JAMB UTME - Mathematics - 2010

Question 1 Report

Evaluate $\begin{vmatrix}2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1\end{vmatrix}$

5y - 2x -18 = 0

102

-102

-42

Answer Details

| \begin{matrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{matrix} |

= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102

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Algebraic Expressions

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Question 2 Report

If $p : q = \frac{2}{3} : \frac{5}{6}$ and $q : r = \frac{3}{4} : \frac{1}{2}$, find $p : q : r$

12 : 15 : 10

12 : 15 : 16

10 : 15 : 24

9 : 10 : 15

Answer Details

If p : q =

\frac{2}{3}

:

\frac{5}{6}

, then the sum S1 of ratio =

\frac{2}{3}

+

\frac{5}{6}

=

\frac{9}{6}

If q : r =

\frac{3}{4}

:

\frac{1}{2}

, then the sum S2 of ratio =

\frac{3}{4}

+

\frac{1}{2}

=

\frac{5}{4}

Let p + q = T1, then
q = (

\frac{5}{6} \div \frac{9}{6}

)T1 = (

\frac{5}{6} \times \frac{6}{9}

)T1 =

\frac{5}{9}

T1
Again, let q + r = T2, then
q = (

\frac{3}{4} \div \frac{5}{4}

)T2 = (

\frac{3}{4} \times \frac{4}{5}

)T2 =

\frac{3}{5}

T2
Using q = q

\frac{5}{9}

T1 =

\frac{3}{5}

T2
5 x 5T1 = 9 x 3T2

\frac{T_{1}}{T_{2}}

=

\frac{9 \times 3}{5 x 5}

=

\frac{27}{5}

Giving that, T1 = 27 and T2 = 25
P = (

\frac{2}{3} \div S_{1}

)T1 = (

\frac{2}{3} \div \frac{9}{6}

)T1
= (

\frac{2}{3} \times \frac{6}{9}

)27 = 12
q = (

\frac{5}{6} \div S_{1}

)T1 = (

\frac{5}{6} \div \frac{9}{6}

)T1
= (

\frac{5}{6} \times \frac{6}{9}

)27 = 15
and r = (

\frac{1}{2} \div S_{2}

)T2 = (

\frac{1}{2} \div \frac{5}{4}

)T2
= (

\frac{1}{2} \times \frac{4}{5}

)25 = 10
Hence p : q : r = 12: 15 : 10

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Positive And Negative Integers, Rational Numbers

Sequence And Series

Ratio, Proportions And Rates

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Question 3 Report

Marks	2	3	4	5	6	7	8
No. of students	3	1	5	2	4	2	3

From the table above, if the pass mark is 5, how many students failed the test?

6

2

9

7

Answer Details

To determine how many students failed the test, we need to add up the frequencies of the students who obtained marks less than 5, since the pass mark is 5. Looking at the table, the marks less than 5 are 2, 3, and 4. Adding up the corresponding frequencies, we get: 3 + 1 + 5 = 9 Therefore, 9 students failed the test. The answer is (C) 9.

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Question 4 Report

Determine the value of x for which (x2 - 1) > 0

x < -1 or x > 1

-1 < x < 1

x > 0

x < -1

Answer Details

We want to solve the inequality (x² - 1) > 0 for x. To do this, we can factor the left-hand side of the inequality: (x² - 1) = (x - 1)(x + 1) Now we have the inequality: (x - 1)(x + 1) > 0 The product of two factors is positive if and only if both factors are positive or both factors are negative. So we can break the inequality into two cases: Case 1: (x - 1) > 0 and (x + 1) > 0 This simplifies to x > 1, which means x is greater than 1. Case 2: (x - 1) < 0 and (x + 1) < 0 This simplifies to x < -1, which means x is less than -1. Therefore, the solution to the inequality (x² - 1) > 0 is: x < -1 or x > 1 So the answer is: x < -1 or x > 1.

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Linear Inequalities

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Question 5 Report

Find the distance between the points $ \left(\frac{1}{2}, -\frac{1}{2}\right) $.

1

o

√3

√2

Answer Details

Let D denote the distance between (

\frac{1}{2}

, -

\frac{1}{2}

) then using
D =

\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

=

\sqrt{(- \frac{1}{2} - \frac{1}{2})^{2} + (- \frac{1}{2} - \frac{1}{2})^{2}}

=

\sqrt{(- 1)^{2} + (- 1)^{2}}

=

\sqrt{1 + 1}

= √2

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Question 6 Report

At what value of x does the function y= -3 – 2x +x2 attain a minimum value?

