WAEC SSCE - General Mathematics - 2016…

Tambaya 1 Rahoto

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 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 2 Rahoto

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

$\frac{5}{36}$

$\frac{1}{6}$ Amsar da ta dace

$\frac{5}{18}$

$\frac{2}{3}$

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 3 Rahoto

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

120^o

108^o

72^o Amsar da ta dace

60^o

Bayanin Amsa

Duba Amsa

Tambaya 4 Rahoto

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

4 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Number Bases

Sets

Modular Arithmetic

Duba Amsa

Tambaya 5 Rahoto

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

$\frac{1}{2}$ Amsar da ta dace

p - 2r

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

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

Bayanin Amsa

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}}.

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Simple Operations On Algebraic Expressions

Quadratic Equations

Duba Amsa

Tambaya 6 Rahoto

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

2 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Sets

Modular Arithmetic

Duba Amsa

Tambaya 7 Rahoto

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 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Angles Of Elevation And Depression (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Angles Of Elevation And Depression

Coordinate Geometry

Duba Amsa

Tambaya 8 Rahoto

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 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Inequalities (JAMB)

Volumes

Inequalities

Duba Amsa

Tambaya 9 Rahoto

Factorize; (2x + 3y)² - (x - 4y)²

(3x - y)(x + 7y) Amsar da ta dace

(3x + y)(2x - 7y)

(3x + y)(x - 7y)

(3x - y)(2x + 7y)

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 10 Rahoto

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

$4\sqrt{11}$

$4\sqrt{10}$

$2\sqrt{17}$

$2\sqrt{13}$ Amsar da ta dace

Bayanin Amsa

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: $2\sqrt{13}$.

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Loci

Circles

Coordinate Geometry

Duba Amsa

Tambaya 11 Rahoto

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

75.00

15.00 Amsar da ta dace

8.66

3.87

Bayanin Amsa

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Sets

Surds (radicals)

Duba Amsa

Tambaya 12 Rahoto

If log₂(3x - 1) = 5, find x.

2.00

3.67

8.67

11 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 13 Rahoto

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

- 8 Amsar da ta dace

- 4

Bayanin Amsa

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.

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Solution Of Linear Equations

Quadratic Equations

Duba Amsa

Tambaya 14 Rahoto

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

21x² - 19x - 12 = 0 Amsar da ta dace

21x² + 37x - 12 = 0

21x² - x + 12 = 0

21x² + 7x - 4 = 0

Bayanin Amsa

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 15 Rahoto

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 Amsar da ta dace

27cm²

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Indices (WAEC)

Karanta bayanin darasi akan Number Bases (JAMB)

Circles

Indices

Number Bases

Duba Amsa

Tambaya 16 Rahoto

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

135^o

105^o

75^o Amsar da ta dace

60^o

Bayanin Amsa

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.

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 17 Rahoto

Express 1975 correct to 2 significant figures

1,900

1,980

2,000 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 18 Rahoto

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 Amsar da ta dace

Joseph plays good football therefore he is a Nigerian footballer

Bayanin Amsa

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.

Karanta bayanin darasi akan Logical Reasoning (WAEC)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Logical Reasoning

Sequence And Series

Duba Amsa

Tambaya 19 Rahoto

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

D8,000.00 Amsar da ta dace

D10,000.00

D12,000.00

D16,000.00

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Representation Of Data

Modular Arithmetic

Duba Amsa

Tambaya 20 Rahoto

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

17.5

19.0 Amsar da ta dace

27.5

28.0

Bayanin Amsa

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.

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Matrices And Determinants

Measures Of Dispersion

Duba Amsa

Tambaya 21 Rahoto

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

7 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 22 Rahoto

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

13 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Linear Inequalities

Measures Of Dispersion

Duba Amsa

Tambaya 23 Rahoto

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

{f, g} Amsar da ta dace

{e}

{e, f}

{e, f, g}

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Sets (WAEC)

Number Bases

Sets

Duba Amsa

Tambaya 24 Rahoto

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 Amsar da ta dace

Bayanin Amsa

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.

