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Tambaya 1 Rahoto
A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30°
Bayanin Amsa
Hint: Make a sketch of the moving points such that the hypotenuse is 8cm and the adjacent x cm.
Cos 30 = x/8
x = 8 cos 30 = 8√3cm
Tambaya 2 Rahoto
Evaluate (0.142 x 0.275) ÷ 7(0.02) to 3 decimal places.
Bayanin Amsa
(0.142 x 0.275) 7(0.02) = (0.14 x 0.14 x 0.275) ÷ (7 x 0.02) = (0.14 x 0.14 x 0.275)/(0.14)
=0.14 x 0.275 = 0.0385 Approx. 0.039
Tambaya 3 Rahoto
Solve the equations
m2 + n2 = 29
m + n = 7
Bayanin Amsa
m2 + n2 = 29 .......(1)
m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)
Tambaya 4 Rahoto
Triangle SPT is the solution of the linear inequalities
Bayanin Amsa
Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1
Tambaya 5 Rahoto
Find the dimensions of a rectangle of greatest area which has a fixed perimeter p.
Bayanin Amsa
Let the rectangle be a square of sides p/4.
So that perimeter of square = 4p
4 x (p/4) = p.
Tambaya 6 Rahoto
The bar chart above shows different colours of passing a particular point of a certain street in two minutes. What fraction of the total number of cars
Bayanin Amsa
The total number of cars that passed the point on the street in two minutes is not given in the chart. Therefore, we cannot determine the fraction of the total number of cars from the information provided. The chart only shows the relative frequency or proportion of cars of different colors that passed the point during the two-minute observation period. For example, if the chart shows that 3 out of 25 cars were blue, then the proportion of blue cars passing the point during the two minutes was 3/25. To find the fraction of the total number of cars, we would need to know the total number of cars that passed the point during the two-minute observation period.
Tambaya 7 Rahoto
Find the number of ways of selecting 8 subjects from 12 subjects for an examination
Bayanin Amsa
This question is asking us to find the number of ways to select 8 subjects out of 12 subjects. This is a combination problem since the order in which the subjects are selected does not matter. The formula for combination is nCr = n!/r!(n-r)!, where n is the total number of items, r is the number of items to be selected, and ! represents the factorial function. Applying this formula to the given problem, we have: 12C8 = 12!/8!(12-8)! = (12x11x10x9)/(4x3x2x1) = 495 Therefore, there are 495 ways of selecting 8 subjects out of 12 subjects. Hence, the correct option is 495.
Tambaya 8 Rahoto
If y = x sinx, find dy/dx when x = ?/2.
Bayanin Amsa
To find dy/dx, we need to take the derivative of y with respect to x. Using the product rule of differentiation, we have: dy/dx = (sinx) + (x)(cosx) So, when x = ?/2, we can substitute this value into the expression to get: dy/dx = (sin(?/2)) + (?/2)(cos(?/2)) We know that sin(?/2) = 1 and cos(?/2) = 0, so we can simplify the expression: dy/dx = (1) + (?/2)(0) dy/dx = 1 Therefore, the answer is 1.
Tambaya 9 Rahoto
What is the rate of change of the volume v of a hemisphere with respect to its radius r when r = 2?
Bayanin Amsa
To solve this problem, we need to use the formula for the volume of a hemisphere, which is V = (2/3)πr^3. We want to find the rate of change of V with respect to r, which is given by dV/dr = 2πr^2. When r = 2, we can substitute this value into the formula to get dV/dr = 2π(2)^2 = 8π. Therefore, the rate of change of the volume of a hemisphere with respect to its radius when r = 2 is 8π.
Tambaya 10 Rahoto
A car dealer bought a second-hand car for ₦250,000 and spent ₦70,000 refurbishing it. He then sold the car for ₦400,000. What is the percentage gain?
Bayanin Amsa
The dealer bought the car for ₦250,000 and spent ₦70,000 refurbishing it, for a total cost of: Total cost = ₦250,000 + ₦70,000 = ₦320,000 The dealer then sold the car for ₦400,000. The profit made by the dealer is the difference between the selling price and the cost price: Profit = Selling price - Cost price Profit = ₦400,000 - ₦320,000 = ₦80,000 The percentage gain is the profit expressed as a percentage of the cost price: Percentage gain = (Profit / Cost price) x 100% Substituting the values we get: Percentage gain = (₦80,000 / ₦320,000) x 100% Percentage gain = 0.25 x 100% Percentage gain = 25% Therefore, the percentage gain made by the dealer is 25%, and the answer is.