4

2

-1

-4

1

Answer Details

To find the minimum value of the function y = -3 - 2x + x^2, we need to determine the value of x that corresponds to the vertex of the parabolic graph. The vertex of a parabolic graph with equation y = ax^2 + bx + c is located at x = -b/2a. In this case, a = 1, b = -2, and c = -3. Therefore, x = -(-2)/(2*1) = 1. So the answer is (E) 1, and that's the value of x at which the function y attains its minimum value.

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Quadratic Equations

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Question 7 Report

A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?

749

π

m2

700

π

m2

350

π

m2

98

π

m2

Answer Details

To calculate the total surface area of the cylindrical pipe, we need to add the surface area of the curved part and the surface area of the two circular ends. The surface area of the curved part can be calculated by multiplying the circumference of the circle (2πr) by the length of the pipe (50cm), which gives us: 2πr x h = 2π x 7m x 50cm = 7π m^2 The surface area of one circular end can be calculated by multiplying the area of the circle (πr^2) by 1, since one end of the pipe is open and has no surface area. Thus, the total surface area of both circular ends is: 2πr^2 = 2π x 7m^2 = 14π m^2 Finally, we add the surface area of the curved part and the surface area of the two circular ends to get the total surface area of the pipe: 7π m^2 + 14π m^2 = 21π m^2 Therefore, the total surface area of the pipe is 21π square meters. The closest option to this answer is 749π, but it is not the correct answer.

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Volumes

Loci

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Question 8 Report

The interior angles of a quadrilateral are (x + 15)o, (2x - 45)o and (x + 10)o. Find the value of the least interior angle.

112o

102o

82o

52o

Answer Details

(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x =

\frac{410^{o}}{5}

= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o

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Algebraic Expressions

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Question 9 Report

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope is 1.25m, what was the percentage error in the measurement?

1.40%

0.01%

0.25%

6.9%

0.80%

Answer Details

The percentage error in measurement is the difference between the measured value and the actual value, divided by the actual value, multiplied by 100. In this case, the measured value is 1.26m, and the actual value is 1.25m. So the difference between the measured value and actual value is: 1.26m - 1.25m = 0.01m The percentage error can be calculated as: (0.01m ÷ 1.25m) × 100% = 0.8% Therefore, the percentage error in the measurement is 0.8%, which corresponds to option E.

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Fractions, Decimals And Approximations

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Question 10 Report

Make Q the subject of formula if $ p = \frac{M}{5}(X + Q) + 1 $

\frac{5 P - M X + 5}{M}

\frac{5 P - M X - 5}{M}

\frac{5 P + M X + 5}{M}

\frac{5 P + M X - 5}{M}

Answer Details

To make Q the subject of the formula if p=M5 in the expression (X+Q)+1, we need to isolate Q on one side of the equation and simplify the expression on the other side. First, we need to remove the parentheses by adding X and 1 together, which gives us X+1. Next, we move M5 to the other side of the equation by subtracting it from both sides, resulting in: (X+Q)+1 - M5 = 0 Then, we can isolate Q by subtracting X and 1 from both sides: (X+Q) - (X+1) - M5 = -1 Simplifying the left-hand side, we get: Q - M5 = -1 Finally, we can solve for Q by adding M5 to both sides: Q = M5 - 1 Therefore, the expression (X+Q)+1 can be simplified to MX+5P-5M-4, and Q is equal to M5-1. The correct answer is (B) 5P-MX-5M.

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Change Of Subject Of A Formula/Relation

Measures Of Location

Simple Operations On Algebraic Expressions

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Question 11 Report

If x is inversely proportional to y and x = $ \frac{1}{2} $ when y = 2, find x if y = 4

4

5

1

\frac{1}{4}

2

\frac{1}{4}

Answer Details

x

α

\frac{1}{y}

.........(1)
x = k x

\frac{1}{y}

.........(2)
When x = 2

\frac{1}{2}

=

\frac{5}{2}

, y = 2
(2) becomes

\frac{5}{2}

= k x

\frac{1}{2}

giving k = 5
from (2), x =

\frac{5}{y}

so when y =4, x =

\frac{5}{y}

= 1

\frac{1}{4}

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Logical Reasoning

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Question 12 Report

Marks	1	2	3	4	5
Frequency	2	2	8	4	4

The table above shows the marks obtained in a given test. How many students took the test?

