WAEC SSCE - General Mathematics - 1994
Question 1 Report
A ladder 6m long leans against a vertical wall so that it makes an angle of 60o with the wall. Calculate the distance of the foot of the ladder from the wall
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Question 2 Report
E = (integers \(\leq\) 20), P = (multiples of 3), Q = (multiples of 4), what are the elements of P'∩Q?
Answer Details
To find P' ∩ Q, we first need to find the complement of P, which consists of all the elements in E that are not in P. Since P is the set of multiples of 3, P' is the set of all integers in E that are not multiples of 3. Therefore, P' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20} Next, we need to find the elements that are common to both P' and Q. Since Q is the set of multiples of 4, we can simply list the multiples of 4 that are in P', which are: 4, 8, 16, 20 Therefore, P' ∩ Q = {4, 8, 16, 20}. So the correct option is (2) (4, 8, 16, 20).
Question 3 Report
A cuboid of base 12.5cm by 20cm holds exactly 1 litre of water. What is the height of the cuboid? (1 litre =1000cm3)
Answer Details
The volume of the cuboid is given by the product of its length, width, and height. Let's call the height of the cuboid "h". Then, we have: Volume of cuboid = length x width x height 1 liter = 1000 cm^3 Substituting the given values, we have: 1000 cm^3 = 12.5 cm x 20 cm x h Simplifying this expression, we get: 1000 cm^3 = 250 cm x h Dividing both sides by 250 cm, we get: h = 1000 cm^3 / 250 cm h = 4 cm Therefore, the height of the cuboid is 4cm. Answer: 4cm.
Question 4 Report
If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
Answer Details
Given sin(\(\theta\)) = K, we need to find tan(\(\theta\)). We know that sin(\(\theta\)) = opposite/hypotenuse and cos(\(\theta\)) = adjacent/hypotenuse in a right-angled triangle with angle \(\theta\). Using Pythagoras theorem, we also know that hypotenuse² = opposite² + adjacent². So, let's assume a right-angled triangle with angle \(\theta\) and opposite side as K. We can find the adjacent side using Pythagoras theorem as hypotenuse² = opposite² + adjacent², which gives us adjacent = \(\sqrt{1 - K^2}\). Now, we can use the definition of tangent, which is tan(\(\theta\)) = opposite/adjacent. Thus, tan(\(\theta\)) = K/\(\sqrt{1 - K^2}\). Therefore, the correct option is \( \frac{k}{\sqrt{1 - k^2}} \).
Question 6 Report
Two ships on the equator are on longitudes 45oW and 45oE respectively. How far are they apart along the equator, correct to 2 significant figures? (Take the radius of earth = 6400km and π = 22/7)
Answer Details
The distance between the two ships along the equator is equal to the length of the arc that connects the two longitudes. We can start by finding the total length of the equator, which is the circumference of the earth, using the formula: C = 2πr where C is the circumference, r is the radius, and π is the mathematical constant pi. Substituting the given values, we get: C = 2 x (22/7) x 6400 C ≈ 40,320 km Since the two ships are on longitudes that are 90 degrees apart, the arc that connects them along the equator is one-fourth of the total circumference of the earth. Therefore, the length of the arc is: (1/4) x C = (1/4) x 40,320 ≈ 10,080 km Rounding this to two significant figures gives us: ≈ 10,000 km Therefore, the answer is, 10,000km.
Question 7 Report
The mean heights of the three groups of students consisting of 20, 16 and 14 students are 1.67m, 1.50m and 1.40m respectively. Find the mean height of the students
Question 8 Report
A house bought for N100,000 was later auctioned for N80,000. Find the loss percent.
Answer Details
The loss is calculated by finding the difference between the selling price (N80,000) and the cost price (N100,000): Loss = Cost price - Selling price = N100,000 - N80,000 = N20,000 To find the loss percentage, we divide the loss by the cost price and then multiply by 100: Loss % = (Loss/Cost price) x 100% = (N20,000/N100,000) x 100% = 20% Therefore, the loss percentage is 20%. Option A is the correct answer.
Question 9 Report
In the diagram ST and QR are parallel. |PS| = 6cm, |SQ| = 8cm and |PR] = 18 2/3. Find |PT|
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Question 10 Report
The diagram above shows a circle PQRS in which ∠PRQ = 54o and ∠SPQ = 97o. Find ∠PQS.
