INTERSECTION MATHS QUIZ CLASS X PRELIMS

The solutions are as follows:

1)) there were many solutions but the easiest ones are
1+2+3+4+5+6+7+(8*9)
(1*2*3*4)*(5+6)+7 - (8+9)

2)) 2^0 , 2^1 , 2^2 , ..................... , 2^9

3)) it is LCM - 1
LCM is 2520 and the number is 2519


4)) the triangle numbers are represented by 1/2(n^2 + n) = nth term
we had to show that nth term + (n+1)th term is a perfect square
s=1/2(n^2 + n + n^2 + 1 + 2n + n+ 1)
=(n + 1)^2


5)) 4 6
7 1 8 2
3 5

6))simple geometry
area of shaded area is 3 times area of sector of angle 60 degrees(eq. triangle) - 2 times area of equilateral triangle
comes out to be 1/2(Pi - sqrt3) = 0.7047 approximately


7)) S= a+ar+ar^2+.................+ar^(n-1) [these are n terms]
rS= ar+ar^2+ar^3+.......................+ar^n [still n terms----multiplyng throughout by r]
Subtracting
S(1-r) = a - ar^n
S=a(1-r^n)/(1-r)

8))i admit this was the toughest one according to your standards
S = 1+3+6+10+15+....................t(n) [t(n) is the nth term]
S= 1+3+ 6+ 10+ ..................t(n-1) + t(n) [shifting and writing again]
Subtracting
0= 1+2+3+4+.........+n - t(n)
t(n)=n(n+1)/2 [these are triangle numbers]

Now the question waxs to find the sum
S=summation[t(n)] summation is given by sigma
= 1/2Sum(n^2) + 1/2Sum(n)
= 1/2{1/6n(n+1)(2n+1) + 1/2n(n+1)}

after simplifying
S= 1/6(n)(n+1)(n+2)


pretty simple!!