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JAMB UTME - Mathematics - 2012

Question 1 Report

The probabilities that a man and his wife live for 80 years are \( \frac{2}{3} \) and \( \frac{3}{5} \) respectively. Find the probability that at least one of them will live up to 80 years

\frac{2}{15}

\frac{3}{15}

\frac{7}{15}

\frac{13}{15}

Answer Details

The probability that the man will not live for 80 years is 1 - 2/3 = 1/3. Similarly, the probability that the woman will not live for 80 years is 1 - 3/5 = 2/5. The probability that both of them will not live for 80 years is the product of their individual probabilities, which is (1/3) * (2/5) = 2/15. Therefore, the probability that at least one of them will live up to 80 years is 1 - 2/15 = 13/15. So, the answer is 13/15.

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Probability

Measures Of Dispersion

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Question 2 Report

If \(y = x^2 - \frac{1}{x}\), find \(\frac{\delta y}{\delta x}\)

2x -

\frac{1}{x^{2}}

2x + x2

2x - x2

2x +

\frac{1}{x^{2}}

Answer Details

To find δyδx (the derivative of y with respect to x), we need to apply the power rule of differentiation, which states that if y = x^n, then δyδx = n*x^(n-1). Applying this rule to y = x^2 - 1/x, we get: δyδx = 2x + 1/x^2 Therefore, the answer is not one of the options provided. Option A (2x - 1/x^2) is close but has a minus sign instead of a plus sign before the second term. Option B (2x + x^2) and option C (2x - x^2) are incorrect because they don't take into account the derivative of the second term (-1/x). Option D (2x + 1/x^2) is the correct answer based on the power rule of differentiation.

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Question 7 Report

The sum to infinity of a geometric progression is \( -\frac{1}{10} \) and the first term is \( -\frac{1}{8} \). Find the common ratio of the progression.

- \frac{1}{4}

- \frac{1}{4}

- \frac{1}{4}

- \frac{1}{2}

Answer Details

Sr =

\frac{a}{1 - r}

- \frac{1}{10}

=

\frac{1}{8} \times \frac{1}{1 - r}

- \frac{1}{10}

=

\frac{1}{8 (1 - r)}

- \frac{1}{10}

=

\frac{1}{8 - 8 r}

cross multiply...
-1(8 - 8r) = -10
-8 + 8r = -10
8r = -2
r = -1/4

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Sequence And Series

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Question 8 Report

If log3x2 = -8, what is x?

\frac{1}{3}

\frac{1}{9}

\frac{1}{27}

\frac{1}{81}

Answer Details

We can start by using the definition of logarithms to rewrite the equation: log3(x^2) = -8 This means that 3 raised to the power of -8 is equal to x^2: 3^(-8) = x^2 To solve for x, we can take the square root of both sides: sqrt(3^(-8)) = sqrt(x^2) On the left side, we can simplify the expression using the rule that says sqrt(a^b) = a^(b/2): 3^(-8/2) = x Simplifying the exponent, we get: 3^(-4) = x Recall that a negative exponent means the reciprocal of the corresponding positive exponent. So: 3^(-4) = 1/3^4 Using the exponent rule that says a^b = a*a*a*...*a (b times), we get: 1/81 = x Therefore, the correct answer is option (D) 181.

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Positive And Negative Integers, Rational Numbers

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Question 10 Report

Simplify \( \frac{3}{x} + \frac{15}{2y} + \frac{6}{xy} \)

\frac{2 y - 5 x}{4}

\frac{9 (2 x - 5 x)}{x^{2} y^{2}}

\frac{5 x - 2 y}{2}

\frac{c^{2} y^{2}}{18 y - 45 x}

\frac{4}{2 y - 5 x}

Solution Of Linear Equations

Algebraic Expressions

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Question 12 Report

In how many ways can the letters of the word TOTALITY be arranged?

6720

6270

6207

6027

Answer Details

\frac{8!}{3!}

\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}

==> 8 x 7 x 6 x 5 x 4 = 6720

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Permutation And Combination

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Question 14 Report

The binary operation on the set of real numbers is defined by \(m*n = \frac{mn}{2}\) for all \(m, n \in R\). If the identity element is 2, find the inverse of -5

- \frac{4}{5}

-

- \frac{2}{5}

4

5

Answer Details

m

\propto

n =

\frac{m n}{2} - a

Identify = e = 2
a

\propto

a-1 = e
a

\propto

a-1 = 2
-5

\propto

a-1 = 2

\frac{- 5 \times a^{- 1}}{2} = 2

a^{- 1} = \frac{2 \times 2}{- 5}

a^{- 1} = - \frac{4}{5}

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Binary Operations

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Question 16 Report

Simplify \( (\sqrt{6}+2)^2-(\sqrt{6}-2)^2 \)

8 \sqrt{6}

8 \sqrt{6}

8 \sqrt{6}

8 \sqrt{6}

Answer Details

We can simplify the given expression using the identity: (a + b)^2 - (a - b)^2 = 4ab where a = √6 and b = 2. Substituting these values, we get: (√6 + 2)^2 - (√6 - 2)^2 = 4(√6 × 2) Simplifying further: (√6 + 2)^2 - (√6 - 2)^2 = 8√6 Therefore, the simplified form of the expression (√6+2)2−(√6−2)2 is 8√6.

