Paper JAMB 2015 General Maths Objective
Objective paper · 34 questions · partial
JAMB 2015 · UTME Topics include Sequences & series (AP, GP), Surds, Expressions, formulae & change of subject, Sine & cosine rules, Angles, triangles & polygons, Commercial arithmetic.
Our copy of this paper is missing questions 5, 7, 12, 18, 23, 40.
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 8 9 10 11 13 14 15 16 17 19 20 21 22 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 The sum of the progression 1 + x + x 2 + … 1 + x + x^2 + \ldots 1 + x + x 2 + … ∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1
A 1 1 − x \dfrac{1}{1 - x} 1 − x 1 B 1 1 + x \dfrac{1}{1 + x} 1 + x 1 C 1 x − 1 \dfrac{1}{x - 1} x − 1 1 D 1 x \dfrac1x x 1
Worked solution (try it first) This is a G.P. with first term
a = 1 a = 1 a = 1 and common ratio
r = x r = x r = x .
Since
∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1 , the sum to infinity exists:
S ∞ = a 1 − r S_\infty = \dfrac{a}{1 - r} S ∞ = 1 − r a .
So the sum is
1 1 − x \dfrac{1}{1 - x} 1 − x 1 , option A.
Watch out
The ratio is + x +x + x 1 − x 1 - x 1 − x 1 1 + x \frac{1}{1 + x} 1 + x 1 1 − x + x 2 − … 1 - x + x^2 - \dots 1 − x + x 2 − … r = − x r = -x r = − x Report a problem with this question
Find a square root of 170 − 20 30 170 - 20\sqrt{30} 170 − 20 30
A 2 10 − 5 2\sqrt{10} - 5 2 10 − 5 B 3 5 − 8 6 3\sqrt5 - 8\sqrt6 3 5 − 8 6 C 2 5 − 5 6 2\sqrt5 - 5\sqrt6 2 5 − 5 6 D 5 5 − 2 6 5\sqrt5 - 2\sqrt6 5 5 − 2 6
Worked solution (try it first) A square root has the form
a 5 − b 6 a\sqrt5 - b\sqrt6 a 5 − b 6 , because
( a 5 − b 6 ) 2 = 5 a 2 + 6 b 2 − 2 a b 30 (a\sqrt5 - b\sqrt6)^2 = 5a^2 + 6b^2 - 2ab\sqrt{30} ( a 5 − b 6 ) 2 = 5 a 2 + 6 b 2 − 2 ab 30 .
Match the surd parts:
2 a b = 20 2ab = 20 2 ab = 20 , so
a b = 10 ab = 10 ab = 10 .
Try
a = 2 a = 2 a = 2 ,
b = 5 b = 5 b = 5 :
5 ( 4 ) + 6 ( 25 ) = 20 + 150 5(4) + 6(25) = 20 + 150 5 ( 4 ) + 6 ( 25 ) = 20 + 150 , which is 170.
That matches, so
( 2 5 − 5 6 ) 2 = 170 − 20 30 (2\sqrt5 - 5\sqrt6)^2 = 170 - 20\sqrt{30} ( 2 5 − 5 6 ) 2 = 170 − 20 30 , and
2 5 − 5 6 2\sqrt5 - 5\sqrt6 2 5 − 5 6 is a square root, option C.
Watch out
Check both parts, not just the 30 \sqrt{30} 30 5 5 − 2 6 5\sqrt5 - 2\sqrt6 5 5 − 2 6 − 20 30 -20\sqrt{30} − 20 30 125 + 24 = 149 125 + 24 = 149 125 + 24 = 149 Report a problem with this question
Multiply ( x + 3 y + 5 ) (x + 3y + 5) ( x + 3 y + 5 ) ( 2 x 2 + 5 y + 2 ) (2x^2 + 5y + 2) ( 2 x 2 + 5 y + 2 )
A 2 x 3 + 3 y x 2 + 10 x y + 15 y 2 + 13 y + 10 x 2 + 2 x + 10 2x^3 + 3yx^2 + 10xy + 15y^2 + 13y + 10x^2 + 2x + 10 2 x 3 + 3 y x 2 + 10 x y + 15 y 2 + 13 y + 10 x 2 + 2 x + 10 B 2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 31 y + 10 x 2 + 2 x + 10 2x^3 + 6yx^2 + 5xy + 15y^2 + 31y + 10x^2 + 2x + 10 2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 31 y + 10 x 2 + 2 x + 10 C 2 x 3 + 3 y x 2 + 5 x y + 10 y 2 + 13 y + 5 x 2 + 2 x + 10 2x^3 + 3yx^2 + 5xy + 10y^2 + 13y + 5x^2 + 2x + 10 2 x 3 + 3 y x 2 + 5 x y + 10 y 2 + 13 y + 5 x 2 + 2 x + 10 D 2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 13 y + 10 x 2 + 2 x + 10 2x^3 + 6yx^2 + 5xy + 15y^2 + 13y + 10x^2 + 2x + 10 2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 13 y + 10 x 2 + 2 x + 10
Worked solution (try it first) Multiply each term of
x + 3 y + 5 x + 3y + 5 x + 3 y + 5 by the second bracket:
x x x gives
2 x 3 + 5 x y + 2 x 2x^3 + 5xy + 2x 2 x 3 + 5 x y + 2 x , and
3 y 3y 3 y gives
6 x 2 y + 15 y 2 + 6 y 6x^2y + 15y^2 + 6y 6 x 2 y + 15 y 2 + 6 y .
