Paper JAMB 2014 General Maths Objective
Objective paper · 46 questions · partial
JAMB 2014 · UTME Topics include Commercial arithmetic, Indices & standard form, Logarithms, Surds, Sets & Venn diagrams, Expressions, formulae & change of subject.
Our copy of this paper is missing questions 1, 2, 29, 44.
Sit this paper Answer every question in order, timed if you like (suggested 30 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.
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 46 47 48 49 50 A woman bought a grinder for ₦60,000. She sold it at a loss of 15 % 15\% 15%
A ₦50,000 B ₦53,000 C ₦52,000 D ₦51,000
Worked solution (try it first) A
15 % 15\% 15% loss means she sold for
100 % − 15 % = 85 % 100\% - 15\% = 85\% 100% − 15% = 85% of the cost.
85 % 85\% 85% of ₦60,000 is
0.85 × 60 000 = 51 000 0.85 \times 60\,000 = 51\,000 0.85 × 60 000 = 51 000 .
So she sold it for ₦51,000, option D.
Watch out
The loss is 15 % 15\% 15% Report a problem with this question
Express the product of 0.00043 and 2000 in standard form.
A 8.6 × 10 8.6 \times 10 8.6 × 10 B 8.3 × 10 − 3 8.3 \times 10^{-3} 8.3 × 1 0 − 3 C 8.6 × 10 − 2 8.6 \times 10^{-2} 8.6 × 1 0 − 2 D 8.6 × 10 − 1 8.6 \times 10^{-1} 8.6 × 1 0 − 1
Worked solution (try it first) 2000 = 2 × 1000 2000 = 2 \times 1000 2000 = 2 × 1000 , so first double:
0.00043 × 2 = 0.00086 0.00043 \times 2 = 0.00086 0.00043 × 2 = 0.00086 .
Multiplying by 1000 moves the point three places right:
0.86 0.86 0.86 .
0.86 is
8.6 × 10 − 1 8.6 \times 10^{-1} 8.6 × 1 0 − 1 , option D.
Watch out
Standard form needs a number from 1 to 10: 0.86 becomes 8.6 by moving the point one place right, so the power is − 1 -1 − 1 8.6 × 10 − 2 8.6 \times 10^{-2} 8.6 × 1 0 − 2 Report a problem with this question
A man donates 10 % 10\% 10%
A ₦62,500 B ₦40,500 C ₦45,000 D ₦52,500
Worked solution (try it first) 10 % 10\% 10% of his net income
E E E is ₦4,500, so
0.1 E = 4500 0.1E = 4500 0.1 E = 4500 .
Divide both sides by 0.1 (multiply by 10):
E = 45 000 E = 45\,000 E = 45 000 .
So his net monthly income is ₦45,000, option C.
Watch out
₦4,500 is the 10 % 10\% 10% Report a problem with this question
If log 7.5 = 0.8751 \log 7.5 = 0.8751 log 7.5 = 0.8751 2 log 75 + log 750 2\log 75 + \log 750 2 log 75 + log 750
A 66.253 B 6.6252 C 6.6253 D 66.252
Worked solution (try it first) 75 = 7.5 × 10 75 = 7.5 \times 10 75 = 7.5 × 10 , so
log 75 = 1 + 0.8751 = 1.8751 \log 75 = 1 + 0.8751 = 1.8751 log 75 = 1 + 0.8751 = 1.8751 .
In the same way,
log 750 = 2 + 0.8751 = 2.8751 \log 750 = 2 + 0.8751 = 2.8751 log 750 = 2 + 0.8751 = 2.8751 .
2 log 75 = 2 × 1.8751 = 3.7502 2\log 75 = 2 \times 1.8751 = 3.7502 2 log 75 = 2 × 1.8751 = 3.7502 .
So the value is
3.7502 + 2.8751 = 6.6253 3.7502 + 2.8751 = 6.6253 3.7502 + 2.8751 = 6.6253 , option C.
Watch out
Double all of log 75 \log 75 log 75 2 × 1.8751 = 3.7502 2 \times 1.8751 = 3.7502 2 × 1.8751 = 3.7502 Report a problem with this question
Solve for x x x 8 x − 2 = 2 25 8x^{-2} = \frac{2}{25} 8 x − 2 = 25 2
Worked solution (try it first) The index
− 2 -2 − 2 applies only to
x x x :
8 x − 2 = 8 x 2 8x^{-2} = \frac{8}{x^2} 8 x − 2 = x 2 8 , so
8 x 2 = 2 25 \frac{8}{x^2} = \frac{2}{25} x 2 8 = 25 2 .
Cross-multiply:
2 x 2 = 200 2x^2 = 200 2 x 2 = 200 , so
x 2 = 100 x^2 = 100 x 2 = 100 .
Take the square root:
x = 10 x = 10 x = 10 , option A.
Watch out
8 x − 2 8x^{-2} 8 x − 2 8 x 2 \frac{8}{x^2} x 2 8 x 2 = 25 16 x^2 = \frac{25}{16} x 2 = 16 25 Report a problem with this question
Simplify 2 2 − 3 2 + 3 \dfrac{2\sqrt2 - \sqrt3}{\sqrt2 + \sqrt3} 2 + 3 2 2 − 3
A 3 6 + 1 3\sqrt6 + 1 3 6 + 1 B 3 6 − 7 3\sqrt6 - 7 3 6 − 7 C 3 6 + 7 3\sqrt6 + 7 3 6 + 7 D 3 6 − 1 3\sqrt6 - 1 3 6 − 1
Worked solution (try it first) Multiply the top and bottom by
3 − 2 \sqrt3 - \sqrt2 3 − 2 , so the bottom becomes
( 3 + 2 ) ( 3 − 2 ) = 3 − 2 (\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2) = 3 - 2 ( 3 + 2 ) ( 3 − 2 ) = 3 − 2 , which is 1.
Top:
( 2 2 − 3 ) ( 3 − 2 ) = 2 6 − 4 − 3 + 6 (2\sqrt2 - \sqrt3)(\sqrt3 - \sqrt2) = 2\sqrt6 - 4 - 3 + \sqrt6 ( 2 2 − 3 ) ( 3 − 2 ) = 2 6 − 4 − 3 + 6 .
Collect like terms:
3 6 − 7 3\sqrt6 - 7 3 6 − 7 , option B.