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Sequence And Series

Surds (radicals)

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Question 17 Report

If angle \(\theta\) is \(135^o\), evaluate \(\cos\theta\)

\frac{1}{2}

\frac{\sqrt{2}}{2}

\frac{\sqrt{2}}{2}

-

\frac{1}{2}

Answer Details

θ

= 135o
Cos 135o = Cos(90 + 45)o
= cos90ocos45o - sin90osin45o
= 0cos45o - (1 x

\frac{\sqrt{2}}{2}

)
=

- \frac{\sqrt{2}}{2}

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Question 21 Report

If \(27^{x+2} \div 9^{x+1} = 3^{2x}\), find x

3

4

5

6

Answer Details

27^{x + 2} \div 9^{x + 1} = 3^{2 x}

Reduce 27 and 9 to have the same base as 3

3^{3 (x + 2)} \div 3^{2 (x + 1)} = 3^{2 x}

Note, - is ÷ and + is × for powers of indices of the same bases.

3^{3 (x + 2) - 2 (x + 1)} = 3^{2 x}

Equating the exponentials

3 (x + 2) - 2 (x + 1) = 2 x

Clear brackets

3 x + 6 - 2 x - 2 = 2 x

Collect like terms

3 x - 2 x - 2 x = - 6 + 2

x - 2 x = - 4

- x = - 4

Divide both sides by

-

x = 4

Answer is B

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Solution Of Linear Equations

Algebraic Expressions

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Question 24 Report

If y varies directly as \( \sqrt{n} \) and y = 4 when n = 4, find y when n = \( \frac{7}{9} \)

√17

\frac{4}{3}

\frac{8}{3}

\frac{2}{3}

Answer Details

y

α \sqrt{n}

y = k

\sqrt{n}

when y = 4, n = 4
4 = k

\sqrt{4}

4 = 2k
k = 2
Therefore,
y = 2

\sqrt{n}

y = 2

\sqrt{\frac{16}{9}}

y = 2

(\frac{4}{3})

y =

\frac{8}{3}

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Sequence And Series

Simple Operations On Algebraic Expressions

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Question 25 Report

Evaluate n+1Cn-2 If n =15

3630

3360

1120

560

Answer Details

\frac{n + 1 (n - 2)}{(n + 1)!}

\frac{(n + 1) + (n - 2)! (n - 2)!}{(n + 1)!}

\frac{(n + 1) (n + 1 - 1) (n + 1 - 2) (n + 1 - 3)!}{3! (n - 2)!}

\frac{(n + 1) (n) (n - 1) (n - 2)!}{3! (n - 2)!}

\frac{(n + 1) (n) (n - 1)}{3!}

Since n = 15

\frac{(15 + 1) (15) (15 - 1)}{3!}

\frac{16 \times 15 \times 14}{3 \times 2 \times 1}

= 560

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Solution Of Linear Equations

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Question 29 Report

The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is

3 \sqrt{2}

3 \sqrt{2}

3 \sqrt{2}

3 \sqrt{2}

Answer Details

The first step is to find the intersection point of the two lines y = 2x + 4 and y = 7 - x. To do this, we can set the two equations equal to each other and solve for x: 2x + 4 = 7 - x 3x = 3 x = 1 So, the intersection point is (1, 2). Next, we can use the distance formula to find the distance between the point (4, 3) and the intersection point (1, 2): distance = √((4 - 1)^2 + (3 - 2)^2) = √((3^2) + (1^2)) = √(9 + 1) = √10 = √(2 * 5) = √2 * √5 = 3√2. So, the distance between the two points is 3√2.

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Coordinate Geometry

Coordinate Geometry

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Question 30 Report

Simply \( \frac{2\frac{2}{3}\times1\frac{1}{2}}{4\frac{4}{5}} \)

1 \frac{1}{4}

1 \frac{1}{6}

\frac{5}{6}

\frac{4}{5}

Answer Details

\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}

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Fractions, Decimals And Approximations

Number Bases

Modular Arithmetic

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Question 34 Report

If \( \begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix} = \begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix} \), find the value of x

3

4

5

7

Answer Details

| \begin{matrix} 5 & 3 \\ x & 2 \end{matrix} |

=

| \begin{matrix} 3 & 5 \\ 4 & 5 \end{matrix} |

10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5

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Sequence And Series

Lengths And Perimeters

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Question 36 Report

Evaluate \( \int_{1}^{3}(X^2 - 1)\,dx \)

