KVS Junior Mathematics Olympiad (JMO) – 2001
M.M. 100 Time : 3 hours
Note : (i) Please check that there are two printed pages and ten question in all.
(ii) Attempt all questions. All questions carry equal marks.
1. 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 a
……..
(e) The sum of the digits of the number 2200052004 is ….
2. (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 (shaded) in cm2.
3. (a) Solve :
log 10 (35 x 3 )
3
log 10 (5 x )
(b) Simplify :
a b b c c a (a b)(b c)(c a )
a b b c c a (a b)(b c)(c a )
4. (a) Factorize :
(x-y)3+(y-z)3+(z-x)3
(b) If x2-x-1=0, then find the value of x3-2x +1
5. ABCD is a square. A line through B intersects CD produced at E, the side
AD at F and the diagonal AC at G.
B A
G
F
C D E
If BG = 3, and GF=1, then find the length of FE,
6. (a) Find all integers n such that (n2-n-1)n+2 = 1
4ab x 2a x 2b
(b) If x = , find the value of
ab x 2a x 2b
7. (a) Find all the positive perfect cubes that divide 99.
(b) Find the integer closest to 100 (12- 143 )
8. In a triangle ABC, BCA=90o. Points E and F lie on the hypotenuse AB
such that AE=AC and BF = BC. Find ECF.
A
F
y E
x
C z B
9. An ant crawls 1 centimetre north, 2 centimetres west, 3 centimetres south, 4
centimetres east, 5 centimetres north and so on, at 1 centimetre per second. Each
segment is 1 centimetre 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 ?
10. Find the lengths of the sides of a triangle with 20, 28 and 35 as the lengths of
its altitudes.
SOLUTION KV JMO 2001
Q1.
(i) 7
(ii) 6 litres
(iii) 20 m
(iv) Friday
(v) 13
Q2.
(a) 25555, 33333, 62222
25 = 32
33 = 27
62 = 36
3 5 2
3 <2 <6
33333 < 25555 < 62222
The required order of the three numbers is :-
33333 , 25555 , 62222
Q2(b)
4
Two rectangles each measuring 3cm x 7 cm are placed in this
manner.
The four triangles formed must be congruent to each other.
EAG BCG
AG = BG
and EG = CG
Let, BG = x
AG = AB – BG
=7–x
CG = AG = 7-x
In BCG, we have,
BC = 3
BG = x
CG = 7 – x
By Pythagoras theorem, In BCG, we have,
CG2 = BC2 + BG2
(7-x)2 = 32 + x2
49 – 14x + x2 = 9 + x2
49 – 9 =14 x
5
40 = 14 x
40
x=
14
20
x=
7
AG = 7 – x
20
=7-
7
49 20
=
7
29
=
7
DC || AB
HC || AG
Similarly AH || GC
AG CH is a parallelogram
Area of the AGCH = (Base) x (Height)
= AG x BC
29
= x3
7
87
=
7
3
= 12 cm 2
7
6
log 10 (35 x 3 )
3.(a) 3
log 10 (5 x )
log 10 (35 – x3) = 3.log10 (5-x)
log 10 (35 – x3) = log10 (5-x)3
35 – x3 = (5-x)3
35 – x3 = 125 – x3 + 3(5) (-x) (5-x)
35 = 125 – 15 x (5-x)
35 = 125 – 75 x + 15x2
15x2 – 75 x + 125- 35 = 0
15x2 – 75 x + 90 = 0
x2 – 5 x + 6 = 0
x2 – 3 x –2x + 6 = 0
x (x - 3) – 2 (x – 3) = 0
(x - 3) (x – 2) = 0
x = 3 or 2
Q3. (b)
a b b c c a (a b)(b c)(c a )
a b b c c a (a b)(b c)(c a )
= (a-b) (b+c) (c+a) + (b-c) (a+b) (c+a)
(c a )(a b)(b c) (a b) (b c)(c a )
=
(a b)(b c)(c a )
7
= (c+a) [(a-b) (b+c) +(b-c) (a+b)]
(c a )[(a b) (b c) (a b)(b c)]
=
(a b)(b c)(c a )
= (c+a) [ab + ac - b2 – bc + ab + b2 – ac - bc]
(c a )[ab ac - b 2 bc ab b 2 - ac bc]
=
(a b)(b c)(c a )
(c a )(2ab - 2bc) (c - a) (2ab 2bc)
=
(a b)(b c)(c a )
2abc 2bc 2 2a 2 b 2bc 2abc 2bc 2 2a 2 b 2abc
=
(a b)(b c)(c a )
0
=
(a b)(b c)(c a )
=0
4. (a)
= (x-y)3 + (y-z)3 + (z-x)3
= (x – y + y - z) [(x-y)2 + (y-z)2 + (x-y) (y-z)] + (z-x)3
= (x –z) [x2+y2-2xy+y2+z2-2yz+xy-xz-y2+yz] + (z-x)3
= (x –z) (x2+y2+z2 –xy-yz-zx) - (x-z)3
= (x –z) [x2+y2+z2 –xy-yz-zx) - (x-z)2]
= (x –z) [x2+y2+z2 –xy-yz-zx – x2-z2 +2xz]
= (x-z) [y2-yz-xy+xz]
= (x-z) [y(y-z)-x(y-z)]
8
= (x-z)(y-x)(y-z)
=(x-y) (y-z) (z-x)
4. (b) x2 – x – 1 = 0
x3 – 2x + 1 = ?
