PQRS is a rhombus. If PR2 + QS2 = kPQ2, determine k.
1
2
3
4
Correct answer is D
PR2 + QS2 = kPQ2
SQ2 = SR2 + RQ2
PR2 + SQ2 = PQ2 + SR2 + 2RQ2
= 2PQ2 + 2RQ2
= 4PQ2
∴ K = 4
A regular polygon of (2k + 1) sides has 140° as the size of each interior angle. Find k
4
412
8
812
Correct answer is A
A regular has all sides and all angles equal. If each interior angle is 140° each exterior angle must be
180° - 140° = 40°
The number of sides must be 360o40o = 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k = 82
= 4
If -8, m, n, 19 are in arithmetic progression, find (m, n)
1, 10
2, 10
3, 13
4, 16
Correct answer is A
-8, m, n, 19 = m + 8
= 19 - n
m + n = 11
i.e. 1, 10
xy
yx
(xy)1n−2
(yx)1n−2
Correct answer is D
Sum of nth term of a G.P = Sn = arn−1r−1
sum of the first two terms = ar2−1r−1
x = a(r + 1)
sum of the last two terms = Sn - Sn - 2
= arn−1r−1 - (arn−1)r−1
= a(rn−1−rn−2+1)r−1 (r2 - 1)
∴ arn−2(r+1)(r−1)1= arn - 2(r + 1) = y
= a(r + 1)r^n - 2
y = xrn - 2
= yrn - 2
yx = r = (yx)1n−2
Simplify x(x+1)12−(x+1)12(x+1)12
−1x+1
1x+1
1x
1x−1
Correct answer is A
x(x+1) - √(x+1)√(x+1)
= xx+1 - 1
x−x−1x+1 = −1x+1