Tambaya 11 Rahoto
A sector of a circle of radius 7.2cm which subtends an angle of 300∘ at the centre is used to form a cone. What is the radius of the base of the cone?
Bayanin Amsa
Tambaya 12 Rahoto
Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3).
Bayanin Amsa
Hint: If two lines a perpendicular, the gradient of one is equal to minus the reciprocal of the other (P, 4) and (6, -2)
Tambaya 13 Rahoto
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode.
Bayanin Amsa
Mode = score with highest frequency = 11.
Square of 11 = 121
Tambaya 14 Rahoto
Find the principal which amounts to ₦5,500 at a simple interest in 5 years at 2% per annum.
Bayanin Amsa
Principal, P = Amount, A - Interest, I.
A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = ₦5,000
Tambaya 15 Rahoto
The mean score is
Bayanin Amsa
Mean ((4*3) + (7*5) + (8*2) + (11*7) + (13*2) + (8*1))/20
(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7
Tambaya 16 Rahoto
Given distribution of color beads: blue, black, yellow, white and brown with frequencies 1, 2, 3, 4, and 5 respectively. Find the probability that a bead picked at random will be blue or white.
Bayanin Amsa
There are a total of 1+2+3+4+5 = 15 beads. The probability of picking a blue bead is 1/15, since there is only one blue bead out of the total of 15 beads. The probability of picking a white bead is 4/15, since there are 4 white beads out of the total of 15 beads. To find the probability of picking a blue or white bead, we add the probabilities of picking a blue bead and a white bead: P(blue or white) = P(blue) + P(white) P(blue or white) = 1/15 + 4/15 P(blue or white) = 5/15 P(blue or white) = 1/3 Therefore, the probability of picking a blue or white bead is 1/3.
Tambaya 17 Rahoto
Find the variance of 2, 6, 8, 6, 2 and 6
Bayanin Amsa
To find the variance of a set of numbers, you need to follow these steps: 1. Find the mean (average) of the set. 2. Subtract the mean from each number in the set. 3. Square each of the differences. 4. Find the mean (average) of the squared differences. Let's apply these steps to the set of numbers given: 2, 6, 8, 6, 2, and 6. 1. The mean of the set is (2+6+8+6+2+6)/6 = 5. 2. Subtracting the mean from each number gives us the following differences: -3, 1, 3, 1, -3, 1. 3. Squaring each difference gives us: 9, 1, 9, 1, 9, 1. 4. Finding the mean of these squared differences gives us (9+1+9+1+9+1)/6 = 5. Therefore, the variance of the set is 5. In summary, variance is a measure of how spread out a set of numbers is. A high variance means that the numbers in the set are far from the mean, while a low variance means that the numbers are closer to the mean.
Tambaya 18 Rahoto
The histogram above shows the distribution of passengers in taxis of a certain motor park. How many taxis have more than 4 passengers
Bayanin Amsa
Looking at the histogram, we can see that the height of the bars represents the number of taxis with a particular number of passengers. We can count the number of taxis with more than 4 passengers by adding up the heights of the bars for 5, 6, 7, and 8 passengers. Counting these bars, we can see that there are a total of 17 taxis with more than 4 passengers. Therefore, the answer is 17.
Tambaya 19 Rahoto
Find the square pf the mode
Tambaya 20 Rahoto
Evaluate 21.05347 - 1.6324 x 0.43 to 3 decimal places
Bayanin Amsa
Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.
