Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

2x2 + 11x2 + 17x + 6 by 2x + 1

**Question 2**
**Report**

find the equation of the curve which passes through by 6x - 5

**Answer Details**

m = dydv
$\frac{dy}{dv}$ = 6x - 5

∫dy = ∫(6x - 5)dx

y = 3x2 - 5x + C

when x = 2, y = 5

∴ 5 = 3(2)2 - 5(2) +C

C = 3

∴ y = 3x2 - 5x + 3

**Question 3**
**Report**

Factorize r2 - r(2p + q) + 2pq

**Answer Details**

To factorize the expression r^{2} - r(2p + q) + 2pq, we need to look for two numbers whose product is 2pq and whose sum is -r(2p + q). Let's try to break up -r(2p + q) into two parts such that their product is 2pq. We can write -r(2p + q) as -2rpq - rq^{2}. Then, we can rewrite the expression as: r^{2} - 2rpq - rq^{2} + 2pq Now, we can group the first two terms and the last two terms together and factor them separately: r(r - 2pq) - q(r - 2pq) We can see that r - 2pq is a common factor, so we can factor it out: (r - 2pq)(r - q) Therefore, the factorization of r^{2} - r(2p + q) + 2pq is (r - 2pq)(r - q). So the correct option is (c) (r - q)(r - 2p).

**Question 4**
**Report**

If y = 243(4x + 5)-2, find dydx $\frac{dy}{dx}$ when x = 1

**Answer Details**

To find dy/dx, we need to differentiate y with respect to x. We can start by using the chain rule, which states that if we have a function of the form f(g(x)), the derivative of that function with respect to x is f'(g(x)) * g'(x). In this case, we have y = 243(4x + 5)-2, which can be written as y = 243/(4x + 5)^2. Using the chain rule, we have: dy/dx = d/dx(243/(4x + 5)^2) = -2 * 243 / (4x + 5)^3 * d/dx(4x + 5) = -2 * 243 / (4x + 5)^3 * 4 = -1944 / (4x + 5)^3 Now, to find the value of dy/dx when x = 1, we just need to substitute x = 1 into the expression we found above: dy/dx = -1944 / (4(1) + 5)^3 = -1944 / 729 = -8/3 Therefore, the answer is option A: -83. To summarize, we used the chain rule to differentiate y with respect to x, which gave us an expression for dy/dx in terms of x. We then substituted x = 1 to find the value of dy/dx at that point.

**Question 5**
**Report**

In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities 25 $\frac{2}{5}$ and 13 $\frac{1}{3}$ respectively of winning the gold in the football event. What is the probability that either team will win the gold?

**Answer Details**

p(x) = 25
$\frac{2}{5}$ p(y) = 13
$\frac{1}{3}$

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= 25
$\frac{2}{5}$ + 13
$\frac{1}{3}$

= 115

**Question 6**
**Report**

Find the value of x in the diagram.

**Answer Details**

30 + x = 100

x = 100 - 30

= 70o

**Question 7**
**Report**

If x, y can take values from the set (1, 2, 3, 4), find the probability that the product of x and y is not greater than 6

**Answer Details**

∣∣ ∣ ∣ ∣ ∣ ∣∣123411234224683369124481216∣∣ ∣ ∣ ∣ ∣ ∣∣
$\left|\overline{)\begin{array}{ccccc}& 1& 2& 3& 4\\ 1& 1& 2& 3& 4\\ 2& 2& 4& 6& 8\\ 3& 3& 6& 9& 12\\ 4& 4& 8& 12& 16\end{array}}\right|$

P (product of x and y not greater than 6) = 1016
$\frac{10}{16}$

= 58

**Question 8**
**Report**

The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?

**Answer Details**

x012345678910f194710879821
$\begin{array}{cccccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ f& 1& 9& 4& 7& 10& 8& 7& 9& 8& 2& 1\end{array}$

no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = 2560 $\frac{25}{60}$ x 100%

= 41.70%

**Question 9**
**Report**

Average hourly earnings(N)5−910−1415−1920−24No. of workers17322524 $\begin{array}{ccccc}\text{Average hourly earnings(N)}& 5-9& 10-14& 15-19& 20-24\\ \text{No. of workers}& 17& 32& 25& 24\end{array}$

