2010 - JAMB Mathematics Past Questions and Answers - page 6

∑f = 2 + 2 + 8 + 4 + 4

= 20

= 3.3

</table></table>

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