Watch out
Both whole-number products are negative: 2 2 × ( − 2 ) = − 4 2\sqrt2 \times (-\sqrt2) = -4 2 2 × ( − 2 ) = − 4 − 3 × 3 = − 3 -\sqrt3 \times \sqrt3 = -3 − 3 × 3 = − 3 − 7 -7 − 7 3 6 − 1 3\sqrt6 - 1 3 6 − 1 Also set as JAMB 2018 · UTME · Q3
Report a problem with this question
Evaluate log 2 8 + log 2 16 − log 2 4 \log_2 8 + \log_2 16 - \log_2 4 log 2 8 + log 2 16 − log 2 4
Worked solution (try it first) 2 3 = 8 2^3 = 8 2 3 = 8 ,
2 4 = 16 2^4 = 16 2 4 = 16 and
2 2 = 4 2^2 = 4 2 2 = 4 , so the logs are 3, 4 and 2.
So the value is
3 + 4 − 2 = 5 3 + 4 - 2 = 5 3 + 4 − 2 = 5 , option D.
Watch out
log 2 4 = 2 \log_2 4 = 2 log 2 4 = 2 2 2 = 4 2^2 = 4 2 2 = 4 3 + 4 − 1 = 6 3 + 4 - 1 = 6 3 + 4 − 1 = 6 Report a problem with this question
If P = { 1 , 2 , 3 , 4 , 5 } P = \{1, 2, 3, 4, 5\} P = { 1 , 2 , 3 , 4 , 5 } P ∪ Q = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } P \cup Q = \{1, 2, 3, 4, 5, 6, 7\} P ∪ Q = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } Q Q Q
A { 5 , 7 } \{5, 7\} { 5 , 7 } B { 6 } \{6\} { 6 } C { 7 } \{7\} { 7 } D { 6 , 7 } \{6, 7\} { 6 , 7 }
Worked solution (try it first) P ∪ Q P \cup Q P ∪ Q holds every element of
P P P or
Q Q Q .
The elements 6 and 7 are in the union but not in
P P P , so both must come from
Q Q Q .
So
Q = { 6 , 7 } Q = \{6, 7\} Q = { 6 , 7 } , option D.
Watch out
Both 6 and 7 have to be in Q Q Q { 6 } \{6\} { 6 } { 7 } \{7\} { 7 } Report a problem with this question
From the Venn diagram, the shaded parts represent
A ( P ∩ Q ) ∩ ( P ∩ R ) (P \cap Q) \cap (P \cap R) ( P ∩ Q ) ∩ ( P ∩ R ) B ( P ∩ Q ) ∪ ( P ∩ R ) (P \cap Q) \cup (P \cap R) ( P ∩ Q ) ∪ ( P ∩ R ) C ( P ∪ Q ) ∩ ( P ∪ R ) (P \cup Q) \cap (P \cup R) ( P ∪ Q ) ∩ ( P ∪ R ) D ( P ∪ Q ) ∪ ( P ∪ R ) (P \cup Q) \cup (P \cup R) ( P ∪ Q ) ∪ ( P ∪ R )
Worked solution (try it first) The shading covers the whole overlap of
P P P and
Q Q Q , which is
P ∩ Q P \cap Q P ∩ Q .
It also covers the whole overlap of
P P P and
R R R , which is
P ∩ R P \cap R P ∩ R .
Together, both pieces make
( P ∩ Q ) ∪ ( P ∩ R ) (P \cap Q) \cup (P \cap R) ( P ∩ Q ) ∪ ( P ∩ R ) , option B.
Watch out
Joining two regions is a union. ( P ∩ Q ) ∩ ( P ∩ R ) (P \cap Q) \cap (P \cap R) ( P ∩ Q ) ∩ ( P ∩ R ) Report a problem with this question
If g t 2 − k − w = 0 gt^2 - k - w = 0 g t 2 − k − w = 0 g g g
A k − w t \dfrac{k - w}{t} t k − w B k + w t 2 \dfrac{k + w}{t^2} t 2 k + w C k − w t 2 \dfrac{k - w}{t^2} t 2 k − w D k + w t \dfrac{k + w}{t} t k + w
Worked solution (try it first) Add
k k k and
w w w to both sides:
g t 2 = k + w gt^2 = k + w g t 2 = k + w .
Divide both sides by
t 2 t^2 t 2 :
g = k + w t 2 g = \dfrac{k + w}{t^2} g = t 2 k + w , option B.
Watch out
Both k k k w w w w w w − w -w − w k − w t 2 \frac{k - w}{t^2} t 2 k − w Report a problem with this question
Factorize 2 y 2 − 15 x y + 18 x 2 2y^2 - 15xy + 18x^2 2 y 2 − 15 x y + 18 x 2
A ( 3 y + 2 x ) ( y − 6 x ) (3y + 2x)(y - 6x) ( 3 y + 2 x ) ( y − 6 x ) B ( 2 y − 3 x ) ( y + 6 x ) (2y - 3x)(y + 6x) ( 2 y − 3 x ) ( y + 6 x ) C ( 2 y − 3 x ) ( y − 6 x ) (2y - 3x)(y - 6x) ( 2 y − 3 x ) ( y − 6 x ) D ( 2 y + 3 x ) ( y − 6 x ) (2y + 3x)(y - 6x) ( 2 y + 3 x ) ( y − 6 x )
Worked solution (try it first) Find two terms with product
2 × 18 x 2 = 36 x 2 2 \times 18x^2 = 36x^2 2 × 18 x 2 = 36 x 2 and sum
− 15 x -15x − 15 x : they are
− 12 x -12x − 12 x and
− 3 x -3x − 3 x .
Split the middle term:
2 y 2 − 12 x y − 3 x y + 18 x 2 2y^2 - 12xy - 3xy + 18x^2 2 y 2 − 12 x y − 3 x y + 18 x 2 .
Group in pairs:
2 y ( y − 6 x ) − 3 x ( y − 6 x ) = ( 2 y − 3 x ) ( y − 6 x ) 2y(y - 6x) - 3x(y - 6x) = (2y - 3x)(y - 6x) 2 y ( y − 6 x ) − 3 x ( y − 6 x ) = ( 2 y − 3 x ) ( y − 6 x ) , option C.
Watch out
The last term + 18 x 2 +18x^2 + 18 x 2 ( 2 y + 3 x ) ( y − 6 x ) (2y + 3x)(y - 6x) ( 2 y + 3 x ) ( y − 6 x ) − 18 x 2 -18x^2 − 18 x 2 Report a problem with this question
Find the value of k k k y − 1 y - 1 y − 1 y 3 + 4 y 2 + k y − 6 y^3 + 4y^2 + ky - 6 y 3 + 4 y 2 + k y − 6
Worked solution (try it first) By the factor theorem,
y − 1 y - 1 y − 1 is a factor, so the expression is 0 at
y = 1 y = 1 y = 1 .
