Paper JAMB 2017 General Maths Objective
Objective paper · 31 questions · partial
JAMB 2017 · UTME Topics include Calculus (JAMB bridge), Matrices & determinants, Coordinate geometry, Dispersion & cumulative frequency, Quadratics & their graphs, Trigonometric ratios.
Our copy of this paper is missing questions 7, 8, 9, 11, 14, 18, 19, 24, 37.
Sit this paper Answer every question in order, timed if you like (suggested 20 min). You're marked when you hand in, then you see where to focus and the working for each question.
Or read it here: every question below has a worked solution.
1 2 3 4 5 6 10 12 13 15 16 17 20 21 22 23 25 26 27 28 29 30 31 32 33 34 35 36 38 39 40 If d y d x = 2 x − 3 \dfrac{dy}{dx} = 2x - 3 d x d y = 2 x − 3 y = 3 y = 3 y = 3 x = 0 x = 0 x = 0 y y y x x x
A 2 x 2 − 3 x 2x^2 - 3x 2 x 2 − 3 x B x 2 − 3 x x^2 - 3x x 2 − 3 x C x 2 − 3 x − 3 x^2 - 3x - 3 x 2 − 3 x − 3 D x 2 − 3 x + 3 x^2 - 3x + 3 x 2 − 3 x + 3
Worked solution (try it first) Integrate:
y = x 2 − 3 x + c y = x^2 - 3x + c y = x 2 − 3 x + c .
Put in
x = 0 x = 0 x = 0 ,
y = 3 y = 3 y = 3 :
3 = 0 − 0 + c 3 = 0 - 0 + c 3 = 0 − 0 + c , so
c = 3 c = 3 c = 3 .
So
y = x 2 − 3 x + 3 y = x^2 - 3x + 3 y = x 2 − 3 x + 3 , option D.
Watch out
2 x 2x 2 x x 2 x^2 x 2 2 x 2 2x^2 2 x 2 2 x 2 − 3 x 2x^2 - 3x 2 x 2 − 3 x 4 x − 3 4x - 3 4 x − 3 y = 0 y = 0 y = 0 x = 0 x = 0 x = 0 Also set as JAMB 2002 · UME · Q48
Report a problem with this question
If P = ( 2 1 − 3 0 ) P = \begin{pmatrix} 2 & 1 \\ -3 & 0 \end{pmatrix} P = ( 2 − 3 1 0 ) I I I 2 × 2 2 \times 2 2 × 2 P 2 − 2 P + 4 I P^2 - 2P + 4I P 2 − 2 P + 4 I
A ( 9 4 − 12 1 ) \begin{pmatrix} 9 & 4 \\ -12 & 1 \end{pmatrix} ( 9 − 12 4 1 ) B ( − 3 0 0 − 3 ) \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} ( − 3 0 0 − 3 ) C ( 1 0 0 1 ) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 1 0 0 1 ) D ( 1 4 4 1 ) \begin{pmatrix} 1 & 4 \\ 4 & 1 \end{pmatrix} ( 1 4 4 1 )
Worked solution (try it first) Square
P P P row by column:
P 2 = ( 4 − 3 2 + 0 − 6 + 0 − 3 + 0 ) P^2 = \begin{pmatrix} 4 - 3 & 2 + 0 \\ -6 + 0 & -3 + 0 \end{pmatrix} P 2 = ( 4 − 3 − 6 + 0 2 + 0 − 3 + 0 ) = ( 1 2 − 6 − 3 ) = \begin{pmatrix} 1 & 2 \\ -6 & -3 \end{pmatrix} = ( 1 − 6 2 − 3 ) .
Subtract
2 P = ( 4 2 − 6 0 ) 2P = \begin{pmatrix} 4 & 2 \\ -6 & 0 \end{pmatrix} 2 P = ( 4 − 6 2 0 ) :
P 2 − 2 P = ( − 3 0 0 − 3 ) P^2 - 2P = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} P 2 − 2 P = ( − 3 0 0 − 3 ) .
Add
4 I 4I 4 I , which adds 4 to each diagonal entry:
( 1 0 0 1 ) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 1 0 0 1 ) , option C.
Watch out
4 I = ( 4 0 0 4 ) 4I = \begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} 4 I = ( 4 0 0 4 ) ( 1 4 4 1 ) \begin{pmatrix} 1 & 4 \\ 4 & 1 \end{pmatrix} ( 1 4 4 1 ) Also set as JAMB 2002 · UME · Q40
Report a problem with this question
Find the value of a a a 2 y − a x + 4 = 0 2y - ax + 4 = 0 2 y − a x + 4 = 0 y + 1 4 x − 1 = 0 y + \frac14x - 1 = 0 y + 4 1 x − 1 = 0
Worked solution (try it first) Rearrange
2 y − a x + 4 = 0 2y - ax + 4 = 0 2 y − a x + 4 = 0 :
2 y = a x − 4 2y = ax - 4 2 y = a x − 4 , so the gradient is
a 2 \frac a2 2 a .
Rearrange
y + 1 4 x − 1 = 0 y + \frac14x - 1 = 0 y + 4 1 x − 1 = 0 :
y = − 1 4 x + 1 y = -\frac14x + 1 y = − 4 1 x + 1 , so the gradient is
− 1 4 -\frac14 − 4 1 .
Perpendicular gradients multiply to
− 1 -1 − 1 :
a 2 × ( − 1 4 ) = − 1 \frac a2 \times \left(-\frac14\right) = -1 2 a × ( − 4 1 ) = − 1 , so
a 8 = 1 \frac a8 = 1 8 a = 1 .
