sector BAD) =1/2 area of one circle

=2

Q9. The square of a number of two digits is four times the number obtained by reversing its

digits . Find the number.

Let Number be 10x+y (10x+y)2 = 4.(10y+x) Number is even and 10y+X IS A PERFECT SQUARE . Possible values=25,49,64 1nd 81 Square root of 4(10y+x) may be 10,14,16 ,18 10 ,14 1nd 16 does not satisfy other conditions Required number is 18

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Q10. An ant wants to go from A to B .It can move only along the grid lines, either horizontally

to the right or vertically upwards.How many differents paths can it take from A to B.

A

B

Ant has to travel 8 steps to reach from A to B

4 in right direction and 4 in upwards direction for each path.

Number of different paths it can take =!4!4

!8 = 70

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KVS JMO 2001 Q1. Fill in the blanks: a.If x+y=1,x3+y3=4 ,then x2+y2=……………… b.After 15 litres of petrol was added to the fuel tank of a car the tank was 75% full. If the capacity of the tank is 28 litres ,then the number of litres in the tank before adding the petrol was…………. c.The perimeter of a rectangle is 56 metres. The ratio of its length to width is 4:3.The length of the diagonal in metres is……….. d.If April 23 falls on Tuesday, then March 23 of the same year was……. e.The sum of the digits of the number 2200052004 is …………… Ans: (a) x3+y3=4 => (x+y)(x2-xy+y2) =4 => ( x+y)2-3xy =4 => xy=-1 Therefore, x2+y2= (x+y)2-2xy = 1+2= 3 (b) Initial quantity of petrol in tank =75%of 28 -15 =6 litres ( c) L+B=56/2=28 L=4*28/7=16m B= 3*28/7=12m Diagonal =(162+122)1/2= 20m ( d) Number of days from March 23 to April 23 =32 = 4 weeks and 4 days Going backward from April 23 we get 23rd march was Saturday ( e) 22000 .52004 = 54.22000.52000 = 625. 102000

Therefore , Sum of digits =6+2+5=13 Q2. ( a) Arrange the following in ascending order: 25555, 33333, 62222. ( b) Two rectangles, each measuring 3 cm x 7 cm, are placed as in the adjoining figure:

Find the area of the overlapping portion in cm2. Ans (a) 25555 =(25)1111 =321111 33333 =(33)1111 =271111

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62222 = (62)1111=361111 Since exponents are equal therefore bases will decide the order of numbers Hence, Ascending order: 33333, 25555, 62222

(b ) Clearly all triangles are congruent

In (7-x)2 +32 =x2 =>14x=58 x=58/14=29/7 Area of shaded region which is a rhombus =29/7x3 sq cm =87/7 sq cm

Q3 ( a) Solve: =3

( b) Simplify :

Ans: ( a) Equation =>

Solving x=3 or x=2

( b) Expression =

In numerator taking a=b Num=0 a-b is a factor By symmetry b-c and c-a will be other two factors Num. =k(a-b)(b-c)(c-a) Comp coeff. of a2b: k=0 Numerator=0 Expression=0 Q4 ( a) Factorise: (x-y)3+(y-z)3+(z-x)3

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(b) If x2-x-1=0 ,then find the value of x3-2x+1 Ans: 4(a) Since (x-y)+(y-z)+(z-x)=0 Therefore, (x-y)3+(y-z)3+(z-x)3= 3(x-y)(y-z)(z-x) (b) x3-2x+1 =(x2-x-1)(x+1)+2 = 2 Q5. ABCD is a square. A line through B intersects CD produced at E, the side AD at F and the diagonal AC at G. If BG=3, and GF=1, then find the length of

FE.

Ans:

And

1+x=9 x =8 Q6. ( a) Find all integers n such that (n2-n-1)n+2=1

( b) If x=

Ans (a) Equation is satisfied if, n+2 =0 => n=-2 and If n2-n-1 =1

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=> n2-n-2=0 =>(n-2)(n+1)=0 n =2 or n=-1 If n is even and n2-n-1=-1 =>n(n-1)=0 =>n=0 n can’t be 1 Hence, Solution={-2,-1,0, 2}

( b) x=

=>

By componendo and dividend

=

Similarly,

Hence,

=

= 2 Q7.( a) Find all the positive perfect cubes that divide 99.

( b) Find all the integers closese to 100(12- ) Ans: (a) 99= (93)3 =7293 Cubes of all factors of 729 will divide 99 Factors are= 1,33,93,273,813,2433, 7293 = 1,33,36,39, 312, 315, 318

(b) Therefore,

=100x.042 = 4.2 Nearest integer =4

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Q8. In a triangle ABC, BCA =900. Points E and F lie on the hypotenuse AB such that AE=AC and BF=BC . Find <ECF.

Ans: AE=AC Triangle ACE is isosceles <CEA=x+y BF=Bc Tri. BCF is isosceles <CFB =x+z In tri CFE X+x+z+x+y=1800 2x=900 x=450

Q9. An ant crawls 1 centimetre north, 2 centimetre west, 3 centimetres south, 4 centimetres east, 5 centimetres north and so on, at 1 cm per second. Each segment is 1 cm longer than the preceding one, and at the end of a segment, the ant makes a left turn. In which direction is the ant moving 1 minute after the start. Ans: Total distance covered in 1 minute = 60 cm Let, there are n segments In consecutive segments length covered are consecutive natural numbers. n(n+1)/2= 60 n2+n-120=0

n=

n is natural number nearest perfect square number near 481=441 n=-1+21/2=10 Number of turns =10 11th segment is incomplete After four segments ant will again be towards north After 9th towards west

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After 10th towards south Q10. Find the lengths of the sides of a triangle with 20, 28 1nd 35 as the lengths of its altitudes. Ans: ratio of sides { Draw figure} a:b:c= 1/20: 1/28: 1/35 { since a= 2A/h} = 7:5:4 Let sides be 7x, 5x and 4x Now, s= (7x+5x+4x)/2 =8x

Area = 7xx20/2 =

7x10x =

x=

lengths of sides

a=

b=

c=

34 LPS, KVRKT

KVS JMO 2002

Q1. Fill in the blanks: a. Yash is carrying 100 hundred rupee notes, 50 fifty rupee notes, 20 twenty rupee notes, 10 ten rupee notes, and 5 five rupee notes. The total amount of money he is carrying, in rupees, is ……. b. In a school, the ratio of boys to girls is 4:3 and the ratio of girls to teachers is 8:1. The ratio of student to teacher is ……….

c.The value of is …………….

d.(123456)2 +123456 +123457 is the square of ………… e.The area of a square is 25 square centimeters .Its perimeter, in centimeters, is ………………. Ans: 1.( a ) Total money Yash has = Rs(1002+502+202+102+52) =Rs13025 (b) Boys: Girls = 4:3 Girls: Teachers = 8:1 =24:3 Student: Girls = 7:3=56: 24 Therefore, Students : Teachers = 56:3 ( c ) (0.5+1/0.5)2 =2.52 = 6 .25 ( d ) (123456)2 +123456 +123457 = (123456)2 +123456 +123456 +1 = (123456)2 +2*123456 *1+12 = (123456+1)2 = 1234572

