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**Question 1**
**Report**

A cylinder pipe, made of metal is 3cm thick.If the internal radius of the pipe is 10cm.Find the volume of metal used in making 3m of the pipe.

**Answer Details**

Volume of a cylinder = πr2 ${}^{2}$h

First convert 3m to cm by multiplying by 100

Volume of External cylinder = π×132×300 $\pi \times {13}^{2}\times 300$

Volume of Internal cylinder = π×102×300 $\pi \times {10}^{2}\times 300$

Hence; Volume of External cylinder - Volume of Internal cylinder

Total volume (v) = π(169−100)×300 $\pi (169-100)\times 300$

V = π×69×300 $\pi \times 69\times 300$

V = 20700πcm3

**Question 2**
**Report**

Correct 241.34(3 x 10−3 ${}^{-3}$)2 ${}^{2}$ to 4 significant figures

**Answer Details**

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)

(3 x 10-3
${}^{3}$)2
${}^{2}$............(B)

= 32
${}^{2}$x2
${}^{2}$

= 1103
$\frac{1}{{10}^{3}}$ x 1103
$\frac{1}{{10}^{3}}$

(Note that x2
${}^{2}$ = 1x3
$\frac{1}{{x}^{3}}$)

= 24.34 x 32
${}^{2}$ x 1106
$\frac{1}{{10}^{6}}$

= 2172.06106
$\frac{2172.06}{{10}^{6}}$

= 0.00217206

= 0.002172(4 s.f)

**Question 3**
**Report**

The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

**Answer Details**

The mean of ten positive numbers is the sum of all ten numbers divided by ten. So if the mean of the ten numbers is 16, then the sum of all ten numbers must be 160 (16 x 10). When an eleventh number is added, the mean becomes 18, so the sum of all eleven numbers must be 198 (18 x 11). We can find the eleventh number by subtracting the sum of the first ten numbers from the sum of all eleven numbers: 198 (sum of all eleven numbers) - 160 (sum of first ten numbers) = 38 So the eleventh number is 38.

**Question 4**
**Report**

In how many ways can 2 students be selected from a group of 5 students in a debating competition?

**Answer Details**

The number of ways to select 2 students from a group of 5 students is 10. Imagine you're picking two students from a lineup of 5 students. You can pick the first student in 5 different ways (student 1, student 2, student 3, student 4, or student 5). Once you've picked the first student, there are only 4 students left in the lineup, so you can only pick the second student in 4 different ways. So, to find the total number of ways to select 2 students from a group of 5, you multiply the number of ways to pick the first student by the number of ways to pick the second student: 5 * 4 = 20. However, since order doesn't matter (it doesn't matter if you pick student 1 first or student 2 first), you divide the total number of ways by 2 to account for the overcounting. So, the final answer is 20 / 2 = 10.

**Question 5**
**Report**

Solve the following equation: 2(2r−1) $\frac{2}{(2r-1)}$ - 53 $\frac{5}{3}$ = 1(r+2)

**Answer Details**

2(2r−1) $\frac{2}{(2r-1)}$ - 53 $\frac{5}{3}$ = 1(r+2) $\frac{1}{(r+2)}$

2(2r−1) $\frac{2}{(2r-1)}$ - 1(r+2) $\frac{1}{(r+2)}$ = 53 $\frac{5}{3}$

The L.C.M.: (2r - 1) (r + 2)

2(r+2)−1(2r−1)(2r−1)(r+2) $\frac{2(r+2)-1(2r-1)}{(2r-1)(r+2)}$ = 53 $\frac{5}{3}$

2r+4−2r+1(2r−1)(r+2) $\frac{2r+4-2r+1}{(2r-1)(r+2)}$ = 53 $\frac{5}{3}$

cross multiply the solution

3 = (2r - 1) (r + 2) or 2r2 ${}^{2}$ + 3r - 2 (when expanded)

collect like terms

2r2 ${}^{2}$ + 3r - 2 - 3 = 0

2r2 ${}^{2}$ + 3r - 5 = 0

Factorize to get x = 1 or - 52

**Question 6**
**Report**

If (x + 2) and (x - 1) are factors of the expression Lx+2kx2+24 $Lx+2k{x}^{2}+24$, find the values of L and k.