16

20

13

15

Answer Details

To determine how many students took the test, we need to sum up the frequencies in the table, since each frequency represents the number of students who obtained the corresponding mark. Adding up the frequencies, we get: 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (B) 20.

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Integration

Linear Inequalities

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Question 13 Report

If two smaller sides of a right angled triangle are 4cm and 5cm, find its area

10 cm2

6 cm2

24 cm2

8 cm2

Answer Details

To find the area of a right angled triangle, we can use the formula: Area = (base x height) / 2 In a right angled triangle, the two smaller sides that form the right angle are the base and height. Therefore, we can substitute 4 cm for the base and 5 cm for the height in the formula: Area = (4 cm x 5 cm) / 2 = 10 cm^2 Therefore, the area of the right angled triangle is 10 cm^2. The answer is (A) 10 cm^2.

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Euclidean Geometry

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Question 14 Report

In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS

55o

45o

35o

25o

Answer Details

Join SR
< PRS = 90

^{\circ}

(Angle in a semicircle)
< PRS = 55

^{\circ}

(Angle between a chord and a tangent = Angle in the alternate segment)
< PSR + < PRS + < RSP = 180

^{\circ}

90v + 55

^{\circ}

+ < RSP = 180

^{\circ}

< RSP = 180

^{\circ}

- 145

^{\circ}

= 35

^{\circ}

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Circles

Coordinate Geometry

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Question 15 Report

Solve the inequality $ (x - 3)(x - 4) \le 0 $

3

\leq

x

\leq

4

3 < x < 4

3

\leq

x < 4

3 < x

\leq

4

Answer Details

(x - 3)(x - 4)

\leq

0
Case 1 (+, -) = x - 3

\geq

0, X - 4

\geq

0
= X

\leq

3, x

\geq

4
= 3 < x

\geq

4 (solution)

Case 2 = (-, +) = x - 3 $\leq$ 0, x - 4 $\geq$ 0
= x $\leq$ 3, x $\geq$ 4
therefore = 3 $\leq$ x $\leq$ 4

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Statistics

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Question 16 Report

At what rate will the interest on ₦400 increases to ₦24 in 3 years reckoning in simple interest?

3%

2%

5%

4%

Answer Details

The formula for simple interest is: I = PRT Where: I = Interest P = Principal R = Rate T = Time We are given the following information: P = ₦400 I = ₦24 T = 3 years Substituting these values into the formula and solving for R: 24 = 400 * R * 3 R = 24 / (400 * 3) = 0.02 = 2% Therefore, the answer is 2%. The rate at which the interest on ₦400 increases to ₦24 in 3 years reckoning in simple interest is 2%.

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Variation

Ratio, Proportions And Rates

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Question 17 Report

Simplify $ \frac{3}{5} \div \left(\frac{2}{7} x \frac{4}{3} \div \frac{4}{9}\right) $

\frac{6}{7}

\frac{4}{5}

\frac{7}{10}

\frac{21}{6}

Answer Details

\frac{3}{5}

\div

(

\frac{2}{7}

x

\frac{4}{3}

\div

\frac{4}{9}

) =

\frac{2}{3}

\div

(

\frac{2}{7}

x

\frac{4}{3}

x

\frac{9}{4}

)
=

\frac{3}{5}

\div

\frac{6}{7}

=

\frac{3}{5}

x

\frac{7}{6}

=

\frac{7}{10}

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Number Bases

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Question 18 Report

actorize completely $ \frac{x^3+3x^2-10x}{2x^2-8} $

\frac{x (x - 5)}{2 (x + 2)}

\frac{x (x + 5)}{2 (x + 2)}

\frac{x (x - 5)}{2 (x - 2)}

\frac{x^{2} + 5}{2 x + 4}

Answer Details

\frac{x^{3} + 3 x^{2} - 10 x}{2 x^{2} - 8}

=

\frac{x (x^{2} + 3 x - 10)}{2 (x^{2} - 4)}

=

\frac{x (x^{2} + 5 x - 2 x - 10)}{2 (x + 2) (x - 2)}

=

\frac{x (x - 2) (x + 5)}{2 (x + 2) (x - 2)}

=

\frac{x (x + 5)}{2 (x + 2)}

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Polynomials

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Question 19 Report

For what range of values of x is $ \frac{1}{2}x + \frac{1}{4} > \frac{1}{3}x + \frac{1}{2} $?

