Paper JAMB 2001 General Maths Objective
Objective paper · 42 questions · partial
JAMB 2001 · UME Topics include Commercial arithmetic, Approximation & error, Surds, Expressions, formulae & change of subject, Linear & simultaneous equations, Quadratics & their graphs.
Our copy of this paper is missing questions 1, 7, 8, 9, 13, 15, 20, 35.
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.
2 3 4 5 6 10 11 12 14 16 17 18 19 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 A car dealer bought a second-hand car for ₦250,000.00 and spent ₦70,000.00 refurbishing it. He then sold the car for ₦400,000.00. What is the percentage gain?
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 B.
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 2017 · UTME (set 2) · Q42
Report a problem with this question
Evaluate 21.05347 − 1.6324 × 0.43 21.05347 - 1.6324 \times 0.43 21.05347 − 1.6324 × 0.43
A 20.351 B 20.352 C 20.980 D 20.981
Worked solution (try it first) Multiply before subtracting:
1.6324 × 0.43 = 0.701932 1.6324 \times 0.43 = 0.701932 1.6324 × 0.43 = 0.701932 .
Subtract:
21.05347 − 0.701932 = 20.351538 21.05347 - 0.701932 = 20.351538 21.05347 − 0.701932 = 20.351538 .
To 3 decimal places, the fourth decimal is 5, so round up: 20.352, option B.
Watch out
Round, don't chop: 20.3515… has 5 in the fourth decimal place, so it rounds up to 20.352. Cutting it off gives 20.351 (option A). Report a problem with this question
Evaluate ( 0.14 ) 2 × 0.275 7 × 0.02 \dfrac{(0.14)^2 \times 0.275}{7 \times 0.02} 7 × 0.02 ( 0.14 ) 2 × 0.275
Worked solution (try it first) Square:
( 0.14 ) 2 = 0.0196 (0.14)^2 = 0.0196 ( 0.14 ) 2 = 0.0196 .
Then
0.0196 × 0.275 = 0.00539 0.0196 \times 0.275 = 0.00539 0.0196 × 0.275 = 0.00539 .
Bottom:
7 × 0.02 = 0.14 7 \times 0.02 = 0.14 7 × 0.02 = 0.14 .
Divide:
0.00539 ÷ 0.14 = 0.0385 0.00539 \div 0.14 = 0.0385 0.00539 ÷ 0.14 = 0.0385 .
The fourth decimal is 5, so round up: 0.039, option B.
Watch out
( 0.14 ) 2 = 0.0196 (0.14)^2 = 0.0196 ( 0.14 ) 2 = 0.0196 Report a problem with this question
Given that p = 1 + 2 p = 1 + \sqrt2 p = 1 + 2 q = 1 − 2 q = 1 - \sqrt2 q = 1 − 2 p 2 − q 2 2 p q \dfrac{p^2 - q^2}{2pq} 2 pq p 2 − q 2
A − 2 ( 2 + 2 ) -2(2 + \sqrt2) − 2 ( 2 + 2 ) B 2 ( 2 + 2 ) 2(2 + \sqrt2) 2 ( 2 + 2 ) C − 2 2 -2\sqrt2 − 2 2 D 2 2 2\sqrt2 2 2
Worked solution (try it first) Factorise the top as a difference of two squares:
p 2 − q 2 = ( p − q ) ( p + q ) p^2 - q^2 = (p - q)(p + q) p 2 − q 2 = ( p − q ) ( p + q ) .
p − q = 2 2 p - q = 2\sqrt2 p − q = 2 2 and
p + q = 2 p + q = 2 p + q = 2 , so
p 2 − q 2 = 4 2 p^2 - q^2 = 4\sqrt2 p 2 − q 2 = 4 2 .
p q = ( 1 + 2 ) ( 1 − 2 ) = 1 − 2 pq = (1 + \sqrt2)(1 - \sqrt2) = 1 - 2 pq = ( 1 + 2 ) ( 1 − 2 ) = 1 − 2 , which is
− 1 -1 − 1 , so
2 p q = − 2 2pq = -2 2 pq = − 2 .
So the value is
4 2 − 2 = − 2 2 \dfrac{4\sqrt2}{-2} = -2\sqrt2 − 2 4 2 = − 2 2 , option C.
Watch out
p q = 1 − ( 2 ) 2 = 1 − 2 = − 1 pq = 1 - (\sqrt2)^2 = 1 - 2 = -1 pq = 1 − ( 2 ) 2 = 1 − 2 = − 1 + 1 +1 + 1 2 2 2\sqrt2 2 2 Report a problem with this question
If y 2 = x \frac y2 = x 2 y = x x 3 y 3 + 1 2 1 2 − x 2 y 2 \dfrac{\frac{x^3}{y^3} + \frac12}{\frac12 - \frac{x^2}{y^2}} 2 1 − y 2 x 2 y 3 x 3 + 2 1
A 5 16 \frac5{16} 16 5 B 5 8 \frac58 8 5 C 5 4 \frac54 4 5 D 5 2 \frac52 2 5
Worked solution (try it first) From
y 2 = x \frac y2 = x 2 y = x ,
y = 2 x y = 2x y = 2 x , so
x y = 1 2 \frac xy = \frac12 y x = 2 1 .
Top:
( 1 2 ) 3 + 1 2 = 1 8 + 1 2 \left(\frac12\right)^3 + \frac12 = \frac18 + \frac12 ( 2 1 ) 3 + 2 1 = 8 1 + 2 1 Bottom:
1 2 − ( 1 2 ) 2 = 1 4 \frac12 - \left(\frac12\right)^2 = \frac14 2 1 − ( 2 1 ) 2 = 4 1 .
Divide:
5 8 ÷ 1 4 = 5 8 × 4 \frac58 \div \frac14 = \frac58 \times 4 8 5 ÷ 4 1 = 8 5 × 4 = 5 2 = \frac52 = 2 5 , option D.
