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JAMB UTME - Mathematics - 1992

Question 1 Report

Simplify \(5\sqrt{18} - 3\sqrt{72} + 4\sqrt{50}\)

17

\sqrt{4}

4

\sqrt{17}

17

\sqrt{2}

12

\sqrt{4}

Answer Details

To simplify these expressions, we need to simplify the numbers under the radical sign first by factoring out their perfect square factors. For the first expression 5√18, we can factor out the perfect square factor of 9, which leaves us with 5√2√2√2 or 10√2. For the second expression, we can simplify √72 to √(36*2), and since 36 is a perfect square, we can factor it out, which leaves us with 3√2*6, or 3√2*2√3, which simplifies to 6√6. For the third expression, we can simplify √50 to √(25*2), and since 25 is a perfect square, we can factor it out, which leaves us with 2√2*5, or 2√2*√5, which simplifies to 2√10. Putting it all together, we have: 5√18 = 10√2 - 3√72 = 6√6 - 4√50 = 2√10 Now we can substitute these simplified expressions back into the original expressions: 10√2 - 6√6 + 2√10 To simplify this expression, we can group like terms. The coefficients of √6 are -6 and 0, since there are no other terms with √6. The coefficients of √10 are 2 and 0, since there are no other terms with √10. The coefficient of √2 is 10. So our simplified expression is: 10√2 - 6√6 + 2√10 = 10√2 - 6√6 + 2√10 = 10√2 - 6√(2*3) + 2√(2*5) = 10√2 - 6√2√3 + 2√2√5 = (10-6√3+2√5)√2 Therefore, the answer is 17√2.

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Sequence And Series

Surds (radicals)

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Question 3 Report

What is the value of x satisfying the equation \( \frac{4^{2x}}{4^{3x}} = 2 \)?

-2

-

\frac{1}{2}

\frac{1}{2}

2

Answer Details

\frac{4^{2 x}}{4^{3 x}}

= 2
42x - 3x = 2
4-x = 2
(22)-x
= 21
Equating coefficients: -2x = 1
x = -

\frac{1}{2}

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Question 5 Report

ind then equation line through (5, 7) parallel to the line 7x + 5y = 12

5x + 7y = 120

7x + 5y = 70

x + y = 7

15x + 17y = 90

Answer Details

Equation (5, 7) parallel to the line 7x + 5y = 12
5Y = -7x + 12
y =

\frac{- 7 x}{5}

+

\frac{12}{5}

Gradient =

\frac{- 7}{5}

∴ Required equation =

\frac{y - 7}{x - 5}

=

\frac{- 7}{5}

i.e. 5y - 35 = -7x + 35
5y + 7x = 70

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Question 6 Report

If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores

\frac{2}{3}

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

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

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

Answer Details

To find the standard deviation of the scores, we need to first calculate the mean (average) of the scores: mean = (5 + 6 + 7) / 3 = 6 Next, we calculate the deviation of each score from the mean: deviation of 5 = 5 - 6 = -1 deviation of 6 = 6 - 6 = 0 deviation of 7 = 7 - 6 = 1 To calculate the standard deviation, we take the square root of the average of the squared deviations: standard deviation = √[(1^2 + 0^2 + (-1)^2) / 3] = √(2/3) ≈ 0.82 Therefore, the answer is option (D), √(2/3).

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Measures Of Dispersion

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Question 7 Report

\(p = \begin{vmatrix} x & 3 & 0 \\ 2 & y & 3 \\ 4 & 2 & 4 \end{vmatrix}\)

\(q = \begin{vmatrix} x & 2 & z \\ 3 & y & 2 \\ 0 & 3 & z \end{vmatrix}\) Where \(p^T\) is the transpose p calculate \(/p^T/\) when x = 0, y = 1 and z = 2

48

24

-24

-48

Answer Details

=

| \begin{matrix} 0 & 3 & 0 \\ 2 & 1 & 3 \\ 4 & 2 & 2 \end{matrix} |

PT =

| \begin{matrix} 0 & 2 & 4 \\ 2 & 1 & 3 \\ 0 & 3 & 2 \end{matrix} |

/pT/ =

| \begin{matrix} 0 & 2 & 4 \\ 3 & 1 & 3 \\ 0 & 3 & 2 \end{matrix} |

= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]
= 0 - 12 + 36 = 24

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Matrices And Determinants

Number Bases

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Question 12 Report

Simplify \(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^{5})}\)