So
a = 8 a = 8 a = 8 , option C.
Watch out
Moving − a x -ax − a x + a x +ax + a x + a 2 +\frac a2 + 2 a − a 2 -\frac a2 − 2 a a = − 8 a = -8 a = − 8 Report a problem with this question
Calculate the mean deviation of the numbers 7, 3, 14, 9, 7 and 8.
A 2 1 6 2\frac16 2 6 1 B 2 1 2 2\frac12 2 2 1 C 1 1 6 1\frac16 1 6 1 D 2 1 3 2\frac13 2 3 1
Worked solution (try it first) The numbers add up to 48 and there are 6 of them, so the mean is 8.
The distances from 8 are 1, 5, 6, 1, 1 and 0, which add up to 14.
The mean deviation is
14 6 = 2 1 3 \frac{14}{6} = 2\frac13 6 14 = 2 3 1 , option D.
Watch out
Add the distances carefully: they total 14, not 15. A total of 15 gives 2 1 2 2\frac12 2 2 1 Also set as JAMB 2002 · UME · Q30
Report a problem with this question
The graph of y = x 2 + 4 y = x^2 + 4 y = x 2 + 4 P Q PQ P Q x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0 P Q PQ P Q
A y = 3 x − 2 y = 3x - 2 y = 3 x − 2 B y = 3 x + 2 y = 3x + 2 y = 3 x + 2 C y = 3 x − 4 y = 3x - 4 y = 3 x − 4 D y = 3 x + 4 y = 3x + 4 y = 3 x + 4
Worked solution (try it first) Rearrange the equation to have
x 2 x^2 x 2 alone:
x 2 = 3 x − 2 x^2 = 3x - 2 x 2 = 3 x − 2 .
Add 4 to both sides so the left side matches the curve:
x 2 + 4 = 3 x + 2 x^2 + 4 = 3x + 2 x 2 + 4 = 3 x + 2 .
So the curve
y = x 2 + 4 y = x^2 + 4 y = x 2 + 4 meets the line
y = 3 x + 2 y = 3x + 2 y = 3 x + 2 at the roots.
P Q PQ P Q is
y = 3 x + 2 y = 3x + 2 y = 3 x + 2 , option B.
Watch out
Add 4 to both sides; don't subtract it. x 2 + 4 = 3 x − 4 x^2 + 4 = 3x - 4 x 2 + 4 = 3 x − 4 x 2 − 3 x + 8 = 0 x^2 - 3x + 8 = 0 x 2 − 3 x + 8 = 0 Also set as JAMB 2003 · UME · Q14
Report a problem with this question
If π 2 ≤ θ < 2 π \frac{\pi}{2} \le \theta < 2\pi 2 π ≤ θ < 2 π f ( θ ) = 4 6 + 2 cos θ f(\theta) = \dfrac{4}{6 + 2\cos\theta} f ( θ ) = 6 + 2 cos θ 4
A 4 B 1 C 2 3 \frac23 3 2 D 1 3 \frac13 3 1
Worked solution (try it first) The top is fixed at 4, so
f ( θ ) f(\theta) f ( θ ) is largest when the bottom,
6 + 2 cos θ 6 + 2\cos\theta 6 + 2 cos θ , is smallest.
cos θ \cos\theta cos θ is smallest,
− 1 -1 − 1 , at
θ = π \theta = \pi θ = π , which is inside the range.
So the maximum is
4 6 − 2 = 4 4 = 1 \frac{4}{6 - 2} = \frac44 = 1 6 − 2 4 = 4 4 = 1 , option B.
Watch out
A fraction gets bigger as its bottom gets smaller, so use cos θ = − 1 \cos\theta = -1 cos θ = − 1 θ = π 2 \theta = \frac{\pi}{2} θ = 2 π cos θ = 0 \cos\theta = 0 cos θ = 0 4 6 = 2 3 \frac46 = \frac23 6 4 = 3 2 Also set as JAMB 2003 · UME · Q28
Report a problem with this question
Find x x x y y y
4 2 4 3 − 1 3 x 4 y 3 4 4 \begin{array}{rcccc} & 4 & 2 & 4 & 3 \\ - & 1 & 3 & x & 4 \\ \hline & y & 3 & 4 & 4 \end{array} − 4 1 y 2 3 3 4 x 4 3 4 4
Worked solution (try it first) Units:
3 − 4 3 - 4 3 − 4 needs a borrow of 5, and
8 − 4 = 4 8 - 4 = 4 8 − 4 = 4 , which matches.
Fives: after lending, the 4 is 3, and
3 − x = 4 3 - x = 4 3 − x = 4 needs another borrow.
So
8 − x = 4 8 - x = 4 8 − x = 4 and
x = 4 x = 4 x = 4 .
Twenty-fives: after lending, the 2 is 1, and
1 − 3 1 - 3 1 − 3 needs a borrow:
6 − 3 = 3 6 - 3 = 3 6 − 3 = 3 , which matches.
Hundred-and-twenty-fives: after lending, the 4 is 3, and
3 − 1 = 2 3 - 1 = 2 3 − 1 = 2 , so
y = 2 y = 2 y = 2 .
That is
x = 4 x = 4 x = 4 and
y = 2 y = 2 y = 2 , option C.
Watch out
"Respectively" means x x x y y y Also set as JAMB 2018 · UTME · Q1 · Also set as JAMB 2004 · UME · Q1
Report a problem with this question
P Q R S T V PQRSTV P QR S T V 7 cm 7\text{ cm} 7 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 12 cm 12\text{ cm} 12 cm B 42 cm 42\text{ cm} 42 cm C 44 cm 44\text{ cm} 44 cm D 56 cm 56\text{ cm} 56 cm
Worked solution (try it first) A regular hexagon splits into six equilateral triangles from the centre, so the radius equals the side:
r = 7 r = 7 r = 7 cm.
