2015 - JAMB Mathematics Past Questions and Answers - page 4

His annual salary = N2000

His allowances = N600

So his taxable income = Annual salary − allowance

= N2000 − N600

= N1400

He pay at 5%

Then, his allowance income tax 5/100 × 600 = N30

Nth term of a linear sequence (AP) = a+(n − 1)d

first term = 6, last term = 10 sum − 40

i.e. a = 6, l = 10, S = 40

sn = n ÷ 2(2a + (n − 1)d or Sn = ÷2 (a + b)

Sn = n ÷ 2(a + l)

40 = n ÷ 2(6 + 10)

40 = 8n

8n = 40

8n = 40

N = 40/8

= 5

The nth term = 5

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

2 + t/5(x + 3/5y ) = s

t/5(x + 3/5y ) = s − 2

[(t(5x + 3y) ÷ 25)] = s − 2

t(5x + 3y) = 25 (s − 2)

5x + 3y = [(25(s − 2))÷ t]

5x = [(25(s − 2))÷ t] − 3y

Divide both sides by 5

x = [25(s − 2) − 3ty)]÷ 5t

log₅20 = x

5x = 20(Take log₁₀ of both sides)

log_5x = log₂₀

xlog₅ = log₂₀

x= [log₂₀ ÷ log₅]

[1.30103 ÷ 0.69897]

x = 1.861

I = A − P

= N588 − N525

∴I = N63

I = PRT ÷ 100

R = [100I ÷PT]

R = [(100 × 63 ) ÷ (525 × 3)]

= (6300 ÷ 1575) = 4

∴ The rate = 4 %

2t − 1 \(\leq\) 3 and 2 − t < 5

2t \(\leq\) 3 + 1 and 2 − 5 \(\leq\) t

2t \(\leq\) 4 and − 3 \(\leq\) t

t \(\leq\) 2 and − 3 \(\leq\) t or t \(\geq\) − 3

Combining the two ,starting from the last

-3 \(\leq\) t \(\leq\) 2

{A ∪ B ∪ C}

{A ∪ B} = {1, 3, 5, 7, 9, 12, 15}

∪ C={2,4,6,8}

{A ∪ B} ∪ C = {1, 3, 5, 7, 9, 12, 15} ∪ {2, 4, 6, 8}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

{A ∪ B} ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}