## Mathematics 2012 Study Mode

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Question 1 Mathematics 2012 | JAMB/UTME (2012)
A
2211
B
2121
C
1212
D
1122
Explanation.
First, convert to base 10

726 = (7 x 61) + (2 x 60)

= 42 + 2

= 4410

$$\begin{array}{c|c} 3 & 44 \\ \hline 3 & 14 & r 2 \\ \hline 3 & 4 & r 2 \\ \hline 3 & 1 & r 1 \\ \hline 3 & 0 & r 1 \end{array}$$

Ans = 1122 = D

Question 2 Mathematics 2012 | JAMB/UTME (2012)
A
$$1\frac{1}{4}$$
B
$$1\frac{1}{6}$$
C
$$\frac{5}{6}$$
D
$$\frac{4}{5}$$
Explanation.
$$\frac{2\frac{2}{3} \times 1\frac{1}{2}}{4\frac{4}{5}}$$

$$\frac{8}{3} \times \frac{3}{2} \div \frac{24}{5}$$

$$\frac{8}{3} \times \frac{3}{2} \times \frac{5}{24}$$

$$\frac{5}{6}$$

Question 3 Mathematics 2012 | JAMB/UTME (2012)
A
2.3
B
2.31
C
2.32
D
2.33
Explanation.

No official Explanation yet!

Question 4 Mathematics 2012 | JAMB/UTME (2012)
A
N525
B
N1,025
C
N1,950
D
N2,975
Explanation.
Amount spent on children = $$\frac{15}{100} \times 3500 = N525$$

Amount spent = Amount on children + Additional amount on food

Amount Spent = N525 + N1950 = N2475

Therefore amount left = N3,500 - N2,475 = N1,025

Question 5 Mathematics 2012 | JAMB/UTME (2012)
A
B
C
D
Explanation.

No official Explanation yet!

Question 6 Mathematics 2012 | JAMB/UTME (2012)
A
$$\frac{1}{3}$$
B
$$\frac{1}{9}$$
C
$$\frac{1}{27}$$
D
$$\frac{1}{81}$$
Explanation.
log3x2 = -8

x2 = 3-8

x = $$3^{-8 \times \frac{1}{2}}$$

x = 3-4

x = $$\frac{1}{3^4}$$

x = $$\frac{1}{81}$$

Question 7 Mathematics 2012 | JAMB/UTME (2012)
A
$$2\sqrt{6}$$
B
$$4\sqrt{6}$$
C
$$8\sqrt{6}$$
D
$$16\sqrt{6}$$
Explanation.
$$(\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2$$

= $$[(\sqrt{6} + 2) + (\sqrt{6} - 2)][(\sqrt{6} + 2) - (\sqrt{6} - 2)]$$

= $$(\sqrt{6} + 2 + \sqrt{6} - 2)(\sqrt{6} + 2 - \sqrt{6} + 2)]$$

= $$(2\sqrt{6})(4)$$

= $$8\sqrt{6}$$

Question 8 Mathematics 2012 | JAMB/UTME (2012)
A
{3}
B
{2,3}
C
{2,3,5}
D
{1,2}
Explanation.
P = {2,3,5}

Q = {2,3}

P $$\cap$$ Q = {2,3}

Question 9 Mathematics 2012 | JAMB/UTME (2012)
A
B
C
D
Explanation.
n(f $$\cap$$ v) + n(f) + n(v) + n(f $$\cap$$ v) = 46

3 + 19 + 23 + x = 46

22 + 23 + x = 46

45 + x = 46

x = 46 - 45

x = 1

Question 10 Mathematics 2012 | JAMB/UTME (2012)
A
$$\frac{1}{c}(\frac{w - 2v}{v + w})$$
B
$$\frac{1}{c}(\frac{w - 2v}{v - w})$$
C
$$\frac{1}{c}(\frac{w + 2v}{v - w})$$
D
$$\frac{1}{c}(\frac{w + 2v}{v + w})$$
Explanation.
w = $$\frac{v(2 + cn)}{1 - cn}$$

2v + cnv = w(1 - cn)

2v + cnv = w - cnw

2v - w = -cnv - cnw

Multiply through by negative sign

-2v + w = cnv + cnw

-2v + w = n(cv + cw)

n = $$\frac{-2v + w}{cv + cw}$$

n = $$\frac{1}{c}\frac{-2v + w}{v + w}$$

Re-arrange...

n = $$\frac{1}{c}\frac{w - 2v}{v + w}$$

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