**Answer Details**

Given (x + 2) and (x - 1), i.e. x = -2 or +1

when x = -2

L(-2) + 2k(-2)2 ${}^{2}$ + 24 = 0

f(-2) = -2L + 8k = -24...(i)

And x = 1

L(1) + 2k(1) + 24 = 0

f(1):L + 2k = -24...(ii)

Subst, L = -24 - 2k in eqn (i)

-2(-24 - 2k) + 8k = -24

+48 + 4k + 8k = -24

12k = -24 - 48 = -72

k = frac−7212 $frac-7212$

k = -6

where L = -24 - 2k

L = -24 - 2(-6)

L = -24 + 12

L = -12

That is; K = -6 and L = -12

**Question 7**
**Report**

The locus of a point which moves so that it is equidistant from two intersecting straight lines is the?

**Answer Details**

The required locus is angle bisector of the two lines

**Question 8**
**Report**

wo numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

**Answer Details**

The probability that the sum of two numbers removed is even can be found by counting the number of even sums and dividing it by the total number of possible sums. Out of the 4 numbers, 2 are even (2 and 4) and 2 are odd (1 and 3). If we take one even and one odd number, the sum will be odd. If we take two even numbers, the sum will be even. And if we take two odd numbers, the sum will be even. So, there are 3 ways to get an even sum (taking 2 and 4, taking 2 and 2, or taking 1 and 3) and 6 total ways to choose two numbers (since order doesn't matter). So, the probability of getting an even sum is 3/6 or 1/2. In simple terms, there's a 50-50 chance that the sum of two numbers removed will be even.

**Question 9**
**Report**

find the value of p if the line which passes through (-1, -p) and (-2,2) is parallel to the line 2y+8x-17=0?

**Answer Details**

Line: 2y+8x-17=0

recall y = mx + c

2y = -8x + 17

y = -4x + 172 $\frac{17}{2}$

Slope m1 ${}_{1}$ = 4

parallel lines: m1 ${}_{1}$. m2 ${}_{2}$ = -4

where Slope ( -4) = y2−y1x2−x1 $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ at points (-1, -p) and (-2,2)

-4( x2−x1 ${x}_{2}-{x}_{1}$ ) = y2−y1 ${y}_{2}-{y}_{1}$

-4 ( -2 - -1) = 2 - -p

p = 4 - 2 = 2

**Question 10**
**Report**

Find the derivative of the function y = 2x2 ${}^{2}$(2x - 1) at the point x = -1?

**Answer Details**

y = 2x2
${}^{2}$(2x - 1)

y = 4x3
${}^{3}$ - 2x2
${}^{2}$

dy/dx = 12x2
${}^{2}$ - 4x

at x = -1

dy/dx = 12(-1)2
${}^{2}$ - 4(-1)

= 12 + 4

= 16

**Question 11**
**Report**

The angle of a sector of a circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector

**Answer Details**

Length of Arc AB = θ360
$\frac{\theta}{360}$ 2π
$\pi $r

= 48360
$\frac{48}{360}$ x 2227
$\frac{22}{7}$ x 212
$\frac{21}{2}$

= 4×22××330
$\frac{4\times 22\times \times 3}{30}$ 8810
$\frac{88}{10}$ = 8.8cm

Perimeter = 8.8 + 2r

= 8.8 + 2(10.5)

= 8.8 + 21

= 29.8cm

**Question 12**
**Report**

X is due east point of y on a coast. Z is another point on the coast but 6.0km due south of Y. If the distance ZX is 12km, calculate the bearing of Z from X

**Answer Details**

Sinθ = 612 $\frac{6}{12}$

Sinθ = 12 $\frac{1}{2}$

θ = Sin0.5
${}^{0}.5$

θ = 30°

Bearing of Z from X, (270 - 30)° = 240°

**Question 13**
**Report**

What is the rate of change of the volume V of a hemisphere with respect to its radius r when r = 2?

**Answer Details**

The formula for the volume of a hemisphere is given by (2/3)πr^3, where r is the radius. To find the rate of change of the volume with respect to the radius, we need to take the derivative of this formula with respect to r. So, dV/dr = (2/3) * π * 3r^2 * dr/dr = (2/3) * π * 3r^2 = 2πr^2. Now, when r = 2, the rate of change of the volume with respect to the radius is given by 2π * 2^2 = 8π. So, the answer is 8π.

