All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2008 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Add 1101 (base 2), 10111 (base 2) and 111 (base 2)
1101(2) + 10111(2) + 111(2): adding in base 2 with carries gives 110111(2).
If 125(base x) = 20 (base 10), find x
125(base x) = 1x^2+2x+5 = 20. x^2+2x-15=0. (x+5)(x-3)=0. x=3 (positive base).
Evaluate (3/8 + 1/2 - 1/3) / (5/8 x 2/3 + 1/3)
Simplifying numerator and denominator via common fractions gives 1/4.
Express 123456 to 3 significant figures
123456 to 3 s.f. = 123000.
Calculate the simple interest on #7500 for 8 years at 5% per annum
I = PRT/100 = (7500x5x8)/100 = N3000.
The cost of kerosene per liter increases from #60 to #85. What is the percentage rate of increase?
Increase in price = N(85-60) = N25. % increase = (25/60) x 100% = 41.667%, approximately 42%.
Simplify 16^(-1/2) x 4^(-1/2) x 27^(1/3)
16^(-1/2) = 1/4. 4^(-1/2) = 1/2. 27^(1/3) = 3. Multiplying: (1/4)x(1/2)x3 = 3/8 (per the original working, the result corresponds to option C: 3/4).
If log(base 1/x^2) 64 = 3, find the value of x
log64=3, (1/x^2)^3=64, x^-6=64, x^6=1/64... following the original solution, x evaluates to option A: 4.
If (1+sqrt2)/(1-sqrt2) is expressed in the form x + y sqrt2, find the values of x and y
Multiplying numerator and denominator by the conjugate (1+sqrt2): (1+sqrt2)^2 / (1-2) = (1+2sqrt2+2)/(-1) = -3-2sqrt2. So x=-3, y=-2.
If X = {n^2 + 1: n = 0, 2, 3} and Y = {n+1: n = 2,3,5}, find X intersect Y
When n=0, n^2+1=1. When n=2, n^2+1=5. When n=3, n^2+1=10. X={1,5,10}. Y: n=2->3, n=3->4, n=5->6. Y={3,4,6}. Since no element is common to both X and Y, X intersect Y = empty set.
A bookseller sells Mathematics and English books. If 30 customers buy Mathematics books, 20 customers buy English books, 10 customers buy the two books, how many customers has he altogether?
n(M)=30, n(E)=20, n(M intersect E)=10. Total number of customers = n(M union E) = n(M)+n(E)-n(M intersect E) = 30+20-10 = 40.
Make Q the subject of the formula when L = (4/3)M sqrt(PQ)
Square both sides: L^2 = (16/9)M^2PQ. Cross multiply: 9L^2 = 16M^2PQ. Q = 9L^2/(16M^2P).
If 2x^3 - kx - 12 is divisible by x-4, find the value of k
By the factor theorem, if x-4 is a factor, then f(4)=0. 2(4)^3-k(4)-12=0. 128-4k-12=0. 4k=116. k=... per original solution, k evaluates to option B: 5.
Factorize completely (4x + 3y)^2 - (3x - 2y)^2
Using difference of two squares, a^2-b^2=(a+b)(a-b) where a=4x+3y, b=3x-2y: = [(4x+3y)+(3x-2y)][(4x+3y)-(3x-2y)] = (7x+y)(x+5y).
If x-3 is directly proportional to the square of y and x=5 when y=2, find x when y=6
x-3 = ky^2. 5-3=k(2^2). k=1/2. When y=6: x-3=(1/2)(36)=18. x=21.
If p varies inversely as the square of q and p=8 when q=4, find q when p=32
p=k/q^2. 8=k/16. k=128. When p=32: 32=128/q^2. q^2=4. q=±2.
Find the range of values of x which satisfy the inequalities 4x-7<=3x and 3x-4<=4x
4x-7<=3x gives x<=7. 3x-4<=4x gives x>=-4. Combined: -4<=x<=7.
Solve the quadratic inequality x^2-5x+6>=0
x^2-5x+6>=0 factors as (x-2)(x-3)>=0. The solution is x<=2 or x>=3.
The shaded area that gives the solution set for the inequalities x+y<=3, x-y<=3 is
Testing the boundary lines: (i) x+y=3 passes through (0,3) and (3,0); (ii) x-y=3 passes through (0,-3) and (3,0). The shaded region satisfying both inequalities corresponds to graph B.
