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**Question 1**
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In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects?

**Answer Details**

We can solve this problem using a Venn diagram. Let's draw two circles, one for Economics and one for Geography, with some overlap between them. We know that 65 students study Economics, so we'll put 65 in the circle for Economics. Similarly, we know that 50 students study Geography, so we'll put 50 in the circle for Geography. Now we need to figure out how many students study both subjects. We can call this number "x" for now. We'll put "x" in the overlap between the two circles. We also know that every student in the class studies either Economics or Geography or both. So we need to account for all 80 students in the class. To do this, we can add up the numbers we've put in the circles and subtract the overlap once (since the students in the overlap were counted twice). In other words, Number of students studying Economics + Number of students studying Geography - Number of students studying both = Total number of students in the class Or, 65 + 50 - x = 80 Simplifying, 115 - x = 80 x = 35 So 35 students study both Economics and Geography. Therefore, the answer is option (C) 35.

**Question 2**
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PQRS is a trapezium in which |PS| = 9cm, |QR| = 15cm, |PQ| = \(2\sqrt{3}, \angle PQR = 90^o and \angle QRS = 30^o\). Calculate the area of the trapezium

**Question 3**
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Express 0.0462 in standard form

**Question 4**
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The diagram shows a triangular prism of length 7cm. The right - angled triangle PQR is a cross section of the prism |PR| = 5cm and |RQ| = 3cm. What is the volume of the prism?

**Question 5**
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The variance of a given distribution is 25. What is the standard deviation?

**Answer Details**

The standard deviation is the square root of the variance. Therefore, to find the standard deviation when given the variance, we simply take the square root of the variance. In this case, the variance is given as 25. Taking the square root of 25 gives us: sqrt(25) = 5 So the standard deviation of the distribution is 5. To summarize, the formula for finding the standard deviation from the variance is to take the square root of the variance. In this case, the variance is 25, so the standard deviation is 5.

**Question 6**
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If \(\frac{y-3}{2}<\frac{2y-1}{3}\), which of the following is true?

**Answer Details**

To solve this inequality, we can start by simplifying both sides. First, we can multiply both sides by 6 to eliminate the denominators: 3(y - 3) < 4(2y - 1) Expanding the right side gives: 3(y - 3) < 8y - 4 Simplifying and collecting like terms, 3y - 9 < 8y - 4 Subtracting 3y from both sides, -9 < 5y - 4 Adding 4 to both sides, -5 < 5y Dividing both sides by 5, -1 < y So the inequality is true for all values of y greater than -1. Therefore, the correct answer is: y > -7.

**Question 7**
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Evaluate \((20_{three})^2 - (11_{three})^2\) in base three

**Question 8**
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The bearing S40°E is the same as

**Answer Details**

The bearing S40°E is equivalent to an angle that is 40 degrees east of the southern direction. The direction opposite to south is north, so if we draw a line going north and another line going 40 degrees to the east of south, the angle formed between these two lines is the direction that corresponds to the bearing S40°E. This angle is equal to 180° - 40° = 140°. Therefore, the correct answer is 140^{o}.

**Question 9**
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PQRS is a rhombus of side 16cm. The diagonal |QS| = 20cm. Calculate, correct to the nearest degree, ?PQR

**Question 10**
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A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that the two balls are red?

**Answer Details**

There are 12 balls in total in the box, out of which 5 are red. When the boy takes out the first ball, there are 11 balls left, out of which 4 are red. Therefore, the probability of the first ball being red is 5/12, and the probability of the second ball being red is 4/11. To find the probability of both events happening together, we multiply the probabilities: \(\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\) Therefore, the probability that the two balls taken out are both red is 5/33. Answer: \(\frac{5}{33}\)

**Question 11**
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What is the mode?

**Answer Details**

The mode is the most frequently occurring value in a set of data. In this case, the data set is {22, 23, 24, 25, 27, 30}. The value that appears most frequently in this set is 23 years, which appears twice, making it the mode. Therefore, the answer is 23.0 years.

**Question 12**
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If 2x : (x +1) = 3:2, what is the value of x?

**Answer Details**

To solve the equation 2x : (x + 1) = 3 : 2, we need to cross-multiply the ratios. That means we multiply the numerator of the left-hand side ratio by the denominator of the right-hand side ratio, and we multiply the denominator of the left-hand side ratio by the numerator of the right-hand side ratio. So we have: 2x * 2 = 3 * (x + 1) Simplifying the right-hand side: 4x = 3x + 3 Subtracting 3x from both sides: x = 3 Therefore, the value of x is 3. To check, we can substitute x = 3 back into the original equation and see if both sides are equal: 2(3) : (3 + 1) = 6 : 4 = 3 : 2 So the equation is true when x = 3, and that is our final answer.

