JAMB UTME - Mathematics - 2012

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The first step is to find the intersection point of the two lines y = 2x + 4 and y = 7 - x. To do this, we can set the two equations equal to each other and solve for x: 2x + 4 = 7 - x 3x = 3 x = 1 So, the intersection point is (1, 2). Next, we can use the distance formula to find the distance between the point (4, 3) and the intersection point (1, 2): distance = √((4 - 1)^2 + (3 - 2)^2) = √((3^2) + (1^2)) = √(9 + 1) = √10 = √(2 * 5) = √2 * √5 = 3√2. So, the distance between the two points is 3√2.

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We can simplify the given expression using the identity: (a + b)^2 - (a - b)^2 = 4ab where a = √6 and b = 2. Substituting these values, we get: (√6 + 2)^2 - (√6 - 2)^2 = 4(√6 × 2) Simplifying further: (√6 + 2)^2 - (√6 - 2)^2 = 8√6 Therefore, the simplified form of the expression (√6+2)2−(√6−2)2 is 8√6.

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The locus of a point equidistant from the intersection of two lines is the perpendicular bisector of the line segment joining the intersection point of the lines. Therefore, we need to find the intersection point of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0, and then find the perpendicular bisector passing through that point. To find the intersection point of the two lines, we can solve their simultaneous equations. Multiplying the first equation by 4 and the second equation by 3, we get: 12x - 28y + 28 = 0 12x - 18y + 3 = 0 Subtracting the second equation from the first, we get: -10y + 25 = 0 Solving for y, we get: y = 5/2 Substituting this value of y in either of the original equations, we get: 3x - 7(5/2) + 7 = 0 3x = 17/2 x = 17/6 So, the intersection point of the lines is (17/6, 5/2). Now, we need to find the perpendicular bisector of the line segment joining this point with any other point equidistant from the intersection point. Since the question does not specify any other point, we can assume that we are looking for the perpendicular bisector of the line segment joining the intersection point with itself. This means we need to find the equation of the line passing through the point (17/6, 5/2) and perpendicular to the line joining the intersection point with itself, which is the x-axis. The line passing through (17/6, 5/2) and perpendicular to the x-axis has a slope of 0, so its equation is of the form y = b, where b is the y-coordinate of the point (17/6, 5/2). Therefore, the equation of the perpendicular bisector is: y = 5/2 This is the equation of a line parallel to the x-axis and passing through the point (0, 5/2). Therefore, the locus of a point equidistant from the intersection of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a line parallel to 7x + 13y + 8 = 0, which is.

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To find the remainder when a polynomial is divided by another polynomial, we can use the polynomial long division method. The steps for polynomial long division are as follows: 1. Divide the highest degree term of the dividend (2x^3 in this case) by the highest degree term of the divisor (x+3) and write the result above the long division bracket. 2. Multiply the divisor (x+3) by the quotient obtained in step 1 and write the result below the dividend. 3. Subtract the result obtained in step 2 from the dividend and write the remainder below. 4. Bring down the next term of the dividend and repeat steps 1-3 until there are no more terms to bring down. Using this method, we can divide 2x^3 - 11x^2 + 8x - 1 by x + 3 to find the remainder: 2x^2 - 17x + 51 ______________________ x + 3 | 2x^3 - 11x^2 + 8x - 1 2x^3 + 6x^2 ____________ -17x^2 + 8x -17x^2 - 51x ____________ 59x - 1 59x + 177 ________ -178 Therefore, the remainder when 2x^3 - 11x^2 + 8x - 1 is divided by x + 3 is -178.

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The mean of seven numbers is the sum of all seven numbers divided by 7. If the mean is 96, then the sum of the seven numbers is 96 * 7 = 672. When an eighth number is added, the mean becomes 112, which means that the sum of all eight numbers is 112 * 8 = 896. To find the eighth number, we can subtract the sum of the first seven numbers from the sum of all eight numbers: Eighth number = 896 - 672 = 224. So, the eighth number is 224.

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The probability that the student passes a physics test is 2/3. Let's call this probability p. The probability that he fails a physics test is therefore 1 - p = 1/3. To find the probability that he passes two out of three tests, we need to consider the different ways in which this can happen. He could pass the first two tests and fail the third, or pass the first and third tests but fail the second, or pass the second and third tests but fail the first. Each of these scenarios has the same probability: p * p * (1 - p) + p * (1 - p) * p + (1 - p) * p * p Simplifying this expression, we get: 3p^2(1-p) Substituting p = 2/3, we get: 3(2/3)^2(1-2/3) = 4/27 Therefore, the probability that the student passes two out of three physics tests is 4/27, which corresponds to option (C).

