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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 repeatedLargest number of six digits = 999999
Smallest number of six digits=100000
Difference=899999
(ii) If digits are not repeatedLargest 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.
A
F
D
P
Q
R
B
E
C
E
Triangle ABC is equilateral =>
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
AB=BC=CA
=>
=>
But
<BCD=<ABF
<BCQ=<DBQ (i)
<PQR =<QBC+<BCQ
{Exterior angle}
=<QBC+<DBQ
{From i }
0
= <ABC =60
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)}
= 2{3x6+x3-3x}
{(a+b)3+(a-b)3=2{a3+3ab2}
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Therefore, Degree of expression =6
(b)
=>
=>
=> 6x-12-2x-4=0
Therefore, x=4
since x-1
0
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.
1m
1m
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
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Ans7(a).
A
D
2x
x
B
C
E
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
X
Q
P
M
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.
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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=
Total volume of toy=
=141.3
= 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
G
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
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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
Ans2.
* * *
-----------------------------* * * * *
* * * * *
* * * * *
-------------------------------* * * * 5 4 1
--------------------------------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
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
(b)
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
1400
x
1400
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Ans7(a)
R
8cm
Q
12cm
8cm
S
P Draw RT bisecting <QRS
T
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)
A
0
140
B
E
1400
x
C
F
0
0
140140
Two exterior angles of triangle are 1400.
Therefore, <FBC=3600 –(1400+1400) =800
<BFE =1800-1400=400
Therefore x=1800-(800+400) =600
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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.
18
6
20
8
666
30
(b) What is the sum of the four angles a, b, c, d in the diagram?
a
b
d
c
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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
T
H
A
A
T
H
H
S
S
S
S S S
S
There are 15 words for each A
Hence,
Total number of such words=15x4=60
A
T
H
H
S
S
16
S
H
S
S
H
S
S
S
LPS, KVRKT
KVS-JMO 1999
Q1.Fill in the blanks:
(a)
(b)
(c)
(d)
(e)
(0.4)2-(0.1)2 equals ………..
The angles of a triangle are in the ratio 2:3:4.The greatest angle in degrees is ……………..
If the radius of a circle is increased by 100%, the area is increased by……%.
If log102=a and log103=b, then log1012 equals………
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 placedx+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.
D
A
E
O
C
B
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
=
18
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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=
Height of cylinder=2r
{ Diameter of cylinder will be equal to diagonal of cube}
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.
C
A
X
Y
B
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.
R
S
P
Ans9.Let,
Distance= D
Speed=x km/hr
Q
50 km
20km
(i)
and,
(ii)
20
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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
x
56 a
56 a
48
1
8
72b
72 b
64
84c
8
84 (c
1
76
1)
96 d
8
96 ( d
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
99999
2016 49 1215
Largest multiple of 5 digits = 99999 1215
98784
Required number = 98776
Q2 (a) If
x2
1
x
x
x
x2
8 x 1 0 , find the value of x 3
1
x3
.
8x 1 0
8
1
x3
3
= 8
3
x
=
1
x
3
3 x
1
x
3 8
= 512 + 24
= 536
Q2 ( b) Which is greater:
3111 or17 14 ?
3111 17 14
17
11
31
17
11
17
3
11
17
22
31
17
11
4913
LPS, KVRKT
88
1)
8
=1
31
2
17
11
31
211
17
11
31
2048
17
11
31
4913
0
17
3111 17 14
0
14
17
3111
99
2 99 399 4 99
Q3 (a) Show that 1
199
2 99
199
3 99
4 99
each is divisible by 5
Q 3 ( b)Solve:
x
x
3
3
x
3
x 3
x 3
x 3
x 3
x
x
2
x2
3
3
x 3
599
is exactly divisible by 5.
4 99
2 99
5 99
3 99
5 99
{x n + y n is divisible by x + y when n is odd}
3
2
3
2
2
3
2
9
2
x
2
x
3
3
x2
9
2
2
x
9
8x
x2
8x
9
0
x
9 x
1
0
9 or x
1
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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2
On observation we conclude that the shaded portion is forming a rectangle of
dimensions
6 18
i.e.18=( 2+3+6+3+4)
Total area of shaded region
6 18 108 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
3x 5 when it is divided by
( x 6)( x 7) . What remainder is obtained if this polynomial is divided by x 6 ?
Q5 (b)
An unknown polynomial leaves a remainder
p ( x)
x
6
x
= (x-6 ){(x-7)q(x) + 3} + 13
Hence Required remainder when p(x) is divided by
7
x
q( x)
3x
6 = 13
2 BC . Triangle ABE overlaps the rectangle and is
equilateral, and M is the mid – point of BE. Find (i) CMB (ii) the fraction of the
Q 6. In a rectangle ABCD, AB =
24
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5
rectangle ABCD that is covered by triangle ABE.
