\(\frac{1 - 3n}{n + 1}\)
\(\frac{3n + 1}{n + 1}\)
\(\frac{3n + 1}{n - 1}\)
\(\frac{3n - 1}{n + 1}\)
Correct answer is B
Using Tn = \(\frac{3n + 1}{n + 1}\),
T1 = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)
T2 = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)
T3 = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)
x < 5
-2 < x < 3
-1 < x < 5
x < 1
Correct answer is C
|x - 2| < 3 implies -(x - 2) < 3 .... or .... +(x - 2) < 3 -x + 2 < 3 .... or .... x - 2 < 3 -x < 3 - 2 .... or .... x < 3 + 2 x > -1 .... or .... x < 5 combining the two inequalities results, we get; -1 < x < 5
x < 4
x > -4
x < -4
x > 4
Correct answer is C
3(x + 2) > 6(x + 3)
3x + 6 > 6x + 18
3x - 6x > 18 - 6
-3x > 12
x < -4
If r varies inversely as the square root of s and t, how does s vary with r and t?
s varies inversely as r and t2
s varies inverely as r2 and t
s varies directly as r2 and t2
s varies directly as r and t
Correct answer is B
\(r \propto \frac{1}{\sqrt{s}}, r \propto \frac{1}{\sqrt{t}}\)
\(r \propto \frac{1}{\sqrt{s}}\) ..... (1)
\(r \propto \frac{1}{\sqrt{t}}\) ..... (2)
Combining (1) and (2), we get
\(r = \frac{k}{\sqrt{s} \times \sqrt{t}} = \frac{k}{\sqrt{st}}\)
This gives \(\sqrt{st} = \frac{k}{r}\)
By taking the square of both sides, we get
st = \(\frac{k^2}{r^2}\)
s = \(\frac{k^2}{r^{2}t}\)
12\(\frac{8}{5}\)
15
10
28\(\frac{8}{5}\)
Correct answer is C
P \(\propto\) mu, p \(\propto \frac{1}{q}\)
p = muk ................ (1)
p = \(\frac{1}{q}k\).... (2)
Combining (1) and (2), we get
P = \(\frac{mu}{q}k\)
4 = \(\frac{m \times u}{1}k\)
giving k = \(\frac{4}{6} = \frac{2}{3}\)
So, P = \(\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}\)
Hence, P = \(\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}\)
P = \(\frac{2 \times 6 \times 4 \times 5}{3 \times 8}\)
p = 10