JAMB UTME - Mathematics - 1997

Find the non-zero positive value of x which satisfies the equation $\left|\begin{array}{ccc}x& 1& 0\\ 1& x& x\\ 0& 1& x\end{array}\right|$ = 0

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x(x2 - 1) - x = 0
= x3 - 2x = 0
x(x2 - 2) = 0
x = 2

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

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1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

If y = x(x4 + x + 1), evaluate ∫${}_0^1$ ydx

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The nth term of a sequence is given 31 - n , find the sum of the first terms of the sequence.

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Tn = 31 - n
S3 = 31 - 1 + 31 - 2 + 31 - 3
= 1 + $\frac{1}{3}$ + $\frac{1}{9}$
= $\frac{13}{9}$

The figure shows circles of radii 3cm and 2cm with centres at X and Y respectively. The circles have a transverse common tangent of length 25cm. Calculate XY.

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A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime

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Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors
itself and 1
The prime numbers among the group are 23, 29. Probability of choosing a prime number
= $\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}$
= $\frac{2}{11}$

A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc

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A cone with the sector angle of 45o is cut out of a circle of radius of the cone.

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Find the range of values of x for which 3x - 7 $\le$ 0 and x + 5 > 0

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The first inequality 3x - 7 ≤ 0 can be solved as follows: 3x - 7 ≤ 0 3x ≤ 7 x ≤ 7/3 The second inequality x + 5 > 0 can be solved as follows: x + 5 > 0 x > -5 So the range of values of x that satisfies both inequalities is -5 < x ≤ 7/3. The correct answer is -5 < x ≤ 7/3.

Given that loga2 = 0.693 and loga3 = 1.097, find loga 13.5

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loga 13.5 = loga $\frac{27}{2}$
= 3loga 3 - log2a
= 3 x 1.097 - 0.693
= 2.598

If the function f(fx) = x3 + 2x2 + qx - 6 is divisible by x + 1, find q

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To find the value of q, we can use the fact that the given function f(fx) is divisible by x+1. If a polynomial f(x) is divisible by x-a, then f(a) = 0. Using this fact, we can substitute x = -1 in the given function f(fx) = x^3 + 2x^2 + qx - 6, since x+1 is a factor of f(fx): f(f-1) = (-1)^3 + 2(-1)^2 + q(-1) - 6 f(f-1) = -1 + 2 - q - 6 f(f-1) = -5 - q Since f(f-1) is divisible by x+1, we have: f(f-1) = -5 - q = 0 Therefore, q = -5. Hence, the value of q is -5 when f(fx) = x^3 + 2x^2 + qx - 6 is divisible by x+1.

Find the difference between the range and the variance of the following set of numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7 where $\sum d^2$ = 60

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Find the volume of the prism.

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To find the volume of a prism, we need to multiply the area of the base by the height of the prism. In this case, the base is a triangle with base 11cm and height 15cm, so its area is 1/2 * base * height = 1/2 * 11cm * 15cm = 82.5cm^2. The height of the prism is given as 6cm. Therefore, the volume of the prism is 82.5cm^2 * 6cm = 495cm^3. Hence, the answer is 495cm^3.

Simplify $\frac{2\sqrt{3}+3\sqrt{5}}{3\sqrt{5}?2\sqrt{3}}$

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$\frac{(2\sqrt{3}+3\sqrt{5})(3\sqrt{5}+2\sqrt{3})}{(3\sqrt{5}?2\sqrt{3})(3\sqrt{5}?2\sqrt{3})}$
= $\frac{5+12\sqrt{15}}{33}$
=$\frac{19+4\sqrt{15}}{11}$

Find the standard derivation of the following data -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