**Question 13**
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Express 25° 45' in decimal (Hint: 1° = 60')

**Answer Details**

To convert 25° 45' to decimal, we first note that there are 60 minutes in a degree. So, to convert minutes to degrees, we divide by 60. Thus, 25° 45' = 25 + 45/60 degrees Simplifying, 25° 45' = 25 + 0.75 degrees Combining the whole and fractional parts, we get: 25° 45' = 25.75 degrees Therefore, the answer is: 25.75°.

**Question 14**
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A solid cylinder of radius 7cm is 10 cm long. Find its total surface area.

**Answer Details**

A solid cylinder has two circular bases and a curved lateral surface. To find its total surface area, we need to find the area of both circular bases and the area of the lateral surface, and then add them together. The area of one circular base is \(\pi r^2\), where r is the radius of the cylinder. Since the radius is given as 7 cm, the area of one circular base is: \begin{align*} \text{Area of one circular base} &= \pi \times (7 \text{ cm})^2 \\ &= 49 \pi \text{ cm}^2 \\ \end{align*} Since there are two circular bases, the total area of both circular bases is: \begin{align*} \text{Total area of both circular bases} &= 2 \times \text{Area of one circular base} \\ &= 2 \times 49 \pi \text{ cm}^2 \\ &= 98 \pi \text{ cm}^2 \\ \end{align*} The area of the lateral surface is the curved surface area of the cylinder, which is given by \(2\pi rh\), where r is the radius of the cylinder and h is the height (or length) of the cylinder. Since the radius is given as 7 cm and the length is given as 10 cm, the area of the lateral surface is: \begin{align*} \text{Area of lateral surface} &= 2\pi rh \\ &= 2\pi \times (7 \text{ cm}) \times (10 \text{ cm}) \\ &= 140 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is the sum of the area of both circular bases and the area of the lateral surface: \begin{align*} \text{Total surface area} &= \text{Area of both circular bases} + \text{Area of lateral surface} \\ &= 98 \pi \text{ cm}^2 + 140 \pi \text{ cm}^2 \\ &= 238 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is 238\(\pi\) cm\(^2\).

**Question 15**
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If the probability that an event will occur is p and the probability that it will not occur is q, which of the following is true?

**Answer Details**

The sum of the probabilities of an event occurring and not occurring is always equal to 1. So, if p is the probability that an event will occur, then the probability that it will not occur is 1-p (since the total probability is 1). Therefore, we have: p + (1-p) = 1 which simplifies to: p - p + 1 = 1-p + p 1 = 1 So, the correct option is: p + q = 1.

**Question 16**
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Simplify \(\frac{2x-1}{3}-\frac{x+3}{2}\)

**Answer Details**

To subtract two fractions, they need to have a common denominator. The common denominator of 3 and 2 is 6, so we can write: \begin{align*} \frac{2x-1}{3}-\frac{x+3}{2} &= \frac{2(2x-1)}{2 \cdot 3}-\frac{3(x+3)}{3 \cdot 2} \\ &= \frac{4x-2}{6}-\frac{3x+9}{6} \\ &= \frac{4x-3x-2-9}{6} \\ &= \frac{x-11}{6} \end{align*} Therefore, the answer is \(\frac{x-11}{6}\).

**Question 17**
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A sector of a circle radius 14 cm subtends an angle 135° at the center of the circle. What is the perimeter of the sector? Take \(\pi = \frac{22}{7}\)

**Answer Details**

The perimeter of a sector is given by the sum of the length of the two radii and the arc length between them. In this question, we are given the radius of the sector to be 14cm and the central angle to be 135°. The arc length can be found using the formula for the circumference of a circle, C = 2πr, where r is the radius of the circle. The central angle of 135° is equivalent to \(\frac{135}{360}\) of the full circle, so the arc length of the sector is: \(\frac{135}{360} \times 2\pi \times 14 \approx 32.91cm\) The two radii have the same length and are equal to 14cm each. Therefore, the perimeter of the sector is: 14cm + 14cm + 32.91cm ≈ 60.91cm Rounding this to the nearest whole number gives us 61cm. Hence, the correct answer is 61cm.

**Question 18**
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Subtract (-y + 3x + 5z) from (4y - x - 2z)

**Answer Details**

To subtract (-y + 3x + 5z) from (4y - x - 2z), we can just subtract the corresponding coefficients of each variable. So, - the coefficient of y is 4 - (-1) = 5 - the coefficient of x is -1 - 3 = -4 - the coefficient of z is -2 - 5 = -7 Therefore, the result is: 5y - 4x - 7z. So the correct option is (a) 5y - 4x - 7z.

