The values of three angles at a point are 3y - 45o, y + 25o and yo . Find the value of y.
40o
58o
68o
76o
Correct answer is D
(3y - 450 + (y + 25) + t = 360o(angles at a point)
3y - 45 + y + 25 + y = 360
5y = 360 + 20
5y = 380
y = \(\frac{380}{5}\)
= 76o
{1, 3, 5, 7, 9}
[2, 4, 6, 8, 10}
\(\varnothing\)
{1, 2, 6, 8, 10}
Correct answer is C
P1 = {2, 4, 6, 8, 10}; {1, 3, 5, 7, 9}
P1 \(\cap\) q1 = {} or \(\varnothing\)
140m
220m
280m
440m
Correct answer is D
radius = \(\frac{diameter}{2}\) = \(\frac{1.4}{2}\)
= 0.7m
circumference = 2\(\pi r = 2 \times \frac{22}{7} \times 0.7m = 4.4m\)
distance covered = 4.4m x 100 = 440m
1:4
1:5
3:4
4:5
Correct answer is D
(1 - \(\frac{20}{100}\)) = cash payment fraction= \(\frac{4}{5}\)
The ratio of the cash payment to the marked price = 4:5
If p = [\(\frac{Q(R - T)}{15}\)]\(^ \frac{1}{3}\), make T the subject of the relation
T = R + \(\frac{P^3}{15Q}\)
T = R - \(\frac{15P^3}{Q}\)
T = R + \(\frac{P^3}{15Q}\)
T = 15R - \(\frac{Q}{P^3}\)
Correct answer is B
Cubic both sides; P3 = \(\frac{Q(R - T)}{15}\)
(cross multiplication) Q(R - T) = 15P3
(divide both sides by Q); R - T = 15\(\frac{1}{Q}\)
(subtract r from both sides) - T = \(\frac{15P^3}{Q - R}\)
T = R - \(\frac{15P^3}{Q}\)