Circumference
= 2 π r = 2 × 22 7 × 7 = 2\pi r = 2 \times \frac{22}{7} \times 7 = 2 π r = 2 × 7 22 × 7 .
So the circumference is 44 cm, option C.
Watch out
6 × 7 = 42 6 \times 7 = 42 6 × 7 = 42 Also set as JAMB 2004 · UME · Q25
Report a problem with this question
The shadow of a pole 5 3 m 5\sqrt3\text{ m} 5 3 m 5 m 5\text{ m} 5 m
A 50 ∘ 50^\circ 5 0 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 75 ∘ 75^\circ 7 5 ∘
Worked solution (try it first) The pole is opposite the angle of elevation and the shadow is adjacent, so
tan θ = 5 3 5 = 3 \tan\theta = \frac{5\sqrt3}{5} = \sqrt3 tan θ = 5 5 3 = 3 .
tan 60 ∘ = 3 \tan60^\circ = \sqrt3 tan 6 0 ∘ = 3 , so
θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘ , option C.
Watch out
Put the height on top: tan θ = height shadow \tan\theta = \frac{\text{height}}{\text{shadow}} tan θ = shadow height 1 3 \frac{1}{\sqrt3} 3 1 30 ∘ 30^\circ 3 0 ∘ Also set as JAMB 2004 · UME · Q34
Report a problem with this question
A container has 30 gold medals, 22 silver medals and 18 bronze medals. If one medal is selected at random, what is the probability that it is not a gold medal?
A 4 7 \frac47 7 4 B 3 7 \frac37 7 3 C 11 35 \frac{11}{35} 35 11 D 9 35 \frac{9}{35} 35 9
Worked solution (try it first) There are
30 + 22 + 18 = 70 30 + 22 + 18 = 70 30 + 22 + 18 = 70 medals.
Not gold means silver or bronze:
22 + 18 = 40 22 + 18 = 40 22 + 18 = 40 medals.
So the probability is
40 70 = 4 7 \frac{40}{70} = \frac47 70 40 = 7 4 , option A.
Watch out
30 70 = 3 7 \frac{30}{70} = \frac37 70 30 = 7 3 Also set as JAMB 2004 · UME · Q50
Report a problem with this question
A polynomial in x x x − 2 -2 − 2 − 1 -1 − 1 3 3 3
A x 3 − 7 x + 6 x^3 - 7x + 6 x 3 − 7 x + 6 B x 3 + 7 x − 6 x^3 + 7x - 6 x 3 + 7 x − 6 C x 3 + 7 x + 6 x^3 + 7x + 6 x 3 + 7 x + 6 D x 3 − 7 x − 6 x^3 - 7x - 6 x 3 − 7 x − 6
Worked solution (try it first) A zero at
a a a gives a factor
x − a x - a x − a , so the factors are
( x + 2 ) (x + 2) ( x + 2 ) ,
( x + 1 ) (x + 1) ( x + 1 ) and
( x − 3 ) (x - 3) ( x − 3 ) .
Multiply the first two:
( x + 2 ) ( x + 1 ) = x 2 + 3 x + 2 (x + 2)(x + 1) = x^2 + 3x + 2 ( x + 2 ) ( x + 1 ) = x 2 + 3 x + 2 .
Multiply by
x − 3 x - 3 x − 3 :
x 3 + 3 x 2 + 2 x − 3 x 2 − 9 x − 6 = x 3 − 7 x − 6 x^3 + 3x^2 + 2x - 3x^2 - 9x - 6 = x^3 - 7x - 6 x 3 + 3 x 2 + 2 x − 3 x 2 − 9 x − 6 = x 3 − 7 x − 6 , option D.
Watch out
The zero − 2 -2 − 2 x + 2 x + 2 x + 2 x − 2 x - 2 x − 2 ( x − 2 ) ( x − 1 ) ( x + 3 ) = x 3 − 7 x + 6 (x - 2)(x - 1)(x + 3) = x^3 - 7x + 6 ( x − 2 ) ( x − 1 ) ( x + 3 ) = x 3 − 7 x + 6 Report a problem with this question
If M = ( 1 3 2 4 5 − 1 − 3 2 0 ) M = \begin{pmatrix} 1 & 3 & 2 \\ 4 & 5 & -1 \\ -3 & 2 & 0 \end{pmatrix} M = 1 4 − 3 3 5 2 2 − 1 0 N = ( 1 − 2 3 4 − 1 5 2 − 3 − 1 ) N = \begin{pmatrix} 1 & -2 & 3 \\ 4 & -1 & 5 \\ 2 & -3 & -1 \end{pmatrix} N = 1 4 2 − 2 − 1 − 3 3 5 − 1 2 M − 3 N 2M - 3N 2 M − 3 N
A ( − 1 12 5 4 7 3 0 − 5 − 3 ) \begin{pmatrix} -1 & 12 & 5 \\ 4 & 7 & 3 \\ 0 & -5 & -3 \end{pmatrix} − 1 4 0 12 7 − 5 5 3 − 3 B ( − 1 0 − 5 − 4 7 − 17 0 − 5 3 ) \begin{pmatrix} -1 & 0 & -5 \\ -4 & 7 & -17 \\ 0 & -5 & 3 \end{pmatrix} − 1 − 4 0 0 7 − 5 − 5 − 17 3 C ( − 1 12 − 5 − 4 13 − 17 − 12 13 3 ) \begin{pmatrix} -1 & 12 & -5 \\ -4 & 13 & -17 \\ -12 & 13 & 3 \end{pmatrix} − 1 − 4 − 12 12 13 13 − 5 − 17 3 D ( − 1 12 − 5 4 13 13 − 12 13 3 ) \begin{pmatrix} -1 & 12 & -5 \\ 4 & 13 & 13 \\ -12 & 13 & 3 \end{pmatrix} − 1 4 − 12 12 13 13 − 5 13 3
Worked solution (try it first) Work entry by entry: each entry is
2 × ( entry of M ) − 3 × ( entry of N ) 2 \times (\text{entry of } M) - 3 \times (\text{entry of } N) 2 × ( entry of M ) − 3 × ( entry of N ) .
