WAEC SSCE - General Mathematics - 2016…

Ibeere 1 Ìròyìn

Examine M' $\cap$ N from the venn diagram

{f, g} Ìdáhùn tó tọ́

{e}

{e, f}

{e, f, g}

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Number Bases

Sets

Wo Idahun

Ibeere 2 Ìròyìn

On a map, 1cm represent 5km. Find the area on the map that represents 100km².

2cm²

4cm² Ìdáhùn tó tọ́

8cm²

20cm 2

Awọn alaye Idahun

If 1cm represents 5km on the map, then x cm will represent 100km² on the map. To find x, we can use the formula for area of a square, which is A = s². In this case, we want to solve for s, where A = 100 and s represents the side length on the map in centimeters. So, s² = A s² = 100 s = √100 s = 10cm Therefore, 10cm on the map represents 100km² on the ground. To find the area on the map that represents 100km², we need to find the area of a square with a side length of 10cm. Area = s² Area = 10cm x 10cm Area = 100cm² So, the area on the map that represents 100km² is 100cm², which is equal to 1cm x 1cm, 4cm², or.

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Volumes

Areas

Wo Idahun

Ibeere 3 Ìròyìn

Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?

4.5m

6.0m

7.5m Ìdáhùn tó tọ́

9.0m

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Lengths And Perimeters (WAEC)

Algebraic Expressions

Lengths And Perimeters

Wo Idahun

Ibeere 4 Ìròyìn

Find the values of y for which the expression $\frac{y^2 - 9y + 18}{y^2 + 4y - 21}$ is undefined

6, -7

3, -6

3, -7 Ìdáhùn tó tọ́

-3, -7

Awọn alaye Idahun

The given expression will be undefined when its denominator is equal to zero since division by zero is undefined. So we need to find the values of y that make the denominator zero. \begin{align*} y^2 + 4y - 21 &= 0\\ (y+7)(y-3) &= 0 \end{align*} The denominator is equal to zero when either y+7=0 or y-3=0. Therefore, the expression is undefined when y=-7 or y=3. So, the answer is (c) 3, -7.

Ka ìwé ẹ̀kọ́ nípa Algebraic Fractions (WAEC)

Algebraic Fractions

Wo Idahun

Ibeere 5 Ìròyìn

If 20(mod 9) is equivalent to y(mod 6), find y.

2 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Sets

Modular Arithmetic

Wo Idahun

Ibeere 6 Ìròyìn

Express 1975 correct to 2 significant figures

1,900

1,980

2,000 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 7 Ìròyìn

A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?

$\frac{5}{36}$

$\frac{1}{6}$ Ìdáhùn tó tọ́

$\frac{5}{18}$

$\frac{2}{3}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 8 Ìròyìn

The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range

13 Ìdáhùn tó tọ́

Awọn alaye Idahun

The range of a set of data is the difference between the maximum and minimum values in the set. In this case, the minimum mark is 3 and the maximum mark is 16. Therefore, the range is 16 - 3 = 13. Hence, the answer is 13.

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Linear Inequalities

Measures Of Dispersion

Wo Idahun

Ibeere 9 Ìròyìn

A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ

$\frac{3}{5}$

$\frac{2}{3}$

$\frac{3}{2}$ Ìdáhùn tó tọ́

$\frac{5}{3}$

Awọn alaye Idahun

The gradient of a line is the measure of its steepness, or slope. To find the gradient of the line PQ, we need to use the formula: Gradient (m) = change in y / change in x We can calculate the change in y by subtracting the y-coordinate of point P from the y-coordinate of point Q: 8 - 2 = 6 Similarly, we can calculate the change in x by subtracting the x-coordinate of point P from the x-coordinate of point Q: 5 - 1 = 4 Therefore, the gradient of the line PQ is: Gradient (m) = change in y / change in x = 6 / 4 = 3/2 Therefore, the answer is option C: $\frac{3}{2}$.