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x = $\frac{\sum x}{N}$
= $\frac{0}{11}$
= 0
$\begin{array}{ccc}x& (x-x)& (x-x{)}^2\\ -5& -5& 25\\ -4& -4& 16\\ -3& -3& 9\\ -2& -2& 4\\ -1& -1& 1\\ 0& 0& 0\\ 1& 1& 1\\ 2& 2& 4\\ 3& 3& 9\\ 4& 4& 16\\ 5& 5& 25\\ & & 110\end{array}$
S.D = $\sqrt{\frac{\sum (x-x{)}^2}{\sum f}}$
= $\sqrt{\frac{110}{11}}$
= $\sqrt{10}$

Find the minimum value of X2 - 3x + 2 for all real values of x

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y = X2 - 3x + 2, $\frac{dy}{dx}$ = 2x - 3
at turning pt, $\frac{dy}{dx}$ = 0
∴ 2x - 3 = 0
∴ x = $\frac{3}{2}$
$\frac{d^{2}y}{d{x}^2}$ = $\frac{d}{dx}$($\frac{d}{dx}$)
= 270
∴ ymin = 2$\frac{3}{2}$ - 3$\frac{3}{2}$ + 2
= $\frac{9}{4}$ - $\frac{9}{2}$ + 2
= -$\frac{1}{4}$

What value of g will make the expression 4x2 - 18xy + g a perfect square?

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4x2 - 18xy + g = g $\to$ ($\frac{18y}{4}$)2
= $\frac{18y^2}{4}$

Differentiate $\frac{6x^3-5x^2+1}{3x^2}$ with respect to x

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$\frac{6x^3-5x^2+1}{3x^2}$
let y = 3x2
y = $\frac{6x^3}{3x^2}$ - $\frac{6x^2}{3x^2}$ + $\frac{1}{3x^2}$
Y = 2x - $\frac{5}{3}$ + $\frac{1}{3x^2}$
$\frac{dy}{dx}$ = 2 + $\frac{1}{3}$(-2)x-3
= 2 - $\frac{2}{3x^3}$

Sn is the sum of the first n terms of a series given by Sn = n2n - 1. Find the nth term

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The given series is: 1 + 3 + 5 + 7 + ... + (2n - 1) Notice that each term in the series is an odd number, and the difference between consecutive terms is 2. To find the nth term, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)d where a1 is the first term, d is the common difference, and n is the term number. In this case, a1 = 1 and d = 2, so we have: an = 1 + (n - 1)2 an = 2n - 1 Therefore, the answer is (D) 2n - 1.

The angle between the positive horizontal axis and a given line is 135o. Find the equation of the line if it passes through the point (2,3)

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The problem requires finding the equation of a line that passes through a given point (2,3) and makes an angle of 135 degrees with the positive x-axis. To solve this problem, we need to first find the slope of the line. Since the given angle is measured from the positive x-axis and is 135 degrees, we know that the angle made with the negative x-axis is 45 degrees. Therefore, the slope of the line is the tangent of 45 degrees, which is 1. Now that we have the slope, we can use the point-slope form of the equation of a line to find the equation of the line. The point-slope form is: y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point on the line. Plugging in the values we have, we get: y - 3 = 1(x - 2) Simplifying this equation gives us: y - x + 3 = 0 This equation is in the form of y = mx + b, where m is the slope and b is the y-intercept. We can see that the slope is 1, which we found earlier, and the y-intercept is 3. Therefore, the equation of the line that passes through the point (2,3) and makes an angle of 135 degrees with the positive x-axis is: y - x + 3 = 0 which is equivalent to: y = x - 3 So, the answer is neither (a) nor (b), but is (c) x + y = 5.

Evaluate 64.7642 - 35.2362 correct to 3 significant figures

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To evaluate the expression 64.7642 - 35.2362, we simply subtract the second number from the first number. 64.7642 - 35.2362 = 29.528 To round this answer to three significant figures, we count three digits starting from the left-most nonzero digit. In this case, the left-most nonzero digit is 2, so we round to three digits after the 2. 29.5 Therefore, the answer is 29.5 to 3 significant figures.

Two binary operations $\ast$ and $\oplus$ are defines as m $\ast$ n = mn - n - 1 and m $\oplus$ n = mn + n - 2 for all real numbers m, n.

