KV JMO 2002 SOLUTIONS Q1. (a) 4025

KV JMO 2002 SOLUTIONS
Q1.
(a)
4025
(b)
56 : 3
(c)
6.25
(d)
123457
(e)
20 cm
Q2.
(a)
All the four digit number , which can be formed are :1121, 1112, 1122
2221, 2211, 2212
1221, 1212, 1211, 1222
2121, 2112, 2111, 2122
i.e. a total of 14 numbers
(b)
First
L.C.M. of 2, 3, 4, 5, 6, and 7
=
420
Now the largest four – digit multiple of 420 is :420 x 23 = 9660
 The req. number is
=
9660 – 1
=
9659 Ans.
Q3.
(a)

F (x) = ax7 + bx5 + cx3 - 6
f (-9) =
3
f (-9) =
a (-9)7 + b (-9)5 + c (-9)3 – 6
a (-9)7 + b (-9)5 + c (-9)3 – 6 = 3
 -a (9)7 + b (-9)5 + -c (9)3 = 9
 -a (9)7 - b (9)5 - c (9)3 = 9
 a (9)7 + b (9)5 + c (9)3 =

(b)
f (9) =
- 9
a (9)7 + b (9)5 + c (9)3 – 6
=
-9-6
=
-15 Ans.
(2002)3  (1002)3  (1000)3
3 x (1002) x (1000)
(2002 1002) [( 2002)2  (2002) (1002)  (1002)2 ]  (1000)3
=
3 x (1002) x (1000)
1000 [( 2002 1002 ) 2  3 (2002 ) (1002 )]  (1000 )3
=
3 x (1002 ) x (1000 )
(1000 ) 2  3 (2002 ) (1002 )  (1000 ) 2
=
3 x (1002 )
3 (2002 ) (1002 )
=
3 x (1002 )
= 2002 Ans.
1
x4 +
Q4.(a)
= 47
x4
2
2
2
 1 
 (x ) +  2  + 2. (x2)  12  - (x2)  12  = 47
x 
x 
x 
2

 2 1 
x  2 
x 

2

 2 1 
x  2 
x 

2

x2 

1
x2
- 2 = 47
= 49
=7
2
1

1
 x    2. ( x )  
x


x

1

x 
x

 x
=7
2
=7+2
1
 9
x
=3
 x3 
1
x3
1

1 1 
  x    x 2  ( x ) x    2 
x


x x 
1
1


  x    x 2  2 1
x
x


=
(3) (7 - 1)
=
3x6
=
18 Ans.
82x
=
16 1 – 2x

82x
=
(82)1 – 2x

(8)2x
=
(8)2 – 4x

2x
=
2–4x

2x+4x
=
2

6x
=
2

x
=
1
3

37x
(b)
1
=
=
=
37 x 3
(3)
3x
3
7
3
1
3
1
3
=
3x
x3
3
3 3
(3 x 3 x 3 )
1
3
3x
x3
1
3

5.
=
3 x3 x 3 3
=
93 3
37x
=
A.
9 3 3 Ans.
.
.
C
D
70 km
35 km
Let, the train’s speed
=
y

=
70  x
y
Total time
Case I :Total time
=
.B
70
x

3
y
y
4
=
70
4x

y
3y
=
210  4x
3y

70  x
210  4x
1
=
+ 1
3
3y
y

210  4x
210  3x  3y  y
=
3y
3y
x

210 + 4x = 210 + 3x + 4y

x – 4y = 0
-------------- (i)
Case II :Total time
=
70  35
x  35

3
y
y
4
=
70  35
4x  140

y
3y
=
105 X 3  4x  140
3y
=
315  140  4x
3y
=
175  4x
3y

175  4x
3y
=
70  x
1
y

175  4x
3y
=
70  x  y
y

175 + 4x
=
210 + 3x + 3y

4x – 3x – 3y = 210 - 175

x – 3y = 45
We have,
(i)
 x – 4y = 0
(ii)
(ii)
 x – 3y = 45
x – 4y = 0
x – 3y = 45
-
+
-
-y = -45
 y = 45 km/h

Speed of train = 45 km / h
Q6.
P
Q
60o
60o
30o
U
45o
T
S
R
Considering UQR,
Q
30o
10 cm
105o
U
45o
R
Sin  QUR = sine 105o = sin (60 + 45)
= sin 60 cos 45 + sin 45 cos 60
=
=
=
3 1
1 1
x

x
2
2
2 2
3
1

2 2 2 2
3 1
2 2
In QUR,
By the law of sine, we have,
UQ
QR
UR
=
=
sin URQ sin QUR sin UQR

UQ
10
UR
o =
o =
sin 45
sin 105
sin 30 o

10
UQ
=
sin 45 sin 105
10. sin 45
 UQ =
sin 105
1
2
3 1
2 2
10 x
=
= 10 x
1
2 2
x
2
3 1
=
20
3 1
x
3 1
3 1
20( 3  1) 20( 3  1)

=
2
2
3 1
( 3) 1


 UQ = 10
= 10 3  1

3  1 cm
We know that,
Area of a triangle = ½ bc sin A
 In  QUR,
ar (UQR) = ½ x UQ x QR x sin UQR


= ½ x 10 3  1 x 10 x sin 30


= 50 3  1 x
= 25 (3 = (75 – 25
 ar (UQR)
Q7.
3
3
2
3)
) cm2
= = (75 – 25
3
) cm2
64cm
32cm
32cm
 side of the 1st square = 64 cm
 side of the 2nd square
=
322  322
=
2x322
=
2 x 32
= 32
=
64
2
2x
2
2
32 2
16 2
16 2
 Side of the 3rd square
=
(16 2 ) 2  (16 2 ) 2
2
= 2  (16 2 )
= (16 2 ) x 2
= 32 x
=
2
2
64
64

2 ( 2 )2
 Side of the 1st square
= 64 cm
 Side of the 2rd square
=
64
cm
2
 Side of the 3rd square
=
64
cm
( 2 )2
 In the similar fashion,
side of the 11th square
64
( 2 )10
=
64
25
=
2 x 2 x 2 x 2 x 2 x 2
2 x 2 x 2 x 2 x 2
=
2 cm
Q8.
Let, each side of a cube = a

There are 7 cubes in the given solid,

Total volume of the given solid = 7a3

7a3 = 448
448
7

a3 =

a3 = 64

a = 3 64
= 4 cm

In the given solid,
We are able to see 5 surfaces of 6 cubes

Total S.A. of the solid
= 5a2 x 6
= 30 x 42
= 30 x 16
= 480 cm2
Q9.
Let, us consider two cases, when Anil is a Rabbit and when Anil is a Fox,
CASE I :
When Anil is a Rabbit,
We have,
Anil is a Rabbit/Fox.
Chintoo is a Fox
Eashwar is a Fox
Bhauna is a Rabbit
Dolly is a Rabbit
In this case Anil may be a Rabbit or Fox.
 This is not possible
CASE II :
When Anil is a Fox
We have,
Anil is a Fox
Chintoo is a Rabbit
Eashwar is a Fox
Bhauna is a Fox
Dolly is a Fox
 There are 4 foxes.
Q10.
Q
x
P
By using the principle of Pascal’s triangle, we have,
x
 No. of shortest possible routes = 14