JAMB UTME - Mathematics - 2007

Question 1 Report

In a basket, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random. What is the probability that the fruit is either a grape or a banana

6/30

5/30

17/30

11/30

Answer Details

To find the probability that a fruit chosen at random is either a grape or a banana, we need to first determine the total number of fruits in the basket, which is the sum of grapes and bananas, since we are only interested in these two types of fruits. Total number of grapes and bananas = 6 + 11 = 17 Therefore, the probability of choosing either a grape or a banana is: P(grape or banana) = (number of grapes + number of bananas) / total number of fruits = 17 / (6 + 11 + 13) = 17 / 30 Hence, the probability that the fruit chosen is either a grape or a banana is 17/30. Therefore, the answer is not any of the options provided.

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 2 Report

In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?

260o

130o

100o

80o

Answer Details

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50o

Also, < QSR = < QPR(angles in the segment)

< QPR = xo

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130o

Read lesson note on Trigonometry (JAMB)

Read lesson note on Circles (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

Trigonometry

Circles

Triangles And Polygons

View Answer

Question 3 Report

The volume of a hemispherical bowl is $718\frac{2}{3}\,\text{cm}^3$. Find its radius.

4.0cm

7.0cm

3.8cm

5.6cm[

π

=

\frac{22}{7}

]

Answer Details

Volume of hemisphere =

\frac{2}{3} π r^{3}

\frac{2}{3} π r^{3}

= 718

\frac{2}{3}

\frac{2}{3}

x

\frac{22}{7} r^{3}

=

\frac{2156}{3}

r3 =

\frac{2156}{3}

x

\frac{3}{2}

x

\frac{7}{22}

r3 = 49 x 7
r = 3

\sqrt{343}

= 7cm3

View Answer

Question 4 Report

Find the value of x for which 2(32x-1) = 162

5/2

3/2

2/5

1/2

Answer Details

2(32x-1) = 162
32x-1 = 162/2
32x-1 = 81
32x-1 = 32
2x - 1 = 4 (equating the indices)
2x = 5
5/2
x =

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

View Answer

Question 5 Report

$W \alpha L^2$ and $W = 6$ when $L = 4$. If $L = \sqrt{17}$, find $W$.

6

\frac{3}{8}

\frac{7}{8}

\frac{1}{8}

\frac{5}{8}

Answer Details

W

α

L2,

\frac{W}{L^{2}}

= k(constant)

\frac{6}{42}

= k

\frac{W}{(\sqrt{17})^{2}}

W =

\frac{17 \times 6}{16}

=

\frac{51}{8}

= 6

\frac{3}{8}

Read lesson note on Simple Operations On Algebraic Expressions (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Simple Operations On Algebraic Expressions

Quadratic Equations

View Answer

Question 6 Report

A binary operation $ \oplus $ on real numbers is defined by $ x \oplus y = xy + x + y $ for any two real numbers $ x $ and $ y $. The value of $ \left(-\frac{3}{4}\right) \oplus 6 $ is

-

\frac{3}{4}

\frac{45}{4}

-

\frac{4}{3}

\frac{3}{4}

Answer Details

x

\oplus

y = xy + x + y
(-

\frac{3}{4}

)

\oplus

6 = -

\frac{3}{4.6}

-

\frac{3}{4}

+ 6
= -

\frac{9}{2}

-

\frac{3}{4}

+ 6
= -

\frac{3}{4}

Read lesson note on Binary Operations (JAMB)

Binary Operations

View Answer

Question 7 Report

x10 = 12145 find x.

124

184

121

180

Answer Details

x10 = 1214 5, 1 x 53 + 2 x 52 + 1 x 51 + 4 x 5o
= 125 + 50 + 5 + 4 18410
x = 184

Read lesson note on Number Bases (WAEC)

Number Bases

View Answer

Question 8 Report

If the lines 2y - kx + 2 = 0 and y + x - k/2 = 0 Intersect at (1, -2), find the value of k

-4

-3

-2

-1

Answer Details

The problem gives two equations of two lines and a point of intersection between them. We need to find the value of "k" in one of the equations. The point of intersection (1, -2) lies on both lines, so it must satisfy both equations. Substituting x=1 and y=-2 in the first equation 2y - kx + 2 = 0 gives: 2(-2) - k(1) + 2 = 0 Simplifying this equation: -4 - k + 2 = 0 -2 - k = 0 k = -2 Therefore, the value of k is -2. Option (C) is the correct answer.

