q.18 (pattern question)
notice that the last numeral in each one is 5. for 5^3 it is 125.
for 5^4 it is 625. Hence we can conclude that for every 5 to the power of an odd number, the last three digits will be 125. For every 5 to the power of an even number, the last three digits will be 625
hence the last three digits of 5^15 will be 125
q.48
Let karishma's age be k.
k + 2 = 3(t + 2)
k + 2 = 3t + 6
k = 3t + 4
Q51
Crai - x
Phil - x
terry -x
Dick - 50%
x = other players = (100 - 50)/3 then divide by 100 = 50/3 divide by 100 = 1/6
Q62
a = 110 degrees
notice that a lies on co interior angle with another angle. that angle is equal to 70 due to the fact that it is alternate to 130 degrees. Then by co interior angles are supplementary, 70 + a = 180. a = 110
Q64
3^ 0 = 1
(1/3)^2 = 1/9
3^0 - 3(1/3)^2 = 1 - 3(1/9)
= 1 - 1/3 = 2/3
Q69
the mean = (143x1) + (148 x 2) + (153x1) + (158 x 6) + (163 x 11) + (168 x9) divide by total runners
= 161.5
Q71
angle ACD= angle ADC = DAC = 60 degrees
angle BDC + angle ADC = 180 (straight angle)
angle BDC = 120 degrees
angle BCD = 30
angle ACB = 30 + 60 = 90
Explanation: since the one triangle is double the other, the two triangles ACD and BDC must be equal in size (congruent). Yet angle ADC = 60 and angle BDC = 120. Since the areas are the same, it can be inferred that side BD = CD. This is because sine is positive in the second quadrant. so therefore, angle BCD = angle DBC (angles opposite equal sides)
Its late now. I'll do the rest later on
notice how I skipped every second question, lol.
I'm surprised at how challenging this stuff looks at from now. I blame it on 3 unit maths
