JAMB UTME - Mathematics - 2001

Akọwa Nkọwa

Total number of beads
= 1+2+4+5+3
=15
Number of blue beads = 1
P(Blue beads) = 1/15
Numbers of white beads = 4
P(white beads) = 4/15
∴P(Blue of white beads)
= P(Blue) + P(White)
= 1/15 + 4/15
= 5/15
= 1/3

Akọwa Nkọwa

The notation 6P_r means the number of permutations of r objects selected from a set of 6 distinct objects. Since 6P_r = 6! / (6 - r)!, we know that 6P_r = 6 when r = 1, since 6P₁ = 6! / (6 - 1)! = 6! / 5! = 6. To find the value of 6P_r+1, we can simply multiply 6P_r by (6 - r) since we are selecting one additional object. So, 6P_r+1 = 6P_r * (6 - r) = 6 * (6 - 1) = 30 when r = 1. Therefore, the answer is 30.

Akọwa Nkọwa

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(0.142 x 0.275) 7(0.02) = (0.14 x 0.14 x 0.275) ÷ (7 x 0.02) = (0.14 x 0.14 x 0.275)/(0.14)
=0.14 x 0.275 = 0.0385 Approx. 0.039

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Akọwa Nkọwa

The dealer bought the car for ₦250,000 and spent ₦70,000 refurbishing it, for a total cost of: Total cost = ₦250,000 + ₦70,000 = ₦320,000 The dealer then sold the car for ₦400,000. The profit made by the dealer is the difference between the selling price and the cost price: Profit = Selling price - Cost price Profit = ₦400,000 - ₦320,000 = ₦80,000 The percentage gain is the profit expressed as a percentage of the cost price: Percentage gain = (Profit / Cost price) x 100% Substituting the values we get: Percentage gain = (₦80,000 / ₦320,000) x 100% Percentage gain = 0.25 x 100% Percentage gain = 25% Therefore, the percentage gain made by the dealer is 25%, and the answer is.

Akọwa Nkọwa

The locus of a point which moves such that its distance from the line y = 4 is a constant, k, is the set of all points which are equidistant from the line y = 4. This set of points is a pair of parallel lines, k units above and below the line y = 4. Therefore, the correct answer is: y = k + or - 4.

Akọwa Nkọwa

In ? PQT,
?PTQ = 25${}^{?}$(base ?s of isosceles ?)
In ? QSR,
?RQS = ?QPT + ?QTP
(Extr = sum of interior opposite ?s)
?RQS = 25 + 25
= 50${}^{?}$
Also in ? QSR,
75 + ?RQS + ?QSR = 180${}^{?}$
(sum of ?s of ?)
?75 + 50 + ?QSR = 180
125 + ?QSR = 180
?QSR = 180 - 125
?QSR = 55${}^{?}$
But ?QSR and ?RST are the same
?RST = 55${}^{?}$

Akọwa Nkọwa

Mode = 11
Square of the mode = 112
=121

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To solve this problem, we need to use the formula for the volume of a hemisphere, which is V = (2/3)πr^3. We want to find the rate of change of V with respect to r, which is given by dV/dr = 2πr^2. When r = 2, we can substitute this value into the formula to get dV/dr = 2π(2)^2 = 8π. Therefore, the rate of change of the volume of a hemisphere with respect to its radius when r = 2 is 8π.

Akọwa Nkọwa

Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.

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This question is asking us to find the number of ways to select 8 subjects out of 12 subjects. This is a combination problem since the order in which the subjects are selected does not matter. The formula for combination is nCr = n!/r!(n-r)!, where n is the total number of items, r is the number of items to be selected, and ! represents the factorial function. Applying this formula to the given problem, we have: 12C8 = 12!/8!(12-8)! = (12x11x10x9)/(4x3x2x1) = 495 Therefore, there are 495 ways of selecting 8 subjects out of 12 subjects. Hence, the correct option is 495.

