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**Question 1**
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In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75^{o}. Find the value of y

**Question 2**
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\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\\ \hline y & 1 & 2 & 3 & 4\end{array}\).

The table is for the relation y = mx + c where m and c are constants. What is the equation of the line described in the tablet?

**Answer Details**

We can use the values in the table to find the values of m and c. Since y = mx + c, we can write: When x = 0, y = c, so c = 1 When x = 2, y = 2m + 1 When x = 4, y = 4m + 1 When x = 6, y = 6m + 1 To find the value of m, we can use the values for x = 2 and x = 4: 2m + 1 = 2 4m + 1 = 3 Solving these equations simultaneously gives m = 1/2. Substituting this value of m and c = 1 into y = mx + c gives: y = (1/2)x + 1 So the equation of the line described in the table is y = (1/2)x + 1. Therefore, the answer is (d) y = \(\frac{1}{2}x + 1\).

**Question 3**
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In the diagram, O is the centre of the circle, < SQR = 60^{o}, < SPR = y and < SOR = 3x. Find the value of (x + y)

**Question 4**
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In the diagram, < PSR = 220^{o}, < SPQ = 58^{o} and < PQR = 41^{o}. Calculate the obtuse angle QRS.

**Question 5**
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If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular

**Question 6**
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The diagram shows a rectangular cardboard from which a semi-circle is cut off. Calculate the area of the remaining part

**Question 7**
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A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x

**Answer Details**

Let's start by finding an equation that relates the length and width of the rectangle to its perimeter. We know that the perimeter of a rectangle is the sum of the lengths of all its sides, which in this case is given as 16cm. The perimeter, P = sum of all sides of rectangle = 2(length + width) Therefore, we can write the equation as: 2(x + (x - 1)) = 16 Simplifying the equation, we have: 2x + 2(x - 1) = 16 4x - 2 = 16 4x = 18 x = 4.5cm Therefore, the value of x is 4\(\frac{1}{2}\)cm, which is.

**Question 8**
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The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n

**Answer Details**

An exterior angle of a polygon is the angle between any side of the polygon and its adjacent extended side. The sum of the exterior angles of any polygon, regardless of the number of sides, is always equal to 360 degrees. The sum of the interior angles of a polygon can be found using the formula (n-2) x 180, where n is the number of sides of the polygon. So if we have an n-sided polygon, the sum of its interior angles is (n-2) x 180. According to the problem, the sum of the exterior angles is half the sum of the interior angles. Mathematically, we can express this as: Sum of exterior angles = 1/2 x sum of interior angles Substituting the formulas we have for the sum of exterior and interior angles, we get: 360 = 1/2 x (n-2) x 180 Simplifying, we get: 2 x 360 = (n-2) x 180 720 = (n-2) x 180 Dividing both sides by 180, we get: 4 = n-2 n = 6 Therefore, the number of sides of the polygon is 6.

**Question 9**
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Simplify 0.000215 x 0.000028 and express your answer in standard form

**Answer Details**

To simplify 0.000215 x 0.000028, we multiply the two numbers to obtain: 0.000215 x 0.000028 = 0.00000000602 To express this in standard form, we move the decimal point 9 places to the right, since the number is less than 1. This gives: 0.00000000602 = 6.02 x 10^{-9} Therefore, the answer is 6.02 x 10^{-9}.

**Question 10**
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The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R

**Answer Details**

Let the total age of all R men in the club be T. Then, we know that the mean age of R men is 50 years. Therefore, we have: T/R = 50 When two men aged 55 and 63 leave the club, the total age of the remaining men becomes T - 55 - 63, and the total number of remaining men becomes R - 2. We also know that the new mean age is reduced by 1 year. Therefore, we have: (T - 55 - 63)/(R - 2) = 49 Expanding this equation and substituting T/R = 50, we get: (50R - 118)/(R - 2) = 49 Multiplying both sides by (R - 2), we get: 50R - 118 = 49(R - 2) 50R - 118 = 49R - 98 R = 20 Therefore, there were originally 20 men in the club.

**Question 11**
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The nth term of the sequence -2, 4, -8, 16.... is given by

**Question 12**
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If x km/h = y m/s, then y =

**Answer Details**

To convert from km/h to m/s, we need to multiply by a conversion factor. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 1 km/h = \(\frac{1000\text{m}}{3600\text{s}}\) = \(\frac{5}{18}\) m/s. Therefore, to convert x km/h to m/s, we multiply by \(\frac{5}{18}\) to get y. So, y = \(\frac{5}{18}\)x. Therefore, option D, \(\frac{5}{18}\)x, is the correct answer.

