Paper JAMB 2017 General Maths Objective
Objective paper · 39 questions · partial
JAMB 2017 · UTME (set 2) Topics include Sets & Venn diagrams, Sequences & series (AP, GP), Probability, Solid mensuration, Trigonometric ratios, Calculus (JAMB bridge).
Our copy of this paper is missing questions 5, 9, 12, 24, 26, 28, 36, 41, 44.
Sit this paper Answer every question in order, timed if you like (suggested 25 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 6 7 8 10 11 13 14 15 16 17 18 19 20 21 22 23 25 27 29 30 31 32 33 34 35 37 38 39 40 42 43 45 46 47 48 Given T = { even numbers from 1 to 12 } T = \{\text{even numbers from 1 to 12}\} T = { even numbers from 1 to 12 } N = { common factors of 6, 8 and 12 } N = \{\text{common factors of 6, 8 and 12}\} N = { common factors of 6, 8 and 12 } T ∩ N T \cap N T ∩ N
A { 2 , 3 } \{2, 3\} { 2 , 3 } B { 2 , 3 , 4 } \{2, 3, 4\} { 2 , 3 , 4 } C { 3 , 4 , 6 } \{3, 4, 6\} { 3 , 4 , 6 } D { 2 } \{2\} { 2 }
Worked solution (try it first) T = { 2 , 4 , 6 , 8 , 10 , 12 } T = \{2, 4, 6, 8, 10, 12\} T = { 2 , 4 , 6 , 8 , 10 , 12 } .
The factors of 6 are 1, 2, 3, 6.
Of 8 are 1, 2, 4, 8.
Of 12 are 1, 2, 3, 4, 6, 12.
Common to all three:
N = { 1 , 2 } N = \{1, 2\} N = { 1 , 2 } .
So
T ∩ N = { 2 } T \cap N = \{2\} T ∩ N = { 2 } , option D.
Watch out
A common factor must divide all three numbers. 3 divides 6 and 12 but not 8, so options A and C are out. Report a problem with this question
What is the next number in the series 2 , 1 , 1 2 , 1 4 , … 2, 1, \frac12, \frac14, \dots 2 , 1 , 2 1 , 4 1 , …
A 1 3 \frac13 3 1 B 2 8 \frac28 8 2 C 3 7 \frac37 7 3 D 1 8 \frac18 8 1
Worked solution (try it first) Each term is half the one before:
1 ÷ 2 = 1 2 1 \div 2 = \frac12 1 ÷ 2 = 2 1 and
1 2 ÷ 1 = 1 2 \frac12 \div 1 = \frac12 2 1 ÷ 1 = 2 1 .
So the common ratio is
1 2 \frac12 2 1 .
The next term is
1 4 × 1 2 = 1 8 \frac14 \times \frac12 = \frac18 4 1 × 2 1 = 8 1 , option D.
Watch out
Simplify before you choose: 2 8 \frac28 8 2 1 4 \frac14 4 1 Report a problem with this question
If U = { x : x is an integer and 1 ≤ x ≤ 20 } U = \{x : x \text{ is an integer and } 1 \le x \le 20\} U = { x : x is an integer and 1 ≤ x ≤ 20 } E 1 = { x : x is a multiple of 3 } E_1 = \{x : x \text{ is a multiple of 3}\} E 1 = { x : x is a multiple of 3 } E 2 = { x : x is a multiple of 4 } E_2 = \{x : x \text{ is a multiple of 4}\} E 2 = { x : x is a multiple of 4 } U U U E 2 E_2 E 2
A 3 4 \frac34 4 3 B 3 10 \frac{3}{10} 10 3 C 1 4 \frac14 4 1 D 1 20 \frac{1}{20} 20 1
Worked solution (try it first) E 2 E_2 E 2 , the multiples of 4 up to 20, is
{ 4 , 8 , 12 , 16 , 20 } \{4, 8, 12, 16, 20\} { 4 , 8 , 12 , 16 , 20 } , which has 5 members.
So
P ( in E 2 ) = 5 20 P(\text{in } E_2) = \frac{5}{20} P ( in E 2 ) = 20 5 Not in
E 2 E_2 E 2 is the complement:
1 − 1 4 = 3 4 1 - \frac14 = \frac34 1 − 4 1 = 4 3 , option A.
Watch out
1 4 \frac14 4 1 E 2 E_2 E 2 E 2 E_2 E 2 Also set as JAMB 2000 · UME · Q45
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The curved surface area of a cylinder 5 cm 5\text{ cm} 5 cm 110 cm 2 110\text{ cm}^2 110 cm 2 [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 2.6 cm 2.6\text{ cm} 2.6 cm B 3.5 cm 3.5\text{ cm} 3.5 cm C 3.6 cm 3.6\text{ cm} 3.6 cm D 7.0 cm 7.0\text{ cm} 7.0 cm
Worked solution (try it first) Curved surface area of a cylinder is
2 π r h 2\pi rh 2 π r h :
2 × 22 7 × r × 5 = 110 2 \times \frac{22}{7} \times r \times 5 = 110 2 × 7 22 × r × 5 = 110 .
That is
220 7 r = 110 \frac{220}{7}r = 110 7 220 r = 110 , so
r = 110 × 7 220 = 3.5 r = 110 \times \frac{7}{220} = 3.5 r = 110 × 220 7 = 3.5 cm, option B.
Watch out
The curved surface is 2 π r h 2\pi rh 2 π r h π r h \pi rh π r h r = 7.0 r = 7.0 r = 7.0 Report a problem with this question
Evaluate sin 45 ∘ + sin 30 ∘ \sin 45^\circ + \sin 30^\circ sin 4 5 ∘ + sin 3 0 ∘
A 3 2 3 \frac{\sqrt3}{2\sqrt3} 2 3 3 B 3 − 1 2 \frac{\sqrt3 - 1}{2} 2 3 − 1 C 1 2 2 \frac12\sqrt2 2 1 2 D 1 + 2 2 \frac{1 + \sqrt2}{2} 2 1 + 2
Worked solution (try it first) Use the exact values:
sin 45 ∘ = 2 2 \sin45^\circ = \frac{\sqrt2}{2} sin 4 5 ∘ = 2 2 and
sin 30 ∘ = 1 2 \sin30^\circ = \frac12 sin 3 0 ∘ = 2 1 .