5 5 5 gives
10 x 2 + 25 y + 10 10x^2 + 25y + 10 10 x 2 + 25 y + 10 .
Collect like terms: the only ones to combine are
6 y + 25 y = 31 y 6y + 25y = 31y 6 y + 25 y = 31 y .
So the product is
2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 31 y + 10 x 2 + 2 x + 10 2x^3 + 6yx^2 + 5xy + 15y^2 + 31y + 10x^2 + 2x + 10 2 x 3 + 6 y x 2 + 5 x y + 15 y 2 + 31 y + 10 x 2 + 2 x + 10 , option B.
Watch out
Option D has every term right except the y y y 3 y × 2 = 6 y 3y \times 2 = 6y 3 y × 2 = 6 y 5 × 5 y = 25 y 5 \times 5y = 25y 5 × 5 y = 25 y 31 y 31y 31 y 13 y 13y 13 y Report a problem with this question
A force of 5 units acts on a particle in the direction due east and another force of 4 units acts on the particle in the direction north-east. The resultant of the two forces is
A 3 \sqrt3 3 B 3 3 3 C 41 + 20 2 \sqrt{41 + 20\sqrt2} 41 + 20 2 D 41 + 202 \sqrt{41 + 202} 41 + 202
Worked solution (try it first) East and north-east are
45 ∘ 45^\circ 4 5 ∘ apart.
In the triangle of forces (5 east, then 4 north-east), the angle between them is
180 ∘ − 45 ∘ = 135 ∘ 180^\circ - 45^\circ = 135^\circ 18 0 ∘ − 4 5 ∘ = 13 5 ∘ .
Cosine rule for the resultant:
R 2 = 5 2 + 4 2 − 2 ( 5 ) ( 4 ) cos 135 ∘ R^2 = 5^2 + 4^2 - 2(5)(4)\cos135^\circ R 2 = 5 2 + 4 2 − 2 ( 5 ) ( 4 ) cos 13 5 ∘ .
cos 135 ∘ = − 2 2 \cos135^\circ = -\frac{\sqrt2}{2} cos 13 5 ∘ = − 2 2 , so the last term is
+ 40 × 2 2 = + 20 2 +40 \times \frac{\sqrt2}{2} = +20\sqrt2 + 40 × 2 2 = + 20 2 and
R 2 = 41 + 20 2 R^2 = 41 + 20\sqrt2 R 2 = 41 + 20 2 .
So
R = 41 + 20 2 R = \sqrt{41 + 20\sqrt2} R = 41 + 20 2 units, option C.
Watch out
Use 135 ∘ 135^\circ 13 5 ∘ 45 ∘ 45^\circ 4 5 ∘ 45 ∘ 45^\circ 4 5 ∘ 41 − 20 2 41 - 20\sqrt2 41 − 20 2 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
In the diagram, P Q PQ P Q R S RS R S x x x
A 20 ∘ 20^\circ 2 0 ∘ B 40 ∘ 40^\circ 4 0 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 80 ∘ 80^\circ 8 0 ∘
Worked solution (try it first) Let the crossing point be
X X X .
Angles on the straight line
R X Q RXQ R X Q :
∠ P X Q = 180 ∘ − 100 ∘ \angle PXQ = 180^\circ - 100^\circ ∠ P X Q = 18 0 ∘ − 10 0 ∘ The angles of triangle
P X Q PXQ P X Q add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P Q X = 180 ∘ − 60 ∘ − 80 ∘ \angle PQX = 180^\circ - 60^\circ - 80^\circ ∠ P QX = 18 0 ∘ − 6 0 ∘ − 8 0 ∘ P Q ∥ R S PQ \parallel RS P Q ∥ R S , so
x = ∠ S R Q = ∠ P Q R = 40 ∘ x = \angle SRQ = \angle PQR = 40^\circ x = ∠ S R Q = ∠ P QR = 4 0 ∘ (alternate angles), option B.
Watch out
x x x Q R QR QR ∠ P Q R \angle PQR ∠ P QR ∠ Q P S \angle QPS ∠ QP S 60 ∘ 60^\circ 6 0 ∘ Report a problem with this question
After getting a rise of 15 % 15\% 15%
A ₦350 B ₦396.75 C ₦300 D ₦293.25
Worked solution (try it first) After a
15 % 15\% 15% rise the new salary is
115 % 115\% 115% of the old one
S S S :
1.15 S = 345 1.15S = 345 1.15 S = 345 .
Divide both sides by 1.15:
S = 300 S = 300 S = 300 .
So he earned ₦300 a month before, option C.
Watch out
The 15 % 15\% 15% 15 % 15\% 15% Report a problem with this question
A trader goes to Ghana for y y y Y Y Y x x x X X X
A Y ( y + x ) y − x \dfrac{Y(y + x)}{y - x} y − x Y ( y + x ) B Y y + X x y − x \dfrac{Yy + Xx}{y - x} y − x Y y + X x C Y − x y x − y \dfrac{Y - xy}{x - y} x − y Y − x y D Y − X x y − x \dfrac{Y - Xx}{y - x} y − x Y − X x
Worked solution (try it first) In the first
x x x days he spends
X X X cedis a day, so
X x Xx X x cedis in all.
He has
Y − X x Y - Xx Y − X x cedis left for the remaining
y − x y - x y − x days.
Share it equally:
Y − X x y − x \dfrac{Y - Xx}{y - x} y − x Y − X x cedis a day, option D.