1 + 4 + k − 6 = 0 1 + 4 + k - 6 = 0 1 + 4 + k − 6 = 0 , so
k − 1 = 0 k - 1 = 0 k − 1 = 0 and
k = 1 k = 1 k = 1 , option D.
Watch out
y − 1 y - 1 y − 1 y = + 1 y = +1 y = + 1 y = − 1 y = -1 y = − 1 − 1 + 4 − k − 6 = 0 -1 + 4 - k - 6 = 0 − 1 + 4 − k − 6 = 0 k = − 3 k = -3 k = − 3 Report a problem with this question
y y y w 2 w^2 w 2 y = 8 y = 8 y = 8 w = 2 w = 2 w = 2 y y y w = 3 w = 3 w = 3
Worked solution (try it first) Put in
y = 8 y = 8 y = 8 ,
w = 2 w = 2 w = 2 :
8 = 4 k 8 = 4k 8 = 4 k , so
k = 2 k = 2 k = 2 .
When
w = 3 w = 3 w = 3 :
y = 2 × 9 = 18 y = 2 \times 9 = 18 y = 2 × 9 = 18 , option B.
Watch out
Square w w w y y y w w w 8 × 3 2 8 \times \frac32 8 × 2 3 Report a problem with this question
P P P Q Q Q R R R Q = 36 Q = 36 Q = 36 R = 16 R = 16 R = 16 P = 27 P = 27 P = 27 P P P Q Q Q R R R
A P = 12 Q R P = \dfrac{12}{QR} P = QR 12 B P = Q 12 R P = \dfrac{Q}{12R} P = 12 R Q C P = 12 Q R P = \dfrac{12Q}{R} P = R 12 Q D P = 12 Q R P = 12QR P = 12 QR
Worked solution (try it first) Directly as
Q Q Q (on top), inversely as
R R R (underneath):
P = k Q R P = \dfrac{kQ}{R} P = R k Q .
Put in the values:
27 = 36 k 16 27 = \dfrac{36k}{16} 27 = 16 36 k , so
k = 27 × 16 36 = 12 k = \dfrac{27 \times 16}{36} = 12 k = 36 27 × 16 = 12 .
So
P = 12 Q R P = \dfrac{12Q}{R} P = R 12 Q , option C.
Watch out
Inversely as R R R R R R P = 12 Q R P = 12QR P = 12 QR R R R Q Q Q Report a problem with this question
What is the solution of x − 5 x + 3 < − 1 \dfrac{x - 5}{x + 3} < -1 x + 3 x − 5 < − 1
A x < − 3 x < -3 x < − 3 x > 5 x > 5 x > 5 B − 3 < x < 1 -3 < x < 1 − 3 < x < 1 C x < − 3 x < -3 x < − 3 x > 1 x > 1 x > 1 D − 3 < x < 5 -3 < x < 5 − 3 < x < 5
Worked solution (try it first) Add 1 to both sides to get 0 on the right:
x − 5 x + 3 + 1 < 0 \dfrac{x - 5}{x + 3} + 1 < 0 x + 3 x − 5 + 1 < 0 .
Use the common denominator:
x − 5 + x + 3 x + 3 = 2 x − 2 x + 3 < 0 \dfrac{x - 5 + x + 3}{x + 3} = \dfrac{2x - 2}{x + 3} < 0 x + 3 x − 5 + x + 3 = x + 3 2 x − 2 < 0 .
The critical values are
x = 1 x = 1 x = 1 (top zero) and
x = − 3 x = -3 x = − 3 (bottom zero).
The fraction is negative when the top and bottom have opposite signs, which is between them.
So
− 3 < x < 1 -3 < x < 1 − 3 < x < 1 , option B.
Watch out
Don't multiply both sides by x + 3 x + 3 x + 3 x x x x < 1 x < 1 x < 1 − 3 -3 − 3 x = − 4 x = -4 x = − 4 − 1 -1 − 1 Report a problem with this question
Solve the inequality x 2 + 3 4 ≤ 5 x 6 − 7 12 \frac x2 + \frac34 \le \frac{5x}{6} - \frac{7}{12} 2 x + 4 3 ≤ 6 5 x − 12 7
A x ≥ − 4 x \ge -4 x ≥ − 4 B x ≥ 4 x \ge 4 x ≥ 4 C x ≤ 3 x \le 3 x ≤ 3 D x ≥ − 3 x \ge -3 x ≥ − 3
Worked solution (try it first) Multiply every term by 12, the LCM of 2, 4, 6 and 12:
6 x + 9 ≤ 10 x − 7 6x + 9 \le 10x - 7 6 x + 9 ≤ 10 x − 7 .
Subtract
6 x 6x 6 x and add 7 to both sides:
16 ≤ 4 x 16 \le 4x 16 ≤ 4 x .
Divide by 4:
4 ≤ x 4 \le x 4 ≤ x , that is
x ≥ 4 x \ge 4 x ≥ 4 , option B.
Watch out
Moving − 7 -7 − 7 + 7 +7 + 7 9 + 7 = 16 9 + 7 = 16 9 + 7 = 16 − 4 x ≤ 16 -4x \le 16 − 4 x ≤ 16 x ≥ − 4 x \ge -4 x ≥ − 4 Report a problem with this question
The 4th term of an A.P. is 13 while the 10th term is 31. Find the 24th term.
Worked solution (try it first) From the 4th term to the 10th is 6 steps of
d d d :
6 d = 31 − 13 = 18 6d = 31 - 13 = 18 6 d = 31 − 13 = 18 , so
d = 3 d = 3 d = 3 .
Go back three steps from the 4th term:
a = 13 − 3 × 3 = 4 a = 13 - 3 \times 3 = 4 a = 13 − 3 × 3 = 4 .
The 24th term is
a + 23 d = 4 + 69 = 73 a + 23d = 4 + 69 = 73 a + 23 d = 4 + 69 = 73 , option D.
