JAMB UTME - Mathematics - 2019

If the 3rd and 7th terms of a G.P are 9 and 1/9 respectively. Find the common ratio.

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Let's suppose that the first term of the GP is 'a' and the common ratio is 'r'. Then, using the given information, we can create the following equations: Third term: ar^2 = 9 ---(1) Seventh term: ar^6 = 1/9 ---(2) To find 'r', we can divide equation (2) by equation (1) as follows: (ar^6) / (ar^2) = (1/9) / 9 r^4 = 1/81 r = (1/81)^(1/4) r = 1/3 Therefore, the common ratio is 1/3.

Calculate the volume of the regular three-dimensional figure drawn above, where

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|AC| = |DF| = 13 cm
Using Pythagoras theorem,
|AC|${}^2$ = |AB|${}^2$ + |BC|${}^2$
13${}^2$ = 12${}^2$ + |BC|${}^2$
|BC|${}^2$ = 169 - 144 = 25
|BC| = $\sqrt{25}$
= 5 cm
Volume of triangular prism = $\frac{1}{2}\times base\times length\times height$
= $\frac{1}{2}\times 5\times 12\times 18$
= 540 cm${}^3$

If y = 8x${}^3$ - 3x${}^2$ + 7x - 1, find $\frac{d^{2}y}{d{x}^2}$.

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To find d2y/dx2, we need to take the second derivative of y with respect to x. dy/dx = 24x^2 - 6x + 7 Taking the derivative of dy/dx, we get: d2y/dx2 = 48x - 6 Therefore, the answer is 48x - 6. To understand this, we need to remember that the second derivative of a function gives us the rate of change of the slope of the function. In this case, the first derivative of y gives us the slope of the function, and the second derivative gives us the rate of change of that slope.

Simplify $\frac{0.0839\times 6.381}{5.44}$ to 2 significant figures.

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A bricklayer charges ₦1,500 per day for himself and ₦500 per day for his assistant. If a two bedroom flat was built for ₦95,000 and the bricklayer worked 10 days more than his assistant, how much did the assistant receive?

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Let the number of days worked by the assistant = t
$\therefore$ The bricklayer worked (t + 10) days.
1500(t + 10) + 500(t) = N 95,000
1500t + 15,000 + 500t = N 95,000
2000t = N 95,000 - N 15,000
2000t = N 80,000
t = 40 days
$\therefore$ The assistant worked for 40 days and received N (500 x 40)
= N 20,000

Determine the values for which $x^2-7x+10\le 0$

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To determine the values of x that satisfy x² - 7x + 10 ≤ 0, we need to find the roots of the quadratic equation x² - 7x + 10 = 0 and then analyze the behavior of the quadratic function. To find the roots, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a In this case, a = 1, b = -7, and c = 10, so: x = (-(-7) ± √((-7)² - 4(1)(10))) / 2(1) x = (7 ± √9) / 2 x1 = 5 and x2 = 2 The roots of the equation are x = 5 and x = 2. Now, we need to analyze the behavior of the quadratic function in the intervals between the roots and to the left and right of the roots. We can do this by creating a sign chart: Interval | x² - 7x + 10 --------------------------------- (-∞, 2) | + (2, 5) | - (5, +∞) | + In the interval (-∞, 2), the quadratic function is positive because all the factors are positive. In the interval (2, 5), the function is negative because x - 2 is negative and x - 5 is positive. In the interval (5, +∞), the function is positive again because both factors are positive. Therefore, the solution to the inequality x² - 7x + 10 ≤ 0 is the interval [2, 5], because the function is non-positive in that interval and positive elsewhere. In interval notation, we can write: 2 ≤ x ≤ 5 So the correct answer is: 2 ≤ x ≤ 5.

Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon.

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Since each interior angle = 140°;
Each exterior angle = 180° - 140° = 40°
Number of sides of the polygon = $\frac{360°}{40°}$
= 9 
Sum of angles in the polygon = 140° x 9
= 1260°

A binary operation Δ is defined by a Δ b = a + 3b + 2.