Row 1:
2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 ,
6 + 6 = 12 6 + 6 = 12 6 + 6 = 12 and
4 − 9 = − 5 4 - 9 = -5 4 − 9 = − 5 .
Row 2:
8 − 12 = − 4 8 - 12 = -4 8 − 12 = − 4 ,
10 + 3 = 13 10 + 3 = 13 10 + 3 = 13 and
− 2 − 15 = − 17 -2 - 15 = -17 − 2 − 15 = − 17 .
Row 3:
− 6 − 6 = − 12 -6 - 6 = -12 − 6 − 6 = − 12 ,
4 + 9 = 13 4 + 9 = 13 4 + 9 = 13 and
0 + 3 = 3 0 + 3 = 3 0 + 3 = 3 .
So
2 M − 3 N 2M - 3N 2 M − 3 N is option C.
Watch out
Subtracting a negative adds: 2 ( 3 ) − 3 ( − 2 ) = 6 + 6 = 12 2(3) - 3(-2) = 6 + 6 = 12 2 ( 3 ) − 3 ( − 2 ) = 6 + 6 = 12 6 − 6 6 - 6 6 − 6 Report a problem with this question
What is the probability that an integer x x x 1 ≤ x ≤ 20 1 \le x \le 20 1 ≤ x ≤ 20
A 1 20 \frac{1}{20} 20 1 B 1 3 \frac13 3 1 C 3 20 \frac{3}{20} 20 3 D 7 20 \frac{7}{20} 20 7
Worked solution (try it first) Divisible by both 2 and 3 means divisible by 6.
The multiples of 6 up to 20 are 6, 12, 18, which is 3 numbers.
So the probability is
3 20 \frac{3}{20} 20 3 , option C.
Watch out
"Both" means multiples of 6. Counting multiples of 2 or 3 gives 10 + 6 − 3 = 13 10 + 6 - 3 = 13 10 + 6 − 3 = 13 Report a problem with this question
A trader bought goats for ₦4000 each. He sold them for ₦180,000 at a loss of 25 % 25\% 25%
Worked solution (try it first) He sold at a
25 % 25\% 25% loss, so ₦180,000 is
75 % 75\% 75% of the cost: cost
= 180 000 ÷ 0.75 = = 180\,000 \div 0.75 = = 180 000 ÷ 0.75 = ₦240,000.
Each goat cost ₦4,000, so the number of goats is
240 000 ÷ 4000 = 60 240\,000 \div 4000 = 60 240 000 ÷ 4000 = 60 .
So he bought 60 goats, option A.
Watch out
₦180,000 is what he sold them for, not what he paid. Dividing it by ₦4,000 gives 45 (option C); find the cost price first. Also set as JAMB 2002 · UME · Q1
Report a problem with this question
Evaluate 0.21 × 0.072 × 0.00054 0.006 × 1.68 × 0.063 \dfrac{0.21 \times 0.072 \times 0.00054}{0.006 \times 1.68 \times 0.063} 0.006 × 1.68 × 0.063 0.21 × 0.072 × 0.00054
A 0.01286 B 0.01285 C 0.1286 D 0.1285
Worked solution (try it first) Cancel in pairs:
0.21 0.063 = 10 3 \frac{0.21}{0.063} = \frac{10}{3} 0.063 0.21 = 3 10 ,
0.072 1.68 = 3 70 \frac{0.072}{1.68} = \frac{3}{70} 1.68 0.072 = 70 3 and
0.00054 0.006 = 0.09 \frac{0.00054}{0.006} = 0.09 0.006 0.00054 = 0.09 .
Multiply:
10 3 × 3 70 = 1 7 \frac{10}{3} \times \frac{3}{70} = \frac17 3 10 × 70 3 = 7 1 , and
1 7 × 0.09 = 9 700 \frac17 \times 0.09 = \frac{9}{700} 7 1 × 0.09 = 700 9 .
9 700 = 0.012857 … \frac{9}{700} = 0.012857\ldots 700 9 = 0.012857 … , which rounds to 0.01286, option A.
Watch out
Count the zeros in 0.00054: it is 0.09 of 0.006, not 0.9. Using 0.9 gives 0.1286 (option C). Report a problem with this question
In the diagram, P S T PST P S T P Q = Q S = R S PQ = QS = RS P Q = QS = R S ∠ R S T = 72 ∘ \angle RST = 72^\circ ∠ R S T = 7 2 ∘ x x x
A 36 ∘ 36^\circ 3 6 ∘ B 18 ∘ 18^\circ 1 8 ∘ C 72 ∘ 72^\circ 7 2 ∘ D 24 ∘ 24^\circ 2 4 ∘
Worked solution (try it first) P Q = Q S PQ = QS P Q = QS , so
∠ Q S P = ∠ Q P S = x \angle QSP = \angle QPS = x ∠ QS P = ∠ QP S = x .