Estimate the mode of the above frequency distribution

**Answer Details**

Class intervalsFClass boundary5−7174.5−9.510−14329.5−14.515−192514.5−19.520−242419.5−24.5
$\begin{array}{ccc}\text{Class intervals}& F& \text{Class boundary}\\ 5-7& 17& 4.5-9.5\\ 10-14& 32& 9.5-14.5\\ 15-19& 25& 14.5-19.5\\ 20-24& 24& 19.5-24.5\end{array}$

mode = 9 + D1D2+D1
$\frac{{D}_{1}}{{D}_{2}+{D}_{1}}$ x C

= 9.5 + 5(32−17)2(32)−17−25
$\frac{5(32-17)}{2(32)-17-25}$

= 9.5 + 7527
$\frac{75}{27}$

= 12.27

≈
$\approx $ 2.3

**Question 10**
**Report**

If b3 = a-2 and c13 $\frac{1}{3}$ = a12 $\frac{1}{2}$b, express c in terms of a

**Answer Details**

c13
$\frac{1}{3}$ = a12
$\frac{1}{2}$b

= a12
$\frac{1}{2}$b x a-2

= a-32
$\frac{3}{2}$

= (c13
$\frac{1}{3}$)3

= (a-32
$\frac{3}{2}$)13
$\frac{1}{3}$

c = a-12

**Question 11**
**Report**

Let = (1001) $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ p = (2345) $\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right)$ Q = (u4+u−2vv) $\left(\begin{array}{cc}u& 4+u\\ -2v& v\end{array}\right)$ be 2 x 2 matrices such that PQ = 1. Find (u, v)

**Answer Details**

PQ = (2345)
$\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right)$(u4+u−2vv)
$\left(\begin{array}{cc}u& 4+u\\ -2v& v\end{array}\right)$

= ((2u−6v2(4+u)+3v)4u−10v4(4+u)+5v)
$\left(\begin{array}{cc}(2u-6v& 2(4+u)+3v)\\ 4u-10v& 4(4+u)+5v\end{array}\right)$

= (1001)
$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

2u - 6v = 1.....(i)

4u - 10v = 0.......(ii)

2(4 + u) + 3v = 0......(iii)

4(4 + u) + 5v = 1......(iv)

2u - 6v = 1 .....(i) x 2

4u - 10v = 0......(ii) x 1

4u - 12v = 0-4u - 10v = 0
$\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}$

-2v = 2 = v = -1

2u - 6(-1) = 1 = 2u = 5

u = -52
$\frac{5}{2}$

∴ (U, V) = (-52
$\frac{5}{2}$ - 1)

**Question 12**
**Report**

If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p

**Answer Details**

2p - 10 = p+1+1−4P22
$\frac{p+1+1-4{P}^{2}}{2}$ (Arithmetic mean)

= 2(2p - 100 = p + 2 - 4P2)

= 4p - 20 = p + 2 - 4p2

= 4p2 + 3p - 22 = 0

= (p - 2)(4p + 11) = 0

∴ p = 2 or -411

**Question 13**
**Report**

In the figure, PQST is a parallelogram and TSR is a straight line. If the area of △
$\u25b3$QRS is 20cm2, find the area of the trapezium PQRT.

**Answer Details**

A△ $\u25b3$ = 12 $\frac{1}{2}$ x 8 x h = 20

= 12 $\frac{1}{2}$ x 8 x h = 4h

h = 204 $\frac{20}{4}$

= 5cm

A△ $\u25b3$(PQTS) = L x H

A△ $\u25b3$PQRT = A△ $\u25b3$QSR + A△ $\u25b3$PQTS

20 + 50 = 70cm2

ALTERNATIVE METHOD

A△ $\u25b3$PQRT = 12 $\frac{1}{2}$ x 5 x 28

= 70cm2

**Question 14**
**Report**

Find the value of k if k?3+?2 $\frac{k}{\sqrt{3}+\sqrt{2}}$ = k?3?2

**Answer Details**

k√3+√2
$\frac{k}{\sqrt{3}+\sqrt{2}}$ = k√3−2
$\sqrt{3-2}$

k√3+√2
$\frac{k}{\sqrt{3}+\sqrt{2}}$ x √3−√2√3−√2
$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

= k√3−2
$\sqrt{3-2}$

= k(√3−√2
$\sqrt{3}-\sqrt{2}$)

= k√3−2
$\sqrt{3-2}$

= k√3
$3$ - k√2
$2$

= k√3−2
$\sqrt{3-2}$

k2 = √2
$2$

k = 2√2
$\frac{2}{\sqrt{2}}$

= √2

**Question 15**
**Report**

Find the variance of the numbers k, k+1, k+2.

**Answer Details**

To find the variance of the numbers k, k+1, k+2, we can use the formula for variance which is the average of the squared differences from the mean. First, we need to find the mean of the three numbers. Mean = (k + k + 1 + k + 2) / 3 = (3k + 3) / 3 = k + 1 So, the mean is k + 1. Next, we find the squared differences from the mean for each number: For k, the difference from the mean is k - (k+1) = -1. The squared difference is (-1)^2 = 1. For k+1, the difference from the mean is (k+1) - (k+1) = 0. The squared difference is 0^2 = 0. For k+2, the difference from the mean is (k+2) - (k+1) = 1. The squared difference is 1^2 = 1. Now we can find the variance: Variance = [(1^2 + 0^2 + 1^2) / 3] = 2/3 = 0.67 (rounded to two decimal places) Therefore, the answer is option (A) 2/3 or as a percentage approximately 66.7%.

