-44
-165
165
44
Correct answer is A
3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44
12 : 15 : 10
12 : 15 : 16
10 : 15 : 24
9 : 10 : 15
Correct answer is A
If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)
If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)
Let p + q = T1, then
q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1
Again, let q + r = T2, then
q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2
Using q = q
\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
5 x 5T1 = 9 x 3T2
\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{25}\)
Giving that, T1 = 27 and T2 = 25
P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1
= (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12
q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1
= (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15
and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2
= (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10
Hence p : q : r = 12: 15 : 10
At what rate will the interest on N400 increases to N24 in 3 years reckoning in simple interest?
3%
2%
5%
4%
Correct answer is B
Using simple interest = \(\frac{P \times T \times R}{100}\),
where: P denoted principal = N400
T denotes time = 3 years
R denotes interest rate = ?
24 = \(\frac{400 \times 3 \times R}{100}\)
24 x 100 = 400 x 3 x R
R = \(\frac{24 \times 100}{400 \times3}\)
= 2%
0.25%
0.01%
0.80%
0.40%
Correct answer is C
Actual length of rope = 1.25m
Measured length of rope = 1.26m
error = (1.26 - 1.25)m - 0.01m
Percentage error = \(\frac{error}{\text{actual length}}\) x 100%
= \(\frac{0.01}{1.25}\) x 100%
= 0.8%
Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
\(\frac{1}{5}\)
\(\frac{1}{4}\)
\(\frac{1}{36}\)
\(\frac{1}{25}\)
Correct answer is C
(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
Applying the rule of BODMAS, we have:
(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)
(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)
\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)
= \(\frac{1}{36}\)