A sequence is given by \(2\frac{1}{2}, 5, 7\frac{1}{2}, .....\) if the nth term is 25, find n
9
10
12
15
Correct answer is B
\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\\
25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\\
45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\)
36
27
24.5
13.5
Correct answer is B
\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)
Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\)
-8
-13
13
19
Correct answer is B
\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 8y - 4 - 9y + 3 = 12 \\
\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)
Solve the equation 2x - 3y = 22; 3x + 2y = 7
-5
-4
4
5
Correct answer is B
2x - 3y = 22 ---- eqn I
3x + 2y = 7 ---- eqn II
multiply eqn I by 2
4x - 6y = 44 ---- eqn III
multiply eqn II by 3
9x + 6y = 21 ---- eqn IV
Adding eqn III and IV
=> 13x = 65 => x = 5
Substituting 5 for x in eqn II
3x5 + 2y = 7 => 2y = -8
y = -4
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)]
\(\frac{\sqrt{2}}{7}cm\)
\(3\frac{1}{2}cm\)
7cm
\(7\sqrt{2}cm\)
Correct answer is D
\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\
\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)