Watch out
5 8 \frac58 8 5 1 4 \frac14 4 1 Report a problem with this question
Solve the equations m 2 + n 2 = 29 m^2 + n^2 = 29 m 2 + n 2 = 29 m + n = 7 m + n = 7 m + n = 7
A ( 5 , 2 ) (5, 2) ( 5 , 2 ) ( 5 , 3 ) (5, 3) ( 5 , 3 ) B ( 5 , 3 ) (5, 3) ( 5 , 3 ) ( 3 , 5 ) (3, 5) ( 3 , 5 ) C ( 2 , 3 ) (2, 3) ( 2 , 3 ) ( 3 , 5 ) (3, 5) ( 3 , 5 ) D ( 2 , 5 ) (2, 5) ( 2 , 5 ) ( 5 , 2 ) (5, 2) ( 5 , 2 )
Worked solution (try it first) Square the sum:
( m + n ) 2 = m 2 + n 2 + 2 m n (m + n)^2 = m^2 + n^2 + 2mn ( m + n ) 2 = m 2 + n 2 + 2 mn , so
49 = 29 + 2 m n 49 = 29 + 2mn 49 = 29 + 2 mn .
So
2 m n = 20 2mn = 20 2 mn = 20 and
m n = 10 mn = 10 mn = 10 .
Two numbers that add to 7 and multiply to 10 are 2 and 5.
So
( m , n ) = ( 2 , 5 ) (m, n) = (2, 5) ( m , n ) = ( 2 , 5 ) or
( 5 , 2 ) (5, 2) ( 5 , 2 ) , option D.
Watch out
Test every pair in both equations. ( 5 , 3 ) (5, 3) ( 5 , 3 ) 5 2 + 3 2 = 34 5^2 + 3^2 = 34 5 2 + 3 2 = 34 Report a problem with this question
Divide a 3 x − 26 a 2 x + 156 a x − 216 a^{3x} - 26a^{2x} + 156a^x - 216 a 3 x − 26 a 2 x + 156 a x − 216 a 2 x − 24 a x + 108 a^{2x} - 24a^x + 108 a 2 x − 24 a x + 108
A a x − 18 a^x - 18 a x − 18 B a x − 6 a^x - 6 a x − 6 C a x − 2 a^x - 2 a x − 2 D a x + 2 a^x + 2 a x + 2
Worked solution (try it first) Then
a 2 x = u 2 a^{2x} = u^2 a 2 x = u 2 and
a 3 x = u 3 a^{3x} = u^3 a 3 x = u 3 , so divide
u 3 − 26 u 2 + 156 u − 216 u^3 - 26u^2 + 156u - 216 u 3 − 26 u 2 + 156 u − 216 by
u 2 − 24 u + 108 u^2 - 24u + 108 u 2 − 24 u + 108 .
u 3 ÷ u 2 = u u^3 \div u^2 = u u 3 ÷ u 2 = u .
Subtract
u ( u 2 − 24 u + 108 ) = u 3 − 24 u 2 + 108 u u(u^2 - 24u + 108) = u^3 - 24u^2 + 108u u ( u 2 − 24 u + 108 ) = u 3 − 24 u 2 + 108 u to leave
− 2 u 2 + 48 u − 216 -2u^2 + 48u - 216 − 2 u 2 + 48 u − 216 .
− 2 u 2 ÷ u 2 = − 2 -2u^2 \div u^2 = -2 − 2 u 2 ÷ u 2 = − 2 , and
− 2 ( u 2 − 24 u + 108 ) -2(u^2 - 24u + 108) − 2 ( u 2 − 24 u + 108 ) is exactly that, so the remainder is 0.
The quotient is
u − 2 = a x − 2 u - 2 = a^x - 2 u − 2 = a x − 2 , option C.
Watch out
− 26 u 2 − ( − 24 u 2 ) = − 2 u 2 -26u^2 - (-24u^2) = -2u^2 − 26 u 2 − ( − 24 u 2 ) = − 2 u 2 − 2 -2 − 2 + 2 +2 + 2 Report a problem with this question
Find the integral values of x x x y y y 3 y + 5 x ≤ 15 3y + 5x \le 15 3 y + 5 x ≤ 15 y > 0 y > 0 y > 0 y < 3 y < 3 y < 3 x > 0 x > 0 x > 0
A ( 1 , 1 ) , ( 2 , 1 ) , ( 1 , 3 ) (1, 1), (2, 1), (1, 3) ( 1 , 1 ) , ( 2 , 1 ) , ( 1 , 3 ) B ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) (1, 1), (1, 2), (1, 3) ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) C ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) (1, 1), (1, 2), (2, 1) ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) D ( 1 , 1 ) , ( 3 , 1 ) , ( 2 , 2 ) (1, 1), (3, 1), (2, 2) ( 1 , 1 ) , ( 3 , 1 ) , ( 2 , 2 )
Worked solution (try it first) y y y is a whole number with
0 < y < 3 0 < y < 3 0 < y < 3 , so
y = 1 y = 1 y = 1 or
2 2 2 .
Also
x x x is a whole number with
x > 0 x > 0 x > 0 .
Test the points in
3 y + 5 x 3y + 5x 3 y + 5 x :
( 1 , 1 ) (1, 1) ( 1 , 1 ) gives 8,
( 1 , 2 ) (1, 2) ( 1 , 2 ) gives 11 and
( 2 , 1 ) (2, 1) ( 2 , 1 ) gives 13, all
≤ 15 \le 15 ≤ 15 .
Other points fail:
( 2 , 2 ) (2, 2) ( 2 , 2 ) gives 16 and
( 3 , 1 ) (3, 1) ( 3 , 1 ) gives 18.
So the points are
( 1 , 1 ) (1, 1) ( 1 , 1 ) ,
( 1 , 2 ) (1, 2) ( 1 , 2 ) and
( 2 , 1 ) (2, 1) ( 2 , 1 ) , option C.