Both are over 2, so add the tops:
2 2 + 1 2 = 1 + 2 2 \frac{\sqrt2}{2} + \frac12 = \frac{1 + \sqrt2}{2} 2 2 + 2 1 = 2 1 + 2 , option D.
Watch out
Add both values: 1 2 2 \frac12\sqrt2 2 1 2 sin 45 ∘ \sin45^\circ sin 4 5 ∘ sin 45 ∘ + sin 30 ∘ \sin45^\circ + \sin30^\circ sin 4 5 ∘ + sin 3 0 ∘ sin 75 ∘ \sin75^\circ sin 7 5 ∘ Report a problem with this question
If y = x sin x y = x\sin x y = x sin x d y d x \dfrac{dy}{dx} d x d y x = π 2 x = \dfrac{\pi}{2} x = 2 π
A − π 2 -\frac{\pi}{2} − 2 π B − 1 -1 − 1 C 1 D π 2 \frac{\pi}{2} 2 π
Worked solution (try it first) Product rule:
d y d x = sin x + x cos x \frac{dy}{dx} = \sin x + x\cos x d x d y = sin x + x cos x .
At
x = π 2 x = \frac\pi2 x = 2 π ,
sin x = 1 \sin x = 1 sin x = 1 and
cos x = 0 \cos x = 0 cos x = 0 , so
d y d x = 1 + 0 = 1 \frac{dy}{dx} = 1 + 0 = 1 d x d y = 1 + 0 = 1 , option C.
Watch out
π 2 \frac\pi2 2 π y y y x = π 2 x = \frac\pi2 x = 2 π d y d x \frac{dy}{dx} d x d y Also set as JAMB 2001 · UME · Q34
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If temperature t t t h h h t = 20 ∘ C t = 20^\circ\text{C} t = 2 0 ∘ C h = 50 J h = 50\text{ J} h = 50 J t t t h = 60 J h = 60\text{ J} h = 60 J
A 24 ∘ C 24^\circ\text{C} 2 4 ∘ C B 20 ∘ C 20^\circ\text{C} 2 0 ∘ C C 34 ∘ C 34^\circ\text{C} 3 4 ∘ C D 30 ∘ C 30^\circ\text{C} 3 0 ∘ C
Worked solution (try it first) Put in
t = 20 t = 20 t = 20 ,
h = 50 h = 50 h = 50 :
k = 20 50 = 2 5 k = \frac{20}{50} = \frac25 k = 50 20 = 5 2 .
When
h = 60 h = 60 h = 60 :
t = 2 5 × 60 = 24 ∘ C t = \frac25 \times 60 = 24^\circ\text{C} t = 5 2 × 60 = 2 4 ∘ C , option A.
Watch out
Direct proportion scales by the same factor, not by the same amount. Adding the increase of 10 to 20 gives 30 ∘ C 30^\circ\text{C} 3 0 ∘ C Report a problem with this question
Given M = N S L T M = N\sqrt{\dfrac{SL}{T}} M = N T S L T T T
A N S L M \dfrac{NSL}{M} M N S L B N 2 S L M 2 \dfrac{N^2SL}{M^2} M 2 N 2 S L C N 2 S L M \dfrac{N^2SL}{M} M N 2 S L D N S L M \dfrac{NSL}{M} M N S L
Worked solution (try it first) Divide by
N N N :
M N = S L T \frac MN = \sqrt{\frac{SL}{T}} N M = T S L .
Square both sides:
M 2 N 2 = S L T \frac{M^2}{N^2} = \frac{SL}{T} N 2 M 2 = T S L .
Cross-multiply to get
T T T on top:
T = N 2 S L M 2 T = \dfrac{N^2SL}{M^2} T = M 2 N 2 S L , option B.
Watch out
Squaring affects both sides, so M M M N N N N N N N 2 S L M \frac{N^2SL}{M} M N 2 S L Report a problem with this question
Simplify 3 n − 1 × 27 n + 1 81 n \dfrac{3^{n-1} \times 27^{n+1}}{81^n} 8 1 n 3 n − 1 × 2 7 n + 1
A 3 2 n 3^{2n} 3 2 n B 9 C 3 n 3^n 3 n D 3 n + 1 3^{n+1} 3 n + 1
Worked solution (try it first) Write everything as a power of 3:
27 n + 1 = 3 3 n + 3 27^{n+1} = 3^{3n + 3} 2 7 n + 1 = 3 3 n + 3 and
81 n = 3 4 n 81^n = 3^{4n} 8 1 n = 3 4 n .
Top: add the indices,
( n − 1 ) + ( 3 n + 3 ) = 4 n + 2 (n - 1) + (3n + 3) = 4n + 2 ( n − 1 ) + ( 3 n + 3 ) = 4 n + 2 , so the top is
3 4 n + 2 3^{4n + 2} 3 4 n + 2 .
Divide: subtract the indices,
( 4 n + 2 ) − 4 n = 2 (4n + 2) - 4n = 2 ( 4 n + 2 ) − 4 n = 2 .
So the value is
3 2 = 9 3^2 = 9 3 2 = 9 , option B.
Watch out
Multiply the whole index when changing base: 27 n + 1 = 3 3 ( n + 1 ) = 3 3 n + 3 27^{n+1} = 3^{3(n + 1)} = 3^{3n + 3} 2 7 n + 1 = 3 3 ( n + 1 ) = 3 3 n + 3 3 3 n + 1 3^{3n + 1} 3 3 n + 1 3 n + 1 3n + 1 3 n + 1 3 0 = 1 3^0 = 1 3 0 = 1 Report a problem with this question
In the diagram, R R R S S S T T T O O O ∠ R S T = 60 ∘ \angle RST = 60^\circ ∠ R S T = 6 0 ∘ x x x
A 100 ∘ 100^\circ 10 0 ∘ B 140 ∘ 140^\circ 14 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 10 ∘ 10^\circ 1 0 ∘
Worked solution (try it first) x x x is at the centre and
∠ R S T \angle RST ∠ R S T is at the circumference, both on arc
R T RT R T .
