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Question 1 Report
In the diagram above, EFGH is a cyclic quadrilateral in which EH//FG, EG and FH are chords. If ?FHG = 42?
$?$ and ?EFH = 34?
$?$, calculate ?HEG
Answer Details
∠EFH = ∠EGH(∠s in same segment)
= 34∘
$\circ $
∠HEG = ∠HFG(∠s in same segment)
= X
also ∠HFG = ∠EHF (alternate ∠s)
But ∠EHG + ∠EFG =180(opp sof a cyclic quad)
42 + x + 34 + x = 180
2x + 76 = 180
2x = 180  76
2x = 104
x = 52∘
Question 2 Report
Find the equation of the locus of a point P(x,y) such that PV = PW, where V = (1,1) and W = (3,5)
Answer Details
To find the locus of point P(x,y) such that PV = PW, we need to find the set of all points that are equidistant from V(1,1) and W(3,5). First, we can find the distance between V and W using the distance formula: d = sqrt((31)^2 + (51)^2) = sqrt(20) The midpoint M of line segment VW is: M = ((1+3)/2, (1+5)/2) = (2,3) Using the midpoint formula, we can find the equation of the perpendicular bisector of VW, which is the locus of points equidistant from V and W: (y  3) = (1/2)(x  2) 2y  6 = x + 2 x + 2y = 8 Therefore, the correct option is (D) x + 2y = 8.
Question 3 Report
From a point P, the bearings of two points Q and R are N670W and N230E respectively. If the bearing of R from Q is N680E and PQ = 150m, calculate PR
Question 6 Report
If the maximum value of y=1+hx−3x2 $y=1+hx3{x}^{2}$ is 13 $13$, find h $h$.
Question 7 Report
If 29 x (3Y)9 = 35 x (3Y)5, find the value of Y.
Question 8 Report
The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the G.P if its sum to infinity is 8.
Answer Details
Le the common ratio be r so that the first term is 2r.
Sum, s = a/(1r)
ie. 8 = 2r/(1r)
8(1r) = 2r, r = 8/5.
Sn = a(1rn)/(1r)
Solve further to get 72/25
Question 9 Report
Divide 4x33x+1 by 2x1
Answer Details
To divide 4x³3x+1 by 2x1, we can use polynomial long division.
First, we set up the division like this:
2x² + x  1  2x  1  4x³ + 0x²  3x + 1
Next, we look at the leading term of the dividend (4x³) and the leading term of the divisor (2x) and ask: "How many times does 2x go into 4x³?" The answer is 2x², so we write that above the division line and multiply by the divisor:
2x² + x  1  2x  1  4x³ + 0x²  3x + 1  4x³ + 2x²  2x²  3x
We then subtract the result from the dividend and bring down the next term (1x):
2x² + x  1  2x  1  4x³ + 0x²  3x + 1  4x³ + 2x²  2x²  3x  2x² + x  2x + 1
We repeat the process with the new polynomial (2x+1) and the divisor (2x1):
2x² + x  1  2x  1  4x³ + 0x²  3x + 1  4x³ + 2x²  2x²  3x  2x² + x  2x + 1 2x + 1  0
We end up with a remainder of 0, which means that the division is exact. Therefore, the quotient is:
2x² + x  1
So the answer is (B) 2x²x1.
Question 10 Report
Find the area bounded by the curve y = x(2x). The xaxis, x = 0 and x = 2.
Answer Details
To find the area bounded by the curve y = x(2x), the xaxis, x = 0 and x = 2, we need to integrate the function with respect to x from 0 to 2. The function y = x(2x) is a quadratic equation, which opens downwards and has its vertex at (1, 1). It intersects the xaxis at x = 0 and x = 2, forming a trapezium. To integrate the function, we need to first expand it: y = x(2x) = 2x  x^2 Then, we integrate it with respect to x from 0 to 2: ∫(2x  x^2)dx, limits 0 to 2 = [x^2  (x^3)/3] from 0 to 2 = [(2^2)  (2^3)/3]  [(0^2)  (0^3)/3] = [4  (8/3)]  [0  0] = (4/3) sq units Therefore, the area bounded by the curve y = x(2x), the xaxis, x = 0 and x = 2 is (4/3) sq units.
