WAEC SSCE - General Mathematics - 1994
Tambaya 1 Rahoto
Which of the following equations can be solved by the points of intersection P and Q of the curve and the line PQ?
Bayanin Amsa
Tambaya 3 Rahoto
A house bought for N100,000 was later auctioned for N80,000. Find the loss percent.
Bayanin Amsa
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.
Tambaya 4 Rahoto
Solve the equation log8x - 4log8x = 2
Bayanin Amsa
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.
Tambaya 5 Rahoto
Calculate, correct to 2 significant figures, the length of the arc of a circle of radius 3.5cm which subtends an angle of 75° at the centre of the circle. [Take π = 22/7].
Bayanin Amsa
The formula for the length of an arc of a circle is given by:
L = (θ/360) x 2πr
where θ is the angle subtended at the centre of the circle in degrees, r is the radius of the circle, and π is pi.
Using the given values, we have:
θ = 75° r = 3.5cm π = 22/7
Substituting these values into the formula, we get:
L = (75/360) x (2 x 22/7 x 3.5) L = (5/24) x (44/7) L = 110/24 L ≈ 4.58cm (correct to 2 significant figures)
Therefore, the length of the arc of the circle is approximately 4.58cm. The correct option is (b) 4.6cm.
Tambaya 7 Rahoto
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
Tambaya 8 Rahoto
Find the probability that a number selected from the number 30 to 50 inclusive is a prime
Bayanin Amsa
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.
Tambaya 9 Rahoto
The bearing of Q from P is 122o, what is the bearing of P from Q?
Bayanin Amsa
If the bearing of Q from P is 122o, then the bearing of P from Q is 180o + 122o = 302o. This is because bearings are measured clockwise from north and the bearing of Q from P is 122o which is to the south-west of P. Therefore, the bearing of P from Q will be in the opposite direction, which is to the north-east, and we add 180o to 122o to get the bearing of P from Q, which is 302o.
Tambaya 10 Rahoto
Two solid spheres have volumes 250cmo and 128cmo respectively. Find the ratio of their radii.
Bayanin Amsa
The ratio of the volumes of two spheres is equal to the cube of the ratio of their radii. Therefore, (volume of first sphere) / (volume of second sphere) = (radius of first sphere)3 / (radius of second sphere)3 Given the volumes of the spheres as 250 cm3 and 128 cm3, we have: 250 / 128 = (radius of first sphere)3 / (radius of second sphere)3 Simplifying this equation, we get: (radius of first sphere) / (radius of second sphere) = (250 / 128)1/3 Using a calculator, we can evaluate the cube root of 250/128 to be approximately 1.25. Therefore, the ratio of the radii of the two spheres is: (radius of first sphere) / (radius of second sphere) = 1.25 Simplifying this ratio, we can express it as 5:4. Therefore, the correct answer is 5:4.
Tambaya 12 Rahoto
A string is 4.8m. A boy measured it to be 4.95m. Find the percentage error.
Bayanin Amsa
Tambaya 14 Rahoto
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?
Bayanin Amsa
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.
Tambaya 16 Rahoto
Factorise: 6x\(^2\) + 7xy - 5y\(^2\)
Bayanin Amsa
To factorize the expression 6x\(^2\) + 7xy - 5y\(^2\), we need to find two binomials whose product equals the expression. First, we need to find two numbers that multiply to 6, the coefficient of the x\(^2\) term, and two numbers that multiply to -5, the coefficient of the y\(^2\) term. We also need to find the signs of the two numbers that add up to 7, the coefficient of the xy term. After some trial and error, we can determine that the factors are (3x - y) and (2x + 5y). Multiplying these two binomials using FOIL method, we get: (3x - y)(2x + 5y) = 6x\(^2\) + 7xy - 5y\(^2\) Therefore, the correct answer is option (D): (3x + 5y)(2x - y).
Tambaya 17 Rahoto
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
Tambaya 18 Rahoto
Which of the following is a root of the equation x\(^2\) +6x = 0?
Bayanin Amsa
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.
Tambaya 20 Rahoto
The diagram above shows a circle PQRS in which ∠PRQ = 54o and ∠SPQ = 97o. Find ∠PQS.
Bayanin Amsa
Tambaya 21 Rahoto
Evaluate \( \frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
Bayanin Amsa
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.
Tambaya 23 Rahoto
Given that 2p - m = 6 and 2p + 4m = 1, find the value of (4p + 3m).
Bayanin Amsa
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).
Tambaya 25 Rahoto
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?
Bayanin Amsa
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.
Tambaya 26 Rahoto
simplify; 2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16
Bayanin Amsa
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.
Tambaya 27 Rahoto
For what value of x is the expression \(\frac{x^2 + 15x + 50}{x - 5}\) not defined ?
Bayanin Amsa
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.
Tambaya 28 Rahoto
Which of the following is not a factors of 2p\(^2\) - 2?
Bayanin Amsa
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.
Tambaya 29 Rahoto
The angle subtended at the centre by a chord of a circle radius 6cm is 120°. Find the length of the chord.
Bayanin Amsa
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.
Tambaya 30 Rahoto
Simplify: 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)
Bayanin Amsa
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.
Tambaya 32 Rahoto
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.
Bayanin Amsa
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.
Tambaya 34 Rahoto
A cuboid of base 12.5cm by 20cm holds exactly 1 litre of water. What is the height of the cuboid? (1 litre =1000cm3)
Bayanin Amsa
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.
Tambaya 35 Rahoto
In the diagram |PS| is a diameter of circle PQRS. |PQ| = |QR| and ∠RSP = 74o. Find ∠QPS
Tambaya 36 Rahoto
IF the exterior angles of a quadilateral are yo, (2y + 5)o , (y + 15)o and (3y - 10)o, find yo
Bayanin Amsa
Tambaya 37 Rahoto
The sum of the angles of regular polygon is 2520o. How many sides does the polygon have?