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Question 11 Report
Which of the following is equal to \(\frac{72}{125}\)
Answer Details
To simplify \(\frac{72}{125}\), we can factorize both numbers. The prime factorization of 72 is \(2^3 \times 3^2\) and the prime factorization of 125 is \(5^3\). Therefore, we can write: \[\frac{72}{125} = \frac{2^3 \times 3^2}{5^3}\] Hence, the answer is \( \frac{2^3 \times 3^2}{5^3} \).
Question 12 Report
Solve the equation log8x - 4log8x = 2
Answer Details
We can use the logarithmic identities to simplify the given equation: log8x - 4log8x = log8(x) - log8(x^4) = log8(x / x^4) = log8(1/x^3) Substituting this into the original equation, we have: log8(1/x^3) = 2 Rewriting in exponential form, we have: 8^2 = 1/x^3 Simplifying this expression, we get: 64 = 1/x^3 x^3 = 1/64 Taking the cube root of both sides, we get: x = 1/4 Therefore, the solution to the equation is x = 1/4. Answer: 1/4.
Question 13 Report
The length of an exercise book is given as 20cm correct to the nearest centimeter. In which of the following ranges of possible measurement does the actual length lie?
Answer Details
The length of the exercise book is given as 20cm correct to the nearest centimeter. This means the actual length of the exercise book could be 0.5cm more or less than the given measurement. Therefore, the actual length of the exercise book can lie within the range of 19.5 - 20.5cm, which is option A: 19.5 - 20.4cm. Option B: 19.1 - 20.9cm and option D: 19.95 - 20.05cm are too wide and do not take into account the fact that the measurement is correct to the nearest centimeter. Option C: 19.9 - 20.1cm is too narrow and does not allow for the possible error in measurement. Hence, the correct option is A: 19.5 - 20.4cm.
Question 14 Report
What is the number whose logarithm to base 10 is \(\bar{3}.4771\)?
Answer Details
The logarithm of a number to base 10 is the power to which 10 must be raised to obtain that number. In this case, we are given the logarithm of a number to base 10, which is \(\bar{3}.4771\). This means that the number is equal to 10 raised to the power of \(\bar{3}.4771\). Using a calculator, we can evaluate this expression to get: 10\(\bar{3}.4771\) = 0.0003441 Therefore, the number whose logarithm to base 10 is \(\bar{3}.4771\) is 0.0003441. So the answer is option (D) 0.003.
Question 16 Report
Find the probability that a number selected from the number 30 to 50 inclusive is a prime
Answer Details
We know that prime numbers are numbers that are divisible only by 1 and themselves. To solve this problem, we can simply list the prime numbers between 30 and 50, and count how many there are. The prime numbers between 30 and 50 are: 31, 37, 41, 43, and 47. There are 5 prime numbers in this range.
The total number of integers between 30 and 50 (inclusive) is 21 (50 - 30 + 1 = 21). Therefore, the probability of selecting a prime number from this range is:
5 prime numbers
--------------- = 5/21
21 numbers
So the answer is 5/21.
Question 18 Report
Which of the following equations can be solved by the points of intersection P and Q of the curve and the line PQ?
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Question 19 Report
A circle has radius x cm. What is the area of a sector of the circle with angle 135°, leaving the answer in terms of x and π?
Answer Details
The area of a sector of a circle can be found using the formula A = (θ/360)πr², where θ is the angle of the sector in degrees and r is the radius of the circle. In this case, the radius of the circle is given as x cm and the angle of the sector is 135°. Substituting these values in the formula, we get: A = (135/360)πx² Simplifying this expression, we get: A = (3/8)πx² Therefore, the area of the sector is (3/8)πx², which is.
Question 22 Report
If 8x- 4 = 6x- 10, find the value of 5x,
Answer Details
We can begin solving this equation by isolating x on one side of the equation. We do this by adding 10 to both sides and subtracting 6x from both sides: 8x - 4 = 6x - 10 8x - 6x = -10 + 4 2x = -6 x = -3 Now that we have found the value of x, we can substitute it into the expression 5x to find its value: 5x = 5(-3) = -15 Therefore, the value of 5x is -15. So, the correct option is (b) -15.