The angle at the centre is twice the angle at the circumference:
x = 2 × 60 ∘ = 120 ∘ x = 2 \times 60^\circ = 120^\circ x = 2 × 6 0 ∘ = 12 0 ∘ , option C.
Watch out
Double, don't halve: the angle at the centre is the bigger one. Halving 60 ∘ 60^\circ 6 0 ∘ 30 ∘ 30^\circ 3 0 ∘ Report a problem with this question
Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5.
Worked solution (try it first) The range is highest minus lowest:
10 − 4 = 6 10 - 4 = 6 10 − 4 = 6 .
Count the values: 10 appears six times, more than any other, so the mode is 10.
So the sum is
6 + 10 = 16 6 + 10 = 16 6 + 10 = 16 , option A.
Watch out
The range is a difference, 10 − 4 = 6 10 - 4 = 6 10 − 4 = 6 Also set as JAMB 2000 · UME · Q48
Report a problem with this question
Find the sum to infinity of the series 1 4 , 1 8 , 1 16 , … \frac14, \frac18, \frac1{16}, \dots 4 1 , 8 1 , 16 1 , …
A 1 2 \frac12 2 1 B 3 5 \frac35 5 3 C − 1 5 -\frac15 − 5 1 D 73 12 \frac{73}{12} 12 73
Worked solution (try it first) This is a G.P. with
a = 1 4 a = \frac14 a = 4 1 and
r = 1 8 ÷ 1 4 = 1 2 r = \frac18 \div \frac14 = \frac12 r = 8 1 ÷ 4 1 = 2 1 .
Since
r r r is between
− 1 -1 − 1 and 1,
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a , and
1 − 1 2 = 1 2 1 - \frac12 = \frac12 1 − 2 1 = 2 1 .
So
S ∞ = 1 4 ÷ 1 2 = 1 2 S_\infty = \frac14 \div \frac12 = \frac12 S ∞ = 4 1 ÷ 2 1 = 2 1 , option A.
Watch out
Divide the first term by 1 − r 1 - r 1 − r 1 4 × 1 2 = 1 8 \frac14 \times \frac12 = \frac18 4 1 × 2 1 = 8 1 Report a problem with this question
The addition below was carried out in a certain base.
1101.01 + 1110.11 1011.10 100111.10 \begin{array}{rr} & 1101.01 \\ + & 1110.11 \\ & 1011.10 \\ \hline & 100111.10 \end{array} + 1101.01 1110.11 1011.10 100111.10
The base in which the operation was performed was
Worked solution (try it first) Every digit is 0 or 1, so try base 2 first.
In base 2:
1101.01 = 13.25 1101.01 = 13.25 1101.01 = 13.25 ,
1110.11 = 14.75 1110.11 = 14.75 1110.11 = 14.75 and
1011.10 = 11.5 1011.10 = 11.5 1011.10 = 11.5 .
Their sum is
39.5 39.5 39.5 .
The answer
100111.10 2 100111.10_2 100111.1 0 2 is
32 + 4 + 2 + 1 + 0.5 = 39.5 32 + 4 + 2 + 1 + 0.5 = 39.5 32 + 4 + 2 + 1 + 0.5 = 39.5 , which matches.
So the base is 2, option B.
Watch out
Digits 0 and 1 are allowed in every base, so check a column. In the last column 1 + 1 + 0 = 2 1 + 1 + 0 = 2 1 + 1 + 0 = 2 Report a problem with this question
The value of x + x ( x x ) x + x(x^x) x + x ( x x ) x = 2 x = 2 x = 2
Worked solution (try it first) Work out the power first:
x x = 2 2 = 4 x^x = 2^2 = 4 x x = 2 2 = 4 .
Then multiply:
x ( x x ) = 2 × 4 = 8 x(x^x) = 2 \times 4 = 8 x ( x x ) = 2 × 4 = 8 .
Finally add:
2 + 8 = 10 2 + 8 = 10 2 + 8 = 10 , option B.
Watch out
Multiply before you add. Adding first gives ( 2 + 2 ) × 4 = 16 (2 + 2) \times 4 = 16 ( 2 + 2 ) × 4 = 16 Report a problem with this question
Simplify 4 27 + 5 12 − 3 75 4\sqrt{27} + 5\sqrt{12} - 3\sqrt{75} 4 27 + 5 12 − 3 75
A 7 B − 7 -7 − 7 C − 7 3 -7\sqrt3 − 7 3 D 7 3 7\sqrt3 7 3
Worked solution (try it first) Take out square factors:
4 27 = 4 × 3 3 4\sqrt{27} = 4 \times 3\sqrt3 4 27 = 4 × 3 3 , which is
12 3 12\sqrt3 12 3 .
Likewise
5 12 = 5 × 2 3 = 10 3 5\sqrt{12} = 5 \times 2\sqrt3 = 10\sqrt3 5 12 = 5 × 2 3 = 10 3 and
3 75 = 3 × 5 3 = 15 3 3\sqrt{75} = 3 \times 5\sqrt3 = 15\sqrt3 3 75 = 3 × 5 3 = 15 3 .
Combine:
12 3 + 10 3 − 15 3 = 7 3 12\sqrt3 + 10\sqrt3 - 15\sqrt3 = 7\sqrt3 12 3 + 10 3 − 15 3 = 7 3 , option D.
Watch out
The answer is still a multiple of 3 \sqrt3 3 7 3 7\sqrt3 7 3 Report a problem with this question
A man covered a distance of 50 miles on his first trip. On a later trip he travelled 300 miles while going 3 times as fast. His new time compared with the old time was
A three times as much B the same C twice as much D half as much
Worked solution (try it first) Let the old speed be
v v v .
Time is distance over speed, so the old time is
50 v \dfrac{50}{v} v 50 .