Bayanin Amsa
The sum of the interior angles of any polygon can be found using the formula: (n-2) x 180, where n is the number of sides of the polygon. For a regular polygon, all interior angles are equal, so we can divide the sum of the angles by the number of angles to find the measure of each interior angle. In this case, the sum of the angles is 2520 degrees. Let's plug this value into the formula: (n-2) x 180 = 2520 Simplifying this equation: n - 2 = 14 n = 16 So the polygon has 16 sides.
Tambaya 38 Rahoto
Express in \( \frac{8.75}{0.025} \)standard form
Tambaya 40 Rahoto
In the diagram ST and QR are parallel. |PS| = 6cm, |SQ| = 8cm and |PR] = 18 2/3. Find |PT|
Bayanin Amsa
Tambaya 41 Rahoto
If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term
Bayanin Amsa
Let us denote the first term of the G.P. by a and the common ratio by r. Then the 2nd term is ar, the 3rd term is ar^2, and so on. Therefore, the 5th term is ar^4. We have two pieces of information, namely that ar = 6 and ar^4 = 48. Dividing the second equation by the first, we obtain (ar^4)/(ar) = r^3 = 48/6 = 8. Therefore, r = 2. Using this value of r, we can solve for a by using the equation ar = 6. Substituting r = 2, we obtain 2a = 6, so a = 3. Therefore, the first four terms of the G.P. are 3, 6, 12, 24, and their sum is 45. Thus, the correct answer is 45.
Tambaya 42 Rahoto
If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
Bayanin Amsa
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 \( \frac{k}{\sqrt{1 - k^2}} \).
Tambaya 43 Rahoto
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
Bayanin Amsa
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.
Tambaya 44 Rahoto
If 8x- 4 = 6x- 10, find the value of 5x,
Bayanin Amsa
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.
Tambaya 45 Rahoto
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)
Bayanin Amsa
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.
Tambaya 46 Rahoto
Which of the following is not a property of all parallelogram?
Bayanin Amsa
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".
Tambaya 47 Rahoto
Which of the following is equal to \(\frac{72}{125}\)
Bayanin Amsa
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 \( \frac{2^3 \times 3^2}{5^3} \).
Tambaya 49 Rahoto
E = (integers \(\leq\) 20), P = (multiples of 3), Q = (multiples of 4), what are the elements of P'∩Q?
Bayanin Amsa
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).
Tambaya 50 Rahoto
What is the number whose logarithm to base 10 is \(\bar{3}.4771\)?
Bayanin Amsa
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.
Tambaya 51 Rahoto
(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60° and < XZY = 75°. Measure /XY/; (ii) the locus \(l_{1}\) of points equidistant from Y and Z ; (iii) the locus \(l_{2}\) of points equidistant from XY and YZ.
(b) Measure QY where Q is the point of intersection of \(l_{1}\) and \(l_{2}\).
Tambaya 52 Rahoto
(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.
(b) If \(P = {1, 2, 3, 4}\) and \(Q = {3, 5, 6}\), find (i) \(P \cap Q\) ; (ii) \(P \cup Q\) ; (iii) \((P \cap Q) \cup Q\) ; (iv) \((P \cap Q) \cup P\).
Tambaya 53 Rahoto
(a) Evaluate, without using mathematical tables, \(17.57^{2} - 12.43^{2}\).
(b) Prove that angles in the same segment of a circle are equal.
Bayanin Amsa
None
Tambaya 54 Rahoto
The table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.
| Scores (x) | 15 | 25 | 35 | 45 | 55 | 65 | 75 |
| Freq (f) | 1 | 4 | 12 | 24 | 18 | 8 | 3 |
(a) Calculate, correct to 3 significant figures, the mean mark.
(b) Find the : (i) mode ; (ii) range of the distribution.
Bayanin Amsa
None
Tambaya 55 Rahoto
(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.
(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) Copy and complete the following table of values for \(y = 2x^{2} - 9x - 1\).
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| y | -1 | -8 | -11 | 17 |
(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 9x - 1\).
(c) Use your graph to find the : (i) roots of the equation \(2x^{2} - 9x = 4\), correct to one decimal place ; (ii) gradient of the curve \(y = 2x^{2} - 9x - 1\) at x = 3.
Tambaya 57 Rahoto
The table below gives the ages, to the nearest 5 years of 50 people.
| Age in years | 10 | 15 | 20 | 25 | 30 |
| No of people | 8 | 19 | 10 | 7 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve (Ogive)
(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.
Tambaya 58 Rahoto
(a) Given that \(3 \times 9^{1 + x} = 27^{-x}\), find x.
(b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
Tambaya 59 Rahoto
(a) Divide \(11111111_{two}\) by \(101_{two}\)
(b) A sector of radius 6 cm has an angle of 105° at the centre. Calculate its:
(i) perimeter ; (ii) area . [Take \(\pi = \frac{22}{7}\)]
Tambaya 60 Rahoto
(a) If \(\log_{10} (3x - 1) - \log_{10} 2 = 3\), find the value of x.
(b) Use logarithm tables to evaluate \(\sqrt{\frac{0.897 \times 3.536}{0.00249}}\), correct to 3 significant figures.
Tambaya 61 Rahoto
A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :
(a) all four balls are black ;
(b) exactly one of the four balls is white.
Bayanin Amsa
None
Tambaya 62 Rahoto
(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth's surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \(\pi = \frac{22}{7}\)]
(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.
None
Bayanin Amsa
None