Required number is 123457 ( e ) side= 5 Cm Perimeter= 4*5=20 Cm Q2. ( a) How many four digit numbers can be formed using the digits 1, 2 only so that each of these digits is used at least once ? ( b) Find the greatest number of four digits which when increased by 1 is exactly divisible by 2,3 ,4,5,6 and 7. Ans: 2.( a )Following cases are possible: i. 1 used thrice and 2 once ii. 1 used twice and 2 twice iii. 1 used once and 2 thrice Number of four digit numbers in i and iii case= 4 each ( one different digit can be placed at any of the four places)

Number of four digit numbers in ii case = = 6 ( 4 digits can be arranged in 24 ways(Nr)

but 1 and 2 occur twice so actual number will be half for each) Required number of numbers=14 (b) LCM of 2,3,4,5,6,and 7=420 Largest four digit number which is a multiple of 420= 9999-339=9660 Since required number is 1 less than the exact multiple therefore required number = 9660-1=9659

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Q3. (a ) If f(x) = ax7 + bx5+ cx3-6, and f(-9)=3 , find f(9). ( b) Find the value of :

Ans: 3 ( a) f(-9)=3 a(-9)7+b(-9)5-c(-9)3-6= 3 97a+95b+93c = -9 Therefore, f(9) =97a+95b+93c -6 = -9-6 = -15 ( b)

=

= 2002

Q4. ( a) If x>0 and =47, find the value of .

(b) If , find the value of 37x.

Ans: =47

Therefore,

= 33- 3.3 = 18 Q5. A train, after travelling 70 km from a station A towards a station B, develops a fault in the engine at C, and covers the remaining journey to B at ¾ of its earlier speed and arrives at B 1

36 LPS, KVRKT

hour and 20 minutes late. If the fault had developed at 35 km further on at D, it would have arrived 20 minutes sooner. Find the speed of the train and the distance from A to B. Ans: Let, Normal speed of train = x km/hr Distance AB = D km In I situation:

In II situation:

Solving equations: x=35 and D=210 Speed of train=35km/hr Distance =210km Q6. The adjoining diagram shows a square PQRS with each side of length 10 cm . Triangle PQT is equilateral. Find the area of the triangle UQR. P Q S R Ans: In figure PM is altitude PU bisects < TPM

Also TM=5 TU+UM=5

UM=

UQ=UM+MQ=10(

Ar(tr. PUQ)=1/2.10(

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=25(3- ) cm2

Ar( tr. UQR) = ar(tr. PRQ)-ar(tr. PUQ)

=50-{25(3- ) }cm2 = 25( -1) cm2

Q7. A square of side length 64cm is given. A second square is obtained by connecting the mid-points of the sides of first square. If the process of forming smaller inner squares by connecting the mid -points of the sides of the previous square is continued, what will be the side length of the eleventh square, counting the original square as the first square? Ans: Side of first square =64 cm Here, Side of next square will be half of the side of previous square.

side of second square= =

Side of third square =

Hence, Side of 11th square =

= 2cm Q8. Seven cubes of the same size are glued together face to face ( six cubes on six faces of a cube).What is the surface area in square cm, of the solid if its volume is 448 cubic cm. Ans: 7a3 = 448 cm3 => a= 4cm Required surface area = 6X 5X 42 cm2 = 480 cm2

Q9. Anil, Bhavna, Chintoo , Dolly and Eshwar play a game in which each is either a FOX or a RABBIT. FOXES statements are always false and RABBITS statements are always true. Anil says that Bhavna is a RABBIT. Chintoo says that Dolly is a FOX. Eshwar says that Anil is not a FOX. Bhavna says that Chintoo is not a RABBIT. Dolly says that Eshwar and Anil are different kinds of animals. How many FOXES are there? (Justify your answer) Ans. Let, Anil’s statement is false then Anil is a FOX. Anil says Bhavna is a RABBIT means Bhavna is a FOX Eshwar says that Anil is not a FOX means Eshwar is a FOX Bhavna says that Chintoo is not a RABBIT means Chintoo is a RABBIT Dolly says Eshwar and Anil are different kinds of animals means Dolly is a FOX Chintoo says that Dolly is a FOX is correct. Therefore Anil, Bhavna, Eshwar and Dolly are four FOXES.

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Q10.The accompanying diagram is a road plan of a city. All the roads go east –west or north south, with the exception of one shown. Due to repairs one road is impassable at the point X, of all the possible routes from P to Q, there are several shortest routes. How many such shortest routes are there? Q

P Ans: All shortest routes will pass through AB. Shortest routes are: 1. PCABIJGHQ 2. PCABIJKHQ 3. PCABIJKLQ 4. PCABEFGHQ 5. PCABEJGHQ 6. PCABEJKHQ 7. PCABEJKLQ 8. PDABIJGHQ 9. PDABIJKHQ 10. PDABIJKLQ 11. PDABEFGHQ 12. PDABEJGHQ 13. PDABEJKHQ 14. PDABEJKLQ There are 14 shortest routes.

E

F J

G K

H L

B I

D A c

39 LPS, KVRKT

KVS JMO 2003 Q 1.Fill in the blanks- The digits of the number 2978 are arranged first in descending order and then in ascending order. The difference between the resulting two numbers is…. Yash is riding his bicycle at a constant speed of 12 kilometres per hour. The number of metres he travels each minute is …… The square root of 35 X 65 X 91 is ………… The number 81 is 15% of ……….. A train leaves New Delhi at 9.45 am and reaches Agra at 12.58 pm. The time taken in the journey, in minutes, is ………… Ans. (a) Required difference = 9872-2789 =7083 (b) No. of metres traveled in each minute = 12000/60 =200 metres

(c) =

(d) Number = =

(e) Time taken in journey = 3 hr 13 minutes = 193 minute Q 2. ( a ) Find the largest prime factor of 203203. ( b ) Find the last two (ten’s and unit’s) digit of (2003)2003.

Ans ( a ) 203203 = Therefore, Largest prime factor = 29 ( b ) Last two digits is remainder when number is divided by 100

(2003)2 32 (mod 100 ) 9 ( mod 100 )

(2003)4 92 (mod 100 ) -19 (mod 100 )

(2003)8 (-19)2 (mod 100 ) 61 (mod 100 )

(2003)16 612 (mod 100 ) 21 (mod 100 )

(2003)32 212 (mod 100 ) 41 (mod 100 )

(2003)40 (2003)32 .(2003)8 (mod 100 ) 41.61 (mod 100 ) 1(mod100)

(2003)2000 (200340)50 150 (mod100) 1(mod100)

(2003)2003 20032000.20032.20031(mod100) 1.9.3(mod100) 27(mod100)

Last two digits of 20032003 =27 Q 3. (a) Find the number of perfect cubes between 1 and 1000009 which are exactly divisible by 9.

( b) If x =5+2 , Find the value of

(ii )

40 LPS, KVRKT

Ans: ( a) Perfect cubes divisible by 9 will be cubes of multiples of 3.