The exterior angle of triangle
P Q S PQS P QS at
Q Q Q is
∠ S Q R = x + x = 2 x \angle SQR = x + x = 2x ∠ S QR = x + x = 2 x .
Q S = R S QS = RS QS = R S , so
∠ Q R S = ∠ S Q R = 2 x \angle QRS = \angle SQR = 2x ∠ QR S = ∠ S QR = 2 x .
∠ R S T \angle RST ∠ R S T is an exterior angle of triangle
P R S PRS P R S , so it equals
∠ P + ∠ R \angle P + \angle R ∠ P + ∠ R :
x + 2 x = 72 ∘ x + 2x = 72^\circ x + 2 x = 7 2 ∘ .
So
3 x = 72 ∘ 3x = 72^\circ 3 x = 7 2 ∘ and
x = 24 ∘ x = 24^\circ x = 2 4 ∘ , option D.
Watch out
The exterior angle at S S S x + 2 x x + 2x x + 2 x 2 x = 72 ∘ 2x = 72^\circ 2 x = 7 2 ∘ 36 ∘ 36^\circ 3 6 ∘ Also set as JAMB 2002 · UME · Q14
Report a problem with this question
If N = ( 3 5 − 4 6 − 3 − 5 − 2 2 1 ) N = \begin{pmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{pmatrix} N = 3 6 − 2 5 − 3 2 − 4 − 5 1 ∣ N ∣ |N| ∣ N ∣
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
3 × ( ( − 3 ) ( 1 ) − ( − 5 ) ( 2 ) ) = 3 × 7 3 \times ((-3)(1) - (-5)(2)) = 3 \times 7 3 × (( − 3 ) ( 1 ) − ( − 5 ) ( 2 )) = 3 × 7 The second term is
− 5 × ( 6 × 1 − ( − 5 ) ( − 2 ) ) = − 5 × ( − 4 ) -5 \times (6 \times 1 - (-5)(-2)) = -5 \times (-4) − 5 × ( 6 × 1 − ( − 5 ) ( − 2 )) = − 5 × ( − 4 ) The third term is
− 4 × ( 6 × 2 − ( − 3 ) ( − 2 ) ) = − 4 × 6 -4 \times (6 \times 2 - (-3)(-2)) = -4 \times 6 − 4 × ( 6 × 2 − ( − 3 ) ( − 2 )) = − 4 × 6 Add them:
21 + 20 − 24 = 17 21 + 20 - 24 = 17 21 + 20 − 24 = 17 , option A.
Watch out
The middle term takes a minus sign, and − 5 × ( − 4 ) = + 20 -5 \times (-4) = +20 − 5 × ( − 4 ) = + 20 + 5 +5 + 5 21 − 20 − 24 = − 23 21 - 20 - 24 = -23 21 − 20 − 24 = − 23 Also set as JAMB 2002 · UME · Q36
Report a problem with this question
A man invested ₦5,000 for 9 months at 4 % 4\% 4%
Worked solution (try it first) Write the time in years: 9 months is
9 12 = 3 4 \frac{9}{12} = \frac34 12 9 = 4 3 year.
Use
I = P R T 100 I = \dfrac{PRT}{100} I = 100 P R T :
I = 5000 × 4 100 × 3 4 I = 5000 \times \dfrac{4}{100} \times \dfrac34 I = 5000 × 100 4 × 4 3 .
That is
200 × 3 4 = 150 200 \times \frac34 = 150 200 × 4 3 = 150 .
The interest is ₦150, option A.
Watch out
The rate is per year, so the time must be in years. Using T = 9 T = 9 T = 9 3 4 \frac34 4 3 Also set as JAMB 2011 · UTME · Q4
Report a problem with this question
Rationalize 2 − 5 3 − 5 \dfrac{2 - \sqrt5}{3 - \sqrt5} 3 − 5 2 − 5
A 1 − 5 2 \dfrac{1 - \sqrt5}{2} 2 1 − 5 B 1 − 5 4 \dfrac{1 - \sqrt5}{4} 4 1 − 5 C 5 − 1 2 \dfrac{\sqrt5 - 1}{2} 2 5 − 1 D 1 + 5 4 \dfrac{1 + \sqrt5}{4} 4 1 + 5
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
3 + 5 3 + \sqrt5 3 + 5 .
Bottom:
( 3 − 5 ) ( 3 + 5 ) = 9 − 5 (3 - \sqrt5)(3 + \sqrt5) = 9 - 5 ( 3 − 5 ) ( 3 + 5 ) = 9 − 5 , which is 4.
Top:
( 2 − 5 ) ( 3 + 5 ) = 6 + 2 5 − 3 5 − 5 (2 - \sqrt5)(3 + \sqrt5) = 6 + 2\sqrt5 - 3\sqrt5 - 5 ( 2 − 5 ) ( 3 + 5 ) = 6 + 2 5 − 3 5 − 5 , which is
1 − 5 1 - \sqrt5 1 − 5 .
So the value is
1 − 5 4 \dfrac{1 - \sqrt5}{4} 4 1 − 5 , option B.
