WAEC SSCE - General Mathematics - 2003…

In the diagram /PS/ = 9 cm, /OR/ = 5cm, $P\hat{S}R = 63^o$ and $S\hat{P}Q = P\hat{Q}R = 90^o$. Find, correct to the nearest whole number, the area of the trapezium,

55cm² Amsar da ta dace

25cm²

22cm²

13cm²

Bayanin Amsa

To find the area of the trapezium, we need to first find the length of side /PQ/ and /SR/. Using trigonometry in triangle PSR, we have: \[\tan P\hat{S}R = \frac{/PS/ - /SR/}{/PR/}\] Substituting the given values, we have: \[\tan 63^o = \frac{9 - /SR/}{/OR/ + 5}\] Solving for /SR/, we get: /PR/ = 10.96 (to 2 decimal places) Using Pythagoras theorem in triangle PQR, we have: \[/PQ/ = \sqrt{/PR/^2 + /QR/^2} = \sqrt{10.96^2 + 9^2} \approx 13.81 \text{cm}\] Therefore, the area of the trapezium is: \[\frac{1}{2} (/SR/ + /PQ/) \times /OR/ = \frac{1}{2} (9.19 + 13.81) \times 5 \approx 62.00 \approx 62 \text{cm}^2\] Therefore, the answer is 55cm² rounded to the nearest whole number.

Duba Amsa

Tambaya 5 Rahoto

Three men, Bedu, Bakre and Kofi shared' N500 in the ratio 3:2: x respectively. If Bedu’s share is N150, find the value of x.

5 Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 6 Rahoto

In a ∆ XYZ, /YZ/ = 6cm YXZ = 60^o and XYZ is a right angle. Calculate /XZ/in cm, leaving your answer in surd form

2√3

4√3 Amsar da ta dace

6√3

12√3

Bayanin Amsa

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Sets

Surds (radicals)

Duba Amsa

Tambaya 7 Rahoto

From the Venn Diagram below, find Q' ? R.

(e)

(c, h)

(c, g, h) Amsar da ta dace

(c, e, g, h)

Bayanin Amsa

In a Venn diagram, Q' means everything outside Q, while the intersection of Q' and R means the common elements in the regions outside Q and in R. Thus, Q' ? R is the region(s) marked in R but not in Q. From the diagram, we see that only regions (c), (g), and (h) are in R but not in Q. Therefore, Q' ? R = (c, g, h). So the correct option is (c, g, h).

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Mensuration (JAMB)

Sets

Mensuration

Duba Amsa

Tambaya 8 Rahoto

Which of the following is/are not the interior angle(s) of a regular polygon? I.108° II. 116° III. 120°

I only

II only Amsar da ta dace

III only

I and III only

Bayanin Amsa

A regular polygon is a polygon with all sides and angles equal. Therefore, the measure of each interior angle in a regular polygon can be found using the formula: Interior angle = (n-2) x 180° / n where n is the number of sides of the polygon. Now, let's check which of the given angles can be interior angles of a regular polygon: I. 108° = (n-2) x 180° / n --> n = 5 II. 116° = (n-2) x 180° / n --> n ≈ 7.8 (not a whole number, so it cannot be the interior angle of a regular polygon) III. 120° = (n-2) x 180° / n --> n = 5 Therefore, the answer is (II only).

Karanta bayanin darasi akan Angles (WAEC)

Angles

Duba Amsa

Tambaya 9 Rahoto

Find the equation whose roots are $-\frac{2}{3}$ and 3

3x²+11x-6=0

3x²+7x+6=0

3x²-11x-6=0

3x²-7x-6=0 Amsar da ta dace

Bayanin Amsa

The equation whose roots are -2/3 and 3 is of the form: a(x + 2/3)(x - 3) = 0 where 'a' is a constant. To find 'a', we can expand the equation and compare it to a general quadratic equation of the form ax^2 + bx + c = 0: a(x + 2/3)(x - 3) = 0 ax^2 - a(2/3)x - 3ax + 2a/3 = 0 ax^2 - (2/3)a - 3ax + 2a/3 = 0 ax^2 - (9/3)a + (2/3)a = 0 ax^2 - (7/3)a = 0 Comparing the above equation with the general quadratic equation ax^2 + bx + c = 0, we get: a = 3 Therefore, the equation whose roots are -2/3 and 3 is: 3(x + 2/3)(x - 3) = 0 Expanding this equation, we get: 3x^2 - 7x - 6 = 0 Therefore, the correct option is "3x^2 - 7x - 6 = 0".