Watch out
The amount spent is the daily spend times the days: X × x = X x X \times x = Xx X × x = X x x y xy x y Report a problem with this question
The mean of the numbers 1.2, 1.0, 0.4, 1.4, 0.8, 0.8, 1.2 and 1.1 is
Worked solution (try it first) Add the eight numbers:
1.2 + 1.0 + 0.4 + 1.4 + 0.8 + 0.8 + 1.2 + 1.1 = 7.9 1.2 + 1.0 + 0.4 + 1.4 + 0.8 + 0.8 + 1.2 + 1.1 = 7.9 1.2 + 1.0 + 0.4 + 1.4 + 0.8 + 0.8 + 1.2 + 1.1 = 7.9 .
Divide by 8:
7.9 8 = 0.9875 \frac{7.9}{8} = 0.9875 8 7.9 = 0.9875 .
That is 1.0 to one decimal place, option C.
Watch out
Divide by the number of values, 8. 0.8 (option B) is the mode, not the mean. Report a problem with this question
A solid cylinder of radius 3 cm 3\text{ cm} 3 cm 36 π cm 2 36\pi\text{ cm}^2 36 π cm 2
A 2 cm 2\text{ cm} 2 cm B 3 cm 3\text{ cm} 3 cm C 4 cm 4\text{ cm} 4 cm D 5 cm 5\text{ cm} 5 cm
Worked solution (try it first) Total surface area of a solid cylinder:
2 π r ( r + h ) = 2 π × 3 × ( 3 + h ) 2\pi r(r + h) = 2\pi \times 3 \times (3 + h) 2 π r ( r + h ) = 2 π × 3 × ( 3 + h ) = 6 π ( 3 + h ) = 6\pi(3 + h) = 6 π ( 3 + h ) .
Set it equal to
36 π 36\pi 36 π and divide both sides by
6 π 6\pi 6 π :
3 + h = 6 3 + h = 6 3 + h = 6 .
So
h = 3 h = 3 h = 3 cm, option B.
Watch out
Square the radius in the two ends: 2 π r 2 = 18 π 2\pi r^2 = 18\pi 2 π r 2 = 18 π 2 π r = 6 π 2\pi r = 6\pi 2 π r = 6 π 30 π 30\pi 30 π h = 5 h = 5 h = 5 Report a problem with this question
Which formula represents the general term of the numbers { − 1 , 2 3 , − 1 2 , 2 5 , … } \left\{-1, \frac23, -\frac12, \frac25, \ldots\right\} { − 1 , 3 2 , − 2 1 , 5 2 , … } n = 1 , 2 , 3 , 4 , … n = 1, 2, 3, 4, \ldots n = 1 , 2 , 3 , 4 , …
A 2 n − 1 \dfrac{2}{n - 1} n − 1 2 B ( − 1 ) n + 1 2 n + 1 (-1)^{n + 1}\dfrac{2}{n + 1} ( − 1 ) n + 1 n + 1 2 C ( − 1 ) n 2 n + 1 (-1)^n\dfrac{2}{n + 1} ( − 1 ) n n + 1 2 D n 2 n − 1 \dfrac{n}{2n - 1} 2 n − 1 n
Worked solution (try it first) Write
− 1 -1 − 1 as
− 2 2 -\frac22 − 2 2 and
− 1 2 -\frac12 − 2 1 as
− 2 4 -\frac24 − 4 2 .
Then the sizes are
2 2 , 2 3 , 2 4 , 2 5 \frac22, \frac23, \frac24, \frac25 2 2 , 3 2 , 4 2 , 5 2 , which is
2 n + 1 \frac{2}{n + 1} n + 1 2 .
The signs go
− , + , − , + -, +, -, + − , + , − , + : negative when
n n n is odd.
That is the pattern of
( − 1 ) n (-1)^n ( − 1 ) n .
So the general term is
( − 1 ) n 2 n + 1 (-1)^n\dfrac{2}{n + 1} ( − 1 ) n n + 1 2 , option C.
Watch out
Check the sign at n = 1 n = 1 n = 1 ( − 1 ) 1 + 1 = + 1 (-1)^{1 + 1} = +1 ( − 1 ) 1 + 1 = + 1 + 1 +1 + 1 − 1 -1 − 1 ( − 1 ) n (-1)^n ( − 1 ) n Report a problem with this question
Write the decimal number 39 in base 2.
A 100111 B 110111 C 111001 D 100101
Worked solution (try it first) Write 39 as a sum of powers of two:
39 = 32 + 4 + 2 + 1 39 = 32 + 4 + 2 + 1 39 = 32 + 4 + 2 + 1 .
Put a 1 in the 32, 4, 2 and 1 places and 0 in the 16 and 8 places.
So
39 = 100111 2 39 = 100111_2 39 = 10011 1 2 , option A.
Watch out
If you divide by 2 repeatedly, read the remainders from the bottom up. Reading them top-down gives 111001 (option C). Report a problem with this question
A pentagon has four of its angles equal. If the size of the fifth angle is 60 ∘ 60^\circ 6 0 ∘
A 60 ∘ 60^\circ 6 0 ∘ B 108 ∘ 108^\circ 10 8 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 150 ∘ 150^\circ 15 0 ∘
Worked solution (try it first) The interior angles of a pentagon add up to
( 5 − 2 ) × 180 ∘ = 540 ∘ (5 - 2) \times 180^\circ = 540^\circ ( 5 − 2 ) × 18 0 ∘ = 54 0 ∘ .