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 9 Report

The histogram above represents the weights of students who travelled out to their school for an examination. How many people made the trip.

78

28

29

69

Euclidean Geometry

Mensuration

View Answer

Question 10 Report

Make L the subjects of the formula if $ \sqrt{\frac{42w}{5l}} $

\sqrt{\frac{42 w}{5 d}}

\frac{42 W}{5 d^{2}}

\frac{42}{5 d W}

\frac{1}{d} \sqrt{\frac{42 w}{5}}

Answer Details

\sqrt{\frac{42 w}{5 l}}

Read lesson note on Change Of Subject Of A Formula/Relation (WAEC)

Read lesson note on Algebraic Fractions (WAEC)

Change Of Subject Of A Formula/Relation

Algebraic Fractions

View Answer

Question 11 Report

Given

P = {1, 3, 5, 7, 9, 11}

And Q = {2, 4, 6, 8, 1, 12}. Determine the relationship between P and Q

P∩Q = ∅

P ⊂ Q

Q⊂P

P = Q

Answer Details

The given sets P and Q are two different sets containing some numbers. To determine the relationship between the sets, we need to compare the elements of the sets. By comparing the elements of P and Q, we can see that they have only one element in common, which is 1. The other elements in the sets are unique and do not appear in both sets. Therefore, we can say that P and Q have some elements that are different and some that are the same. Hence, the correct answer is P ∩ Q ≠ ∅.

Read lesson note on Positive And Negative Integers, Rational Numbers (WAEC)

Read lesson note on Sets (WAEC)

Positive And Negative Integers, Rational Numbers

Sets

View Answer

Question 12 Report

Integrate $ \frac{x^2-\sqrt{x}}{x} $ with respect to x

\frac{x^{2}}{2} - 2 \sqrt{x} + K

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

\frac{x^{2}}{2} - \sqrt{x} + K

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

Answer Details

To integrate x² - √x/x with respect to x, we can start by factoring the expression as follows: x² - √x/x = x - 1/√x Then we can integrate each term separately: ∫(x - 1/√x) dx = ∫x dx - ∫(1/√x) dx The first integral is straightforward: ∫x dx = 1/2 x² + C1 For the second integral, we can use the substitution u = √x, du/dx = 1/(2√x), dx = 2√x du: ∫(1/√x) dx = ∫2 du = 2u + C2 = 2√x + C2 Substituting back u = √x, we get: ∫(1/√x) dx = 2√x + C2 Putting everything together, we have: ∫(x² - √x/x) dx = ∫x dx - ∫(1/√x) dx = (1/2 x² + C1) - (2√x + C2) = 1/2 x² - 2√x + C where C = C1 - C2 is the constant of integration. Therefore, the correct option is x²/2 - 2√x + K, where K = C is the constant of integration.

Read lesson note on Integration (JAMB)

Integration

View Answer

Question 13 Report

If log102 = x, express log1012.5 in terms of x

2(1 + x)

2 + 3x

2(1 - x)

2 - 3x

Answer Details

We know that log base 10 of 2 is x, which means that 10 to the power of x is equal to 2. To express log base 10 of 12.5 in terms of x, we need to find a way to write 12.5 in terms of 2 and x. We can write 12.5 as 10 to the power of 1.09691 (approximately) using a calculator. Now, we can use the laws of logarithms to simplify the expression: log base 10 of 12.5 = log base 10 of (10^1.09691) = 1.09691 * log base 10 of 10 = 1.09691 Therefore, we want to find an expression among the given options that equals 1.09691 when x is substituted into it. We can check each option by substituting x into it and simplifying: : 2(1 + x) = 2 + 2x Substituting x = log base 10 of 2 gives 2 + 2(log base 10 of 2), which does not equal 1.09691. : 2 + 3x Substituting x = log base 10 of 2 gives 2 + 3(log base 10 of 2), which also does not equal 1.09691. : 2(1 - x) = 2 - 2x Substituting x = log base 10 of 2 gives 2 - 2(log base 10 of 2), which also does not equal 1.09691. : 2 - 3x Substituting x = log base 10 of 2 gives 2 - 3(log base 10 of 2), which equals 1.09691. Therefore, the correct answer is: 2 - 3x.