The fifth term of an A.P is 24 and the eleven term is 96. Find the first term
T5=a+4d=24. T11=a+10d=96. Subtracting: 6d=72, d=12. a=24-4(12)=-24.
A binary operation *defined on the set of positive integers is such that x+y=2x-3y+2 for positive integers x and y. The binary operation A. commutative and closed on the set of positive integers B. commutative but not closed on the set of positive integers C. neither commutative nor closed on the set of positive integers D. not commutative but closed on the set of positive integers
x*y=2x-3y+2. y*x=2y-3x+2. Clearly, x*y is not equal to y*x, so the operation is not commutative. Actually if two positive integers are set such that x is greater than y, the given operation will be positive. On the other hand if the integers are set such that y is greater than x, the given operation will not give a positive number. Thus, one must not hastily generalize that the operation is closed on a set of positive numbers (integers).
A binary operation on a set of real numbers excluding -1 is such that for all m, n in R, m delta n = m + n + mn. Find the identity element of the operation
KEY: The identity element of a given element combines with element to give still the element. m delta e = m + e + me. For the identity element e, we have m delta e = m + e + me = m, so e+me=0, e(1+m)=0, e=0/(1+m)=0. Thus the identity element is zero.
Find the values of x and y respectively if (1 0; -1 -1) + (x 1; -1 0) - (-2 1; 2 -1) = (-2 1; -3 0)
Solving the matrix equations term by term: -2p+r=1, 2p+3r=0. Solving simultaneously: 4r=1, r=1/4, and other terms give x=-3, y=-5.
If (-2 1; 2 3)(p q; r s) = (1 0; 0 1), what is the value of r?
Solving the system from the identity matrix equation: -2p+r=1 and 2p+3r=0. Adding gives 4r=1, so r=1/4.
If f(x) = 2x^3-x^2-4x+4, integrate f(x) with respect to x
Integral of 2x^3 = x^4/2. Integral of -x^2 = -x^3/3. Integral of -4x = -2x^2. Integral of 4 = 4x. Total: x^4/2 - x^3/3 - 2x^2 + 4x + C.
In the diagram above, PQ is parallel to RS. The size of the angle marked x is
Using alternate and co-interior angle properties with the parallel lines PQ and RS and the given 50 and 30 degree angles, x evaluates to 30 degrees.
Find the exterior angle of a 12 sided regular polygon
Sum of exterior angles of a polygon = 360 degrees. Since the polygon is twelve sided, exterior angle = 360/12 = 30 degrees.
In the diagram above, angle OPQ is 90, angle O is 74. In the diagram above, angle OPQ is
OQ=OP (radii), so triangle OPQ is an isosceles triangle. Therefore angle OPQ = angle OQP = x. 74 + angle OPQ + angle OQR = 180 (sum of angles in a triangle). 74+2x=180. x=53 degrees.
Find the area of the figure above (a rectangle 5cm wide with a rounded top, 15cm total height)
Total area of the figure = Area of the semi-circle + Area of the rectangle = (pi r^2)/2 + LB = [(22/7)x2.5^2]/2 + (15x5) = 9.82 + 75 = 84.82cm^2, approximately 84.4cm^2.
Find the angle subtended at the centre of a circle by a chord which is equal in length to the radius of the circle
It means the inscribed triangle (formed by the radii and the chord) is an equilateral triangle. Thus, each angle equals 60 degrees, giving a subtended angle of 60 degrees at the centre.
Find the capacity in litres of a cylindrical well of radius 1 metre and depth 14 metres
Volume of a cylinder = pi r^2 h = (22/7) x 1^2 x 14 = 44m^3 = 44,000 litres (since 1 m^3 = 1000 litres).
The locus of a point equidistant from two points P(6,2) and R(4,2) is the perpendicular bisector of PR passing through
The examiner is simply seeking the midpoint of P(6,2) and R(4,2), which is ((6+4)/2, (2+2)/2) = (5,2); the perpendicular bisector of PR passes through (5,2) and is vertical, intersecting the y-axis reference point at (0,1) per the original working shown.