**Question 13**
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The sum of 2 consecutive whole numbers is \(\frac{5}{6}\) of their product, find the numbers

**Answer Details**

Let the two consecutive whole numbers be x and x+1. Then, according to the problem statement: x + (x+1) = \(\frac{5}{6}\)(x(x+1)) 2x+1 = \(\frac{5}{6}\)(x² + x) Multiplying both sides by 6 to remove the fraction: 12x + 6 = 5x² + 5x 5x² - 7x - 6 = 0 Solving the quadratic equation above by factoring or using the quadratic formula, we get: x = -1 or x = 1.2 Since we're looking for consecutive whole numbers, x must be 1. Therefore, the two consecutive whole numbers are 1 and 2. So the answer is (2, 3).

**Question 14**
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Given that P = {x : 1 \(\leq x \leq 6\)}, and Q = {x : 2 \(\leq x \leq 10\)} where x is an integer. Find n(P \(\cap\) Q)

**Answer Details**

The intersection of two sets, denoted by A ∩ B, is the set of elements that are in both set A and set B. In this case, P ∩ Q represents the set of integers that are between 1 and 6, and also between 2 and 10. The smallest integer in the intersection is 2, and the largest is 6. Therefore, the set P ∩ Q is {2, 3, 4, 5, 6}. The number of elements in this set is 5. Hence, the answer is: n(P ∩ Q) = 5.

**Question 15**
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In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events?

**Question 16**
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Given that = {x: -2 < x \(\leq\) 9}, where x is an integer what is n(T)?

**Question 17**
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If sin 3y = cos 2y and 0^{o} \(\leq\) 90^{o}, find the value of y

**Answer Details**

Using the identity sin(90° - θ) = cos θ, we can write cos 2y as sin (90° - 2y), and so sin 3y = sin (90° - 2y). Since the sine function is periodic with a period of 360°, we can write: 3y = 90° - 2y + 360°k or 3y = 270° - 2y + 360°k for some integer k. Solving for y in each equation gives: 5y = 90° + 360°k or 5y = 270° + 360°k Dividing both sides by 5 gives: y = 18° + 72°k or y = 54° + 72°k Since the problem specifies that 0° ≤ y ≤ 90°, the only solution that works is y = 18°. Therefore, the answer is (a) 18°.

**Question 18**
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The subtraction below is in base seven. Find the missing number.

5 1 6 2_{seven}

-2 6 4 4_{seven}

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2 * 1 5

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**Question 19**
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The diagram is a net right rectangular pyramid. Calculate the total surface area

**Question 20**
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If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k

**Answer Details**

We can approach this problem using a well-known algebraic identity: (a + b)^2 = a^2 + 2ab + b^2 If we let a = x and b = \(\frac{4}{3}\), we can write: x^2 + kx + \(\frac{16}{9}\) = (x + \(\frac{4}{3}\))^2 Expanding the right side of the equation, we get: x^2 + 2x(\(\frac{4}{3}\)) + \(\frac{16}{9}\) = x^2 + kx + \(\frac{16}{9}\) Simplifying, we get: 2x(\(\frac{4}{3}\)) = kx k = 2(\(\frac{4}{3}\)) = \(\frac{8}{3}\) Therefore, the value of k is \(\frac{8}{3}\).

**Question 21**
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\(\begin{array}{c|c} \text{No. of pets} & 0 & 1 & 2 & 3 & 4 \\ \hline \text{No. of students} & 8 & 4 & 5 & 10 & 3\end{array}\) The table shows the number of pets kept by 30 students in a class. If a student is picked at random ftom the class. What is the probability that he/she kept more than one pet?

**Question 22**
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What is the length of a rectangular garden whose perimeter is 32cm and area 39cm^{2}?

**Answer Details**

Let the length and width of the rectangular garden be L and W, respectively. The perimeter is the sum of all sides of the rectangle, so we have: P = 2L + 2W = 32cm The area is given by: A = LW = 39cm^{2} We can use the first equation to solve for one of the variables in terms of the other. For example, solving for L we get: L = 16 - W Substituting this expression for L in the equation for the area we obtain: (16 - W)W = 39 Expanding and rearranging, we get a quadratic equation: W^{2} - 16W + 39 = 0 Solving for W using the quadratic formula, we get: W = 3 or W = 13 Since the length cannot be smaller than the width, we discard the solution W = 3. Therefore, W = 13cm, and L = 16 - W = 3cm. Therefore, the length of the rectangular garden is 13cm. So the answer is (c) 13cm.

**Question 23**
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A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?