Akọwa Nkọwa

To divide the given expression, we can use the long division method.

                  ax^2 - 26x + 156a - 216b
     ___________________________________
ax | ax^3 + 3x^2 - 26x^2 + 156ax - 216bx^2 - 24ax + 108
      -ax^3
      -----
           3x^2 - 216bx^2
           3x^2 - 3ax^2
           --------------
                     -216bx^2 + 3ax^2
                     -216bx^2 + 36ax^2
                     ------------------
                                    -33ax^2 + 156a - 216bx^2 - 24ax + 108
                                    -33ax^2 + 33bx^2
                                    ------------------
                                                156a - 33ax^2 - 24ax + 108
                                                156a - 33ax^2 - 156ax + 33ax
                                                -------------------------
                                                             -24ax + 108
                                                             -24ax + 24bx
                                                             -------------
                                                                       108 - 24bx
                                                                       108 - 2ax
                                                                       --------
                                                                                2ax - 24bx + 108

Therefore, the quotient is ax - 2 with a remainder of 2ax - 24bx + 108.

Hence, the answer is ax - 2.

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To find the variance of a set of numbers, you need to follow these steps: 1. Find the mean (average) of the set. 2. Subtract the mean from each number in the set. 3. Square each of the differences. 4. Find the mean (average) of the squared differences. Let's apply these steps to the set of numbers given: 2, 6, 8, 6, 2, and 6. 1. The mean of the set is (2+6+8+6+2+6)/6 = 5. 2. Subtracting the mean from each number gives us the following differences: -3, 1, 3, 1, -3, 1. 3. Squaring each difference gives us: 9, 1, 9, 1, 9, 1. 4. Finding the mean of these squared differences gives us (9+1+9+1+9+1)/6 = 5. Therefore, the variance of the set is 5. In summary, variance is a measure of how spread out a set of numbers is. A high variance means that the numbers in the set are far from the mean, while a low variance means that the numbers are closer to the mean.

Akọwa Nkọwa

Mode = score with highest frequency = 11.
Square of 11 = 121

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Hint: Make a sketch of the moving points such that the hypotenuse is 8cm and the adjacent x cm.
Cos 30 = x/8
x = 8 cos 30 = 8√3cm

Akọwa Nkọwa

Principal, P = Amount, A - Interest, I.
A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = ₦5,000

Akọwa Nkọwa

Given: 4x2 - 9y2 + 20x + 25
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)

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To evaluate the definite integral ∫2(2x-3)^2/3 dx, we can use the following formula: ∫uⁿ du = (uⁿ⁺¹)/(n+1) + C where C is the constant of integration. Using this formula, we can rewrite the integral as: ∫2(2x-3)^2/3 dx = 2 ∫(2x-3)^2/3 dx Let u = 2x - 3, then du/dx = 2 and dx = du/2. Substituting this into the integral, we get: 2 ∫(2x-3)^2/3 dx = 2 ∫u^2/3 (du/2) = ∫u^2/3 du Using the formula, we get: ∫u^2/3 du = (u^5/3)/(5/3) + C Substituting back for u, we get: ∫2(2x-3)^2/3 dx = (2(2x-3)^5/3)/(5/3) + C Simplifying, we get: ∫2(2x-3)^2/3 dx = (6(2x-3)^5/3)/5 + C Therefore, the correct answer is option A: 3/5(2x-3)^5/3 + k.

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The given curve is y = 2kx² + x + 1, and we are asked to find the value of k when the gradient at x = 1 is 9. The gradient of a curve at a particular point is the slope of the tangent to the curve at that point. To find the gradient of this curve at x = 1, we need to differentiate it with respect to x: dy/dx = 4kx + 1 Now we can substitute x = 1 and set the result equal to 9, which gives: 4k(1) + 1 = 9 Simplifying this equation, we get: 4k = 8 Dividing both sides by 4, we obtain: k = 2 Therefore, the value of k that satisfies the given conditions is 2.

Akọwa Nkọwa

To find the mean score, we need to calculate the sum of all the scores, and then divide it by the total number of scores. The scores are given as 4, 7, 8, 11, 13, 8 with corresponding frequencies of 3, 5, 2, 7, 2, 1 respectively. So, the sum of all the scores is: 4 × 3 + 7 × 5 + 8 × 2 + 11 × 7 + 13 × 2 + 8 × 1 = 12 + 35 + 16 + 77 + 26 + 8 = 174 The total number of scores is: 3 + 5 + 2 + 7 + 2 + 1 = 20 Therefore, the mean score is: mean = sum of scores / total number of scores = 174 / 20 = 8.7 So, the answer is.