Watch out
Watch the signs in the top: − 5 × 5 = − 5 -\sqrt5 \times \sqrt5 = -5 − 5 × 5 = − 5 2 5 − 3 5 = − 5 2\sqrt5 - 3\sqrt5 = -\sqrt5 2 5 − 3 5 = − 5 + 5 +\sqrt5 + 5 1 + 5 4 \frac{1 + \sqrt5}{4} 4 1 + 5 Also set as JAMB 2011 · UTME · Q8
Report a problem with this question
Factorize completely 9 y 2 − 16 x 2 9y^2 - 16x^2 9 y 2 − 16 x 2
A ( 3 y − 2 x ) ( 3 y + 4 x ) (3y - 2x)(3y + 4x) ( 3 y − 2 x ) ( 3 y + 4 x ) B ( 3 y + 4 x ) ( 3 y + 4 x ) (3y + 4x)(3y + 4x) ( 3 y + 4 x ) ( 3 y + 4 x ) C ( 3 y + 2 x ) ( 3 y − 4 x ) (3y + 2x)(3y - 4x) ( 3 y + 2 x ) ( 3 y − 4 x ) D ( 3 y + 4 x ) ( 3 y − 4 x ) (3y + 4x)(3y - 4x) ( 3 y + 4 x ) ( 3 y − 4 x )
Worked solution (try it first) Both terms are squares:
9 y 2 = ( 3 y ) 2 9y^2 = (3y)^2 9 y 2 = ( 3 y ) 2 and
16 x 2 = ( 4 x ) 2 16x^2 = (4x)^2 16 x 2 = ( 4 x ) 2 .
Use the difference of two squares:
A 2 − B 2 = ( A + B ) ( A − B ) A^2 - B^2 = (A + B)(A - B) A 2 − B 2 = ( A + B ) ( A − B ) .
So
9 y 2 − 16 x 2 = ( 3 y + 4 x ) ( 3 y − 4 x ) 9y^2 - 16x^2 = (3y + 4x)(3y - 4x) 9 y 2 − 16 x 2 = ( 3 y + 4 x ) ( 3 y − 4 x ) , option D.
Watch out
The square root of 16 x 2 16x^2 16 x 2 4 x 4x 4 x 2 x 2x 2 x 4 x 2 4x^2 4 x 2 Also set as JAMB 2011 · UTME · Q14
Report a problem with this question
Solve the inequality x 2 + 2 x > 15 x^2 + 2x > 15 x 2 + 2 x > 15
A x < − 3 x < -3 x < − 3 x > 5 x > 5 x > 5 B − 5 < x < 3 -5 < x < 3 − 5 < x < 3 C x < 3 x < 3 x < 3 x > 5 x > 5 x > 5 D x > 3 x > 3 x > 3 x < − 5 x < -5 x < − 5
Worked solution (try it first) Bring everything to one side:
x 2 + 2 x − 15 > 0 x^2 + 2x - 15 > 0 x 2 + 2 x − 15 > 0 .
Factorise:
( x + 5 ) ( x − 3 ) > 0 (x + 5)(x - 3) > 0 ( x + 5 ) ( x − 3 ) > 0 , so the roots are
x = − 5 x = -5 x = − 5 and
x = 3 x = 3 x = 3 .
"Greater than 0" means outside the roots:
x < − 5 x < -5 x < − 5 or
x > 3 x > 3 x > 3 , option D.
Watch out
Option B, − 5 < x < 3 -5 < x < 3 − 5 < x < 3 x 2 + 2 x − 15 x^2 + 2x - 15 x 2 + 2 x − 15 x = 0 x = 0 x = 0 0 > 15 0 > 15 0 > 15 Also set as JAMB 2011 · UTME · Q20
Report a problem with this question
A circle of perimeter 28 cm 28\text{ cm} 28 cm
A 56 cm 2 56\text{ cm}^2 56 cm 2 B 49 cm 2 49\text{ cm}^2 49 cm 2 C 98 cm 2 98\text{ cm}^2 98 cm 2 D 28 cm 2 28\text{ cm}^2 28 cm 2
Worked solution (try it first) The length of wire does not change, so the square's perimeter is 28 cm.
Each side is
28 ÷ 4 = 7 28 \div 4 = 7 28 ÷ 4 = 7 cm, so the area is
7 2 = 49 cm 2 7^2 = 49\text{ cm}^2 7 2 = 49 cm 2 , option B.
Watch out
28 cm is the perimeter, not the area (option D). Find the side first, 28 ÷ 4 = 7 28 \div 4 = 7 28 ÷ 4 = 7 Also set as JAMB 2011 · UTME · Q28
Report a problem with this question
If the graphs y = p x 2 + q y = px^2 + q y = p x 2 + q y = 2 x 2 − 1 y = 2x^2 - 1 y = 2 x 2 − 1 x = 2 x = 2 x = 2 p p p q q q
A 7 + q 8 \dfrac{7 + q}{8} 8 7 + q B 8 − q 2 \dfrac{8 - q}{2} 2 8 − q C q − 8 7 \dfrac{q - 8}{7} 7 q − 8 D 7 − q 4 \dfrac{7 - q}{4} 4 7 − q
Worked solution (try it first) Find
y y y on the known curve at
x = 2 x = 2 x = 2 :
y = 2 ( 2 ) 2 − 1 = 7 y = 2(2)^2 - 1 = 7 y = 2 ( 2 ) 2 − 1 = 7 .
The other curve passes through the same point, so
p ( 2 ) 2 + q = 7 p(2)^2 + q = 7 p ( 2 ) 2 + q = 7 , which is
4 p + q = 7 4p + q = 7 4 p + q = 7 .
Subtract
q q q and divide by 4:
p = 7 − q 4 p = \dfrac{7 - q}{4} p = 4 7 − q , option D.