The new speed is
3 v 3v 3 v , so the new time is
300 3 v = 100 v \dfrac{300}{3v} = \dfrac{100}{v} 3 v 300 = v 100 .
100 v \dfrac{100}{v} v 100 is twice
50 v \dfrac{50}{v} v 50 , so the new time is twice as much, option C.
Watch out
Both the distance and the speed change. The distance is 6 times as far, but going 3 times as fast divides the time by 3, so the time is 6 ÷ 3 = 2 6 \div 3 = 2 6 ÷ 3 = 2 Report a problem with this question
A 40 ∘ 40^\circ 4 0 ∘ B 55 ∘ 55^\circ 5 5 ∘ C 50 ∘ 50^\circ 5 0 ∘ D 60 ∘ 60^\circ 6 0 ∘
Worked solution (try it first) The three angles go all the way round the point, and angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ :
2 x + 3 x + 4 x = 360 ∘ 2x + 3x + 4x = 360^\circ 2 x + 3 x + 4 x = 36 0 ∘ .
So
9 x = 360 ∘ 9x = 360^\circ 9 x = 36 0 ∘ and
x = 40 ∘ x = 40^\circ x = 4 0 ∘ , option A.
Watch out
Angles at a point add up to 360 ∘ 360^\circ 36 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 180 ∘ 180^\circ 18 0 ∘ x = 20 ∘ x = 20^\circ x = 2 0 ∘ Report a problem with this question
Evaluate 0.00000231 0.007 \dfrac{0.00000231}{0.007} 0.007 0.00000231
A 3.3 × 10 4 3.3 \times 10^4 3.3 × 1 0 4 B 3.3 × 10 − 3 3.3 \times 10^{-3} 3.3 × 1 0 − 3 C 3.3 × 10 − 4 3.3 \times 10^{-4} 3.3 × 1 0 − 4 D 3.3 × 10 − 8 3.3 \times 10^{-8} 3.3 × 1 0 − 8
Worked solution (try it first) Write both in standard form:
0.00000231 = 2.31 × 10 − 6 0.00000231 = 2.31 \times 10^{-6} 0.00000231 = 2.31 × 1 0 − 6 and
0.007 = 7 × 10 − 3 0.007 = 7 \times 10^{-3} 0.007 = 7 × 1 0 − 3 .
Divide the numbers,
2.31 ÷ 7 = 0.33 2.31 \div 7 = 0.33 2.31 ÷ 7 = 0.33 , and subtract the powers,
10 − 6 − ( − 3 ) = 10 − 3 10^{-6 - (-3)} = 10^{-3} 1 0 − 6 − ( − 3 ) = 1 0 − 3 .
0.33 × 10 − 3 = 3.3 × 10 − 4 0.33 \times 10^{-3} = 3.3 \times 10^{-4} 0.33 × 1 0 − 3 = 3.3 × 1 0 − 4 , option C.
Watch out
0.33 is not in standard form: making it 3.3 lowers the power by one more, to 10 − 4 10^{-4} 1 0 − 4 10 − 3 10^{-3} 1 0 − 3 Report a problem with this question
If a rod 10 cm 10\text{ cm} 10 cm 10.5 cm 10.5\text{ cm} 10.5 cm
A 5 % 5\% 5% B 10 % 10\% 10% C 8 % 8\% 8% D 7 % 7\% 7%
Worked solution (try it first) The error is
10.5 − 10 = 0.5 10.5 - 10 = 0.5 10.5 − 10 = 0.5 cm.
Divide by the true length and multiply by 100:
0.5 10 × 100 % = 5 % \frac{0.5}{10} \times 100\% = 5\% 10 0.5 × 100% = 5% , option A.
Watch out
The error is 0.5 cm, not 1 cm. Using 1 cm gives 10% (option B). Report a problem with this question
The pie chart shows the allocation of money to each sector of a farm. The total amount allocated to the farm is ₦80,000. Find the amount allocated to fertilizer.
A ₦35,000 B ₦40,000 C ₦25,000 D ₦20,000
Worked solution (try it first) The angles add up to
360 ∘ 360^\circ 36 0 ∘ , so Fertilizer is
360 ∘ − ( 40 ∘ + 50 ∘ + 80 ∘ + 70 ∘ + 30 ∘ ) = 90 ∘ 360^\circ - (40^\circ + 50^\circ + 80^\circ + 70^\circ + 30^\circ) = 90^\circ 36 0 ∘ − ( 4 0 ∘ + 5 0 ∘ + 8 0 ∘ + 7 0 ∘ + 3 0 ∘ ) = 9 0 ∘ .
90 ∘ 90^\circ 9 0 ∘ is a quarter of the circle.
So Fertilizer gets
1 4 × 80 000 = 20 000 \frac14 \times 80\,000 = 20\,000 4 1 × 80 000 = 20 000 naira: ₦20,000, option D.
Watch out
Find the missing angle before using the total. Using the Farm implements sector, 80 ∘ 80^\circ 8 0 ∘ Report a problem with this question
y y y x x x y = 6 y = 6 y = 6 x = 7 x = 7 x = 7
Worked solution (try it first) y = k x y = \dfrac{k}{x} y = x k , so the constant is
k = x y k = xy k = x y .
k = 7 × 6 = 42 k = 7 \times 6 = 42 k = 7 × 6 = 42 , option B.