Watch out
y < 3 y < 3 y < 3 y = 3 y = 3 y = 3 ( 1 , 3 ) (1, 3) ( 1 , 3 ) 3 y + 5 x ≤ 15 3y + 5x \le 15 3 y + 5 x ≤ 15 y < 3 y < 3 y < 3 Report a problem with this question
The sixth term of an arithmetic progression is half of its twelfth term. The first term is equal to
A half of the common difference B double the common difference C the common difference D zero
Worked solution (try it first) The 6th term is
a + 5 d a + 5d a + 5 d and the 12th term is
a + 11 d a + 11d a + 11 d .
The 6th is half the 12th, so twice the 6th equals the 12th:
2 a + 10 d = a + 11 d 2a + 10d = a + 11d 2 a + 10 d = a + 11 d .
Take
a + 10 d a + 10d a + 10 d from both sides:
a = d a = d a = d .
The first term equals the common difference, option C.
Watch out
The n n n a + ( n − 1 ) d a + (n - 1)d a + ( n − 1 ) d a + 6 d a + 6d a + 6 d a + 12 d a + 12d a + 12 d a = 0 a = 0 a = 0 Report a problem with this question
An operation ∗ * ∗ a ∗ b = a + b + 1 a * b = a + b + 1 a ∗ b = a + b + 1 − 1 -1 − 1
Worked solution (try it first) The inverse
b b b of 2 combines with 2 to give the identity
− 1 -1 − 1 :
2 ∗ b = − 1 2 * b = -1 2 ∗ b = − 1 .
Use the rule:
2 + b + 1 = − 1 2 + b + 1 = -1 2 + b + 1 = − 1 , so
b + 3 = − 1 b + 3 = -1 b + 3 = − 1 .
Take 3 from both sides:
b = − 4 b = -4 b = − 4 , option A.
Watch out
Set the result equal to the identity, − 1 -1 − 1 2 + b + 1 = 0 2 + b + 1 = 0 2 + b + 1 = 0 b = − 3 b = -3 b = − 3 Report a problem with this question
The identity element with respect to the multiplication shown in the table is
⊗ \otimes ⊗ k k k l l l m m m
k k k l l l m m m k k k
l l l m m m k k k l l l
m m m k k k l l l m m m
Worked solution (try it first) The identity leaves every element unchanged, so its row and column must repeat the headings
k , l , m k, l, m k , l , m .
The row for
m m m reads
k , l , m k, l, m k , l , m , and so does the column under
m m m .
So the identity is
m m m , option C.
Watch out
Don't pick the first element by habit: the row for k k k l , m , k l, m, k l , m , k k ⊗ k = l k \otimes k = l k ⊗ k = l k k k Report a problem with this question
Given the matrix K = ( 2 1 3 4 ) K = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} K = ( 2 3 1 4 ) K 2 + K + I K^2 + K + I K 2 + K + I I I I 2 × 2 2 \times 2 2 × 2
A ( 9 8 22 23 ) \begin{pmatrix} 9 & 8 \\ 22 & 23 \end{pmatrix} ( 9 22 8 23 ) B ( 10 7 21 24 ) \begin{pmatrix} 10 & 7 \\ 21 & 24 \end{pmatrix} ( 10 21 7 24 ) C ( 7 2 12 21 ) \begin{pmatrix} 7 & 2 \\ 12 & 21 \end{pmatrix} ( 7 12 2 21 ) D ( 6 3 13 20 ) \begin{pmatrix} 6 & 3 \\ 13 & 20 \end{pmatrix} ( 6 13 3 20 )
Worked solution (try it first) K 2 K^2 K 2 means
K × K K \times K K × K , row by column:
K 2 = ( 4 + 3 2 + 4 6 + 12 3 + 16 ) K^2 = \begin{pmatrix} 4 + 3 & 2 + 4 \\ 6 + 12 & 3 + 16 \end{pmatrix} K 2 = ( 4 + 3 6 + 12 2 + 4 3 + 16 ) = ( 7 6 18 19 ) = \begin{pmatrix} 7 & 6 \\ 18 & 19 \end{pmatrix} = ( 7 18 6 19 ) .
Add
K K K entry by entry:
( 9 7 21 23 ) \begin{pmatrix} 9 & 7 \\ 21 & 23 \end{pmatrix} ( 9 21 7 23 ) .
I I I adds 1 to each diagonal entry only:
( 10 7 21 24 ) \begin{pmatrix} 10 & 7 \\ 21 & 24 \end{pmatrix} ( 10 21 7 24 ) , option B.
Watch out
K 2 K^2 K 2 ( 4 1 9 16 ) \begin{pmatrix} 4 & 1 \\ 9 & 16 \end{pmatrix} ( 4 9 1 16 ) Report a problem with this question
Evaluate ∣ − 1 − 1 − 1 3 1 1 1 2 1 ∣ \begin{vmatrix} -1 & -1 & -1 \\ 3 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} − 1 3 1 − 1 1 2 − 1 1 1
A 4 B − 2 -2 − 2 C − 4 -4 − 4 D − 12 -12 − 12
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The minor of the first entry is
1 × 1 − 1 × 2 = − 1 1 \times 1 - 1 \times 2 = -1 1 × 1 − 1 × 2 = − 1 , so the first term is
( − 1 ) ( − 1 ) = 1 (-1)(-1) = 1 ( − 1 ) ( − 1 ) = 1 .
The minor of the second entry is
3 × 1 − 1 × 1 = 2 3 \times 1 - 1 \times 1 = 2 3 × 1 − 1 × 1 = 2 .
With the minus sign, the term is
− ( − 1 ) ( 2 ) = 2 -(-1)(2) = 2 − ( − 1 ) ( 2 ) = 2 .
The third term is
− 1 × ( 3 × 2 − 1 × 1 ) = − 5 -1 \times (3 \times 2 - 1 \times 1) = -5 − 1 × ( 3 × 2 − 1 × 1 ) = − 5 .
Add them:
1 + 2 − 5 = − 2 1 + 2 - 5 = -2 1 + 2 − 5 = − 2 , option B.