Watch out
The 24th term has 23 d 23d 23 d 24 d 24d 24 d 24 d 24d 24 d 4 + 72 = 76 4 + 72 = 76 4 + 72 = 76 Report a problem with this question
What is the common ratio of the G.P. ( 10 + 5 ) , ( 10 + 2 5 ) , … (\sqrt{10} + \sqrt5), (\sqrt{10} + 2\sqrt5), \ldots ( 10 + 5 ) , ( 10 + 2 5 ) , …
Worked solution (try it first) The common ratio is the second term divided by the first:
r = 10 + 2 5 10 + 5 r = \dfrac{\sqrt{10} + 2\sqrt5}{\sqrt{10} + \sqrt5} r = 10 + 5 10 + 2 5 .
Since
10 = 2 5 \sqrt{10} = \sqrt2\sqrt5 10 = 2 5 , take
5 \sqrt5 5 out of the top and bottom:
r = 2 + 2 2 + 1 r = \dfrac{\sqrt2 + 2}{\sqrt2 + 1} r = 2 + 1 2 + 2 .
Write
2 = 2 × 2 2 = \sqrt2 \times \sqrt2 2 = 2 × 2 , so the top is
2 ( 1 + 2 ) \sqrt2(1 + \sqrt2) 2 ( 1 + 2 ) and cancels with the bottom.
So
r = 2 r = \sqrt2 r = 2 , option B.
Watch out
You can only cancel factors, not terms added together. Crossing out 10 \sqrt{10} 10 2 5 5 = 2 \frac{2\sqrt5}{\sqrt5} = 2 5 2 5 = 2 Report a problem with this question
A binary operation ∗ * ∗ x ∗ y = x y x * y = xy x ∗ y = x y x ∗ x = 12 − x x * x = 12 - x x ∗ x = 12 − x x x x
A − 3 , − 4 -3, -4 − 3 , − 4 B 3 , 4 3, 4 3 , 4 C 3 , − 4 3, -4 3 , − 4 D − 3 , 4 -3, 4 − 3 , 4
Worked solution (try it first) x ∗ x = x × x = x 2 x * x = x \times x = x^2 x ∗ x = x × x = x 2 , so the equation is
x 2 = 12 − x x^2 = 12 - x x 2 = 12 − x .
Bring everything to one side:
x 2 + x − 12 = 0 x^2 + x - 12 = 0 x 2 + x − 12 = 0 , which factorises as
( x − 3 ) ( x + 4 ) = 0 (x - 3)(x + 4) = 0 ( x − 3 ) ( x + 4 ) = 0 .
So
x = 3 x = 3 x = 3 or
x = − 4 x = -4 x = − 4 , option C.
Watch out
Each factor gives the root with the opposite sign: ( x − 3 ) (x - 3) ( x − 3 ) 3 3 3 ( x + 4 ) (x + 4) ( x + 4 ) − 4 -4 − 4 − 3 , 4 -3, 4 − 3 , 4 Report a problem with this question
Find y y y ( 5 − 6 2 − 7 ) ( x y ) = ( 7 − 11 ) \begin{pmatrix} 5 & -6 \\ 2 & -7 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \end{pmatrix} ( 5 2 − 6 − 7 ) ( x y ) = ( 7 − 11 )
Worked solution (try it first) Multiply out the left side to get two equations:
5 x − 6 y = 7 5x - 6y = 7 5 x − 6 y = 7 and
2 x − 7 y = − 11 2x - 7y = -11 2 x − 7 y = − 11 .
Make the
x x x terms match: multiply the first by 2 and the second by 5, giving
10 x − 12 y = 14 10x - 12y = 14 10 x − 12 y = 14 and
10 x − 35 y = − 55 10x - 35y = -55 10 x − 35 y = − 55 .
Subtract the second from the first:
23 y = 69 23y = 69 23 y = 69 , so
y = 3 y = 3 y = 3 , option D.
Watch out
The question asks for y y y x = 5 x = 5 x = 5 5 ( 5 ) − 6 ( 3 ) = 7 5(5) - 6(3) = 7 5 ( 5 ) − 6 ( 3 ) = 7 Report a problem with this question
If ∣ − x 12 − 1 4 ∣ = − 12 \begin{vmatrix} -x & 12 \\ -1 & 4 \end{vmatrix} = -12 − x − 1 12 4 = − 12 x x x
Worked solution (try it first) A
2 × 2 2 \times 2 2 × 2 determinant is
a d − b c ad - bc a d − b c :
( − x ) ( 4 ) − ( 12 ) ( − 1 ) = − 4 x + 12 (-x)(4) - (12)(-1) = -4x + 12 ( − x ) ( 4 ) − ( 12 ) ( − 1 ) = − 4 x + 12 .
So
− 4 x + 12 = − 12 -4x + 12 = -12 − 4 x + 12 = − 12 .
Take 12 from both sides:
− 4 x = − 24 -4x = -24 − 4 x = − 24 .
Divide by
− 4 -4 − 4 :
x = 6 x = 6 x = 6 , option A.
Watch out
Dividing a negative by a negative gives a positive: − 24 ÷ ( − 4 ) = 6 -24 \div (-4) = 6 − 24 ÷ ( − 4 ) = 6 − 6 -6 − 6 Report a problem with this question
Find the value of ∣ 0 3 2 1 7 8 0 5 4 ∣ \begin{vmatrix} 0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4 \end{vmatrix} 0 1 0 3 7 5 2 8 4
Worked solution (try it first) The first column is
( 0 , 1 , 0 ) (0, 1, 0) ( 0 , 1 , 0 ) , so expand down it: only the middle entry, 1, gives a term.
Its place sign is
− - − (row 2, column 1).
Its minor, crossing out row 2 and column 1, is
3 × 4 − 2 × 5 = 2 3 \times 4 - 2 \times 5 = 2 3 × 4 − 2 × 5 = 2 .
So the determinant is
− 1 × 2 = − 2 -1 \times 2 = -2 − 1 × 2 = − 2 , option A.
Watch out
The minor is 3 × 4 − 2 × 5 3 \times 4 - 2 \times 5 3 × 4 − 2 × 5 Report a problem with this question
How many sides has a regular polygon whose interior angles are 135 ∘ 135^\circ 13 5 ∘
Worked solution (try it first) An interior angle and its exterior angle add up to
180 ∘ 180^\circ 18 0 ∘ , so each exterior angle is
180 ∘ − 135 ∘ = 45 ∘ 180^\circ - 135^\circ = 45^\circ 18 0 ∘ − 13 5 ∘ = 4 5 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 45 = 8 360 \div 45 = 8 360 ÷ 45 = 8 , option A.