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Arrange all the values in ascending order,
2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

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Sr = $\frac{a}{1-r}$
$-\frac{1}{10}$ = $\frac{1}{8}\times \frac{1}{1-r}$
$-\frac{1}{10}$ = $\frac{1}{8(1-r)}$
$-\frac{1}{10}$ = $\frac{1}{8-8r}$
cross multiply...
-1(8 - 8r) = -10
-8 + 8r = -10
8r = -2
r = -1/4

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$\frac{2\frac{2}{3}\times 1\frac{1}{2}}{4\frac{4}{5}}$
$\frac{8}{3}\times \frac{3}{2}÷\frac{24}{5}$
$\frac{8}{3}\times \frac{3}{2}\times \frac{5}{24}$
$\frac{5}{6}$

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The set P contains all the prime factors of 30, which are 2, 3, and 5. The set Q contains all the factors of 18 less than 10, which are 1, 2, and 3. Therefore, the intersection of sets P and Q (denoted by the symbol ∩) is the set of factors that are both prime factors of 30 and less than 10, which is {2, 3}. Thus, the answer is: {2, 3}.

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PQ = $\frac{y_1-y_0}{x_1-x_0}$ = $\frac{-3-(-7)}{-2-5}$ = $\frac{-3+7}{-2-5}$ = $\frac{4}{-7}$

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Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o

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cos(x + 40)o = 0.0872

x + 40 = cos-10.0872

x + 40o = 84.99o

x = 84.99o - 40o

x = 44.99

x = 45o

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$\frac{y-y_1}{x+x_1}$ = $\frac{y_2-y_1}{x-x_1}$
$\frac{y-1}{x+2}$ = $\frac{4-1}{-\frac{1}{2}+2}$
= $\frac{y-1}{x+2}$ = $\frac{3}{\frac{3}{2}}$
y = 2x + 5

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To find the number of students who play neither football nor volleyball, we need to subtract the total number of students who play either football or volleyball (including those who play both) from the total number of students in the class. First, we add the number of students who play football and the number of students who play volleyball, then subtract the number of students who play both (since they were counted twice). 22 + 26 - 3 = 45 So there are 45 students who play either football or volleyball or both. To find the number of students who play neither, we subtract this from the total number of students in the class: 46 - 45 = 1 Therefore, only 1 student plays neither football nor volleyball. The correct answer is 1.

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${27}^{x+2}÷9^{x+1}=3^{2x}$
Reduce 27 and 9 to have the same base as 3
$3^{3(x+2)}÷3^{2(x+1)}=3^{2x}$
Note, - is ÷ and + is × for powers of indices of the same bases.
$3^{3(x+2)-2(x+1)}=3^{2x}$
Equating the exponentials
$3(x+2)-2(x+1)=2x$
Clear brackets
$3x+6-2x-2=2x$
Collect like terms
$3x-2x-2x=-6+2$
$x-2x=-4$
$-x=-4$
Divide both sides by $-$
$x=4$
Answer is B

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We can solve for x and y by using substitution and elimination methods. Starting with the first equation: x^2 - y^2 = 4 We can rewrite this as: x^2 = 4 + y^2 Next, we substitute this expression for x^2 into the second equation: x + y = 2 x^2 + 2xy + y^2 = 4 + y^2 + 2xy + y^2 x^2 + 2xy + y^2 = 4 + 2xy + 2y^2 x^2 = 4 x = ±2 Since x cannot be negative, we can conclude that x = 2. Finally, we can substitute x = 2 into the second equation to find y: 2 + y = 2 y = 0 So, the solution is x = 2, y = 0

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y $\alpha \sqrt{n}$
y = k$\sqrt{n}$
when y = 4, n = 4
4 = k$\sqrt{4}$
4 = 2k
k = 2
Therefore,
y = 2$\sqrt{n}$
y = 2$\sqrt{\frac{16}{9}}$
y = 2$\left(\frac{4}{3}\right)$
y = $\frac{8}{3}$

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Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70${}^{\circ }$
If 360${}^{\circ }$ represents 36 students
1${}^{\circ }$ will represent 36/360
50${}^{\circ }$ will represent 36/360 * 70/1
= 7

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The man earns ₦3,500 per month and spends 15% of it on his children's education. We can calculate the amount he spends on his children's education as: 15% of ₦3,500 = (15/100) x ₦3,500 = ₦525 So, his total expenses on education and food are: ₦525 + ₦1,950 = ₦2,475 Subtracting his total expenses from his monthly income, we get his savings: ₦3,500 - ₦2,475 = ₦1,025 Therefore, the man has ₦1,025 left after spending on his children's education and food. Hence, the answer to the question is ₦1,025.

