JAMB UTME - Mathematics - 1997

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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%

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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

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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.

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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

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

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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.

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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$\theta$ = $\frac{2}{10}$

= $\frac{1}{5}$

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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

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

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

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

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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.

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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
= -$\frac{1}{4}$

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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.

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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.

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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.

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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.

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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.

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

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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 = $\frac{3}{2}$

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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 = $\frac{g{t}^2}{gv-t^2}$

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

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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

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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
= $\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}$
= $\frac{2}{11}$

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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.

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

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

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3 + 5 + 8 + 1 + 2 = 19

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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 - $\frac{2}{3x^3}$

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

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

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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 = $\sqrt{\frac{\sum (x-x{)}^2}{\sum f}}$
= $\sqrt{\frac{110}{11}}$
= $\sqrt{10}$

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