Find (3 Δ 2) Δ 5

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The binary operation Δ is defined as a Δ b = a + 3b + 2. To find (3 Δ 2) Δ 5, we need to evaluate the expression step by step. First, we need to find 3 Δ 2: 3 Δ 2 = 3 + 3(2) + 2 = 11 Next, we need to find (3 Δ 2) Δ 5: (3 Δ 2) Δ 5 = 11 Δ 5 = 11 + 3(5) + 2 = 28 Therefore, the answer is 28.

Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

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Equation: 4y = 7x + 3
$⟹y=\frac{7}{4}x+\frac{3}{4}$
Slope = coefficient of x = $\frac{7}{4}$
Slope of perpendicular line = $\frac{-1}{\frac{7}{4}}$
= $\frac{-4}{7}$
The perpendicular line passes (-3, 1)
$\therefore$ Using the equation of line $y=mx+b$
m = slope and b = intercept.
$y=\frac{-4}{7}x+b$
To find the intercept, substitute y = 1 and x = -3 in the equation.
$1=\frac{-4}{7}(-3)+b$
$1=\frac{12}{7}+b$
$b=\frac{-5}{7}$
$\therefore y=\frac{-4}{7}x-\frac{5}{7}$
$7y+4x+5=0$

Evaluate $\frac{2{\log }_{3}9\times {\log }_3{81}^{-2}}{{\log }_{5}625}$

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We can simplify the expression using the properties of logarithms. First, we can rewrite the expression as: 2log3(3^2) * log3(3^4) - 2log5(5^4) Using the power rule of logarithms, we can simplify the first term: 2log3(3^2) * log3(3^4) = 2(2) * 4 = 16 Using the power rule of logarithms again, we can simplify the second term: 2log5(5^4) = 8log5(5) = 8 Substituting these simplified terms back into the original expression, we get: 16 - 8 = 8 Therefore, the value of the expression is 8, which is option (C) in the given choices.

Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)

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To find the locus of a point A that is equidistant from two given points B and C, we need to find the perpendicular bisector of the line segment joining B and C. The locus of point A will be the line that is perpendicular to the line segment BC and passes through the midpoint of BC. First, we need to find the midpoint of the line segment BC: Midpoint = ((0+2)/2, (2+1)/2) = (1, 1.5) The slope of the line segment BC is (1-2)/(2-0) = -1/2. The slope of a line perpendicular to this line is the negative reciprocal of the slope, which is 2. Using the point-slope form of a line with the midpoint (1, 1.5) and slope 2, we get: y - 1.5 = 2(x - 1) Simplifying the equation, we get: y = 2x - 2 + 1.5 y = 2x - 0.5 Therefore, the equation of the locus of point A that is equidistant from B(0, 2) and C(2, 1) is 4x - 2y = 1, which is option (C) in the given choices.

Find the value of x for $\frac{2+2x}{3}-2\ge \frac{4x-6}{5}$

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$\frac{2+2x}{3}-2\ge \frac{4x-6}{5}$
$\frac{2+2x-6}{3}\ge \frac{4x-6}{5}$
$\frac{2x-4}{3}\ge \frac{4x-6}{5}$
$5(2x-4)\ge 3(4x-6)$
$10x-20\ge 12x-18$
$10x-12x\ge -18+20$
$-2x\ge 2$
$x\le -1$

A factory worker earns ₦50,000 per month out of which he spends 15% on his children's education, ₦13,600 on Food, 3% on electricity and uses the rest for his personal purpose. How much does he have left?

