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KVS Junior Mathematics Olympiad (JMO) – 2002 Solution

kvs Junior Mathematics Solution 2002

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

Þa (-9)7 + b (-9)5 + c (-9)3 – 6 = 3

Þ -a (9)7 + b (-9)5 + -c (9)3 = 9

Þ -a (9)7  –  b (9)5  – c (9)3  =  9

Þ a (9)7  +  b (9)5  + c (9)3  =  –  9

f (9)   =       a (9)7 + b (9)5 + c (9)3 – 6

=       – 9 – 6

=       -15 Ans.

(b)

=

=

=

=

=  2002 Ans.

Q4.(a)         x4 +   = 47

Þ (x2) +  + 2. (x2)  – (x2)  = 47

Þ       – 2 = 47

Þ       = 49

= 7

Þ  = 7

= 7 + 2

Þ

= 3

=       (3)  (7 – 1)

=       3 x 6

=       18 Ans.

(b)               82x     =       16 1 – 2x

Þ      82x     =       (82)1 – 2x

Þ      (8)2x            =       (8)2 – 4x

                Þ      2 x              =       2 – 4 x

Þ      2 x + 4 x     =       2

Þ      6x               =       2

Þ      x                 =

37x

=

=          =       (33 x 33 x 33)

=        x  x

=       3 x3 x

=       9

37x     =       9  Ans.

5.

C

D

A.                 .                .                                    .B

Let, the train’s speed     =       y

Total time             =

Case I :-

Total time   =

=

=

=  +

Þ       =

Þ      210 + 4x = 210 + 3x + 4y

Þ      x – 4y = 0                                                             ————– (i)

Case II :-

Total time   =

=

=

=

=

=

Þ           =

Þ      175 + 4x     =       210 + 3x + 3y

Þ      4x – 3x – 3y = 210 – 175

Þ      x – 3y = 45                                                                    (ii)

We have,

(i)                Þ x – 4y = 0

(ii)             Þ x – 3y = 45

x – 4y = 0

x – 3y = 45

–                   +       –

-y = -45

Þ y = 45 km/h

Speed of train = 45 km / h

Q6.                       P                                         Q

60o                        60o

30o

45o

T

S                                    R

Considering DUQR,                          Q

30o

10 cm

U                     45o

R

Sin QUR = sine 105o = sin (60 + 45)

= sin 60 cos 45 + sin 45 cos 60

=

=

=

In DQUR,

By the law of sine, we have,

= =

Þ =  =

=

Þ UQ =

=

=

=

=

=

UQ =  cm

We know that,

Area of a triangle = ½ bc sin A

In D QUR,

ar (DUQR) = ½ x UQ x QR x sin ÐUQR

= ½ x x 10 x sin 30

=

= 25 (3 – )

= (75 – 25 ) cm2

ar (DUQR)  = = (75 – 25 ) cm2

Q7.

64cm

32cm

32cm

side of the 1st square = 64 cm

side of the 2nd square          =

=

=

= 32

=

Side of the 3rd square          =

=

=

=

=

Side of the 1st square          = 64 cm

Side of the 2rd square          = cm

Side of the 3rd square          = cm

In the similar fashion,

side of the 11th square

=

=

=       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

Þ               a3 =

Þ               a3 = 64

Þ               a =

= 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

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

Þa (-9)7 + b (-9)5 + c (-9)3 – 6 = 3

Þ -a (9)7 + b (-9)5 + -c (9)3 = 9

Þ -a (9)7  –  b (9)5  – c (9)3  =  9

Þ a (9)7  +  b (9)5  + c (9)3  =  –  9

f (9)   =       a (9)7 + b (9)5 + c (9)3 – 6

=       – 9 – 6

=       -15 Ans.

(b)

=

=

=

=

=  2002 Ans.

Q4.(a)         x4 +   = 47

Þ (x2) +  + 2. (x2)  – (x2)  = 47

Þ       – 2 = 47

Þ       = 49

= 7

Þ  = 7

= 7 + 2

Þ

= 3

=       (3)  (7 – 1)

=       3 x 6

=       18 Ans.

(b)               82x     =       16 1 – 2x

Þ      82x     =       (82)1 – 2x

Þ      (8)2x            =       (8)2 – 4x

                Þ      2 x              =       2 – 4 x

Þ      2 x + 4 x     =       2

Þ      6x               =       2

Þ      x                 =

37x

=

=          =       (33 x 33 x 33)

=        x  x

=       3 x3 x

=       9

37x     =       9  Ans.

5.

C

D

A.                 .                .                                    .B

Let, the train’s speed     =       y

Total time             =

Case I :-

Total time   =

=

=

=  +

Þ       =

Þ      210 + 4x = 210 + 3x + 4y

Þ      x – 4y = 0                                                             ————– (i)

Case II :-

Total time   =

=

=

=

=

=

Þ           =

Þ      175 + 4x     =       210 + 3x + 3y

Þ      4x – 3x – 3y = 210 – 175

Þ      x – 3y = 45                                                                    (ii)

We have,

(i)                Þ x – 4y = 0

(ii)             Þ x – 3y = 45

x – 4y = 0

x – 3y = 45

–                   +       –

-y = -45

Þ y = 45 km/h

Speed of train = 45 km / h

Q6.                       P                                         Q

60o                        60o

30o

45o

T

S                                    R

Considering DUQR,                          Q

30o

10 cm

U                     45o

R

Sin QUR = sine 105o = sin (60 + 45)

= sin 60 cos 45 + sin 45 cos 60

=

=

=

In DQUR,

By the law of sine, we have,

= =

Þ =  =

=

Þ UQ =

=

=

=

=

=

UQ =  cm

We know that,

Area of a triangle = ½ bc sin A

In D QUR,

ar (DUQR) = ½ x UQ x QR x sin ÐUQR

= ½ x x 10 x sin 30

=

= 25 (3 – )

= (75 – 25 ) cm2

ar (DUQR)  = = (75 – 25 ) cm2

Q7.

64cm

32cm

32cm

side of the 1st square = 64 cm

side of the 2nd square          =

=

=

= 32

=

Side of the 3rd square          =

=

=

=

=

Side of the 1st square          = 64 cm

Side of the 2rd square          = cm

Side of the 3rd square          = cm

In the similar fashion,

side of the 11th square

=

=

=       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

Þ               a3 =

Þ               a3 = 64

Þ               a =

= 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

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Filed under: KVS Junior Mathematics Olympiad (JMO) Model question Papers

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