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**Question 2**
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Rationalize 2√3+3√23√2−2√3

**Answer Details**

2√3+3√23√2−2√3
$\frac{2\sqrt{3}+3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}}$

= 2√3+3√23√2−2√3
$\frac{2\sqrt{3}+3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}}$ x 3√2+2√33√2−2√3
$\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$

4(3)+9(2)+2(6)√69(2)−4(3)
$\frac{4(3)+9(2)+2(6)\sqrt{6}}{9(2)-4(3)}$

12+18+12√61‘8−12
$\frac{12+18+12\sqrt{6}}{1\u20188-12}$

= 30+12√66
$\frac{30+12\sqrt{6}}{6}$

= 5 + 2√6

**Question 3**
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Multiply (x2 - 3x + 1) by (x - a)

**Answer Details**

To multiply the expression (x^2 - 3x + 1) by (x - a), we need to use the distributive property of multiplication over addition. We start by multiplying each term in the first expression (x^2 - 3x + 1) by (x - a) and then add the resulting terms. (x^2 - 3x + 1) * (x - a) = x^3 - ax^2 - 3x^2 + 3ax + x - a = x^3 - (a+3)x^2 + (3a+1)x - a Therefore, the correct option is: - x^3 - (a+3)x^2 + (3a+1)x - a

**Question 4**
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If the exterior angles of a pentagon are xo, (x + 5)o, (x + 10), (x + 15)o and (x + 20)o, find x

**Question 5**
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The area of a square is 144 sq cm. Find the length of its diagonal

**Answer Details**

The area of a square is given as 144 sq cm. To find the length of its diagonal, we need to use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides. In the case of a square, we know that all sides are equal in length, so we can use this fact to simplify the equation. Let's call the length of one side of the square "s". Then, we know that the area of the square is s^2 = 144 sq cm. Solving for "s", we get s = sqrt(144) = 12 cm. Now, let's draw a diagonal line across the square, splitting it into two right-angled triangles. We can label the hypotenuse of each triangle as "d" (which is the length of the diagonal we want to find), and the other two sides as "s". Using the Pythagorean theorem, we get: d^2 = s^2 + s^2 d^2 = 2s^2 d^2 = 2(12^2) d^2 = 288 d = sqrt(288) = 12sqrt(2) So the length of the diagonal is 12sqrt(2) cm.

**Question 6**
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Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result:

No. defective per box456789No. of boxes27171086 $\begin{array}{ccccccc}\text{No. defective per box}& 4& 5& 6& 7& 8& 9\\ \text{No. of boxes}& 2& 7& 17& 10& 8& 6\end{array}$

Find the percentage of boxes containing at least 5 defective bolts each.

**Answer Details**

To find the percentage of boxes containing at least 5 defective bolts each, we need to first add up the number of boxes that have 5, 6, 7, 8, or 9 defective bolts. These are the boxes that contain at least 5 defective bolts. From the table, we can see that there are: - 7 boxes with 5 defective bolts - 17 boxes with 6 defective bolts - 10 boxes with 7 defective bolts - 8 boxes with 8 defective bolts - 6 boxes with 9 defective bolts Adding these up, we get a total of 7+17+10+8+6 = 48 boxes. Therefore, out of the 50 boxes that were inspected, 48 of them contain at least 5 defective bolts each. To find the percentage, we can use the formula: percentage = (part/whole) x 100 In this case, the "part" is the number of boxes containing at least 5 defective bolts (48), and the "whole" is the total number of boxes inspected (50). Substituting in the values, we get: percentage = (48/50) x 100 = 96% Therefore, the answer is 96%, option (A).

**Question 7**
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Evaluate x2(x2 - 1)-12 $\frac{1}{2}$ - (x2 - 1)12

**Answer Details**

x2(x2 - 1)-12
$\frac{1}{2}$ - (x2 - 1)12
$\frac{1}{2}$ = x2(x2?1)12
$\frac{{x}^{2}}{({x}^{2}?1{)}^{\frac{1}{2}}}$ - (x2?1)121
$\frac{({x}^{2}?1{)}^{\frac{1}{2}}}{1}$

= x2?(x2?1)(x2?1)12
$\frac{{x}^{2}?({x}^{2}?1)}{({x}^{2}?1{)}^{\frac{1}{2}}}$

= x2?x2+1(x2?1)12
$\frac{{x}^{2}?{x}^{2}+1}{({x}^{2}?1{)}^{\frac{1}{2}}}$

= (x2 - 1)-12

**Question 8**
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Which of the following is a factor of rs + tr - pt - ps?

