JAMB UTME - Mathematics - 2006

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P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5

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Area of rectangle PQRS
= 10 x 7 = 70 cm2
Area of semi circle
QR = Dπ = 7 x 22/7 = 22cm2
∴Area of the figure = 70 + 22
= 92cm2
= 90cm2

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Let's call the smallest of the five consecutive integers "x". Since the numbers are consecutive, we know that the next four integers will be x + 1, x + 2, x + 3, and x + 4. To find the mean of these five numbers, we add them up and divide by 5: (x + (x+1) + (x+2) + (x+3) + (x+4)) / 5 = 30 Simplifying the left side: (5x + 10) / 5 = 30 Canceling the 5's on the left side: x + 2 = 30 Subtracting 2 from both sides: x = 28 Therefore, the five consecutive integers are 28, 29, 30, 31, and 32, and the largest of these numbers is 32. So the correct option is: 32.

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To find the percentage of the population that speaks both the local language and pidgin English, we need to use the formula: Percentage = (Number of people who speak both / Total population) x 100% We are given that there are 500 people in the village, 350 speak the local language, and 200 speak pidgin English. However, we do not know the number of people who speak both. To find the number of people who speak both, we can add the number of people who speak the local language and the number of people who speak pidgin English and subtract it from the total population. Number of people who speak both = Total population - (Number of people who speak local language + Number of people who speak pidgin English) Number of people who speak both = 500 - (350 + 200) Number of people who speak both = 50 So, 50 people speak both languages. Now we can substitute this value in the formula to find the percentage. Percentage = (Number of people who speak both / Total population) x 100% Percentage = (50 / 500) x 100% Percentage = 0.1 x 100% Percentage = 10% Therefore, the percentage of the population that speaks both the local language and pidgin English is 10%. Answer is correct.

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3, -6, +12, -24
a = 3, r = -2
$8n=\frac{a(1-r^n)}{1-r}\therefore 1-2^8=\frac{3(1-(-2^{n-1}\left)\right)}{1-(-2)}1-2^8=\frac{3(1-(-2^{n-1}\left)\right)}{3}$
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9

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Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50o

Also, < QSR = < QPR(angles in the segment)

< QPR = xo

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130o

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To find the equation of the curve, we need to integrate the gradient. Since the gradient is 2x + 7, the equation of the curve will be of the form y = x^2 + 7x + C, where C is a constant of integration. We can find the value of C by using the fact that the curve passes through the point (2, 0). Substituting x = 2 and y = 0 into the equation, we get: 0 = 2^2 + 7(2) + C 0 = 4 + 14 + C C = -18 Therefore, the equation of the curve is y = x^2 + 7x - 18. The answer is.

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Draw a line perpendicular to |SR| to form |PT| in $△$ PST, cos 60 = \(\frac{|ST|}{8}\(

|PT| = 8 cos 6 = 4cm

Since |TR| = |PQ|

SR = ST + TR

= (4 + 10)cm

= 14cm

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Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90${}^{\circ }$
Expenditure on transport = 45${}^{\circ }$
x + 2x + 90 + 45 = 360${}^{\circ }$
3x + 135 = 360${}^{\circ }$
3x = 360 - 135
3x = 225
x = 225/3 = 75${}^{\circ }$
∴∠ for food = 2x = 2 * 75
= 150
360${}^{\circ }$ = N30,000
1${}^{\circ }$ = ?
1${}^{\circ }$ = 30,000/360
150${}^{\circ }$ = (30000/360) * (150/1)
= N12500

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To evaluate -5 * 3 using the binary operation * on the set of rational numbers, we substitute x = -5 and y = 3 in the given expression: -5 * 3 = (-5)^2 - 3^2 / 2(-5)(3) Simplifying the numerator, we get: (-5)^2 - 3^2 = 25 - 9 = 16 Simplifying the denominator, we get: 2(-5)(3) = -30 Substituting these values back into the expression, we get: -5 * 3 = 16 / -30 Simplifying the fraction, we get: -5 * 3 = -8/15 Therefore, the value of -5 * 3 using the binary operation * on the set of rational numbers is -8/15.

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y = x2 - x - 12
= (x - 4)(x - 3)
∴ x = 4 or x = -3
-3 < x < 4

But y $\ge$ 0
∴ -3 < x $\le$ 4

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The given equation of a line can be rewritten in the standard form of a line, which is y = mx + b, where m is the slope and b is the y-intercept. To rewrite the equation, we can solve for y: ax + bc + c = 0 ax + bc = -c y = -(a/b)x - (c/b) The locus of points equidistant from the lines y = -(a/b)x - (c/b) is a line perpendicular to the given lines, with the distance from the origin equal to the absolute value of c/b. This is because the perpendicular bisector of a line segment is equidistant from the endpoints of the segment, and in this case the "endpoints" are the given lines. Since the slope of the given lines is -(a/b), the slope of the perpendicular bisector is b/a. The perpendicular bisector passes through the origin (0,0), so we can find its equation by using the point-slope form of a line: y - 0 = (b/a)x y = (b/a)x This is equivalent to the equation y = mx, where m = b/a, which is the equation of a line passing through the origin with slope b/a. Therefore, the answer is: a line ax - ay + q = 0.