**Question 16**
**Report**

The locus of all points at a distance 8cm from a point N passes through points T and S. If S is equidistant from T and N, find the area of triangle STN.

**Question 17**
**Report**

a cylindrical drum of diameter 56 cm contains 123.2 litres of oil when full. Find the height of the drum in centimeters

**Answer Details**

To solve the problem, we need to use the formula for the volume of a cylinder, which is V = πr²h, where V is the volume, r is the radius, and h is the height of the cylinder. We are given the diameter of the drum, which is 56 cm. To find the radius, we need to divide the diameter by 2: radius (r) = diameter / 2 = 56 cm / 2 = 28 cm We are also given that the drum contains 123.2 litres of oil when full. To convert litres to cubic centimeters, we need to multiply by 1000: 123.2 litres * 1000 = 123200 cubic centimeters Now we can use the formula for the volume of a cylinder to find the height (h): V = πr²h h = V / (πr²) h = 123200 / (π28²) h ≈ 123200 / 2463.47 h ≈ 50.00 cm Therefore, the height of the drum is approximately 50.00 cm.

**Question 18**
**Report**

If the distance between the points (x, 3) and (-x, 2) is 5. Find x

**Answer Details**

To solve the problem, we need to use the distance formula between two points in a coordinate plane. The distance formula is given by: d = sqrt[(x2 - x1)^2 + (y2 - y1)^2] where d is the distance between the two points (x1, y1) and (x2, y2). Using the given points, we have: (x2, y2) = (-x, 2) (x1, y1) = (x, 3) Substituting these values into the distance formula, we get: d = sqrt[(-x - x)^2 + (2 - 3)^2] Simplifying the expression inside the square root: d = sqrt[4x^2 + 1] We are given that d = 5, so we can substitute that into the equation and solve for x: 5 = sqrt[4x^2 + 1] 25 = 4x^2 + 1 24 = 4x^2 6 = x^2 x = sqrt(6) or x = -sqrt(6) Since we are only interested in the positive value of x, the answer is x = sqrt(6). Therefore, the correct option is: √

**Question 19**
**Report**

In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35o, find ∠BAO.

**Question 20**
**Report**

The pie chart shows the monthly expenditure pf a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7200?

**Answer Details**

Based on the information given in the pie chart, the monthly expenditure on school fees can be represented by the fraction 1/10 (since it is one-tenth of the total expenditure) and the monthly expenditure on housing can be represented by the fraction 2/10 (since it is twice the amount of school fees). To find out how much the worker spends on housing, we need to first calculate the total amount of money he spends on all his monthly expenses. From the pie chart, we see that the total monthly expenditure is represented by the fraction 7/10. If we let x represent the amount of money the worker spends on housing each month, we can set up the following equation to solve for x: x + (1/10)*7200 = (2/10)*7200 Simplifying the equation, we get: x + 720 = 1440 Subtracting 720 from both sides, we get: x = 720 Therefore, the worker spends ₦7200/10 = ₦720 on school fees each month, and ₦720 * 2 = ₦1440 on housing each month. So the answer is ₦2000.

**Question 21**
**Report**

A man is paid r naira per hour for normal work and double rate for overtime. if he does a 35-hour week which includes q hours of overtime, what is his weekly earning in naira?

**Answer Details**

The cost of normal work = 35r

The cost of overtime = q x 2r = 2qr

The man's total weekly earning = 35r + 2qr

= r(35 + 2q)

**Question 22**
**Report**

The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is

**Answer Details**

To find the midpoint of a line segment, we need to find the average of the endpoints. The x-intercept of the line y = 4x + 3 is found by setting y = 0 and solving for x: 0 = 4x + 3 x = -3/4 So the x-coordinate of the midpoint is the average of -3/4 and 0: x = (-3/4 + 0)/2 = -3/8 To find the y-coordinate of the midpoint, we plug in x = -3/8 to the equation of the line: y = 4(-3/8) + 3 = -3/2 + 3 = 3/2 So the midpoint is (-3/8, 3/2). Therefore, the answer is (-3/8, 3/2).

**Question 23**
**Report**

Express in partial fractions 11+26x2−x−1

**Answer Details**

11+26x2−x−1
$\frac{11+2}{6{x}^{2}-x-1}$ = 11+23x+1
$\frac{11+2}{3x+1}$

= A3x+1
$\frac{A}{3x+1}$ + B2x−1
$\frac{B}{2x-1}$

11x = 2 = A(2x - 1) + B(3x + 1)

put x = 12
$\frac{1}{2}$

= -−53
$\frac{-5}{3}$

= -−53
$\frac{-5}{3}$A →
$\to $ A = 1

∴ 11x+26x2−x−1
$\frac{11x+2}{6{x}^{2}-x-1}$ =