Question 23 Report
Which of the following is not a property of all parallelogram?
Answer Details
The property that is not true for all parallelograms is: "the diagonal cut at right angles". Explanation: A parallelogram is a four-sided figure with opposite sides parallel. Its diagonals (lines connecting opposite corners) bisect each other, meaning they divide each other into two equal parts. Also, each diagonal bisects the area of the parallelogram, and the opposite sides and angles are equal. However, the diagonals do not always cut at right angles. In fact, they only cut at right angles if and only if the parallelogram is a rectangle (a special type of parallelogram with all angles equal to 90 degrees). Therefore, the correct answer is "the diagonal cut at right angles".
Question 25 Report
The sum of the 1st and 2nd terms of an A.P. is 4 and the 10th term is 19. Find the sum of the 5th and 6th terms.
Answer Details
Let's start by finding the common difference of the arithmetic progression using the 1st and 2nd terms. Let the first term be "a" and the common difference be "d". The first term is "a", the second term is "a + d", and the 10th term is "a + 9d". Given that the sum of the 1st and 2nd terms is 4, we have: a + (a + d) = 4 2a + d = 4 -----(1) Also, the 10th term is 19, so: a + 9d = 19 -----(2) We now have two equations with two variables, which we can solve to find the values of "a" and "d". From equation (1), we get: d = 4 - 2a Substituting into equation (2), we get: a + 9(4 - 2a) = 19 Solving for "a", we get: a = 1 Substituting this value of "a" into equation (1), we get: 2 + d = 4 d = 2 So the common difference is 2. Now, we can find the 5th and 6th terms of the arithmetic progression: The 5th term = a + 4d = 1 + 4(2) = 9 The 6th term = a + 5d = 1 + 5(2) = 11 Therefore, the sum of the 5th and 6th terms is: 9 + 11 = 20 So the answer is 20.
Question 27 Report
Find the quadratic equation whose roots are 3 and \(\frac{2}{3}\).
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Question 28 Report
Which of the following is a point on the curve y = x\(^2\) - 4x + 7?
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Question 30 Report
The angle subtended at the centre by a chord of a circle radius 6cm is 120°. Find the length of the chord.
Answer Details
In a circle, the angle subtended by a chord at the centre of the circle is twice the angle subtended by the chord at any point on the circumference. Therefore, if the angle subtended by a chord at the centre is 120°, then the angle subtended by the chord at any point on the circumference is 60°.
Consider the triangle formed by joining the endpoints of the chord to the centre of the circle. We know that the angle at the centre of the circle is 120°, and we know that the radius of the circle is 6 cm.
Now we can use trigonometry to find the length of the chord. Let the length of the chord be 2x. Then, in the triangle we have:
\[\sin 60^\circ = \frac{x}{6}\]
\[\Rightarrow x = 6\sin 60^\circ = 6\cdot\frac{\sqrt{3}}{2} = 3\sqrt{3}\]
Therefore, the length of the chord is 2x, which is equal to:
\[2\cdot3\sqrt{3} = 6\sqrt{3}\]
Hence, the answer is (E) \(6\sqrt{3}\) cm.
Question 31 Report
Given that 2p - m = 6 and 2p + 4m = 1, find the value of (4p + 3m).
Answer Details
We can solve for the values of p and m by adding the two given equations:
2p - m = 6
2p + 4m = 1
-----------
4p + 3m = 7
So the value of (4p + 3m) is 7. Therefore, the correct option is (7).
Question 32 Report
Which of the following is not a factors of 2p\(^2\) - 2?
Answer Details
To determine which of the options is not a factor of 2p\(^2\) - 2, we can factorize the expression first. We can start by taking out a common factor of 2: 2p\(^2\) - 2 = 2(p\(^2\) - 1) We can then further factorize the expression inside the parentheses using the difference of squares: 2p\(^2\) - 2 = 2(p - 1)(p + 1) Now we can check each of the options to see if they are factors of the expression. - 2 is a factor since it is a common factor of the original expression. - p - 1 is a factor since it is one of the factors obtained from the factorization. - p + 1 is a factor since it is the other factor obtained from the factorization. - 2p - 2 is a factor since it is equivalent to 2(p - 1). - 2p + 1 is not a factor of the expression since it cannot be obtained by any combination of the factors we found earlier. Therefore, the answer is 2p + 1.