Tambaya 23 Rahoto
The bar chart shows different colours of cars passing a particular point of a certain street in two minutes. What fraction of the cars is yellow
Bayanin Amsa
colour of carsNumber (frequency)yellow3white4red8green2blue6black225
Thus, the fraction of the total numbers that are yellow is 325
Tambaya 24 Rahoto
Evaluate ∫2(2x-3)2/3dx
Bayanin Amsa
To evaluate the definite integral ∫2(2x-3)2/3 dx, we can use the following formula: ∫un du = (un+1)/(n+1) + C where C is the constant of integration. Using this formula, we can rewrite the integral as: ∫2(2x-3)2/3 dx = 2 ∫(2x-3)2/3 dx Let u = 2x - 3, then du/dx = 2 and dx = du/2. Substituting this into the integral, we get: 2 ∫(2x-3)2/3 dx = 2 ∫u2/3 (du/2) = ∫u2/3 du Using the formula, we get: ∫u2/3 du = (u5/3)/(5/3) + C Substituting back for u, we get: ∫2(2x-3)2/3 dx = (2(2x-3)5/3)/(5/3) + C Simplifying, we get: ∫2(2x-3)2/3 dx = (6(2x-3)5/3)/5 + C Therefore, the correct answer is option A: 3/5(2x-3)5/3 + k.
Tambaya 25 Rahoto
Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k.
Bayanin Amsa
The locus of a point which moves such that its distance from the line y = 4 is a constant, k, is the set of all points which are equidistant from the line y = 4. This set of points is a pair of parallel lines, k units above and below the line y = 4. Therefore, the correct answer is: y = k + or - 4.
Tambaya 26 Rahoto
The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST.
Bayanin Amsa
In a circle, the length of an arc is proportional to the angle it subtends at the center of the circle. Since ST is a chord and its length is equal to the radius of the circle, then angle SOT, where O is the center of the circle, is 60 degrees (because it subtends an equilateral triangle). Therefore, the length of arc ST is 1/6 of the circumference of the circle (because 60 degrees is 1/6 of a full revolution, which is 360 degrees). The circumference of the circle is 2πr, so the length of arc ST is: Length of arc ST = 1/6 * 2πr = πr/3 Hence, the answer is option D: πr/3.
Tambaya 27 Rahoto
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
Bayanin Amsa
Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6
Tambaya 28 Rahoto
If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k.
Bayanin Amsa
The given curve is y = 2kx² + x + 1, and we are asked to find the value of k when the gradient at x = 1 is 9. The gradient of a curve at a particular point is the slope of the tangent to the curve at that point. To find the gradient of this curve at x = 1, we need to differentiate it with respect to x: dy/dx = 4kx + 1 Now we can substitute x = 1 and set the result equal to 9, which gives: 4k(1) + 1 = 9 Simplifying this equation, we get: 4k = 8 Dividing both sides by 4, we obtain: k = 2 Therefore, the value of k that satisfies the given conditions is 2.
Tambaya 29 Rahoto
A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.
Bayanin Amsa
Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
Tambaya 30 Rahoto
The sixth term of an A.P is half of its twelfth term. The first term of the A.P is equal to
Bayanin Amsa
Let's assume that the first term of the A.P is 'a', and the common difference is 'd'. We know that for an A.P, the nth term is given by: a + (n-1)d According to the problem, the sixth term is equal to half of the twelfth term. So, we can write: a + 5d = 1/2 (a + 11d) Multiplying both sides by 2 to eliminate the fraction: 2a + 10d = a + 11d Simplifying, we get: a = d Therefore, the first term of the A.P is equal to the common difference. So, the correct answer is "the common difference."
Tambaya 31 Rahoto
The graph shows the cumulative frequency of the distribution of masses of fertilizer for 48 workers in one institution. Which of the following gives the inter-quartile range?
Tambaya 32 Rahoto
Divide: ax3x - 26x2x + 156ax - 216 by a2x - 24ax + 108
Bayanin Amsa
To divide the given expression, we can use the long division method.
ax^2 - 26x + 156a - 216b ___________________________________ ax | ax^3 + 3x^2 - 26x^2 + 156ax - 216bx^2 - 24ax + 108 -ax^3 ----- 3x^2 - 216bx^2 3x^2 - 3ax^2 -------------- -216bx^2 + 3ax^2 -216bx^2 + 36ax^2 ------------------ -33ax^2 + 156a - 216bx^2 - 24ax + 108 -33ax^2 + 33bx^2 ------------------ 156a - 33ax^2 - 24ax + 108 156a - 33ax^2 - 156ax + 33ax ------------------------- -24ax + 108 -24ax + 24bx ------------- 108 - 24bx 108 - 2ax -------- 2ax - 24bx + 108
Therefore, the quotient is ax - 2 with a remainder of 2ax - 24bx + 108.
Hence, the answer is ax - 2.
Tambaya 33 Rahoto
Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units.