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The factory worker earns ₦50,000 per month. He spends 15% of his income on his children's education. 15% of ₦50,000 = (15/100) x ₦50,000 = ₦7,500 He spends ₦13,600 on food. He spends 3% of his income on electricity. 3% of ₦50,000 = (3/100) x ₦50,000 = ₦1,500 Therefore, the total amount he spends on education, food and electricity is: ₦7,500 + ₦13,600 + ₦1,500 = ₦22,600 To find out how much he has left for personal purposes, we subtract the total amount he spends from his income. ₦50,000 - ₦22,600 = ₦27,400 Therefore, he has ₦27,400 left for his personal purpose. So the answer is N27,400.

Evaluate $\frac{2\sin 30+5\tan 60}{\sin 60}$, leaving your answer in surd form.

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$\frac{2\sin 30+5\tan 60}{\sin 60}$
$\sin 30=\frac{1}{2};\tan 60=\sqrt{3};\sin 60=\frac{\sqrt{3}}{2}$
$\therefore \frac{2\sin 30+5\tan 60}{\sin 60}=\frac{2\left(\frac{1}{2}\right)+5\left(\sqrt{3}\right)}{\frac{\sqrt{3}}{2}}$
= $\frac{1+5\sqrt{3}}{\frac{\sqrt{3}}{2}}$
= $\frac{2(1+5\sqrt{3})}{\sqrt{3}}$
= $\frac{2+10\sqrt{3}}{\sqrt{3}}$
Rationalizing, we get
= $\frac{2\sqrt{3}+30}{3}$
= $\frac{2}{3}\sqrt{3}+10$

This table below gives the scores of a group of students in a Further Mathematics Test.

Find the mode of the distribution.

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Mode = Score with the highest frequency
= 5

In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC.

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< BAC = $\frac{130}{2}$ (angle subtended at the centre)
< BAC = 65°
Also, x = 26° (theorem)
y = 65° - 26° = 39°
< AOC = 180° - (39° + 39°)
= 102°

Differentiate $\frac{2x}{\sin x}$ with respect to x.

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To differentiate 2xsinx with respect to x, we use the product rule of differentiation, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. Applying this rule, we get: (2xsinx)' = 2(x)'sinx + 2x(sin x)' Simplifying this expression, we have: (2xsinx)' = 2sinx + 2xcosx Therefore, the answer is option (D) 2cscx(1−xcotx). None of the other options match the simplified expression we obtained using the product rule.

Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30

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To solve this system of equations, we can use substitution or elimination. Here's how to use substitution: From the first equation, we can rearrange it to solve for y in terms of x: 3x + y = 21 y = 21 - 3x Then we can substitute this expression for y into the second equation: xy = 30 x(21 - 3x) = 30 Expanding the left side: 21x - 3x^2 = 30 Rearranging and factoring: 3x^2 - 21x + 30 = 0 Dividing by 3: x^2 - 7x + 10 = 0 This quadratic equation can be factored as: (x - 2)(x - 5) = 0 So the possible values of x are 2 and 5. To find the corresponding values of y, we can substitute each value of x into the expression we found for y: When x = 2: y = 21 - 3(2) = 15 When x = 5: y = 21 - 3(5) = 6 Therefore, the solution to the system of equations is: x = 2 or 5, y = 15 or 6 So the correct option is: - x = 2 or 5, y = 15 or 6

In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females?

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To select a committee of 5 members, including 2 females, we can break down the problem into the following steps: Step 1: Select 2 females from the 3 available females. This can be done in $\binom{3}{2} = 3$ ways. Here, $\binom{n}{r}$ denotes the number of ways to choose r items from a set of n items, also known as "n choose r". Step 2: Select 3 members from the remaining 4 males and 1 female. This can be done in $\binom{4}{3} \cdot \binom{1}{0} = 4$ ways. Here, we choose 3 males from the 4 available males, and 0 females from the 1 remaining female. Step 3: Multiply the results of Steps 1 and 2 to obtain the total number of ways to choose the committee. Therefore, the total number of ways to select a committee of 5 members, including 2 females, is: $\binom{3}{2} \cdot \binom{4}{3} \cdot \binom{1}{0} = 3 \cdot 4 \cdot 1 = 12$ Hence, there are 12 ways to choose the members of the committee. Therefore, the answer is 12 ways.