BM = BC
also
BMC
BCM
MBC 30
CMB 75
BC
BC 2
Area of two triangles not included =
.BC
3
3
2
BC
2 BC 2
2 3 1
3
Fraction of triangle in rectangle =
=
2 BC 2
2 3
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
26
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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 =
8!
4!4!
27
= 70
LPS, KVRKT
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
28
LPS, KVRKT
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
29
LPS, KVRKT
(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
30
LPS, KVRKT
=> 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
31
LPS, KVRKT
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
32
LPS, KVRKT
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
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
{ Draw figure}
Area = 7xx20/2 =
7x10x =
x=
lengths of sides
a=
b=
c=
33
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
34
LPS, KVRKT
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
7
5
a(-9) +b(-9) -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
.
, find the value of 37x.
(b) If
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
35
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(
36
LPS, KVRKT
=25(3) cm2
Ar( tr. UQR) = ar(tr. PRQ)-ar(tr. PUQ)
=50-{25(3-
) }cm2
-1) cm2
= 25(
Q7. A square of side length 64cm is given. A second square is obtained by connecting the midpoints 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.
37
LPS, KVRKT
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
F
G
H
E
J
B
D
K
L
I
A
c
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.
38
LPS, KVRKT
KVS JMO 2003
Q 1.Fill in the blanksThe 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 )
(2003)2000 (200340)50 150 (mod100)
41.61 (mod 100 )
1(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 )
39
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)
=
40
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.
41
LPS, KVRKT
(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
42
LPS, KVRKT
. 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 =
Shaded area is forming regular hexagon of side
Ratio of shaded area to the total area =
a/3
a/3
=
. 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)
}
2
one face painted = 24 { 6(n-2)
}
3
no face painted = 8 { (n-2)
}
. 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.
43
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
=
44
LPS, KVRKT
=
Hence,
45
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
3x 2 34
x 2 4
x 6
(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 =
=
1
CD AC sq. units
2
1
7 4 sq. units = 14 sq. units
2
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
=
100
=100
3
46
LPS, KVRKT
300
= 60
5
300
No. of multiples of 3 and 5 both =
= 20
15
300
No. of multiples of 10
=
= 30
10
300
N0 of multiples of 15
=
= 20
15
300
No. of multiples of 10nd 15both =
=10
30
No. of multiples of 5
x denotes
greatest
=
int eger less
than / equal x
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:
x2
x2
xy
xy
y2
y2
19
49
(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)
Adding
x 2 xy y 2 49
x 2 xy y 2 19
2(x2 + y2) = 68
x2 + y2 = 34
Subtracting
Hence, (x – y)2 = 64
x–y = 8
2 xy = - 30
47
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
Alternate method:
and c= -3
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.
48
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
x
y
2
5
u y = 25
5
y
x
2
5
u
x 25
2
49
LPS, KVRKT
xu
25 5
10
2
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
2
Side of square =
units
Area of four semicircles
50
LPS, KVRKT
2
1
4
2
=
4
1
2
2
2
2
4
sq. units
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
Similarly,
In triangle DCQ
BQ = PQ
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?
51
LPS, KVRKT
Solution: Number of routes through AXB = 2
Number of routes through AXYB = 2
Number of routes through AYB = 1
Total number of routes
=2
= 11
1
1 3
3
1+2
1
3+1
3
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
2
= 104 cm
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.
52
LPS, KVRKT
Solution:
Number of triangles that are not right triangle
6 5 4
1 1 12
1 2 3
= 20 – 14
=6
i.e. ABF , ACE , BCD, DEC, DFB, EFA
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
150 1
36 2
150 14 x
3x
50
6
53
LPS, KVRKT
Average speed
200 3 x
150 14 x
x 5
7x2 – 190x + 375 = 0
(x – 25) (7x – 15) = 0
x = 25 or x = 15/7
But
50
x
14
3
10
, x = 15/7 does not satisfy it.
Therefore,
x = 25
Speed for next part = 25 km/hr
54
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 ……………..
4 4
2
b.If 1
0, then equals ………………………
2
x x
x
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
………………………………………….
4
SOLUTION1: (a)
2r
r 2 1, Area=1 sq.unit
2 r
2
4 4
2
2
0
1
0
1
2
x x
x
x
(c) (a+b+c-d)+(a+b-c+d)+(a-b+c+d)+(-a+b+c+d)
=2(a+b+c+d)
= 2x1111
=2222
1
10
10
(d) New area= .(1
)b.(1
)h
2
100
100
99 1
bh
=
100 2
1 1
bh
Change in area=
100 2
% change
= 1% decrease
(b) 1
(e) x3+y3=(x+y)3-3xy(x+y)
=13-3.1.1
= -2
2.(a) If x= (log82)log28, find the value of log3x.