**Question 9**
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Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result:

No. defective per box456789No. of boxes27171086 $\begin{array}{ccccccc}\text{No. defective per box}& 4& 5& 6& 7& 8& 9\\ \text{No. of boxes}& 2& 7& 17& 10& 8& 6\end{array}$

The mean and the median of the distribution are respectively

**Answer Details**

To find the mean of the distribution, we need to calculate the sum of all the defective bolts in the boxes and divide it by the total number of boxes: Mean = (4x2 + 5x7 + 6x17 + 7x10 + 8x8 + 9x6) / 50 = 335 / 50 = 6.7 To find the median of the distribution, we need to arrange the data in order of increasing magnitude and find the middle value. In this case, we have an even number of observations (50), so the median will be the average of the two middle values, which are the 25th and 26th observations. From the table, we can see that the 25th and 26th observations are both 6, so the median is: Median = (6 + 6) / 2 = 6 Therefore, the mean and the median of the distribution are 6.7 and 6, respectively.

**Question 10**
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Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe?

**Answer Details**

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

= 10×2×4100
$\frac{10\times 2\times 4}{100}$

= 45
$\frac{4}{5}$

= 0.8

Total amount = N10.80

He pays N8.00

Remainder = 10.80 - 8.00

= N2.80

**Question 11**
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Given that 1/2 log10 P = 1, find the value Of P

**Question 12**
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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 w km per hour. Find its average speed from Calabar to Benin

**Answer Details**

Average speed = total DistanceTotal Time
$\frac{\text{total Distance}}{\text{Total Time}}$

from Calabar to Enugu in time t1, hence

t1 = pu
$\frac{p}{u}$ also from Enugu to Benin

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

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

= p+qpu+qw
$\frac{p+q}{\frac{p}{u}+\frac{q}{w}}$

= p + q × uwpw+qu
$\frac{uw}{pw+qu}$

= uw(p+q)pw+qu

**Question 13**
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Obi borrows ₦10.00 at 2% per month simple interest and repays ₦8.00 after 4 months, how much does he still owe?

**Question 14**
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Express 62+3 $62+3$ as decimal correct to 3 significant figures.

**Question 15**
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If w varies inversely as uru+r $\frac{ur}{u+r}$ and is equal to 8 when u = 2 and r = 6, find a relationship between u, v, w.

**Answer Details**

W α
$\alpha $ 1uvu+v
$\frac{\frac{1}{uv}}{u+v}$

∴ w = kuvu+v
$\frac{\frac{k}{uv}}{u+v}$

= k(u+v)uv
$\frac{k(u+v)}{uv}$

w = k(u+v)uv
$\frac{k(u+v)}{uv}$

w = 8, u = 2 and v = 6

8 = k(2+6)2(6)
$\frac{k(2+6)}{2(6)}$

= k(8)12
$\frac{k(8)}{12}$

k = 12

**Question 16**
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If 2257 is the result of subtracting 4577 from 7056 in base n, find n

**Answer Details**

7056- 45772257
$\begin{array}{c}7056\\ \text{- 4577}\\ 2257\end{array}$

By trial and error method

Let the base to 8

i.e. Let n = 8 and it is easily verified that the subtraction holds.

The subtraction does not hold when other values of n are tried

n = 8

**Question 17**
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What is the nth term of the progression 27, 9, 3,......?

**Answer Details**

The given progression is decreasing by a factor of 3 in each term, starting from 27. Therefore, the nth term can be calculated by multiplying 27 by 1/3 raised to the power of (n-1). This is because in the nth term, the factor of 3 has been applied (n-1) times to the first term 27. Therefore, the answer is: 27 x (1/3)^(n-1)

**Question 18**
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Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x2 + xy = 6

**Answer Details**

We can solve this system of equations by substitution or elimination method. Here, we will use substitution method. First, we solve the first equation for y in terms of x: 3x + y = 8 y = 8 - 3x Now we substitute this expression for y into the second equation and solve for x: x^2 + xy = 6 x^2 + x(8 - 3x) = 6 x^2 + 8x - 3x^2 = 6 -2x^2 + 8x - 6 = 0 x^2 - 4x + 3 = 0 (x - 3)(x - 1) = 0 x = 3 or x = 1 Now we substitute each of these values of x into the expression we found for y earlier: y = 8 - 3x When x = 3, y = -1 When x = 1, y = 5 Therefore, the two values of y that satisfy the system of equations are -1 and 5. So, the correct answer is: - -1 and 5

**Question 19**
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Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm

**Question 21**
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Simplify 313 $\frac{1}{3}$ - 114 $\frac{1}{4}$ x 23 $\frac{2}{3}$ + 125

**Answer Details**

3 - 13
$\frac{1}{3}$ - (54
$\frac{5}{4}$ x 23
$\frac{2}{3}$) + 125
$\frac{2}{5}$

= 103
$\frac{10}{3}$ - 56
$\frac{5}{6}$ + 75
$\frac{7}{5}$

= 100−25+4230
$\frac{100-25+42}{30}$

= 11730
$\frac{117}{30}$

= 3.9

≈
$\approx $ 4

**Question 22**
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From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X.

