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Question 1 Report
Rationalize 2√3+3√23√2−2√3
Answer Details
2√3+3√23√2−2√3
= 2√3+3√23√2−2√3
x 3√2+2√33√2−2√3
4(3)+9(2)+2(6)√69(2)−4(3)
12+18+12√61‘8−12
= 30+12√66
= 5 + 2√6
Question 3 Report
Factorize 1 - (a - b)2
Answer Details
1 - (a - b)2 = [1 + (a - b)][1 - a + b]
= (1 + a - b)(1 - a + b)
Question 4 Report
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm
Question 5 Report
Obi borrows ₦10.00 at 2% per month simple interest and repays ₦8.00 after 4 months, how much does he still owe?
Question 6 Report
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
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 8 Report
If 2257 is the result of subtracting 4577 from 7056 in base n, find n
Answer Details
7056- 45772257
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 9 Report
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
= 10×2×4100
= 45
= 0.8
Total amount = N10.80
He pays N8.00
Remainder = 10.80 - 8.00
= N2.80
Question 10 Report
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water?
Answer Details
% of water in the mixture
= Total Amount of waterTotal quantity of spirit
x 1001
3(20100)+5(15100)3+5
x 1001
= 610+751008
x 1001
= 0.6+0.758
x 1001
= 1.358
x 1001
= 33.752
= 16.875
= 1678
Question 11 Report
Simplify 2log 25 - log72125 + log 9
Answer Details
2log 25
- log72125
+ log 9
[25
)2 x 9] = log 425
x 91
x 12572
= log 72125
= log 52
= log 104
= log 10 - log 4
= log10 10 - log10 22
= 1 - 2 log2
Question 12 Report
If the exterior angles of a pentagon are xo, (x + 5)o, (x + 10), (x + 15)o and (x + 20)o, find x
Answer Details
Question 13 Report
If cos x = √ab
find coses x
Answer Details
cosx = √ab
y2 + √(a)2
= √(b)2
by pythagoras
y2 = b - a
∴ y = b - a
cosec x = 1sinx
= 1y
by
= √b√b−a
= √bb−a
Question 14 Report
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 15 Report
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 16 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 w km per hour. Find its average speed from Calabar to Benin
Answer Details
Average speed = total DistanceTotal Time
from Calabar to Enugu in time t1, hence
t1 = pu
also from Enugu to Benin
t2 = qw
Average speed = p+qt1+t2
= p+qpu+qw
= p + q × uwpw+qu
= uw(p+q)pw+qu
Question 17 Report
Given that 1/2 log10 P = 1, find the value Of P
Question 19 Report
Find the positive number n such that thrice its square is equal to 12 times the number.
Answer Details
The problem can be translated into an equation using algebra. Let's start by translating the words into math symbols. Let n be the positive number we are looking for. The phrase "thrice its square" means 3 times n^2 or 3n^2. The phrase "12 times the number" means 12 times n or 12n. So we can translate the sentence "thrice its square is equal to 12 times the number" into the equation: 3n^2 = 12n Now we can solve for n. First, we can simplify the equation by dividing both sides by 3: n^2 = 4n Next, we can rearrange the equation by subtracting 4n from both sides: n^2 - 4n = 0 Now we can factor out an n: n(n - 4) = 0 So the solutions to this equation are n = 0 and n - 4 = 0, which gives n = 4. However, we are looking for a positive value of n, so we can discard the solution n = 0. Therefore, the positive number n such that thrice its square is equal to 12 times the number is n = 4. Therefore, the answer is 4.
Question 20 Report
Simplify cos2 x (sec2x + sec2 x tan2x)
Answer Details
To simplify cos2 x (sec2x + sec2 x tan2x), we can use the trigonometric identity: 1 + tan2 x = sec2 x. First, we can simplify sec2 x + sec2 x tan2 x: sec2 x + sec2 x tan2 x = sec2 x (1 + tan2 x) Next, we can substitute sec2 x (1 + tan2 x) into the original equation: cos2 x (sec2 x + sec2 x tan2 x) = cos2 x (sec2 x (1 + tan2 x)) Using the identity 1 + tan2 x = sec2 x, we can simplify: cos2 x (sec2 x (1 + tan2 x)) = cos2 x (sec2 x)(sec2 x) Finally, using the identity sec2 x = 1/cos2 x, we can simplify: cos2 x (sec2 x)(sec2 x) = cos2 x (1/cos2 x)(1/cos2 x) = 1/cos2 x = sec2 x Therefore, the answer is 1+tan2xsec2x.
Question 21 Report
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 22 Report
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.
Answer Details
oin PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o
Question 23 Report
One angle of a rhombus is 60o. The shorter of the two diagonals is 8cm long. Find the length of the longer one.
Answer Details
In a rhombus, all sides are congruent, and opposite angles are equal. If one angle of the rhombus is 60 degrees, then the opposite angle is also 60 degrees. Let the shorter diagonal of the rhombus be divided into two equal halves. Let the half of the shorter diagonal be `a`. Using the properties of a 30-60-90 degree triangle, we know that the length of the longer diagonal of the rhombus is `2a√3`. Since the shorter diagonal is 8cm, `a = 4cm`. Therefore, the length of the longer diagonal is: `2a√3 = 2(4cm)√3 = 8√3 cm` Therefore, the length of the longer diagonal of the rhombus is `8√3 cm`. So the correct answer is: 8√3.
Question 24 Report
Simplify 313 - 114 x 23 + 125
Answer Details
3 - 13
- (54
x 23
) + 125
= 103
- 56
+ 75
= 100−25+4230
= 11730
= 3.9
≈
4
Question 25 Report
What is the product of 2751 (3)-3 and (1)−15 ?
Answer Details
2751
(3)-3 x (1)−15
= 275
x 133
x 115
= 275
x 127
x 51
= 1
Question 26 Report
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 27 Report
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 29 Report
In this figure, PQ = PR = PS and SRT = 68∘
. Find QPS.
Answer Details
Since PQRS is quadrilateral 2y + 2x = QPS = 360∘
i.e. 2(y + x) + QPS = 360∘
QPS = 360∘ - 2(y + x)
But x + y + 68