Since,

Also x is a multiple of 3 But, 101=3x33 + 2 Between 1 and 101 there are 33 multiples of 3 . Required number of perfect cubes =33 ( b)

x =5+2 ,

Therefore,

+

-

Therefore,

And,

= 103 -3. 10 = 970 Q 4.( a) Solve:

(b) Find the remainder when x81+x49 +x25+x9 +x is divided by x3 –x. Ans: ( a)

=

41 LPS, KVRKT

x81+ x49+x25+x9+x = x ( x80 + x48+ x24 + x8 + 1)

x3-x = x( x2 -1) Therefore, When x81+ x49+x25+x9+x is divided by x3-x remainder will be same as when x80 + x48+ x24 + x8 + 1 is divided by x2 -1. Taking y= x2

x80 + x48+ x24 + x8 + 1 = y40+y24+ y12+y4 +1 is divided by y-1

Remainder = 140+124+112+14 +1 = 5

OPQ is a quadrant of a circle and semicircles are drawn on OP and OQ . Areas a and b

are shaded. Find a/b.

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(b) Assuming all vertical lines are parallel , all angles are right angles and all the horizontal lines are equally spaced, what fraction of the figure is shaded?

Ans: ( a) Let OP =OQ =R Area of region which is not shaded = area of quadrant –(a + b) = area of two semicircles – 2a

(b) Shaded area on horizontal shifting to one row makes one complete row.

Area of shaded portion = ¼ of total area of rectangle

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. Alternate vertices of a regular hexagon are joined as shown. What fraction of the total area of the hexagon is shaded? ( Justify your answer)

Ans: Let side of given regular hexagon =a Base of isosceles triangle formed by sides of hexagon and line segment joining the alternate

vertices = a

Side of equilateral triangles formed = one third of diagonals = a/3

Shaded area is forming regular hexagon of side a/3

Ratio of shaded area to the total area = =

. Question is incomplete.

. A cube with edge of length 4 units is painted green on all the faces. The cube is then cut into 64 unit cubes. How many of these small cubes have 3 faces painted (ii) 2 faces painted (iii) one face painted (iv) no face painted Ans: Number of cubes having 3 faces painted = 8 { At 8 corners } 2 faces painted = 24 { 12(n-2) } one face painted = 24 { 6(n-2)2 } no face painted = 8 { (n-2)3 }

. Let PQR be an equilateral triangle with each side of length 3 units. Let U,V, W,X,Y, and Z divide the sides into unit lengths. Find the ratio of the area UWXY (shaded) to the whole triangle PQR.

44 LPS, KVRKT

Ans: Length of altitude on PY from U =

Length of altitude on QW from U =

Shaded area =

=

= sq units

Area of

fraction = =

. Five houses P,Q,R,S and T are situated on the opposite side of a street from five other

houses U,V,W,X and Y as shown in the diagram:

Houses on the same side of the street are 20 metres apart. A postman is trying to decide whether to deliver the letters using route PQRSTYXWUV or route PUQVRWSXTY, and finds that the total distance is the same in each case. Find the total distance in metres. Ans: Let PU=x Then PQRSTYXWUV = (20+20+20+20+x+20+20+20+20) = 160+x And PUQVRWSXTY

=

45 LPS, KVRKT

= Hence,

46 LPS, KVRKT

KVS (JMO) 2004 Q1. Fill in the blanks: a.The number of hours from 8 p.m. Tuesday until 5 am Friday of the same week is …………… b.If 3 x - 2 = 81 then x equals………….. c.In a school the ratio of boys to girls is 3:5 and the ratio of girls to teachers is 6:1. The ratio of boys to teachers is …………. . d.If 7n + 9 > 100 and n is an integer the smallest possible value of n is ……… e.In the diagram, AC = 4, BC = 3, and BD = 10. The area of the shaded triangle is …………. .

Solution: 1. (a) Tuesday 8 pm to Thursday 8 pm = 24 * 2 = 48 hrs Thursday 8 pm to Friday 5 am = 9 hrs Total hours = 57 hrs (b) 3 x – 2 = 81

42 33x

42x

6x (c) Boys : Girls = 3:5 = 18:30 Girls : Teacher = 6:1 = 30:5 Boys : Teacher = 18:5 (d) 7n +9 > 100

7n >91

n > 13

Smallest value of n=14Required Area = ACCD2

1 sq. units

= 472

1 sq. units = 14 sq. units

Q2. (a) Find the number of positive integers less than or equal to 300 that are multiples of 3 or 5, but are not multiples of 10 or 15. (b) The product of the digits of each of the three – digit numbers 138, 262 and 432 is 24. Write down all three digit numbers having 24 as the product of the digits. Solution:

2. (a) No. of multiples of 3 = 3

100 =100

47 LPS, KVRKT

No. of multiples of 5 = 5

300= 60

No. of multiples of 3 and 5 both = 15

300= 20

No. of multiples of 10 = 10

300= 30

N0 of multiples of 15 = 15

300= 20

No. of multiples of 10nd 15both = 30

300 =10

xequalthanlessegergreatestdenotesx /int

Therefore, Required number of numbers = (100 +60-20) - (30+20-10) = 140-40 = 100 (b) 24 can be written as a product of three numerals as -

1 3 8

1 6 4

2 4 3

2 6 2 For three different numerals there are 6 arrangements of each possible product and for fourth product having 2 two’s number of arrangements will be 6/2=3 All three digit numbers having product of their digits 24 = 138, 183, 318, 381, 813, 831, 164, 146, 461, 416, 614, 641, 243, 234, 342, 324, 432, 423, 262, 226, 622 Q3. (a) Solve:

1922 yxyx

4922 yxyx

(b) The quadratic polynomial p ( x ) = a ( x – 3 ) 2 + bx +1 and q (x) = 2 x2 + c(x – 2) + 13 are equal for all values of x. Find the value of a, b, c.

Solution: 3. (a) 4922 yxyx

1922 yxyx

Adding 2(x2 + y2) = 68

x2 + y2 = 34 Subtracting 2 xy = - 30 Hence, (x – y)2 = 64 x – y = 8

48 LPS, KVRKT

(x + y)2 = 4 x + y = 2 Solving equations possible values of x and y are: x = 5 , y = - 3 x = -5 , y = 3 x = 3 , y = - 5 x = - 3 , y = 5 (b) p(x) = a(x – 3)2 + bx + 1 q(x) = 2x2 + c(x – 2) +13 P(x) = q(x)

a(x2 – 6x + 9) + bx +1 = 2x2 + cx – 2c + 13

a(x2 – 6x +9) + bx + 1 = 2x2 + cx – 2c + 13

(a – 2)x2 + ( - 6a +b – c)x + 9a + 2c – 12 = 0 a – 2=0, - 6a + b – c = 0 , 9a+2c-12=0

a = 2, b = 9 and c= -3 Alternate method:

x=3 3b+1=18+c+13 3b-c=30

x=0 9a+1=-2c+13 9a+2c=12

x=2 a+2b+1=8+13 a+2b=20 Solving equations: a=2, b=9 and c=-3 Q4.(a) Two squares, each with side length 5 cm, overlap as shown. The shape of their overlap is a square, which has an area of4 cm 2. Find the perimeter, in centimeters, of the shaded figure.

(b) A rectangle is divided into four smaller rectangles. The areas of three of these rectangles are 6, 15 and 25, as shown. Find the area of the

shaded rectangle.