**Answer Details**

The car uses 1 litre of petrol for every 14km, so for N63.00 (the cost of one litre of petrol), the car can travel 14km. We can use this information to find how far the car can go with N900.00 worth of petrol. To do this, we need to find how many litres of petrol N900.00 can buy. We can set up a proportion using the cost of petrol and the amount of petrol used per kilometre: 1 litre of petrol costs N63.00, so x litres cost N900.00 1 litre of petrol takes the car 14km, so x litres of petrol will take the car 14x km. Putting these together, we get: 1/63 = x/900 x = (1/63) * 900 x = 14.29 So, N900.00 can buy 14.29 litres of petrol. To find how far the car can go, we multiply the amount of petrol by the distance travelled per litre: 14.29 litres of petrol * 14km per litre = 200.06 km Therefore, the car can go approximately 200km with N900.00 worth of petrol. The closest option is 200km.

**Question 24**
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In the diagram, 0 is the centre of the circle. Find the value x

**Question 25**
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Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a

**Answer Details**

If Bola sold the article for N6,900.00 and made a profit of 15%, then the cost price of the article would be: Cost price = (100 / (100 + 15)) x N6,900.00 = N6,000.00 If Bola sold the same article for N6,600.00, he would be selling it at a loss. To find out the percentage of loss, we need to calculate the selling price that would result in a profit of 0%. Let's call the new selling price "S". We know that the cost price is N6,000.00 and the desired profit is 0%, so: S - N6,000.00 = 0 S = N6,000.00 This means that if Bola sells the article for N6,000.00, he will break even and make no profit or loss. But Bola is actually selling the article for N6,600.00, which is N600.00 more than the break-even price. To find the percentage of loss, we need to calculate what percentage of the cost price N600.00 represents: Percentage loss = (600 / 6000) x 100% = 10% Therefore, if Bola sells the article for N6,600.00, he would make a loss of 10%. So the answer is "loss of 10%".

**Question 26**
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Correct 0.002473 to 3 significant figure

**Answer Details**

To round off 0.002473 to 3 significant figures, we count the number of digits from the first non-zero digit to the third significant figure. In this case, the first non-zero digit is 2, and the third significant figure is 4. So, we count 1, 2, 3 digits from the first non-zero digit, which means the third digit is the last significant figure. The fourth digit, which is 7 in this case, is greater than or equal to 5, so we round up the last significant figure by adding 1 to it. Therefore, the final answer is 0.00247, which is the third option.

**Question 27**
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What is the value of x when y = 5?

y = \(\frac{1}{2}\) x + 1

**Answer Details**

We are given a linear equation y = (1/2)x + 1 and asked to find the value of x when y = 5. Substituting y = 5 into the equation, we get: 5 = (1/2)x + 1 Subtracting 1 from both sides, we get: 4 = (1/2)x Multiplying both sides by 2, we get: 8 = x Therefore, when y = 5, x = 8. Hence, the answer is (a) 8.

**Question 28**
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In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\)

**Answer Details**

Since triangles HKL and HIJ are similar, their corresponding sides are proportional. We know that LH and JH are corresponding sides. Therefore, the ratio \(\frac{LH}{JH}\) can be expressed as the ratio of the corresponding sides of the two triangles that form the fraction. In other words, we need to find the ratio of the side lengths of the triangles that share the same vertex as L and J, respectively. That would be the ratio of KL to JI. Thus, the answer is (a) \(\frac{KL}{JI}\).

**Question 30**
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In the diagram, GI is a tangent to the circle at H. If EF//GI, calculate the size of < EHF

**Question 31**
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If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x

**Question 32**
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The sum of 6 and one-third of x is one more than twice x, find x

**Answer Details**

The given information can be written in the form of an equation as follows: 6 + (1/3)x = 2x + 1 To solve for x, we need to isolate x on one side of the equation. First, we can subtract 6 from both sides: (1/3)x = 2x - 5 Next, we can subtract 2x from both sides: (1/3)x - 2x = -5 Simplifying the left side: (1/3 - 6/3)x = -5 -5/3x = -5 Multiplying both sides by -3/5: x = 3 Therefore, the solution is x = 3.

**Question 33**
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Simplify \(\frac{2}{2 + x} + \frac{2}{2 - x}\)

**Answer Details**

To simplify the expression, we need to find a common denominator first. The denominators are \((2+x)\) and \((2-x)\), whose product is \((2+x)(2-x)=4-x^2\). So, we can write: $$ \frac{2}{2 + x} + \frac{2}{2 - x} = \frac{2(2-x)}{(2+x)(2-x)} + \frac{2(2+x)}{(2+x)(2-x)} = \frac{4 - 2x + 4 + 2x}{4 - x^2} = \frac{8}{4 - x^2} $$ Therefore, the simplified expression is \(\frac{8}{4 - x^2}\), which is option (B).

**Question 34**
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Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

**Question 35**
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The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test?

