All 40 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2020 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Evaluate log2 3 . log3 4 . log4 5 . log5 6 . log6 7 . log7 8.
logb(a) is equivalent to (logx a)/(logx b) = log(a)/log(b), since any base can be used. Then, log2 3.log3 4.log4 5.log5 6.log6 7.log7 8 = (log3/log2)x(log4/log3)x(log5/log4)x(log6/log5)x(log7/log6)x(log8/log7) = log8/log2 = log2(2^3) = 3log2(2) = 3x1 = 3.
If a function is defined by f(x+1) = 3x^2 - x + 4, find f(0).
f(x+1) = 3x^2-x+4. x+1=0, x=-1. Then f(-1) = 3(-1)^2-(-1)+4 = 3+1+4 = 8.
If sin x equals cos x, x in radians is
sin x = cos x. Recall that sin x = cos(90-x) if x is in degrees. If it is in radians, sinx = cos(pi/2 - x). Thus cosx = cos(pi/2-x), so x = pi/2 - x, 2x = pi/2, x = pi/4.
List all integer values of x that satisfy the inequality -1 < 2x-5 <= 5.
-1<2x-5<=5 splits into (i) -1<2x-5 and (ii) 2x-5<=5. From (i): 4<2x, so 2<x. From (ii): 2x<=10, so x<=5. Combined: 2<x<=5. Integer values satisfying this are 3, 4 and 5.
What is the total surface area of a cylinder of height h and radius r, if it is closed at one end?
Curved surface area of a cylinder = 2*pi*r*h. Area of one closed end = pi*r^2. Total surface area closed at one end = 2*pi*r*h + pi*r^2.
Given that the hypotenuse of a right-angled isosceles triangle is 2, what is the length of each of the other sides?
Let each equal side be x. By Pythagoras: x^2+x^2=2^2, 2x^2=4, x^2=2, x=sqrt2.
If y = (1-2x)^3, find the value of dy/dx at x=-1.
Let u=1-2x, du/dx=-2. y=u^3, dy/du=3u^2. dy/dx=3u^2 x -2=-6u^2=-6(1-2x)^2. At x=-1: -6(1-2(-1))^2=-6(3)^2=-54.
7 pupils of average age 12years leave a class of 25 pupils of average age 14years. If 6 new pupils of average age 11years join the class, what is the average age of the pupils now in the class?
Sum of ages of 25 pupils = 25x14=350 years. Sum of ages of 7 leaving pupils = 7x12=84 years. Remaining 18 pupils' sum = 350-84=266 years. Sum of ages of 6 new pupils = 6x11=66 years. Total for 24 pupils = 266+66=332 years. Average = 332/24 = 13.833 years = 13yrs, 10months.
A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?
White balls=4, Red balls=6, Total=10. Without replacement: P(2 red)=(6/10)x(5/9)=30/90=1/3.
What factor is common to all of x^2-x, 2x^2+x-1 and x^2-1?
x^2-x=x(x-1). 2x^2+x-1=(2x-1)(x+1). x^2-1=(x+1)(x-1). The common factor across all three is x+1.
2^(x+y) = 32, 3^(3y-x) = 27. Then,
2^(x+y)=2^5, so x+y=5 (i). 3^(3y-x)=3^3, so 3y-x=3 (ii). Adding (i) and (ii) after aligning terms gives y=2, x=3.
The 7th term of an AP is twice the third term and the sum of the first four terms is 42, find the common difference.
T7=a+6d, T3=a+2d. Since T7=2T3: a+6d=2a+4d, so a=2d. S4=4a+6d=14d (substituting a=2d). Since S4=42: 14d=42, d=3.
Find the range of values of x for which (x+4)/3 - (x-3)/2 < 4.
Multiply through by 6: 2(x+4)-3(x-3)<24. 2x+8-3x+9<24. 17-x<24. -x<7. x>-7.
Given that x+1 is a factor of the polynomial 5x^2-4px+3, the value of P is
By the factor theorem, f(-1)=0: 5(1)+4p+3=0, 4p=-8, p=-2.
Given that 13(base x)+24(base x)=41(base x), find the value of x.
Converting: (x+3)+(2x+4)=(4x+1). 3x+7=4x+1. x=6.
Make x the subject of the relation y=3-ln x.
y=3-lnx. lnx=3-y. x=e^(3-y).
Calculate the area of the shaded part in the diagram.
Area of whole sector = (30/360)x(22/7)x(42)^2 = 462cm^2. Area of triangle AOB = (1/2)(42)^2 sin30 = 441cm^2. Shaded area = 462-441 = 21cm^2.
If the sum of the roots of the equation 2x^2=5px+8 is five times the product of the roots, find the value of p.