**Answer Details**

To find the bearing of Y from X, we need to determine the angle between the direction of due north and the direction of XY. First, we can draw a diagram to represent the problem. Let X be the origin, and let Y be the point that is 60√3 m east and some distance north of X. Let Z be the point that is 60 m due north of X, and let W be the point that is due west of Y and due south of Z. [Diagram not included as plain text] We can see that △XYZ is a right-angled triangle, with XY as the hypotenuse, XZ as the adjacent side, and YZ as the opposite side. We know that XZ has a length of 60 m, and YZ has a length of 60√3 m. Therefore, we can use the tangent ratio to find the angle between XZ and XY: tan θ = YZ/XZ tan θ = (60√3)/60 tan θ = √3/1 θ = tan⁻¹(√3) θ = 60° Therefore, the bearing of Y from X is 060°. Option (C) is the correct answer.

**Question 23**
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Solve the equation (x - 2) (x - 3) = 12

**Answer Details**

(x - 2) (x - 3) = 12

x2 - 3x - 2x + 6 = 12

x2 - 5x - 6 = 0

(x - 1)(x - 6) = 0

x = -1 or 6

**Question 24**
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What is the product of 2751 $\frac{27}{{5}^{1}}$(3)-3 and (1)−15 $\frac{(1{)}^{-1}}{5}$?

**Answer Details**

2751
$\frac{27}{{5}^{1}}$(3)-3 x (1)−15
$\frac{(1{)}^{-1}}{5}$ = 275
$\frac{27}{5}$ x 133
$\frac{1}{{3}^{3}}$ x 115
$\frac{1}{\frac{1}{5}}$

= 275
$\frac{27}{5}$ x 127
$\frac{1}{27}$ x 51
$\frac{5}{1}$

= 1

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

**Answer Details**

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

[25
$\frac{2}{5}$)2 x 9] = log 425
$\frac{4}{25}$ x 91
$\frac{9}{1}$ x 12572
$\frac{125}{72}$

= log 72125
$\frac{72}{125}$

= log 52
$\frac{5}{2}$

= log 104
$\frac{10}{4}$

= log 10 - log 4

= log10 10 - log10 22

= 1 - 2 log2

**Question 26**
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PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters

**Answer Details**

In this problem, we are given a circle MQTRN and two secants PMN and PQR intersecting at point T and a tangent PT. First, we can use the theorem that states that the product of the lengths of the two segments of each secant is equal. So we have: PM * PN = PT * PR Substituting the given values, we get: 5cm * 12cm = PT * PR 60cm² = PT * PR Next, we need to find the respective lengths of PR and PT. To do this, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. In triangle PRT, PT is the hypotenuse, so we have: PT² = PR² + RT² We can also use the fact that RT is perpendicular to PT, which means that triangle PRT is a right triangle. To find RT, we can use the theorem that states that the product of the segments of a secant and its external part is equal. So we have: PT * TP = QT * TR Substituting the given values, we get: PT * PT = 4.8cm * TR PT² = 4.8cm * TR TR = PT² / 4.8cm Now we can substitute RT in the Pythagorean theorem and simplify: PT² = PR² + (PT² / 4.8cm)² PT⁴ / (4.8cm)² = PR² PR = sqrt(PT⁴ / (4.8cm)²) = PT² / 4.8cm Finally, we can substitute the value of PR in the first equation we obtained and solve for PT: 60cm² = PT * (PT² / 4.8cm) PT³ = 288cm³ PT = cube root of 288cm³ ≈ 6.67cm Now we can substitute the value of PT in the equation we obtained for PR: PR = PT² / 4.8cm ≈ 7.7cm Therefore, the respective lengths of PR and PT are approximately 7.7cm and 6.67cm, respectively. So the answer is 7.7, 12.5.

**Question 27**
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Simplify √1+x+√x√1+x−√x

**Question 28**
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If cos x = √ab
$\sqrt{\frac{a}{b}}$ find coses x

**Answer Details**

cosx = √ab
$\sqrt{\frac{a}{b}}$

y2 + √(a)2
$\sqrt{(a{)}^{2}}$ = √(b)2
$\sqrt{(b{)}^{2}}$ by pythagoras

y2 = b - a

∴ y = b - a

cosec x = 1sinx
$\frac{1}{sinx}$ = 1y
$\frac{1}{y}$

by
$\frac{b}{y}$ = √b√b−a
$\frac{\sqrt{b}}{\sqrt{b-a}}$

= √bb−a

**Question 29**
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Simplify 11+?5 $\frac{1}{1+\sqrt{5}}$ - 11??5

**Answer Details**

11+√5
$\frac{1}{1+\sqrt{5}}$ - 11−√5
$\frac{1}{1-\sqrt{5}}$

= 3−√5−3−√5(3+√5)(3−√5
$\frac{3-\sqrt{5}-3-\sqrt{5}}{(3+\sqrt{5})(3-\sqrt{5}}$

= −2√