[hi.shivani]

Originally Posted by masoom

i haven't given ans for this question..

just elaborated the approach

post has been edited...the key might be wrong..

[dare2]

Originally Posted by hi.shivani

i solved totally logicaly..so might be wrong neways,

the total number of cases possible =11*2*10*2=440

and for the quetion the cases possible=66
so i got 66/440
=3/20

well nice try..
but i think i should give the solution before u start thinking more sophistically.. this was a simple question and required simple approach..

either A throws more heads than B, or A throws more tails than B, but not both (since A has only one extra coin) .
by symmetry, these two mutually exclusive possibilities occur with equal probability.

therefore the probability that A obtains more heads than B is ½.

[hi.shivani]

New Question Set:
1) For how many values of x
(x-1)(x+4)(x+2)/(x-3)(x+8](x+3)<0

2) For real values of x,the expression(x-c)(x-b)/(x-a) will assume all real values provided ;
a) a=<c=<b
b) b=<a=<c
c) b=<c=<a
d) a=<b=<c

3) If a^2+b^2+c^2=1;then ab+bc+ca lies in between (include)
a) -2 & 1 b) -1 & -2 c) 0 &1 d) NOT

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

5)The number of points (x,y) satisfies i) 3y-4x=20,ii)x^2+y^2=<16
is/are?

[shivam_01]

3) The cordinates of the diagonals of a square are (2,0) and (0,5) .What is the area of the square?

see take the distance betwwe the points that is sqrt(29) so the side of the sqaure is
sqrt(29)/ sqrt(2) so the area is

( sqrt(29)/ sqrt(2))^2 = 14.5 sq units

[masoom]

1) For how many values of x

(x-1)(x+4)(x+2)/(x-3)(x+(x+3)

[masoom]

1) For how many values of x

(x-1)(x+4)(x+2)/(x-3)(x+(x+3) 0

These type of questions are best solved by number line concepts..

first find the critical points..ie the points where the factors change their sign

now plot them on number line..

--------(-)8-------(-)4---------(-3)------(-2)--------1-------3-----------


+ve -ve +ve -ve +ve -ve +vesince we
are looking for -ve sign..

solution set is :

x in (-8,-4) U (-3,-2) U (1,3)

[romy4allcse]

Originally Posted by hi.shivani

Few Simple co-ordinate geometry questions:
1) If each of the points(x1,4),(-2,y1) lies on the line joining the point (2,-1),(5,-3) then the point (x1,y1) lies on the line.
a) 6(x+y)-25=0
b) 2x+3y-6=0
c) 2x+6y+1=0
d) 6(x+y)+25=0

2) If the straight line 5x+12y-43=0 and 10x+24y+57=0,both tangent to a certain circle,then what is the approx. circumference of the circle?

a) 17.3 b) 20.8 c) 15.6 d) 19.1

3) The cordinates of the diagonals of a square are (2,0) and (0,5) .What is the area of the square?

soln1 : eq. of the line joining the point (2,-1),(5,-3)
y+1 = {2/(-3)}(x-2)
2x + 3y -1 = 0
As points(x1,4),(-2,y1) lies on this line hence x1 = -11/2 and y1 = 5/3
as (x1,y1) lies on some line they will satisfy the eq. of that line
put these values of x1,y1 in all the option and u vl get the ans
option (3) 2x+6y+1=0

soln2: 5x+12y-43=0 and 10x+24y+57=0 are parallel lines hence diameter of the circle will be the distance between these two parallel lines hence distance D = |-43-57/2|/sqrt(169) = 5.5
hence radius = 5.5/2 = 2.75
perimeter = 2*3.14*2.75 = 17.27 = 17.3 hence option (1) ans.

soln3: coordinates of the diagonal are (2,0) and (0,5)
length of this diagonal = sqrt(29)
hence length of the side of the square = sqrt(29)/sqrt(2)
area = 29/2 = 14.5 ans.

[hi.shivani]

Originally Posted by masoom

1) For how many values of x

(x-1)(x+4)(x+2)/(x-3)(x+(x+3)

arrey yaar ye 8 with small bracket smiley ka code hain..have changed the bracket plz recheck the edited question..

[shivam_01]

Originally Posted by masoom

option 2 is the ans i think..

as i am getting x1 as -5/2 and y1 as 11/3..

so eqn 2 is satisying

eq. of the line joining the point (2,-1),(5,-3)
2x + 3y -1 = 0
(x1,4),(-2,y1)
x1 = -11/2 and y1 = 5/3

option (3) 2x+6y+1=0 satisfies i think u are wrong some where dude

[romy4allcse]

3) If a^2+b^2+c^2=1;then ab+bc+ca lies in between (include)
a) -2 & 1 b) -1 & -2 c) 0 &1 d) NOT

am > gm

a^2 + b^2 >= 2ab
b^2 + c^2 >= 2bc
c^2 + a^2 >= 2ac
adding these three
2(a^2+b^2+c^2)>=2(ab+bc+ca)
(a^2+b^2+c^2)>=(ab+bc+ca)
hence
0<(ab+bc+ca) <= 1
option (c)

5)The number of points (x,y) satisfies i) 3y-4x=20,ii)x^2+y^2=<16
is/are?

x^2+y^2=<16 this is a circle whose radius is 4
3y-4x=20
draw the diagram of these two eq.
distance from origin to this line = 20/5 = 4
hence point (x,y) satisfies 3y-4x=20 and x^2+y^2=<16 vl be one and it would be when line is tangent to the circle and circle eq is x^2+y^2=16
hence ans is 1 point

soln 2 :
let (x-c)(x-b)/(x-a) = k
now solve this quadratic eq and make the descriminant >=0
we vl get b+c>=2a
hence a<= (b+c)/2
hence option 2 satisfies as a vl lie in between b and c

b) b=<a=<c