Watch out
The first entry is negative and its minor is 1 − 2 = − 1 1 - 2 = -1 1 − 2 = − 1 + 1 +1 + 1 − 1 -1 − 1 − 1 + 2 − 5 = − 4 -1 + 2 - 5 = -4 − 1 + 2 − 5 = − 4 Report a problem with this question
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
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 C.
Watch out
It is the interior angle that is twice the exterior. The other way round gives e = 120 ∘ e = 120^\circ e = 12 0 ∘ Also set as JAMB 2016 · UTME · Q40
Report a problem with this question
In the figure, P Q R PQR P QR P Q = Q T PQ = QT P Q = QT P Q T PQT P QT ∠ S R Q = 75 ∘ \angle SRQ = 75^\circ ∠ S R Q = 7 5 ∘ ∠ Q P T = 25 ∘ \angle QPT = 25^\circ ∠ QP T = 2 5 ∘ ∠ R S T \angle RST ∠ R S T
A 25 ∘ 25^\circ 2 5 ∘ B 45 ∘ 45^\circ 4 5 ∘ C 50 ∘ 50^\circ 5 0 ∘ D 55 ∘ 55^\circ 5 5 ∘
Worked solution (try it first) P Q = Q T PQ = QT P Q = QT , so the base angles of triangle
P Q T PQT P QT are equal:
∠ Q T P = 25 ∘ \angle QTP = 25^\circ ∠ QT P = 2 5 ∘ .
The exterior angle at
Q Q Q equals the sum of the two interior opposite angles:
∠ T Q R = 25 ∘ + 25 ∘ \angle TQR = 25^\circ + 25^\circ ∠ T QR = 2 5 ∘ + 2 5 ∘ Q Q Q ,
T T T and
S S S are in a straight line, so triangle
Q R S QRS QR S has angles
50 ∘ 50^\circ 5 0 ∘ at
Q Q Q and
75 ∘ 75^\circ 7 5 ∘ at
R R R .
Angles in a triangle add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ R S T = 180 ∘ − 50 ∘ − 75 ∘ \angle RST = 180^\circ - 50^\circ - 75^\circ ∠ R S T = 18 0 ∘ − 5 0 ∘ − 7 5 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option D.
Watch out
50 ∘ 50^\circ 5 0 ∘ ∠ T Q R \angle TQR ∠ T QR Q R S QRS QR S S S S Report a problem with this question
A cylindrical tank has a capacity of 3080 m 3 3080\text{ m}^3 3080 m 3 [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
Worked solution (try it first) Radius
= 14 ÷ 2 = 7 = 14 \div 2 = 7 = 14 ÷ 2 = 7 m, so the base area is
22 7 × 49 = 154 m 2 \frac{22}{7} \times 49 = 154\text{ m}^2 7 22 × 49 = 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 A.
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 2017 · UTME (set 2) · Q40
Report a problem with this question
A sector of a circle of radius 7.2 cm which subtends an angle of 300 ∘ 300^\circ 30 0 ∘
Worked solution (try it first) The arc of the sector becomes the circumference of the base:
300 360 × 2 π × 7.2 = 2 π r \frac{300}{360} \times 2\pi \times 7.2 = 2\pi r 360 300 × 2 π × 7.2 = 2 π r .
Divide both sides by
2 π 2\pi 2 π :
r = 7.2 × 5 6 = 6 r = 7.2 \times \frac56 = 6 r = 7.2 × 6 5 = 6 cm, option A.
Watch out
The 7.2 cm becomes the slant height of the cone, not its base radius. The base must be smaller, so multiply by 300 360 \frac{300}{360} 360 300 360 300 \frac{360}{300} 300 360 Report a problem with this question
The chord S T ST S T r r r S T ST S T
A π r 2 \frac{\pi r}{2} 2 π r B π r 3 \frac{\pi r}{3} 3 π r C π r 6 \frac{\pi r}{6} 6 π r D π r 12 \frac{\pi r}{12} 12 π r
Worked solution (try it first) The chord and the two radii to
S S S and
T T T are all
r r r , so that triangle is equilateral and the arc makes
60 ∘ 60^\circ 6 0 ∘ at the centre.
Arc length is
θ 360 ∘ × 2 π r \frac{\theta}{360^\circ} \times 2\pi r 36 0 ∘ θ × 2 π r , so the arc is
60 360 × 2 π r \frac{60}{360} \times 2\pi r 360 60 × 2 π r .
That simplifies to
π r 3 \frac{\pi r}{3} 3 π r , option B.
Watch out
Use the whole circumference 2 π r 2\pi r 2 π r 60 360 \frac{60}{360} 360 60 π r \pi r π r π r 6 \frac{\pi r}{6} 6 π r Report a problem with this question
A point P P P Q Q Q R R R Q R QR QR P R = 8 PR = 8 P R = 8 ∠ P R Q = 30 ∘ \angle PRQ = 30^\circ ∠ P R Q = 3 0 ∘
A 4 cm B 4 3 4\sqrt3 4 3 C 8 cm D 8 3 8\sqrt3 8 3
Worked solution (try it first) P P P is equidistant from
Q Q Q and
R R R , so
P Q = P R = 8 PQ = PR = 8 P Q = P R = 8 cm and triangle
P Q R PQR P QR is isosceles with
∠ P Q R = ∠ P R Q = 30 ∘ \angle PQR = \angle PRQ = 30^\circ ∠ P QR = ∠ P R Q = 3 0 ∘ .
The perpendicular from
P P P meets
Q R QR QR at its mid-point.
Half of
Q R QR QR is
8 cos 30 ∘ = 8 × 3 2 8\cos30^\circ = 8 \times \frac{\sqrt3}{2} 8 cos 3 0 ∘ = 8 × 2 3 So
Q R = 2 × 4 3 = 8 3 QR = 2 \times 4\sqrt3 = 8\sqrt3 QR = 2 × 4 3 = 8 3 cm, option D.