Question 11 Report
In the figure, PQRS is a circle with STRQ. Find the value of x if PT = PS
Answer Details
< PST = < RQT = x(corresponding angle)
< PST = < PTS = < PTS x (PS = PT0
< SRQ = < SPT = 180∘ $\circ $ (sum of < on straight line)
< SPT = 180∘ $\circ $  110∘ $\circ $ = 70∘ $\circ $
in < SPT, < PST = PTS = < PSt = 180∘ $\circ $
2x + 70 = 180∘ $\circ $
2x = 180∘ $\circ $  70∘ $\circ $ = 110∘ $\circ $
x = 110o2 $\frac{{110}^{o}}{2}$ = 55∘
Question 12 Report
A man 1.7m tall observes a bird on top of a tree at an angle of 300. if the distance between the man's head and the bird is 25m, what is the height of the tree?
Answer Details
Hint: Make a sketch forming a right angled triangle. Let X = height required. Such that x/25 = tan 30.
Question 13 Report
Find the volume of solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the xaxis.
Answer Details
To find the volume of the solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the xaxis, we need to use the formula for the volume of a solid of revolution: V = π∫[a,b]f(x)^2dx where f(x) is the function defining the curve being rotated, and a and b are the limits of integration. In this case, the limits of integration are 0 and 3, and the function defining the curve is f(x) = 2x. Substituting into the formula, we get: V = π∫[0,3](2x)^2dx = π∫[0,3]4x^2dx = π[4x^3/3] from 0 to 3 = π[(4(3)^3/3)  (4(0)^3/3)] = π[36] = 36π Therefore, the volume of the solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the xaxis is 36π cubic units. The correct option is 36 π cubic units.
Question 14 Report
In the figure above, PQRS is a circle with ST//RQ. Find the value of x PT = PS.
Answer Details
PQRS is a cyclic quad
^P = 180  110 (opp ∠s of a cyclic quad)
^P = 70
In ΔPTS, ^S = ^T (base ∠s of Isc Δ)
∴^T = (18070) / 2
= 110/2
= 55∘
$\circ $
But ^T x(corr ∠)
∴x =55
Question 15 Report
The shaded portion in the graph is represented by
Question 16 Report
The table above shows the frequency distribution of the ages (in years) of pupils in a certain secondary school. What percentage of the total number of pupils is over 15 years but less than 21 years?
Answer Details
% of those above 15 but less than 21 years
age = 25/50 x 100/1
= 50%
Question 17 Report
Find the tangent to the acute angle between the lines 2x+y = 3 and 3x2y = 5.
Question 18 Report
The shaded portion in the graph above is represented by
Answer Details
y + x  x2 ≥ 0
y = x3  x
on x axis, y = 0
∴x2  x = 0
x(x2  1) = 0
x = 0 or x2  1 = 0
x2 = 1
x = +/1
∴ x = 1, 0 and 1 which are the roots of the equation y + x  x2 ≥ 0
Also y  x ≤ 0
=> y ≤ 0 + x
∴ the region where y ≤ 0
Question 19 Report
Find the length XY in the triangle above.
Answer Details
XY2 = XY2 + Y2  2XY.YZ cos 120
XZ2 = 22 + 12  2 x 2 1 cos 120
= 4 + 1 2 x 2 x 1 x cos(180  120)
= 4 + 1 + 4 cos 60
= 5 + 4 x 1/2
= 5 + 2
= 7
XY = √7 m
Question 20 Report
Express 1X3−1 $\frac{1}{{X}^{3}1}$ in partial fractions
Question 21 Report
If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X.
Answer Details
We can use the change of base formula to evaluate log 5 to base 8 in terms of X: log 5 to base 8 = log 5 / log 8 We know that log 10 to base 8 = X, which means: 8^X = 10 We can rewrite this as: 2^3X = 10 Taking the logarithm of both sides of the equation with base 2, we get: log 2 (2^3X) = log 2 10 3X = log 2 10 X = (1/3) log 2 10 Substituting this value of X into the expression for log 5 to base 8, we get: log 5 to base 8 = log 5 / log 8 = (log 5 / log 2) / (log 8 / log 2) = log 2 5 / log 2 8 = log 2 5 / (3 log 2 2) = (1/3) log 2 5 Therefore, log 5 to base 8 in terms of X is: X  (1/3) log 2 5.