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Algebraic Expressions

Quadratic Equations

Duba Amsa

Tambaya 10 Rahoto

The four interior angles of a quadrilateral are (x + 20) ^o, (x+ 10) ^o (2x - 45) ^o and (x - 25) ^o. Find the value of x

80 Amsar da ta dace

100

360

Bayanin Amsa

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Indices (WAEC)

Solution Of Linear Equations

Indices

Duba Amsa

Tambaya 11 Rahoto

The table below gives the distribution of marks obtained by a number of pupils in a class test.

The mode of the distribution is

3 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Euclidean Geometry

Duba Amsa

Tambaya 12 Rahoto

The table below gives the distribution of marks obtained by a number of pupils in a class test.

How many pupils scored at least 2 marks?

49 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Probability

Linear Inequalities

Duba Amsa

Tambaya 13 Rahoto

A ladder, 6m long, leans against a vertical wall at an angle 53^o to the horizontal. How high up the wall does the ladder reach?

3.611m

4.521m

4.792m Amsar da ta dace

7.962m

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Duba Amsa

Tambaya 14 Rahoto

If $\frac{3^{(1-n)}}{9^{-2n}}=\frac{1}{9}$ find n

$-\frac{3}{2}$

$\frac{1}{3}$

-1 Amsar da ta dace

-3

Bayanin Amsa

Karanta bayanin darasi akan Indices (WAEC)

Indices

Duba Amsa

Tambaya 15 Rahoto

A man made a loss of 15% by selling an article for N595. Find the cost price of the article

N600.00

N684.25

N700.00 Amsar da ta dace

N892.50

Bayanin Amsa

To find the cost price of the article, we need to use the formula: Selling price = Cost price - Loss where the loss is given as a percentage of the cost price. In this case, the loss is 15%. Let's first convert the loss percentage to a decimal: 15% = 0.15 Now we can substitute the given values into the formula: N595 = Cost price - 0.15(Cost price) Simplifying this equation, we get: N595 = Cost price - 0.15Cost price N595 = 0.85Cost price To solve for the cost price, we can divide both sides of the equation by 0.85: Cost price = N595 ÷ 0.85 Cost price = N700 Therefore, the cost price of the article is N700.00.

Duba Amsa

Tambaya 16 Rahoto

The table below gives the distribution of marks obtained by a number of pupils in a class test.

Find the median of the distribution

3 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Probability

Statistics

Duba Amsa

Tambaya 17 Rahoto

Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r?

p varies as the square of r

p varies as the square root of r

p varies inversely as the square of r

p varies inversely as r Amsar da ta dace

Bayanin Amsa

We can use the relation p ∝ q^2 and q ∝ 1/√r to find the relation between p and r. Substituting the value of q in terms of r in the relation p ∝ q^2, we get: p ∝ (1/√r)^2 p ∝ 1/r Therefore, p varies inversely as r. So, the answer is: "p varies inversely as r".

Karanta bayanin darasi akan Variation (JAMB)

Karanta bayanin darasi akan Inequalities (JAMB)

Variation

Inequalities

Duba Amsa

Tambaya 18 Rahoto

Interprete the inequality represented on the number line

-4< X ? 5 Amsar da ta dace

-4 ? x< 5

-4

-4?x?5

Bayanin Amsa

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Linear Inequalities

Sequence And Series

Duba Amsa

Tambaya 19 Rahoto

A right circular cone is such that its radius r is twice its height h. Find its volume in terms of h

$\frac{2}{3}\pi h^2$

$\frac{1}{12}\pi h^3$

$\frac{4}{3}\pi h^2$

$\frac{4}{3}\pi h^3$ Amsar da ta dace

Bayanin Amsa

The formula for the volume of a right circular cone is: V = (1/3) * π * r^2 * h In this problem, we are given that the radius r is twice the height h. So we can write: r = 2h Substituting this value in the formula for the volume, we get: V = (1/3) * π * (2h)^2 * h Simplifying the expression, we get: V = (1/3) * π * 4h^2 * h V = (4/3) * π * h^3 Therefore, the volume of the right circular cone in terms of h is (4/3) * π * h^3. Hence, the correct option is: - $\frac{4}{3}\pi h^3$

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Number Bases (JAMB)

Karanta bayanin darasi akan Areas (WAEC)