x <

\frac{3}{2}

x >

\frac{3}{2}

x < -

\frac{3}{2}

x > -

\frac{3}{2}

Answer Details

\frac{1}{2}

x +

\frac{1}{4}

>

\frac{1}{3}

x +

\frac{1}{2}

Multiply through by through by the LCM of 2, 3 and 4
12 x

\frac{1}{2}

x + 12 x

\frac{1}{4}

> 12 x

\frac{1}{3}

x + 12 x

\frac{1}{2}

6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3

\frac{2 x}{2}

>

\frac{3}{2}

x >

\frac{3}{2}

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Solution Of Linear Equations

Linear Inequalities

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Question 20 Report

If $ P = \begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix} $

(\begin{matrix} - \frac{1}{5} & - \frac{3}{5} \\ - \frac{1}{5} & - \frac{2}{5} \end{matrix})

(\begin{matrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & \frac{2}{5} \end{matrix})

(\begin{matrix} - \frac{1}{5} & \frac{3}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix})

(\begin{matrix} \frac{1}{5} & \frac{3}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix})

Answer Details

P =

(\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix})

|P| = 2 - 1 x -3 = 5
P-1 =

\frac{1}{5}

(\begin{matrix} 1 & 3 \\ - 1 & 2 \end{matrix})

=

(\begin{matrix} \frac{1}{5} & \frac{3}{5} \\ - \frac{1}{5} & \frac{2}{5} \end{matrix})

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Number Bases

Modular Arithmetic

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Question 21 Report

Evaluate $ \left(\frac{81}{16}\right)^{-\frac{1}{4}} \times 2^{-1} $

1/3

3

6

1/6

Answer Details

{(\frac{81}{16})}^{\frac{- 1}{4}} \times 2^{- 1} = \frac{1}{{(\frac{81}{16})}^{\frac{1}{4}}} \times \frac{1}{2} = {(\frac{16}{81})}^{\frac{1}{4}} \times \frac{1}{2} = {(\frac{2}{3})}^{4 \times \frac{1}{4}} = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}

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Question 22 Report

Marks	1	2	3	4	5
Frequency	2	2	8	4	4

The table above show the marks obtained in a given test.

Find the mean mark

3.1

3.0

3.2

3.3

Answer Details

To find the mean mark, we need to calculate the sum of all the marks obtained and divide it by the total number of students. The sum of all the marks obtained can be found by multiplying each mark by its corresponding frequency and adding up the results. So, sum of all marks = (1 x 2) + (2 x 2) + (3 x 8) + (4 x 4) + (5 x 4) = 2 + 4 + 24 + 16 + 20 = 66 The total number of students can be found by adding up all the frequencies. So, total number of students = 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark = (sum of all marks) / (total number of students) = 66 / 20 = 3.3 Hence, the answer is 3.3.

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Euclidean Geometry

Statistics

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Question 23 Report

If three unbiased coins are tossed, find the probability that they are all heads

\frac{1}{2}

\frac{1}{3}

\frac{1}{9}

\frac{1}{8}

Answer Details

P(H) =

\frac{1}{2}

and P(T) =

\frac{1}{2}

Using the binomial prob. distribution,
(H + T)3 = H3 + 3H2T1 + 3HT2 + T3
Hence the probability that three heads show in a toss of the three coins is H3
= (

\frac{1}{2}

)3
=

\frac{1}{8}

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Probability

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Question 24 Report

If $ \begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15 $, find the value of x

4

5

2

3

Answer Details

The expression |x327| means the absolute value of x to the power of 327. The given equation |x327| = 15 means that the absolute value of x to the power of 327 is equal to 15. To solve for x, we can take the 327th root of both sides of the equation. Thus, we have: |x327| = 15 Taking the 327th root of both sides: |x| = 15^(1/327) Since x can be positive or negative, we have two solutions: x = 15^(1/327) or x = -15^(1/327) Using a calculator, we can approximate the value of x as approximately 2.905 or -2.905. However, only one of these values is among the answer choices, which is x = 3. Therefore, the correct answer is 3.