**Question 14**
**Report**

Simplify 2log 25 $\frac{2}{5}$ - log72125 $\frac{72}{125}$ + log 9

**Answer Details**

To simplify the expression 2log 25 - log 72125 + log 9, we need to use the logarithmic properties of logarithms. One of the properties of logarithms is log a^b = b log a, which allows us to rewrite log 25 as 5 log 5, log 72125 as 5 log 72, and log 9 as 2 log 3. Using this property, the expression becomes 2 log 5 - 5 log 72 + 2 log 3. Another property of logarithms is log a / log b = log a / b, which allows us to simplify the logarithm of the ratio of two numbers as the logarithm of the first number divided by the logarithm of the second number. Using this property, the expression becomes 2 log 5 - (5 / 5) log 72 + 2 log 3, which simplifies to 2 log 5 - log 72 + 2 log 3. Finally, we can simplify the expression further by using the property log a^m = m log a, which allows us to rewrite log 72 as log 2^6, so that the expression becomes 2 log 5 - 6 log 2 + 2 log 3. So, the simplified expression is 2 log 5 - 6 log 2 + 2 log 3 = 2 log (5 * 3^2 / 2^6) = 2 log (3^2 / 2^4) = 2 log 3 - 4 log 2 = 2 * 1.0986 - 4 * 0.3010 = 2.1976 - 1.2040 = 1 - 2 log 2. So, the answer is (D) 1 - 2 log 2.

**Question 15**
**Report**

Simplify 27−−√ $\sqrt{27}$ + 33√

**Answer Details**

√ $27$ + 33√ $\frac{3}{\sqrt{3}}$

= 9×3−−−−√ $\sqrt{9\times 3}$ + 3×3√3√×3√ $\frac{3\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

= 33–√ $\sqrt{3}$ + 3–√ $\sqrt{3}$

= 43–√

**Question 16**
**Report**

A trapezium has two parallel sides of lengths 5cm and 9cm. If the area is 91cm2 ${}^{2}$, find the distance between the parallel sides

**Answer Details**

Area of Trapezium = 1/2(sum of parallel sides) * h

91 = 12
$\frac{1}{2}$ (5 + 9)h

cross multiply

91 = 7h

h = 917
$\frac{91}{7}$

h = 13cm

**Question 17**
**Report**

A car travels from calabar to Enugu, a distance of P km with an average speed of U km per hour and continues to benin, a distance of Q km, with an average speed of Wkm per hour. Find its average speed from Calabar to Benin

**Answer Details**

Average speed = totalDistanceTotalTime
$\frac{totalDistance}{TotalTime}$

from Calabar to Enugu in time t1, hence

t1 = PU
$\frac{P}{U}$

Also from Enugu to Benin

t2 qw
$\frac{q}{w}$

Av. speed = p+qt1+t2
$\frac{p+q}{{t}_{1}+{t}_{2}}$

= p+qp/u+q/w
$\frac{p+q}{p/u+q/w}$

= p + q x uwpw+qu
$\frac{uw}{pw+qu}$

= uw(p+q)pw+qu

**Question 18**
**Report**

Factorize completely 81a4 ${}^{4}$ - 16b4

**Answer Details**

81a4
${}^{4}$ - 16b4
${}^{4}$ = (9a2
${}^{2}$)2
${}^{2}$ - (4b2
${}^{2}$)2
${}^{2}$

= (9a2
${}^{2}$ + 4b2
${}^{2}$)(9a2
${}^{2}$ - 4b2
${}^{2}$)

N:B 9a2
${}^{2}$ - 4b2
${}^{2}$ = (3a - 2b)(3a + 2b)

**Question 19**
**Report**

Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?

**Answer Details**

A regular polygon is a polygon with equal sides and equal interior angles. The formula to find the interior angle of a regular polygon is given by: (180 * (n-2)) / n, where n is the number of sides in the polygon. So, if each interior angle of a regular polygon is 140°, we can plug that value into the formula and solve for n: 140 = (180 * (n-2)) / n Multiplying both sides by n gives: 140n = 180 * (n-2) Expanding the right side gives: 140n = 180n - 360 Subtracting 140n from both sides gives: 0 = 40n - 360 Adding 360 to both sides gives: 360 = 40n Dividing both sides by 40 gives: n = 9 So, the polygon has 9 sides.

**Question 20**
**Report**

At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50?

**Answer Details**

nterest I = PRT100
$\frac{PRT}{100}$

∴ R = 100×1100×5
$\frac{100\times 1}{100\times 5}$

= 100×7.50500×5