49 LPS, KVRKT

Solution: 4. (a)

From the figure:

Perimeter of shaded portion = (4 5 + 4 3 ) = 32 cm. 4. ( b )

From figure areas of given regions xz = 6 yz = 15

5

2

y

x

u y = 25

xy

2

5

25

2

5xu

50 LPS, KVRKT

10

2

525xu

Hence, Area of fourth rectangle = 10 sq. units Q5. (a) A square ABCD is inscribed in a circle of unit radius. Semicircles are described on each side as a diameter. Find the area of the region bounded by the four semi-circles and the circle.

(b) In a parallelogram ABCD, H is the mid-pointof AB and M is the mid-point of CD. Show that AM and CH divide the diagonal DB in three equal parts.

Solution: 5. (a) Diagonal of square = 2 units

Side of square = 2 units Area of four semicircles

51 LPS, KVRKT

2

2

2

2

14 sq. units

= 4

2

2

14 sq. units

= sq. units Therefore, Area of four segments = ( - 2) sq. units Area of required region = { -( -2)}sq. units = 2 sq. units.

(b) AH = CM AH CM

Therefore, Quadrilateral AMCH is a parallelogram In triangle ABP, H is mid point

HQ AP

Therefore,

Q is mid point of BP BQ = PQ Similarly,

In triangle DCQ DP = PQ

BQ = PQ = DP Q6. A two-digit number has the property that the square of its tens digit plus ten times its units digit is equal to the square of its units digit plus ten times its tens digit. Find all two digit numbers which have this property, and are prime numbers. Solution: 6. Let, Ten’s digit = x and one’s digit = y x2 + 10y = y2 + 10x

x2 – y2 = 10(x – y)

(x – y) (x + y – 10) = 0 Either x – y = 0 or x + y = 10 x = y Only 11 is such prime number For x + y = 10 Numbers may be 19, 28, 37, 46, 55, 64, 73, 82, 91 All two digit prime numbers having the property = 11, 19, 37, 73 Q7. In the diagram, it is possible to travel only in the direction indicated by the arrow. How many different routes from A to B are there in all?

52 LPS, KVRKT

Solution: Number of routes through AXB = 2 1

Number of routes through AXYB = 2 1 3

Number of routes through AYB = 1 3

Total number of routes = 2 1 + 2 1 3 + 1 3 = 11 Q8.The Object shown in the diagram is made by gluing together the adjacent faces of six wooden cubes, each having edges of length 2 cm. Find the total surface area of the object in square centimeters.

Solution:

Surface area of object = Number of visible faces after gluing area of one face

Total surface area of object = (5+4+4+4+4+5) 22 cm2 = 104 cm2 Or Faces visible after gluing = Total faces before gluing – faces glued

=6 6-2 5 =26

Surface area = 26 4 cm2

= 104 cm2

Q9. Six points A, B, C, 0, E, and F are placed on a square grid, as shown. How many triangles that are not right-angled can be drawn by using 3 of these 6 points as vertices.

53 LPS, KVRKT

Solution: Number of triangles that are not right triangle

1211321

456

= 20 – 14 = 6

i.e. EFADFBDECBCDACEABF ,,,,,

Or For making triangles that are not right triangle we must select two points from one row and the third from the other which is not directly opposite to these two.

It can be done in 2 (3 1)=6 ways Q 10. A distance of 200 km is to be covered by car in less than 10 hours. Yash does it in two parts. He first drives for 150 km at an average speed of 36 km/hr, without stopping. After taking rest for 30 minutes, he starts again and covers the remaining distance non-stop. His average for the entire journey (including the period of rest) exceeds that for the second part by 5 km/hr. Find the speed at which he covers the second part. Solution:

Let, Speed in II part = x km/hr Average speed = (x + 5) km/h

Total time taken 6

50

2

1

36

150

x

x

3

14150

54 LPS, KVRKT

Average speed x

x

14150

3200 5x

7x2 – 190x + 375 = 0

(x – 25) (7x – 15) = 0

x = 25 or x = 15/7

But 103

1450

x , x = 15/7 does not satisfy it.

Therefore, x = 25 Speed for next part = 25 km/hr

55 LPS, KVRKT

KVS (JMO) 2005 Q1.Fill in the blanks: a.If four times the reciprocal of the circumference of a circle equals the diameter, then the area of the circle is ……………..

b.If x

thenxx

2,0

441

2 equals ………………………

c.If a=1000, b=100, c=10 and d=1, then (a+b+c-d)+(a+b-c+d)+(a-b+c+d)+(-a+b+c+d) is equal to ……………… d.When the base of a triangle is increased by 10% and the altitude to the base is decreased by 10%, the change in area is …………………….. e.If the sum of two numbers is 1, and their product is 1,then the sum of their cubes is ………………………………………….

SOLUTION1: (a) ,122

4 2rrr

Area=1 sq.unit

(b) 12

02

1044

1

2

2 xxxx

(c) (a+b+c-d)+(a+b-c+d)+(a-b+c+d)+(-a+b+c+d) =2(a+b+c+d) = 2x1111 =2222

(d) New area= hb )100

101.()

100

101.(

2

1

= bh2

1

100

99

Change in area= bh2

1

100

1

% change = 1% decrease (e) x3+y3=(x+y)3-3xy(x+y) =13-3.1.1 = -2 2.(a) If x= (log82)log

28, find the value of log3x.

(b) If ,2433

98

2

45 y

yx

yx

x

and find the value of x-y.

SOLUTION:

( a) x=

8log

2

2

8log

1

ba

a

blog

1log

56 LPS, KVRKT

=

32 2log

3

2 2log

1

=

2log3

2

2

2log3

1 mnm a

n

a loglog

= 3

3

1 1log aa

log3x=log33-3=-3log33=-3

(b) 32282

4 3 yxyx

yx

x

Other condition will be required to find values of x and y. 3. (a) Find the number of digits in the number 22005 .52000 when written in full. (b) Find the remainder when 22005 is divided by 13. SOLUTION: 22005.52000 = 25.22000.52000 = 32. 102000 Number of digits = 2(non zero digits 2&3) +2000(zeros) =2002 22005=22000.25

)13(mod625 (Note:Go through article on congruences in the project)

210= (25)2 62(mod13) 10(mod13) 220= (210)2 102(mod13) 9(mod13) 240= (220)2 92(mod13) 3(mod13) 2200= (240)5 35(mod13) 9(mod13) 2400=(2200)2 92(mod13) 3(mod13) 22000=(2400)5 35(mod13) 9(mod13) 22005=22000.25 6.9 (mod13) 2(mod13) Remainder is 2 when 22005 is divided by 13.