The four equal angles share
540 ∘ − 60 ∘ = 480 ∘ 540^\circ - 60^\circ = 480^\circ 54 0 ∘ − 6 0 ∘ = 48 0 ∘ .
So each is
480 ∘ ÷ 4 = 120 ∘ 480^\circ \div 4 = 120^\circ 48 0 ∘ ÷ 4 = 12 0 ∘ , option C.
Watch out
108 ∘ 108^\circ 10 8 ∘ 60 ∘ 60^\circ 6 0 ∘ Report a problem with this question
In the figure, P Q ∥ S R PQ \parallel SR P Q ∥ S R S T ∥ R Q ST \parallel RQ S T ∥ R Q P S = 7 cm PS = 7\text{ cm} P S = 7 cm P T = 7 cm PT = 7\text{ cm} P T = 7 cm S R = 4 cm SR = 4\text{ cm} S R = 4 cm Q R S T QRST QR S T P Q R S PQRS P QR S
A 56 : 77 56 : 77 56 : 77 B 56 : 105 56 : 105 56 : 105 C 28 : 105 28 : 105 28 : 105 D 28 : 49 28 : 49 28 : 49
Worked solution (try it first) T Q ∥ S R TQ \parallel SR T Q ∥ S R and
S T ∥ R Q ST \parallel RQ S T ∥ R Q , so
Q R S T QRST QR S T is a parallelogram and
T Q = S R = 4 TQ = SR = 4 T Q = S R = 4 cm.
So
P Q = 7 + 4 = 11 PQ = 7 + 4 = 11 P Q = 7 + 4 = 11 cm.
Area of
Q R S T QRST QR S T = base × height
= 4 × 7 = 28 cm 2 = 4 \times 7 = 28\text{ cm}^2 = 4 × 7 = 28 cm 2 .
Area of trapezium
P Q R S PQRS P QR S = 1 2 ( 11 + 4 ) × 7 = \frac12(11 + 4) \times 7 = 2 1 ( 11 + 4 ) × 7 = 52.5 cm 2 = 52.5\text{ cm}^2 = 52.5 cm 2 .
The ratio is
28 : 52.5 28 : 52.5 28 : 52.5 .
Double both to clear the decimal:
56 : 105 56 : 105 56 : 105 , option B.
Watch out
P Q PQ P Q P T + T Q = 11 PT + TQ = 11 P T + T Q = 11 38.5 cm 2 38.5\text{ cm}^2 38.5 cm 2 56 : 77 56 : 77 56 : 77 Report a problem with this question
Find a two-digit number such that three times the tens digit is 2 less than twice the units digit, and twice the number is 20 greater than the number obtained by reversing the digits.
Worked solution (try it first) Let the tens digit be
t t t and the units digit
u u u , so the number is
10 t + u 10t + u 10 t + u .
The first fact gives
3 t = 2 u − 2 3t = 2u - 2 3 t = 2 u − 2 , or
3 t − 2 u = − 2 3t - 2u = -2 3 t − 2 u = − 2 .
Reversed, the number is
10 u + t 10u + t 10 u + t .
The second fact gives
2 ( 10 t + u ) = 10 u + t + 20 2(10t + u) = 10u + t + 20 2 ( 10 t + u ) = 10 u + t + 20 , which simplifies to
19 t − 8 u = 20 19t - 8u = 20 19 t − 8 u = 20 .
Multiply the first equation by 4, giving
12 t − 8 u = − 8 12t - 8u = -8 12 t − 8 u = − 8 , and subtract it from the second:
7 t = 28 7t = 28 7 t = 28 , so
t = 4 t = 4 t = 4 .
Then
2 u = 3 t + 2 = 14 2u = 3t + 2 = 14 2 u = 3 t + 2 = 14 , so
u = 7 u = 7 u = 7 .
The number is 47, option D.
Watch out
The tens digit is 4, so the number is 10 × 4 + 7 = 47 10 \times 4 + 7 = 47 10 × 4 + 7 = 47 Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z X Y = 3 cm XY = 3\text{ cm} X Y = 3 cm X Z = 5 cm XZ = 5\text{ cm} X Z = 5 cm Y Z = 7 cm YZ = 7\text{ cm} Y Z = 7 cm ∠ X Y Z \angle XYZ ∠ X Y Z X Z XZ X Z W W W X W XW X W
A 1.5 cm 1.5\text{ cm} 1.5 cm B 2.5 cm 2.5\text{ cm} 2.5 cm C 3 cm 3\text{ cm} 3 cm D 4 cm 4\text{ cm} 4 cm
Worked solution (try it first) The bisector of an angle of a triangle cuts the opposite side in the ratio of the two sides next to that angle.
So
X W : W Z = Y X : Y Z = 3 : 7 XW : WZ = YX : YZ = 3 : 7 X W : W Z = Y X : Y Z = 3 : 7 .
X Z = 5 XZ = 5 X Z = 5 cm is split into
3 + 7 = 10 3 + 7 = 10 3 + 7 = 10 parts, so
X W = 3 10 × 5 XW = \frac{3}{10} \times 5 X W = 10 3 × 5 .
So
X W = 1.5 XW = 1.5 X W = 1.5 cm, option A.
Watch out
An angle bisector does not cut the opposite side in half unless the two sides next to the angle are equal. Halving X Z XZ X Z Report a problem with this question
Marks scored by some children in an arithmetic test are 5, 3, 6, 9, 4, 7, 8, 6, 2, 7, 8, 4, 3, 2, 1, 0, 6, 9, 0, 8. The arithmetic mean of the marks is
Worked solution (try it first) Add the 20 marks: the total is 98.