Find the polynomial if given q(x) = x${}^2$ - x - 5, d(x) = 3x - 1 and r(x) = 7.

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Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder.
i.e In this case, the polynomial = (x${}^2$ - x - 5)(3x - 1) + 7.
= (3x${}^3$ - x${}^2$ - 3x${}^2$ + x - 15x + 5) + 7
= (3x${}^3$ - 4x${}^2$ - 14x + 5) + 7
= 3x${}^3$ - 4x${}^2$ - 14x + 12

If $\sin x=\frac{4}{5}$, find $\frac{1+{\cot }^{2}x}{{\csc }^{2}x-1}$.

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$\sin x=\frac{opp}{Hyp}=\frac{4}{5}$
5${}^2$ = 4${}^2$ + adj${}^2$
adj${}^2$ = 25 - 16 = 9
adj = $\sqrt{9}$ = 3
$\tan x=\frac{4}{3}$
$\cot x=\frac{1}{\frac{4}{3}}=\frac{3}{4}$
${\cot }^{2}x=(\frac{3}{4}{)}^2=\frac{9}{16}$
$\csc x=\frac{1}{\sin x}$
= $\frac{1}{\frac{4}{5}}=\frac{5}{4}$
${\csc }^{2}x=(\frac{5}{4}{)}^2=\frac{25}{16}$
$\therefore \frac{1+{\cot }^{2}x}{{\csc }^{2}x-1}=\frac{1+\frac{9}{16}}{\frac{25}{16}-1}$
= $\frac{25}{16}÷\frac{9}{16}$
= $\frac{25}{9}$

Express $(0.0439÷3.62)$ as a fraction.

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$(0.0439÷3.62)$
= 0.01213
$\approxeq$ 0.012
= $\frac{12}{1000}$

Given the matrix $A=\left|\begin{array}{cc}3& -2\\ 1& 6\end{array}\right|$. Find the inverse of matrix A.

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$A=\left|\begin{array}{cc}3& -2\\ 1& 6\end{array}\right|$
|A| = (3 x 6) - (-2 x 1)
= 18 + 2
= 20.
A${}^{-1}$ = $\frac{1}{20}\left|\begin{array}{cc}6& 2\\ -1& 3\end{array}\right|$
= $\left|\begin{array}{cc}\frac{6}{20}& \frac{2}{20}\\ \frac{-1}{20}& \frac{3}{20}\end{array}\right|$
= $\left|\begin{array}{cc}\frac{3}{10}& \frac{1}{10}\\ \frac{-1}{20}& \frac{3}{20}\end{array}\right|$

A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man?

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The man bought the car for ₦1,250,000 and later sold it at the rate of ₦1,085,000, which means he incurred a loss of ₦165,000 (1,250,000 - 1,085,000). To calculate the percentage loss, we divide the loss amount by the original cost and multiply by 100: ₦165,000 ÷ ₦1,250,000 * 100 = 13.2% So, the man incurred a loss of 13.2%.

In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°.

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The length of an arc in a circle can be found using the formula: Arc Length = (Angle Measure / 360) * (Circumference of Circle) The circumference of the circle can be found using the formula: Circumference = 2 * pi * radius So, first find the circumference of the circle: Circumference = 2 * pi * 7 = 14 * pi Next, find the angle measure in degrees in terms of pi: 57 degrees = 57 * pi / 180 radians Now, plug in the values into the formula for arc length: Arc Length = (57 * pi / 180) * (14 * pi) = (57 / 360) * (14 * pi) = (57 / 360) * (43.98) = 43.98 * (57 / 360) = (43.98 * 57) / 360 = 2493.86 / 360 = 6.97 cm So, the length of the arc AB is 6.97 cm.

Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree.