Q4.(a) A polynomial p(x) leaves a remainder three when divided by x-1 and a remainder five when divided by x-3. Find the remainder when p(x) is divided by (x-1)(x-3). (b)Find two numbers both lying between 60 and 70, each of which divides 248-1. SOLUTION 4. (a) Let p(x)=(x-1)(x-3)q+ax+b p(x) when divided by x-1 leaves remainder 3. p(1)=3 a+b=3 p(x) when divided by x-3 leaves remainder 5. p(3)=5 3a+b=5

57 LPS, KVRKT

Solving we get a=1,b=2 Remainder= x+2

b. 248-1=(26-1)(26+1)(212+1)(224+1) 212+1 and 224+1 are greater than 70 Therefore, Numbers between 60 and 70 are 26-1 and 26+1 i.e. 63 and 65

5. In a triangle ABC the medians AM and CN to the sides BC and AB respectively intersect in the point O. P is the mid point of side AC and MP intersects CN in Q.If the area of triangle OMQ is 24cm2, find the area of triangle ABC. SOLUTION:

Area( OMQ)=24Cm2 M and P are mid points, Hence MP║AB

AON~ MOQ

42

2

MO

AO

MOQar

AONar { O is centroid}

ar( AON)=4.24Cm2=96cm2 ar( AOB)=2ar( AON)=2.96cm2 =192cm2 { ON is median} hence, ar( ABC)=3.ar( AOB)=3.192cm2 =576 cm2 { O is centroid,ar( AOB)=ar( BOC)=ar( AOC)} 6. The base of a pyramid is an equilateral triangle of side length 6 cm.The other edges of the

pyramid are each of length 15 cm. Find the volume of the pyramid.

SOLUTION:

58 LPS, KVRKT

BM= Cm3362

3 { altitude of equilateral triangle}

BG= CmCmBM 32333

2

3

2

Hence, PG2 =BP2-BG2 =15-12=3 Therefore,

h=PG= 3

Area of base= 2264

3Cm

Volume of pyramid= 33 93364

3

3

1CmCm { v=1/3area of base .height}

7. Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED are of lengths 2cm, 6cm and 3cm respectively. Find the length of the diameter of the circle. SOLUTION:

CE.DE=AE.BE AE=2Cm, DE=3Cm, EB=6Cm Therefore, CE=4Cm

59 LPS, KVRKT

Let O be centre then DN=3.5 Cm NE=0.5Cm=OM Also BM= 4Cm Hence , OB2=BM2+OM2=16+1/4=65/4

Radius=2

65Cm, diameter= Cm65

8. Three men A,B,C working together do a job in 6 hours less time than A alone, in 1 hour less time than B alone and in one half the time needed by C when working alone. How many hours will be needed by A and B working together to do the job? SOLUTION: Let, A,B &C together do job in x hours A completes work in x+6 hrs B completes work in x+1 hrs C completes work in 2x hrs A&B’s 1 hrs work = A,B,C’s 1 hrs work-C’s 1 hrs work

xxxx 2

11

1

1

6

1

0673 2 xx (x+3)(3x-2)=0

x=2/3 or x=-2 not possible. A& B complete work in 1/2xhrs =3/4 hrs

9. Pegs are put on a board 1 unit apart both horizontally and vertically .A rubber band is stretched over 4 pegs as shown in the figure, forming a quadrilateral . Find the area of the quadrilateral in square units.

SOLUTION:

Area of quad ABCD = ar( )() ABDarBCD

= squnits242

114

2

1

60 LPS, KVRKT

= 6 sq units Q10.The odd positive numbers 1,3,5,7…….. are arranged in five columns counting eith the pattern shown on the right.Counting from the left, in which column( I,II,III,IV orV) does the number 2005 appear ? (Justify your answer) I II III IV V 1 3 5 7 15 13 11 9 17 19 21 23 31 29 27 25 SOLUTION: 2005 =2X1002+1 i.e. It is 1003rd odd number starting from 1 There is a cycle of 8 numbers ,spread in two rows . 1003=8X125+3 Therefore 2005 is 3rd number in 126th cycle Hence, 2005 will lie in IV column.

61 LPS, KVRKT

KVS JMO 2006 Q1. a, b, c are three distinct real numbers and there are real numbers x, y such that a3 + ax

+ y = 0, b3 + bx + y = 0 and c3 + cx + y = 0. Show that a + b + c = 0.

Ans: a3 + ax + y = 0 => y = - (a3 + ax)

b3 + bx + y = 0 => y = -(b3 + bx)

c3 + cx + y = 0 => y = -(c3 + cx)

Hence,

- (a3 + ax) = - (b3 + bx) = - (c3 + cx)

=> (a – b)x = - (a3 – b3)

x = - (a2 + ab + b2)

=> a2 – c2 + ab – bc = 0

(a – c) (a + c) + b(a – c) = 0

(a – c) (a + c + b) = 0

a – c ≠ 0

Therefore,

a + b + c = 0

62 LPS, KVRKT

Q2. The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2·PD. Ans:

Const: PD = DE

ΔCDP ≅ ΔCDE

<CAB = <CPB {Angles in the same segment}

<CBA = <CPA {Angles in the same segment}

But CA = CB

=> <CAB = <CBA

<CPB = <CPA

<CPB = <CPA

<CPB = <CEB {cpct}

In ΔCAP and ΔCEB

CA = CB

<CPA = <CEB

<CAB = <CBE

ΔCBE ≌ΔCAP

Therefore BE = AP

Hence,

C

A

P

D

E

B

63 LPS, KVRKT

AP + PB = BE + BD + PD

= DE + PD

= PD + PD

= 2PD

Q 3 Given reals x, y with (x2 + y2) /(x2 - y2) + (x2 - y2)/(x2 + y2) = k,

find (x8 + y8)/(x8 - y8) + (x8 - y8)/(x8 + y8) in terms of k.

Ans:

Now

Divide Numerator and Denominator by x16

64 LPS, KVRKT

Value

=

=

Q4. In a right triangle ABC right angled at B, a point P is taken on the side AB such that AP = h and PB = b. If BC = d and AC = y such that h + y = b + d. Prove that h = bd/(2b+d) Ans:

Given, h+y=b+d

y=b+d-h

In ΔABC, By Pythagoras Theorem

y2=(h+b)2+d2

Hence, (h+b)2+d2= (b+d-h)2

d

y

A

P

B C

h

b

65 LPS, KVRKT

h(4b+2d)=2bd

h=bd/2b+d

Q5. P is a point inside the triangle ABC. Lines are drawn through P Parallel to the sides of the triangle. The areas of the three resulting triangles with a vertex at P, have areas 4, 9 and 49. What is the area of triangle ABC?

Ans: Since, QR ll BC, QU ll AB and SV ll AC hence, let BC=k

but, quadTPUV is a parallelogram

1

Similarly,

2

3

Adding eq. 1,2 1nd 3

ar(

S

T

U V

Q

R

66 LPS, KVRKT

Q6. A lotus plant in a pool of water is ½ cubit above water level. When propelled by air, the lotus sinks in the pool 2 cubits away from its position. Find the depth of water in the pool. Ans:

………………2………………………….