Divide by 20:
98 20 = 4.9 \frac{98}{20} = 4.9 20 98 = 4.9 .
To the nearest whole number the mean is 5, option B.
Watch out
Count the zeros: they add nothing to the total but they still count as marks, so divide by 20, not 18. Report a problem with this question
The graphical method of solving the equation x 3 + 3 x 2 + 4 x − 28 = 0 x^3 + 3x^2 + 4x - 28 = 0 x 3 + 3 x 2 + 4 x − 28 = 0
A y = x 3 y = x^3 y = x 3 y = 3 x 2 + 4 x − 48 y = 3x^2 + 4x - 48 y = 3 x 2 + 4 x − 48 B y = x 3 + 3 x 2 + 4 x − 28 y = x^3 + 3x^2 + 4x - 28 y = x 3 + 3 x 2 + 4 x − 28 y = 1 y = 1 y = 1 C y = x 3 + 3 x 2 + 4 x y = x^3 + 3x^2 + 4x y = x 3 + 3 x 2 + 4 x y = 28 x y = \frac{28}{x} y = x 28 D y = x 2 + 3 x + 4 y = x^2 + 3x + 4 y = x 2 + 3 x + 4 y = 28 x y = \frac{28}{x} y = x 28
Worked solution (try it first) Move the constant to the right:
x 3 + 3 x 2 + 4 x = 28 x^3 + 3x^2 + 4x = 28 x 3 + 3 x 2 + 4 x = 28 .
Divide every term by
x x x :
x 2 + 3 x + 4 = 28 x x^2 + 3x + 4 = \dfrac{28}{x} x 2 + 3 x + 4 = x 28 .
So the roots are where
y = x 2 + 3 x + 4 y = x^2 + 3x + 4 y = x 2 + 3 x + 4 meets
y = 28 x y = \dfrac{28}{x} y = x 28 , option D.
Watch out
Divide both sides by x x x x 3 + 3 x 2 + 4 x x^3 + 3x^2 + 4x x 3 + 3 x 2 + 4 x 28 x \frac{28}{x} x 28 Report a problem with this question
A sector of a circle is bounded by two radii 7 cm 7\text{ cm} 7 cm 6 cm 6\text{ cm} 6 cm
A 42 cm 2 42\text{ cm}^2 42 cm 2 B 3 cm 2 3\text{ cm}^2 3 cm 2 C 21 cm 2 21\text{ cm}^2 21 cm 2 D 24 cm 2 24\text{ cm}^2 24 cm 2
Worked solution (try it first) The area of a sector is
1 2 × \frac12 \times 2 1 × radius
× \times × arc length.
So the area is
1 2 × 7 × 6 = 21 cm 2 \frac12 \times 7 \times 6 = 21\text{ cm}^2 2 1 × 7 × 6 = 21 cm 2 , option C.
Watch out
Don't drop the 1 2 \frac12 2 1 7 × 6 = 42 cm 2 7 \times 6 = 42\text{ cm}^2 7 × 6 = 42 cm 2 Report a problem with this question
The mean age of 30 pupils in a class is 15.3 years. One boy leaves the class and one girl is enrolled, and the new mean age of the 30 pupils becomes 15.2 years. How much older is the boy than the girl?
A 30 years B 6 years C 9 years D 3 years
Worked solution (try it first) The old total age is
30 × 15.3 = 459 30 \times 15.3 = 459 30 × 15.3 = 459 years.
The new total is
30 × 15.2 = 456 30 \times 15.2 = 456 30 × 15.2 = 456 years.
The number of pupils is unchanged, so the drop of
459 − 456 = 3 459 - 456 = 3 459 − 456 = 3 years is the boy's age minus the girl's.
So the boy is 3 years older, option D.
Watch out
The mean fell by only 0.1 year, but that is spread over 30 pupils: the change in total is 30 × 0.1 = 3 30 \times 0.1 = 3 30 × 0.1 = 3 Report a problem with this question
A world congress of mathematicians was held in Nice in 1970 with 800 people participating: 300 from Europe, 200 from America, 150 from Asia, 45 from Africa and 105 from Australia. On a pie chart, the angle of the sector representing Asia is
A 150 ∘ 150^\circ 15 0 ∘ B 67.5 ∘ 67.5^\circ 67. 5 ∘ C 67 ∘ 67^\circ 6 7 ∘ D 135 ∘ 135^\circ 13 5 ∘
Worked solution (try it first) There are 800 people, so each person gets
360 ∘ 800 = 0.45 ∘ \frac{360^\circ}{800} = 0.45^\circ 800 36 0 ∘ = 0.4 5 ∘ .
Asia has 150 people:
150 × 0.45 ∘ = 67.5 ∘ 150 \times 0.45^\circ = 67.5^\circ 150 × 0.4 5 ∘ = 67. 5 ∘ , option B.
Watch out
Give the exact angle, 67.5 ∘ 67.5^\circ 67. 5 ∘ 67 ∘ 67^\circ 6 7 ∘ Report a problem with this question
Find the sum to infinity of the sequence 1 , 9 10 , ( 9 10 ) 2 , ( 9 10 ) 3 , … 1, \frac{9}{10}, \left(\frac{9}{10}\right)^2, \left(\frac{9}{10}\right)^3, \ldots 1 , 10 9 , ( 10 9 ) 2 , ( 10 9 ) 3 , …
A 1 10 \frac{1}{10} 10 1 B 9 10 \frac{9}{10} 10 9 C 10 9 \frac{10}{9} 9 10 D 10
Worked solution (try it first) This is a G.P. with
a = 1 a = 1 a = 1 and
r = 9 10 r = \frac{9}{10} r = 10 9 .
