JAMB Mathematics Past Questions & Answers - Page 97

Convert the binary to base 10 and they convert back to base two

10011₂ + xxxxx₂ + 11100₂ + 101₂ = 1001111₂


(1 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰) + xxxxx₂ +(1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰) + (1 × 2² + 0 × 2¹ + 1 × 2⁰)

= (16 + 0 + 0 + 2 + 1) + xxxxx₂ + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)

=(64 + 0 + 0 + 8 + 4 + 2 + 1)

19 + xxxxx₂ + 33 = 79

xxxxx₂ + 52 = 79

xxxxx₂ = 79 − 52

xxxxx₂ = 27₁₀

$\begin{array}{c|c} 2 & 27 \\ \hline 2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow$

27₁₀ = 11011₂

Therefore xxxxx₂ = 27₁₀ = 11011₂

log₇17

= [log 17 ÷ log7]

= [1.2304 ÷ 0.8451]

[10^0.0899 ÷ 10^1.9270]

= 1.455(antilog)

[Return ÷ Investment] as a ratio ;

i.e The Ratio is Return : Investmen

[(Return1÷ Investment1 ) = (Return2÷ Investment2)]

R1 = N 25000

R2 =?

I1 = N450,000,

I1 = N 700000

(25000 ÷ 450000) = (R2 ÷ 700000)

R2 = [(25000 × 700000 ) ÷ 450000]

= N38,888.90K

∴ The return on a investment of Y = N38888.90K

S = {0, 1, 2, 3, 4, 5}

T = {− 1, 0, 1, 2}

S $\cup$ T = {− 1, 0, 1, 2, 3, 4, 5 }

⇒ − 2 < x ≤ 5

1st G.P. = a =2

2nd G.P. = ar$^{2 −1}$ = 54

2(r) = 54

r = 54/2 = 27

r = 27

3rd term = ar$^2$ = (2) (27)$^2$

2 × 27 × 27

= 1458