4x
9x y
243, find the value of x-y.
(b) If x y 8and 5 y
2
3
SOLUTION:
1
( a) x=
log 2 8
log 2 8
log b a
55
1
log a b
LPS, KVRKT
1
=
log 2 2 3
1
=
3 log 2 2
log 2 23
3 log 2 2
log a m n
n log a m
3
1
=
log a a 1
3
log3x=log33-3=-3log33=-3
4x
x y 3
(b) x y 8 2 x y 2 3
2
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
25 6(mod 13) (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)
200
40 5
2 = (2 )
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 2 48-1.
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
56
LPS, KVRKT
Solving we get a=1,b=2
Remainder= x+2
48
6
6
12
24
b. 2 -1=(2 -1)(2 +1)(2 +1)(2 +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
ar AON
AO 2
{ O is centroid}
4
ar MOQ MO 2
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:
57
LPS, KVRKT
BM=
3
6
2
3 3Cm { altitude of equilateral triangle}
2
2
BG= BM
3 3Cm
3
3
Hence,
PG2 =BP2-BG2 =15-12=3
Therefore,
h=PG= 3
2 3Cm
3 2
6 Cm 2
4
1
3
36 3Cm 3 9Cm 3 { v=1/3area of base .height}
Volume of pyramid=
3 4
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:
Area of base=
CE.DE=AE.BE
AE=2Cm, DE=3Cm,
Therefore, CE=4Cm
EB=6Cm
58
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
65
Radius=
Cm, diameter= 65Cm
2
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
1
1
1 1
x 6 x 1 x 2x
3x 2 7 x 6 0
(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( BCD) ar ( ABD )
1
1
4 1
4 2 squnits
=
2
2
59
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.
60
LPS, KVRKT
KVS JMO 2006
Q1. a, b, c are three distinct real numbers and there are real numbers x, y such that a 3 + 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
61
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.
C
Ans:
E
A
B
D
Const: PD = DE
ΔCDP ≅ ΔCDE
P
<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,
62
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
63
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)
A
Ans:
h
y
P
b
B
Given,
C
d
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
64
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?
Q
S
R
T
Ans:
hence,
U
Since, QR ll BC, QU ll AB and SV ll AC
V
let BC=k
but, quadTPUV is a parallelogram
1
Similarly,
2
3
Adding eq. 1,2 1nd 3
ar(
65
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.
1/2
Ans:
………………2………………………….
h
h+1/2
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:
A
C1
B
A1
C
B1
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
66
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
67
LPS, KVRKT
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
-1
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
0
1
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
68
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.
D
R
C
S
Q
A
P
B
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 llgm 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
69
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.
A
M
x
√117
B
C
30 – x
Ans: In ΔABC, <B = 90
0
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+x) – √(a-x)
Ans: By Componendo & Dividendo
=
a
x
Squaring both sides
Again by Componendo & Dividendo
=>
x = 0 is not possible
=> x2 = a2
70
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
71
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?
P
Q
Ans: Total steps = 6
Forward steps = 3
Downward steps = 3
Therefore,
No. of ways =
= 20
72
LPS, KVRKT
11th KVS Maths Olympiad -2008
1. Find the value of S
12
22
32
4 2 .......... 98 2
99 2.
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
N ( N 1)( 2 N 1)
.
6
This can also be expressed using the Σ-notation as:
N
n2
n 1
12
22
32  N 2
N ( N 1)( 2 N 1)
6
Solution:
S
12
22
32
(12
49
2 2 ) (3 2
42
.......... 98 2
99 2.
4 2 ) ......... (97 2
(2n 1) 2 (2n) 2
n 1
49
4n 1
99 2
n 1
49
4n 1
99 2
n 1
49
99 2
4n 1
n 1
4 49 50
99 2
49
2
99 2 49 99
99 2
98 2 ) 99 2.
(first no in each bracket is odd, 2nd no is even)
(( 2n 1) 2
(2n) 2 is simplified )
99 50
4950 .
73
LPS, KVRKT
Alternate method I:
12 2 2 3 2 4 2 .......... 98 2 99 2.
(12 2 2 32 4 2 .......... 98 2 99 2 ) 2(2 2
99
49
n 2 2 ( 2n) 2
n 1
n 1
4 2  98 2 )
sum of the squares of first 99 nat.nos - 8 sum of the squares of first 49 nat.nos
99 100 199 8 49 50 99
6
6
99
100 199 196
6
99 50
4950 .