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Positive And Negative Integers, Rational Numbers

Number Bases

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Question 25 Report

Solve the inequality $ -6(x + 3) \le 4(x - 2) $

x

\leq

2

x

\geq

-1

x

\geq

-2

x

\leq

-1

Answer Details

-6(x + 3)

\leq

4(x - 2)
-6(x +3)

\leq

4(x - 2)
-6x -18

\leq

4x - 8
-18 + 8

\leq

4x +6x
-10x

\leq

10x
10x

\leq

-10
x

\leq

1

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Linear Inequalities

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Question 26 Report

If $x \ast y = x + y^2$, find the value of $(2 \ast 3) \ast 5$

25

11

55

36

Answer Details

Given that,
X

*

y = X + y2
(2

*

3)

*

5 = (2 + 32)

*

5
= (2 + 9)

*

5 = 11

*

5
Hence 11

*

5 = 11 + 52
= 11 + 25 = 36

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Surds (radicals)

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Question 27 Report

Find the equation of a line parallel to y = -4x + 2 passing through (2,3)

y + 4x + 11 = 0

y - 4x - 11 = 0

y + 4x - 11 = 0

y - 4x + 11 = 0

Answer Details

By comparing y = mx + c with y = -4x + 2, the gradient of y = -4x + 2 is m1 = -4
Let the gradient of the line parallel to the given line be m2,
then, m2 = m1 = -4 (condition for parallelism)
Using: y - y1 = m2(x - x1)
Hence the equation of the parallel line is
y - 3 = -4(x-2)
y - 3 = -4 x + 8
y + 4x = 8 + 3
y + 4x = 11
y + 4x - 11 = 0

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Algebraic Expressions

Graphs Of Linear And Quadratic Functions

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Question 28 Report

The 3rd term of an arithmetic progression is -9 and the 7th term is -29. Find the 10th term of the progression

-44

-165

165

44

Answer Details

An arithmetic progression is a sequence of numbers where each term is obtained by adding a fixed value to the previous term. Let's call this fixed value "d". Then, the nth term of an arithmetic progression can be expressed as: an = a1 + (n-1)d where "an" is the nth term, "a1" is the first term and "n" is the position of the term. In this problem, we are given the 3rd and 7th terms, which are -9 and -29 respectively. Using the formula above, we can write two equations: a3 = a1 + 2d = -9 a7 = a1 + 6d = -29 We can solve this system of equations to find "a1" and "d". First, we can subtract the first equation from the second equation: 4d = -20 This gives us d = -5. Substituting this value of "d" into the first equation, we get: a1 + 2(-5) = -9 a1 = 1 So the first term is 1, and the common difference is -5. Now we can use the formula to find the 10th term: a10 = a1 + 9d a10 = 1 + 9(-5) a10 = -44 Therefore, the 10th term of the arithmetic progression is -44. Option A is the correct answer.

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Sequence And Series

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Question 29 Report

Cos 65o has the same value as

Sin 65o

Cos 25o

Cos 115o

Cos 205o

Cos 295o

Trigonometry

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Question 30 Report

If y = (2x + 1)3 find dy/dx

3(2x+1)2

6(2x+1)

3(2x+1)

6(2x+1)2

Answer Details

y = (2x + 1)3
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2

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Differentiation

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Question 31 Report

Find the standard deviation of 2, 3, 5 and 6

√6

√10

√

\frac{2}{5}

√

\frac{5}{2}

Answer Details

\begin{matrix} x & x - \bar{x} & (x - \bar{x})^{2} \\ 2 & - 2 & 4 \\ 3 & - 1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4 \\ \sum x = 16 & \sum (x - {\bar{x}}^{2}) = 0 \end{matrix}

___________________________________

\bar{x}

=

\frac{\sum x}{N}

=

\frac{16}{4}

= 4
S =

\sqrt{\frac{(x - \bar{x})^{2}}{N}}

=

\sqrt{\frac{(10)}{4}}

=

\sqrt{\frac{(5)}{2}}

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Measures Of Dispersion

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Question 32 Report

An arc subtends an angle of $50^\circ$ at the center of circle of radius 6cm. Calculate the area of the sector formed

100/7

80/7

110/7

90/7

Answer Details

Area of a sector =	θ	x πr2
	360

= 50/360 x 22/7 x 36/1
= 110/7

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Circles

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Question 33 Report

If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.