Watch out
8 cos 30 ∘ 8\cos30^\circ 8 cos 3 0 ∘ Q R QR QR 4 3 4\sqrt3 4 3 Report a problem with this question
Find the locus of a point which moves such that its distance from the line y = 4 y = 4 y = 4 k k k
A y = 4 + k y = 4 + k y = 4 + k B y = k − 4 y = k - 4 y = k − 4 C y = k ± 4 y = k \pm 4 y = k ± 4 D y = 4 ± k y = 4 \pm k y = 4 ± k
Worked solution (try it first) y = 4 y = 4 y = 4 is a horizontal line.
Points a distance
k k k from it lie on two horizontal lines, one above and one below.
The line above is
y = 4 + k y = 4 + k y = 4 + k and the line below is
y = 4 − k y = 4 - k y = 4 − k .
So the locus is
y = 4 ± k y = 4 \pm k y = 4 ± k , option D.
Watch out
Remember the line below as well as the line above. y = 4 + k y = 4 + k y = 4 + k 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 = \frac1{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 2017 · UTME · Q32
Report a problem with this question
P ( − 6 , 1 ) P(-6, 1) P ( − 6 , 1 ) Q ( 6 , 6 ) Q(6, 6) Q ( 6 , 6 )
A 3.5 units B 6.5 units C 7.0 units D 13.0 units
Worked solution (try it first) The changes from
P P P to
Q Q Q are
6 − ( − 6 ) = 12 6 - (-6) = 12 6 − ( − 6 ) = 12 and
6 − 1 = 5 6 - 1 = 5 6 − 1 = 5 .
The diameter is
P Q = 12 2 + 5 2 = 169 = 13 PQ = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 P Q = 1 2 2 + 5 2 = 169 = 13 .
The radius is half the diameter:
6.5 6.5 6.5 units, option B.
Watch out
P Q PQ P Q Report a problem with this question
Find the value of p p p ( p , 4 ) (p, 4) ( p , 4 ) ( 6 , − 2 ) (6, -2) ( 6 , − 2 ) ( 2 , p ) (2, p) ( 2 , p ) ( − 1 , 3 ) (-1, 3) ( − 1 , 3 )
Worked solution (try it first) Gradient of the first line:
− 2 − 4 6 − p = − 6 6 − p \dfrac{-2 - 4}{6 - p} = \dfrac{-6}{6 - p} 6 − p − 2 − 4 = 6 − p − 6 .
Gradient of the second line:
3 − p − 1 − 2 = p − 3 3 \dfrac{3 - p}{-1 - 2} = \dfrac{p - 3}{3} − 1 − 2 3 − p = 3 p − 3 .
Perpendicular gradients multiply to
− 1 -1 − 1 :
− 6 ( p − 3 ) 3 ( 6 − p ) = − 1 \dfrac{-6(p - 3)}{3(6 - p)} = -1 3 ( 6 − p ) − 6 ( p − 3 ) = − 1 , so
2 ( p − 3 ) = 6 − p 2(p - 3) = 6 - p 2 ( p − 3 ) = 6 − p .
Then
3 p = 12 3p = 12 3 p = 12 , so
p = 4 p = 4 p = 4 , option C.
Watch out
For perpendicular lines the product of the gradients is − 1 -1 − 1 p 2 − 9 p = 0 p^2 - 9p = 0 p 2 − 9 p = 0 p = 0 p = 0 p = 0 Also set as JAMB 2016 · UTME · Q39
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The bearings of P P P Q Q Q N N N 020 ∘ 020^\circ 02 0 ∘ 300 ∘ 300^\circ 30 0 ∘ P P P Q Q Q N N N P P P Q Q Q
A 320 ∘ 320^\circ 32 0 ∘ B 280 ∘ 280^\circ 28 0 ∘ C 070 ∘ 070^\circ 07 0 ∘ D 040 ∘ 040^\circ 04 0 ∘
Worked solution (try it first) The angle between the bearings
300 ∘ 300^\circ 30 0 ∘ and
020 ∘ 020^\circ 02 0 ∘ (through north) is
60 ∘ + 20 ∘ = 80 ∘ 60^\circ + 20^\circ = 80^\circ 6 0 ∘ + 2 0 ∘ = 8 0 ∘ .
N P = N Q NP = NQ N P = N Q , so the base angles of triangle
N P Q NPQ N P Q are
180 ∘ − 80 ∘ 2 = 50 ∘ \frac{180^\circ - 80^\circ}{2} = 50^\circ 2 18 0 ∘ − 8 0 ∘ = 5 0 ∘ .
The bearing of
N N N from
Q Q Q is the back bearing
300 ∘ − 180 ∘ = 120 ∘ 300^\circ - 180^\circ = 120^\circ 30 0 ∘ − 18 0 ∘ = 12 0 ∘ .
P P P is
50 ∘ 50^\circ 5 0 ∘ anticlockwise from that direction, so the bearing of
P P P from
Q Q Q is
120 ∘ − 50 ∘ = 070 ∘ 120^\circ - 50^\circ = 070^\circ 12 0 ∘ − 5 0 ∘ = 07 0 ∘ , option C.
Watch out
∠ P N Q \angle PNQ ∠ P N Q 80 ∘ 80^\circ 8 0 ∘ 300 ∘ − 20 ∘ = 280 ∘ 300^\circ - 20^\circ = 280^\circ 30 0 ∘ − 2 0 ∘ = 28 0 ∘ Report a problem with this question
Find the value of θ \theta θ t t t t t t 3 t \sqrt3t 3 t
A 30 ∘ 30^\circ 3 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 100 ∘ 100^\circ 10 0 ∘ D 120 ∘ 120^\circ 12 0 ∘
Worked solution (try it first) θ \theta θ is between the two sides
t t t , and the base
3 t \sqrt3t 3 t faces it.