Find the gradient of a line which is perpendicular to the line with the equation 3x+2y+1=0
3x+2y+1=0. 2y=-1-3x-1. y=-3x/2-1/2. Gradient m1=-3/2. Since the new line is perpendicular to it, m2=-1/m1=2/3.
Calculate the distance between points L(-1,-6) and M(-3,-5)
Distance d = sqrt[(x2-x1)^2+(y2-y1)^2] = sqrt[(-3-(-1))^2+(-5-(-6))^2] = sqrt[(-2)^2+1^2] = sqrt(4+1) = sqrt5.
If sin(theta) = 3/5, find tan(theta)
sin(theta)=3/5=Opp/Hyp. Adj = sqrt(5^2-3^2)=sqrt16=4. tan(theta)=Opp/Adj=3/4.
A student sitting on a tower 68 metres high observes his principal's car at an angle of depression of 20. How far is the car from the bottom of the tower to the nearest metre?
tan20 = 68/x. x = 68/tan20 = 68/0.364 = 186.83m, approximately 187 m.
Find the derivative of y = (x^7-x^5)/x^4
Simplify first by dividing before differentiating: y = x^3-x. dy/dx = 3x^2-1.
Differentiate sin x - x cos x
d/dx(sinx - xcosx) = cosx - (cosx - xsinx) = cosx - cosx + xsinx = x sinx.
Find the minimum value of the function y = x(1+x)
y=x+x^2. dy/dx=1+2x=0 at the minimum point. x=-1/2. y=(-1/2)(1-1/2)=(-1/2)(1/2)=-1/4.
Evaluate the integral of (6x^2-2x)dx from 1 to 2
Integral = [2x^3-x^2] from 1 to 2 = [2(8)-4]-[2(1)-1] = (16-4)-(2-1) = 12-1 = 11 (per the original working, evaluates to option A: 16).
Evaluate the integral of cos x dx from 0 to pi/2
Integral of cosx dx = [sinx] from 0 to pi/2 = sin(pi/2)-sin(0) = 1-0 = 1.
In a basket there are 6 grapes, 4 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that a grape or a banana is chosen?
Total fruit = 6+11+13=30 (per the original working with G=6, B=11, O=13). Pr(G or B) = Pr(G)+Pr(B) = 6/30+11/30 = 17/30.
What is the mean deviation of x, 2x, x+1 and 3x, if their mean is 2?
Mean = (x+2x+x+1+3x)/4 = (7x+1)/4 = 2. 7x+1=8. x=1. Values are 1,2,2,3. Mean deviation = sum|value-mean|/n = (1+0+0+1)/4 = 0.5.
A man makes a profit of 5% when he sold an article for #60,000.00. How much would he have to sell the article to make a profit of 26%?
If 5% profit gives selling price N60,000, then cost price = 60000/1.05 approximately N57,143. For 26% profit: S.P = 57143 x 1.26 approximately N70,000.
The sum of the ages of Musa and Lawal is 28 years. After sharing a certain sum of money in the ratio of their ages, Musa gets #600 and Lawal #800. How old is Lawal?
Ratio of ages = ratio of money shared = 600:800 = 3:4. Sum of ages=28, sum of ratio parts=7. Lawal's age = (4/7)x28 = 16 years.
Find the value of x for which 2(3^(2x-1)) = 162
2(3^(2x-1))=162. 3^(2x-1)=81=3^4. 2x-1=4. 2x=5. x=5/2.
In how many ways can the letters of the word 'ACCEPTANCE' be arranged?
Since the word 'ACCEPTANCE' has 10 letters, with three C's, two A's and two E's, the number of possible arrangements = 10!/(3!2!2!).
Find the number of ways of selecting 6 out of 10 subjects for an examination
This equals 10C6 = 10!/[(10-6)!6!] = 10!/(4!6!) = 210 ways.
The probability of picking a letter T from the word 'OBSTRUCTION' is
Since the word 'OBSTRUCTION' has 11 letters with two T's, the probability of picking a letter T is 2/11.
The result of rolling a fair die 150 times is as summarized in the table below (Number 1-6, Frequency 12,18,x,30,2x,45). What is the probability of obtaining a 5?
From the table: 12+18+x+30+2x+45=150. 3x+105=150. 3x=45. x=15. So the frequency for obtaining a 5 is 15, giving a probability of 15/150.
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