Akọwa Nkọwa

Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6

Akọwa Nkọwa

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P(Games ends in draw)
This implies that Team P and Q wins
∴ P(P wins) = 1/2
P(Q wins) = 1/2
∴ P(game ends in a draw) = 1/2*1/2 = 1/4

Akọwa Nkọwa

Akọwa Nkọwa

First, let's evaluate p^2 and q^2: p^2 = (1+√2)^2 = 1 + 2√2 + 2 = 3 + 2√2 q^2 = (1-√2)^2 = 1 - 2√2 + 2 = 3 - 2√2 Next, let's substitute the values of p^2 and q^2 into the expression (p^2 - q^2)/(2pq): (p^2 - q^2)/(2pq) = ((3 + 2√2) - (3 - 2√2))/(2(1+√2)(1-√2)) = (4√2)/(2(1-2)) = (4√2)/(-2) = -2√2 Therefore, the answer is -2√2.

Akọwa Nkọwa

To find the value of p in terms of q, we can use the given information that the two graphs intersect at x = 2. This means that at x = 2, the values of y for both graphs are the same. So, we can substitute x = 2 into both equations and set them equal to each other: px^2 + q = 2x^2 - 1 Substituting x = 2: p(2)^2 + q = 2(2)^2 - 1 4p + q = 7 Solving for p in terms of q, we can isolate p: 4p = 7 - q p = (7 - q) / 4 Therefore, the value of p in terms of q is (7 - q) / 4.

Akọwa Nkọwa

Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1

Akọwa Nkọwa

Hint: If two lines a perpendicular, the gradient of one is equal to minus the reciprocal of the other (P, 4) and (6, -2)

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Using cosine formula (t$\sqrt{3}$)2 = t2 + t2 - 2t2 cos$?$

3t2 = 2t2 - 2t2 cos$?$ = 2t2(1 - cos$?$)

1 - cos$?$ = $\frac{3t^2}{2t^2}$ = $\frac{3}{2}$

cos = 1 - $\frac{3}{2}=?\frac{1}{2}$

$?$ = cos-1(-$\frac{1}{2}$) = 120${}^{?}$ and 240${}^{?}$

N.B 0 $?$ $?$ 360

Akọwa Nkọwa

Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.

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In a circle, the length of an arc is proportional to the angle it subtends at the center of the circle. Since ST is a chord and its length is equal to the radius of the circle, then angle SOT, where O is the center of the circle, is 60 degrees (because it subtends an equilateral triangle). Therefore, the length of arc ST is 1/6 of the circumference of the circle (because 60 degrees is 1/6 of a full revolution, which is 360 degrees). The circumference of the circle is 2πr, so the length of arc ST is: Length of arc ST = 1/6 * 2πr = πr/3 Hence, the answer is option D: πr/3.

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To find dy/dx, we need to take the derivative of y with respect to x. Using the product rule of differentiation, we have: dy/dx = (sinx) + (x)(cosx) So, when x = ?/2, we can substitute this value into the expression to get: dy/dx = (sin(?/2)) + (?/2)(cos(?/2)) We know that sin(?/2) = 1 and cos(?/2) = 0, so we can simplify the expression: dy/dx = (1) + (?/2)(0) dy/dx = 1 Therefore, the answer is 1.

Akọwa Nkọwa

Looking at the histogram, we can see that the height of the bars represents the number of taxis with a particular number of passengers. We can count the number of taxis with more than 4 passengers by adding up the heights of the bars for 5, 6, 7, and 8 passengers. Counting these bars, we can see that there are a total of 17 taxis with more than 4 passengers. Therefore, the answer is 17.

Akọwa Nkọwa

m2 + n2 = 29 .......(1)
m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)

Akọwa Nkọwa

The total number of cars that passed the point on the street in two minutes is not given in the chart. Therefore, we cannot determine the fraction of the total number of cars from the information provided. The chart only shows the relative frequency or proportion of cars of different colors that passed the point during the two-minute observation period. For example, if the chart shows that 3 out of 25 cars were blue, then the proportion of blue cars passing the point during the two minutes was 3/25. To find the fraction of the total number of cars, we would need to know the total number of cars that passed the point during the two-minute observation period.