Dividing (x3 – 2x + 1) by (x2 – x - 1), we get,
x 1
x 2 x 1 x 2x 1
3
x3 x 2 x
x2 x 1
x 2 x 1
2
x3 – 2x + 1 = (x2-x-1) (x+1) + 2
x3 - 2x + 1 = 0 x (x + 1) + 2
x3 - 2x + 1 = 0 + 2
x3 - 2x + 1 = 2
x3 – 2x + 1 = 2
Q5
9
.
Given :
ABCD is a square
BG = 3
GF = 1
BE = ?
Let, AB = BC=CD = DA = a
Now,
In BCG and FAG,
BCG = FAG (Each = 45o)
BGC = AGF (Vertically opposite angles)
BCG ~ FAG by AA rule
BC BG
AF GF
a 3
AF 1
10
a
AF
3
a 2
DF
= AD – AF = a a
3 3
Now,
In ABF and DEF,
A = D = 90o
AFB = DFE (Vertically opposite angles)
ABF ~ DEF by AA rule
BF EF
AF DF
BF EF
a 2
a
3 3
2.BF EF
2 X 4 = EF
EF = 8
BE = BF + FE
= 4+ 8
= 12 units.
11
Q6.(a) (n2-n-1) n+2 = 1
(n2-n-1) n+2 = (1)n+2
n2 – n – 1 = 1
n2 – n – 2 = 0
n2 - 2n + n – 2 = 0
n (n - 2) + 1 (n – 2) = 0
(n - 2) (n + 1) = 0
n = 2 or -1
4ab
Q6. (b) x=
ab
x 2a x 2b
?
x 2a x 2b
x 2a x 2b
x 2a x 2b
x 2a x 2b
1 1
x 2a x 2b
x 2a x 2a x 2b x 2b
x 2a x 2b
2x 4b
x 2a x 2b
2x 4b
2 2
x 2a x 2b
12
2x 2x 4a 4b 2x 4b
x 2a x 2b
4a 2x
x 2a x 2b
4ab
2x
4a
ab
4ab 4ab
2b
ab ab
4a (a b) 8ab
4ab 2a 2 2ab 4ab 2ab 2b 2
4a 2 4ab 8ab
2ab 2a 2 2ab 2b 2
2a 2b 4a
ba ab
2a 2b 4a
ab
2a 2b
ab
2(a b)
(a b )
=2
Q7. (a) 99
= (33)9
= (3)27
13
The positive perfect cubes that divide 99 are :
13, 33, (32)3, (33)3, (34)3, (35)3, (36)3, (37)3, (39)3.
i.e. 10 numbers
(b) 100 (12- 143 )
143 144
i.e. 143 < 12
and 143 is less than 12 by a very small margin.
The closest integer to 100 (12 - 143 )
is , 100.
8.
A
F
x+z
y x+y E
x
C z B
Given :
BCA = 90o
AE = AC
BF = BC
ECF = ?
BCA = 90
x + y + z = 90o
In ACE
14
AE = AC
AEC = ACE = x + y
In BCF,
BF = CF
BCF = BFC = x + z
In CFE,
FCE + CFE + CEF = 180
x + x + z + x + y = 180
2x + x + y + z = 180
2x + 90 = 180
2x = 90
x = 45o
ECF = x
= 45o
ECF = 45o
9. 2cm (W)
3 cm (s) 1 cm (N)
4 cm (E)
15
So, we get that,
The ant always travels (4k+1) cm North,
The ant always travels (4k+2) cm West,
The ant always travels (4k+3) cm South,
The ant always travels (4k+4) cm East,
In 1 min the ant travels, distance = 60 x 1 cm
= 60 cm
We get the following AP,
1, 2, 3, …….
a=1
d=1
n
Sn = 2a + (n-1)d
2
n
60 = 2 x 1 + (n-1) x 1
2
120 = n2 + n-1
120 = nn+1
120 = n2n
n2 + n
1 12 4(1)(120)
n =
2x1
16
1 1 480
2
1 481
2
1 22
2
21
Taking the positive value
2
1
10
2
But we’ll have to take, n = 11
a11 = 1 + 10 x 1
= 1 + 10
= 11
=4x2+3
The ant was traveling towards South.
Q10. A
B C
17
ha = 20
hb = 28
hc = 35
1 2
= a x ha a =
2 ha
2
=
20 10
1 2
= b x hb b =
2 hb
2
=
28 14
1 2 2
= c x hc c = =
2 hc 35
2
a =
20
2
b =
28
2
c =
35
a b c 1 2 2 2
S =
28 35
=
2 2 20
1 1 1
=
20 28 35
18
1 1 1 2
S-a = - 20
20 28 35
1 1 1
= -
28 35 20
35x 20 28x 20 28x35
=
20x 28x35
700 560 980
=
20x 28x35
10
280
= 20 x 28x35
2
1
=
70
=
70
1 1 1 2
=
20 28 35 28
S-b -
1 1 1
= -
20 35 28
35x 28 28x 20 20x35
=
20x 28x35
980 560 700
=
20x 28x35
19
42
63
840
= 20 x 28 x 35
14 5
3
=
70
3
=
70
1 1 1 2
S-c = - 35
20 28 35
1 1 1
= -
20 28 35
35x 28 20x35 20x 28
=
20x 28x35
980 700 560
=
20x 28x35
56
2
1120
= 20x 28x35
2
=
35
2
=
35
= s(s a )(s b)(s c)
20
1 1 1 3 2
=
20 28 35 70 70 35
2240 1 3 2
= 2 x x x
20x 28x35 70 70 35
16x3
= 2
28x35x10x35x35
6
= 2
35x35x35x35
6
= 2
35x 35
6
2 =
35x 35
35x 35
6
2
a=
20
245 6
a=
6
2
b=
28
175 6
b=
12
2 35 6
c= = units.
35 3
21