Watch out
For inverse proportion the constant is the product x y xy x y y x = 6 7 \frac{y}{x} = \frac67 x y = 7 6 Report a problem with this question
Find ∫ ( x 2 + 3 x − 5 ) d x \displaystyle\int (x^2 + 3x - 5)\,dx ∫ ( x 2 + 3 x − 5 ) d x
A x 3 3 − 3 x 2 2 − 5 x + k \frac{x^3}{3} - \frac{3x^2}{2} - 5x + k 3 x 3 − 2 3 x 2 − 5 x + k B x 3 3 − 3 x 2 2 + 5 x + k \frac{x^3}{3} - \frac{3x^2}{2} + 5x + k 3 x 3 − 2 3 x 2 + 5 x + k C x 3 3 + 3 x 2 2 − 5 x + k \frac{x^3}{3} + \frac{3x^2}{2} - 5x + k 3 x 3 + 2 3 x 2 − 5 x + k D x 3 3 + 3 x 2 2 + 5 x + k \frac{x^3}{3} + \frac{3x^2}{2} + 5x + k 3 x 3 + 2 3 x 2 + 5 x + k
Worked solution (try it first) Integrate each term: add one to the power and divide by the new power.
x 2 x^2 x 2 gives
x 3 3 \frac{x^3}{3} 3 x 3 ,
3 x 3x 3 x gives
3 x 2 2 \frac{3x^2}{2} 2 3 x 2 , and
− 5 -5 − 5 gives
− 5 x -5x − 5 x .
So the integral is
x 3 3 + 3 x 2 2 − 5 x + k \frac{x^3}{3} + \frac{3x^2}{2} - 5x + k 3 x 3 + 2 3 x 2 − 5 x + k , option C.
Watch out
Each term keeps its sign when you integrate. Changing the signs gives options A, B and D. Report a problem with this question
Make s s s p = s + s m 2 n r p = s + \dfrac{sm^2}{nr} p = s + n r s m 2
A s = n r p n r + m 2 s = \dfrac{nrp}{nr + m^2} s = n r + m 2 n r p B s = n r + m 2 m r p s = nr + \dfrac{m^2}{mrp} s = n r + m r p m 2 C s = n r p m r + m 2 s = \dfrac{nrp}{mr} + m^2 s = m r n r p + m 2 D s = n r p n r + m 2 s = \dfrac{nrp}{nr} + m^2 s = n r n r p + m 2
Worked solution (try it first) Multiply every term by
n r nr n r to clear the fraction:
n r p = n r s + s m 2 nrp = nrs + sm^2 n r p = n r s + s m 2 .
Take out
s s s as a common factor:
n r p = s ( n r + m 2 ) nrp = s(nr + m^2) n r p = s ( n r + m 2 ) .
Divide by
n r + m 2 nr + m^2 n r + m 2 :
s = n r p n r + m 2 s = \dfrac{nrp}{nr + m^2} s = n r + m 2 n r p , option A.
Watch out
s s s s s s Report a problem with this question
The operation ∗ * ∗ R \mathbb{R} R x ∗ y = 3 x + 2 y − 1 x * y = 3x + 2y - 1 x ∗ y = 3 x + 2 y − 1 3 ∗ ( − 1 ) 3 * (-1) 3 ∗ ( − 1 )
Worked solution (try it first) Put
x = 3 x = 3 x = 3 and
y = − 1 y = -1 y = − 1 in the rule:
3 ∗ ( − 1 ) = 3 ( 3 ) + 2 ( − 1 ) − 1 3 * (-1) = 3(3) + 2(-1) - 1 3 ∗ ( − 1 ) = 3 ( 3 ) + 2 ( − 1 ) − 1 .
That is
9 − 2 − 1 = 6 9 - 2 - 1 = 6 9 − 2 − 1 = 6 , option C.
Watch out
Keep the order: x = 3 x = 3 x = 3 y = − 1 y = -1 y = − 1 3 ( − 1 ) + 2 ( 3 ) − 1 = 2 3(-1) + 2(3) - 1 = 2 3 ( − 1 ) + 2 ( 3 ) − 1 = 2 Report a problem with this question
Find the gradient of the line joining the points ( 3 , 2 ) (3, 2) ( 3 , 2 ) ( 1 , 4 ) (1, 4) ( 1 , 4 )
A 3 2 \frac32 2 3 B 2 2 2 C − 1 -1 − 1 D 1 1 1
Worked solution (try it first) Gradient
= y 2 − y 1 x 2 − x 1 = \dfrac{y_2 - y_1}{x_2 - x_1} = x 2 − x 1 y 2 − y 1 .
With
( 3 , 2 ) (3, 2) ( 3 , 2 ) first and
( 1 , 4 ) (1, 4) ( 1 , 4 ) second, this is
4 − 2 1 − 3 \dfrac{4 - 2}{1 - 3} 1 − 3 4 − 2 .
That is
2 − 2 = − 1 \frac{2}{-2} = -1 − 2 2 = − 1 , option C.
Watch out
Take both differences in the same order. The bottom is 1 − 3 = − 2 1 - 3 = -2 1 − 3 = − 2 1 1 1 Report a problem with this question
Simplify ( 64 a 3 3 ) − 1 \left(\sqrt[3]{64a^3}\right)^{-1} ( 3 64 a 3 ) − 1
A 4 B 1 8 a \frac{1}{8a} 8 a 1 C 8 a 8a 8 a D 1 4 a \frac{1}{4a} 4 a 1
Worked solution (try it first) Take the cube root of each part:
64 3 = 4 \sqrt[3]{64} = 4 3 64 = 4 and
a 3 3 = a \sqrt[3]{a^3} = a 3 a 3 = a , so
64 a 3 3 = 4 a \sqrt[3]{64a^3} = 4a 3 64 a 3 = 4 a .
The index
− 1 -1 − 1 means the reciprocal:
( 4 a ) − 1 = 1 4 a (4a)^{-1} = \frac{1}{4a} ( 4 a ) − 1 = 4 a 1 , option D.
Watch out
It is a cube root: 4 3 = 64 4^3 = 64 4 3 = 64 64 3 = 4 \sqrt[3]{64} = 4 3 64 = 4 1 8 a \frac{1}{8a} 8 a 1 Report a problem with this question
If 2 3 − 2 3 + 2 2 = m + n 6 \dfrac{2\sqrt3 - \sqrt2}{\sqrt3 + 2\sqrt2} = m + n\sqrt6 3 + 2 2 2 3 − 2 = m + n 6 m m m n n n
A 1 , − 2 1, -2 1 , − 2 B − 2 , 1 -2, 1 − 2 , 1 C − 2 5 , 1 -\frac25, 1 − 5 2 , 1 D 2 3 \frac23 3 2
Worked solution (try it first) Multiply the top and bottom by the conjugate of the bottom,
3 − 2 2 \sqrt3 - 2\sqrt2 3 − 2 2 .