By Pythagoras theorem: (h+1/2)2=h2+22

Solving we get, h=15/4cubits

Q7. Let C1 be any point on side AB of a triangle ABC. Join C1C .The lines through A and B parallel to CC1 meet BC and AC produced at A1and B1 respectively. Prove that 1/AA1 + 1/BB1 =1/CC1

Ans:

Triangle AC1C is similar Triangle ABB1 Hence AC1/AB=AC/AB1=CC1/BB1 (1)

Triangle BC1C is Similar to triangle BAA1

BC1/AB= CC1/AA1==BC/A1B (2) From (1) & (2) AC1/AB + BC1/AB =CC1/BB1+ CC1/AA1

AC1+BC1=AB CC1( 1/BB1+1/AA1)=1 => 1/CC1=1/BB1+ 1/AA1

B

A

C A1

C1

B1

1/2

h

h+1/2

67 LPS, KVRKT

Q8. The triangle ABC has angle B = 90o. When it is rotated about AB it gives a cone of volume 800π. When it is rotated about BC it gives a cone of volume 1920π. Find the length AC. Ans: Let , In Right Triangle ABC , angle B=900 The sides are a,b and c . According to given condition

Which on solving gives a=12k, b=5K, Using 1/3πb2a =800π we get , a=24 and b=10 Hence,

AC= =26 units Q9. A number when divided by 7,11 and 13(the prime factor of 1001) successively leave the remainders 6,10 and 12 respectively. Find the remainder if the number is divided by 1001. Ans: Let, X= 7q1+6 = 7(q1+1)-1 X= 11q2+10 = 11(q2+1)-1 X= 13q3+12 = 13(q3+1)-1 Hence number is 1 less than common multiple of 7,11 and 13 LCM of 7,11 1nd 13=1001 Hence, X=1001q-1 =1001(q-1)+1000 Therefore when X is divided by 1001 will leave remainder 1000. Q10. Two candles of the same height are lighted together. First one gets burnt up completely in 3 hours while the second in 4 hours. At what point of time, the length of second candle will be double the length of the first candle. Ans: I candle burns completely in 3 hours II candle burns completely in 4 hours Let, Rate of burning of I candle=x/3 unit/hr Rate of burning of II candle=x/4 unit/hr Suppose after time t length of II candle is double of the I. Then, (x-xt/4)=2(x-xt/3)

t=12/5 hrs

68 LPS, KVRKT

-1 0 1

KVS (JMO) 2007

1. Solve |x-1| + |x| + |x+1| = x+2

Ans: Critical points: -1, 0, 1

|x-1| + |x| + |x+1| = x+2

Case I: if x< -1

then

- (x-1) – x – (x + 1) = x + 2

=> - x + 1 – x -1 – x = 2

=> - 4x = 2

=> x = -1/2

-1/2 is not less than 1

Case II: -1≤ x ≤ 0

-(x – 1) – x + x + 1 = x + 2

=> - 2x = 0

=> x = 0

Case III: 0 ≤ x ≤ 1

-(x – 1) + x + x + 1 = x + 2

=> 2 = 2

Always true

Case IV: x ≥ 1

x – 1 + x + x + 1 = x + 2

2x = 2

x = 1

Hence,

x = 1 or 0

2. Find the greatest number of four digits which when divided by 3, 5, 7, 9

leaves remainders 1, 3, 5, 7 respectively.

Ans: Let,

Number = 3x + 1 = 5y + 3 = 7z + 5 = 9u + 7 = 3(x + 1) – 2 = 5(y+1) – 2 = 7(z+1) – 2 = 9(u + 1) – 2 i.e. Number is 2 less than common multiple of 3,5,7 and 9. L.C.M. of 3,5,7 and 9 = 315 Greatest no. of 4 digits = 9999 = 315×31+ 234. Greatest number of 4 digits which is a multiple of 315 = 10000 – 235=9765 Therefore, required number = 9765-2= 9763

69 LPS, KVRKT

3. A printer numbers the pages of a book starting with 1. He uses 3189 digits in

all. How many pages does the book have?

Ans: Total number of digits used in 1 digit number = 9×1 = 9

Total number of digits used in 2 digit number = 90×2 = 180 Total number of digits used in 3 digit number = 900×3 = 2700 Total digits used till three digit numbers = 9 + 180 + 2700 = 2889 Remaining digits used for 4 digit numbers = 3189 – 2889 = 300 Therefore, number of 4 digit numbers = 300/4 = 75 Number of pages = 1074

4. ABCD is a parallelogram. P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the parallelogram is 16 cm2, find the area of the quadrilateral PQRS.

Ans: Since, AP= DR and AP || DR Therefore, APRD and PBCR are also parallelograms Therefore ar(ΔPRS) = ½ ar(IIgm APRD) {Area of triangle is half the area of ll

gm on the same base and between same parallel lines}

ar(ΔPRQ) = ½ ar(IIgm PBCR) {Area of triangle is half the area of llgm

on the same base and between same parallel lines}

Therefore area of quadrilateral PQRS = ½ [ar(APRD) + ar(PBCR)] = ½ ar(IIgm ABCD) = ½ × 16cm2 = 8 cm2

D

A B P

Q

R C

S

70 LPS, KVRKT

5. ABC is a right triangle with B= 900.

M is the midpoint of AC and BM =

√117 cm. Sum of the lengths of sides AB and BC is 30 cm. Find the area of the

triangle ABC.

Ans: In ΔABC, <B = 900

M is midpoint of AC

Therefore, BM = AM = AC

=> AC = 2√117

Now x2 + (30 – x)

2 = (2√117)

2

=> x2 – 30x + 216 = 0

(x – 18) (x – 12) = 0

x = 18 or x = 12

Therefore ar(ΔABC)

= ½ × 18 × 12 cm2

= 108 cm2

6. Solve √(a+x) + √(a-x) = a

√(a+x) – √(a-x) x

Ans: By Componendo & Dividendo

Squaring both sides

Again by Componendo & Dividendo

=>

x = 0 is not possible

=> x2 = a

2

x M

√117

A

C B

30 – x

71 LPS, KVRKT

=> x = ±a

7. Without actually calculating, find which is greater: 3111 or 1714. Ans: 3111

< 3211

3111

<(25)

11

3111

<255

AND

1714

>1614

1714

>(24)

14

1714

>256

Hence

3111

<255

<256

<1714

=> 3111

<1714

8. Show that there do not exist any distinct natural numbers a, b, c, d such that a3 + b3 = c3 + d3 and a + b = c + d. Ans: a3 + b3 = c3 + d3

a + b = c + d => (a + b)3 = (c + d)3 => a3 + b3+3ab(a+b) = c3+d3+3cd(c+d) => ab = cd = n => a and b are roots of quadratic equation x2 – mx + n = 0 Also c and d are roots of same quadratic equation But A quadratic equation has at most two distinct roots Either a = c or d or b = c or d => a,b,c and d are not distinct.

9. Find the largest prime factor of 312 +212 – 2.66. Ans: 312 + 212 - 2.66

= (36)2 + (26)2 – 2.36.26

= (36 – 26)2 = {(33 – 23) (33 + 23)}2 = {(3 – 2) (32 + 3.2 + 22). (3 + 2) (32 – 3.2 + 22) = {19.5.7}2 Therefore, Largest Prime Factor = 19

72 LPS, KVRKT

10. If only downward motion along lines is allowed, what is the total number of

paths from point P to point Q in the figure below?

Ans: Total steps = 6

Forward steps = 3 Downward steps = 3 Therefore,

No. of ways = = 20

P

Q

73 LPS, KVRKT

11th KVS Maths Olympiad -2008

Remember: (i) Any odd number can be written in the form 2n -1 or 2n +1 Any even number can be written in the form 2n.