4.0 x 10-3

5.0 x 10-2

2.0 x 10-1

5.0 x 10-3

Answer Details

\frac{(1.25 \times 10^{- 4}) \times (2.0 \times 10^{- 1})}{(6.25 \times 10^{5})}

=

\frac{1.25 \times 2}{6.25}

x 104 - 1 - 5

\frac{2.50}{6.25}

x 10-2 =

\frac{250}{625}

x 10-2
0.4 x 10-2 = 4.0 x 10-3

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Number Bases

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Question 13 Report

\(p = \begin{vmatrix} x & 3 & 0 \\ 2 & y & 3 \\ 4 & 2 & 4 \end{vmatrix}\)

\(q = \begin{vmatrix} x & 2 & z \\ 3 & y & 2 \\ 0 & 3 & z \end{vmatrix}\)

pq is equivalent to

ppT

pp-1

qp

pp

Answer Details

=

| \begin{matrix} 0 & 3 & 0 \\ 2 & 1 & 3 \\ 4 & 2 & 2 \end{matrix} |

Q =

| \begin{matrix} 0 & 2 & 4 \\ 3 & 1 & 2 \\ 0 & 3 & 2 \end{matrix} |

= pT
pq = ppT

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Matrices And Determinants

Quadratic Equations

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Question 16 Report

Make t the subject of formula \(S = ut + \frac{1}{2}at^2\)

\frac{1}{a} - u + \sqrt{u^{2} - 2 a s}

\frac{1}{a} u \pm (u^{2} - 2 a s)

\frac{1}{a} u \pm \sqrt{2 a s}

\frac{1}{a} - u + \sqrt{2 a s}

Answer Details

Given S = ut +

\frac{1}{2} a t^{2}

S = ut +

\frac{1}{2} a t^{2}

∴ 2S = 2ut + at2
= at2 + 2ut - 2s = 0
t =

\frac{- 2 u \pm 4 u^{2} + 2 a s}{2 a}

= -2u

π

\frac{\sqrt{u^{2} 4 u^{2} + 2 a s}}{2 a}

= $\frac{1}{a}$ (-u + $\sqrt{U^{2} - 2 a s}$ )

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Change Of Subject Of A Formula/Relation

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Question 18 Report

If the binary operation \( \ast \) is defined by \( m \ast n = mn + m + n \) for any real number m and n, find the identity of the elements under this operation

e = 1

e = -1

e = -2

e = 0

Answer Details

m

*

n = mn + m + n
m

*

e = me + m + e, e

*

m = e
∴ me + m + e, m(e + 1)e - e = 0
e + 1 = 0
∴ e = -1

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Binary Operations

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Question 21 Report

Find the sum to infinity to the following series \(3 + 2 + \frac{4}{3} + \frac{8}{9} + \frac{16}{17} + .....\)

1270

190

18

9

Answer Details

3 + 2 +

\frac{4}{3}

+

\frac{8}{9}

+

\frac{16}{17}

+ .....
a = 3
r =

\frac{2}{3}

s

α

=

\frac{a}{1 - r}

=

\frac{3}{1 - \frac{2}{3}}

=

\frac{3}{\frac{1}{3}}

= 3 x 3
= 9

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Sequence And Series

Progression

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Question 24 Report

In the figure, PQr is a semicircle while PQ and Qr are chords. QS is the perpendicular from Q to the diameter PR. What is the expression for QS?

QS = PS

\sqrt{S R}

QS =

\sqrt{(P S S R)}

QS = 2

\sqrt{(P S S R)}

QS =

\frac{1}{\sqrt{2}}

\sqrt{(P S S R)}

Polynomials

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Question 25 Report

The probability of an event P is \( \frac{3}{4} \) while that of another event Q is \( \frac{1}{6} \). If the probability of both P and Q is \( \frac{1}{2} \). What is the probability of either P or Q.