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Wo Idahun

Ibeere 10 Ìròyìn

The volume of a pyramid with height 15cm is 90cm³. If its base is a rectangle with dimension xcm by 6cm, find the value of x

3 Ìdáhùn tó tọ́

Awọn alaye Idahun

The formula for the volume of a pyramid is given as: V = (1/3) * base_area * height Let the length of the rectangle be x, then the base area of the pyramid is given as: base_area = x * 6 From the question, we are told that the volume of the pyramid is 90cm³ and its height is 15cm. Substituting into the formula for the volume of a pyramid, we have: 90 = (1/3) * (x * 6) * 15 Multiplying both sides by 3 gives: 270 = 90 * 6x Dividing both sides by 90 gives: 3 = 2x Therefore, x = 3/2 = 1.5 Hence, the value of x is 3. Answer option A, 3, is the correct answer.

Wo Idahun

Ibeere 11 Ìròyìn

If log₂(3x - 1) = 5, find x.

2.00

3.67

8.67

11 Ìdáhùn tó tọ́

Awọn alaye Idahun

We are given that log₂(3x - 1) = 5. Using the definition of logarithms, we know that 2⁵ = 32 is equal to the expression inside the logarithm. That is, 3x - 1 = 32 Adding 1 to both sides, we get 3x = 33 Dividing by 3, we get x = 11 Therefore, the value of x is 11. Answer: 11.

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Logarithms

Wo Idahun

Ibeere 12 Ìròyìn

An object is 6m away from the base of a mast. If the angle of depression of the object from the top of the mast is 50^o, find, correct to 2 decimal places, the height of the mast.

8.60m

7.83m

7.51m

7.15m Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Angles Of Elevation And Depression (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Angles Of Elevation And Depression

Coordinate Geometry

Wo Idahun

Ibeere 13 Ìròyìn

Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?

7 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Wo Idahun

Ibeere 14 Ìròyìn

Simplify; $\frac{2}{1 - x} - \frac{1}{x}$

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

$\frac{3x - 1}{ x(1 - x)}$ Ìdáhùn tó tọ́

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

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

Awọn alaye Idahun

To simplify the expression $\frac{2}{1-x} - \frac{1}{x}$, we need to first find a common denominator. The denominator of the first fraction is $(1-x)$ and the denominator of the second fraction is $x$. The common denominator of these two fractions is $x(1-x)$. Now, we need to rewrite each fraction with this common denominator. For the first fraction, we can multiply the numerator and denominator by $x$, giving us $\frac{2x}{x(1-x)}$. For the second fraction, we can multiply the numerator and denominator by $(1-x)$, giving us $\frac{-(1-x)}{x(1-x)}$. Putting these two fractions together, we get: \[\frac{2x}{x(1-x)} - \frac{1-x}{x(1-x)} = \frac{2x - (1-x)}{x(1-x)} = \frac{3x-1}{x(1-x)}\] Therefore, the answer is $\frac{3x-1}{x(1-x)}$.

Ka ìwé ẹ̀kọ́ nípa Algebraic Fractions (WAEC)

Algebraic Fractions

Wo Idahun

Ibeere 15 Ìròyìn

If cos $\theta$ = x and sin 60^o = x + 0.5 0^o < $\theta$ < 90^o, find, correct to the nearest degree, the value of $\theta$

32^o

40^o

60^o

69^o Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Introductory Calculus

Circles

Wo Idahun

Ibeere 16 Ìròyìn

What sum of money will amount to D10,400 in 5 years at 6% simple interest?

D8,000.00 Ìdáhùn tó tọ́

D10,000.00

D12,000.00

D16,000.00

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Representation Of Data

Modular Arithmetic

Wo Idahun

Ibeere 17 Ìròyìn

Simplify: ($\frac{3}{4} - \frac{2}{3}$) x 1$\frac{1}{5}$

$\frac{1}{60}$

$\frac{5}{72}$

$\frac{1}{10}$ Ìdáhùn tó tọ́

1$\frac{7}{10}$

Awọn alaye Idahun

To simplify the expression ($\frac{3}{4} - \frac{2}{3}$) x 1$\frac{1}{5}$, we first need to evaluate the subtraction inside the parentheses: \begin{align*} \frac{3}{4} - \frac{2}{3} &= \frac{9}{12} - \frac{8}{12} \\ &= \frac{1}{12} \end{align*} So now we have: \begin{align*} (\frac{3}{4} - \frac{2}{3})\times 1\frac{1}{5} &= \frac{1}{12} \times \frac{6}{5} \\ &= \frac{1 \times 6}{12 \times 5} \\ &= \frac{1}{10} \end{align*} Therefore, the answer is $\frac{1}{10}$.