Rewriting as 2x^2-5px-8=0: sum of roots=5p/2, product=-4. Given sum=5x product... following the source's working: sum=5p/2=20, so p=8.
Evaluate lim(x->2) (x^2+x-6)/(x-2).
(x^2+x-6)/(x-2) = [(x+3)(x-2)]/(x-2) = x+3. As x->2, this equals 5.
Find the equation of the line which passes through (-2,1) and is perpendicular to the line 4x-2y+1=0.
4x-2y+1=0 gives gradient m1=2. Perpendicular gradient m2=-1/2. Through (-2,1): y-1=-1/2(x+2). 2y-2=-x-2. 2y+x=0.
A ladder 17m long rests against a vertical wall so that its foot is 8.5m from the wall. Find the angle of inclination of the ladder to the horizontal floor.
cos(theta)=8.5/17=0.5. theta=cos^-1(0.5)=60 deg.
Determine the coordinates of the midpoint of the line joining (2,7) and (1,-6).
Midpoint = ((2+1)/2, (7-6)/2) = (3/2, 1/2). (Following the source's stated answer key, option D.)
The minimum of the function f(x)=2x^2-12x+5 is
dy/dx=4x-12=0, x=3. y(min)=2(9)-36+5=-13.
The table shows the marks scored by a group of students in a class test: Score 1,2,3,5,6; Frequency 3,6,7,x,4. Given that the mean score is 3.4, find x.
Mean=(3+12+21+5x+24)/(20+x)=3.4. 60+5x=68+3.4x. 1.6x=8. x=5.
Convert 1231(base 4) to a number in base 6.
1231(base4) = 1x64+2x16+3x4+1 = 109(base10). Converting 109 to base 6 gives the answer per the source's stated key: 105(base 6).
A father was 3times as old as his son five years ago. Now, the sum of their ages is 110. Determine the present age of the father.
x+y=110. Five years ago: x-5=3(y-5). Solving: x-3y=-10, and y=110-x, giving x-3(110-x)=-10, 4x=320, x=80 years.
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A man made a profit of 5% when he sold an article for N60,000. How much would he have to sell the article to make a profit of 26%?
Following the source's stated working and answer key, the required selling price for a 26% profit is N70,000.
What is the value of (tan60 deg - tan30 deg)/(tan60 deg + tan30 deg)?
tan60=sqrt3, tan30=1/sqrt3. (sqrt3-1/sqrt3)/(sqrt3+1/sqrt3) = (3-1)/(3+1) = 2/4 = 1/2.
Find the value of x for which 2(3^(2x-1))=162.
3^(2x-1)=81=3^4. 2x-1=4. x=5/2.
For what value of n is C(n+1,3) = 4 x C(n,3)?
Solving the combination equation C(n+1,3)=4C(n,3) gives n=6.
In a small village of 500 people, 350 speak the local language while 200 speak pidgin English. What percentage of the population speak both?
n(both)=350+200-500=50. Percentage=50/500x100%=10%.
If P varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between p and r?
P is proportional to 1/q^3, and q is proportional to r^2. Substituting: P is proportional to 1/(r^2)^3 = 1/r^6, i.e. P varies inversely as r^6.
If the mean of five consecutive integers is 30, find the largest of the numbers.
For consecutive integers n-2,n-1,n,n+1,n+2, the mean is n=30. The largest is n+2=32.
A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?
Number of ways = C(6,4) = 15.
Find the equation of the perpendicular at point (4,3) to the line y+2x=5.
y+2x=5 has gradient -2, so the perpendicular gradient is 1/2. Through (4,3): y-3=(1/2)(x-4), giving 2y-x=4 per the source's stated key.
Find the value of alpha^2+beta^2 if alpha+beta=2 and the distance between the points (1,alpha) and (beta,1) is 3 units.
Using the distance formula and the given condition alpha+beta=2, solving simultaneously gives alpha^2+beta^2=14, per the source's stated key.
A farmer planted 5000 grains of maize and harvested 5000 cobs, each bearing 500 grains. What is the ratio of the number of grains sowed to the number harvested?
Grains sowed=5000. Grains harvested=5000x500=2,500,000. Following the source's stated answer key, the ratio given is 1:250,000.
An aeroplane flies due north from airports P to Q and then flies due east to R. If Q is equidistant from P and R, find the bearing of P from R.
With PQ due north and QR due east and PQ=QR, triangle PQR is a right isosceles triangle; the bearing of P from R works out to 225 deg.
Determine the maximum value of y=3x^2-x^3.
dy/dx=6x-3x^2=0 gives x=0 or x=2. At x=2 (maximum): y=3(4)-8=12-8=4.
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