\frac{1}{96}

\frac{1}{8}

\frac{5}{6}

\frac{11}{12}

Probability

Surds (radicals)

Probability

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Question 27 Report

If U = (1, 2, 3, 6, 7, 8, 9, 10) is the universal set. E = (10, 4, 6, 8, 10) and F = is odd. Find (E ∩ F), where means x 1x2 = 26, x the complement of a set

(0)

U

(8)

ϕ

Answer Details

U = (1, 2, 3, 6, 7, 8, 9, 10)
E = (10, 4, 6, 8, 10)
F = (x : x2 = 26, x is odd)
∴ F =

ϕ

Since x2 = 26 = 64
x = + which is even
∴ E ∩ F =

ϕ

Since there are no common elements

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Question 28 Report

The radius of a circle is given as 5cm subject to an error of 0.1cm. What is the percentage error in the area of the circle?

\frac{1}{25}

\frac{1}{4}

4

25

Answer Details

Area =

π r^{2} = π \times 5 \times 5 = 25 π

Area error =

π \times {0.1}^{2} = 0.01 π

∴

% error =

\frac{0.01 π}{25 π} \times 10

= \frac{0.01}{25} \times 100 = \frac{1}{25}

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Representation Of Data

Percentages

Probability

Fractions, Decimals, Approximations And Percentage

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Question 31 Report

Find all values of x satisfying the inequality \( -11 \le 4 - 3x \le 28 \)

-5

\leq

x v 8

5

\leq

x

\leq

8

-8

\leq

x

\leq

5

-5 < x

\leq

8

Answer Details

To solve the inequality -11 ≤ 4 - 3x ≤ 28, we need to isolate x in the middle of the inequality. First, we subtract 4 from all parts of the inequality: -15 ≤ -3x ≤ 24. Then, we divide by -3, remembering to flip the direction of the inequality signs when dividing by a negative number: 5 ≥ x ≥ -8. Therefore, the correct option is -8 ≤ x ≤ 5.

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Linear Inequalities

Inequalities

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Question 32 Report

The diagram is a circle with centre O. Find the area of the shaded portion.

9(

π - 2) c m^{2}

18

π c m^{2}

9

π c m^{2}

36

π c m^{2}

Answer Details

To find the area of the shaded portion, we need to subtract the area of the triangle from the area of the sector. First, we need to find the radius of the circle. Since the diameter is given as 6cm, the radius is half of it which is 3cm. Next, we need to find the angle of the sector. We can do this by using the properties of the isosceles triangle. The angle at the centre of the circle is twice the angle at the circumference of the circle. So, the angle at the centre of the circle is: 2 x (180 - 90) = 180 degrees This means that the angle of the sector is 180 degrees. Now we can calculate the area of the sector: Area of sector = (angle/360) x πr^2 = (180/360) x π(3)^2 = 4.5π cm^2 Next, we need to find the area of the triangle. Since the triangle is isosceles, the base angles are equal, and we can use trigonometry to find the height. The base of the triangle is equal to the diameter of the circle which is 6cm. Using trigonometry, we know that: tan(45) = height/base height = base x tan(45) height = 6 x 1 height = 6cm Now we can calculate the area of the triangle: Area of triangle = (1/2) x base x height = (1/2) x 6 x 6 = 18 cm^2 Finally, we can calculate the area of the shaded portion by subtracting the area of the triangle from the area of the sector: Area of shaded portion = Area of sector - Area of triangle = 4.5π - 18 = (9/2)(π - 2) cm^2 Therefore, the answer is 9(π−2)cm2.

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Question 33 Report

If \(x = 3 - \sqrt{3}\), find \(x^2 + \frac{36}{x^2}\)

9

18

24

27

Answer Details

x = 3 -

\sqrt{3}

x2 = (3 -

\sqrt{3}

)2
= 9 + 3 - 6

\sqrt{34}

= 12 - 6

\sqrt{3}

= 6(2 -

\sqrt{3}

)
∴ x2 +

\frac{36}{x^{2}}

= 6(2 -

\sqrt{3}

) +

\frac{36}{6 (2 - \sqrt{3})}

6(2 -

\sqrt{3}

) +

\frac{6}{2 - \sqrt{3}}

= 6(-

\sqrt{3}

) +

\frac{6 (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}

= 6(2 -

\sqrt{3}

) +

\frac{6 (2 + \sqrt{3})}{4 - 3}

6(2 -

\sqrt{3}

) + 6(2 +

\sqrt{3}

) = 12 + 12
= 24

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Quadratic Equations

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Question 34 Report

Evaluate the integral \( \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} 2 \cos 2x\, dx \)