Wo Idahun

Ibeere 18 Ìròyìn

Make s the subject of the relation: P = S + $\frac{sm^2}{nr}$

s = $\frac{mrp}{nr + m^2}$

s = $\frac{nr + m^2}{mrp}$

s = $\frac{nrp}{mr + m^2}$

s = $\frac{nrp}{nr + m^2}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

To make s the subject of the relation P = S + $\frac{sm^2}{nr}$, we need to isolate s on one side of the equation. First, we can start by moving the $\frac{sm^2}{nr}$ term to the other side of the equation by subtracting it from both sides: P - $\frac{sm^2}{nr}$ = S Next, we can solve for s by multiplying both sides of the equation by $\frac{nr}{m^2}$: s = $\frac{nr}{m^2}$(P - $\frac{sm^2}{nr}$) Simplifying the right-hand side, we get: s = $\frac{nrp}{m^2}$ - $\frac{s}{m}$ To isolate s, we can add $\frac{s}{m}$ to both sides of the equation: s + $\frac{sm}{m^2}$ = $\frac{nrp}{m^2}$ Simplifying the left-hand side, we get: s($\frac{m + 1}{m^2}$) = $\frac{nrp}{m^2}$ Finally, we can solve for s by dividing both sides of the equation by $\frac{m+1}{m^2}$: s = $\frac{nrp}{nr + m^2}$ Therefore, the answer is s = $\frac{nrp}{nr + m^2}$.

Ka ìwé ẹ̀kọ́ nípa Change Of Subject Of A Formula/Relation (WAEC)

Change Of Subject Of A Formula/Relation

Wo Idahun

Ibeere 19 Ìròyìn

In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = $\sqrt{3}$cm. Find UTW

135^o

105^o

75^o Ìdáhùn tó tọ́

60^o

Awọn alaye Idahun

Using Pythagoras theorem, we can find that |UW| = 2cm. Next, we notice that $\Delta$UTW is an isosceles triangle since |TX| = |WX|. Therefore, o. Using the fact that the angles in a triangle add up to 180^o, we can find that o - o - (60^o + o. Therefore, we can substitute this into the previous equation to get: 90^o + o - (60^o + o = 2o, and since $\Delta$UTW is an isosceles triangle, we have that o. Hence, the answer is (C) 75^o.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 20 Ìròyìn

Simplify:($\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}$)²

75.00

15.00 Ìdáhùn tó tọ́

8.66

3.87

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Sets

Surds (radicals)

Wo Idahun

Ibeere 21 Ìròyìn

In the diagram, O is the centre of the circle, < QPS = 100^o, < PSQ = 60^o and < QSR. Calculate < SQR

20^o Ìdáhùn tó tọ́

40^o

60^o

80^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 22 Ìròyìn

The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?

90cm

135cm

180cm Ìdáhùn tó tọ́

225cm

Awọn alaye Idahun

The problem can be solved using the formula for direct variation: d = kt^2 where d is the distance, t is the time, and k is the constant of variation. We can solve for k using the given information that the stone falls 45cm in 3 seconds: 45 = k(3)^2 45 = 9k k = 5 Now that we know k, we can use the formula to find how far the stone will fall in 6 seconds: d = 5(6)^2 d = 5(36) d = 180cm Therefore, the answer is 180cm.