Watch out
Find the exterior angle carefully: 180 − 135 = 45 180 - 135 = 45 180 − 135 = 45 360 ÷ 45 = 8 360 \div 45 = 8 360 ÷ 45 = 8 140 ∘ 140^\circ 14 0 ∘ Report a problem with this question
In the figure, K L ∥ N M KL \parallel NM K L ∥ N M L N LN L N ∠ K N M \angle KNM ∠ K N M ∠ K L N = 54 ∘ \angle KLN = 54^\circ ∠ K L N = 5 4 ∘ ∠ M K N = 35 ∘ \angle MKN = 35^\circ ∠ M K N = 3 5 ∘ ∠ K M N \angle KMN ∠ K M N
A 19 ∘ 19^\circ 1 9 ∘ B 91 ∘ 91^\circ 9 1 ∘ C 89 ∘ 89^\circ 8 9 ∘ D 37 ∘ 37^\circ 3 7 ∘
Worked solution (try it first) K L ∥ N M KL \parallel NM K L ∥ N M , so
∠ L N M = ∠ K L N = 54 ∘ \angle LNM = \angle KLN = 54^\circ ∠ L N M = ∠ K L N = 5 4 ∘ (alternate angles).
L N LN L N bisects
∠ K N M \angle KNM ∠ K N M , so
∠ K N M = 2 × 54 ∘ \angle KNM = 2 \times 54^\circ ∠ K N M = 2 × 5 4 ∘ The angles of triangle
K M N KMN K M N add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ K M N = 180 ∘ − 108 ∘ − 35 ∘ \angle KMN = 180^\circ - 108^\circ - 35^\circ ∠ K M N = 18 0 ∘ − 10 8 ∘ − 3 5 ∘ = 37 ∘ = 37^\circ = 3 7 ∘ , option D.
Watch out
54 ∘ 54^\circ 5 4 ∘ N N N 180 ∘ − 54 ∘ − 35 ∘ = 91 ∘ 180^\circ - 54^\circ - 35^\circ = 91^\circ 18 0 ∘ − 5 4 ∘ − 3 5 ∘ = 9 1 ∘ Report a problem with this question
From the figure, what is the value of p p p
A 135 ∘ 135^\circ 13 5 ∘ B 90 ∘ 90^\circ 9 0 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 45 ∘ 45^\circ 4 5 ∘
Worked solution (try it first) Vertically opposite angles are equal, so
q = 30 ∘ q = 30^\circ q = 3 0 ∘ .
The angle
( p + 2 q ) ∘ (p + 2q)^\circ ( p + 2 q ) ∘ and the
30 ∘ 30^\circ 3 0 ∘ angle lie on a straight line, so
p + 2 q + 30 = 180 p + 2q + 30 = 180 p + 2 q + 30 = 180 .
Put in
q = 30 q = 30 q = 30 :
p + 60 + 30 = 180 p + 60 + 30 = 180 p + 60 + 30 = 180 , so
p = 90 ∘ p = 90^\circ p = 9 0 ∘ , option B.
Watch out
Replace both q q q 2 q = 60 2q = 60 2 q = 60 p p p 180 ∘ 180^\circ 18 0 ∘ p = 60 ∘ p = 60^\circ p = 6 0 ∘ 150 ∘ 150^\circ 15 0 ∘ Report a problem with this question
Find the value of x x x
A 4 3 cm 4\sqrt3\text{ cm} 4 3 cm B 120 3 cm 120\sqrt3\text{ cm} 120 3 cm C 10 3 cm 10\sqrt3\text{ cm} 10 3 cm D 5 3 cm 5\sqrt3\text{ cm} 5 3 cm
Worked solution (try it first) The angles add up to
180 ∘ 180^\circ 18 0 ∘ , so the third angle is
90 ∘ 90^\circ 9 0 ∘ .
The longest side is the hypotenuse.
At the
60 ∘ 60^\circ 6 0 ∘ corner,
x x x is the opposite side and the 10 cm side is the adjacent side, so
tan 60 ∘ = x 10 \tan60^\circ = \dfrac{x}{10} tan 6 0 ∘ = 10 x .
So
x = 10 tan 60 ∘ = 10 3 x = 10\tan60^\circ = 10\sqrt3 x = 10 tan 6 0 ∘ = 10 3 cm, option C.
Watch out
10 cm is a shorter side, not the hypotenuse. Treating it as the hypotenuse gives x = 10 sin 60 ∘ = 5 3 x = 10\sin60^\circ = 5\sqrt3 x = 10 sin 6 0 ∘ = 5 3 Report a problem with this question
A cylindrical tank has a capacity of 6160 m 3 6160\text{ m}^3 6160 m 3 28 m 28\text{ m} 28 m [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 8.0 m 8.0\text{ m} 8.0 m B 7.5 m 7.5\text{ m} 7.5 m C 5.0 m 5.0\text{ m} 5.0 m D 2.5 m 2.5\text{ m} 2.5 m
Worked solution (try it first) Base area:
π r 2 = 22 7 × 28 2 \pi r^2 = \frac{22}{7} \times 28^2 π r 2 = 7 22 × 2 8 2 = 2464 m 2 = 2464\text{ m}^2 = 2464 m 2 .
Volume = base area × depth, so the depth is
6160 ÷ 2464 = 2.5 6160 \div 2464 = 2.5 6160 ÷ 2464 = 2.5 m, option D.
Watch out
Check by multiplying back: 2464 × 2.5 = 6160 2464 \times 2.5 = 6160 2464 × 2.5 = 6160 2464 × 8 = 19 712 m 3 2464 \times 8 = 19\,712\text{ m}^3 2464 × 8 = 19 712 m 3 Report a problem with this question
The locus of a dog tethered to a pole with a rope of 4 m 4\text{ m} 4 m
A semi-circle with radius 4 m 4\text{ m} 4 m B circle with diameter 4 m 4\text{ m} 4 m C circle with radius 4 m 4\text{ m} 4 m D semi-circle with diameter 4 m 4\text{ m} 4 m
Worked solution (try it first) With the rope pulled tight, the dog is always 4 m from the pole.
Points at a fixed distance from a point form a circle, so the locus is a circle with radius 4 m, option C.