Read lesson note on Logarithms (WAEC)

Logarithms

View Answer

Question 14 Report

If x10 = 12145 find x

124

121

184

180

Answer Details

x10 = 12145
= x10 = 1 x 53 + 2 x 52 + 1 * 51 + 4 x 50
= 1 x 125 + 2 x 25 + 1 x 5 + 4 x 1
= 125 + 50 + 5 + 4
= 184

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

View Answer

Question 15 Report

Factorize 2t2 + t - 15

(2t - 3)(t + 5)

(t + 3)(2t - 5)

(t + 3)(t - 5)

(2t + 3)(t - 5)

Answer Details

To factorize 2t² + t - 15, we need to find two binomials that multiply to give us 2t² + t - 15. To do this, we can use the factoring method called "AC method." First, we need to find two numbers whose product is 2(-15) = -30 and whose sum is 1. These numbers are 6 and -5. Next, we replace the middle term t with 6t - 5t: 2t² + 6t - 5t - 15 Then we group the terms: (2t² + 6t) - (5t + 15) We factor out the greatest common factor from each group: 2t(t + 3) - 5(t + 3) We notice that we have a common binomial factor of (t + 3), so we can factor it out: (t + 3)(2t - 5) Therefore, the factored form of 2t² + t - 15 is (t + 3)(2t - 5). So, the correct option is: (t + 3)(2t - 5).

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 16 Report

The area of a square is 144 sqcm. Find the length of the diagonal.

13 cm

12√2 cm

12 cm

11√3 cm

Answer Details

To find the length of the diagonal of a square, we need to use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides (legs) is equal to the square of the longest side (hypotenuse). In this case, the two legs of the right triangle are the sides of the square, and the hypotenuse is the length of the diagonal. Let's call the length of one side of the square "x". Then, we know that the area of the square is given by: x^2 = 144 sqcm Taking the square root of both sides, we get: x = 12 cm Now, using the Pythagorean theorem, we can find the length of the diagonal: d^2 = x^2 + x^2 = 2x^2 d = sqrt(2x^2) = x * sqrt(2) = 12 * sqrt(2) cm Therefore, the length of the diagonal is 12√2 cm.

Read lesson note on Number Bases (JAMB)

Read lesson note on Areas (WAEC)

Number Bases

Areas

View Answer

Question 17 Report

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

\frac{4}{5}

\frac{7}{10}

\frac{6}{6}

\frac{21}{6}

Answer Details

\frac{3}{5} \div (\frac{2}{7} \times \frac{4}{3} \div \frac{4}{9}) = \frac{3}{5} \div (\frac{2}{7} \times \frac{4}{3} \times \frac{9}{4}) = \frac{3}{5} \div \frac{6}{7} = \frac{3}{5} \times \frac{7}{6} = \frac{7}{10}

Read lesson note on Fractions, Decimals And Approximations (WAEC)

Read lesson note on Number Bases (WAEC)

Read lesson note on Sets (WAEC)

Fractions, Decimals And Approximations

Number Bases

Sets

View Answer

Question 18 Report

If y = x cos x, find dy/dx

sin x - x cos x

sin x + x cos x

cos x + x sin x

cos x - x sin x

Answer Details

To find dy/dx of y = x cos x, we can use the product rule of differentiation, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. In this case, we have: y = x cos x Using the product rule, we get: dy/dx = cos x - x sin x Therefore, the correct option is: cos x - x sin x. To explain it in simple terms, the derivative of x cos x is equal to cos x minus x times the derivative of cos x, which is -sin x. This gives us cos x - x sin x as the answer.

Read lesson note on Differentiation (JAMB)

Differentiation

View Answer

Question 19 Report

Find the sum to infinity of the series $2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \ldots$

1

2

8

4

Answer Details

a = 2 r = \frac{3}{4} S = \frac{a}{1 - r} S = \frac{2}{1 - \frac{3}{4}} = \frac{2}{\frac{1}{4}} S = \frac{2}{1} \times \frac{4}{1} = 8

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 20 Report

Integrate $ \frac{x^2-\sqrt{x}}{x} $ with respect to x

\frac{x^{2} - \sqrt{x}}{x^{2}}

+ k

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

-

\sqrt{x}

+ k

\frac{2 (x^{2} - x)}{3 x}

+ k

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

- 2

\sqrt{x}

+ k

Answer Details

Integrate

\frac{x^{2} - \sqrt{x}}{x}

with respect to x
∫

\frac{x^{2} - \sqrt{x}}{x}

dx = ∫(

\frac{x^{2}}{x} - \frac{\sqrt{x}}{x}

)dx
= ∫(x - x

\frac{1}{2}

)dx
=

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

-

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

+ k
=

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

- 2

\sqrt{x}

+ k

Read lesson note on Integration (JAMB)

Integration

View Answer

Question 21 Report

Each of the interior angles of a regular polygon is 140o. How many sides has the polygon?

