All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2007 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Evaluate 1011(base 2)^2 - 101(base 2)^2
1011(2) = 11 and 101(2) = 5 in base 10. 11^2 - 5^2 = 121 - 25 = 96 (base 10), which converts to 1100000 in base 2.
If X10 = 1214(base 5), find X.
1214(base5) = 1x5^3 + 2x5^2 + 1x5^1 + 4x5^0 = 125+50+5+4 = 184.
Evaluate 3/5 + (2/7 x 4/3 + 4/7)
Following order of operations on the bracketed expression and adding to 3/5 gives 7/10.
Evaluate [(0.5625)^2 - (0.4375)^2] / 0.04, correct to 3 significant figures.
Using difference of two squares: (0.5625+0.4375)(0.5625-0.4375)/0.04 = 1 x 0.125/0.04 = 3.125, which rounds to 3.13 (3 s.f.).
A man made 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%?
Cost price = 60,000/1.05 = #57,142.86. Selling price for 26% profit = 1.26 x 57,142.86 = #72,000.
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. Lawal's age = 28 x 4/7 = 16 years.
Find the value of x for which 2(3^(2x-1)) = 162.
3^(2x-1) = 81 = 3^4, so 2x-1=4, giving x = 5/2.
If log10 2 = x, express log10 12.5 in terms of x.
log10 12.5 = log10(100/8) = log10 10^2 - log10 2^3 = 2 - 3x.
Find y, if sqrt(12) - sqrt(147) + y.sqrt(3) = 0.
sqrt(12)=2sqrt(3), sqrt(147)=7sqrt(3). So 2sqrt(3)-7sqrt(3)+y.sqrt(3)=0, giving sqrt(3)(y-5)=0, so y=5.
Given: P={1,3,5,7,9,11} and Q={2,4,6,8,10,12}. Determine the relationship between P and Q.
P and Q share no common elements, so P and Q are disjoint sets: P intersect Q = empty set.
If X = {all perfect squares less than 40} and Y = {all odd numbers from 1 to 15}, find X intersect Y.
X = {1,4,9,16,25,36}, Y = {1,3,5,7,9,11,13,15}. X intersect Y = {1,9}.
Make L the subject of the formula d = sqrt(42w/(5L)).
Squaring both sides: d^2 = 42w/(5L). Rearranging: L = 42w/(5d^2), which corresponds to the given option for L in terms of d and w.
Factorize 2t^2 + t - 15.
2t^2+t-15 = 2t^2+6t-5t-15 = 2t(t+3)-5(t+3) = (t+3)(2t-5).
The graph above (a cubic curve crossing the x-axis at x=-2, -1 and 1) is represented by
The curve's roots are x=-2,-1,1, so y=(x+2)(x+1)(x-1) = x^3+2x^2-x-2.
W is proportional to L^2, and W=6 when L=4. If L = sqrt(17), find W.
K = W/L^2 = 6/16 = 3/8. W = (3/8) x (sqrt(17))^2 = (3/8) x 17 = 51/8 = 6 3/8.
The solution set of the quadratic inequality x^2+x-12 >= 0 is
x^2+x-12=(x-3)(x+4)>=0, giving x>=3 or x<=-4.
Solve the inequality -3(x-2) < -2(x+3).
-3x+6<-2x-6 gives -x<-12; dividing by -1 (reversing the inequality) gives x>12.
Find the nth term of the sequence 3/2, 3, 7, 16, 35, 74, ...
Checking the formula Tn=5(2^(n-2))-n against n=1,2,3 confirms it generates 3/2, 3, 7 - matching the sequence.
Find the sum to infinity of the series 2 + 3/2 + 9/8 + 27/32 + ...
a=2, r=(3/2)/2=3/4. Sum to infinity = a/(1-r) = 2/(1/4) = 8.
A binary operation Delta is defined by a Delta b = a+b+1 for any real numbers a and b. Find the inverse of the real number 7 under the operation Delta, if the identity element is -1.
Combining an element with its inverse gives the identity: 7+l+1=-1, so l=-1-8=-9.
Evaluate the matrix expression (3,-7/2,5) + 2(-2,4/3,-1).
(3,-7/2,5) + (-4,8/6,-2) = (3-4, -7+8/2+6, 5-2) = (-1,6/-1,3).
If f(x)=3x-2, P = (2,1/-1,0), and I is a 2x2 identity matrix, evaluate f(P).
f(P)=3P-2I = 3(2,1/-1,0) - 2(1,0/0,1) = (6,3/-3,0) - (2,0/0,2) = (4,3/-3,-2).
Using the same matrix setup as the previous question (f(x)=3x-2, P and the 2x2 identity matrix I), evaluate f(P) again.
As in question 22, f(P) = 3P - 2I = (4,3/-3,-2). Note: the source scan shows this working printed twice; presented here as given.
If the lines 2y-kx+2=0 and y+x-k/2=0 intersect at (1,2), find the value of k.
Substituting x=1, y=2 into k(-x+1/2) = -y+x-2 and solving gives k = -2.
Find the size of each exterior angle of a regular octagon.
Exterior angle = 360/n = 360/8 = 45 degrees.
In the parallelogram PQRS above (with angle QRS=30 deg and QSR=50 deg marked), find angle SQR.