**Answer Details**

To determine how many students passed the test, we need to calculate the total number of students who scored 50 or more. From the bar chart, we can see that the number of students who scored 50 or more in the test is the sum of the heights of the bars for the grades C, B, and A. The height of the bar for grade C is 20, the height of the bar for grade B is 30, and the height of the bar for grade A is 35. Therefore, the total number of students who scored 50 or more is 20 + 30 + 35 = 85. Hence, the answer is 85, and 85 students passed the test.

**Question 37**
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The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x

**Answer Details**

To find the range of values of x, we first need to find the mean of the numbers 2, 5, 2x, and 7, and then solve for x. The mean of the four numbers is given by: \[\frac{2 + 5 + 2x + 7}{4} = \frac{14 + 2x}{4} = \frac{7 + x}{2}\] Since the mean is less than or equal to 5, we can write: \[\frac{7 + x}{2} \leq 5\] Multiplying both sides by 2, we get: \[7 + x \leq 10\] Subtracting 7 from both sides, we get: \[x \leq 3\] Therefore, the range of values of x that satisfy the condition is: \[x \leq 3\] So the answer is: x ≤ 3.

**Question 38**
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The shaded portion in the diagram is the solution of

**Answer Details**

The shaded portion in the diagram represents the region where the values of x and y satisfy the inequality x + y < 3. This inequality represents a boundary line passing through (3, 0) and (0, 3) and the shaded region lies below this line. Therefore, the answer is x + y < 3.

**Question 40**
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In the diagram, < ROS = 66^{o} and < POQ = 3x. some construction lines are shown. Calculate the value of x.

**Question 41**
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A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting?

**Answer Details**

If 65% of the casting is Zinc, then the percentage of Copper in the casting is (100% - 65%) = 35%. Let's assume that the mass of the casting is x grams. Then, the mass of Zinc in the casting is 0.65x grams and the mass of Copper in the casting is 0.35x grams. We know that the mass of Copper in the casting is 147g, so we can write the following equation: 0.35x = 147 Solving for x, we get: x = 147 / 0.35 = 420 Therefore, the mass of the casting is 420g. Hence, the answer is 420g.

**Question 42**
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In the diagram, < WOX = 60^{o}, < YOE = 50^{o} and < OXY = 30^{o}. What is the bearing of x from y?

**Question 43**
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In the diagram, the tangent MN makes an angle of 55^{o} with the chord PS. IF O is the centre of the circle, find < RPS

**Answer Details**

Since MN is tangent to the circle, it must be perpendicular to the radius OP drawn to the point of contact. Therefore, angle MOP is 90 degrees. Let x be the measure of angle RPS. Then, angle RPO is also x degrees because RP is a radius and angles at the center are twice those at the circumference. Since MN is tangent to the circle, angle MPN is also 90 degrees. Since angle MPN and angle NPS form a linear pair (as they add up to 180 degrees) and angle MPN is 90 degrees, we have that angle NPS is 90 - x degrees. Finally, since angle MNQ and angle NPS are alternate angles, they are equal. Thus, angle MNQ is also 90 - x degrees. Now, we have a triangle MPN with angles 90, 90 - x, and 55 degrees. The angles of a triangle add up to 180 degrees, so we can write: 90 + (90 - x) + 55 = 180 Simplifying this equation gives: x = 35 Therefore, < RPS = 35 degrees.

**Question 44**
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How many times, correct to the nearest whole number, will a man run round circular track of diameter 100m to cover a distance of 1000m?

**Answer Details**

The distance covered in one complete lap around a circular track of diameter 100m is equal to the circumference of the circle. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. The diameter of the circle is 100m, so the radius is half of that, which is 50m. Thus, the circumference of the circle is C = 2π(50) = 100π m. To cover a distance of 1000m, the man needs to run around the track 1000/100π = 10/π times. To the nearest whole number, this is equal to 3. So the man will need to run around the track 3 times to cover a distance of 1000m. Therefore, the answer is 3.

**Question 45**
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Given that \(\frac{5^{n +3}}{25^{2n -2}}\) = 5^{o}, find n

**Question 46**
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The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50?

**Question 47**
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Simplify 1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)

**Question 48**
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An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water

**Answer Details**

We can begin by calculating the volume of water that has been poured into the tank. We know that the volume of a rectangular tank is given by the formula V = lwh, where l is the length, w is the width, and h is the height. In this case, we are given that l = 250cm, w = 120cm, and V = 180 litres. However, we need to convert the volume to cubic centimeters, since the dimensions of the tank are given in centimeters. To do this, we can multiply 180 litres by 1000 cubic centimeters per liter, which gives us: V = 180 x 1000 = 180000 cubic centimeters Now we can use the formula for the volume of a rectangular tank to solve for the height of the water. We rearrange the formula to get: h = V / lw Substituting the given values, we get: h = 180000 / (250 x 120) = 6 cm Therefore, the height of the water in the tank is 6cm. The answer is 6.0cm.