9

8

7

5

Answer Details

The sum of the interior angles of a polygon can be found using the formula: S = (n - 2) × 180o where S is the sum of the interior angles, and n is the number of sides in the polygon. For a regular polygon, all interior angles have the same measure. In this case, the interior angle of the polygon is given as 140o. Therefore, we can use the formula: S = n × 140o Substituting this into the formula for the sum of interior angles, we get: n × 140o = (n - 2) × 180o Simplifying this equation, we get: 140n = 180n - 360 Solving for n, we get: 40n = 360 n = 9 Therefore, the polygon has 9 sides. The answer is 9.

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

View Answer

Question 22 Report

The nth term of the sequence 3/2, 3, 7, 16, 35, 74 ..... is

5 . 2n-2 - n

5 . 2n-2 - (n+1) / 2

3 . 2n-2

3/2 n

Answer Details

The nth term of the sequence is 5 . 2^n-2 - n. To understand why, we can look at how the sequence is generated. The first term is 3/2, the second term is 3, and each subsequent term is generated by doubling the previous term and subtracting its position in the sequence. For example, to get the third term, we double the second term (which is 3) to get 6, and then subtract the position of the term (which is 3) to get 3+3=6. Similarly, to get the fourth term, we double the third term (which is 6) to get 12, and then subtract the position of the term (which is 4) to get 12-4=8. Using this pattern, we can derive the general formula for the nth term: 5 . 2^n-2 - n.

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 23 Report

Find the value of x for which 2(32x - 1) = 162

\frac{3}{2}

\frac{1}{2}

\frac{5}{2}

\frac{2}{5}

Answer Details

2(32x - 1) = 162, 32x - 1 = 81
= 34
2x - 1 = 4 , 2x = 5
x =

\frac{5}{2}

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Areas (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Algebraic Expressions

Areas

Quadratic Equations

View Answer

Question 24 Report

The graph above is represented by

y = x3 - 4x2 + 5x - 2

y = x3 - 3x - 2

y = x3 - 4x + 2

y = x3 + 2x2 - x - 2

Answer Details

The roots of the graph are -2, -1 and 1

y = (x + 2)(x + 1)(x - 1) = (x + 2)(x2 - 1)

= x3 + 2x2 - x - 2

Read lesson note on Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

View Answer

Question 25 Report

A binary operation ⊕ on real numbers is defined by x⊕y = xy + x + y for any two real numbers x and y. The value of (-3/4)⊕6 is

3/4

-9/2

45/4

-3/4

Answer Details

The given binary operation ⊕ on real numbers is defined as x⊕y = xy + x + y for any two real numbers x and y. Substituting the values x = -3/4 and y = 6 in the given expression, we get: (-3/4)⊕6 = (-3/4)×(6) + (-3/4) + (6) = (-9/2) + (21/4) = (-18/4) + (21/4) = 3/4 Therefore, the value of (-3/4)⊕6 is 3/4. Hence, option (A) is the correct answer.

Read lesson note on Binary Operations (JAMB)

Binary Operations

View Answer

Question 26 Report

A particle p moves between points S and T such that angle SPT is always constant. Find the locus of P.

it is a straight line perpendicular to ST.

it is a quadrant of a circle with ST as diameter

it is a perpendicular bisector of ST

it is a semi-circle with ST as diameter

Answer Details

The locus of P is a semi-circle with ST as diameter.