**Question 19**
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A chord of length 30cm is 8cm away from the center of the circle. What is the radius of the circle. What is the perimeter of the sector? Take \(\pi =\frac{22}{7}\)

**Question 20**
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Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\)

**Question 21**
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If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24

**Question 22**
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Factorize \(6x^2 + 7x - 20\)

**Answer Details**

To factorize \(6x^2 + 7x - 20\), we need to find two numbers that multiply to -120 (the product of the leading coefficient, 6, and the constant, -20) and add up to the coefficient of the middle term, 7. These numbers are 15 and -8. Therefore, we can write: \begin{align*} 6x^2 + 7x - 20 &= 6x^2 + 15x - 8x - 20 \\ &= 3x(2x+5) - 4(2x+5) \\ &= (3x-4)(2x+5) \end{align*} So the factorization of \(6x^2 + 7x - 20\) is \((3x-4)(2x+5)\).

**Question 23**
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Which of the following is not the the size of an exterior angle of a regular polygon?

**Question 24**
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Evaluate \(log_{10}5 + log_{10}20\)

**Answer Details**

To evaluate \(log_{10}5 + log_{10}20\), we can use the logarithmic rule that states: $log_{a}x + log_{a}y = log_{a}(xy)$ Applying this rule, we get: $log_{10}5 + log_{10}20 = log_{10}(5 \times 20)$ Simplifying the right-hand side: $log_{10}(5 \times 20) = log_{10}100$ Finally, we know that \(log_{10}100 = 2\), so: \(log_{10}5 + log_{10}20 = log_{10}100 = 2\) Therefore, the answer is 2.

**Question 26**
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In the diagram, PQ is a diameter of the circle and ∠PRS = 58°. Find ∠STQ.

**Question 27**
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Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1

**Answer Details**

Substituting x = -1, we get: \[\frac{(-1)^2 + (-1) - 2}{2(-1)^2 + (-1) - 3}\] \[=\frac{1 - 1 - 2}{2 - 1 - 3}\] \[=\frac{-2}{-2}\] \[=1\] Therefore, the value of the expression is 1 when x = -1. Hence the correct option is (d) 1.

**Question 28**
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which of the following is not quadratic expression?

**Answer Details**

A quadratic expression is a polynomial of degree two. This means that the highest exponent in the expression is 2. Therefore, the expression that is not quadratic is y = 5(x-1) which is a linear expression of degree 1.

**Question 29**
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Three observation posts P,Q and R are such that Q is due east of P and R is due north of Q. If |PQ| = 5km and |PR| = 10km, find |QR|

**Question 31**
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Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p

**Answer Details**

We know that \(\sin p = \dfrac{5}{13}\) and $p$ is acute. Using Pythagorean identity, we can find $\cos p$: \begin{align*} \cos^2 p &= 1 - \sin^2 p \\ \cos^2 p &= 1 - \left(\dfrac{5}{13}\right)^2 \\ \cos^2 p &= \dfrac{144}{169} \\ \cos p &= \dfrac{12}{13} \end{align*} Using the definition of tangent, we can find $\tan p$: \begin{align*} \tan p &= \dfrac{\sin p}{\cos p} \\ \tan p &= \dfrac{\frac{5}{13}}{\frac{12}{13}} \\ \tan p &= \dfrac{5}{12} \end{align*} Therefore, \begin{align*} \cos p - \tan p &= \dfrac{12}{13} - \dfrac{5}{12} \\ &= \dfrac{144}{156} - \dfrac{65}{156} \\ &= \dfrac{79}{156} \end{align*} Hence, the answer is \(\frac{79}{156}\).

**Question 32**
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A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate

**Answer Details**

The error in the boy's estimate is the difference between what he estimated and the actual cost of the journey, which is N200 - N190 = N10. To find the percentage error, we divide the error by the actual cost and then multiply by 100%. Percentage error = (error / actual cost) x 100% = (10 / 200) x 100% = 5% Therefore, the percentage error in the boy's estimate is 5%. Option (d) is the correct answer.

**Question 33**
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Given that y = px + q and y = 5 when x = 3, while y = 4 when x = 2, find the value of p and q.

**Answer Details**

To find the values of p and q, we need to use the information given in the problem. We are given that: y = px + q When x = 3, y = 5 5 = 3p + q --- (Equation 1) When x = 2, y = 4 4 = 2p + q --- (Equation 2) We can solve these two equations simultaneously to get the values of p and q. We can do this by subtracting equation 2 from equation 1: 5 = 3p + q - (4 = 2p + q) 1 = p Substituting p = 1 into equation 2, we get: 4 = 2(1) + q 4 = 2 + q q = 2 Therefore, the values of p and q are p = 1 and q = 2. Therefore, the answer is (B) p = 1, q = 2.