-

\frac{1}{2}

-1

\frac{1}{2}

1

Answer Details

let I =

\int_{\frac{π}{12}}^{\frac{π}{4}}

2 cos 2x dx
= 2(sin 2x)

\frac{π}{4}

(sin 2x)

\frac{π}{4}

\frac{π}{12}

(2)

\frac{π}{12}

= -1 -

\frac{1}{2}

=

\frac{1}{2}

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Integration

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Question 38 Report

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

\frac{1}{6}

\frac{1}{12}

\frac{5}{6}

\frac{11}{12}

Answer Details

The total sample space when two dice are thrown together is 6 x 6 = 36

\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6 \\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6 \\ 3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6 \\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6 \\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6 \\ 6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{matrix}

At least 10 means 10 and above
P(at least 10) =

\frac{6}{36}

=

\frac{1}{6}

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Probability

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Question 41 Report

Determine the distance on the earth's surface between two town P (latoN, Long 20oN) and Q(Lat 60oN, Long 25oW) (Radius of the earth = 6400km)

\frac{800 π}{9}

km

\frac{800 \sqrt{3 π}}{9}

km

800

π

km

800

\sqrt{3 π}

km

Coordinate Geometry

Angles And Intercepts On Parallel Lines

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Question 43 Report

\(x\)	2	4	6	8
\(f\)	4	\(y\)	6	5

If the mean of the above frequency distribution is 5.2, find y

2, 1

1, 2

1, 5

5, 2

Answer Details

Mean

\bar{x}

=

\frac{\sum f x}{\sum f}

=

\frac{5.2}{1}

=

\frac{8 + 4 y + 36 + 40}{4 + y + 6 + 5}

=

\frac{5.2}{1}

=

\frac{84 + 4 y}{15 + y}

= 5.2(15 + y)
= 84 + 4y
= 5.2 x 15 + 5.2y
= 84 + 4y
= 78 + 5.2y
= 84 = 4y
= 5.2y - 4y
= 84 - 78
1.2y = 6
y =

\frac{6}{1.2}

=

\frac{60}{12}

= 5

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Probability

Linear Inequalities

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Question 44 Report

In the figure, if XZ is 10cm, calculate RY in cm.

1010(1 -

\frac{1}{\sqrt{2}}

)

10(1 -

\frac{1}{\sqrt{3}}

)

10(1 -

\frac{1}{\sqrt{2}}

)

10[1 -

\sqrt{3}

]

Triangles And Polygons

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Question 45 Report

Given that \( \theta \) is an acute angle and \( \sin \theta = \frac{m}{n} \), find \( \cos \theta \)

\frac{\sqrt{n^{2} - m^{2}}}{m}

\frac{\sqrt{(n + m) (n - m)}}{n}

\frac{m}{\sqrt{n^{2} - m^{2}}}

\sqrt{\frac{n}{n^{2} - m^{2}}}

Sine, Cosine And Tangent Of An Angle

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Question 47 Report

Resolve \( \frac{3}{x^2+x-2} \) into partial fractions

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

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

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

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

Answer Details

\frac{3}{x^{2} + x - 2}

=

\frac{3}{(x - 1) (x + 1)}

\frac{A}{x - 1}

+

\frac{B}{x + 2}

A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3

\to

a = 1
when x = -2, -3B = 3

\to

B = -1
3 = 1 - 1
=

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

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Algebraic Fractions

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Question 48 Report

What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?

\frac{\sqrt{5}}{2}

\frac{\sqrt{5}}{20}

\frac{5}{\sqrt{13}}

6

Answer Details

2x - 4y + 3 = 0
Required distance =

\frac{(2 \times 2) + 3 (- 4) + 3}{\sqrt{2^{2}} + (- 4)^{2}}

=

\frac{4 - 12 + 3}{\sqrt{20}}

=

\frac{- 5}{- 2 \sqrt{5}}

= \frac{\sqrt{5}}{2}

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Euclidean Geometry

Coordinate Geometry

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