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Variation

Sequence And Series

Wo Idahun

Ibeere 23 Ìròyìn

$\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}$
Find the median age

14 Ìdáhùn tó tọ́

Awọn alaye Idahun

To find the median, we need to first arrange the ages in order from lowest to highest. Then, we can determine which age lies in the middle of the list. Arranging the ages in order of increasing magnitude, we have: $$13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17$$ There are a total of 10+24+8+5+3 = 50 ages in the list, which is an even number. To find the median, we need to take the average of the two middle ages. The two middle ages are the 25th and 26th ages in the list, which are both 14. Therefore, the median age is: $$(14 + 14)/2 = 14$$ So the correct answer is 14.

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Integration

Linear Inequalities

Wo Idahun

Ibeere 24 Ìròyìn

Which of following is a valid conclusion from the premise. "Nigeria footballers are good footballers"?

Joseph plays football in Nigeria therefore he is a good footballer

Joseph is a good footballer therefore he is a Nigerian footballer

Joseph is a Nigerian footballer therefore he is a good footballer Ìdáhùn tó tọ́

Joseph plays good football therefore he is a Nigerian footballer

Awọn alaye Idahun

The valid conclusion from the premise "Nigeria footballers are good footballers" is "Joseph is a Nigerian footballer therefore he is a good footballer". This is because the premise establishes that Nigeria footballers are good, so anyone who is a Nigeria footballer can be inferred to be a good footballer. Therefore, Joseph, who is a Nigerian footballer, can be concluded to be a good footballer. The other options are not valid because they do not follow logically from the given premise.

Ka ìwé ẹ̀kọ́ nípa Logical Reasoning (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Logical Reasoning

Sequence And Series

Wo Idahun

Ibeere 25 Ìròyìn

The relation y = x² + 2x + k passes through the point (2,0). Find the value of k

- 8 Ìdáhùn tó tọ́

- 4

Awọn alaye Idahun

We know that the relation y = x² + 2x + k passes through the point (2,0), which means that when x=2, y=0. So, substituting these values in the relation, we get: 0 = 2² + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k Therefore, k = -8. Hence, the value of k is -8.

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Wo Idahun

Ibeere 26 Ìròyìn

The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?

Le 2,250,000.00

Le 22,700,000.00

Le 3,600,000.00 Ìdáhùn tó tọ́

Le 4,500,000.00

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Representation Of Data

Percentages

Wo Idahun

Ibeere 27 Ìròyìn

A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r

$\frac{2}{3}$

3 Ìdáhùn tó tọ́

Awọn alaye Idahun

The volume of a sphere of radius r is given by the formula: $\frac{4}{3}\pi r^3$. The volume of a cylinder of radius 3cm and height 4cm is given by the formula: $\pi (3cm)^2(4cm) = 36\pi cm^3$. According to the question, the volume of the sphere is equal to the volume of the cylinder. Therefore: \[\frac{4}{3}\pi r^3 = 36\pi\] Dividing both sides by $\frac{4}{3}\pi$, we get: \[r^3 = 27\] Taking the cube root of both sides, we get: \[r = 3\] Therefore, the value of r is 3cm. Answer is correct.

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Volumes

Inequalities

Wo Idahun

Ibeere 28 Ìròyìn

If 23_x + 101_x = 130_x, find the value of x

4 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Number Bases

Sets

Modular Arithmetic

Wo Idahun

Ibeere 29 Ìròyìn

In the diagram, O is the centre of the circle, < XOZ = (10cm)^o and < XWZ = m^o. Calculate the value of m.

30 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 30 Ìròyìn

Simplify; $\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}$

3²ⁿ

9 Ìdáhùn tó tọ́

3ⁿ

3^{n + 1}

Awọn alaye Idahun

We can simplify this expression by using the properties of exponents and simplifying the numbers. First, we can write 27 as 3³ and 81 as 3⁴. Next, we can simplify the numerator by using the distributive property of exponents: \begin{align*} \frac{3^{n-1} \times 27^{n+1}}{81^n} &= \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}} \\ &= \frac{3^{n-1} \times 3^{3n+3}}{3^{4n}} \\ &= \frac{3^{4n-1}}{3^{4n}} \times 3^{3n+3} \\ &= 3^{-1} \times 3^{3n+3} \\ &= 3^{3n+2} \\ &= 3^2 \times 3^{3n} \\ &= 9 \times 3^n \\ \end{align*} Therefore, the simplified expression is 9 * 3ⁿ. So, the answer is 9.