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$\tan \theta =\frac{9}{20}=0.45$
$\theta ={\tan }^{-1}\left(0.45\right)$
= 24.23°
$\therefore$ The bearing of Y from X = 180° - 24.23°
= 155.77°
= 156° (to the nearest degree)

If M varies directly as N and inversely as the root of P. Given that M = 3, N = 5 and P = 25. Find the value of P when M = 2 and N = 6.

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The problem tells us that "M varies directly as N and inversely as the root of P". This means that M is directly proportional to N and inversely proportional to the square root of P, which can be written mathematically as: M = k(N / sqrt(P)) where k is a constant of proportionality. We can solve for k by using the given values of M, N, and P. When M = 3, N = 5, and P = 25, we can plug these values into the equation above and solve for k: 3 = k(5 / sqrt(25)) 3 = k(5 / 5) 3 = k So, we have found that k = 3. Now, we can use this value of k to find P when M = 2 and N = 6. We can rearrange the equation above to solve for P: M = k(N / sqrt(P)) sqrt(P) = k(N / M) P = (k(N / M))^2 Plugging in k = 3, N = 6, and M = 2, we get: P = (3(6 / 2))^2 P = 9(3)^2 P = 81 Therefore, the value of P when M = 2 and N = 6 is 81. So, option D is the correct answer.

If the volume of a frustrum is given as $V=\frac{\pi h}{3}(R^2+Rr+r^2)$, find $\frac{dV}{dR}$.

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To find dV/dR, we need to take the derivative of V with respect to R, while treating all other variables as constants. Using the product rule of differentiation, we get: dV/dR = πh/3 [2R + r] Therefore, the answer is πh/3 [2R + r]. To understand this, we need to remember that the derivative of a function gives us the rate of change of the function. In this case, the volume of a frustrum is a function of its height (h) and radii (R and r). By taking the derivative of the volume with respect to one of the radii (R), we get the rate of change of the volume with respect to that radius. In other words, dV/dR tells us how much the volume changes for a small change in R, while holding h and r constant.

Find the probability that a number selected at random from 21 to 34 is a multiple of 3

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S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34}
n(S) = 14
multiples of 3 = {21, 24, 27, 30, 33}
n(multiples of 3) = 5
Prob( picking a multiple of 3) = 5/14

If $\left|\begin{array}{ccc}2& -5& 3\\ x& 1& 4\\ 0& 3& 2\end{array}\right|=132$, find the value of x.

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Explanation:

This table below gives the scores of a group of students in a Further Mathematics Test.

Calculate the mean deviation for the distribution

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Mean = $\frac{\sum fx}{\sum f}$

= $\frac{156}{40}$

= 3.9

M.D = $\frac{\sum f|x–\overline{x}|}{\sum f}$

= $\frac{60.4}{40}$

= 1.51

Find the value of k in the equation: $\sqrt{28}+\sqrt{112}-\sqrt{k}=\sqrt{175}$

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To find the value of k, we need to isolate the variable on one side of the equation. Let's start by simplifying each of the square roots: √28 = √4 × √7 = 2√7 √112 = √16 × √7 = 4√7 √175 is already simplified, so we leave it as is. Now we can substitute these values into the original equation: 2√7 + 4√7 - √k = √175 Simplifying the left side, we get: 6√7 - √k = √175 Next, we'll isolate the radical term by moving the non-radical term to the other side: 6√7 = √k + √175 Squaring both sides to eliminate the radicals: (6√7)² = (√k + √175)² 36 × 7 = k + 2√k × √175 + 175 252 = k + 5√7√35 252 = k + 5√(7 × 35) 252 = k + 5√(5² × 7) 252 = k + 5 × 5√7 252 = k + 25√7 k = 252 - 25√7 Therefore, the value of k is 252 - 25√7.

The histogram above represents the number of candidates who did Further Mathematics examination in a school. How many candidates scored more than 40?

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Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5
= 150 students.

If 2x${}^2$ + x - 3 divides x - 2, find the remainder.