Watch out
The rope is the radius, not the diameter. A circle of diameter 4 m (option B) would keep the dog only 2 m from the pole. Report a problem with this question
Find the midpoint of S ( − 5 , 4 ) S(-5, 4) S ( − 5 , 4 ) T ( − 3 , − 2 ) T(-3, -2) T ( − 3 , − 2 )
A ( 4 , − 1 ) (4, -1) ( 4 , − 1 ) B ( − 4 , 2 ) (-4, 2) ( − 4 , 2 ) C ( 4 , − 2 ) (4, -2) ( 4 , − 2 ) D ( − 4 , 1 ) (-4, 1) ( − 4 , 1 )
Worked solution (try it first) The midpoint is the average of the ends.
x x x :
− 5 + ( − 3 ) 2 = − 8 2 = − 4 \frac{-5 + (-3)}{2} = \frac{-8}{2} = -4 2 − 5 + ( − 3 ) = 2 − 8 = − 4 .
y y y :
4 + ( − 2 ) 2 = 2 2 = 1 \frac{4 + (-2)}{2} = \frac22 = 1 2 4 + ( − 2 ) = 2 2 = 1 .
So the midpoint is
( − 4 , 1 ) (-4, 1) ( − 4 , 1 ) , option D.
Watch out
Halve both sums. Forgetting to halve the y y y ( − 4 , 2 ) (-4, 2) ( − 4 , 2 ) Report a problem with this question
The gradient of a line joining ( x , 4 ) (x, 4) ( x , 4 ) ( 1 , 2 ) (1, 2) ( 1 , 2 ) 1 2 \frac12 2 1 x x x
Worked solution (try it first) Gradient:
4 − 2 x − 1 = 1 2 \dfrac{4 - 2}{x - 1} = \frac12 x − 1 4 − 2 = 2 1 .
Cross-multiply:
x − 1 = 2 × 2 = 4 x - 1 = 2 \times 2 = 4 x − 1 = 2 × 2 = 4 .
So
x = 5 x = 5 x = 5 , option B.
Watch out
Subtract in the same order top and bottom. Writing the bottom as 1 − x 1 - x 1 − x 1 − x = 4 1 - x = 4 1 − x = 4 x = − 3 x = -3 x = − 3 Report a problem with this question
What is the equation of the line that passes through the y y y ( 0 , 5 ) (0, 5) ( 0 , 5 ) x x x ( 5 , 0 ) (5, 0) ( 5 , 0 )
A y = − x − 5 y = -x - 5 y = − x − 5 B y = x + 5 y = x + 5 y = x + 5 C y = − x + 5 y = -x + 5 y = − x + 5 D y = x − 5 y = x - 5 y = x − 5
Try it on a graph The line through (0, 5) and (5, 0).
Open the interactive graph Worked solution (try it first) Gradient:
0 − 5 5 − 0 = − 1 \dfrac{0 - 5}{5 - 0} = -1 5 − 0 0 − 5 = − 1 .
The line cuts the
y y y -axis at 5, so
c = 5 c = 5 c = 5 .
So
y = − x + 5 y = -x + 5 y = − x + 5 , option C.
Watch out
The line falls from ( 0 , 5 ) (0, 5) ( 0 , 5 ) ( 5 , 0 ) (5, 0) ( 5 , 0 ) y = x + 5 y = x + 5 y = x + 5 x x x − 5 -5 − 5 Also set as JAMB 2018 · UTME · Q9
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Calculate the midpoint of the segment of the line y − 4 x + 3 = 0 y - 4x + 3 = 0 y − 4 x + 3 = 0 x x x y y y
A ( − 2 3 , 3 2 ) \left(-\frac23, \frac32\right) ( − 3 2 , 2 3 ) B ( 3 8 , − 3 2 ) \left(\frac38, -\frac32\right) ( 8 3 , − 2 3 ) C ( 3 8 , 3 2 ) \left(\frac38, \frac32\right) ( 8 3 , 2 3 ) D ( − 3 2 , 3 2 ) \left(-\frac32, \frac32\right) ( − 2 3 , 2 3 )
Worked solution (try it first) Write the line as
y = 4 x − 3 y = 4x - 3 y = 4 x − 3 .
It meets the
x x x -axis where
y = 0 y = 0 y = 0 :
4 x = 3 4x = 3 4 x = 3 , so
x = 3 4 x = \frac34 x = 4 3 , giving
( 3 4 , 0 ) \left(\frac34, 0\right) ( 4 3 , 0 ) .
It meets the
y y y -axis at
( 0 , − 3 ) (0, -3) ( 0 , − 3 ) .
Average the two points:
( 3 8 , − 3 2 ) \left(\frac38, -\frac32\right) ( 8 3 , − 2 3 ) , option B.
Watch out
The y y y − 3 -3 − 3 ( 3 8 , 3 2 ) \left(\frac38, \frac32\right) ( 8 3 , 2 3 ) − 3 2 − 4 ( 3 8 ) + 3 = 0 -\frac32 - 4\left(\frac38\right) + 3 = 0 − 2 3 − 4 ( 8 3 ) + 3 = 0 Similar: JAMB 1998 · UME · Q33
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Find the equation of the straight line through ( − 2 , 3 ) (-2, 3) ( − 2 , 3 ) 4 x + 3 y − 5 = 0 4x + 3y - 5 = 0 4 x + 3 y − 5 = 0
A 5 x − 2 y − 11 = 0 5x - 2y - 11 = 0 5 x − 2 y − 11 = 0 B 3 x − 4 y + 18 = 0 3x - 4y + 18 = 0 3 x − 4 y + 18 = 0 C 3 x + 2 y − 18 = 0 3x + 2y - 18 = 0 3 x + 2 y − 18 = 0 D 4 x + 5 y + 3 = 0 4x + 5y + 3 = 0 4 x + 5 y + 3 = 0
Worked solution (try it first) Rearrange
4 x + 3 y − 5 = 0 4x + 3y - 5 = 0 4 x + 3 y − 5 = 0 :
y = − 4 3 x + 5 3 y = -\frac43x + \frac53 y = − 3 4 x + 3 5 , so its gradient is
− 4 3 -\frac43 − 3 4 .
The perpendicular gradient is the negative reciprocal,
3 4 \frac34 4 3 .
Through
( − 2 , 3 ) (-2, 3) ( − 2 , 3 ) :
y − 3 = 3 4 ( x + 2 ) y - 3 = \frac34(x + 2) y − 3 = 4 3 ( x + 2 ) .
Multiply by 4:
4 y − 12 = 3 x + 6 4y - 12 = 3x + 6 4 y − 12 = 3 x + 6 .
Collect terms:
3 x − 4 y + 18 = 0 3x - 4y + 18 = 0 3 x − 4 y + 18 = 0 , option B.
