Download Solution_pf_13th_JMo_Q.Paper 2010

13th KVS Mathematics Olympiad Solution
1.
Let a, b, c R, a  o, such that a and 4a + 3 b + 2c have
the same sign. Show that the equation a x 2 + b x + c = 0
can not have both roots in the interval (1,2).
As a, 4a +3a +2c have the same sign
≥0
[2 marks]

 4+ 3(-( X1+X2))+2.x1.x2 > = 0 where x1, x2 are the roots of
equation a x 2 + b x + c = 0
[3 marks]
 4-3 (x1+x2 ) + 2x1,x2 > 0
(x1-1) (x2-2) +(x1-2)(x2-1) > 0
[3 marks]
Thus, if x1,x2 € (1,2) then each term of the sum should be
strictly negative ,which is a contradiction
So both the roots of given equation can not belong to [1,2]
[2 marks]
Q.2 (a) Find all the integers which are equal to 11 times the
sum of their digits.
Sol.
(1) There is no such number in 1 digit number .
(2) let a and b be the ten’s and one’s digits of 2 digit number .
So 10 a+ b = 11 (a+b)
10 a+b =11 a+11 b
-a= 10b
Not possible.
There also exists no such number.
[2 marks]
(3) let a,b and c be the hundred’s ten’s and one’s digit of a 3 digit
number.
So 100 a+10 b +c =11 (a+ b+c)
100a +10 b+c =11 a +11 b+11 c
89 a= b+10 c
Which must be less then 100
89 a<100
a<100/89
a= 1
Implies 89=b +10 c
B= 9 and c= 8
[2 marks]
∴ one possible such number is 198. [1 marks]
Q. 2 (b) Prove that 3 (2+ 5) + 3 2- 5 is a rational number
Let
+
x3 2 5
=>
-
=0
By using identity if a+ b+c = 0
we get ,x3 + {3.x(-
}3 +{)
x3 – (2+√5 ) – (2-√5) = 3x
[1 marks]
a3 +b3 +c3 = 3abc
}3 =
[2 marks]
[1 marks]
= 3xCube root ( -1) == -3x
x3 – 4 = 3x
3
+3x -4 = 0
Which gives x= 1 which is rational. [1 marks]
Q3. A circle centered at A with radius 1 unit and another
circle centered at B with radius 4 units touch each other
externally. A third circle is drawn to touch the first two
circles and one of their common external tangents as shown
in the figure. What is the radius of the third circle?
[1 mark]
Let x be the radius of third circle.
In ∆ BED, BD = 4+x
BE = 4-x
∴ DE2 = (4+ X)2 –(4-X)2
= 16 X
DE = 4 √x
In = ∆ ADE , AD = 1+x
AF = 1-x
FD2 = (1+X)2-(1-X)2
= 4x
∴ FD=2 √x
In ∆ ABC, AB = 5 , BC = 3
∴ AC = 4 = FD +DE
= 2 √x + 4 √x = 6 √x
[2 marks]
[3 marks]
√x = =
x=
[4 marks]
Q4. Given three non collinear points A,H and G. Construct a triangle with A
as vertex, H as orthocenter and G as centroid.
Steps :
1) Join AG ,Produce to D, Such that GD = AG
2) Produce AH and draw DN AH to meet AH at N,Extend DN
both Side .
3) Draw DO DN to meet HG produced at o.
4) On DN produced cut off OB and OC equal to OA ( Taking O as
Centre ) On opposite side of D.
5) Join AB and AC
6) ∆ ABC is required triangle.
[two marks each for 5 steps]
Q5.
In Triangle ABC angle A is twice the angle B . Prove that
a2=b(b+c) where a,b and c are the sides opposite to
angles A,B and C respectively
[1 mark]
Draw bisector AD,
So that ,
L BAD = L DAC = LB
∆ ABC ∆ DAC
∴
=
∴
=
---------(i)
---------(ii)
=
=
=
=
[3 mark]
=
a2 = b(b+c)
[3 mark]
------------------Proved
[3 mark]
Q6. The equation x2+px+q=0, where p and q are integers, has rational
roots. Prove that the roots are integers.