Question 22 Report
Find the value of l in the frustrum above
Answer Details
ΔABE and ΔACD are similar
∴  x 


x+4 
Question 23 Report
In the diagram, EFGH is a cyclic quadrilateral in which EH  FG, EG and FH are chords. If < FHG = 424∘ $\circ $ and < EFH = 34∘
Answer Details
< EFH = < EGH = 34∘ $\circ $ (angles n the same segment)
< GHF = < GEF = 42∘ $\circ $(angles in the same segment)
< FOG = 42 + 34 = 76(exterior angle)
< FOG = < EOH = 76(vertically opposite angle)
< EDO = 90∘ $\circ $, < DOE = 762 $\frac{76}{2}$ = 38∘ $\circ $
< HEG = 90∘ $\circ $  38∘ $\circ $ = 52∘
Question 24 Report
What is the derivative of t2 sin (3t  5) with respect to t?
Answer Details
To find the derivative of the function t^2 sin(3t5), we will use the product rule of differentiation. The product rule states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. So, applying the product rule to the given function, we get: d/dt [t^2 sin(3t5)] = t^2 d/dt[sin(3t5)] + sin(3t5) d/dt[t^2] Now, we need to find the derivatives of sin(3t5) and t^2. The derivative of sin(3t5) is cos(3t5) times the derivative of the inside function (3t5), which is simply 3. So, d/dt[sin(3t5)] = 3cos(3t5). The derivative of t^2 is 2t. So, d/dt[t^2] = 2t. Now, we can substitute these values back into the original equation: d/dt [t^2 sin(3t5)] = t^2 * 3cos(3t5) + sin(3t5) * 2t Simplifying this expression, we get: d/dt [t^2 sin(3t5)] = 3t^2 cos(3t5) + 2t sin(3t5) Therefore, the correct option is: 2t sin (3t  5) + 3t^2 cos (3t  5).
Question 25 Report
Find the value of x for which the function y = x3  x has a minimum value.
Question 26 Report
If m∗n=(mn−nm $m\ast n=(\frac{m}{n}\frac{n}{m}$) for m, n belong to R, evaluate 3*4
Answer Details
Using the formula given, we have: 3 * 4 = (3*4  4*3) / (3 + 4) = 0 / 7 = 0 Therefore, the answer is 0. is the closest approximation to this value, which is 7/12.
Question 27 Report
A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?
Answer Details
Let the three items be M, Y and P.
n{M ∩ Y} only = 43 = 1
n{M ∩ P) only = 53 = 2
n{ Y ∩ P} only = 2
n{M} only = 12(1+3+2) = 6
n{Y} only = 10(1+2+3) = 4
n{P} only = 14(2+3+2) = 7
Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.
Question 28 Report
A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss
Answer Details
First, let's calculate how much the trader spent on the oranges. The trader bought 100 oranges at 5 for N1.20, which means that the cost of each orange was (N1.20 ÷ 5) = N0.24. Therefore, the total cost of 100 oranges was N0.24 × 100 = N24. Now, 20 of the oranges got spoilt and were not sold. So the trader had 80 oranges left to sell. The remaining 80 oranges were sold at 4 for N1.50, which means that the selling price of each orange was (N1.50 ÷ 4) = N0.375. Therefore, the total revenue from selling 80 oranges was N0.375 × 80 = N30. To calculate the profit or loss, we need to compare the cost and revenue. The cost of buying 100 oranges was N24, but the trader only earned N30 from selling the remaining 80 oranges. So, the trader earned a profit of N30  N24 = N6. To calculate the percentage gain or loss, we need to use the formula: Percentage gain or loss = (Profit or loss / Cost price) × 100% In this case, the profit was N6, and the cost was N24. Percentage gain or loss = (6/24) × 100% = 25% Therefore, the answer is "25% gain."
Question 29 Report
If {(a2b3c)3/4/a1b4c5} = apbqcr; what is the value of p+2q?
Answer Details
Hint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = 25/4 and r = 9/2.
Then substitute the values of p and q into p+2q to get 10.
Question 30 Report
Simplify √(0.0023∗750)(0.00345∗1.25)
Answer Details
To simplify the given expression: $$\sqrt{\frac{(0.0023\cdot750)}{(0.00345\cdot1.25)}}$$ We can first simplify the expression inside the square root by multiplying the numerators and denominators: $$\sqrt{\frac{0.0023\cdot750}{0.00345\cdot1.25}} = \sqrt{\frac{1.725}{0.0043125}}$$ Then we can divide the numerator and denominator by 0.0043125 to get: $$\sqrt{\frac{1.725}{0.0043125}} = \sqrt{400} = 20$$ Therefore, the simplified expression is 20.