2y + 3x

2y - 3x

3x + 2y

3x - 2y

Answer Details

27x3 - 8y3 = (3x - 2y)3
But 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y

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Algebraic Expressions

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Question 34 Report

Simplify $ \left(\frac{16}{81}\right)^{\frac{1}{4}} \div \left(\frac{9}{16}\right)^{-\frac{1}{2}} $

$\frac{2}{3}$

\frac{1}{2}

$\frac{8}{9}$

$\frac{1}{3}$

Answer Details

$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{- \frac{1}{2}}$
$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}$
$(\frac{2^{4}}{3^{4}})^{\frac{1}{4}} \div (\frac{4^{2}}{3^{2}})^{\frac{1}{2}}$
$\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}$
$\frac{2}{3} \div \frac{4}{3}$
$\frac{2}{3} \times \frac{3}{4}$
$\frac{2}{4}$
$\frac{1}{2}$

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Number Bases

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Question 35 Report

If the area of $ \triangle PQR $ above is $12\sqrt{3}\text{ cm}^2$, find the value of q?

7cm

8cm

5cm

6cm

Answer Details

Let A denote the area of

△

PQR, then A =

\frac{1}{2} b h

Using Sin 60 $^{\circ}$ = $\frac{h}{q}$

h = q sin 60 $^{\circ}$

So A = $\frac{1}{2} b (q \sin 60^{o})$

12 $\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}$

12 $\sqrt{3}$ - 2q $\sqrt{3}$

q = $\frac{12}{2} = 6$ cm

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Question 36 Report

Find $ \int $(sin x + 2) dx.

-cosx + 2x + k

cosx + 2x + k

-cosx + x2 + k

cosx + x2 + k

Answer Details

\int

(Sin x + 2)dx = -cos x + 2x + k

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Question 37 Report

If $y = x \sin x$, Find $\frac{d^2y}{d^2x}$

2 cosx - x - x sinx

sinx + x cosx

sinx - x cosx

x sinx - 2 cosx

Answer Details

To find the second derivative of the given function, we need to differentiate it twice with respect to x. First, we differentiate y with respect to x using the product rule: y = x sin x y' = x cos x + sin x Then, we differentiate y' with respect to x using the product rule again: y' = x cos x + sin x y'' = cos x - x sin x + cos x Simplifying the expression: y'' = 2cos x - x sin x Therefore, the second derivative of y = x sin x is y'' = 2cos x - x sin x.

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Differentiation

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Question 38 Report

Solve for x and y if x - y = 2 and x2 - y2 = 8

(-1, 3)

(3, 1)

(-3, 1)

(1, 3)

Answer Details

x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x =

\frac{12}{4}

= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1

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Solution Of Linear Equations

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Question 39 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 40 Report

From the cyclic quadrilateral TUVW above, find the value of x

26

^{\circ}

24

^{\circ}

20

^{\circ}

23

^{\circ}

Answer Details

TUVW is a cyclic quad
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3χ = 24

^{\circ}

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Question 41 Report

If $ \cos \theta = \frac{12}{13} $. Find $ \theta + \cos^2 \theta $

\frac{169}{25}

\frac{25}{169}

\frac{169}{144}

\frac{144}{169}

Answer Details

Cos

θ

=

\frac{12}{13}

x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan

θ

=

\frac{5}{12}

and cos

θ

=

\frac{12}{13}

If cos2

θ

= 1 +

\frac{1}{t a n^{2} θ}

= 1 +

\frac{1}{\frac{(5)^{2}}{12}}

= 1 +

\frac{1}{\frac{25}{144}}

= 1 +

\frac{144}{25}

=

\frac{25 + 144}{25}

=

\frac{169}{25}

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Positive And Negative Integers, Rational Numbers

Angles Of Elevation And Depression

Modular Arithmetic

Fractions, Decimals, Approximations And Percentage

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Question 42 Report

Rationalise	2√3+√5
	√5-√3

(3 √ 15 +11) / 2

3 √ 15 -11

(3 √ 15 -11) / 2

3 √ 15 +11

Answer Details

To rationalize the given expression, we need to eliminate the radical from the denominator. To do that, we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of √5-√3 is √5+√3. Therefore, we have: (2√3+√5) / (√5-√3) x (√5+√3) / (√5+√3) Simplifying the numerator and the denominator using FOIL (First, Outer, Inner, Last) method, we get: = [2√3(√5) + 2√3(√3) + √5(√5) + √5(√3)] / [(√5)(√5) - (√3)(√5) + (√5)(√3) - (√3)(√3)] = [2√15 + 6 + 5 + √15] / [5 - 3 + √15 - 3] = [3√15 + 11] / 2 Therefore, the answer is (3√15 + 11) / 2.