Sol. Given equation x2+px+q=0, where p and q are integers, has
rational roots.
∴
[1 mark]
Given that roots are rational,
∴ P2 -4q should be perfect square
[2 mark]
(I) If P is even , P2 and 4q are even and hence p2 -4q ,is an even
integer and hence
is an even integer and hence
is an integer .
[3 mark]
(II) If p is odd, (P2 -4q) is odd and
even integer and hence,
Hence the result.
is
is integer.
[4 mark]
Q7. Triangles ABC is a right triangle with  ACB as its right
angle, m  ABC = 60 0, and AB = 10 units. Let P be point
chosen randomly inside  ABC. Extend BP to meet AC at D.
What is the probability that BD > 5 2.
[1 mark]
AB= 10, LABC = 60
BC =5, AC = 5
Choose E on AC so that CE =5
Then, BE = 5 .
[2 mark]
[2 mark]
For BD to be greater then = 5 , P has to be inside ∆ ABE. The
probability that P is inside ∆ ABE is
=
=
[3 mark]
=
=
Answer
[2 mark]
Q8.
Prove that sum of the hypotenuse and the attitude of the right angled
∆ dropped on the hypotenuse exceeds the half perimeter of the
triangle.
Solution :
[1 mark]
Let p be the length of perpendicular dropped from C on
hypotenuse BC.
ar(∆ ABC ) =1/2 ab = (AB)p
ab = cp ----------------------------(i)
[2 mark]
also , c2 = a2 +b2 ----------- (ii)
and a + b > c (Sum of two side grater then third side )
∴ (a+ b) (a + b) > c (a + b)
a2 + 2ab + b2 > c ( a + b)
a2 + b2 + 2ab > c ( a + b)
[3 mark]
c2 +2cp >c( a + b) ( using a2+b2=c2 and ab = cp )
c + 2p > a + b
as c > 0
[2 mark]
c +2 p +c > (a + b) +c
2c + 2p > a + b+ c
p+c>
c + p > half the perimeter of the triangle. [2 mark]
Q9. How many times is digit 0 written when listing all numbers from
1 to 3333 ?
Solution : we have to consider integer t , such ,
that I < t < 3333
The gratest number having 0 at units place is 3330.
(1)
There are 333 numbers ‘t’ having 0 at units place like
10,20,30,_______3330
We can express it as x.0
Where x is any one out of 1,2,3,_____ ,333
[3 mark]
(2)
Similarary , when 0 comes at ten’s place , the number is of
the from xoy ,when x can be any one from 1,2,_____,33 and y
can be any one out of 0,1,2,_____,9 Hence total number of such
number =33 x 10
= 330.
[3 mark]
(3)
When ,0 place is at hundred’s place the number is of the
from xo y z where x can be any one out of 1,2,3 and y & z can
be any one out of 0,1,2,____ 9 .
Hence total number of numbers = 3 x 10 x10 =300 [3 mark]
Hence total number of time 0 is written = 333+330+300
= 963---------Answer
[1 mark]
Q10.
Find out the remainder when x +x 9 +x 25 +x 49+x81 is divided by
x3-x .
Solution :Given expression can be written as
x +x9 +x25 +x49 +x81
=x[1+x8+x24+x48+x80]
[1mark]
= x[(x80 -1)+ (x48-1)+(x24-1)+(x8-1)+5]
[3 mark]
= x(x80-1)+x(x48-1)+x(x24-1)+ x(x8-1) +5x----------------(i) [2 mark]
Divisor is x3-x i.e x (x2-1) ,hence each term of (i) is divisible by
x(x2-1)except last term 5x, hence remainder = 5x --Ans [4 mark]