6.0
3.0
2.4
1.8
Correct answer is C
\(\begin{array}{c|c} x & f & fx\\ \hline 1 & 6 & 6 \\
2 & 8 & 16\\ 3 & 8 & 18\\ 4 & 5 & 20\end{array}\)
mean x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{60}{25}\)
x = 2.4
10
24
25
30
Correct answer is C
6 + 8 + 6 + 5 = 25
155o
80o
158o
91o
Correct answer is C
< NPQ = < PQB = 50o(alt. < s)
< PQB = 50o
< PQR = < PQR < PQB + < QBR = 72o
< QBR = < PQR - < PQB = 72o - 50o
= 22o
< NQR = 180 - < QBR = 180o - 22 = 158o, the bearing of R from Q = 158o
i and iv
ii
iii
iv
Correct answer is C
m + y + x = 180o(sum of < s on straight line)
y = n(vertically opp. angle)
m + n + x = 180o
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
150o
110o
100o
95o
Correct answer is C
L + 30o - 180o(Sum of < s on straight line)
L = 180o - 30o = 150o
L = q = 150o(opposite < s are equal)
y = b = 30o(alt. < s)
b + c = 180o(sum of < s on str. line)
30o + c 180
c = 180 - 30
c = 150o
b = a = 30o (opp < s are equal)
c = d = 150o (opp < s are equal)
a + k + 70o = 180o (sum of < s on \(\bigtriangleup\))
30o + k + 70o = 180
k + 100o = 180
k = 180 - 100
k = 80o
x + 80o = 180(sum of < s on straight line)
x = 180o - 80o
x = 100o