Paper JAMB 1983 General Maths Objective
Objective paper · 40 questions · partial
JAMB 1983 · UME Topics include Statistics: data & averages, Commercial arithmetic, Angles, triangles & polygons, Linear & simultaneous equations, Polynomials & algebraic division, Variation.
Our copy of this paper is missing questions 8, 11, 13, 21, 22, 34, 38, 43, 44, 48.
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 5 6 7 9 10 12 14 15 16 17 18 19 20 23 24 25 26 27 28 29 30 31 32 33 35 36 37 39 40 41 42 45 46 47 49 50 If M M M D D D ( M , D ) (M, D) ( M , D )
A ( 6 , 5 ) (6, 5) ( 6 , 5 ) B ( 5 , 8 ) (5, 8) ( 5 , 8 ) C ( 5 , 7 ) (5, 7) ( 5 , 7 ) D ( 5 , 5 ) (5, 5) ( 5 , 5 ) E ( 7 , 5 ) (7, 5) ( 7 , 5 )
Worked solution (try it first) Put the measurements in order: 3, 5, 5, 8, 9.
There are 5 values, so the median is the 3rd one:
M = 5 M = 5 M = 5 .
The mode is the value that occurs most often: 5 appears twice, so
D = 5 D = 5 D = 5 .
So
( M , D ) = ( 5 , 5 ) (M, D) = (5, 5) ( M , D ) = ( 5 , 5 ) , option D.
Watch out
Order the list before you pick the middle. The middle of 5, 9, 3, 5, 8 as written is 3, and taking the mean (30 ÷ 5 = 6 30 \div 5 = 6 30 ÷ 5 = 6 ( 6 , 5 ) (6, 5) ( 6 , 5 ) Report a problem with this question
A construction company is owned by two partners X X X Y Y Y 4 : 5 4 : 5 4 : 5 Y Y Y X X X
A ₦20,000.00 B ₦25,000.00 C ₦30,000.00 D ₦15,000.00 E ₦45,000.00
Worked solution (try it first) The ratio
4 : 5 4 : 5 4 : 5 splits the profit into
4 + 5 = 9 4 + 5 = 9 4 + 5 = 9 equal shares.
Y Y Y gets
5 − 4 = 1 5 - 4 = 1 5 − 4 = 1 share more than
X X X , so one share is ₦5,000.
So the total profit is
9 × 5000 = 9 \times 5000 = 9 × 5000 = ₦45,000, option E.
Watch out
The question asks for the total. X X X Y Y Y Also set as JAMB 2018 · UTME · Q10
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Given a regular hexagon, calculate each interior angle of the hexagon.
A 60 ∘ 60^\circ 6 0 ∘ B 30 ∘ 30^\circ 3 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 45 ∘ 45^\circ 4 5 ∘ E 135 ∘ 135^\circ 13 5 ∘
Worked solution (try it first) The exterior angles of any polygon add up to
360 ∘ 360^\circ 36 0 ∘ , so each exterior angle of a regular hexagon is
360 ∘ ÷ 6 = 60 ∘ 360^\circ \div 6 = 60^\circ 36 0 ∘ ÷ 6 = 6 0 ∘ .
An interior angle and its exterior angle lie on a straight line, so the interior angle is
180 ∘ − 60 ∘ = 120 ∘ 180^\circ - 60^\circ = 120^\circ 18 0 ∘ − 6 0 ∘ = 12 0 ∘ , option C.
Watch out
60 ∘ 60^\circ 6 0 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
Solve the equations 4 x − 3 = 3 x + y = 2 y + 5 x − 12 4x - 3 = 3x + y = 2y + 5x - 12 4 x − 3 = 3 x + y = 2 y + 5 x − 12
A x = 5 , y = 2 x = 5, y = 2 x = 5 , y = 2 B x = 2 , y = 5 x = 2, y = 5 x = 2 , y = 5 C x = − 2 , y = − 5 x = -2, y = -5 x = − 2 , y = − 5 D x = 5 , y = − 2 x = 5, y = -2 x = 5 , y = − 2 E x = − 5 , y = − 2 x = -5, y = -2 x = − 5 , y = − 2
Worked solution (try it first) Split the chain into two equations.