Question 31 Report
If (a2b−3c)34a−1b4c5=apbqcr $\frac{({a}^{2}{b}^{3}c{)}^{\frac{3}{4}}}{{a}^{1}{b}^{4}{c}^{5}}={a}^{p}{b}^{q}{c}^{r}$ What is the value of p+2q?
Answer Details
Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at p =(5/2), q = (25/4) and r = (9/2).
Then p+2q will give you 52+2(−254)=−10
Question 32 Report
Evaluate: ∫z0(sinx−cosx)dx
${\int}_{0}^{z}(sinxcosx)dx$
Where z=π4.(π=pi)
Answer Details
To solve the definite integral ∫z0(sinx−cosx)dx, we need to find the antiderivative of the given expression first, and then evaluate it from 0 to z. The antiderivative of sin(x) is cos(x), and the antiderivative of cos(x) is sin(x). Therefore, the antiderivative of (sin(x)  cos(x)) is (cos(x)  sin(x)). Thus, ∫z0(sinx−cosx)dx = [cos(x)  sin(x)] evaluated from 0 to z. Plugging in the values of z and 0, we get: [cos(z)  sin(z)]  [cos(0)  sin(0)] = [cos(π/4)  sin(π/4)]  [cos(0)  sin(0)] = [(1/√2)  (1/√2)]  [1  0] = √2 + 1 Therefore, the value of the definite integral is √2 + 1. Hence, the option that represents this result is √21.
Question 33 Report
In the figure above, TZ is tangent to the circle QPZ. Find x if TZ = 6 units and PQ = 9 units
Answer Details
62 = 9 x χ
36 = 9χχ = 36/9χ = 4
Question 34 Report
The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers
Answer Details
Let's call the two numbers x and y. From the problem, we can write two equations: x + y = 2(x  y) (the sum of the two numbers is twice their difference) x  y = P (the difference of the numbers is P) To solve for x and y, we can use the first equation to eliminate one of the variables. x + y = 2(x  y) x + y = 2x  2y 3y = x Now we can substitute 3y for x in the second equation: 3y  y = P 2y = P y = P/2 Since x = 3y, we have: x = 3(P/2) = 3P/2 Therefore, the larger of the two numbers is 3P/2. So, the answer is: 3P/2.
Question 35 Report
Three consecutive positive integers k, l and m are such that l2 = 3(k+m). Find the value of m
Answer Details
Let's start by translating the given problem into equations. First, we know that the three integers are consecutive, so we can express them as k, k+1, and k+2. Second, we know that l^2 = 3(k+m). We can substitute k+1 for l to get (k+1)^2 = 3(k+m). Expanding (k+1)^2, we get k^2 + 2k + 1 = 3(k+m). Simplifying, we get k^2  k(63m) + 3m  1 = 0. We can use the quadratic formula to solve for k: k = [(63m) ± sqrt((63m)^2  4(3m1))]/2. Simplifying under the square root, we get sqrt(9m^2  36m + 37). Since k is a positive integer, the expression under the square root must be a perfect square. We can check the perfect squares between 1 and 37 and find that only 4, 16, and 25 are possibilities. Solving for m in each case gives us three potential solutions: 4, 5, and 7. However, only one of these solutions satisfies the original equation l^2 = 3(k+m), and that is m=4. Therefore, the answer to the problem is m=4.
Question 36 Report
The grades of 36 students in a test are shown in the pie chart above. How many students had excellent?
Answer Details
Angle of Excellent
= 360  (120+80+90)
= 360  290
= 70∘
$\circ $
If 360∘
$\circ $ represents 36 students
1∘
$\circ $ will represent 36/360
50∘
$\circ $ will represent 36/360 * 70/1
= 7
Question 37 Report
Find the value of X if √2x+√2=1x−√2
Answer Details
Start your solution by crossmultiplying, then collect like terms and factorize accordingly to get the unknown.
Question 38 Report
In ∆MNO, MN = 6 units, MO = 4 units and NO = 12 units. If the bisector of and M meets NO at P, calculate NP.
Answer Details
Hint: start with making a sketch of the figure described.
Question 39 Report
What is the answer when 24346 is divided by 426?
Answer Details
Hint: Two methods can be used;
1. Direct division (if you know division in number bases)
2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356
Question 40 Report
A binary operation * is defined by a*b = ab+a+b for any real number a and b. if the identity element is zero, find the inverse of 2 under this operation.
Answer Details
a*a1 = aa1 + a + a1 = e
if e = 0
2.21 + 2 + 21 = 0
3.21 + 2 = 0
= 21 = 23
Question 41 Report
Find the length XZ in the triangle