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 − 9 10 = 1 10 1 - \frac{9}{10} = \frac{1}{10} 1 − 10 9 = 10 1 .
So
S ∞ = 1 ÷ 1 10 = 10 S_\infty = 1 \div \frac{1}{10} = 10 S ∞ = 1 ÷ 10 1 = 10 , option D.
Watch out
Divide by 1 − r 1 - r 1 − r r r r 9 10 \frac{9}{10} 10 9 10 9 \frac{10}{9} 9 10 Also set as JAMB 1995 · UME · Q23
Report a problem with this question
Which of the following is a sketch of y = 3 sin x y = 3\sin x y = 3 sin x
Worked solution (try it first) At
x = 0 x = 0 x = 0 ,
y = 3 sin 0 = 0 y = 3\sin 0 = 0 y = 3 sin 0 = 0 .
At
x = π 2 x = \frac{\pi}{2} x = 2 π ,
y = 3 y = 3 y = 3 .
At
x = − π 2 x = -\frac{\pi}{2} x = − 2 π ,
y = − 3 y = -3 y = − 3 .
Sketches C and D do not pass through the origin, so they are out.
Sketch B passes through the origin, rises to 3 at
π 2 \frac{\pi}{2} 2 π and falls to
− 3 -3 − 3 at
− π 2 -\frac{\pi}{2} − 2 π , so the answer is option B.
Watch out
Sketch A also goes through the origin, but it is − 3 -3 − 3 π 2 \frac{\pi}{2} 2 π y = − 3 sin x y = -3\sin x y = − 3 sin x x = π 2 x = \frac{\pi}{2} x = 2 π Report a problem with this question
Given that log a 2 = 0.693 \log_a 2 = 0.693 log a 2 = 0.693 log a 3 = 1.097 \log_a 3 = 1.097 log a 3 = 1.097 log a 13.5 \log_a 13.5 log a 13.5
Worked solution (try it first) Write 13.5 using 2 and 3:
13.5 = 27 2 = 3 3 2 13.5 = \frac{27}{2} = \frac{3^3}{2} 13.5 = 2 27 = 2 3 3 .
So
log a 13.5 = 3 log a 3 − log a 2 \log_a 13.5 = 3\log_a 3 - \log_a 2 log a 13.5 = 3 log a 3 − log a 2 .
That is
3 ( 1.097 ) − 0.693 = 3.291 − 0.693 = 2.598 3(1.097) - 0.693 = 3.291 - 0.693 = 2.598 3 ( 1.097 ) − 0.693 = 3.291 − 0.693 = 2.598 , option C.
Watch out
27 is 3 3 3^3 3 3 log a 3 \log_a 3 log a 3 log a 6 = 1.790 \log_a 6 = 1.790 log a 6 = 1.790 Also set as JAMB 1997 · UME · Q4
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If f ( x ) = x 3 + 2 x 2 + q x − 6 f(x) = x^3 + 2x^2 + qx - 6 f ( x ) = x 3 + 2 x 2 + q x − 6 x + 1 x + 1 x + 1 q q q
Worked solution (try it first) Divisible by
x + 1 x + 1 x + 1 means
f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 , by the factor theorem.
f ( − 1 ) = − 1 + 2 − q − 6 = − 5 − q f(-1) = -1 + 2 - q - 6 = -5 - q f ( − 1 ) = − 1 + 2 − q − 6 = − 5 − q .
Set it to 0:
q = − 5 q = -5 q = − 5 , option A.
Watch out
q × ( − 1 ) = − q q \times (-1) = -q q × ( − 1 ) = − q − 5 − q = 0 -5 - q = 0 − 5 − q = 0 q = − 5 q = -5 q = − 5 q = 5 q = 5 q = 5 Also set as JAMB 1997 · UME · Q11
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What value of g g g 4 x 2 − 18 x y + g 4x^2 - 18xy + g 4 x 2 − 18 x y + g
A 9 B 9 y 2 4 \dfrac{9y^2}{4} 4 9 y 2 C 81 y 2 81y^2 81 y 2 D 81 y 2 4 \dfrac{81y^2}{4} 4 81 y 2
Worked solution (try it first) A perfect square starting
4 x 2 4x^2 4 x 2 has the form
( 2 x − k y ) 2 = 4 x 2 − 4 k x y + k 2 y 2 (2x - ky)^2 = 4x^2 - 4kxy + k^2y^2 ( 2 x − k y ) 2 = 4 x 2 − 4 k x y + k 2 y 2 .
Match the middle terms:
4 k = 18 4k = 18 4 k = 18 , so
k = 9 2 k = \frac92 k = 2 9 .
So
g = k 2 y 2 = 81 y 2 4 g = k^2y^2 = \dfrac{81y^2}{4} g = k 2 y 2 = 4 81 y 2 , option D.
Watch out
The middle term is 2 × 2 x × k y = 4 k x y 2 \times 2x \times ky = 4kxy 2 × 2 x × k y = 4 k x y k = 18 4 k = \frac{18}{4} k = 4 18 k = 9 k = 9 k = 9 81 y 2 81y^2 81 y 2 Also set as JAMB 1997 · UME · Q15
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An arc of a circle subtends an angle of 70 ∘ 70^\circ 7 0 ∘ 6 cm 6\text{ cm} 6 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 22 cm 2 22\text{ cm}^2 22 cm 2 B 44 cm 2 44\text{ cm}^2 44 cm 2 C 66 cm 2 66\text{ cm}^2 66 cm 2 D 88 cm 2 88\text{ cm}^2 88 cm 2
Worked solution (try it first) Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 , with
r 2 = 36 r^2 = 36 r 2 = 36 .