Read lesson note on Loci (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

Loci

Triangles And Polygons

View Answer

Question 27 Report

Determine the value of $ \displaystyle \int_{0}^{\frac{\pi}{2}}(-2\cos x)\,dx $

-2

-1/2

-3

-3/2

Answer Details

\int_{0}^{\frac{π}{2}} (- 2 c o s x) d x = [- 2 s i n x + c]_{0}^{\frac{π}{2}} = (- 2 s i n \frac{π}{2} + c + 2 s i n 0 - c) = - 2 s i n 90 + c + 2 s i n 0 - c = - 2 (1) + 2 (0) = - 2

Read lesson note on Introductory Calculus (WAEC)

Read lesson note on Trigonometry (JAMB)

Introductory Calculus

Trigonometry

View Answer

Question 28 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

To find the sum to infinity of this series, we need to determine if it is a converging or diverging series. We can do this by finding the common ratio between each term. The common ratio between the second and first term is 2/3. The common ratio between the third and second term is 4/3. The common ratio between the fourth and third term is 8/9, and so on. We can see that the common ratio is less than 1, so the series is converging. Therefore, we can use the formula for the sum of an infinite geometric series: S = a/(1 - r) where S is the sum, a is the first term, and r is the common ratio. In this case, the first term is 3 and the common ratio is 2/3. So, plugging these values into the formula, we get: S = 3/(1 - 2/3) = 3/(1/3) = 9 Therefore, the sum to infinity of this series is 9. So, the answer to the question is option (D) 9.

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 29 Report

The solution of the quadratic inequality (x3 + x - 12) ≥ 0 is

x ≥ -3 or x ≤ 4

x ≥ 3 or x ≥ -4

x ≤ 3 or x v -4

x ≥ 3 or x ≤ -4

Answer Details

(x3 + x - 12) ≥ 0
(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

View Answer

Question 30 Report

Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value

\frac{2}{3}

1

-1

-

\frac{2}{3}

Answer Details

f(x) = 2x3 - x2 - 4x + 4
f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -

\frac{2}{3}

or 1
f11(x) = 12x - 2,
when x =

\frac{2}{3}

, f11(x) = 12(-

\frac{2}{3}

) - 2 = -10 < 0

$\to$ f(x) is maximum @ x = - $\frac{2}{3}$
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
$\to$ f(x) is maximum @ x = 1

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

View Answer

Question 31 Report

If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1 respectively, find the product of the modal and medial number.

36

48

38

40

Answer Details

To find the modal and medial numbers from the given frequency distribution, we need to first determine the mode and median of the dataset. The mode is the number that occurs most frequently, and the median is the middle number when the data is arranged in order. In this case, the mode is 6, which occurs with a frequency of 4. The median can be found by arranging the numbers in order: 2, 5, 5, 5, 6, 6, 6, 6, 8, 8 The median is the middle number, which is also 6. Therefore, the modal number is 6, and the medial number is also 6. The product of the modal and medial numbers is: 6 × 6 = 36 Therefore, the answer is 36.

Read lesson note on Probability (WAEC)

Read lesson note on Sets (WAEC)

Probability

Sets

View Answer

Question 32 Report

Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value

2/3

1

- 2/3

- 1

Answer Details

f(x) = 2x3 - x2 - 4x – 4
f’(x) = 6x2 - 2x – 4
As f’(x) = 0
Implies 6x2 - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -2/3
Or x + 1 = 0 implies x = -1
f’(x) = 6x2 - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0

∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1

Read lesson note on Differentiation (JAMB)

Read lesson note on Quadratic Equations (WAEC)

Differentiation

Quadratic Equations

View Answer

Question 33 Report

The graph above is represented by

y = x3 - 3x - 2

y = x3 + 2x2 - x - 2

y = x3 - 4x2 + 5x - 2

y = x3 - 4x + 2

Answer Details

x = -2, x = -1 and x = 1
then the factors; x+2, x+1 and x-1
Product of the factors; (x+2)(x+1)(x-1)
= y = (x + 2)(x2 - x + x - 1)
= y = (x+2)(x2-1)
x3 - x + 2x2 - 2 = y
x3 + 2x2 - x - 2 = y

Read lesson note on Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

View Answer

Question 34 Report

If the lines 3y = 4x - 1 and qy = x + 3 are parallel to each other, the value of q is

-

\frac{4}{3}

-

\frac{3}{4}

\frac{3}{4}

\frac{4}{3}

Answer Details

Two lines are parallel if their slopes are equal. Slope is defined as the ratio of the change in the y-coordinates to the change in the x-coordinates between two points on the line. The given lines are 3y = 4x - 1 and qy = x + 3. To find the slope of the first line, we can rearrange the equation in the slope-intercept form y = mx + b where m is the slope and b is the y-intercept. 3y = 4x - 1 y = (4/3)x - 1/3 So the slope of the first line is 4/3. To find the slope of the second line, we can rearrange the equation in the same way: qy = x + 3 y = (1/q)x + 3/q So the slope of the second line is 1/q. If the two lines are parallel, their slopes are equal. Therefore: 4/3 = 1/q We can solve for q by cross-multiplying: 4q = 3 q = 3/4 Therefore, the value of q is 3/4.