Sum of the squares of first N natural numbers is 6

)12)(1( NNN.

This can also be expressed using the Σ-notation as:

6

)12)(1(2232221

1

2 NNNN

N

n

n

Solution:

.299298..........24232221 of value theFind .1 S

.4950

5099

9949299

492

50494299

49

1

14299

29949

1

14

)2)2(2)12((29949

1

14

even) is no 2nd odd, isbracket each in nofirst (29949

1

2)2(2)12(

.299)298297(.........)2423()2221(

.299298..........24232221

n

n

n

n

simplifiedisnn

n

n

n

nn

S

74 LPS, KVRKT

Alternate method I:

Alternate method II:

S= 12-22+32-42+……………-982+992

=(12-22 )+(32-42)+……………+(972-982 )+(992-1002) +1002

= ( -3-7-11…………….-199) +10000 { n2-(n+1)2 =-(2n+1) }

= -50/2[ 2*3+49*4] +10000

= 4950

2. Find the smallest multiple of 15 such that each digit of the multiple is either ‘0’ or ‘8’. Prime factors of 15 are 3 ,5. Therefore any multiple of 15 must be divisible by 3 and 5. As the required no has to be divisible by 5, it should end in zero (the option 5 is not applicable here) Also, the given no must be divisible by 3. Therefore if you put one 8 or two eights or one 8 and zero before zero i.e. 80 or 880 or 800 or 8080 are not divisible by 3. Also, we want the smallest multiple of 15 and therefore the only possibility is 8880. The required no is 8880.

.4950

5099

1961991006

99

6

9950498

6

19910099

nat.nos 49first of squares theof sum8 - nat.nos 99first of squares theof sum

99

1

49

1

2)2(22

)2982422(2)299298..........24232221(

.299298..........24232221

n n

nn

75 LPS, KVRKT

3. At the end of year 2002, Ram was half as old as his grandfather. The sum of years in which they were born is 3854. What is the age of Ram at the end of year 2003? Let grand father’s age at the end of 2002 be ‘x’ years Therefore Ram’s age at the end of 2002 will be x/2 years. Accordingly, the year in which they were born will be (2002- x), (2002- x/2) (2002- x) + (2002- x/2) = 3854. Solving this simple eqn gives x = 100. Therefore age of Ram at the end of 2002 will be 50 and his age at the end of 2003 will be 51 years. 4. Find the area of the largest square, which can be inscribed in a right triangle with legs ‘4’ and ‘8’ units.

Here inscribed square can be formed in two ways

I.

Let, AB=4 units BC=8 units

Side of square be x, BF=y,BG=z

GBFis similar to FEC

A

B C

D

E

F

G

76 LPS, KVRKT

ADGGBFalso

zyxx

z

y

x)8(

8

2

)4(4

2 zyxx

y

z

x

Equating and solving we get

y=2z

but x2=y2+z2

z=8/7

Area of square=5z2 =320/49=649

26sq units

A

B

C

D E

G

x

x x

x

8-x

4-x

y

√80 - y

sq.units9

64squaretheofArea

3

8 x gives solving and plyingCrossmulti

x-4

x

x

x-8)(&)(

)(80

x-4

x ABEG //

(i) 80

x

x-8

EC

AE

DB

AD BC // DE

units. y'' EC and units x'' be square theof sideeach Let

808016642AC 4, BC 8,AB

iii

iiy

y

y

y

AC

77 LPS, KVRKT

Area is smallest in the second case. 5. In a triangle the length of an altitude is 4 unit and this altitude divides the opposite side in two parts in the ratio 1:8. Find the length of a segment parallel to altitude which bisects the area of the given triangle.

6. A number ‘X’ leaves the same remainder while dividing 5814, 5430, 5958. What is the largest possible value of ‘X’. According to the given condition, 5814 = aX + r, 5430 = bX + r, 5958 = cX +r and this implies the difference of any of the above 3 numbers is divisible by X. 5814 – 5430 = 384, 5958 – 5430 = 528, 5958 – 5814 = 144. The required number is H.C.F of 384, 528, 144.

A

B D

C

E

F

x 8x

y 4

.3

92

492

12

2

1

2

1

2

1ABC.ar

2

1CEFar,

284

AD // EF

y. i.e. EF, find tohave We

y. EF 8x, DC x, BDLet

(given) 8:1 DC:BD

ABC.ar2

1CEFar i.e.

parts. equal twointo ABC triangle thedivides and AD // EF 4,AD

unitsy

y

xxyy

ADBCCFEFNow

xyCFx

CFy

CD

CF

AD

EFCADCEF

78 LPS, KVRKT

Find the H.C.F as 48. The required number here is 48. Note that it is enough even if you find the H.C.F of any 2 of the numbers 384, 528, 144. A sports meet was organized for 4 days. On each day, half of existing total medals and one more medal was awarded. Find the number of medals awarded on each day.

Alternate solution: Let,

Existing medals on fourth day=x

Distributed=x/2+1

Remaining=x-(x/2+1)=x/2-1

But x/2-1=0

x=2

Existing medals on third day=y

Distributed=y/2+1

Remaining=y-(y/2+1)=y/2-1

2

3

42

1

41

2medals Remaining

)....(..........2

1

411

22

1day 2ndon awarded medals of No

12

)12

(medals Remaining

)....(..........12

x day 1st on awarded medals of No

x.be medals totalLet the

xxx

iixx

xxx

i

30. as x of valueget the andeqn above theSolve

8

1

164

1

82

1

41

2.,.

)()()()(,

)......(..........8

1

161

4

7

82

1day4th on awarded medals of No

4

7

84

1

82

3

4 medals Remaining

)....(..........4

1

81

2

3

42

1day 3rdon awarded medals of No

xxxxx

ei

xiviiiiiiNow

ivxx

xxx

iiixx

79 LPS, KVRKT

But y/2-1=2

y= 6

Similarly, z/2-1=6

Z=14

u/2-1=14

u=30

So, Medals distributed

I day=16

II day=8,

III day=4

Iv day=2

result.your

justify and BED Find .51CBE and 24BCDsuch that ly respective ACand AB

sideson points thebe E and DLet .78ACBABC with isosceles be ABCLet 8.