Bottom:
( 3 + 2 2 ) ( 3 − 2 2 ) = 3 − 8 (\sqrt3 + 2\sqrt2)(\sqrt3 - 2\sqrt2) = 3 - 8 ( 3 + 2 2 ) ( 3 − 2 2 ) = 3 − 8 , which is
− 5 -5 − 5 .
Top:
( 2 3 − 2 ) ( 3 − 2 2 ) = 6 − 4 6 − 6 + 4 (2\sqrt3 - \sqrt2)(\sqrt3 - 2\sqrt2) = 6 - 4\sqrt6 - \sqrt6 + 4 ( 2 3 − 2 ) ( 3 − 2 2 ) = 6 − 4 6 − 6 + 4 , which is
10 − 5 6 10 - 5\sqrt6 10 − 5 6 .
Divide by
− 5 -5 − 5 :
− 2 + 6 -2 + \sqrt6 − 2 + 6 .
So
m = − 2 m = -2 m = − 2 and
n = 1 n = 1 n = 1 , option B.
Watch out
"Respectively" means m m m n n n − 2 + 1 6 -2 + 1\sqrt6 − 2 + 1 6 − 2 , 1 -2, 1 − 2 , 1 1 , − 2 1, -2 1 , − 2 Also set as JAMB 2000 · UME · Q3
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If α \alpha α β \beta β 3 x 2 + 5 x − 2 = 0 3x^2 + 5x - 2 = 0 3 x 2 + 5 x − 2 = 0 1 α + 1 β \dfrac1\alpha + \dfrac1\beta α 1 + β 1
A − 5 3 -\frac53 − 3 5 B − 2 3 -\frac23 − 3 2 C 1 2 \frac12 2 1 D 5 2 \frac52 2 5
Worked solution (try it first) For
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 , the sum of the roots is
− b a -\frac ba − a b and the product is
c a \frac ca a c .
So
α + β = − 5 3 \alpha + \beta = -\frac53 α + β = − 3 5 and
α β = − 2 3 \alpha\beta = -\frac23 α β = − 3 2 .
Add the fractions:
1 α + 1 β = α + β α β \dfrac1\alpha + \dfrac1\beta = \dfrac{\alpha + \beta}{\alpha\beta} α 1 + β 1 = α β α + β .
So the value is
− 5 3 ÷ ( − 2 3 ) = 5 2 -\frac53 \div \left(-\frac23\right) = \frac52 − 3 5 ÷ ( − 3 2 ) = 2 5 , option D.
Watch out
− 5 3 -\frac53 − 3 5 α + β \alpha + \beta α + β α β = − 2 3 \alpha\beta = -\frac23 α β = − 3 2 Also set as JAMB 2000 · UME · Q22
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Find the range of the following set of numbers: 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2.
Worked solution (try it first) The largest number is 0.47.
The smallest is the most negative,
− 0.53 -0.53 − 0.53 .
The range is
0.47 − ( − 0.53 ) 0.47 - (-0.53) 0.47 − ( − 0.53 ) , which is
0.47 + 0.53 = 1.0 0.47 + 0.53 = 1.0 0.47 + 0.53 = 1.0 , option D.
Watch out
Subtracting − 0.53 -0.53 − 0.53 0.53 − 0.47 0.53 - 0.47 0.53 − 0.47 Report a problem with this question
What is the product of 2 x 2 − x + 1 2x^2 - x + 1 2 x 2 − x + 1 3 − 2 x 3 - 2x 3 − 2 x
A 4 x 3 − 8 x 2 + 5 x + 3 4x^3 - 8x^2 + 5x + 3 4 x 3 − 8 x 2 + 5 x + 3 B − 4 x 3 + 8 x 2 − 5 x + 3 -4x^3 + 8x^2 - 5x + 3 − 4 x 3 + 8 x 2 − 5 x + 3 C − 4 x 3 − 8 x 2 + 5 x + 3 -4x^3 - 8x^2 + 5x + 3 − 4 x 3 − 8 x 2 + 5 x + 3 D 4 x 3 + 8 x 2 − 5 x + 3 4x^3 + 8x^2 - 5x + 3 4 x 3 + 8 x 2 − 5 x + 3
Worked solution (try it first) Multiply each term of
2 x 2 − x + 1 2x^2 - x + 1 2 x 2 − x + 1 by 3:
6 x 2 − 3 x + 3 6x^2 - 3x + 3 6 x 2 − 3 x + 3 .
Multiply each term by
− 2 x -2x − 2 x :
− 4 x 3 + 2 x 2 − 2 x -4x^3 + 2x^2 - 2x − 4 x 3 + 2 x 2 − 2 x .
Add and collect like terms:
− 4 x 3 + 8 x 2 − 5 x + 3 -4x^3 + 8x^2 - 5x + 3 − 4 x 3 + 8 x 2 − 5 x + 3 , option B.
Watch out
2 x 2 × ( − 2 x ) = − 4 x 3 2x^2 \times (-2x) = -4x^3 2 x 2 × ( − 2 x ) = − 4 x 3 Report a problem with this question
Find the equation of the locus of a point P ( x , y ) P(x, y) P ( x , y ) P V = P W PV = PW P V = P W V = ( 1 , 1 ) V = (1, 1) V = ( 1 , 1 ) W = ( 3 , 5 ) W = (3, 5) W = ( 3 , 5 )
A 2 x + 2 y = 9 2x + 2y = 9 2 x + 2 y = 9 B 2 x + 3 y = 8 2x + 3y = 8 2 x + 3 y = 8 C 2 x + y = 9 2x + y = 9 2 x + y = 9 D x + 2 y = 8 x + 2y = 8 x + 2 y = 8
Worked solution (try it first) P V = P W PV = PW P V = P W , so square both distances:
( x − 1 ) 2 + ( y − 1 ) 2 = ( x − 3 ) 2 + ( y − 5 ) 2 (x - 1)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 ( x − 1 ) 2 + ( y − 1 ) 2 = ( x − 3 ) 2 + ( y − 5 ) 2 .