Question 33 Report
In the diagram above, PQ is the tangent to the circle RST at T.|ST| = ST and ?RTQ = 68o. Find ?PTS
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Question 34 Report
In the diagram |PS| is a diameter of circle PQRS. |PQ| = |QR| and ∠RSP = 74o. Find ∠QPS
Question 35 Report
Simplify: 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)
Answer Details
To simplify this expression, we need to use the rules of exponents: - 16\(^{\frac{5}{4}}\) can be written as (2\(^4\))\(^{\frac{5}{4}}\) since 16 is equal to 2\(^4\). - Using the power of a power rule, we can multiply the exponents 4 and 5/4 to get 2\(^5\). - 2\(^{-3}\) is the same as 1/2\(^3\) or 1/8 since we have a negative exponent. - 3\(^0\) is equal to 1 since any number raised to the power of 0 is 1. Therefore, 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\) simplifies to: (2\(^4\))\(^{\frac{5}{4}}\) x 1/8 x 1 = 2\(^5\) x 1/8 = 32/8 = 4 So the answer is 4.
Question 36 Report
Express in \( \frac{8.75}{0.025} \)standard form
Question 37 Report
IF the exterior angles of a quadilateral are yo, (2y + 5)o , (y + 15)o and (3y - 10)o, find yo
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Question 38 Report
Which of the following is a root of the equation x\(^2\) +6x = 0?
Answer Details
To find the roots of the given equation, we need to solve for x when x\(^2\) + 6x = 0. Factorizing the equation, we get x(x+6) = 0 Hence, the roots are x = 0 and x = -6. So, the correct answer is 0 or -6.
Question 39 Report
In the diagram, PQRS is a trapezium with PQ||SR. STR is a straight line, PSHQT, PTHQR. If |PQ| = 12cm and the height of the trapezium is 9cm, find its area.
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Question 40 Report
simplify; 2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16
Answer Details
We can use logarithmic rules to simplify the expression: 2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16 = log\(_{3}\) (6\(^{2}\)) + log\(_{3}\) 12 - log\(_{3}\) 16 (using the rule log\(_{b}\) (x) + log\(_{b}\) (y) = log\(_{b}\) (xy)) = log\(_{3}\) (36) + log\(_{3}\) 12 - log\(_{3}\) 16 = log\(_{3}\) (36 x 12) - log\(_{3}\) 16 (using the rule log\(_{b}\) (x) - log\(_{b}\) (y) = log\(_{b}\) (x/y)) = log\(_{3}\) (432/16) = log\(_{3}\) 27 = 3 Therefore, the answer is 3.
Question 41 Report
Evaluate \( \frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
Answer Details
To evaluate the expression, we can use the laws of exponents, which state that \(a^{m/n} = (a^m)^{1/n}\) and \(a^{-n} = \frac{1}{a^n}\). Applying these laws, we have: \(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} = \frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}} = \frac{3^{3\cdot\frac{1}{3}}}{2^{4\cdot(-\frac{1}{4})}} = \frac{3}{2^{-1}} = 3 \cdot 2 = 6\) Therefore, the answer is 6.
Question 42 Report
Two cards are drawn one after the other with replacement from a well-shuffled ordinary deck of 52 cards containing four Aces. Find the probability that they are both Aces
Answer Details
When a card is drawn from a well-shuffled deck of 52 cards, the probability of getting an Ace is 4/52 since there are 4 Aces in the deck. Since we're drawing two cards with replacement, the probability of getting two Aces is the product of the probability of getting an Ace on the first draw and the probability of getting an Ace on the second draw. So: Probability of getting two Aces = (4/52) x (4/52) = 16/2704 = 1/169 Therefore, the probability that the two cards drawn are both Aces is 1/169.
Question 43 Report
For what value of x is the expression \(\frac{x^2 + 15x + 50}{x - 5}\) not defined ?
Answer Details
The expression \(\frac{x^2 + 15x + 50}{x - 5}\) is not defined when the denominator, which is \(x - 5\), equals zero. In other words, the expression is undefined when \(x = 5\). Therefore, the value of x that makes the expression undefined is 5.
Question 44 Report
The bearing of Q from P is 122o, what is the bearing of P from Q?