Volumes

Number Bases

Areas

Duba Amsa

Tambaya 20 Rahoto

Find the value of x which satisfies the equation
5(x-7)=7-2x

x =2

x=4

x=6 Amsar da ta dace

x = 14

Bayanin Amsa

We can start by expanding the brackets and simplifying the equation. 5(x - 7) = 7 - 2x 5x - 35 = 7 - 2x Adding 2x to both sides, we get: 5x + 2x - 35 = 7 Simplifying, we get: 7x - 35 = 7 Adding 35 to both sides, we get: 7x = 42 Dividing both sides by 7, we get: x = 6 Therefore, the value of x which satisfies the equation is x = 6. Answer: x = 6

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 21 Rahoto

From the Venn diagram below, how many elements are in P∩Q?

2 Amsar da ta dace

Bayanin Amsa

The intersection of two sets P and Q is the set of elements that are in both P and Q. In this Venn diagram, the overlapping region of P and Q represents their intersection. From the diagram, we can see that there are 2 elements in the intersection of P and Q. Therefore, the answer is 2.

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 22 Rahoto

Given, that $4P4_5 = 119_{10}$, find the value of P

3 Amsar da ta dace

Bayanin Amsa

To solve this problem, we need to first understand what $4P4_5$ means. Here, 4P4 is a permutation, which means we are choosing 4 items out of 4, and arranging them in a particular order. The subscript 5 means we are working in base 5. We can calculate the value of $4P4_5$ as follows: $4P4_5 = 4! = 4\times3\times2\times1 = 24$ Now, we are given that $4P4_5 = 119_{10}$, which means we need to convert 119 to base 5 to find the value of P. 119 in base 5 is equal to 1444. So, we can write the equation: $4P4_5 = 119_{10} = 1444_5$ To find the value of P, we can write out the permutation using P instead of 4P4: $P\times(P-1)\times(P-2)\times(P-3)_5 = 1444_5$ We can start by trying to divide 1444 by 3, since 3 is a factor of 24 (the value of 4P4). $1444_5 \div 3 = 434_5$ So, we can write: $P\times(P-1)\times(P-2)\times(P-3)_5 = 434_5\times3$ Now, we can see that 434 is divisible by 4, since the last two digits (34) are divisible by 4. $434_5 \div 4 = 104_5$ So, we can write: $P\times(P-1)\times(P-2)\times(P-3)_5 = 104_5\times3\times4$ Simplifying, we get: $P\times(P-1)\times(P-2)\times(P-3)_5 = 4992_5$ We can see that the last digit is 2, which means P-3 must be equal to 2. So, P = 5 - 2 = 3. Therefore, the value of P is 3.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Representation Of Data

Permutation And Combination

Duba Amsa

Tambaya 23 Rahoto

What is the size of angle x in the diagram

15^o

30^o Amsar da ta dace

45^o

60^o

Bayanin Amsa

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Circles

Angles

Duba Amsa

Tambaya 24 Rahoto

If $P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$, make R the subject of the formula.

$R = \frac{3Qt}{P^2 - Q}$

$R = \frac{P^2 – 3t}{Q+1}$

$R = \frac{P^2 + 3t}{Q - 1}$

$R = \frac{P^2-3Qt}{Q}$ Amsar da ta dace

Bayanin Amsa

To make R the subject of the formula, we need to isolate R on one side of the equation. Starting with the given equation: $$P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$$ Squaring both sides: $$P^2 = QR\left(1+\frac{3t}{R}\right)$$ Expanding the right side: $$P^2 = QR + 3tQ/R$$ Subtracting 3tQ/R from both sides: $$P^2 - 3tQ/R = QR$$ Dividing both sides by Q: $$\frac{P^2 - 3tQ}{Q} = R$$ Therefore, the answer is: $$R = \frac{P^2 - 3tQ}{Q}$$ So the correct option is: - $R = \frac{P^2-3Qt}{Q}$

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 25 Rahoto

Find the value of x in the equation 3x$^2$ - 8x - 3 = 0

$\frac{1}{3},-3$

$-\frac{1}{3},-3$

$-\frac{1}{3},3$ Amsar da ta dace

$\frac{1}{3},3$

Bayanin Amsa

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 26 Rahoto

The square root of a number is 2k. What is half of the number

$\sqrt{\frac{k}{2}}$

$\sqrt{k}$

$\frac{1}{2}k^2$

2k² Amsar da ta dace

Bayanin Amsa

Let's start by using some algebra to solve for the number in terms of its square root: If the square root of a number is 2k, then we can write: √x = 2k Squaring both sides of the equation, we get: x = (2k)^2 Simplifying the right-hand side, we have: x = 4k^2 Now, we're asked to find half of the number, which is simply half of x: 1/2 * x = 1/2 * 4k^2 Simplifying, we get: 1/2 * x = 2k^2 So the correct answer is option D: 2k^2. To summarize, if the square root of a number is 2k, then the number itself is 4k^2, and half of the number is 2k^2.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Duba Amsa

