2010 - JAMB Mathematics Past Questions and Answers - page 6

∑f = 2 + 2 + 8 + 4 + 4
= 20

Mark(χ)	f	fχ
1	2	2
2	2	4
3	8	24
4	4	16
5	4	20
	∑f = 20	∑fχ = 66

Mean =	∑fχ
∑f

=	66
20

= 3.3

6r7₈ = 511₉

6 x 8² + r x 8¹ + 7 x 8^o = 5 x 9² + 1 x 9¹ + 1 x 9^o

6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1

384 + 8r + 7 = 405 + 9 + 1

391 + 8r = 24

r = \(\frac{24}{8}\)

= 3

(\(\frac{3}{4}\) 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 \(\frac{4}{9}\) \(\div\)\(\frac{24}{19}\)) \(\div\)\(\frac{24}{19}\)

(\(\frac{3}{9}\) \(\div\) \(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\) = (\(\frac{3}{9}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)

(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)

\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)

= \(\frac{1}{36}\)

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%

= 0.08%

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%

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

\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{5}\)

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

log 112 = log(16 x 7)

= log 16 + log7

= log 2⁴ + log7

= 4log2 + log7

= 4(0.3010) + 0.8451

= 1.204 + 0.8451

= 2.0491

Given that,

X \(\ast\) y = X + y²

(2 \(\ast\) 3) \(\ast\) 5 = (2 + 3²)\(\ast\)5

= (2 + 9)\(\ast\)5 = 11 \(\ast\)5

Hence 11 \(\ast\) 5 = 11 + 5²

= 11 + 25 = 36

18(p+q) = (18+p)q
18p + 18q = 18q + pq
By comparison
18p = 18q or p = q
Again, 18q = pq or 18 = p which is required