So the area is
70 360 × 22 7 × 36 \frac{70}{360} \times \frac{22}{7} \times 36 360 70 × 7 22 × 36 .
Cancel the 7 into 70 and the 36 into 360:
10 × 22 10 = 22 cm 2 \frac{10 \times 22}{10} = 22\text{ cm}^2 10 10 × 22 = 22 cm 2 , option A.
Watch out
6 cm is the radius, not the diameter. Taking the radius as 12 gives r 2 = 144 r^2 = 144 r 2 = 144 88 cm 2 88\text{ cm}^2 88 cm 2 Also set as JAMB 1997 · UME · Q29
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The angle of elevation of a building from a measuring instrument placed on the ground is 30 ∘ 30^\circ 3 0 ∘ 40 m 40\text{ m} 40 m
A 20 3 m \frac{20}{\sqrt3}\text{ m} 3 20 m B 40 3 m \frac{40}{\sqrt3}\text{ m} 3 40 m C 20 3 m 20\sqrt3\text{ m} 20 3 m D 40 3 m 40\sqrt3\text{ m} 40 3 m
Worked solution (try it first) The height is opposite the
30 ∘ 30^\circ 3 0 ∘ angle and the distance
d d d is adjacent, so
tan 30 ∘ = 40 d \tan30^\circ = \frac{40}{d} tan 3 0 ∘ = d 40 .
Rearrange:
d = 40 tan 30 ∘ d = \frac{40}{\tan30^\circ} d = t a n 3 0 ∘ 40 .
Dividing by
1 3 \frac{1}{\sqrt3} 3 1 multiplies by
3 \sqrt3 3 :
d = 40 3 d = 40\sqrt3 d = 40 3 m, option D.
Watch out
Divide the height by tan 30 ∘ \tan30^\circ tan 3 0 ∘ 40 tan 30 ∘ = 40 3 40\tan30^\circ = \frac{40}{\sqrt3} 40 tan 3 0 ∘ = 3 40 Report a problem with this question
Integrate 1 x + cos x \dfrac1x + \cos x x 1 + cos x x x x
A − 1 x + sin x + k -\frac1x + \sin x + k − x 1 + sin x + k B ln x + sin x + k \ln x + \sin x + k ln x + sin x + k C ln x − sin x + k \ln x - \sin x + k ln x − sin x + k D − 1 8 sin x + k -\frac18\sin x + k − 8 1 sin x + k
Worked solution (try it first) 1 x \frac1x x 1 integrates to
ln x \ln x ln x , and
cos x \cos x cos x integrates to
sin x \sin x sin x .
So the integral is
ln x + sin x + k \ln x + \sin x + k ln x + sin x + k , option B.
Watch out
cos x \cos x cos x + sin x +\sin x + sin x ∫ sin x = − cos x \int \sin x = -\cos x ∫ sin x = − cos x Also set as JAMB 1997 · UME · Q41
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Find d d x cos ( 3 x 2 − 2 x ) \dfrac{d}{dx}\cos(3x^2 - 2x) d x d cos ( 3 x 2 − 2 x )
A − sin ( 6 x − 2 ) -\sin(6x - 2) − sin ( 6 x − 2 ) B − sin ( 3 x 2 − 2 ) -\sin(3x^2 - 2) − sin ( 3 x 2 − 2 ) C ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) (6x - 2)\sin(3x^2 - 2x) ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) D − ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) -(6x - 2)\sin(3x^2 - 2x) − ( 6 x − 2 ) sin ( 3 x 2 − 2 x )
Worked solution (try it first) Chain rule:
cos u \cos u cos u differentiates to
− sin u -\sin u − sin u times
d u d x \frac{du}{dx} d x d u .
Here
u = 3 x 2 − 2 x u = 3x^2 - 2x u = 3 x 2 − 2 x , so
d u d x = 6 x − 2 \frac{du}{dx} = 6x - 2 d x d u = 6 x − 2 .
So the derivative is
− ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) -(6x - 2)\sin(3x^2 - 2x) − ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) , option D.
Watch out
cos differentiates to − sin -\sin − sin ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) (6x - 2)\sin(3x^2 - 2x) ( 6 x − 2 ) sin ( 3 x 2 − 2 x ) Report a problem with this question
If log 8 10 = x \log_8 10 = x log 8 10 = x log 8 5 \log_8 5 log 8 5 x x x
A 1 2 x \frac{1}{2x} 2 x 1 B x − 1 4 x - \frac14 x − 4 1 C x − 1 3 x - \frac13 x − 3 1 D x − 1 2 x - \frac12 x − 2 1
Worked solution (try it first) Write 5 as
10 2 \frac{10}{2} 2 10 :
log 8 5 = log 8 10 − log 8 2 \log_8 5 = \log_8 10 - \log_8 2 log 8 5 = log 8 10 − log 8 2 .
8 1 3 = 2 8^{\frac13} = 2 8 3 1 = 2 (the cube root of 8 is 2), so
log 8 2 = 1 3 \log_8 2 = \frac13 log 8 2 = 3 1 .
So
log 8 5 = x − 1 3 \log_8 5 = x - \frac13 log 8 5 = x − 3 1 , option C.