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The probability that the man will not live for 80 years is 1 - 2/3 = 1/3. Similarly, the probability that the woman will not live for 80 years is 1 - 3/5 = 2/5. The probability that both of them will not live for 80 years is the product of their individual probabilities, which is (1/3) * (2/5) = 2/15. Therefore, the probability that at least one of them will live up to 80 years is 1 - 2/15 = 13/15. So, the answer is 13/15.

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To find 2 * (3 * 4), we need to evaluate the expression in parentheses first, i.e., 3 * 4. Using the given binary operation *, we have: 3 * 4 = 3(4) + 3 - 4 = 11 Now we can substitute this value into the original expression: 2 * (3 * 4) = 2 * 11 = 2(11) + 2 - 11 = 13 Therefore, the answer is 13.

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$\frac{8!}{3!}$
$\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}$
==> 8 x 7 x 6 x 5 x 4 = 6720

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m $\propto$ n = $\frac{mn}{2}-a$
Identify = e = 2
a $\propto$ a-1 = e
a $\propto$ a-1 = 2
-5 $\propto$ a-1 = 2
$\frac{-5\times a^{-1}}{2}=2$
$a^{-1}=\frac{2\times 2}{-5}$
$a^{-1}=-\frac{4}{5}$

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The nth term of the given sequence is n^2 - 6n - 4. To find the sum of the 3rd and 4th terms, we need to substitute n = 3 and n = 4 respectively, and then add the resulting values. So, the third term of the sequence is: 3^2 - 6(3) - 4 = 1 And, the fourth term of the sequence is: 4^2 - 6(4) - 4 = -12 Therefore, the sum of the 3rd and 4th terms is: 1 + (-12) = -11 Hence, the correct answer is option (D) -25 is not correct because -11 is the sum of the 3rd and 4th terms, not the 4th term itself.

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We can start by using the definition of logarithms to rewrite the equation: log3(x^2) = -8 This means that 3 raised to the power of -8 is equal to x^2: 3^(-8) = x^2 To solve for x, we can take the square root of both sides: sqrt(3^(-8)) = sqrt(x^2) On the left side, we can simplify the expression using the rule that says sqrt(a^b) = a^(b/2): 3^(-8/2) = x Simplifying the exponent, we get: 3^(-4) = x Recall that a negative exponent means the reciprocal of the corresponding positive exponent. So: 3^(-4) = 1/3^4 Using the exponent rule that says a^b = a*a*a*...*a (b times), we get: 1/81 = x Therefore, the correct answer is option (D) 181.

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$\left|\begin{array}{cc}5& 3\\ x& 2\end{array}\right|$ = $\left|\begin{array}{cc}3& 5\\ 4& 5\end{array}\right|$
10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5

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To convert 726 to base three, we need to find the largest power of three that is less than or equal to 726. In this case, 3^6 = 729, which is greater than 726, so we need to use powers of three lower than 729. First, we divide 726 by 243, which is 3^5. The quotient is 2 and the remainder is 240. Next, we divide the remainder 240 by 81, which is 3^4. The quotient is 2 and the remainder is 78. Continuing in this way, we divide 78 by 27, which is 3^3. The quotient is 2 and the remainder is 24. Then, we divide 24 by 9, which is 3^2. The quotient is 2 and the remainder is 6. Finally, we divide 6 by 3, which is 3^1. The quotient is 2 and the remainder is 0. Therefore, the base-three representation of 726 is 222202, which corresponds to the last option (1122).

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Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3

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$\frac{n+1(n-2)}{(n+1)!}$
$\frac{(n+1)+(n-2)!(n-2)!}{(n+1)!}$
$\frac{(n+1)(n+1-1)(n+1-2)(n+1-3)!}{3!(n-2)!}$
$\frac{(n+1)\left(n\right)(n-1)(n-2)!}{3!(n-2)!}$
$\frac{(n+1)\left(n\right)(n-1)}{3!}$
Since n = 15
$\frac{(15+1)\left(15\right)(15-1)}{3!}$
$\frac{16\times 15\times 14}{3\times 2\times 1}$
= 560

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A unit matrix is a square matrix in which all the diagonal elements are equal to 1, and all the other elements are equal to 0. The symbol I3 represents the unit matrix of order 3, which is a 3x3 matrix. So, the matrix I3 can be written as: |1 0 0| |0 1 0| |0 0 1| To find the determinant of I3, we can use the formula for the determinant of a 3x3 matrix: |a b c| |d e f| |g h i| = a(ei - fh) - b(di - fg) + c(dh - eg) Applying this formula to I3, we get: |1 0 0| |0 1 0| |0 0 1| = 1(1*1 - 0*0) - 0(0*1 - 0*0) + 0(0*1 - 0*0) = 1 Therefore, the determinant of I3 is 1, and the correct answer is option (C) 1. Options (A) -1, (B) 0, and (D) 2 are incorrect.

