13 cm
4 cm
6 cm
7 cm
Correct answer is A
Area of Trapezium = 1/2(sum of parallel sides) * h
91 = \(\frac{1}{2}\) (5 + 9)h
cross multiply
91 = 7h
h = \(\frac{91}{7}\)
h = 13cm
Determine the maximum value of y=3x\(^2\) + 5x - 3
6
0
2
No correct option
Correct answer is D
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option
Solve the following equation: \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
( -1,\(\frac{5}{2}\) )
( 1, - \(\frac{5}{2}\) )
( \(\frac{5}{2}\), 1 )
(2,1)
Correct answer is B
\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\)
The L.C.M.: (2r - 1) (r + 2)
\(\frac{2(r + 2) - 1(2r - 1)}{(2r - 1) (r + 2)}\) = \(\frac{5}{3}\)
\(\frac{2r + 4 - 2r + 1}{ (2r - 1) (r + 2)}\) = \(\frac{5}{3}\)
cross multiply the solution
3 * 5 = (2r - 1) (r + 2) * 5
divide both sides 5
3 = 2r\(^2\) + 3r - 2 (when expanded)
collect like terms
2r\(^2\) + 3r - 2 - 3 = 0
2r\(^2\) + 3r - 5 = 0
Factors are -2r and +5r
2r\(^2\) -2r + 5r - 5 = 0
[2r\(^2\) -2r] [+ 5r - 5] = 0
2r(r-1) + 5(r-1) = 0
(2r+5) (r-1) = 0
r = 1 or - \(\frac{5}{2}\)
3
57
19
17
Correct answer is C
a + 3d = 37
-20 + 3d = 37
3d = 37 + 20 = 57
d = \(\frac{57}{3}\)
= 19
If 7 + y = 4 (mod 8), find the least value of y, 10 \(\leq y \leq 30\)
11
13
19
21
Correct answer is B
7 + y = 4 (mod 8)
y = 4 - 7 (mod 8)
y = -3 + 8 (mod 8)
y = 5 + 8 (mod 8)
y = 13
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