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3−4×92×81−1=log9(3−4×92×81−1)=log9(134×92×181)=log9(181×811×181)=log9181=log9192=log99−2=−2log99−2×1=−2

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1100112 + 111112 = 10100102

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To calculate the tax on an income of N20,000 using the given tax rules, we need to break down the income into three parts: 1. The first N10,000: No tax is paid on this amount. 2. The next N5,000: Tax is paid at N50 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N50 = N250. 3. The remainder of N5,000: Tax is paid at N55 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N55 = N275. Therefore, the total tax on an income of N20,000 would be N0 + N250 + N275 = N525. In summary, the tax on an income of N20,000 using the given tax rules is N525, which is option (D).

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Given that PQ and RS are two parallel lines, we can observe that their slopes are equal. The slope of PQ can be found using the coordinates of points P and Q: slope of PQ = (2 - q)/(3 - 1) = (2 - q)/2 Similarly, the slope of RS can be found using the coordinates of points R and S: slope of RS = (2q - 4)/(5 - 3) = (2q - 4)/2 Since PQ and RS are parallel, their slopes are equal: (2 - q)/2 = (2q - 4)/2 Simplifying the equation above, we get: 2 - q = q - 2 2q = 4 q = 2 Therefore, the value of q is 2, which corresponds to.

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($\frac{52}{160}=\frac{36}{160}$)% = $\frac{88}{160}$%

= 55%

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P ∝ 1/q3
P = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6
∴ r varies inversely as 6√P

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The cost of renovating a room is directly proportional to the area of the room. If the cost of renovating a 6 m square room is N540, then we can find the cost of renovating a 9 m square room using the proportionality constant. Let C be the cost of renovating a room and A be the area of the room. Then, we have: C = kA where k is the proportionality constant. We can find k by using the given information that the cost of renovating a 6 m square room is N540: 540 = k(6) k = 90 Now, we can use this value of k to find the cost of renovating a 9 m square room: C = kA = 90(9) = N810 Therefore, the cost of renovating a 9 m square room is N810.

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Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)

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We need to integrate the given expression to find the value of y. Integrating both sides with respect to x, we get: ∫dy = ∫(x + cos x)dx y = 1/2 * x^2 + sin x + c where c is the constant of integration. Therefore, the correct answer is: - x^2/2 + sin x + c

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∫0−4(1−2x)dx=[x−x2]0−4=(0−0+C)−(−4(−4)2+C)=C−(−4−16+C)=C−(−20+C)=C+20−C=20

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In ΔPST; cos 60 = ST/8
ST = 8cos 60
ST = 8 x 1/2 = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm

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Let's start by using the total number of marbles in the bag which is the sum of the number of black, white, and red marbles. Total number of marbles = 5 (black) + 4 (white) + x (red) = 9 + x Now, we are given that the probability of picking a red marble is 2/3. This means that out of all the marbles in the bag, 2/3 of them are red. So, we can write an equation using this information: Number of red marbles / Total number of marbles = 2/3 Simplifying this equation, we get: x / (9 + x) = 2/3 Cross-multiplying, we get: 3x = 2(9 + x) Simplifying this equation, we get: 3x = 18 + 2x Subtracting 2x from both sides, we get: x = 18 Therefore, the value of x is 6. So, there are 6 red marbles in the bag.

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To calculate the number of ways a student can answer any 4 out of 6 questions, we can use the combination formula, which is: nCk = n! / (k! * (n-k)!) where n is the total number of items, k is the number of items to be selected, and ! represents the factorial function. In this case, we have n = 6 (the total number of questions on the exam) and k = 4 (the number of questions the student must answer). So we can calculate the number of ways to select 4 questions out of 6 using the combination formula as follows: 6C4 = 6! / (4! * (6-4)!) = (6 * 5 * 4 * 3) / (4 * 3 * 2 * 1) = 15 Therefore, the answer is 15, which means that a student can answer any 4 out of 6 questions in 15 different ways.

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The given expression is (x^2 - 1)/x^2. To differentiate this expression with respect to x, we can use the quotient rule of differentiation. The quotient rule states that the derivative of (f(x)/g(x)) = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2. Applying the quotient rule, we get: [(x^2)*2x - (x^2 - 1)*2x]/(x^2)^2 Simplifying the expression further, we get: [2x^3 - 2x^3 + 2x]/x^4 = 2/x^3 Therefore, the answer is option C: 4x^2 - 2 - 2/x^3.