Find the value of 3 $\oplus$ (4 $\ast$ 50)

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m $\ast$ n = mn - n - 1, m $\oplus$ n = mn + n - 2
3 $\oplus$ (4 $\ast$ 5) = 3 $\oplus$ (4 x 5 - 5 - 1) = 3 $\oplus$ 14
3 $\oplus$ 14 = 3 x 14 + 14 - 2
= 54

The table above shows that the scores of a group of students in a test. If the average score is 3.5, find the value of x

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7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2

Make F the subject of the formula t = $\sqrt{\frac{v}{\frac{1}{f}+\frac{1}{g}}}$

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t = $\sqrt{\frac{v}{\frac{1}{f}+\frac{1}{g}}}$
t2 = $\frac{v}{\frac{1}{f}+\frac{1}{g}}$
= $\frac{vfg}{ftg}$
$\frac{1}{f}+\frac{1}{g}$ = $\frac{v}{t^2}$
= (g + f)t2 = vfg
gt2 = vfg - ft2
gt2 = f(vg - t2)
f = $\frac{g{t}^2}{gv-t^2}$

Determine x + y if $\left(\begin{array}{cc}2& ?3\\ ?1& 4\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$ = $\left(\begin{array}{c}?1\\ 8\end{array}\right)$

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The pie chart shows the income of a civil servant in a month. If his monthly income is N600, find his monthly basic salary

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$\frac{d}{dx}$ cos(3x2 - 2x) is equal to

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Find the value of k if $\frac{5+2r}{(r+1)(r-2)}$ expressed in partial fraction is $\frac{k}{r-2}$ + $\frac{L}{r+1}$ where K and L are constants

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5 + 2r = k(r + 1) + L(r - 2)
but r - 2 = 0 and r = 2
9 = 3k
k = 3

Find the value of (0.006)3 + (0.004)3 in standard form

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To solve this problem, we simply need to evaluate the expressions inside the parentheses first, then add the two results. (0.006)³ = 0.000000216, which can be written in standard form as 2.16 x 10^-7. (0.004)³ = 0.000000064, which can be written in standard form as 6.4 x 10^-8. Adding these two values gives: 2.16 x 10^-7 + 6.4 x 10^-8 = 2.76 x 10^-7 Therefore, the answer is 2.8 x 10-7.

An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.($\pi$ = $\frac{22}{7}$)

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Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2

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To find the distance between point Q and the point common to the lines 2x - y = 4 and x + y = 2, we need to first find the coordinates of the point of intersection of the two lines. We can solve the system of equations: 2x - y = 4 x + y = 2 by either substitution or elimination to get the coordinates of the point of intersection, which is (1, 1). Then, we can use the distance formula to find the distance between point Q (4, 3) and (1, 1): d = √[(4 - 1)² + (3 - 1)²] = √[9 + 4] = √13. Therefore, the answer is √13.

$\begin{array}{ccccccc}\text{Age}& 20& 25& 30& 35& 40& 45\\ \text{Number of people}& 3& 5& 1& 1& 2& 3\end{array}$

Find the median age of the frequency distribution in the table above.

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To find the median age, we need to arrange the ages in ascending order and then find the middle value. However, since the table only gives us frequency counts for each age, we first need to construct a cumulative frequency distribution. Age | Number of People | Cumulative Frequency --- | --- | --- 20 | 3 | 3 25 | 5 | 8 30 | 1 | 9 35 | 1 | 10 40 | 2 | 12 45 | 3 | 15 The total number of people is 15, which is an odd number, so the median is simply the value that corresponds to the middle person. Since the cumulative frequency for 8 people is at age 25 and the cumulative frequency for 9 people is at age 30, the median age is 25. Therefore, the answer is: - 25

Cos 65o has the same value as

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A survey of 100 students in an institution shows that 80 students speak Hausa and 20 students speak Igbo, while only 9 students speak both language. How many students speak neither Hausa nor Igbo?