Bayanin Amsa
In an isosceles triangle, the two equal sides are opposite to the two equal angles. Therefore, the angle opposite to the side of length √3t would be equal to the other equal angle of the triangle. Let's call this angle x. Then, using the fact that the sum of the angles in a triangle is 180°, we can write: x + x + θ = 180 Simplifying this equation, we get: 2x + θ = 180 Now, we know that the two equal sides of the triangle have length t, so we can use the cosine formula to write: cos(x) = (t/2) / √(3t^2/4) Simplifying this expression, we get: cos(x) = 1/√3 Taking the inverse cosine of both sides, we get: x = 30° Substituting this value of x into the earlier equation, we get: 2(30°) + θ = 180° θ = 120° Therefore, the answer is option (B) 120°.
Tambaya 34 Rahoto
If two graphs y = px2 + q and y = 2x2 -1 intersect at x = 2, find the value of p in terms q.
Bayanin Amsa
To find the value of p in terms of q, we can use the given information that the two graphs intersect at x = 2. This means that at x = 2, the values of y for both graphs are the same. So, we can substitute x = 2 into both equations and set them equal to each other: px^2 + q = 2x^2 - 1 Substituting x = 2: p(2)^2 + q = 2(2)^2 - 1 4p + q = 7 Solving for p in terms of q, we can isolate p: 4p = 7 - q p = (7 - q) / 4 Therefore, the value of p in terms of q is (7 - q) / 4.
Tambaya 35 Rahoto
In the figure above, PQR is a straight line segment, PQ = QT. Triangle PQT is an isosceles triangle, ?SQR is 75?
and ?QPT is 25?
. Calculate the value of ?RST.
Bayanin Amsa
In ? PQT,
?PTQ = 25?
(base ?s of isosceles ?)
In ? QSR,
?RQS = ?QPT + ?QTP
(Extr = sum of interior opposite ?s)
?RQS = 25 + 25
= 50?
Also in ? QSR,
75 + ?RQS + ?QSR = 180?
(sum of ?s of ?)
?75 + 50 + ?QSR = 180
125 + ?QSR = 180
?QSR = 180 - 125
?QSR = 55?
But ?QSR and ?RST are the same
?RST = 55?
Tambaya 36 Rahoto
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is
Bayanin Amsa
To find the mean score, we need to calculate the sum of all the scores, and then divide it by the total number of scores. The scores are given as 4, 7, 8, 11, 13, 8 with corresponding frequencies of 3, 5, 2, 7, 2, 1 respectively. So, the sum of all the scores is: 4 × 3 + 7 × 5 + 8 × 2 + 11 × 7 + 13 × 2 + 8 × 1 = 12 + 35 + 16 + 77 + 26 + 8 = 174 The total number of scores is: 3 + 5 + 2 + 7 + 2 + 1 = 20 Therefore, the mean score is: mean = sum of scores / total number of scores = 174 / 20 = 8.7 So, the answer is.
Tambaya 37 Rahoto
Find the integral values of x and y satisfying the inequality 3y + 5x ≤ 15, given that y > 0, y < 3 and x > 0.
Bayanin Amsa
Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.
Tambaya 38 Rahoto
An operation * is defined on the set of real numbers by a*b = a + b + 1. If the identity elements is -1, find the inverse of the element 2 under *.
Bayanin Amsa
First, let's find the inverse of the element 2 under the operation * on the set of real numbers. The inverse of an element "a" under the operation "*" is another element "b" such that a*b = b*a = identity element. The identity element under the operation "*" is given as -1, so we have: 2 * b = b * 2 = -1 Substituting the value of the identity element, we get: 2 * b = b * 2 + 1 = -1 Simplifying the equation, we get: 2b = -3 b = -3/2 Therefore, the inverse of the element 2 under the operation * is -3/2. To check if this is correct, we can substitute the values of 2 and -3/2 into the operation * and see if we get the identity element: 2 * (-3/2) = -3/2 * 2 = -1 Since we get the identity element, we can confirm that the inverse of the element 2 under the operation * is -3/2. So, the answer to the question is option (C) -2.
Tambaya 39 Rahoto
Find the value of ?
in the diagram
Bayanin Amsa
Using cosine formula (t?3 )2 = t2 + t2 - 2t2 cos?