Watch out
Flip the gradient and change its sign: − 4 3 -\frac43 − 3 4 + 3 4 +\frac34 + 4 3 3 ( − 2 ) − 4 ( 3 ) + 18 = 0 3(-2) - 4(3) + 18 = 0 3 ( − 2 ) − 4 ( 3 ) + 18 = 0 Report a problem with this question
If sin θ = 12 13 \sin\theta = \frac{12}{13} sin θ = 13 12 θ \theta θ 1 + cos θ 1 + \cos\theta 1 + cos θ
A 5 13 \frac{5}{13} 13 5 B 25 13 \frac{25}{13} 13 25 C 18 13 \frac{18}{13} 13 18 D 8 13 \frac{8}{13} 13 8
Worked solution (try it first) sin θ = 12 13 \sin\theta = \frac{12}{13} sin θ = 13 12 , so draw a right-angled triangle with opposite side 12 and hypotenuse 13.
Pythagoras gives the adjacent side
169 − 144 = 5 \sqrt{169 - 144} = 5 169 − 144 = 5 .
So
cos θ = 5 13 \cos\theta = \frac{5}{13} cos θ = 13 5 .
Add 1:
1 + 5 13 = 18 13 1 + \frac{5}{13} = \frac{18}{13} 1 + 13 5 = 13 18 , option C.
Watch out
Remember the 1: cos θ \cos\theta cos θ 5 13 \frac{5}{13} 13 5 Report a problem with this question
If y = 4 x 3 − 2 x 2 + x y = 4x^3 - 2x^2 + x y = 4 x 3 − 2 x 2 + x d y d x \dfrac{dy}{dx} d x d y
A 12 x 2 − 4 x + 1 12x^2 - 4x + 1 12 x 2 − 4 x + 1 B 8 x 2 − 2 x + 1 8x^2 - 2x + 1 8 x 2 − 2 x + 1 C 8 x 2 − 4 x + 1 8x^2 - 4x + 1 8 x 2 − 4 x + 1 D 12 x 2 − 2 x + 1 12x^2 - 2x + 1 12 x 2 − 2 x + 1
Worked solution (try it first) Differentiate each term: bring the power down and reduce it by one.
4 x 3 4x^3 4 x 3 gives
12 x 2 12x^2 12 x 2 ,
− 2 x 2 -2x^2 − 2 x 2 gives
− 4 x -4x − 4 x , and
x x x gives 1.
So
d y d x = 12 x 2 − 4 x + 1 \frac{dy}{dx} = 12x^2 - 4x + 1 d x d y = 12 x 2 − 4 x + 1 , option A.
Watch out
Multiply every term by its power: − 2 x 2 -2x^2 − 2 x 2 − 4 x -4x − 4 x − 2 x -2x − 2 x Report a problem with this question
If y = cos 3 x y = \cos 3x y = cos 3 x d y d x \dfrac{dy}{dx} d x d y
A − 3 sin 3 x -3\sin 3x − 3 sin 3 x B 1 3 sin 3 x \frac13\sin 3x 3 1 sin 3 x C − 1 3 sin 3 x -\frac13\sin 3x − 3 1 sin 3 x D 3 sin 3 x 3\sin 3x 3 sin 3 x
Worked solution (try it first) Chain rule:
cos 3 x \cos3x cos 3 x differentiates to
− sin 3 x -\sin3x − sin 3 x times the derivative of
3 x 3x 3 x , which is 3.
So
d y d x = − 3 sin 3 x \frac{dy}{dx} = -3\sin3x d x d y = − 3 sin 3 x , option A.
Watch out
Differentiating multiplies by the 3; dividing by 3 is what you do when integrating. That mix-up gives − 1 3 sin 3 x -\frac13\sin3x − 3 1 sin 3 x Report a problem with this question
Find the minimum value of y = x 2 − 2 x − 3 y = x^2 - 2x - 3 y = x 2 − 2 x − 3
Worked solution (try it first) At the minimum
d y d x = 2 x − 2 = 0 \frac{dy}{dx} = 2x - 2 = 0 d x d y = 2 x − 2 = 0 , so
x = 1 x = 1 x = 1 .
Put
x = 1 x = 1 x = 1 into
y y y :
1 − 2 − 3 = − 4 1 - 2 - 3 = -4 1 − 2 − 3 = − 4 .
So the minimum value is
− 4 -4 − 4 , option A.
Watch out
The question asks for the minimum value of y y y x = 1 x = 1 x = 1 Also set as NECO 2023 · Paper 1 · Q23
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Evaluate ∫ sin 2 x d x \displaystyle\int \sin 2x\,dx ∫ sin 2 x d x
A − cos 2 x + k -\cos 2x + k − cos 2 x + k B cos 2 x + k \cos 2x + k cos 2 x + k C 1 2 cos 2 x + k \frac12\cos 2x + k 2 1 cos 2 x + k D − 1 2 cos 2 x + k -\frac12\cos 2x + k − 2 1 cos 2 x + k
Worked solution (try it first) sin \sin sin integrates to
− cos -\cos − cos .
For
sin 2 x \sin2x sin 2 x , also divide by 2, the coefficient of
x x x .
So
∫ sin 2 x d x = − 1 2 cos 2 x + k \int\sin2x\,dx = -\frac12\cos2x + k ∫ sin 2 x d x = − 2 1 cos 2 x + k , option D.
Watch out
Divide by the 2 inside. − cos 2 x -\cos2x − cos 2 x 2 sin 2 x 2\sin2x 2 sin 2 x Report a problem with this question
Evaluate ∫ ( 2 x + 3 ) 1 2 d x \displaystyle\int (2x + 3)^{\frac12}\,dx ∫ ( 2 x + 3 ) 2 1 d x
A − 1 12 ( 2 x + 3 ) 3 4 + k -\frac{1}{12}(2x + 3)^{\frac34} + k − 12 1 ( 2 x + 3 ) 4 3 + k B − 1 12 ( 2 x + 3 ) 6 + k -\frac{1}{12}(2x + 3)^6 + k − 12 1 ( 2 x + 3 ) 6 + k C 1 3 ( 2 x + 3 ) 1 42 + k \frac13(2x + 3)^{\frac{1}{42}} + k 3 1 ( 2 x + 3 ) 42 1 + k D 1 3 ( 2 x + 3 ) 3 2 + k \frac13(2x + 3)^{\frac32} + k 3 1 ( 2 x + 3 ) 2 3 + k
Worked solution (try it first) Add one to the power:
1 2 + 1 = 3 2 \frac12 + 1 = \frac32 2 1 + 1 = 2 3 .
