Solve the simultaneous linear equations: 2x + 5y = 11, 7x + 4y = 2
x = -8, y = 1
x = -2, y = 4
x = \(\frac{-34}{27}\), y = \(\frac{73}{27}\)
x = 7, y = -9
Correct answer is C
2x + 5y = 11.......(i)
7x + 4y = 2.......(ii)
(i) x 7 \(\to\) 14x + 35y = 77.........(iii)
(ii) x 2 \(\to\) 14x + 8y = 4........(iv)
(iii) - (iv)
27y = 73
y = \(\frac{73}{27}\)
1\(\frac{2}{3}\)
2
5
6
10
Correct answer is C
Let x rept. the usual speed = \(\frac{Distance}{Time}\)
= \(\frac{20}{1}\) x \(\frac{1}{2}\)
= 5
(a), (b), (c)
(c)
None of the choices
All of the above
Correct answer is B
0 < \(\frac{x + 3}{x - 1}\) < 2
Put x = 0, -3 and 9
0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2
i.e. 0 < 1.5 \(\leq\) 2 (true)
but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2
i.e. 0 < -3 \(\leq\) 2 (not true)
-3 \(\leq\) 2
-3 is not greater than 0
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0
c = \(\frac{2d - 5 - b}{ad}\)
c = \(\frac{2d - 5 - abd}{ad}\)
c = \(\frac{2d - 5 - ab}{ad}\)
c = \(\frac{5 - 2d - b}{ad}\)
Correct answer is B
a(b + c) + \(\frac{5}{d}\) - 2 = 0
ab + ac + \(\frac{5}{d}\) - 2 = 0
abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d
acd = 2d - 5 - abd
c = \(\frac{2d - 5 - abd}{ad}\)
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If < PQS = 15o find < QRS
75o
37\(\frac{1}{2}\)o
127\(\frac{1}{2}\)o
105o
Correct answer is D
PSO = 90o(angle in a semicircle)
SPO = 180o - (90o + 15o) = 75o
< QRS = 108o - 75o
= 105o(opposite angles in cyclic quadrilateral are supplementary)