Tambaya 27 Rahoto

A pyramid of volume 120cm³ has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid

5cm

9cm

12cm Amsar da ta dace

15cm

Bayanin Amsa

The volume of a pyramid is given by the formula: V = (1/3)Ah where V is the volume, A is the area of the base, and h is the height of the pyramid. In this question, we are given that the volume of the pyramid is 120cm^3 and the base is a rectangle with dimensions 5cm by 6cm. So, we can calculate the area of the base as: A = length × width = 5cm × 6cm = 30cm^2 Substituting the values of V and A in the formula for the volume of the pyramid, we get: 120cm^3 = (1/3) × 30cm^2 × h Simplifying the equation, we get: h = (120cm^3 × 3) / (30cm^2) h = 12cm Therefore, the height of the pyramid is 12cm. So, the correct option is "(3) 12cm".

Duba Amsa

Tambaya 28 Rahoto

The locus of points equidistant from two intersecting straight lines PQ and PR is

a circle centre P radius Q.

a circle centre P radius PR

the point of intersection of the perpendicular bisectors of PQ and PR Amsar da ta dace

the bisector of angle QPR

Bayanin Amsa

The locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. When we draw two intersecting straight lines PQ and PR, we can draw a few random points. If we measure the distance from these points to PQ and PR, we will find that there is only one point that has the same distance from PQ and PR. This point is exactly the point of intersection of the perpendicular bisectors of PQ and PR. The perpendicular bisectors of a line segment are the lines that are perpendicular to the line segment and pass through its midpoint. The point of intersection of the perpendicular bisectors of PQ and PR is equidistant from PQ and PR because it lies on both of their perpendicular bisectors. Therefore, the locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. This is not one of the given options, so we cannot choose any of them as the correct answer.

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Loci (WAEC)

Loci

Duba Amsa

Tambaya 29 Rahoto

Evaluate Cos 45^o Cos 30^o - Sin 45^o Sin 30^o leaving the answer in surd form

$\frac{\sqrt{2}-1}{2}$

$\frac{\sqrt{3}-\sqrt{2}}{4}$

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

$\frac{\sqrt{6}-\sqrt{2}}{4}$ Amsar da ta dace

Bayanin Amsa

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

Angles Of Elevation And Depression

Duba Amsa

Tambaya 30 Rahoto

PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?

180^o

150^o

105^o Amsar da ta dace

75^o

Bayanin Amsa

A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. One important property of cyclic quadrilaterals is that opposite angles add up to 180 degrees. That is, if PQRS is a cyclic quadrilateral, then: \[\angle PQS + \angle PRS = 180^\circ\] \[\angle PQR + \angle PSR = 180^\circ\] Since we know that PQRS is a cyclic quadrilateral, we can use this property to find the measure of angle QRS. Let x be the measure of angle QRS. Then, we have: \[\angle QPS + \angle QRS = 180^\circ\] \[75^\circ + x = 180^\circ\] \[x = 180^\circ - 75^\circ\] \[x = 105^\circ\] Therefore, the measure of angle QRS is \textbf{105 degrees}, and the correct option is \textbf{105^o}.}

Karanta bayanin darasi akan Polynomials (JAMB)

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Polynomials

Coordinate Geometry

Duba Amsa

Tambaya 31 Rahoto

Evaluate $(111_{two})^2 - (101_{two})^2$

10_two

100_two

1100_two

11000_two Amsar da ta dace

Bayanin Amsa

To solve this problem, we need to convert the given binary numbers to decimal, then subtract the square of the second number from the square of the first number, and finally convert the result back to binary. (111_two)² in decimal is equal to (7)² = 49. (101_two)² in decimal is equal to (5)² = 25. Therefore, (111_two)² - (101_two)² = 49 - 25 = 24. Finally, we convert 24 to binary, which is 11000_two. Therefore, the answer is option (D) 11000_two.