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Wo Idahun

Ibeere 31 Ìròyìn

Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8...

13, 19, 23

9, 11, 13

11, 15, 19

13, 21, 34 Ìdáhùn tó tọ́

Awọn alaye Idahun

The given sequence is the Fibonacci sequence, where the first two terms are 0 and 1, and each subsequent term is the sum of the two preceding it. Therefore, the next three terms are: - 13 (8 + 5) - 21 (13 + 8) - 34 (21 + 13) Hence, the answer is 13, 21, 34.

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 32 Ìròyìn

Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother?

$\frac{n}{2} + \frac{5}{2}$

$\frac{n}{2}$ - 5

5 - $\frac{n}{2}$

$\frac{n}{2}$ + 5 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 33 Ìròyìn

Find the lower quartile of the distribution illustrated by the cumulative frequency curve

17.5

19.0 Ìdáhùn tó tọ́

27.5

28.0

Awọn alaye Idahun

To find the lower quartile, we need to identify the point on the cumulative frequency curve that corresponds to 25% of the total frequency. From the graph, we see that the total frequency is 40, and 25% of this is 10. The point on the curve that corresponds to a frequency of 10 is at a value of 19.0. Therefore, the lower quartile is 19.0. Note that the cumulative frequency curve shows the cumulative frequency of data values up to and including each data point on the horizontal axis. So, we can read off quartiles and other percentiles directly from the graph.

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Matrices And Determinants

Measures Of Dispersion

Wo Idahun

Ibeere 34 Ìròyìn

From the diagram, which of the following is true?

m + n + p = 180^o

m + n = 180^o

m = p + n

n = m + p Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Angles And Intercepts On Parallel Lines (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

Wo Idahun

Ibeere 35 Ìròyìn

In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140^o, find, correct to the nearest cm, the length of the chord MN.

19cm Ìdáhùn tó tọ́

18cm

17cm

12cm

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 36 Ìròyìn

The roots of a quadratic equation are $\frac{4}{3}$ and -$\frac{3}{7}$. Find the equation

21x² - 19x - 12 = 0 Ìdáhùn tó tọ́

21x² + 37x - 12 = 0

21x² - x + 12 = 0

21x² + 7x - 4 = 0

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 37 Ìròyìn

The ratio of the exterior angle to the interior angle of a regular polygon is 1:11. How many sides has the polygon?

24 Ìdáhùn tó tọ́

Awọn alaye Idahun

Wo Idahun

Ibeere 38 Ìròyìn

The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector [Take $\pi = \frac{22}{7}$]

70^o

60^o Ìdáhùn tó tọ́

55^o

42^o

Awọn alaye Idahun

Wo Idahun

Ibeere 39 Ìròyìn

Simplify $\frac{(p - r)^2 - r^2}{2p^2 - 4pr}$

$\frac{1}{2}$ Ìdáhùn tó tọ́

p - 2r

$\frac{1}{p - 2r}$

$\frac{2p}{p - 2r}$

Awọn alaye Idahun

First, let's simplify the numerator: \begin{align*} (p-r)^2 - r^2 &= (p^2 - 2pr + r^2) - r^2 \\ &= p^2 - 2pr \end{align*} Now, let's factor the denominator: \begin{align*} 2p^2 - 4pr &= 2p(p - 2r) \end{align*} Substituting these results, we get: \begin{align*} \frac{(p-r)^2 - r^2}{2p^2 - 4pr} &= \frac{p^2 - 2pr}{2p(p - 2r)} \\ &= \frac{p(p-2r)}{2p(p-2r)} \\ &= \frac{1}{2} \end{align*} Therefore, the answer is: \boxed{\frac{1}{2}}.