Alternate method II:
S= 12-22+32-42+……………-982+992
=(12-22 )+(32-42)+……………+(972-982 )+(992-1002) +1002
{ n2-(n+1)2 =-(2n+1) }
= ( -3-7-11…………….-199) +10000
= -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.
74
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
A
D
E
G
B
F
C
I.
Let, AB=4 units BC=8 units
Side of square be x, BF=y,BG=z
GBFis similar to
FEC
75
LPS, KVRKT
x
z
8 y x
also GBF
x
4 z
y
x
x2
(8
y) z
ADG
x2
y (4 z )
Equating and solving we get
y=2z
but x2=y2+z2
z=8/7
Area of square=5z2 =320/49=6
26
sq units
49
A
√80 - y
8-x
x
D
x
E
y
x
x
B
C
4-x
G
AB 8, BC 4, AC 2 64 16 80
AC
80
Let each side of the square be ' x' units and EC ' y' units.
DE // BC
EG // AB
(i ) & (ii )
AD
DB
AE
EC
8-x
x
x
4-x
8-x
x
80 y
y
x
4-x
80 y
y
Crossmulti plying and solving gives x
Area of the square
(i)
(ii )
8
3
64
sq.units
9
76
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.
A
E
4
B
y
x
8x
D
F
C
AD
4, EF // AD and divides the triangle ABC into two equal parts.
1
i.e. ar CEF
ar ABC.
2
BD : DC 1 : 8 (given)
Let BD x, DC 8x, EF y.
We have to find EF, i.e. y.
EF CF
y CF
EF // AD
CEF
CAD
CF 2 xy
AD CD
4 8x
1
1
1 1
Now, ar CEF
ar ABC.
EF CF
BC AD
2
2
2 2
1
y 2 xy
9x 4
2
y2 9
y
3 units.
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.
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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.
Let the total medals be x.
x
1..............(i )
2
x
x
Remaining medals x (
1)
1
2
2
1 x
x
No of medals awarded on 2nd day
1 1
2 2
4
No of medals awarded on 1st day
1
..............(ii )
2
x
x 1
x 3
Remaining medals
1
1 x 43 2
x 1
No of medals awarded on 3rd day 2
1 4 2 ..............(iii )
2 4 2
8 4
Remaining medals
x
4
3
2
x
8
No of medals awarded on 4th day
Now, (i ) (ii ) (iii ) (iv )
x
x 1 x 1 x
i.e.,
1
2
4 2 8 4 16
Solve the above eqn and get the
1
4
x
8
7
4
1 x
2 8
7
4
1
x
16
1
................(iv )
8
x
1
x
8
value of x as 30.
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
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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
8. Let ABC be isosceles with
ABC
AB andAC respective ly such that
78 . Let D and E be the points on sides
24  and CBE 51. Find BED and justify
ACB
BCD
your result.
A
0
24
E
D
0
27
0
B
51
0
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0
24
C
51
BC CE.......(i )
From BDC, BDC 78 
BC CD.....(ii )
(i ) & (ii )
BC
CD CE
From BEC, BEC
In
CDE, if
Now, from
BED
CDE
x
CED then
CED, x x 54 180
x - 51 12 
BED
CED - CEB
x - 51
x 63.
9. If , , and are the roots of the equation (x - a)(x - b)(x - c) 1 0,
then show that a, b and c are the roots of the equation
( - x)( - x)( - x) 1 0.
Let p(x) (x - a)(x - b)(x - c) 1 , q(x) ( - x)( - x)( - x) 1
Since , , and are the roots of p(x) we can write p(x) (x - )(x - )(x - ).
Now, q(x) ( - x)( - x)( - x) 1 -(x - )(x - )(x - ) 1 1 - p(x)
q(x) 1 - p(x) 1 - (x - a)(x - b)(x - c) 1
(x - a)(x - b)(x - c)
roots of q(x) are a, b, c.
Alternate Solution:
, , are roots of
x3-(a+b+c)x2+(ab+bc+ca)-abc+1=0
a b c
ab bc ca
abc 1
(
x)(
x3
(
x)(
x) 1 0
)x2
(
)x
1 0
Substituting from above,
x3
(a b c) x 2
(ab bc ca ) x abc 1 1 0
( x a)( x b)( x c)
a, b, c
0
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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
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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
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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)
2
2
2x – 3xy + 2y = 14
(ii)
Subtracting (i) from (ii)
2
2
y –x =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)
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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:
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∠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:
N
Let x be the radius of the required circle
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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)
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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 :-
[email protected]
[email protected]
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