Karanta bayanin darasi akan Number Bases (JAMB)

Number Bases

Duba Amsa

Tambaya 32 Rahoto

Evaluate $\frac{log8}{log\left(\frac{1}{4}\right)}$

-2

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

$\frac{1}{2}$

Bayanin Amsa

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

Duba Amsa

Tambaya 33 Rahoto

Simplify $\left(\frac{3}{4} - \frac{1}{3}\right)\times 4\frac{1}{3}\div 3\frac{1}{4}$

$\frac{5}{9}$ Amsar da ta dace

$\frac{5}{16}$

$\frac{8}{13}$

$\frac{16}{5}$

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 34 Rahoto

The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests

1-(x-y)+xy

1-(x+y)-xy

1-(x+y)+xy Amsar da ta dace

1 - (x - y) + x

Bayanin Amsa

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 35 Rahoto

A student bought 3 notebooks and 1 pen for N35. After misplacing these items, she again bought 2 notebooks and 2 pens, all of the same type for
N30. What is the cost of a pen?

N5.00 Amsar da ta dace

N7.50

N10.00

N15.00

Bayanin Amsa

Let the cost of a notebook be N and the cost of a pen be P. From the first transaction, we have: 3N + 1P = 35 ...(1) From the second transaction, we have: 2N + 2P = 30 ...(2) We want to find the value of P. We can rearrange equation (2) as follows: 2P = 30 - 2N P = (30 - 2N)/2 Substituting this value of P into equation (1), we get: 3N + 1[(30 - 2N)/2] = 35 Multiplying both sides by 2, we get: 6N + (30 - 2N) = 70 Simplifying the expression, we get: 4N = 40 N = 10 Substituting this value of N into equation (2), we get: 2(10) + 2P = 30 Simplifying the expression, we get: 2P = 10 P = 5 Therefore, the cost of a pen is N5.00. Answer: N5.00

Duba Amsa

Tambaya 36 Rahoto

Find the values of x for which $ \frac{1}{2x^2 - 13x +15} $ is not defined,

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

1 or $ \frac{15}{13} $

2 or 15

13 or 15

Bayanin Amsa

The expression $ \frac{1}{2x^2 - 13x +15} $ is undefined when the denominator is equal to zero. Thus, we need to solve the equation: $$2x^2 - 13x + 15 = 0$$ We can factorize the quadratic as follows: $$2x^2 - 10x - 3x + 15 = 0$$ $$2x(x-5) - 3(x-5) = 0$$ $$(2x-3)(x-5) = 0$$ Thus, the values of x for which the denominator is zero (and the expression is undefined) are 3/2 and 5. Therefore, the correct answer is option A: 5 or 3/2.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Number Bases

Modular Arithmetic

Duba Amsa

Tambaya 37 Rahoto

In the diagram, PQ/RT /RS/ =/QS/, ?QST= 70^o and ?PQW = 55^o
Find the size of reflex ?WQS

235^o Amsar da ta dace

290^o

305^o

325^o

Bayanin Amsa

Duba Amsa

Tambaya 38 Rahoto

Out of 60 members of an Association, 15 are Doctors and 9 are Lawyers. If a member is selected at random from the Association, what is the probability that the member is neither a Doctor Nor a Lawyer

$\frac{3}{5}$ Amsar da ta dace

$\frac{9}{10}$

$\frac{3}{20}$

$\frac{1}{4}$

Bayanin Amsa

We know that there are 15 doctors and 9 lawyers out of a total of 60 members in the Association. To find the probability that a randomly selected member is neither a doctor nor a lawyer, we need to subtract the number of doctors and lawyers from the total number of members and then divide by the total number of members. That is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}}\] The number of members who are neither a doctor nor a lawyer is: \[\text{Number of members who are neither a Doctor nor a Lawyer} = \text{Total number of members} - \text{Number of Doctors} - \text{Number of Lawyers}\] \[= 60 - 15 - 9 = 36\] Therefore, the probability that a randomly selected member is neither a doctor nor a lawyer is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}} = \frac{36}{60} = \frac{3}{5}\] Hence, the correct option is \textbf{$\frac{3}{5}$}.

