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) = ax^{7} + bx^{5} + cx^{3} – 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) x^{4} + = 47
Þ (x^{2}) + + 2. (x^{2}) – (x^{2}) = 47
Þ – 2 = 47
Þ = 49
= 7
Þ = 7
= 7 + 2
Þ
= 3
= (3) (7 – 1)
= 3 x 6
= 18 Ans.
(b) 8^{2x} = 16 ^{1 – 2x}
Þ 8^{2x} = (8^{2})^{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 =
3^{7x}
=
= = (3^{3} x 3^{3} x 3^{3})
= x x
= 3 x3 x
= 9
3^{7x} = 9 Ans.
5.


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
60^{o } 60^{o}


T
S R
Considering DUQR, Q
30^{o}


U 45^{o}
R
Sin QUR = sine 105^{o} = 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 ) cm^{2}
ar (DUQR) = = (75 – 25 ) cm^{2}
Q7.
64cm
32cm
32cm
side of the 1^{st} square = 64 cm
side of the 2^{nd} square =
=
=
= 32
=
Side of the 3^{rd} square =
=
=
=
=
Side of the 1^{st} square = 64 cm
Side of the 2^{rd} square = cm
Side of the 3^{rd} square = cm
In the similar fashion,
side of the 11^{th} 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 = 7a^{3}
7a^{3} = 448
Þ a^{3} =
Þ a^{3} = 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 = 5a^{2} x 6
= 30 x 4^{2}
= 30 x 16
= 480 cm^{2}
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


P
By using the principle of Pascal’s triangle, we have,


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) = ax^{7} + bx^{5} + cx^{3} – 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) x^{4} + = 47
Þ (x^{2}) + + 2. (x^{2}) – (x^{2}) = 47
Þ – 2 = 47
Þ = 49
= 7
Þ = 7
= 7 + 2
Þ
= 3
= (3) (7 – 1)
= 3 x 6
= 18 Ans.
(b) 8^{2x} = 16 ^{1 – 2x}
Þ 8^{2x} = (8^{2})^{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 =
3^{7x}
=
= = (3^{3} x 3^{3} x 3^{3})
= x x
= 3 x3 x
= 9
3^{7x} = 9 Ans.
5.


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
60^{o } 60^{o}


T
S R
Considering DUQR, Q
30^{o}


U 45^{o}
R
Sin QUR = sine 105^{o} = 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 ) cm^{2}
ar (DUQR) = = (75 – 25 ) cm^{2}
Q7.
64cm
32cm
32cm
side of the 1^{st} square = 64 cm
side of the 2^{nd} square =
=
=
= 32
=
Side of the 3^{rd} square =
=
=
=
=
Side of the 1^{st} square = 64 cm
Side of the 2^{rd} square = cm
Side of the 3^{rd} square = cm
In the similar fashion,
side of the 11^{th} 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 = 7a^{3}
7a^{3} = 448
Þ a^{3} =
Þ a^{3} = 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 = 5a^{2} x 6
= 30 x 4^{2}
= 30 x 16
= 480 cm^{2}
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


P
By using the principle of Pascal’s triangle, we have,


No. of shortest possible routes = 14
Filed under: KVS Junior Mathematics Olympiad (JMO) Model question Papers