From
4 x − 3 = 3 x + y 4x - 3 = 3x + y 4 x − 3 = 3 x + y , take
3 x 3x 3 x from both sides:
x − y = 3 x - y = 3 x − y = 3 .
From
3 x + y = 2 y + 5 x − 12 3x + y = 2y + 5x - 12 3 x + y = 2 y + 5 x − 12 , take
2 y 2y 2 y and
5 x 5x 5 x from both sides:
− 2 x − y = − 12 -2x - y = -12 − 2 x − y = − 12 .
Multiply by
− 1 -1 − 1 :
2 x + y = 12 2x + y = 12 2 x + y = 12 .
Add the two equations so
y y y cancels:
3 x = 15 3x = 15 3 x = 15 , so
x = 5 x = 5 x = 5 .
Then
y = x − 3 = 2 y = x - 3 = 2 y = x − 3 = 2 .
So
x = 5 x = 5 x = 5 ,
y = 2 y = 2 y = 2 , option A.
Watch out
Check both equations with your pair. Option B swaps the values: x = 2 x = 2 x = 2 y = 5 y = 5 y = 5 4 x − 3 = 5 4x - 3 = 5 4 x − 3 = 5 3 x + y = 11 3x + y = 11 3 x + y = 11 Report a problem with this question
If x = 1 x = 1 x = 1 x 3 − 2 x 2 − 5 x + 6 = 0 x^3 - 2x^2 - 5x + 6 = 0 x 3 − 2 x 2 − 5 x + 6 = 0
A − 3 -3 − 3 B − 2 -2 − 2 C 3 and − 2 -2 − 2 D 1 and 3 E − 3 -3 − 3
Worked solution (try it first) x = 1 x = 1 x = 1 is a root, so
x − 1 x - 1 x − 1 is a factor.
Divide it out:
x 3 − 2 x 2 − 5 x + 6 = ( x − 1 ) ( x 2 − x − 6 ) x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6) x 3 − 2 x 2 − 5 x + 6 = ( x − 1 ) ( x 2 − x − 6 ) .
Factorise the quadratic: two numbers that multiply to
− 6 -6 − 6 and add to
− 1 -1 − 1 are
− 3 -3 − 3 and 2, so
x 2 − x − 6 = ( x − 3 ) ( x + 2 ) x^2 - x - 6 = (x - 3)(x + 2) x 2 − x − 6 = ( x − 3 ) ( x + 2 ) .
Each factor gives a root:
x = 3 x = 3 x = 3 and
x = − 2 x = -2 x = − 2 , option C.
Watch out
The factor x − 3 x - 3 x − 3 + 3 +3 + 3 x + 2 x + 2 x + 2 − 2 -2 − 2 − 3 -3 − 3 Also set as JAMB 2018 · UTME · Q11
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If x x x y y y z z z x x x y y y z z z
A 4 : 1 increase B 2 : 1 increase C 1 : 4 decrease D 1 : 1 no change E 3 : 4 decrease
Worked solution (try it first) Jointly proportional means
x = k y 3 z 4 x = ky^3z^4 x = k y 3 z 4 .
Halving
y y y multiplies
y 3 y^3 y 3 by
( 1 2 ) 3 = 1 8 \left(\frac12\right)^3 = \frac18 ( 2 1 ) 3 = 8 1 .
Doubling
z z z multiplies
z 4 z^4 z 4 by
2 4 = 16 2^4 = 16 2 4 = 16 .
So
x x x is multiplied by
1 8 × 16 = 2 \frac18 \times 16 = 2 8 1 × 16 = 2 : a
2 : 1 2 : 1 2 : 1 increase, option B.