Expand, and cancel
x 2 x^2 x 2 and
y 2 y^2 y 2 :
− 2 x + 1 − 2 y + 1 = − 6 x + 9 − 10 y + 25 -2x + 1 - 2y + 1 = -6x + 9 - 10y + 25 − 2 x + 1 − 2 y + 1 = − 6 x + 9 − 10 y + 25 .
Collect terms:
4 x + 8 y = 32 4x + 8y = 32 4 x + 8 y = 32 .
Divide by 4:
x + 2 y = 8 x + 2y = 8 x + 2 y = 8 , option D.
Watch out
Expand the brackets fully: ( y − 5 ) 2 = y 2 − 10 y + 25 (y - 5)^2 = y^2 - 10y + 25 ( y − 5 ) 2 = y 2 − 10 y + 25 y 2 − 5 y + 25 y^2 - 5y + 25 y 2 − 5 y + 25 ( 2 , 3 ) (2, 3) ( 2 , 3 ) V W VW V W 2 + 6 = 8 2 + 6 = 8 2 + 6 = 8 Also set as JAMB 1999 · UME · Q23
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In a regular polygon, each interior angle is double its corresponding exterior angle. Find the number of sides of the polygon.
Worked solution (try it first) Let the exterior angle be
e e e .
The interior angle is
2 e 2e 2 e , and the two add up to
180 ∘ 180^\circ 18 0 ∘ :
3 e = 180 ∘ 3e = 180^\circ 3 e = 18 0 ∘ , so
e = 60 ∘ e = 60^\circ e = 6 0 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 60 = 6 360 \div 60 = 6 360 ÷ 60 = 6 , option B.
Watch out
It is the interior angle that is double. Making the exterior angle double gives e = 120 ∘ e = 120^\circ e = 12 0 ∘ Also set as JAMB 2000 · UME · Q32
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A cylindrical tank has a capacity of 3080 m 3 3080\text{ m}^3 3080 m 3 14 m 14\text{ m} 14 m [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 23 m 23\text{ m} 23 m B 25 m 25\text{ m} 25 m C 20 m 20\text{ m} 20 m D 22 m 22\text{ m} 22 m
Worked solution (try it first) Radius
= 14 ÷ 2 = 7 = 14 \div 2 = 7 = 14 ÷ 2 = 7 m, so the base area is
22 7 × 7 2 = 154 m 2 \frac{22}{7} \times 7^2 = 154\text{ m}^2 7 22 × 7 2 = 154 m 2 .
Volume = base area × depth:
154 h = 3080 154h = 3080 154 h = 3080 .
So
h = 3080 ÷ 154 = 20 h = 3080 \div 154 = 20 h = 3080 ÷ 154 = 20 m, option C.
Watch out
Halve the diameter. Using 14 m as the radius gives a base of 616 m 2 616\text{ m}^2 616 m 2 Also set as JAMB 2001 · UME · Q23
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A car dealer bought a second-hand car for ₦250,000 and spent ₦70,000 refurbishing it. He then sold the car for ₦400,000. What is the percentage gain?
A 60 % 60\% 60% B 32 % 32\% 32% C 25 % 25\% 25% D 20 % 20\% 20%
Worked solution (try it first) The cost includes the refurbishing:
250 000 + 70 000 = 250\,000 + 70\,000 = 250 000 + 70 000 = ₦320,000.
The gain is
400 000 − 320 000 = 400\,000 - 320\,000 = 400 000 − 320 000 = ₦80,000.
As a percentage of the cost:
80 000 320 000 × 100 % = 25 % \dfrac{80\,000}{320\,000} \times 100\% = 25\% 320 000 80 000 × 100% = 25% , option C.
Watch out
The ₦70,000 spent on the car is part of its cost. Leaving it out and dividing ₦80,000 by ₦250,000 gives 32 % 32\% 32% Also set as JAMB 2001 · UME · Q2
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Find the principal which amounts to ₦5,500 at simple interest in 5 years at 2 % 2\% 2%
A ₦4,900 B ₦5,000 C ₦4,700 D ₦4,800
Worked solution (try it first) Simple interest for 5 years at
2 % 2\% 2% is
5 × 2 % = 10 % 5 \times 2\% = 10\% 5 × 2% = 10% of the principal
P P P .
The amount is principal plus interest, so
1.1 P = 5500 1.1P = 5500 1.1 P = 5500 .
Divide both sides by 1.1:
P = 5000 P = 5000 P = 5000 .
The principal is ₦5,000, option B.
Watch out
Interest is 10 % 10\% 10% 10 % 10\% 10% Report a problem with this question
In how many ways can the letters of the word ACCEPTANCE be arranged?
A 10 ! 3 ! \dfrac{10!}{3!} 3 ! 10 ! B 10 ! 2 ! 3 ! \dfrac{10!}{2!\,3!} 2 ! 3 ! 10 ! C 10 ! 2 ! 2 ! \dfrac{10!}{2!\,2!} 2 ! 2 ! 10 ! D 10 ! 3 ! 2 ! 2 ! \dfrac{10!}{3!\,2!\,2!} 3 ! 2 ! 2 ! 10 !
Worked solution (try it first) ACCEPTANCE has 10 letters: C appears 3 times, A twice and E twice.
P, T and N appear once.
Divide
10 ! 10! 10 ! by the factorial of each repeat count, so that swapping identical letters is not counted as new.
So the number of arrangements is
10 ! 3 ! 2 ! 2 ! \dfrac{10!}{3!\,2!\,2!} 3 ! 2 ! 2 ! 10 ! , option D.
