Mathematics 2007 Jamb Examination

Operation: aΔb = a+b+1. The identity element e satisfies aΔe=a ⇒ a+e+1=a ⇒ e=-1.
Inverse x of 7 satisfies 7Δx=e= -1 ⇒ 7+x+1=-1 ⇒ x+8=-1 ⇒ x=-9.

Check option A: For n=2 ⇒ 5·2^0 - 2 = 3 ✓, n=3 ⇒ 10-3=7 ✓, n=4 ⇒ 20-4=16 ✓, n=5 ⇒ 40-5=35 ✓, n=6 ⇒ 80-6=74 ✓. Matches all terms, so the nth term is $5·2^{n-2} - n$.

It is a geometric series with first term a=2 and common ratio r=(3/2)/2=3/4.
Sum to infinity = a / (1-r) = 2 / (1-3/4) = 2 / (1/4) = 8.

$\sqrt{12} = 2\sqrt{3}$, $\sqrt{147} = 7\sqrt{3}$.
So equation becomes 2√3 - 7√3 + y√3 = 0 ⇒ (-5+y)√3=0 ⇒ y=5.

$(1214)_5 = 1·5^3 + 2·5^2 + 1·5^1 + 4·5^0 = 125+50+5+4=184$.
Therefore x=184.

Use the difference of squares: \(a^2-b^2=(a-b)(a+b)\). Let \(a=0.5625\) and \(b=0.4375\).
\((0.5625)^2-(0.4375)^2\)
\(=(0.5625-0.4375)(0.5625+0.4375)\)
\(=0.125\times1.000=0.125.\)
Now divide by \(0.04\): \(\dfrac{0.125}{0.04}=3.125.\)
Rounded to three significant figures: \(3.12\).

Divide both sides by 2: \(3^{2x-1}=81\).
Recognize \(81=3^4\), so \(3^{2x-1}=3^4\Rightarrow 2x-1=4\Rightarrow 2x=5\Rightarrow x=\tfrac{5}{2}.\)

Compute inside the parentheses first: \(\dfrac{2}{7}\times\dfrac{4}{3}=\dfrac{8}{21}.\)
Then \(\dfrac{8}{21}\div\dfrac{4}{9}=\dfrac{8}{21}\times\dfrac{9}{4}=\dfrac{72}{84}=\dfrac{6}{7}.\)
Now \(\dfrac{3}{5}\div\dfrac{6}{7}=\dfrac{3}{5}\times\dfrac{7}{6}=\dfrac{21}{30}=\dfrac{7}{10}.\)

Write \(12.5=\dfrac{25}{2}\).
\(\log_{10}12.5=\log_{10}25-\log_{10}2=2\log_{10}5 - x.\)
But \(\log_{10}5=\log_{10}\left(\dfrac{10}{2}\right)=1-x.\) So \(2\log_{10}5 - x = 2(1-x)-x = 2-3x.\)

Let CP be the cost price. From 5% profit: \(60000=1.05\times CP\Rightarrow CP=\dfrac{60000}{1.05}=57142.857...\).
To make 26% profit: \(SP=1.26\times CP=1.26\times57142.857...=72000.\)