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The sum of the interior angles of a polygon can be found using the formula: (n - 2) * 180°, where n is the number of sides of the polygon. In this case, the polygon has 5 sides, so the sum of its interior angles is (5 - 2) * 180° = 540°. Let's call the value of x "a". The sum of the angles of the polygon is given by: x + 2x + 3x + 4x + 5x = 15x Since the sum of the angles is 540°, we have: 15a = 540 Dividing both sides by 15, we get: a = 36 So, the value of x, or the common difference between the angles of the polygon, is 36°

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$\begin{array}{ccc}x& (x-\varkappa )& (x-\varkappa {)}^2\\ 2& -5& 25\\ 3& -4& 16\\ 8& 1& 1\\ 10& 3& 9\\ 12& 5& 25\\ & & 76\end{array}$
S.D = $\sqrt{\frac{(x-\varkappa {)}^2}{n}}$
S.D = $\sqrt{\frac{76}{5}}$
S.D = 3.9

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To find δyδx (the derivative of y with respect to x), we need to apply the power rule of differentiation, which states that if y = x^n, then δyδx = n*x^(n-1). Applying this rule to y = x^2 - 1/x, we get: δyδx = 2x + 1/x^2 Therefore, the answer is not one of the options provided. Option A (2x - 1/x^2) is close but has a minus sign instead of a plus sign before the second term. Option B (2x + x^2) and option C (2x - x^2) are incorrect because they don't take into account the derivative of the second term (-1/x). Option D (2x + 1/x^2) is the correct answer based on the power rule of differentiation.

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When two quantities, U and V, are inversely proportional, it means that as one of them increases, the other decreases, and vice versa, in such a way that their product remains constant. In mathematical terms, we can express this relationship as U*V^3 = k, where k is a constant. In this problem, we are told that U is inversely proportional to the cube of V. Therefore, we can write U*V^3 = k, where k is a constant of proportionality that we need to find. We are also given that U equals 81 when V equals 2. Substituting these values into the equation, we get: 81*2^3 = k k = 648 Now that we have the constant of proportionality, we can use the equation U*V^3 = 648 to find U when V equals 3: U*3^3 = 648 U*27 = 648 U = 648/27 Simplifying this expression, we get U = 24. Therefore, the answer is 24.

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To evaluate 21/9 to 3 significant figures, we need to round the result to the nearest thousandth. The result of 21 divided by 9 is 2.33333333333... The third digit after the decimal point is 3, which is greater than or equal to 5. Therefore, we need to round up the second digit after the decimal point, which is 3. Therefore, rounding 2.33333333333... to 3 significant figures gives us 2.33. Therefore, the correct option is 2.33.

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For a regular polygon with n sides, each interior angle is given by: interior angle = (n-2) x 180 / n We are given that each interior angle of the polygon is 140 degrees. Substituting this into the formula, we get: 140 = (n-2) x 180 / n Multiplying both sides by n, we get: 140n = (n-2) x 180 Expanding the right-hand side, we get: 140n = 180n - 360 Simplifying and isolating n, we get: n = 360 / (180 - 140) n = 360 / 40 n = 9 Therefore, the regular polygon has 9 sides. Hence, the answer to the question is 9.

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The range of a set of numbers is the difference between the largest and smallest values in the set. To find the range of the set {4, 9, 6, 3, 2, 8, 10, 11}, we need to find the largest and smallest values in the set and then subtract the smallest value from the largest value. The smallest value in the set is 2 and the largest value is 11. So, the range is 11 - 2 = 9. So, the range of the set {4, 9, 6, 3, 2, 8, 10, 11} is 9.

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$\theta$ = 135o
Cos 135o = Cos(90 + 45)o
= cos90ocos45o - sin90osin45o
= 0cos45o - (1 x $\frac{\sqrt{2}}{2}$)
= $-\frac{\sqrt{2}}{2}$

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w = $\frac{v(2+cn)}{1-cn}$
2v + cnv = w(1 - cn)
2v + cnv = w - cnw
2v - w = -cnv - cnw
Multiply through by negative sign
-2v + w = cnv + cnw
-2v + w = n(cv + cw)
n = $\frac{-2v+w}{cv+cw}$
n = $\frac{1}{c}\frac{-2v+w}{v+w}$
Re-arrange...
n = $\frac{1}{c}\frac{w-2v}{v+w}$

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$\frac{1}{p}$ - $\frac{1}{q}$ + $\frac{p}{q}$ - $\frac{q}{p}$ = $\frac{q-p}{pq}$ + $\frac{p^2-q^2}{pq}$
$\frac{q-p}{pq}$ x $\frac{pq}{p^2{q}^2}$ = $\frac{q-p}{p^2{q}^2}$
$\frac{-(p-q)}{(p+q)(p-q)}$
= $\frac{-1}{p+q}$

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