Watch out
Count every repeated letter: A and E each appear twice. Allowing for only one of them gives 10 ! 2 ! 3 ! \frac{10!}{2!\,3!} 2 ! 3 ! 10 ! Report a problem with this question
Factorize completely x 2 + 2 x y + y 2 + 3 x + 3 y − 18 x^2 + 2xy + y^2 + 3x + 3y - 18 x 2 + 2 x y + y 2 + 3 x + 3 y − 18
A ( x + y + 6 ) ( x + y − 3 ) (x + y + 6)(x + y - 3) ( x + y + 6 ) ( x + y − 3 ) B ( x − y − 6 ) ( x − y + 3 ) (x - y - 6)(x - y + 3) ( x − y − 6 ) ( x − y + 3 ) C ( x − y + 6 ) ( x − y − 3 ) (x - y + 6)(x - y - 3) ( x − y + 6 ) ( x − y − 3 ) D ( x + y − 6 ) ( x + y + 3 ) (x + y - 6)(x + y + 3) ( x + y − 6 ) ( x + y + 3 )
Worked solution (try it first) The first three terms are a perfect square:
x 2 + 2 x y + y 2 = ( x + y ) 2 x^2 + 2xy + y^2 = (x + y)^2 x 2 + 2 x y + y 2 = ( x + y ) 2 .
The next two are
3 ( x + y ) 3(x + y) 3 ( x + y ) .
The expression is
u 2 + 3 u − 18 u^2 + 3u - 18 u 2 + 3 u − 18 .
Find two numbers with product
− 18 -18 − 18 and sum 3: they are 6 and
− 3 -3 − 3 .
So
u 2 + 3 u − 18 = ( u + 6 ) ( u − 3 ) u^2 + 3u - 18 = (u + 6)(u - 3) u 2 + 3 u − 18 = ( u + 6 ) ( u − 3 ) .
Put back
u = x + y u = x + y u = x + y :
( x + y + 6 ) ( x + y − 3 ) (x + y + 6)(x + y - 3) ( x + y + 6 ) ( x + y − 3 ) , option A.
Watch out
The middle term is + 3 u +3u + 3 u ( u − 6 ) ( u + 3 ) (u - 6)(u + 3) ( u − 6 ) ( u + 3 ) − 3 u -3u − 3 u Also set as JAMB 1999 · UME · Q19
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Divide 4 x 3 − 3 x + 1 4x^3 - 3x + 1 4 x 3 − 3 x + 1 2 x − 1 2x - 1 2 x − 1
A 2 x 2 − x + 1 2x^2 - x + 1 2 x 2 − x + 1 B 2 x 2 − x − 1 2x^2 - x - 1 2 x 2 − x − 1 C 2 x 2 + x + 1 2x^2 + x + 1 2 x 2 + x + 1 D 2 x 2 + x − 1 2x^2 + x - 1 2 x 2 + x − 1
Worked solution (try it first) Write the missing
x 2 x^2 x 2 term:
4 x 3 + 0 x 2 − 3 x + 1 4x^3 + 0x^2 - 3x + 1 4 x 3 + 0 x 2 − 3 x + 1 .
Then
4 x 3 ÷ 2 x = 2 x 2 4x^3 \div 2x = 2x^2 4 x 3 ÷ 2 x = 2 x 2 .
Subtract
4 x 3 − 2 x 2 4x^3 - 2x^2 4 x 3 − 2 x 2 to leave
2 x 2 − 3 x + 1 2x^2 - 3x + 1 2 x 2 − 3 x + 1 .
2 x 2 ÷ 2 x = x 2x^2 \div 2x = x 2 x 2 ÷ 2 x = x .
Subtract
2 x 2 − x 2x^2 - x 2 x 2 − x to leave
− 2 x + 1 -2x + 1 − 2 x + 1 .
− 2 x ÷ 2 x = − 1 -2x \div 2x = -1 − 2 x ÷ 2 x = − 1 , and
− 1 ( 2 x − 1 ) = − 2 x + 1 -1(2x - 1) = -2x + 1 − 1 ( 2 x − 1 ) = − 2 x + 1 leaves 0.
So the quotient is
2 x 2 + x − 1 2x^2 + x - 1 2 x 2 + x − 1 , option D.
Watch out
Put in 0 x 2 0x^2 0 x 2 − 2 x 2 -2x^2 − 2 x 2 0 x 2 0x^2 0 x 2 + 2 x 2 +2x^2 + 2 x 2 2 x 2 − x − 1 2x^2 - x - 1 2 x 2 − x − 1 ( 2 x − 1 ) ( 2 x 2 − x − 1 ) = 4 x 3 − 4 x 2 − x + 1 (2x - 1)(2x^2 - x - 1) = 4x^3 - 4x^2 - x + 1 ( 2 x − 1 ) ( 2 x 2 − x − 1 ) = 4 x 3 − 4 x 2 − x + 1 Also set as JAMB 1999 · UME · Q16
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In the diagram, M N MN M N K M N KMN K M N O O O 10 cm 10\text{ cm} 10 cm ∠ M O N = 140 ∘ \angle MON = 140^\circ ∠ M O N = 14 0 ∘ M N MN M N
A 19 cm 19\text{ cm} 19 cm B 18 cm 18\text{ cm} 18 cm C 17 cm 17\text{ cm} 17 cm D 12 cm 12\text{ cm} 12 cm
Worked solution (try it first) O M = O N = 10 OM = ON = 10 O M = O N = 10 cm.
The perpendicular from
O O O to
M N MN M N bisects the chord and the
140 ∘ 140^\circ 14 0 ∘ angle, giving two right-angled triangles with a
70 ∘ 70^\circ 7 0 ∘ angle at
O O O .
In one of them, half the chord is opposite the
70 ∘ 70^\circ 7 0 ∘ :
M N 2 = 10 sin 70 ∘ = 9.397 \frac{MN}{2} = 10\sin70^\circ = 9.397 2 M N = 10 sin 7 0 ∘ = 9.397 cm.
So
M N = 2 × 9.397 = 18.79 MN = 2 \times 9.397 = 18.79 M N = 2 × 9.397 = 18.79 cm, which is 19 cm to the nearest cm, option A.
Watch out
Round, don't cut off: 18.79 cm is nearer 19 than 18. Dropping the decimals gives 18 cm (option B). Report a problem with this question