Angle QSR=50 (corresponding angles). In triangle QRS: 50+30+SQR=180, so SQR=100 degrees.
A square has an area of 144cm^2. Find the length of its diagonal.
Side L = sqrt(144) = 12cm. Diagonal d = sqrt(12^2+12^2) = sqrt(288) = 12.sqrt(2) cm.
Calculate the length of an arc of a circle of diameter 14cm, which subtends an angle of 90 degrees at the centre of the circle.
Radius=7cm. Arc length = (90/360) x 2 x pi x 7 = 7pi/2 cm.
The volume of a hemispheric bowl is 718 2/3 cm^3. Find its radius. [pi=22/7]
Volume of hemisphere = (2/3)pi.r^3. Solving (2/3)(22/7)r^3 = 718(2/3) gives r^3=343, so r=7cm.
Find the locus of points equidistant from two straight lines y-5=0 and y-3=0.
The equidistant locus lies midway between y=5 and y=3, i.e. y=(5+3)/2=4, giving the locus y-4=0.
A particle P moves between points S and T such that angle SPT is always constant. Find the locus of P.
Since the angle subtended at any point on the circumference by a diameter is constant (90 degrees), the locus of P is a semi-circle with ST as diameter.
What is the value of k if the mid-point of the line joining (1-k,-4) and (2,k+1) is (-k,k)?
Equating the x-coordinate: -k = (1-k+2)/2 = (3-k)/2. Solving gives -2k=3-k, so k=-3.
If the lines 3y=4x-1 and qy=x+3 are parallel to each other, the value of q is
Gradient of first line = 4/3. Gradient of second line = 1/q. Since parallel lines have equal gradients: 1/q=4/3, so q=3/4.
Find the value of (cos60 - tan30)/(cos60 + tan30).
Using tan60=sqrt(3), tan30=sqrt(3)/3 (and comparable cos values as given in the source working), the expression simplifies to 1/2.
A man 40m from the foot of a tower observes the angle of elevation of the tower to be 30 degrees. Determine the height of the tower.
tan30 = h/40, so h = 40.tan30 = 40 x (1/sqrt(3)) = 40sqrt(3)/3 m.
If y=x.cos(x), find dy/dx.
Using the product rule: dy/dx = x(-sin x) + cos(x)(1) = cos(x) - x.sin(x).
If y=(1+x)^2, find dy/dx.
Let u=1+x. y=u^2, dy/du=2u. dy/dx = 2u x 1 = 2(1+x) = 2+2x.
Find the value of x for which the function f(x)=2x^3-x^2-4x+4 has a maximum value.
f'(x)=6x^2-2x-4=0 simplifies to 3x^2-x-2=0, giving x=-2/3 or x=1. Testing the second derivative shows x=-2/3 gives the maximum.
Integrate (x^2 - sqrt(x))/x with respect to x.
(x^2-sqrt(x))/x = x - x^(-1/2). Integrating: x^2/2 - 2sqrt(x) + k.
Determine the value of the integral of -2cos(x) dx from 0 to pi/2.
Integral of -2cos(x) dx = -2sin(x). Evaluating from 0 to pi/2: -2sin(90) - (-2sin(0)) = -2 - 0 = -2.
Mark: 3,4,5,6,7,8; Frequency: 5,y-1,y,9,4,1. The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, calculate the value of y.
Total mark = sum of (frequency x mark) = 3(5)+4(y-1)+5(y)+6(9)+7(4)+8(1) = 9y+101 = 200, giving y=11.
The histogram above represents the weight of students who travelled out of their school for an excursion. How many people made the trip?
The number of people who made the trip is the sum of the heights (frequencies) of all the bars: 2+3+5+6+8+4+1=29.
The pie chart above illustrates the amount of private time a student spends in a week studying various subjects (Maths 105 deg, Social 75 deg, Science k, Others 3k, English 2k). Find the value of k.
Sum of all sectoral angles = 360 deg: 105+75+3k+k+2k=360, so 180+6k=360, giving k=30 degrees.
The cumulative frequency for 5<=x<=12 from the distribution above is
Summing the frequencies for values from 5 to 12 (10,12,10,8,6,8,10,8) gives a cumulative frequency of 72.
If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1 respectively, find the product of the modal and the median number.
Arranged in order: 2,5,5,5,6,6,6,6,8,8. Modal number (highest frequency) = 6. Median (middle value) = 6. Product = 6x6=36.
What is the mean deviation of 3, 5, 8, 11, 12 and 21?
A senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered?
Number of possible itineraries = 7P4 = 7!/(7-4)! = 7x6x5x4 = 840.
In how many ways can 6 subjects be selected from 10 subjects for an examination?
Number of ways = 10C6 = 10!/(4!6!) = 210.
In a basket there are 6 grapes, 11 bananas, and 13 oranges. If one fruit is chosen at random, what is the probability that it is either a grape or a banana?
P(grape or banana) = P(grape) + P(banana) = 6/30 + 11/30 = 17/30.
Age (years): 10, 11, 12; No. of pupils: 6, 27, 7. The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
Total pupils = 6+27+7 = 40. Pupils at least 11 years old = 27+7 = 34. Probability = 34/40 = 17/20.
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