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Tan θ = 5/4
x2 = 52 + 42
= 25 + 16
= 41
x = √41
sin2θ - cos2θ = (5/√41)2 - (4/√41)2
= 25/41 - 16/41
= 9/41

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X ⊕ y = xy/6 = xe/6 = x
Xe = 6x
e = 6
X ⊕ y = xx1/6 where x1 = 20
Xx20/6 = 6
X = 36/20
X = 9/5

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∠yxz = 180 - (45 + 15)

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Y = x implies y ≤ x
Y + x = 4 implies y = 4 – x
∴y = 4 – x
∴y ≤ x, y + x ≤ 4 and y ≥ 0

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To solve this problem, we need to use the fact that the angle between a tangent and a chord of a circle is equal to the angle formed by the chord in the opposite segment. Let's call the point where the tangent MN touches the circle point T. Then, angle MOT is 90 degrees because OT is a radius of the circle and MT is a tangent, and these two lines are perpendicular at point T. Also, angle MTS is 55 degrees because MN is tangent to the circle and makes an angle of 55 degrees with PS. Therefore, angle MTN is 180 - 90 - 55 = 35 degrees because the angles in a triangle add up to 180 degrees. Now, let's consider the triangle OTN. We know that angle OTN is 90 degrees because OT is a radius of the circle and TN is a tangent, and these two lines are perpendicular at point T. We also know that angle MTN is 35 degrees. Therefore, angle OTM is 180 - 90 - 35 = 55 degrees because the angles in a triangle add up to 180 degrees. Finally, we can see that angle PTS is equal to angle OTM because they both intercept the same arc PS. Therefore, angle PTS is also 55 degrees. Since angle PTS is half of angle POS (which is equal to 110 degrees because it intercepts a semicircle), angle POS is 2 * 110 = 220 degrees. Finally, the angle between the tangent MN and the chord PS is equal to angle MTS, which is 55 degrees. Therefore, the answer is option (C) 35o.

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The answer is NQ. By definition, the locus of points that are equidistant from two points P and Q is the perpendicular bisector of the line segment PQ. Therefore, if this locus intersects line PQ at point N, then N must be the midpoint of the segment PQ. So, PN and NQ are equal, and since N is the midpoint of PQ, PN is half of PQ. Thus, PN is equal to NQ.

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To solve the given equation, we can use the formula for combinations, which is: nCr = n! / r!(n-r)! Substituting n+1 for n and 3 for r, we get: (n+1)C3 = (n+1)! / 3!(n+1-3)! (n+1)C3 = (n+1)! / 6n! Simplifying the right side of the equation using the given information, we get: 4(nC3) = 4n! / 6n! (nC3) = n! / 6n! Now we have: (n+1)C3 = 4(nC3) Substituting the formulas we derived earlier, we get: (n+1)! / 3!(n+1-3)! = 4n! / 3!(n-3)! Simplifying and canceling the terms, we get: (n+1)(n)(n-1) = 4(n)(n-1)(n-2) Expanding and simplifying the equation, we get: n^3 - 3n^2 - 10n + 12 = 0 Factoring the equation, we get: (n-4)(n^2 - n - 3) = 0 The roots of the equation are n = 4, (1+sqrt(13))/2, and (1-sqrt(13))/2. Since n represents a counting number, the only possible solution is n = 4. Therefore, the answer is 4.

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< QRS = 30${}^{\circ }$(corresponding < ), < QSR = 50${}^{\circ }$

< SQR = 180${}^{\circ }$ - (30 + 50)${}^{\circ }$ (Sum of < S in a $△$)

180${}^{\circ }$ - 80${}^{\circ }$ = 100${}^{\circ }$

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T = 2 π √1/gT/2π = √1/g
(T/2π)2 = (√1/g)2T2/4π2 = 1/g
T2 x g = 4π2 x l
G = (4π2 x l) / T2

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1st convert to base 10
22324 = 2x43 + 2x42 + 3x41 + 2x40
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 4506

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F(x) = Q x D + R
Kx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10

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To find the variance of a set of numbers, we first need to find the mean of the set. The mean of the set {2x, 2x-1, 2x+1} is: (2x + 2x-1 + 2x+1)/3 = 2x Next, we need to find the difference between each number and the mean, square these differences, and take the average of the squares. This gives us the variance. The differences between each number and the mean are: 2x - 2x = 0 2x-1 - 2x = -1 2x+1 - 2x = 1 Squaring these differences, we get: 0^2 = 0 (-1)^2 = 1 1^2 = 1 Taking the average of these squares, we get: (0 + 1 + 1)/3 = 2/3 Therefore, the variance is 2/3. Thus, the answer is option A: 2/3.

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To find the values of n and m, we need to first determine the mode and the number of students who scored 4 or more. The mode is the most frequently occurring score in the set. In this case, the score of 4 appears most frequently, with a total of 12 students achieving that score. Therefore, the mode is 4. To find the number of students who scored 4 or more, we need to add up the frequencies of all scores that are 4 or greater. Adding the frequencies of scores 4, 5, 6, and 7 gives a total of 33 students. Therefore, the values of (n, m) are (33, 4).

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(K2)6 * 36 = 35(K4)5
k*61 + 2*60 + 3*50 = 3*50 + K*51 + 4*50
6K + 2 = 5K + 4
6K - 5k = 4 - 2
K = 2