Divide by the new power
3 2 \frac32 2 3 and by 2, the derivative of
2 x + 3 2x + 3 2 x + 3 :
( 2 x + 3 ) 3 / 2 3 2 × 2 = 1 3 ( 2 x + 3 ) 3 2 \dfrac{(2x + 3)^{3/2}}{\frac32 \times 2} = \frac13(2x + 3)^{\frac32} 2 3 × 2 ( 2 x + 3 ) 3/2 = 3 1 ( 2 x + 3 ) 2 3 .
So the integral is
1 3 ( 2 x + 3 ) 3 2 + k \frac13(2x + 3)^{\frac32} + k 3 1 ( 2 x + 3 ) 2 3 + k , option D.
Watch out
When integrating, the power goes up by one (1 2 → 3 2 \frac12 \to \frac32 2 1 → 2 3 ( 2 x + 3 ) 1 2 (2x + 3)^{\frac12} ( 2 x + 3 ) 2 1 Report a problem with this question
The pie chart shows the monthly distribution of a man's salary on food items. If he spent ₦8,000 on rice, how much did he spend on yam?
A ₦42,000 B ₦18,000 C ₦16,000 D ₦12,000
Worked solution (try it first) The angles add up to
360 ∘ 360^\circ 36 0 ∘ , so Yam is
360 ∘ − ( 70 ∘ + 80 ∘ + 50 ∘ ) = 160 ∘ 360^\circ - (70^\circ + 80^\circ + 50^\circ) = 160^\circ 36 0 ∘ − ( 7 0 ∘ + 8 0 ∘ + 5 0 ∘ ) = 16 0 ∘ .
Rice is
80 ∘ 80^\circ 8 0 ∘ for ₦8,000, so each degree is ₦100.
Yam:
160 × 100 = 16 000 160 \times 100 = 16\,000 160 × 100 = 16 000 , so he spent ₦16,000, option C.
Watch out
Find the Yam angle correctly: 360 − 200 = 160 ∘ 360 - 200 = 160^\circ 360 − 200 = 16 0 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question
Values
0
1
2
3
4
Frequency
1
2
2
1
9
Find the mode of the distribution.
Worked solution (try it first) The mode is the value with the highest frequency.
Value 4 has frequency 9, more than any other, so the mode is 4, option A.
Watch out
Give the value, not its frequency: the mode is 4, not 9. And don't pick the value with the lowest frequency. Report a problem with this question
Find the median of 5, 9, 1, 10, 3, 8, 9, 2, 4, 5, 5, 5, 7, 3 and 6.
Worked solution (try it first) Put the 15 numbers in order: 1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 10.
With 15 numbers the median is the
15 + 1 2 = 8 \frac{15 + 1}{2} = 8 2 15 + 1 = 8 th.
The 8th number is 5, so the median is 5, option C.
Watch out
Order the list first: the 8th number as written is 2. Count carefully in the ordered list; 6 (option B) is the 10th. Report a problem with this question
Find the standard deviation of 5, 4, 3, 2, 1.
A 10 \sqrt{10} 10 B 2 \sqrt2 2 C 3 \sqrt3 3 D 6 \sqrt6 6
Worked solution (try it first) The mean of 5, 4, 3, 2, 1 is 3.
The squared deviations are 4, 1, 0, 1, 4, which add up to 10.
The variance is
10 5 = 2 \frac{10}{5} = 2 5 10 = 2 , so the standard deviation is
2 \sqrt2 2 , option B.
Watch out
Divide the sum of squares by 5 before taking the root. 10 \sqrt{10} 10 Report a problem with this question
In how many ways can a team of 3 girls be selected from 7 girls?
A 7 ! 2 ! 5 ! \dfrac{7!}{2!5!} 2 ! 5 ! 7 ! B 7 ! 3 ! \dfrac{7!}{3!} 3 ! 7 ! C 7 ! 4 ! \dfrac{7!}{4!} 4 ! 7 ! D 7 ! 3 ! 4 ! \dfrac{7!}{3!4!} 3 ! 4 ! 7 !
Worked solution (try it first) A team is a selection, so the order doesn't matter: this is
7 C 3 ^7C_3 7 C 3 .
n C r = n ! r ! ( n − r ) ! ^nC_r = \dfrac{n!}{r!\,(n - r)!} n C r = r ! ( n − r )! n ! , so
7 C 3 = 7 ! 3 ! 4 ! ^7C_3 = \dfrac{7!}{3!\,4!} 7 C 3 = 3 ! 4 ! 7 ! , option D.
Watch out
7 ! 4 ! \frac{7!}{4!} 4 ! 7 ! 7 P 3 ^7P_3 7 P 3 3 ! 3! 3 ! Report a problem with this question
Number
1
2
3
4
5
6
Frequency
18
22
20
16
10
14
The table represents the outcome of throwing a die 100 times. What is the probability of obtaining at least a 4?
A 3 4 \frac34 4 3 B 1 5 \frac15 5 1 C 1 2 \frac12 2 1 D 2 5 \frac25 5 2
Worked solution (try it first) The die was thrown 100 times.
At least 4 means 4, 5 or 6:
16 + 10 + 14 = 40 16 + 10 + 14 = 40 16 + 10 + 14 = 40 throws.
So the probability is
40 100 = 2 5 \frac{40}{100} = \frac25 100 40 = 5 2 , option D.
Watch out
Use the frequencies in the table, not the fair-die value 3 6 = 1 2 \frac36 = \frac12 6 3 = 2 1 Report a problem with this question
A number is chosen at random from 10 to 30, both inclusive. What is the probability that the number is divisible by 3?
A 3 5 \frac35 5 3 B 2 15 \frac{2}{15} 15 2 C 1 10 \frac{1}{10} 10 1 D 1 3 \frac13 3 1
Worked solution (try it first) From 10 to 30 inclusive there are
30 − 10 + 1 = 21 30 - 10 + 1 = 21 30 − 10 + 1 = 21 numbers.
The multiples of 3 are 12, 15, 18, 21, 24, 27, 30, which is 7 numbers.
So the probability is
7 21 = 1 3 \frac{7}{21} = \frac13 21 7 = 3 1 , option D.
Watch out
Include both ends: there are 21 numbers, not 20. Using 20 gives 7 20 \frac{7}{20} 20 7 Report a problem with this question