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Surds (radicals)

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Question 43 Report

Find r, if 6r78 = 5119

3

2

6

5

Answer Details

6r78 = 5119
6 x 82 + r x 81 + 7 x 8o = 5 x 92 + 1 x 91 + 1 x 9o
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
391 + 8r = 24
r =

\frac{24}{8}

= 3

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Number Bases

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Question 44 Report

If the area of ΔPQR above is 12√3 cm2, find the value of q?

6 cm

7 cm

8 cm

5 cm

Answer Details

Area of a triangle = 1/2 ab Sinθ
12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3

q =	12√3
	2√3

q = 6

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Volumes

Triangles And Polygons

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Question 45 Report

In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

100

200

30

50

Answer Details

To determine how many ways a committee of 2 women and 3 men can be chosen from 6 men and 5 women, we can use the combination formula. The number of combinations of k objects that can be chosen from a set of n objects is given by: nCk = n! / (k! * (n - k)!) where n! denotes n factorial, which is the product of all positive integers up to n. So, in this case, the number of ways to choose 2 women from 5 is 5C2 = 5! / (2! * (5-2)!) = 10. Similarly, the number of ways to choose 3 men from 6 is 6C3 = 6! / (3! * (6-3)!) = 20. Using the multiplication principle, we can multiply these two numbers together to find the total number of ways to choose 2 women and 3 men: 10 * 20 = 200. Therefore, there are 200 ways to choose a committee of 2 women and 3 men from 6 men and 5 women. The answer is (B) 200.

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Permutation And Combination

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Question 46 Report

If $ y = (2x + 1)^3 $, find $ \frac{dy}{dx} $

6(2x + 1)

3(2x + 1)

6(2x + 1)2

2(2x + 1)2

Answer Details

If y = (2x + 1)3, then
Let u = 2x + 1 so that, y = u3

\frac{d y}{d u}

= 3u2 and

\frac{d y}{d x}

= 2
Hence by the chain rule,

\frac{d y}{d x}

=

\frac{d y}{d u}

x

\frac{d u}{d x}

= 3u2 x 2
= 6u2
= 6(2x + 1)2

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Algebraic Expressions

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Question 47 Report

Marks	1	2	3	4	5
Frequency	2	2	8	4	4

The table above show the marks obtained in a given test.

How many student too the test

16

13

20

15

Answer Details

To find the total number of students who took the test, we need to add up the frequency of all the marks. 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (C) 20.

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Read lesson note on Statistics (WAEC)

Integration

Euclidean Geometry

Statistics

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Question 48 Report

If x * y = x + y2, find then value of (2*3)*5

36

25

11

55

Answer Details

x * y = x + y2
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

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Question 49 Report

Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + .....

\frac{5}{8}

\frac{5}{7}

\frac{5}{11}

\frac{5}{9}

Answer Details

Using S

\infty

=

\frac{a}{1 - r}

r =

\frac{0.05}{0.5}

=

\frac{1}{10}

S

\infty

=

\frac{0.5}{\frac{1}{10}}

= $\frac{0.5}{(\frac{9}{10})}$
= $\frac{0.5 \times 10}{9}$
= $\frac{5}{9}$

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

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Question 50 Report

In a survey of 50 newspaper readers, 40 read Champion and 30 read Guardian, how many read both papers?

10

5

15

20

Answer Details

To find out how many people read both Champion and Guardian, we need to use a concept called "intersection" from mathematics. Out of 50 readers, 40 read Champion and 30 read Guardian. We need to find out how many people are reading both Champion and Guardian. To do this, we can draw two circles to represent the readers who read Champion and those who read Guardian. Then, we can see how much they overlap, which is the number of people who read both. So, if we draw two circles, one for Champion and one for Guardian, we can see that the overlapping region represents the people who read both newspapers. Since we don't have a visual representation, we can use a formula to find the answer. We can use the formula: Number of people who read both = Number of people who read Champion + Number of people who read Guardian - Total number of people Substituting the given values, we get: Number of people who read both = 40 + 30 - 50 Number of people who read both = 20 Therefore, the answer is 20.

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Read lesson note on Inequalities (JAMB)

Matrices And Determinants

Inequalities

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