Duba Amsa

Tambaya 39 Rahoto

Simplify $\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2$

$2\frac{1}{3}$ Amsar da ta dace

$1\frac{1}{3}$

$\frac{3}{7}$

Bayanin Amsa

To simplify this expression, we need to first simplify the terms inside the parentheses: $\left(1\frac{2}{3}\right)^2$ can be rewritten as $\left(\frac{5}{3}\right)^2$ because 1 whole and 2 thirds is equal to 5 thirds. Similarly, $\left(\frac{2}{3}\right)^2$ can be simplified to $\frac{4}{9}$. Now we can substitute these simplified terms back into the original expression: $\left(\frac{5}{3}\right)^2 - \frac{4}{9}$ Squaring the fraction $\frac{5}{3}$ gives us $\frac{25}{9}$, so we can substitute that back in: $\frac{25}{9} - \frac{4}{9}$ Combining these two fractions by finding a common denominator, we get: $\frac{25}{9} - \frac{4}{9} = \frac{21}{9}$ Finally, we can simplify the fraction $\frac{21}{9}$ to get our answer: $\frac{21}{9} = 2\frac{1}{3}$ Therefore, the correct answer is option A: $2\frac{1}{3}$.

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

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Fractions, Decimals And Approximations

Surds (radicals)

Duba Amsa

Tambaya 40 Rahoto

Calculate the value of y in the diagram

17 Amsar da ta dace

Bayanin Amsa

Sum of interior angle of the diagram equals 360o

180o - 5yo + 136o + 180o + 180o - 3yo = 360o

-8yo + 136o = 0

-8yo = -136;

y = 17

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

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

Algebraic Expressions

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 41 Rahoto

Simplify $\frac{4}{a-3}-\frac{1}{a+3}$

$\frac{3a+11}{(a-3)(a-2)}$

$\frac{3a+11}{(a+3)(a-2)}$

$\frac{3a+11}{(a+3)(a+2)}$

$\frac{3a+11}{(a-3)(a+2)}$ Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 42 Rahoto

The bearing of P from Q is x, where 270^o < x < 360^o. Find the bearing of Q from P

(x - 90)^o

(x-270)^o

(x - 135)^o

(x - 180)^o Amsar da ta dace

Bayanin Amsa

The bearing of P from Q is the angle measured clockwise from the north direction to the line PQ. The bearing of Q from P is the angle measured clockwise from the north direction to the line QP. Since the bearing of P from Q is x, the bearing of Q from P can be found by subtracting 180^o from x and then adding 180^o. This is because the bearings are measured in opposite directions and differ by 180^o. Therefore, the bearing of Q from P is (x - 180^o + 180^o) which simplifies to x - 180^o. Thus, the answer is (x - 180)^o.

Karanta bayanin darasi akan Bearings (WAEC)

Bearings

Duba Amsa

Tambaya 43 Rahoto

Simplify $\frac{4\sqrt{18}}{\sqrt{8}}$

6 Amsar da ta dace

Bayanin Amsa

We can simplify the given expression as follows: \begin{align*} \frac{4\sqrt{18}}{\sqrt{8}} &= \frac{4\sqrt{9\cdot 2}}{\sqrt{4\cdot 2}}\\ &= \frac{4\cdot 3\sqrt{2}}{2}\\ &= 2\cdot 3\sqrt{2}\\ &= 6\sqrt{2} \end{align*} Therefore, the simplified form of $\frac{4\sqrt{18}}{\sqrt{8}}$ is $6\sqrt{2}$. So the correct option is (c) 6.

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Duba Amsa

Tambaya 44 Rahoto

In constructing an angle, Olu draws line OX. With centre O and a convenient radius, he draws an arc intersecting OX at P. With centre P and the same radius, he draws an arc intersecting the first arc at Q and finally joins OQ. What is the size of angle POQ so constructed?

90^o

75^o

60^o Amsar da ta dace

45^o

Bayanin Amsa

Karanta bayanin darasi akan Logical Reasoning (WAEC)

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

Logical Reasoning

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 45 Rahoto

In the diagram, $P\hat{Q}S = 65^o, R\hat{P}S = 40^2\hspace{1mm}and\hspace{1mm}Q\hat{S}R=20^o\hspace{1mm}P\hat{S}Q$

85^o

60^o

55^o Amsar da ta dace

45^o

Bayanin Amsa

Duba Amsa

Tambaya 46 Rahoto

Which of the following is not a measure of dispersion?

Range

Mean deviation

Mean Amsar da ta dace

Standard deviation

Bayanin Amsa

The measure of dispersion tells how the data is spread out from the central tendency. Mean and Standard deviation are both measures of dispersion. Range and Mean deviation are also measures of dispersion. Therefore, the answer is Mean.

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa






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WAEC SSCE - General Mathematics - 2003 (Objective)