Cosine rule:
cos θ = t 2 + t 2 − ( 3 t ) 2 2 × t × t \cos\theta = \dfrac{t^2 + t^2 - (\sqrt3t)^2}{2 \times t \times t} cos θ = 2 × t × t t 2 + t 2 − ( 3 t ) 2 .
( 3 t ) 2 = 3 t 2 (\sqrt3t)^2 = 3t^2 ( 3 t ) 2 = 3 t 2 , so
cos θ = − t 2 2 t 2 \cos\theta = \dfrac{-t^2}{2t^2} cos θ = 2 t 2 − t 2 The cosine is negative, so
θ \theta θ is obtuse:
θ = 180 ∘ − 60 ∘ = 120 ∘ \theta = 180^\circ - 60^\circ = 120^\circ θ = 18 0 ∘ − 6 0 ∘ = 12 0 ∘ , option D.
Watch out
Keep the minus sign: cos θ = − 1 2 \cos\theta = -\frac12 cos θ = − 2 1 120 ∘ 120^\circ 12 0 ∘ 60 ∘ 60^\circ 6 0 ∘ t t t 3 t \sqrt3t 3 t Report a problem with this question
Differentiate ( 2 x + 5 ) 2 ( x − 4 ) (2x + 5)^2(x - 4) ( 2 x + 5 ) 2 ( x − 4 ) x x x
A ( 2 x + 5 ) ( 6 x − 11 ) (2x + 5)(6x - 11) ( 2 x + 5 ) ( 6 x − 11 ) B ( 2 x + 5 ) ( 2 x − 13 ) (2x + 5)(2x - 13) ( 2 x + 5 ) ( 2 x − 13 ) C 4 ( 2 x + 5 ) ( x − 4 ) 4(2x + 5)(x - 4) 4 ( 2 x + 5 ) ( x − 4 ) D 4 ( 2 x + 5 ) ( 4 x − 3 ) 4(2x + 5)(4x - 3) 4 ( 2 x + 5 ) ( 4 x − 3 )
Worked solution (try it first) Product rule with
u = ( 2 x + 5 ) 2 u = (2x + 5)^2 u = ( 2 x + 5 ) 2 and
v = x − 4 v = x - 4 v = x − 4 .
By the chain rule,
u ′ = 2 ( 2 x + 5 ) × 2 = 4 ( 2 x + 5 ) u' = 2(2x + 5) \times 2 = 4(2x + 5) u ′ = 2 ( 2 x + 5 ) × 2 = 4 ( 2 x + 5 ) , and
v ′ = 1 v' = 1 v ′ = 1 .
So
d y d x = 4 ( 2 x + 5 ) ( x − 4 ) + ( 2 x + 5 ) 2 \frac{dy}{dx} = 4(2x + 5)(x - 4) + (2x + 5)^2 d x d y = 4 ( 2 x + 5 ) ( x − 4 ) + ( 2 x + 5 ) 2 .
Take out the common factor
2 x + 5 2x + 5 2 x + 5 :
( 2 x + 5 ) ( 4 x − 16 + 2 x + 5 ) = ( 2 x + 5 ) ( 6 x − 11 ) (2x + 5)(4x - 16 + 2x + 5) = (2x + 5)(6x - 11) ( 2 x + 5 ) ( 4 x − 16 + 2 x + 5 ) = ( 2 x + 5 ) ( 6 x − 11 ) , option A.
Watch out
The product rule has two terms. 4 ( 2 x + 5 ) ( x − 4 ) 4(2x + 5)(x - 4) 4 ( 2 x + 5 ) ( x − 4 ) u v ′ = ( 2 x + 5 ) 2 u v' = (2x + 5)^2 u v ′ = ( 2 x + 5 ) 2 Report a problem with this question
If y = x sin x y = x\sin x y = x sin x d y d x \frac{dy}{dx} d x d y x = π 2 x = \frac\pi2 x = 2 π
A π 2 \frac\pi2 2 π B 1 C − 1 -1 − 1 D π 2 − 2 \frac\pi2 - 2 2 π − 2
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 B.
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 2017 · UTME (set 2) · Q7
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Find the rate of change of the volume V V V r r r r = 1 r = 1 r = 1
A 4 π 4\pi 4 π B 8 π 8\pi 8 π C 12 π 12\pi 12 π D 24 π 24\pi 24 π
Worked solution (try it first) The volume of a sphere is
V = 4 3 π r 3 V = \frac43\pi r^3 V = 3 4 π r 3 .
Differentiate:
d V d r = 3 × 4 3 π r 2 \frac{dV}{dr} = 3 \times \frac43\pi r^2 d r d V = 3 × 3 4 π r 2 At
r = 1 r = 1 r = 1 :
d V d r = 4 π \frac{dV}{dr} = 4\pi d r d V = 4 π , option A.
Watch out
Differentiate the volume, not the surface area. 4 π r 2 4\pi r^2 4 π r 2 8 π r 8\pi r 8 π r 8 π 8\pi 8 π Report a problem with this question
Find the dimensions of the rectangle of greatest area which has a fixed perimeter p p p
A Square of sides p 4 \frac p4 4 p B Square of sides p 2 \frac p2 2 p C Square of sides p p p D Square of sides 2 p 2p 2 p
Worked solution (try it first) Let the sides be
x x x and
y y y .
The perimeter is
2 x + 2 y = p 2x + 2y = p 2 x + 2 y = p , so
y = p 2 − x y = \frac p2 - x y = 2 p − x and the area is
A = x ( p 2 − x ) A = x\left(\frac p2 - x\right) A = x ( 2 p − x ) = p 2 x − x 2 = \frac p2x - x^2 = 2 p x − x 2 .
At the maximum
d A d x = p 2 − 2 x = 0 \frac{dA}{dx} = \frac p2 - 2x = 0 d x d A = 2 p − 2 x = 0 , so
x = p 4 x = \frac p4 x = 4 p .
Then
y = p 2 − p 4 = p 4 y = \frac p2 - \frac p4 = \frac p4 y = 2 p − 4 p = 4 p too: a square of sides
p 4 \frac p4 4 p , option A.