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When you divide a polynomial p(x) by (x - a), the remainder = p(a)
i.e. In the case of 2x${}^2$ + x - 3 $÷$ (x - 2), the remainder = p(2).
= 2(2)${}^2$ + 2 - 3
= 8 + 2 - 3
= 7.

If the universal set μ = {x : 1 ≤ x ≤ 20} and

A = {y : multiple of 3}

B = |z : odd numbers}

Find A ∩ B

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To find A ∩ B, we need to find the elements that are common to both A and B. A = {y : multiple of 3} contains all the numbers that are multiples of 3 between 1 and 20. So, A = {3, 6, 9, 12, 15, 18}. B = {z : odd numbers} contains all the odd numbers between 1 and 20. So, B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. To find A ∩ B, we need to find the common elements of A and B. The common elements of A and B are 3, 9, and 15. Therefore, A ∩ B = {3, 9, 15}. So, option C - {3, 9, 15} is the correct answer.

If ${25}^{1-x}\times 5^{x+2}÷(\frac{1}{125}{)}^x={625}^{-1}$, find the value of x.

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${25}^{1-x}\times 5^{x+2}÷(\frac{1}{125}{)}^x={625}^{-1}$
$(5^2{)}^{(1-x)}\times 5^{(x+2)}÷(5^{-3}{)}^x=(5^4{)}^{-1}$
$5^{2-2x}\times 5^{x+2}÷5^{-3x}=5^{-4}$
$5^{(2-2x)+(x+2)-(-3x)}=5^{-4}$
Equating bases, we have
$2-2x+x+2+3x=-4$
$4+2x=-4⟹2x=-4-4$
$2x=-8$
$x=-4$

Integrate $\int (4x^{-3}-7x^2+5x-6)dx$.

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The correct answer is  &#x222B;(4x&#x2212;3&#x2212;7x2+5x&#x2212;6)dx = -2x - 2 - 73x^3 + 52x^2 - 6x. The symbol &#x222B; represents an integral, which is the area under a curve. To find the integral of a polynomial, we use antiderivatives, which are the reverse of derivatives. In this case, the polynomial is 4x - 3 - 7x^2 + 5x - 6, and we find its antiderivative by adding a constant of integration to each term, which is why there is a " - 2" at the end of the answer.

The table above is the distribution of data with mean equals to 3. Find the value of y.

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Mean = $\frac{\sum fx}{\sum f}$
$3=\frac{26-y}{3y+2}$
$3(3y+2)=26-y$
$9y+6=26-y$
$9y+y=26-6$
$10y=20⟹y=2$

If $P=(\frac{Q(R-T)}{15}{)}^{\frac{1}{3}}$, make T the subject of the formula.

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To isolate T on one side of the equation, we need to undo each operation that has been performed on T. To do this, we have to work backwards through the equation. Starting with the equation: P=(Q(R−T)15)13, - We will first undo the exponent operation (13), by taking the 13th root of both sides: P^(1/13) = (Q(R−T)15)^(1/13) - Then, we will undo the multiplication operation (15), by dividing both sides by 15: P^(1/13)/15 = (Q(R−T))^(1/13) - Next, we will undo the parenthesis operation, by multiplying both sides by (R−T): (P^(1/13)/15)(R−T) = Q - Finally, we will undo the subtraction operation (R−T), by adding T to both sides: T = R − (Q * 15 * (P^(1/13))) So the final answer is: T=R−15P3Q

Find the distance between the points C(2, 2) and D(5, 6).

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To find the distance between two points, we can use the distance formula which is: distance = √((x2 - x1)^2 + (y2 - y1)^2) Where (x1, y1) and (x2, y2) are the coordinates of the two points. Using this formula, let's find the distance between points C(2, 2) and D(5, 6). distance = √((5 - 2)^2 + (6 - 2)^2) distance = √(3^2 + 4^2) distance = √(9 + 16) distance = √25 distance = 5 units Therefore, the distance between points C and D is 5 units.