Watch out
log 8 2 = 1 3 \log_8 2 = \frac13 log 8 2 = 3 1 1 2 \frac12 2 1 8 1 2 ≈ 2.83 8^{\frac12} \approx 2.83 8 2 1 ≈ 2.83 1 2 \frac12 2 1 x − 1 2 x - \frac12 x − 2 1 Also set as JAMB 1999 · UME · Q8
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Simplify 0.0023 × 750 0.00345 × 1.25 \sqrt{\dfrac{0.0023 \times 750}{0.00345 \times 1.25}} 0.00345 × 1.25 0.0023 × 750
Worked solution (try it first) Pair the numbers to cancel:
0.0023 0.00345 = 2300 3450 \frac{0.0023}{0.00345} = \frac{2300}{3450} 0.00345 0.0023 = 3450 2300 750 1.25 = 600 \frac{750}{1.25} = 600 1.25 750 = 600 .
So the fraction is
2 3 × 600 = 400 \frac23 \times 600 = 400 3 2 × 600 = 400 , and
400 = 20 \sqrt{400} = 20 400 = 20 , option B.
Watch out
Take the square root at the end. The fraction inside is 400, and the answer is its square root, 20. Halving 400 by mistake leads nowhere near the options. Also set as JAMB 1999 · UME · Q7
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Find the matrix T T T S T = I ST = I S T = I S = ( − 1 1 1 − 2 ) S = \begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix} S = ( − 1 1 1 − 2 ) I I I
A ( − 2 1 − 1 1 ) \begin{pmatrix} -2 & 1 \\ -1 & 1 \end{pmatrix} ( − 2 − 1 1 1 ) B ( − 2 − 1 − 1 − 1 ) \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} ( − 2 − 1 − 1 − 1 ) C ( − 1 − 1 0 − 1 ) \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix} ( − 1 0 − 1 − 1 ) D ( − 1 1 0 1 ) \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix} ( − 1 0 1 1 )
Worked solution (try it first) S T = I ST = I S T = I means
T T T is the inverse of
S S S .
The determinant is
∣ S ∣ = ( − 1 ) ( − 2 ) − ( 1 ) ( 1 ) = 1 |S| = (-1)(-2) - (1)(1) = 1 ∣ S ∣ = ( − 1 ) ( − 2 ) − ( 1 ) ( 1 ) = 1 .
Swap the two diagonal entries and change the signs of the other two:
T = ( − 2 − 1 − 1 − 1 ) T = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} T = ( − 2 − 1 − 1 − 1 ) .
Check:
S T = ( 2 − 1 1 − 1 − 2 + 2 − 1 + 2 ) ST = \begin{pmatrix} 2 - 1 & 1 - 1 \\ -2 + 2 & -1 + 2 \end{pmatrix} S T = ( 2 − 1 − 2 + 2 1 − 1 − 1 + 2 ) , which is
I I I .
So the answer is option B.
Watch out
Check your choice by multiplying. S S S ( 1 0 − 1 1 ) \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} ( 1 − 1 0 1 ) I I I Also set as JAMB 1999 · UME · Q15
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The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8.
A 8 5 \frac85 5 8 B 8 3 \frac83 3 8 C 72 25 \frac{72}{25} 25 72 D 56 9 \frac{56}{9} 9 56
Worked solution (try it first) Put
a = 2 r a = 2r a = 2 r into
S ∞ = a 1 − r = 8 S_\infty = \dfrac{a}{1 - r} = 8 S ∞ = 1 − r a = 8 :
2 r = 8 ( 1 − r ) 2r = 8(1 - r) 2 r = 8 ( 1 − r ) .
So
10 r = 8 10r = 8 10 r = 8 , giving
r = 4 5 r = \frac45 r = 5 4 and
a = 8 5 a = \frac85 a = 5 8 .
The second term is
a r = 8 5 × 4 5 ar = \frac85 \times \frac45 a r = 5 8 × 5 4 = 32 25 = \frac{32}{25} = 25 32 .
So the first two terms add up to
40 25 + 32 25 = 72 25 \frac{40}{25} + \frac{32}{25} = \frac{72}{25} 25 40 + 25 32 = 25 72 , option C.
Watch out
8 5 \frac85 5 8 a + a r a + ar a + a r Also set as JAMB 1999 · UME · Q12
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In △ M N O \triangle MNO △ M N O M N = 9 MN = 9 M N = 9 M O = 6 MO = 6 M O = 6 N O = 12 NO = 12 N O = 12 M M M N O NO N O P P P N P NP N P
A 4.8 units B 7.2 units C 8.0 units D 18.0 units
Worked solution (try it first) The bisector of
∠ M \angle M ∠ M cuts
N O NO N O in the ratio of the sides next to
M M M :
N P : P O = M N : M O = 9 : 6 NP : PO = MN : MO = 9 : 6 N P : P O = M N : M O = 9 : 6 , which is
3 : 2 3 : 2 3 : 2 .
N O = 12 NO = 12 N O = 12 is split into
3 + 2 = 5 3 + 2 = 5 3 + 2 = 5 parts, so
N P = 3 5 × 12 NP = \frac35 \times 12 N P = 5 3 × 12 .
So
N P = 7.2 NP = 7.2 N P = 7.2 units, option B.
Watch out
N P NP N P N N N M N = 9 MN = 9 M N = 9 2 5 × 12 = 4.8 \frac25 \times 12 = 4.8 5 2 × 12 = 4.8 P O PO P O Report a problem with this question