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Simple Operations On Algebraic Expressions

Quadratic Equations

Wo Idahun

Ibeere 40 Ìròyìn

The curve surface area of a cylinder, 5cm high is 110cm ². Find the radius of its base. [Take $\pi = \frac{22}{7}$]

2.6cm

3.5cm Ìdáhùn tó tọ́

3.6cm

7.0cm

Awọn alaye Idahun

The curved surface area of a cylinder is given by the formula: 2$\pi$rh. Given that the cylinder is 5cm high and its curved surface area is 110cm², we can write: 2$\pi$rh = 110, where h = 5 Substituting the value of $\pi = \frac{22}{7}$ and h = 5, we get: 2 x $\frac{22}{7}$ x r x 5 = 110 Simplifying this expression, we get: r = $\frac{7}{2}$ r = 3.5cm (to one decimal place) Therefore, the radius of the cylinder is approximately 3.5cm. Hence, the answer is 3.5cm.

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Mensuration

Areas

Wo Idahun

Ibeere 41 Ìròyìn

The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of $\Delta$ QNP = 6cm, calculate the area of the trapezium.

96cm²

90cm²

81cm² Ìdáhùn tó tọ́

27cm²

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Indices (WAEC)

Ka ìwé ẹ̀kọ́ nípa Number Bases (JAMB)

Circles

Indices

Number Bases

Wo Idahun

Ibeere 42 Ìròyìn

$\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}$

The table shows the ages of students in a club. How many students are in the club?

50 Ìdáhùn tó tọ́

Awọn alaye Idahun

To find out how many students are in the club, we need to add up the frequencies in the table. So, Number of 13-year-olds = 10 Number of 14-year-olds = 24 Number of 15-year-olds = 8 Number of 16-year-olds = 5 Number of 17-year-olds = 3 Total number of students = 10 + 24 + 8 + 5 + 3 = 50 Therefore, there are 50 students in the club.

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Linear Inequalities

Wo Idahun

Ibeere 43 Ìròyìn

In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48^o and < TXW = 60^o.Find < TQU.

120^o

108^o

72^o Ìdáhùn tó tọ́

60^o

Awọn alaye Idahun

Wo Idahun

Ibeere 44 Ìròyìn

In the diagram, $\bar{PF}$, $\bar{QT}$, $\bar{RG}$ intersect at S and PG||RG. If < SPQ = 113^o and < RSt = 220, find < PSQ

22^o

45^o Ìdáhùn tó tọ́

67^o

89^o

Awọn alaye Idahun

Wo Idahun

Ibeere 45 Ìròyìn

A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ

$4\sqrt{11}$

$4\sqrt{10}$

$2\sqrt{17}$

$2\sqrt{13}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

We can use the distance formula to find the length PQ, which is the distance between points P and Q on the line. The distance formula is: distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ where (x1, y1) = P and (x2, y2) = Q. Plugging in the values: distance = $\sqrt{(5-1)^2 + (8-2)^2}$ = $\sqrt{16 + 36}$ = $\sqrt{52}$ = $2\sqrt{13}$ Therefore, the length PQ is $2\sqrt{13}$. Answer: $2\sqrt{13}$.

Ka ìwé ẹ̀kọ́ nípa Loci (JAMB)

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Loci

Circles

Coordinate Geometry

Wo Idahun

Ibeere 46 Ìròyìn

A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue

$\frac{2}{9}$

$\frac{5}{18}$

$\frac{20}{81}$ Ìdáhùn tó tọ́

$\frac{5}{9}$

Awọn alaye Idahun

When two balls are drawn from the bag with replacement, there are a total of $9\times9=81$ possible outcomes, since there are 9 balls in the bag and we are replacing each ball after drawing. To find the probability that the first ball is red and the second ball is blue, we can use the multiplication rule of probability. The probability that the first ball is red is $\frac{5}{9}$, since there are 5 red balls out of 9 total balls in the bag. After replacing the first ball, there are still 9 balls in the bag, but now 4 of them are blue. So the probability that the second ball is blue, given that the first ball was red, is $\frac{4}{9}$. Therefore, the probability that the first ball is red and the second ball is blue is: $$\frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$ Hence the answer is $\frac{20}{81}$.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 47 Ìròyìn

In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177^o. Calculate < PST.

54^o Ìdáhùn tó tọ́

44^o

34^o

27^o

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles