JAMB UTME - Mathematics - 2010

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If p : q = $\frac{2}{3}$ : $\frac{5}{6}$, then the sum S1 of ratio = $\frac{2}{3}$ + $\frac{5}{6}$ = $\frac{9}{6}$
If q : r = $\frac{3}{4}$ : $\frac{1}{2}$, then the sum S2 of ratio = $\frac{3}{4}$ + $\frac{1}{2}$ = $\frac{5}{4}$
Let p + q = T1, then
q = ($\frac{5}{6}÷\frac{9}{6}$)T1 = ($\frac{5}{6}\times \frac{6}{9}$)T1 = $\frac{5}{9}$T1
Again, let q + r = T2, then
q = ($\frac{3}{4}÷\frac{5}{4}$)T2 = ($\frac{3}{4}\times \frac{4}{5}$)T2 = $\frac{3}{5}$T2
Using q = q
$\frac{5}{9}$T1 = $\frac{3}{5}$T2
5 x 5T1 = 9 x 3T2
$\frac{T_1}{T_2}$ = $\frac{9\times 3}{5x5}$ = $\frac{27}{5}$
Giving that, T1 = 27 and T2 = 25
P = ($\frac{2}{3}÷S_1$)T1 = ($\frac{2}{3}÷\frac{9}{6}$)T1
= ($\frac{2}{3}\times \frac{6}{9}$)27 = 12
q = ($\frac{5}{6}÷S_1$)T1 = ($\frac{5}{6}÷\frac{9}{6}$)T1
= ($\frac{5}{6}\times \frac{6}{9}$)27 = 15
and r = ($\frac{1}{2}÷S_2$)T2 = ($\frac{1}{2}÷\frac{5}{4}$)T2
= ($\frac{1}{2}\times \frac{4}{5}$)25 = 10
Hence p : q : r = 12: 15 : 10

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-6(x + 3) $\le$ 4(x - 2)
-6(x +3) $\le$ 4(x - 2)
-6x -18 $\le$ 4x - 8
-18 + 8 $\le$ 4x +6x
-10x $\le$ 10x
10x $\le$ -10
x $\le$ 1

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$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{9}{16}{)}^{-\frac{1}{2}}$
$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{16}{9}{)}^{\frac{1}{2}}$
$(\frac{2^4}{3^4}{)}^{\frac{1}{4}}÷(\frac{4^2}{3^2}{)}^{\frac{1}{2}}$
$\frac{2^{4\times \frac{1}{4}}}{3^{4\times \frac{1}{4}}}÷\frac{4^{2\times \frac{1}{2}}}{3^{2\times \frac{1}{2}}}$
$\frac{2}{3}÷\frac{4}{3}$
$\frac{2}{3}\times \frac{3}{4}$
$\frac{2}{4}$
$\frac{1}{2}$

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Let A denote the area of $△$PQR, then A = $\frac{1}{2}bh$

Using Sin 60${}^{\circ }$ = $\frac{h}{q}$

h = q sin 60${}^{\circ }$

So A = $\frac{1}{2}b\left(q\sin {60}^o\right)$

12$\sqrt{3}=\frac{1}{2}\times 8\times q\times \frac{\sqrt{3}}{3}$

12$\sqrt{3}$ - 2q$\sqrt{3}$

q = $\frac{12}{2}=6$cm

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27x3 - 8y3 = (3x - 2y)3
But 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y

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P = $\left(\begin{array}{cc}2& ?3\\ 1& 1\end{array}\right)$
|P| = 2 - 1 x -3 = 5
P-1 = $\frac{1}{5}$$\left(\begin{array}{cc}1& 3\\ ?1& 2\end{array}\right)$
= $\left(\begin{array}{cc}\frac{1}{5}& \frac{3}{5}\\ ?\frac{1}{5}& \frac{2}{5}\end{array}\right)$

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To rationalize the given expression, we need to eliminate the radical from the denominator. To do that, we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of √5-√3 is √5+√3. Therefore, we have: (2√3+√5) / (√5-√3) x (√5+√3) / (√5+√3) Simplifying the numerator and the denominator using FOIL (First, Outer, Inner, Last) method, we get: = [2√3(√5) + 2√3(√3) + √5(√5) + √5(√3)] / [(√5)(√5) - (√3)(√5) + (√5)(√3) - (√3)(√3)] = [2√15 + 6 + 5 + √15] / [5 - 3 + √15 - 3] = [3√15 + 11] / 2 Therefore, the answer is (3√15 + 11) / 2.

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To find out how many people read both Champion and Guardian, we need to use a concept called "intersection" from mathematics. Out of 50 readers, 40 read Champion and 30 read Guardian. We need to find out how many people are reading both Champion and Guardian. To do this, we can draw two circles to represent the readers who read Champion and those who read Guardian. Then, we can see how much they overlap, which is the number of people who read both. So, if we draw two circles, one for Champion and one for Guardian, we can see that the overlapping region represents the people who read both newspapers. Since we don't have a visual representation, we can use a formula to find the answer. We can use the formula: Number of people who read both = Number of people who read Champion + Number of people who read Guardian - Total number of people Substituting the given values, we get: Number of people who read both = 40 + 30 - 50 Number of people who read both = 20 Therefore, the answer is 20.

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The expression |x327| means the absolute value of x to the power of 327. The given equation |x327| = 15 means that the absolute value of x to the power of 327 is equal to 15. To solve for x, we can take the 327th root of both sides of the equation. Thus, we have: |x327| = 15 Taking the 327th root of both sides: |x| = 15^(1/327) Since x can be positive or negative, we have two solutions: x = 15^(1/327) or x = -15^(1/327) Using a calculator, we can approximate the value of x as approximately 2.905 or -2.905. However, only one of these values is among the answer choices, which is x = 3. Therefore, the correct answer is 3.

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(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = $\frac{{410}^o}{5}$
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o

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An arithmetic progression is a sequence of numbers where each term is obtained by adding a fixed value to the previous term. Let's call this fixed value "d". Then, the nth term of an arithmetic progression can be expressed as: an = a1 + (n-1)d where "an" is the nth term, "a1" is the first term and "n" is the position of the term. In this problem, we are given the 3rd and 7th terms, which are -9 and -29 respectively. Using the formula above, we can write two equations: a3 = a1 + 2d = -9 a7 = a1 + 6d = -29 We can solve this system of equations to find "a1" and "d". First, we can subtract the first equation from the second equation: 4d = -20 This gives us d = -5. Substituting this value of "d" into the first equation, we get: a1 + 2(-5) = -9 a1 = 1 So the first term is 1, and the common difference is -5. Now we can use the formula to find the 10th term: a10 = a1 + 9d a10 = 1 + 9(-5) a10 = -44 Therefore, the 10th term of the arithmetic progression is -44. Option A is the correct answer.

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x * y = x + y2
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36

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$\int$(Sin x + 2)dx = -cos x + 2x + k

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To find the second derivative of the given function, we need to differentiate it twice with respect to x. First, we differentiate y with respect to x using the product rule: y = x sin x y' = x cos x + sin x Then, we differentiate y' with respect to x using the product rule again: y' = x cos x + sin x y'' = cos x - x sin x + cos x Simplifying the expression: y'' = 2cos x - x sin x Therefore, the second derivative of y = x sin x is y'' = 2cos x - x sin x.

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TUVW is a cyclic quad
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3χ = 24${}^{\circ }$

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To find the mean mark, we need to calculate the sum of all the marks obtained and divide it by the total number of students. The sum of all the marks obtained can be found by multiplying each mark by its corresponding frequency and adding up the results. So, sum of all marks = (1 x 2) + (2 x 2) + (3 x 8) + (4 x 4) + (5 x 4) = 2 + 4 + 24 + 16 + 20 = 66 The total number of students can be found by adding up all the frequencies. So, total number of students = 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark = (sum of all marks) / (total number of students) = 66 / 20 = 3.3 Hence, the answer is 3.3.

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Area of a triangle = 1/2 ab Sinθ
12√3 = 1/2 x 8 x q sin 60
12√3 = 4 x q x √3/2
12√3 = 2q√3

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y = (2x + 1)3
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2

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By comparing y = mx + c with y = -4x + 2, the gradient of y = -4x + 2 is m1 = -4
Let the gradient of the line parallel to the given line be m2,
then, m2 = m1 = -4 (condition for parallelism)
Using: y - y1 = m2(x - x1)
Hence the equation of the parallel line is
y - 3 = -4(x-2)
y - 3 = -4 x + 8
y + 4x = 8 + 3
y + 4x = 11
y + 4x - 11 = 0

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If y = (2x + 1)3, then
Let u = 2x + 1 so that, y = u3
$\frac{dy}{du}$ = 3u2 and $\frac{dy}{dx}$ = 2
Hence by the chain rule,
$\frac{dy}{dx}$ = $\frac{dy}{du}$ x $\frac{du}{dx}$
= 3u2 x 2
= 6u2
= 6(2x + 1)2

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The formula for simple interest is: I = PRT Where: I = Interest P = Principal R = Rate T = Time We are given the following information: P = ₦400 I = ₦24 T = 3 years Substituting these values into the formula and solving for R: 24 = 400 * R * 3 R = 24 / (400 * 3) = 0.02 = 2% Therefore, the answer is 2%. The rate at which the interest on ₦400 increases to ₦24 in 3 years reckoning in simple interest is 2%.

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To make Q the subject of the formula if p=M5 in the expression (X+Q)+1, we need to isolate Q on one side of the equation and simplify the expression on the other side. First, we need to remove the parentheses by adding X and 1 together, which gives us X+1. Next, we move M5 to the other side of the equation by subtracting it from both sides, resulting in: (X+Q)+1 - M5 = 0 Then, we can isolate Q by subtracting X and 1 from both sides: (X+Q) - (X+1) - M5 = -1 Simplifying the left-hand side, we get: Q - M5 = -1 Finally, we can solve for Q by adding M5 to both sides: Q = M5 - 1 Therefore, the expression (X+Q)+1 can be simplified to MX+5P-5M-4, and Q is equal to M5-1. The correct answer is (B) 5P-MX-5M.

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To calculate the total surface area of the cylindrical pipe, we need to add the surface area of the curved part and the surface area of the two circular ends. The surface area of the curved part can be calculated by multiplying the circumference of the circle (2πr) by the length of the pipe (50cm), which gives us: 2πr x h = 2π x 7m x 50cm = 7π m^2 The surface area of one circular end can be calculated by multiplying the area of the circle (πr^2) by 1, since one end of the pipe is open and has no surface area. Thus, the total surface area of both circular ends is: 2πr^2 = 2π x 7m^2 = 14π m^2 Finally, we add the surface area of the curved part and the surface area of the two circular ends to get the total surface area of the pipe: 7π m^2 + 14π m^2 = 21π m^2 Therefore, the total surface area of the pipe is 21π square meters. The closest option to this answer is 749π, but it is not the correct answer.

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To determine how many ways a committee of 2 women and 3 men can be chosen from 6 men and 5 women, we can use the combination formula. The number of combinations of k objects that can be chosen from a set of n objects is given by: nCk = n! / (k! * (n - k)!) where n! denotes n factorial, which is the product of all positive integers up to n. So, in this case, the number of ways to choose 2 women from 5 is 5C2 = 5! / (2! * (5-2)!) = 10. Similarly, the number of ways to choose 3 men from 6 is 6C3 = 6! / (3! * (6-3)!) = 20. Using the multiplication principle, we can multiply these two numbers together to find the total number of ways to choose 2 women and 3 men: 10 * 20 = 200. Therefore, there are 200 ways to choose a committee of 2 women and 3 men from 6 men and 5 women. The answer is (B) 200.

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To determine how many students failed the test, we need to add up the frequencies of the students who obtained marks less than 5, since the pass mark is 5. Looking at the table, the marks less than 5 are 2, 3, and 4. Adding up the corresponding frequencies, we get: 3 + 1 + 5 = 9 Therefore, 9 students failed the test. The answer is (C) 9.

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x $\alpha$ $\frac{1}{y}$ .........(1)
x = k x $\frac{1}{y}$ .........(2)
When x = 2$\frac{1}{2}$
= $\frac{5}{2}$, y = 2
(2) becomes $\frac{5}{2}$ = k x $\frac{1}{2}$
giving k = 5
from (2), x = $\frac{5}{y}$
so when y =4, x = $\frac{5}{y}$ = 1$\frac{1}{4}$

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(x - 3)(x - 4) $\le$ 0
Case 1 (+, -) = x - 3 $\ge$ 0, X - 4 $\ge$ 0
= X $\le$ 3, x $\ge$ 4
= 3 < x $\ge$ 4 (solution)

Case 2 = (-, +) = x - 3 $\le$ 0, x - 4 $\ge$ 0
= x $\le$ 3, x $\ge$ 4
therefore = 3 $\le$ x $\le$ 4

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Join SR
< PRS = 90${}^{?}$(Angle in a semicircle)
< PRS = 55${}^{?}$ (Angle between a chord and a tangent = Angle in the alternate segment)
< PSR + < PRS + < RSP = 180${}^{?}$
90v + 55${}^{?}$ + < RSP = 180${}^{?}$
< RSP = 180${}^{?}$ - 145${}^{?}$
= 35${}^{?}$

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$\frac{x^3+3x^2-10x}{2x^2-8}$ = $\frac{x(x^2+3x-10)}{2(x^2-4)}$
= $\frac{x(x^2+5x-2x-10)}{2(x+2)(x-2)}$
= $\frac{x(x-2)(x+5)}{2(x+2)(x-2)}$
= $\frac{x(x+5)}{2(x+2)}$

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Using S$\infty$ = $\frac{a}{1-r}$
r = $\frac{0.05}{0.5}$ = $\frac{1}{10}$
S$\infty$ = $\frac{0.5}{\frac{1}{10}}$

= $\frac{0.5}{\left(\frac{9}{10}\right)}$
= $\frac{0.5\times 10}{9}$
= $\frac{5}{9}$

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$\left|\begin{array}{ccc}2& 0& 5\\ 4& 6& 3\\ 8& 9& 1\end{array}\right|$
= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102

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6r78 = 5119
6 x 82 + r x 81 + 7 x 8o = 5 x 92 + 1 x 91 + 1 x 9o
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
391 + 8r = 24
r = $\frac{24}{8}$
= 3

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x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = $\frac{12}{4}$
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1

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We want to solve the inequality (x² - 1) > 0 for x. To do this, we can factor the left-hand side of the inequality: (x² - 1) = (x - 1)(x + 1) Now we have the inequality: (x - 1)(x + 1) > 0 The product of two factors is positive if and only if both factors are positive or both factors are negative. So we can break the inequality into two cases: Case 1: (x - 1) > 0 and (x + 1) > 0 This simplifies to x > 1, which means x is greater than 1. Case 2: (x - 1) < 0 and (x + 1) < 0 This simplifies to x < -1, which means x is less than -1. Therefore, the solution to the inequality (x² - 1) > 0 is: x < -1 or x > 1 So the answer is: x < -1 or x > 1.

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$\begin{array}{ccc}x& x-\overline{x}& (x-\overline{x}{)}^2\\ 2& -2& 4\\ 3& -1& 1\\ 5& 1& 1\\ 6& 2& 4\\ \sum x=16& & \sum (x-{\overline{x}}^2)=0\end{array}$
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$\overline{x}$ = $\frac{\sum x}{N}$
= $\frac{16}{4}$
= 4
S = $\sqrt{\frac{(x-\overline{x}{)}^2}{N}}$
= $\sqrt{\frac{\left(10\right)}{4}}$
= $\sqrt{\frac{\left(5\right)}{2}}$

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To find the area of a right angled triangle, we can use the formula: Area = (base x height) / 2 In a right angled triangle, the two smaller sides that form the right angle are the base and height. Therefore, we can substitute 4 cm for the base and 5 cm for the height in the formula: Area = (4 cm x 5 cm) / 2 = 10 cm^2 Therefore, the area of the right angled triangle is 10 cm^2. The answer is (A) 10 cm^2.

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$\frac{3}{5}$ $÷$ ($\frac{2}{7}$ x $\frac{4}{3}$ $÷$ $\frac{4}{9}$) = $\frac{2}{3}$ $÷$ ($\frac{2}{7}$ x $\frac{4}{3}$ x $\frac{9}{4}$)
= $\frac{3}{5}$ $÷$ $\frac{6}{7}$
= $\frac{3}{5}$ x $\frac{7}{6}$
= $\frac{7}{10}$

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To find the minimum value of the function y = -3 - 2x + x^2, we need to determine the value of x that corresponds to the vertex of the parabolic graph. The vertex of a parabolic graph with equation y = ax^2 + bx + c is located at x = -b/2a. In this case, a = 1, b = -2, and c = -3. Therefore, x = -(-2)/(2*1) = 1. So the answer is (E) 1, and that's the value of x at which the function y attains its minimum value.

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$\frac{1}{2}$x + $\frac{1}{4}$ > $\frac{1}{3}$x + $\frac{1}{2}$
Multiply through by through by the LCM of 2, 3 and 4
12 x $\frac{1}{2}$x + 12 x $\frac{1}{4}$ > 12 x $\frac{1}{3}$x + 12 x $\frac{1}{2}$
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
$\frac{2x}{2}$ > $\frac{3}{2}$
x > $\frac{3}{2}$

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Let D denote the distance between ($\frac{1}{2}$, -$\frac{1}{2}$) then using
D = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
= $\sqrt{(-\frac{1}{2}-\frac{1}{2}{)}^2+(-\frac{1}{2}-\frac{1}{2}{)}^2}$
= $\sqrt{(-1{)}^2+(-1{)}^2}$
= $\sqrt{1+1}$
= √2

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Cos $\theta$ = $\frac{12}{13}$
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan$\theta$ = $\frac{5}{12}$ and cos$\theta$ = $\frac{12}{13}$
If cos2$\theta$ = 1 + $\frac{1}{ta{n}^2\theta }$
= 1 + $\frac{1}{\frac{(5{)}^2}{12}}$
= 1 + $\frac{1}{\frac{25}{144}}$
= 1 + $\frac{144}{25}$
= $\frac{25+144}{25}$
= $\frac{169}{25}$

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The percentage error in measurement is the difference between the measured value and the actual value, divided by the actual value, multiplied by 100. In this case, the measured value is 1.26m, and the actual value is 1.25m. So the difference between the measured value and actual value is: 1.26m - 1.25m = 0.01m The percentage error can be calculated as: (0.01m ÷ 1.25m) × 100% = 0.8% Therefore, the percentage error in the measurement is 0.8%, which corresponds to option E.

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To find the total number of students who took the test, we need to add up the frequency of all the marks. 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (C) 20.

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Given that,
X $\ast$ y = X + y2
(2 $\ast$ 3) $\ast$ 5 = (2 + 32)$\ast$5
= (2 + 9)$\ast$5 = 11 $\ast$5
Hence 11 $\ast$ 5 = 11 + 52
= 11 + 25 = 36

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To determine how many students took the test, we need to sum up the frequencies in the table, since each frequency represents the number of students who obtained the corresponding mark. Adding up the frequencies, we get: 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (B) 20.