Akọwa Nkọwa

Hint:
y = 4 - x2 and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = -3 or x = 1.
Integrating x2 + 2x - 3 from (-3, to 1) w.r.t x will give 31/3 = 10(1/3)

Akọwa Nkọwa

$\begin{array}{cc}\text{colour of cars}& \text{Number (frequency)}\\ yellow& 3\\ white& 4\\ red& 8\\ green& 2\\ blue& 6\\ black& 2\\ & 25\end{array}$

Thus, the fraction of the total numbers that are yellow is $\frac{3}{25}$

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First, let's find the inverse of the element 2 under the operation * on the set of real numbers. The inverse of an element "a" under the operation "*" is another element "b" such that a*b = b*a = identity element. The identity element under the operation "*" is given as -1, so we have: 2 * b = b * 2 = -1 Substituting the value of the identity element, we get: 2 * b = b * 2 + 1 = -1 Simplifying the equation, we get: 2b = -3 b = -3/2 Therefore, the inverse of the element 2 under the operation * is -3/2. To check if this is correct, we can substitute the values of 2 and -3/2 into the operation * and see if we get the identity element: 2 * (-3/2) = -3/2 * 2 = -1 Since we get the identity element, we can confirm that the inverse of the element 2 under the operation * is -3/2. So, the answer to the question is option (C) -2.

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Let's assume that the first term of the A.P is 'a', and the common difference is 'd'. We know that for an A.P, the nth term is given by: a + (n-1)d According to the problem, the sixth term is equal to half of the twelfth term. So, we can write: a + 5d = 1/2 (a + 11d) Multiplying both sides by 2 to eliminate the fraction: 2a + 10d = a + 11d Simplifying, we get: a = d Therefore, the first term of the A.P is equal to the common difference. So, the correct answer is "the common difference."

Akọwa Nkọwa

Let the rectangle be a square of sides p/4.
So that perimeter of square = 4p
4 x (p/4) = p.

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Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3

Akọwa Nkọwa

The formula for the volume of a cylindrical tank is V = πr^2h, where V is the volume, r is the radius, and h is the height or depth of the tank. We are given that the capacity or volume of the tank is 3080 m^3, and we can find the radius by dividing the diameter by 2: d = 14 m r = d/2 = 7 m Substituting the given values in the formula, we get: 3080 = (22/7) x 7^2 x h Simplifying the expression, we get: 3080 = 22 x 7 x 7 x h/7 3080 = 22 x 7 x h 3080 = 154h h = 3080/154 h = 20 Therefore, the depth of the tank is 20 meters. Answer: 20 m

Akọwa Nkọwa

To find the range of a set of numbers, we need to find the difference between the largest and smallest numbers in the set. The given set of numbers are: 1/6, 1/3, 3/2, 2/3, 8/9, and 4/3. The smallest number in the set is 1/6 and the largest number in the set is 3/2. So, the range of the set is: 3/2 - 1/6 = 9/6 - 1/6 = 8/6 = 4/3 Therefore, the answer is  4/3.

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In an isosceles triangle, the two equal sides are opposite to the two equal angles. Therefore, the angle opposite to the side of length √3t would be equal to the other equal angle of the triangle. Let's call this angle x. Then, using the fact that the sum of the angles in a triangle is 180°, we can write: x + x + θ = 180 Simplifying this equation, we get: 2x + θ = 180 Now, we know that the two equal sides of the triangle have length t, so we can use the cosine formula to write: cos(x) = (t/2) / √(3t^2/4) Simplifying this expression, we get: cos(x) = 1/√3 Taking the inverse cosine of both sides, we get: x = 30° Substituting this value of x into the earlier equation, we get: 2(30°) + θ = 180° θ = 120° Therefore, the answer is option (B) 120°.

Akọwa Nkọwa

Mean ((4*3) + (7*5) + (8*2) + (11*7) + (13*2) + (8*1))/20
(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7

Akọwa Nkọwa

There are a total of 1+2+3+4+5 = 15 beads. The probability of picking a blue bead is 1/15, since there is only one blue bead out of the total of 15 beads. The probability of picking a white bead is 4/15, since there are 4 white beads out of the total of 15 beads. To find the probability of picking a blue or white bead, we add the probabilities of picking a blue bead and a white bead: P(blue or white) = P(blue) + P(white) P(blue or white) = 1/15 + 4/15 P(blue or white) = 5/15 P(blue or white) = 1/3 Therefore, the probability of picking a blue or white bead is 1/3.