Duba Amsa

Tambaya 25 Rahoto

In the diagram, $\bar{YW}$ is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY.

70^o

80^o Amsar da ta dace

105^o

110^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 26 Rahoto

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

54^o Amsar da ta dace

44^o

34^o

27^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 27 Rahoto

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 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 28 Rahoto

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

14 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Integration

Linear Inequalities

Duba Amsa

Tambaya 29 Rahoto

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

30 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 30 Rahoto

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

24 Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 31 Rahoto

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

$\frac{1}{60}$

$\frac{5}{72}$

$\frac{1}{10}$ Amsar da ta dace

1$\frac{7}{10}$

Bayanin Amsa

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 $\frac{1}{10}$.

Duba Amsa

Tambaya 32 Rahoto

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

$\frac{2}{9}$

$\frac{5}{18}$

$\frac{20}{81}$ Amsar da ta dace

$\frac{5}{9}$

Bayanin Amsa

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 $\frac{20}{81}$.

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 33 Rahoto

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 Amsar da ta dace

67^o

89^o

Bayanin Amsa

Duba Amsa

Tambaya 34 Rahoto

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 Amsar da ta dace

Le 4,500,000.00

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Percentages (WAEC)

Representation Of Data

Percentages

Duba Amsa

Tambaya 35 Rahoto

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 Amsar da ta dace

3.6cm

7.0cm

Bayanin Amsa

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.

Karanta bayanin darasi akan Mensuration (JAMB)

Karanta bayanin darasi akan Areas (WAEC)

Mensuration

Areas

Duba Amsa

Tambaya 36 Rahoto

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 = $\frac{nrp}{nr + m^2}$ Amsar da ta dace

Bayanin Amsa

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 = $\frac{nrp}{nr + m^2}$.

Karanta bayanin darasi akan Change Of Subject Of A Formula/Relation (WAEC)

Change Of Subject Of A Formula/Relation

Duba Amsa

Tambaya 37 Rahoto

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

$\frac{3}{5}$

$\frac{2}{3}$

$\frac{3}{2}$ Amsar da ta dace

$\frac{5}{3}$

Bayanin Amsa

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: $\frac{3}{2}$.

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 38 Rahoto

From the diagram, which of the following is true?

m + n + p = 180^o

m + n = 180^o

m = p + n

n = m + p Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Angles And Intercepts On Parallel Lines

Angles

Duba Amsa

Tambaya 39 Rahoto

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 Amsar da ta dace

225cm

Bayanin Amsa

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.

Karanta bayanin darasi akan Variation (WAEC)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Variation

Sequence And Series

Duba Amsa

Tambaya 40 Rahoto

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 Amsar da ta dace

-3, -7

Bayanin Amsa

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.

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 41 Rahoto

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 Amsar da ta dace

55^o

42^o

Bayanin Amsa

Duba Amsa

Tambaya 42 Rahoto

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 Amsar da ta dace

18cm

17cm

12cm

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 43 Rahoto

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

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

$\frac{3x - 1}{ x(1 - x)}$ Amsar da ta dace

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

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 44 Rahoto

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

2cm²

4cm² Amsar da ta dace

8cm²

20cm 2

Bayanin Amsa

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.

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Volumes

Areas

Duba Amsa

Tambaya 45 Rahoto

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 Amsar da ta dace

9.0m

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Algebraic Expressions

Lengths And Perimeters

Duba Amsa

Tambaya 46 Rahoto

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

3²ⁿ

9 Amsar da ta dace

3ⁿ

3^{n + 1}

Bayanin Amsa

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.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Duba Amsa

Tambaya 47 Rahoto

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

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

50 Amsar da ta dace

Bayanin Amsa

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.

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Linear Inequalities

Duba Amsa

Tambaya 48 Rahoto

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

20^o Amsar da ta dace

40^o

60^o

80^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Circles