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Angles And Intercepts On Parallel Lines (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Algebraic Expressions

Angles And Intercepts On Parallel Lines

Quadratic Equations

View Answer

Question 35 Report

Find the locus of point equidistant from two straight lines y - 5 = 0 and y - 3 = 0

y - 2 = 0

y - 4 = 0

y - 1 = 0

y - 3 = 0

Answer Details

Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0

Read lesson note on Loci (JAMB)

Read lesson note on Statistics (WAEC)

Loci

Statistics

View Answer

Question 36 Report

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

x - 1

2 + 2x

1 + 2x

2x - 1

Answer Details

To find dy/dx for y = (1 + x)^2, we can use the chain rule of differentiation. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x), where f'(g(x)) is the derivative of the outer function evaluated at g(x), and g'(x) is the derivative of the inner function evaluated at x. In this case, we can let f(x) = x^2 and g(x) = 1 + x, so that y = f(g(x)) = (1 + x)^2. Then, we can find the derivatives of f(x) and g(x) as follows: f'(x) = 2x g'(x) = 1 Using the chain rule, we can now find dy/dx: dy/dx = f'(g(x)) * g'(x) = 2(1 + x) * 1 = 2 + 2x Therefore, dy/dx = 2 + 2x.

View Answer

Question 37 Report

The solution of the quadratic inequality $ (x^2 + x - 12) \ge 0 $ is

x

\geq

3 or x

\geq

-4

x

\leq

3 or x

\leq

-4

x

\geq

3 or x

\leq

-4

x

\geq

-3 or x

\leq

4

Answer Details

(x2 + x - 12)

\geq

0 , (x - 3)(x + 4)

\geq

0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3

\geq

0 and x + 4

\geq

0
or x - 3

\leq

0 and x + 4

\leq

0
when x

\geq

3, the condition is satisfied
when x

\geq

-4, the condition is not satisfied.
when x

\leq

3, the condition is not satisfied
when x

\leq

-4 , the condition is not satisfied. Thus, the solution of the inequality is x

\geq

3 or x

\leq

-4 ,

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

View Answer

Question 38 Report

The nth term of the sequence $ \frac{3}{2} $, 3, 7, 16, 35, 74, ..., is

3.2n - 2

5.2n - 2 - n

\frac{3}{2}

n

\frac{3}{2}

n - 2

Answer Details

In the given sequence, each term is obtained by adding 1 to twice the previous term, starting from 32/3, which is the same as (96/3) + (32/3) = 128/3. So, the second term of the sequence is 2(128/3) + 1 = 257/3. Similarly, the third term is 2(257/3) + 1 = 515/3. Continuing this pattern, we can find the nth term by performing the following operation: Tn = 2(Tn-1) + 1 Therefore, the nth term of the sequence is 2^n * (128/3) - (2^n + 1)/3. Thus, the correct option is 5.2n - 2 - n.

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 39 Report

The volume of a hemispherical bowl is $718\frac{2}{3}$. Find its radius .

4.0 cm

5.6 cm

7.0 cm

3.8 cm

Answer Details

Volume of bowl

\frac{2}{3} π r^{2} 718 \frac{2}{3} = \frac{2}{3} π r^{2} \frac{2156}{3} = \frac{2}{3} \times \frac{22}{7} \times r^{3} ∴ r^{3} = \frac{2156 \times 3 \times 7}{3 \times 2 \times 22} r^{3} = 343 r = \sqrt[3]{343} r = 7.0 c m

Read lesson note on Volumes (WAEC)

Volumes

View Answer

Question 40 Report

A senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered?

640

840

520

920

Answer Details

Number of itineraries = 7P4
=

\frac{7!}{(7 - 4)!} = \frac{7!}{3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840

Read lesson note on Permutation And Combination (JAMB)

Permutation And Combination

View Answer

Question 41 Report

If y = x cosx, find $ \frac{dy}{dx} $.

cos x - x sinx

sinx - x cosx

sin x + x cosx

cosx + x sin x

Answer Details

If y = x cosx, find

\frac{d y}{d x}

.
Using product rule

\frac{d y}{d x}

= cosx - x sinx

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

View Answer

Question 42 Report

The sum of the ages of Musa and Lawal is 28 years. After sharing a certain sum of money in the ratio of their ages, Musa gets ₦600 and Lawal ₦800. How old is Lawal?

14 years

20 years

12 years

16 years

Answer Details

Let the age of Musa be x, and the age of Lawal be y. According to the first statement, we know that: x + y = 28 ---- (Equation 1) Let the common ratio be k, so the ratio of Musa to Lawal's share is x:y. Therefore, we can write: x:y = k:k = 1:1 (since the ratio is not given) So, the total amount of money shared can be represented as: x + y = 2k ---- (Equation 2) From the second statement, we know that: x/y = 6/8 = 3/4 Multiplying both sides by 4y, we get: 4x = 3y Now we have two equations with two variables (Equations 1 and 2), which can be solved to find the values of x and y: x + y = 28 ---- (Equation 1) 2x = 3y ---- (Equation 3) Solving the equations 1 and 3 simultaneously, we get: x = 12 y = 16 Therefore, Lawal's age is 16 years.

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

View Answer

Question 43 Report

A man made a profit of 5% when he sold an article for N60,000.00. How much would he have sell the article to make a profit of 26%

N68,000

N72,000

N65,000

N70,000

Answer Details

5% profit = 100 + 5 = 105%
26% profit = 100 + 26 = 126%
∴ 105% → N60,000
1% → 60000/15
126% = 1000/105 x 126/1
=N72,000

Read lesson note on Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

View Answer

Question 44 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}

View Answer

Question 45 Report

If the lines $2y - kx + 2 = 0$ and $y + x - \frac{k}{2} = 0$ intersect at (1, 2), find the value of k

-2

-4

-1

-3

Answer Details

2y - kx + 2 = 0, @ x = 1, y = -2
2(-2) - k(1) + 2 = 0
-4 - k + 2 = 0
k = -2
Or y + x -

\frac{k}{2}

= 0, @ x= 1, y = -2
-2 + 1 -

\frac{k}{2}

= 0
-1 =

\frac{k}{2}

k = -1 x 2
= -2

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 46 Report

The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 1 years old?

27/40

17/20

3/30

33/40

Answer Details

P(At east 11 yrs) = P(11yrs) + P(12yrs)
= 27/40 + 7/40
= 34/40
= 17/20

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 47 Report

The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, calculate the value of y

15

13

11

8

Answer Details

To solve this problem, we need to use the formula for the mean of a frequency distribution: mean = (sum of (value × frequency))/total frequency From the table, we can see that the total frequency is 25, and the sum of (value × frequency) is: 5 × 4 + 6 × 6 + 7 × 7 + 8 × 5 + 9 × 3 = 20 + 36 + 49 + 40 + 27 = 172 So, the mean mark is: mean = 172/25 = 6.88 We also know that the total mark scored is 200, so we can set up an equation: total mark = mean × total frequency + y Substituting in the values we know, we get: 200 = 6.88 × 25 + y Solving for y, we get: y = 200 - 6.88 × 25 = 11 Therefore, the value of y is 11. Answer: 11.

Read lesson note on Integration (JAMB)

Read lesson note on Loci (WAEC)

Integration

Loci

View Answer

Question 48 Report

Marks	3	4	5	6	7	8
Frequency	5	$y - 1$	$y$	9	4	1

The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, the value of y is

11

15

9

3

Answer Details

(3 x 5) + (4(y - 1)) + (5y) + (6 x 9) + (7 x 4) + 8 = 200
15 + 4y - 4 + 5y + 54 + 28 + 8 = 200
9y + 105 - 4 = 200
9y + 101 = 200
9y = 200 - 101
9y = 99
y = 11

Read lesson note on Probability (WAEC)

Read lesson note on Statistics (WAEC)

Probability

Statistics

View Answer

Question 49 Report

If X = {all the perfect squares less than 40}

Y = {all the odd numbers fro, 1 to 15}. Find X ∩ Y.

{3, 9}

{9}

{9, 25}

{1, 9}