Watch out
Square x x x x = 2 x = 2 x = 2 p x 2 = 4 p px^2 = 4p p x 2 = 4 p 2 p 2p 2 p 2 x 2 − 1 = 7 2x^2 - 1 = 7 2 x 2 − 1 = 7 8 − q 2 \frac{8 - q}{2} 2 8 − q Report a problem with this question
A straight line makes an angle of 30 ∘ 30^\circ 3 0 ∘ x x x y y y y = 5 y = 5 y = 5
A 3 y = x + 5 3 \sqrt3y = x + 5\sqrt3 3 y = x + 5 3 B 3 y = − x + 5 3 \sqrt3y = -x + 5\sqrt3 3 y = − x + 5 3 C y = x + 5 y = x + 5 y = x + 5 D y = 1 10 x + 5 y = \frac{1}{10}x + 5 y = 10 1 x + 5
Worked solution (try it first) The gradient is the tangent of the angle:
tan 30 ∘ = 1 3 \tan 30^\circ = \frac{1}{\sqrt3} tan 3 0 ∘ = 3 1 .
The
y y y -intercept is 5, so
y = 1 3 x + 5 y = \frac{1}{\sqrt3}x + 5 y = 3 1 x + 5 .
Multiply every term by
3 \sqrt3 3 :
3 y = x + 5 3 \sqrt3y = x + 5\sqrt3 3 y = x + 5 3 , option A.
Watch out
The gradient is tan 30 ∘ = 1 3 \tan 30^\circ = \frac{1}{\sqrt3} tan 3 0 ∘ = 3 1 45 ∘ 45^\circ 4 5 ∘ y = x + 5 y = x + 5 y = x + 5 Also set as JAMB 2001 · UME · Q28
Report a problem with this question
Find the area bounded by the curves y = 4 − x 2 y = 4 - x^2 y = 4 − x 2 y = 2 x + 1 y = 2x + 1 y = 2 x + 1
A 10 1 3 10\frac13 10 3 1 B 10 2 3 10\frac23 10 3 2 C 20 1 3 20\frac13 20 3 1 D 20 2 3 20\frac23 20 3 2
Worked solution (try it first) The curves meet where
4 − x 2 = 2 x + 1 4 - x^2 = 2x + 1 4 − x 2 = 2 x + 1 , that is
x 2 + 2 x − 3 = 0 x^2 + 2x - 3 = 0 x 2 + 2 x − 3 = 0 , so
( x + 3 ) ( x − 1 ) = 0 (x + 3)(x - 1) = 0 ( x + 3 ) ( x − 1 ) = 0 and
x = − 3 x = -3 x = − 3 or
x = 1 x = 1 x = 1 .
Between them the parabola is on top, so integrate top minus bottom:
( 4 − x 2 ) − ( 2 x + 1 ) = 3 − 2 x − x 2 (4 - x^2) - (2x + 1) = 3 - 2x - x^2 ( 4 − x 2 ) − ( 2 x + 1 ) = 3 − 2 x − x 2 .
[ 3 x − x 2 − x 3 3 ] − 3 1 \left[3x - x^2 - \frac{x^3}{3}\right]_{-3}^{1} [ 3 x − x 2 − 3 x 3 ] − 3 1 : at
x = 1 x = 1 x = 1 it is
5 3 \frac53 3 5 , and at
x = − 3 x = -3 x = − 3 it is
− 9 -9 − 9 .
Subtract:
5 3 + 9 = 32 3 = 10 2 3 \frac53 + 9 = \frac{32}{3} = 10\frac23 3 5 + 9 = 3 32 = 10 3 2 square units, option B.
Watch out
Use the points where the two curves meet, x = − 3 x = -3 x = − 3 x = 1 x = 1 x = 1 x x x Similar: JAMB 2001 · UME · Q39
Report a problem with this question
Teams P P P Q Q Q
A 1 4 \frac14 4 1 B 1 3 \frac13 3 1 C 1 2 \frac12 2 1 D 2 3 \frac23 3 2
Worked solution (try it first) A football game has three possible results:
P P P wins,
Q Q Q wins, or a draw.
Taking the three results as equally likely, a draw is 1 of 3.
So the probability is
1 3 \frac13 3 1 , option B.
Watch out
Don't forget the draw itself as an outcome. Thinking of only win or lose gives 1 2 \frac12 2 1 Also set as JAMB 2001 · UME · Q50
Report a problem with this question
Find, without using logarithm tables, the value of log 3 27 − log 1 4 64 log 1 3 81 \dfrac{\log_3 27 - \log_{\frac14} 64}{\log_{\frac13} 81} log 3 1 81 log 3 27 − log 4 1 64
A 7 4 \frac74 4 7 B − 7 4 -\frac74 − 4 7 C − 3 2 -\frac32 − 2 3 D 7 3 \frac73 3 7
Worked solution (try it first) 27 = 3 3 27 = 3^3 27 = 3 3 , so
log 3 27 = 3 \log_3 27 = 3 log 3 27 = 3 .
64 = ( 1 4 ) − 3 64 = \left(\frac14\right)^{-3} 64 = ( 4 1 ) − 3 , so
log 1 4 64 = − 3 \log_{\frac14} 64 = -3 log 4 1 64 = − 3 .
81 = ( 1 3 ) − 4 81 = \left(\frac13\right)^{-4} 81 = ( 3 1 ) − 4 , so
log 1 3 81 = − 4 \log_{\frac13} 81 = -4 log 3 1 81 = − 4 .
The value is
3 − ( − 3 ) − 4 = 6 − 4 \frac{3 - (-3)}{-4} = \frac{6}{-4} − 4 3 − ( − 3 ) = − 4 6 = − 3 2 = -\frac32 = − 2 3 , option C.
