How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
-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 \(\frac{19}{24}\)
= \(\frac{1}{36}\)