Watch out
A rectangle has two lengths and two widths, so one length and one width add up to p 2 \frac p2 2 p p p p p 2 \frac p2 2 p 2 p 2p 2 p Report a problem with this question
Evaluate ∫ 2 ( 2 x − 3 ) 2 3 d x \displaystyle\int 2(2x - 3)^{\frac23}\,dx ∫ 2 ( 2 x − 3 ) 3 2 d x
A 2 x − 3 + k 2x - 3 + k 2 x − 3 + k B 2 ( 2 x − 3 ) + k 2(2x - 3) + k 2 ( 2 x − 3 ) + k C 6 5 ( 2 x − 3 ) 5 3 + k \frac65(2x - 3)^{\frac53} + k 5 6 ( 2 x − 3 ) 3 5 + k D 3 5 ( 2 x − 3 ) 5 3 + k \frac35(2x - 3)^{\frac53} + k 5 3 ( 2 x − 3 ) 3 5 + k
Worked solution (try it first) Add one to the power:
2 3 + 1 = 5 3 \frac23 + 1 = \frac53 3 2 + 1 = 3 5 .
Divide by the new power
5 3 \frac53 3 5 and by 2, the derivative of
2 x − 3 2x - 3 2 x − 3 .
So the integral is
2 × ( 2 x − 3 ) 5 / 3 5 3 × 2 = 3 5 ( 2 x − 3 ) 5 3 + k 2 \times \dfrac{(2x - 3)^{5/3}}{\frac53 \times 2} = \frac35(2x - 3)^{\frac53} + k 2 × 3 5 × 2 ( 2 x − 3 ) 5/3 = 5 3 ( 2 x − 3 ) 3 5 + k , option D.
Watch out
Divide by the 2 from the inside bracket as well as by 5 3 \frac53 3 5 6 5 ( 2 x − 3 ) 5 3 \frac65(2x - 3)^{\frac53} 5 6 ( 2 x − 3 ) 3 5 Report a problem with this question
Find the area bounded by the curve y = 4 − x 2 y = 4 - x^2 y = 4 − x 2 x x x
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 curve meets the
x x x -axis where
4 − x 2 = 0 4 - x^2 = 0 4 − x 2 = 0 , at
x = − 2 x = -2 x = − 2 and
x = 2 x = 2 x = 2 .
These are the limits.
∫ − 2 2 ( 4 − x 2 ) d x = [ 4 x − x 3 3 ] − 2 2 \int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} ∫ − 2 2 ( 4 − x 2 ) d x = [ 4 x − 3 x 3 ] − 2 2 .
At
x = 2 x = 2 x = 2 this is
8 − 8 3 = 16 3 8 - \frac83 = \frac{16}{3} 8 − 3 8 = 3 16 , and at
x = − 2 x = -2 x = − 2 it is
− 16 3 -\frac{16}{3} − 3 16 .
Subtract:
16 3 + 16 3 = 32 3 \frac{16}{3} + \frac{16}{3} = \frac{32}{3} 3 16 + 3 16 = 3 32 = 10 2 3 = 10\frac23 = 10 3 2 square units, option B.
Watch out
Find the limits from where the curve crosses the axis, x = ± 2 x = \pm2 x = ± 2 5 1 3 5\frac13 5 3 1 Similar: JAMB 2017 · UTME · Q33
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The bar chart shows the colours of cars passing a point on a street in two minutes. What fraction of the total number of cars is yellow?
A 4 15 \frac4{15} 15 4 B 1 5 \frac15 5 1 C 3 25 \frac3{25} 25 3 D 2 25 \frac2{25} 25 2
Worked solution (try it first) Read the bars and add:
3 + 4 + 8 + 2 + 6 + 2 = 25 3 + 4 + 8 + 2 + 6 + 2 = 25 3 + 4 + 8 + 2 + 6 + 2 = 25 cars.
The yellow bar is 3, so the fraction is
3 25 \frac{3}{25} 25 3 , option C.
Watch out
Divide by all the cars, 25, including the yellow ones. Also read the yellow bar itself, not the tallest one. Report a problem with this question
The histogram shows the distribution of passengers in taxis of a certain motor park. How many taxis have more than 4 passengers?
Worked solution (try it first) The class boundaries 0.5, 2.5, 4.5, … mean the classes are 1–2, 3–4, 5–6, 7–8, 9–10 and 11–12 passengers.
More than 4 passengers means the bars from 4.5 upwards, with heights 7, 5, 4 and 1.
Add them:
7 + 5 + 4 + 1 = 17 7 + 5 + 4 + 1 = 17 7 + 5 + 4 + 1 = 17 taxis, option D.
Watch out
Include the short last bar (10.5 to 12.5), which is 1 taxi. Leaving it out gives 16 (option C). Report a problem with this question
Using the table below, find the square of the mode.
Score
4
7
8
11
13
8
Frequency
3
5
2
7
2
1
Worked solution (try it first) The mode is the score with the highest frequency.
Even with both 8s together (
2 + 1 = 3 2 + 1 = 3 2 + 1 = 3 ), score 11, with frequency 7, is the most common.
So the mode is 11 and its square is
11 2 = 121 11^2 = 121 1 1 2 = 121 , option D.
Watch out
Look at the frequency row, not at which score is printed twice: 8 appears twice in the table but only 3 times in the data, giving 64 (option C). Report a problem with this question
Using the same table, the mean score is
Score
4
7
8
11
13
8
Frequency
3
5
2
7
2
1
Worked solution (try it first) Multiply each score by its frequency and add:
12 + 35 + 16 + 77 + 26 + 8 = 174 12 + 35 + 16 + 77 + 26 + 8 = 174 12 + 35 + 16 + 77 + 26 + 8 = 174 .
Total frequency:
3 + 5 + 2 + 7 + 2 + 1 = 20 3 + 5 + 2 + 7 + 2 + 1 = 20 3 + 5 + 2 + 7 + 2 + 1 = 20 .
