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
The remainder when 6p3 - p2 - 47p + 30 is divided by p - 3 is
21
42
63
18
Correct answer is B
Let f(p) = 6p3 - p2 - 47p + 30
Then by the remainder theorem,
(p - 3): f(3) = remainder R,
i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R
162 - 9 - 141 + 30 = R
192 - 150 = R
R = 42
If x - 4 is a factor of x2 - x - k, then k is
4
12
20
2
Correct answer is B
Let f(x) = x2 - x - k
Then by the factor theorem,
(x - 4): f(4) = (4)2 - (4) - k = 0
16 - 4 - k = 0
12 - k = 0
k = 12