\frac{2}{3}

- \frac{2}{3}

- 6 \frac{2}{3}

6 \frac{2}{3}

Answer Details

\int_{1}^{3} (x^{2} - 1) d x = {[\frac{1}{3} x^{2} - x]}_{1}^{2} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (- \frac{2}{3}) = 6 + \frac{2}{3} = 6 \frac{2}{3}

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Integration

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Question 37 Report

Find the standard deviation of 2,3,8,10 and 12

3.9

4.9

5.9

6.9

Answer Details

\begin{matrix} x & (x - ϰ) & (x - ϰ)^{2} \\ 2 & - 5 & 25 \\ 3 & - 4 & 16 \\ 8 & 1 & 1 \\ 10 & 3 & 9 \\ 12 & 5 & 25 \\ 76 \end{matrix}

S.D =

\sqrt{\frac{(x - ϰ)^{2}}{n}}

S.D =

\sqrt{\frac{76}{5}}

S.D = 3.9

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Measures Of Dispersion

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Question 40 Report

Simplify \( \frac{1}{p} - \frac{1}{q} + \frac{p}{q} - \frac{q}{p} \)

\frac{1}{p - q}

\frac{- 1}{p + q}

\frac{1}{p q}

\frac{1}{p q (p - q)}

Answer Details

\frac{1}{p}

-

\frac{1}{q}

+

\frac{p}{q}

-

\frac{q}{p}

=

\frac{q - p}{p q}

+

\frac{p^{2} - q^{2}}{p q}

\frac{q - p}{p q}

x

\frac{p q}{p^{2} q^{2}}

=

\frac{q - p}{p^{2} q^{2}}

\frac{- (p - q)}{(p + q) (p - q)}

=

\frac{- 1}{p + q}

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Algebraic Expressions

Algebraic Fractions

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Question 41 Report

The grades of 36 students in a test are shown in the pie chart above. How many students had excellent?

7

8

9

12

Answer Details

Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70

^{\circ}

If 360

^{\circ}

represents 36 students
1

^{\circ}

will represent 36/360
50

^{\circ}

will represent 36/360 * 70/1
= 7

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Representation Of Data

Linear Inequalities

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Question 42 Report

Find the equation of the line through the points \((-2, 1)\) and \(\left(-\frac{1}{2}, 4\right)\)

y = 2x - 3

y = 2x + 5

y = 3x - 2

y = 2x + 1

Answer Details

\frac{y - y_{1}}{x + x_{1}}

=

\frac{y_{2} - y_{1}}{x - x_{1}}

\frac{y - 1}{x + 2}

=

\frac{4 - 1}{- \frac{1}{2} + 2}

=

\frac{y - 1}{x + 2}

=

\frac{3}{\frac{3}{2}}

y = 2x + 5

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Coordinate Geometry

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Question 43 Report

Make 'n' the subject of the formula if \( w = \frac{v(2+cn)}{1-cn} \)

\frac{1}{c} (\frac{w - 2 v}{v + w})

\frac{1}{c} (\frac{w - 2 v}{v - w})

\frac{1}{c} (\frac{w + 2 v}{v - w})

\frac{1}{c} (\frac{w + 2 v}{v + w})

Answer Details

w =

\frac{v (2 + c n)}{1 - c n}

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{- 2 v + w}{c v + c w}

n =

\frac{1}{c} \frac{- 2 v + w}{v + w}

Re-arrange...
n =

\frac{1}{c} \frac{w - 2 v}{v + w}

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Question 46 Report

The probability that a student passes a physics test is \( \frac{2}{3} \). If he takes three physics tests, what is the probability that he passes two of the tests?

\frac{2}{27}

\frac{3}{27}

\frac{4}{27}

\frac{5}{3}

Answer Details

The probability that the student passes a physics test is 2/3. Let's call this probability p. The probability that he fails a physics test is therefore 1 - p = 1/3. To find the probability that he passes two out of three tests, we need to consider the different ways in which this can happen. He could pass the first two tests and fail the third, or pass the first and third tests but fail the second, or pass the second and third tests but fail the first. Each of these scenarios has the same probability: p * p * (1 - p) + p * (1 - p) * p + (1 - p) * p * p Simplifying this expression, we get: 3p^2(1-p) Substituting p = 2/3, we get: 3(2/3)^2(1-2/3) = 4/27 Therefore, the probability that the student passes two out of three physics tests is 4/27, which corresponds to option (C).

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Representation Of Data

Probability

Probability

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Question 47 Report

The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is

\frac{1}{2}

\frac{2}{5}

- \frac{4}{7}

- \frac{2}{3}

Answer Details

PQ =

\frac{y_{1} - y_{0}}{x_{1} - x_{0}}

=

\frac{- 3 - (- 7)}{- 2 - 5}

=

\frac{- 3 + 7}{- 2 - 5}

=

\frac{4}{- 7}

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Coordinate Geometry

Graphs Of Linear And Quadratic Functions

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