B

A

C

D

E

510 24

0

240

540 27

0

80 LPS, KVRKT

Alternate Solution: ,, are roots of

x3-(a+b+c)x2+(ab+bc+ca)-abc+1=0

1abc

cabcab

cba

01))()(( xxx

01)()( 23 xxx

Substituting from above,

cba

cxbxax

abcxcabcabxcbax

,,

0))()((

011)()( 23

1251-xBED

63. x 180 54 x x CED, fromNow,

51-xCEB-CEDBED then CEDxCDE if CDE,In

)(&)(

).....(78BDC BDC, From

).......(51BEC BEC, From

CECDBCiii

iiCDBC

iCEBC

c.b,a, are q(x) of roots c)-b)(x-a)(x-(x1 c)-b)(x-a)(x-(x-1 p(x) - 1 q(x)

p(x) - 1 1 )-)(x-)(x--(x1 x)-x)(-x)(-( q(x) Now,

).-)(x-)(x-(x p(x) can write wep(x) of roots theare and ,, Since

1 x)-x)(-x)(-( q(x) , 1 c)-b)(x-a)(x-(x p(x)Let

0. 1 x)-x)(-x)(-(

equation theof roots theare c and b a, that showthen

0, 1 c)-b)(x-a)(x-(xequation theof roots theare and ,, If.9

81 LPS, KVRKT

are roots of second equation

10. A 4 X 4 wooden cube is painted so that one pair of opposite faces is blue, one pair green and one pair red. The cube is now sliced into 64 cubes of side 1 unit each. How many of the smaller cubes have no painted face? How many of the smaller cubes have one painted face? How many of the smaller cubes have exactly two painted faces? How many of the smaller cubes have exactly 3 painted faces? How many of the smaller cubes have exactly one face painted blue and One face painted green? Ans: No. of cubes with no painted face = (n – 2)3 = (4 – 2)3 = 8 No. of cubes with one face painted = 6(n – 2)2 = 6 (4 – 2)2 = 24 No. of cubes with exactly two painted faces = 12(n – 2) = 12 (4 – 2) = 24 No. of cubes with exactly 3 painted faces = 8 No. of cubes with exactly one face painted blue and one face painted green = 8

82 LPS, KVRKT

XII KVS JMO 2009 Problems And Solutions

Q1. Consider the following multiplication in decimal notations (999).(abc) = def132, determine the digits a,b,c,d,e,f. Answer 1: 999 X abc = def132 LHS = (1000 – 1) abc = abc000-abc 10 – c =2 ⇒c = 8 9 – b = 3 ⇒ b = 6 9 – a = 1 ⇒ a = 8 c – 1 = f ⇒ f = 7 d=a=8 e=b=6 Q2. Find the greatest number of 4 digits, which when divided by 3,5,7 and 9 leaves remainder 1,3,5 and 7 respectively. Answer 2: Greatest four digit number = 9999 LCM of 3,5,7 and 9 = 315Highest four digit multiple of 315 = 9765 Required Number = 9765 – 2 = 9763 As 3 - 1 = 5 - 3 = 7 – 5 = 9 – 7 = 2 Q3. If n is a positive integer such that n/810 = 0.d25d25… where d is a single digit in decimal base. Find ‘n’. Answer 3: Let,N=.d25d25… Solving we get N = d25/999 d25/999 = n/810

83 LPS, KVRKT

Now, 925 = 37 X 25 n/30 = 37 X 25/37 n = 25 X 30 n = 750 Q4. Solve the integers 3x2 – 3xy + y2 = 7 and 2x2 – 3xy + 2y2 = 14 Answer 4: 3x2 – 3xy + y2 = 7 (i) 2x2 – 3xy + 2y2 = 14 (ii) Subtracting (i) from (ii) y2 – x2 = 7 (y-x)(y+x) = 7 ⇒ y – x = 1 or -1 y + x = 7 or -7 ⇒ y = ± 4 and x = ± 3 Q5. Let x be the LCM of 32002-1 and 32002+1. Find the last digit of x. Answer 5: 32002=(34)500 X 32 As 34= 81 = (Unit digit 1) X 9 = unit digit of 32002 is 9 Unit place digit of (32002-1) = 8 Unit place digit of (32002+1 )= 0 5 & 2 are the factors of their LCM

Factors of LCM must be 2X5 = 10 If 10 is factor of LCM then it’s unit digit will be 0 Q6. Let f0(X)=1/(1-X) and fn(x) = f0(fn-1(x)) Where n = 1,2,3….Calculate f2009(2009) Answer 6: fX(x) = 1/ 1-fn-1(x) f1(x) = f0(f0(x)) f1(x)=1/(1-1/x) f1(x) = (x-1)/x f2(x)=x f3(x) = 1/(1-x)

84 LPS, KVRKT

f3(x) = f0(x) Similarly fo(x) = f3(x) = f6(x) = ……f2007(x) = 1/(1-x) f2008(x) = 1/(1- f2007(x)) f2008(x) = 1/1-(1/(1-x)) = (x – 1)/x f2009(x) = 1/1-((x-1)/x) = x f2009 (2009) = 2009 Q7. ∆ABC and ∆DAC are two isosceles triangles with ∠BAC=20˚, ∠ADC=100˚.Show that AB=BC+CD. Answer 7. Construction:

Proof: Produce BC to E so that CE = CD. ∴ ∠DCE = 60˚. Then ∆DCE is isosceles( Equilateral too) and so ∠CDE = 60˚. Since DA = DE, we have that ∠DAE = ∠DEA = 10˚. Therefore, ∠BAE = 60˚ − 10˚ = 50˚ and ∠BEA = 60˚ - 10˚ = 50˚, hence AB = BE. Q8. Two intersecting circles E1 and E2 have a common tangent which touches E1 at P and E2 at Q. These two circles meet at point M and N where N is nearer to PQ than M. The line PN meets the circle E2 at R. Prove that MQ bisects ∠PMR. Answer 8: Proof:

85 LPS, KVRKT

∠1=∠2 {Angles in the alternate segment} ∠3 = ∠4 (-do-) ⇒ ∠1 + ∠3 = ∠2 + ∠4 ⇒ ∠1 + ∠3 = ∠PMQ (i) ∠5 = ∠1 + ∠3 (ii) {Exterior angle property} ∠5 = ∠6 {Angles in the same segment} ∠5 = ∠QMR (iii) From (i), (ii) and (iii) ∠PMQ = ∠QMR Proved Q9. AB is a line segment of length 24cm and C is its middle point. On AB, AC and CB semi circles are described. Determine the radius of the circle which touches all the three semi circles. Answer 9:

Let x be the radius of the required circle

N

86 LPS, KVRKT

CO2 = (6 + x)2 – 62 CO = 12 – x

(12-x)2 = (12+x).x

144 + x2-24x-12x-x2 = 0

36x = 144

x = 144/36

= 4

Therefore required answer is 4cm.

Q10. Prove that a4+b4+c4 abc (a + b + c)

Answer 10: Without any loss in generality we may assume that a< b<c.

Applying Tchebychef’s inequality to the 3 sets of number same as (a,b,c)

(a3+b3+c3) / 3 ( a+ b+ c)/3.( a+ b+ c)/3 .( a +b +c)/3

a3+b3+c3>(a+b+c)3/9 (i)

Since Arithmetic Mean exceeds Geometric Mean

(a+b+c/3)3>abc (ii)

From (i) and (ii)

a3+b3+c3>(a+b+c)3/9>3abc (iii)

Since a<b<c, therefore a3 < b3 < c3

Applying Tchebychef’s inequality to the sets of number (a,b,c);( a3 , b3 , c3), we get

(a4+b4+c4)/3 > (a3+b3+c3)/ 3. (a+b+c)/3 (iv)

87 LPS, KVRKT

From (iii)

(a3+b3+c3 )/ 3>abc (v)

From (iv) and (v)

a4+b4+c4> abc (a + b + c)

For corrections and suggestions e – mail at :- lpsah2008@rediffmail.com

sahlp@yahoo.com