3t2 = 2t2 - 2t2 cos? = 2t2(1 - cos? )
1 - cos? = 3t22t2 = 32
cos = 1 - 32=?12
? = cos-1(-12 ) = 120? and 240?
N.B 0 ? ? 360
Tambaya 40 Rahoto
Given that p=1+√2 and q=1−√2, evaluate (p2−q2)2pq .
Bayanin Amsa
First, let's evaluate p^2 and q^2: p^2 = (1+√2)^2 = 1 + 2√2 + 2 = 3 + 2√2 q^2 = (1-√2)^2 = 1 - 2√2 + 2 = 3 - 2√2 Next, let's substitute the values of p^2 and q^2 into the expression (p^2 - q^2)/(2pq): (p^2 - q^2)/(2pq) = ((3 + 2√2) - (3 - 2√2))/(2(1+√2)(1-√2)) = (4√2)/(2(1-2)) = (4√2)/(-2) = -2√2 Therefore, the answer is -2√2.
Tambaya 41 Rahoto
Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3
Bayanin Amsa
To find the range of a set of numbers, we need to find the difference between the largest and smallest numbers in the set. The given set of numbers are: 1/6, 1/3, 3/2, 2/3, 8/9, and 4/3. The smallest number in the set is 1/6 and the largest number in the set is 3/2. So, the range of the set is: 3/2 - 1/6 = 9/6 - 1/6 = 8/6 = 4/3 Therefore, the answer is 4/3.
Tambaya 43 Rahoto
The distribution of colours of beads in a bowl is given above. What is the probability that a bead selected at random will be blue or white?
Bayanin Amsa
Total number of beads
= 1+2+4+5+3
=15
Number of blue beads = 1
P(Blue beads) = 1/15
Numbers of white beads = 4
P(white beads) = 4/15
∴P(Blue of white beads)
= P(Blue) + P(White)
= 1/15 + 4/15
= 5/15
= 1/3
Tambaya 44 Rahoto
Find the area bounded by the curves y = 4 - x2 and y = 2x + 1
Bayanin Amsa
Hint:
y = 4 - x2 and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = -3 or x = 1.
Integrating x2 + 2x - 3 from (-3, to 1) w.r.t x will give 31/3 = 10(1/3)
Tambaya 46 Rahoto
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?
(Take pi = 22/7)
Bayanin Amsa
The formula for the volume of a cylindrical tank is V = πr^2h, where V is the volume, r is the radius, and h is the height or depth of the tank. We are given that the capacity or volume of the tank is 3080 m^3, and we can find the radius by dividing the diameter by 2: d = 14 m r = d/2 = 7 m Substituting the given values in the formula, we get: 3080 = (22/7) x 7^2 x h Simplifying the expression, we get: 3080 = 22 x 7 x 7 x h/7 3080 = 22 x 7 x h 3080 = 154h h = 3080/154 h = 20 Therefore, the depth of the tank is 20 meters. Answer: 20 m
Tambaya 47 Rahoto
Team P and Q are involved in a game of football.What is the probability that the game ends in a draw?
Bayanin Amsa
P(Games ends in draw)
This implies that Team P and Q wins
∴ P(P wins) = 1/2
P(Q wins) = 1/2
∴ P(game ends in a draw) = 1/2*1/2 = 1/4
Tambaya 48 Rahoto
If 6Pr = 6, find the value of 6Pr+1
Bayanin Amsa
The notation 6Pr means the number of permutations of r objects selected from a set of 6 distinct objects. Since 6Pr = 6! / (6 - r)!, we know that 6Pr = 6 when r = 1, since 6P1 = 6! / (6 - 1)! = 6! / 5! = 6. To find the value of 6Pr+1, we can simply multiply 6Pr by (6 - r) since we are selecting one additional object. So, 6Pr+1 = 6Pr * (6 - r) = 6 * (6 - 1) = 30 when r = 1. Therefore, the answer is 30.
Tambaya 49 Rahoto
⊗ | k | l | m |
k | l | m | k |
l | m | k | l |
m | k | l | m |
The identity element with respect to the multiplication shown in the table above is
Bayanin Amsa
Tambaya 50 Rahoto
Factorize 4x2 - 9y2 + 20x + 25
Bayanin Amsa
Given: 4x2 - 9y2 + 20x + 25
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
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