Mean
= 174 20 = 8.7 = \frac{174}{20} = 8.7 = 20 174 = 8.7 , option C.
Watch out
Divide by the total frequency, 20, not by the 6 columns. Averaging the scores in the top row gives 8.5, which ignores how often each occurs. Report a problem with this question
Find the range of 1 6 , 1 3 , 3 2 , 2 3 , 8 9 \frac16, \frac13, \frac32, \frac23, \frac89 6 1 , 3 1 , 2 3 , 3 2 , 9 8 4 3 \frac43 3 4
A 4 3 \frac43 3 4 B 7 6 \frac76 6 7 C 5 6 \frac56 6 5 D 3 4 \frac34 4 3
Worked solution (try it first) Compare the fractions as decimals:
1 6 ≈ 0.17 \frac16 \approx 0.17 6 1 ≈ 0.17 ,
1 3 ≈ 0.33 \frac13 \approx 0.33 3 1 ≈ 0.33 ,
3 2 = 1.5 \frac32 = 1.5 2 3 = 1.5 ,
2 3 ≈ 0.67 \frac23 \approx 0.67 3 2 ≈ 0.67 ,
8 9 ≈ 0.89 \frac89 \approx 0.89 9 8 ≈ 0.89 and
4 3 ≈ 1.33 \frac43 \approx 1.33 3 4 ≈ 1.33 .
The largest is
3 2 \frac32 2 3 and the smallest is
1 6 \frac16 6 1 .
Over the common denominator 6, the range is
9 6 − 1 6 = 8 6 \frac96 - \frac16 = \frac86 6 9 − 6 1 = 6 8 , which simplifies to
4 3 \frac43 3 4 , option A.
Watch out
3 2 = 1.5 \frac32 = 1.5 2 3 = 1.5 4 3 ≈ 1.33 \frac43 \approx 1.33 3 4 ≈ 1.33 4 3 \frac43 3 4 4 3 − 1 6 = 7 6 \frac43 - \frac16 = \frac76 3 4 − 6 1 = 6 7 Report a problem with this question
Find the variance of 2, 6, 8, 6, 2 and 6.
Worked solution (try it first) The six numbers add up to 30, so the mean is 5.
The squared deviations are 9, 1, 9, 1, 9, 1, which add up to 30.
The variance is
30 6 = 5 \frac{30}{6} = 5 6 30 = 5 , option C.
Watch out
Don't take the square root: 5 \sqrt5 5 Report a problem with this question
A cumulative frequency graph shows the distribution of masses of fertilizer for 48 workers. Which of the following gives the interquartile range?
A Q 3 − Q 1 Q_3 - Q_1 Q 3 − Q 1 B Q 3 − Q 2 Q_3 - Q_2 Q 3 − Q 2 C Q 2 − Q 1 Q_2 - Q_1 Q 2 − Q 1 D 1 2 ( Q 3 − Q 1 ) \frac12(Q_3 - Q_1) 2 1 ( Q 3 − Q 1 )
Worked solution (try it first) The interquartile range is the upper quartile minus the lower quartile.
So it is
Q 3 − Q 1 Q_3 - Q_1 Q 3 − Q 1 , option A.
Watch out
1 2 ( Q 3 − Q 1 ) \frac12(Q_3 - Q_1) 2 1 ( Q 3 − Q 1 ) Report a problem with this question
Find the number of ways of selecting 8 subjects from 12 subjects for an examination.
Worked solution (try it first) The order of the subjects doesn't matter, so this is
12 C 8 ^{12}C_8 12 C 8 , which is the same as
12 C 4 ^{12}C_4 12 C 4 .
12 C 4 = 12 × 11 × 10 × 9 4 × 3 × 2 × 1 ^{12}C_4 = \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} 12 C 4 = 4 × 3 × 2 × 1 12 × 11 × 10 × 9 = 11 880 24 = \dfrac{11\,880}{24} = 24 11 880 .
That is 495, option C.
Watch out
The options are close together, so work carefully. Choosing 8 to take is the same as choosing 4 to leave out, and 12 C 4 ^{12}C_4 12 C 4 12 ! 8 ! 4 ! \frac{12!}{8!\,4!} 8 ! 4 ! 12 ! Report a problem with this question
If 6 P r = 6 ^6P_r = 6 6 P r = 6 6 P r + 1 ^6P_{r + 1} 6 P r + 1
Worked solution (try it first) 6 P 1 = 6 ^6P_1 = 6 6 P 1 = 6 , so
r = 1 r = 1 r = 1 .
Then
6 P r + 1 = 6 P 2 = 6 × 5 = 30 ^6P_{r + 1} = {^6P_2} = 6 \times 5 = 30 6 P r + 1 = 6 P 2 = 6 × 5 = 30 , option B.
Watch out
P P P 6 C 2 = 15 ^6C_2 = 15 6 C 2 = 15 2 ! 2! 2 ! 6 P 2 ^6P_2 6 P 2 Report a problem with this question
The distribution of colours of beads in a bowl is: blue 1, black 2, yellow 4, white 5, brown 3. What is the probability that a bead selected at random will be blue or white?
A 1 15 \frac1{15} 15 1 B 1 3 \frac13 3 1 C 2 5 \frac25 5 2 D 7 15 \frac7{15} 15 7
Worked solution (try it first) There are
1 + 2 + 4 + 5 + 3 = 15 1 + 2 + 4 + 5 + 3 = 15 1 + 2 + 4 + 5 + 3 = 15 beads.
Blue or white is
1 + 5 = 6 1 + 5 = 6 1 + 5 = 6 beads.
So the probability is
6 15 = 2 5 \frac{6}{15} = \frac25 15 6 = 5 2 , option C.
Watch out
"Blue or white" includes both colours. Using only the white beads gives 5 15 = 1 3 \frac{5}{15} = \frac13 15 5 = 3 1 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 2017 · UTME · Q34
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