Watch out
With a base less than 1, a number bigger than 1 has a negative log. Taking log 1 4 64 \log_{\frac14} 64 log 4 1 64 + 3 +3 + 3 log 1 3 81 \log_{\frac13} 81 log 3 1 81 + 4 +4 + 4 + 3 2 +\frac32 + 2 3 Also set as JAMB 1984 · UME · Q10
Report a problem with this question
In the figure, determine the angle marked y y y
A 66 ∘ 66^\circ 6 6 ∘ B 110 ∘ 110^\circ 11 0 ∘ C 26 ∘ 26^\circ 2 6 ∘ D 70 ∘ 70^\circ 7 0 ∘
Worked solution (try it first) Angles in triangle
P Q T PQT P QT add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P T Q = 180 ∘ − 44 ∘ − 70 ∘ \angle PTQ = 180^\circ - 44^\circ - 70^\circ ∠ P T Q = 18 0 ∘ − 4 4 ∘ − 7 0 ∘ P T S PTS P T S is a straight line, so
∠ P T Q \angle PTQ ∠ P T Q is the exterior angle of the cyclic quadrilateral
Q R S T QRST QR S T at
T T T .
The exterior angle of a cyclic quadrilateral equals the interior opposite angle, which is at
R R R .
So
y = 66 ∘ y = 66^\circ y = 6 6 ∘ , option A.
Watch out
The vertex opposite R R R T T T Q Q Q 70 ∘ 70^\circ 7 0 ∘ Q Q Q 70 ∘ 70^\circ 7 0 ∘ 110 ∘ 110^\circ 11 0 ∘ Report a problem with this question
A right triangular prism is 8 cm 8\text{ cm} 8 cm 5 cm 5\text{ cm} 5 cm 4 cm 4\text{ cm} 4 cm
A 160 cm 3 160\text{ cm}^3 160 cm 3 B 48 cm 3 48\text{ cm}^3 48 cm 3 C 96 cm 3 96\text{ cm}^3 96 cm 3 D 120 cm 3 120\text{ cm}^3 120 cm 3
Worked solution (try it first) The third side of the triangle, by Pythagoras:
5 2 − 4 2 = 9 = 3 \sqrt{5^2 - 4^2} = \sqrt9 = 3 5 2 − 4 2 = 9 = 3 cm.
The two shorter sides are at right angles, so the end has area
1 2 × 3 × 4 = 6 cm 2 \frac12 \times 3 \times 4 = 6\text{ cm}^2 2 1 × 3 × 4 = 6 cm 2 .
Volume = end area × length:
6 × 8 = 48 cm 3 6 \times 8 = 48\text{ cm}^3 6 × 8 = 48 cm 3 , option B.
Watch out
A triangle's area has a 1 2 \frac12 2 1 12 × 8 = 96 cm 3 12 \times 8 = 96\text{ cm}^3 12 × 8 = 96 cm 3 Report a problem with this question
Find the remainder when x 3 − 2 x 2 + 3 x − 3 x^3 - 2x^2 + 3x - 3 x 3 − 2 x 2 + 3 x − 3 x 2 + 1 x^2 + 1 x 2 + 1
A 2 x − 1 2x - 1 2 x − 1 B x + 3 x + 3 x + 3 C 2 x + 1 2x + 1 2 x + 1 D x − 3 x - 3 x − 3
Worked solution (try it first) Long division:
x 3 ÷ x 2 = x x^3 \div x^2 = x x 3 ÷ x 2 = x .
Subtract
x ( x 2 + 1 ) = x 3 + x x(x^2 + 1) = x^3 + x x ( x 2 + 1 ) = x 3 + x to leave
− 2 x 2 + 2 x − 3 -2x^2 + 2x - 3 − 2 x 2 + 2 x − 3 .
− 2 x 2 ÷ x 2 = − 2 -2x^2 \div x^2 = -2 − 2 x 2 ÷ x 2 = − 2 .
Subtract
− 2 ( x 2 + 1 ) = − 2 x 2 − 2 -2(x^2 + 1) = -2x^2 - 2 − 2 ( x 2 + 1 ) = − 2 x 2 − 2 to leave
2 x − 1 2x - 1 2 x − 1 .
2 x − 1 2x - 1 2 x − 1 has a lower power than
x 2 + 1 x^2 + 1 x 2 + 1 , so it is the remainder, option A.
Watch out
Subtracting − 2 -2 − 2 − 3 -3 − 3 − 3 + 2 = − 1 -3 + 2 = -1 − 3 + 2 = − 1 2 x + 1 2x + 1 2 x + 1 Also set as JAMB 2011 · UTME · Q13
Report a problem with this question
In a class of 60 students, 30 offer physics and 40 offer chemistry. If a student is picked at random from the class, what is the probability that the student offers both physics and chemistry?
A 1 3 \frac13 3 1 B 1 4 \frac14 4 1 C 1 2 \frac12 2 1 D 1 6 \frac16 6 1
Worked solution (try it first) Each student offers at least one of the subjects, so those doing both are counted twice in
30 + 40 = 70 30 + 40 = 70 30 + 40 = 70 .
Both
= 70 − 60 = 10 = 70 - 60 = 10 = 70 − 60 = 10 students.
So the probability is
10 60 = 1 6 \frac{10}{60} = \frac16 60 10 = 6 1 , option D.
Watch out
The overlap is the excess over the class size: 30 + 40 − 60 = 10 30 + 40 - 60 = 10 30 + 40 − 60 = 10 30 60 = 1 2 \frac{30}{60} = \frac12 60 30 = 2 1 Also set as JAMB 2011 · UTME · Q50
Report a problem with this question