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In the survey, 80 students speak Hausa, 20 students speak Igbo and 9 students speak both Hausa and Igbo. If we add the number of students speaking only Hausa and the number of students speaking only Igbo, we get 80 + 20 - 9 = 91. Therefore, there are 100 - 91 = 9 students who speak neither Hausa nor Igbo. So, the answer is 9.

The angle of elevation of a building from a measuring instrument placed on the ground is 30o. If the building is 40m high, how far is the instrument from the foot of the building?

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$\frac{40}{x}$ = tan 30o
x = $\frac{40}{tan36}$
= $\frac{40}{1\sqrt{3}}$
= 40√3m

If U = (s, p, i, e, n, d, o, u, r), X = (s, p, e, n, d) Y = (s, e, n, o, u), Z = (p, n, o, u, r) find X ∩ Y ∪ Z

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Find the simple interest rate percent per annum at which ₦1,000 accumulates to ₦1,240 in 3 years

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The formula for simple interest is: I = P * r * t Where: I = Interest P = Principal (initial amount borrowed or invested) r = Interest rate per annum (as a decimal) t = Time in years From the given information: P = ₦1,000 I = ₦1,240 - ₦1,000 = ₦240 t = 3 years Substituting these values into the formula, we get: 240 = 1000 * r * 3 Simplifying the equation, we get: r = 240 / (1000 * 3) = 0.08 Converting to a percentage, we get: r = 0.08 * 100% = 8% Therefore, the simple interest rate percent per annum is 8%. Answer: 8%

Solve the simultaneous equations $\frac{2}{x}-\frac{2}{x}$ = 2, $\frac{4}{x}+\frac{3}{y}$ = 10

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$\frac{2}{x}-\frac{2}{x}$ = 2.....(1)
$\frac{4}{x}+\frac{3}{y}$ = 10
$\frac{6}{x}$ = 12 $\to$ x = $\frac{6}{12}$
x = $\frac{1}{2}$
put x = $\frac{1}{2}$ in equation (i)
= 4 - $\frac{3}{y}$ = 2
= 4 - 2
= $\frac{3}{y}$
therefore y = $\frac{3}{2}$

In an examination, the result of a certain school is as shown in the histogram above. How many candidates did the school present?

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3 + 5 + 8 + 1 + 2 = 19

Integrate $\frac{1}{x}$ + cos x with respect to x

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A point P moves so that is equidistant from point L and M. If LM is 6cm, find the distance of P from LM when P is 10cm from L

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p from LM = $\sqrt{{10}^2}$ - 82
= $\sqrt{36}$ = 6cm

If X $\ast$ Y = X + Y - XY, find x when (x $\ast$ 2) + (x $\ast$ 3) = 68

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x $\ast$ y = x + y - xy
(x $\ast$ 2) + (x $\ast$ 3) = 68
= x + 2 - 2x + x + 3 - 3x
= 86
3x = 63
x = -21

In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that the fruit is either a grape or a banana?

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There are a total of 6 + 11 + 13 = 30 fruits in the basket. The probability of selecting a grape is 6/30 and the probability of selecting a banana is 11/30. To find the probability of selecting either a grape or a banana, we add these probabilities: 6/30 + 11/30 = 17/30 So the probability of selecting either a grape or a banana is 17/30. Therefore, the answer is: - 17/30

Each of the base angles of a isosceles triangle is 59o and the verticles of the triangle lie on a circle. Determine rhe angle which the base of the triangle subtends at the centre of the circle.

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In the figure, XYZ is a triangle with XY = 5cm, XZ = 2cm and XZ is produced to E making the angle < YZE = 150${}^{\circ }$. If the angle XYZ = $\theta$, calculate the value of sin$\theta$

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$\frac{5}{\sin {36}^o}=\frac{2}{\sin \theta }=\frac{5}{\frac{1}{2}}=\frac{2}{\sin \theta }$

10 = $\frac{2}{\sin \theta }$

sin$\theta$ = $\frac{2}{10}$

= $\frac{1}{5}$