Watch out
Cube the half: ( 1 2 ) 3 = 1 8 \left(\frac12\right)^3 = \frac18 ( 2 1 ) 3 = 8 1 1 4 \frac14 4 1 Report a problem with this question
In the figure, ∠ P Q R = 60 ∘ \angle PQR = 60^\circ ∠ P QR = 6 0 ∘ ∠ Q P R = 90 ∘ \angle QPR = 90^\circ ∠ QP R = 9 0 ∘ ∠ P R S = 90 ∘ \angle PRS = 90^\circ ∠ P R S = 9 0 ∘ ∠ R P S = 45 ∘ \angle RPS = 45^\circ ∠ R P S = 4 5 ∘ Q R = 8 cm QR = 8\text{ cm} QR = 8 cm P S PS P S
A 2 3 2\sqrt3 2 3 B 4 6 4\sqrt6 4 6 C 2 6 2\sqrt6 2 6 D 8 6 8\sqrt6 8 6 E 8 cm
Worked solution (try it first) In triangle
P Q R PQR P QR the right angle is at
P P P , so
Q R = 8 QR = 8 QR = 8 cm is the hypotenuse and
P R PR P R is opposite the
60 ∘ 60^\circ 6 0 ∘ angle at
Q Q Q .
So
P R = 8 sin 60 ∘ PR = 8\sin60^\circ P R = 8 sin 6 0 ∘ = 8 × 3 2 = 8 \times \frac{\sqrt3}{2} = 8 × 2 3 In triangle
P R S PRS P R S the right angle is at
R R R , so
P S PS P S is the hypotenuse and
P R PR P R is adjacent to the
45 ∘ 45^\circ 4 5 ∘ angle:
P S = P R cos 45 ∘ PS = \dfrac{PR}{\cos45^\circ} P S = cos 4 5 ∘ P R = 4 3 × 2 = 4\sqrt3 \times \sqrt2 = 4 3 × 2 .
So
P S = 4 6 PS = 4\sqrt6 P S = 4 6 cm, option B.
Watch out
P S PS P S P R S PRS P R S cos 45 ∘ \cos45^\circ cos 4 5 ∘ 4 3 × 2 2 = 2 6 4\sqrt3 \times \frac{\sqrt2}{2} = 2\sqrt6 4 3 × 2 2 = 2 6 P R PR P R Report a problem with this question
If 0.0000152 × 0.00042 = A × 10 B 0.0000152 \times 0.00042 = A \times 10^B 0.0000152 × 0.00042 = A × 1 0 B 1 ≤ A < 10 1 \le A < 10 1 ≤ A < 10 A A A B B B
A A = 9 A = 9 A = 9 B = 6.38 B = 6.38 B = 6.38 B A = 6.38 A = 6.38 A = 6.38 B = − 9 B = -9 B = − 9 C A = 6.38 A = 6.38 A = 6.38 B = 9 B = 9 B = 9 D A = 6.38 A = 6.38 A = 6.38 B = − 1 B = -1 B = − 1 E A = 6.38 A = 6.38 A = 6.38 B = 1 B = 1 B = 1
Worked solution (try it first) Write each number in standard form:
0.0000152 = 1.52 × 10 − 5 0.0000152 = 1.52 \times 10^{-5} 0.0000152 = 1.52 × 1 0 − 5 and
0.00042 = 4.2 × 10 − 4 0.00042 = 4.2 \times 10^{-4} 0.00042 = 4.2 × 1 0 − 4 .
Multiply the numbers:
1.52 × 4.2 = 6.384 1.52 \times 4.2 = 6.384 1.52 × 4.2 = 6.384 .
Add the powers of 10:
10 − 5 × 10 − 4 = 10 − 9 10^{-5} \times 10^{-4} = 10^{-9} 1 0 − 5 × 1 0 − 4 = 1 0 − 9 .
So the product is
6.384 × 10 − 9 6.384 \times 10^{-9} 6.384 × 1 0 − 9 :
A = 6.38 A = 6.38 A = 6.38 and
B = − 9 B = -9 B = − 9 , option B.
Watch out
Both numbers are less than 1, so the power of 10 must be negative: − 5 + ( − 4 ) = − 9 -5 + (-4) = -9 − 5 + ( − 4 ) = − 9 B = 9 B = 9 B = 9 Report a problem with this question
If x + 2 x + 2 x + 2 x − 1 x - 1 x − 1 l x 3 + 2 k x 2 + 24 lx^3 + 2kx^2 + 24 l x 3 + 2 k x 2 + 24 l l l k k k
A l = − 6 l = -6 l = − 6 k = − 9 k = -9 k = − 9 B l = − 2 l = -2 l = − 2 k = 1 k = 1 k = 1 C l = − 2 l = -2 l = − 2 k = − 1 k = -1 k = − 1 D l = 0 l = 0 l = 0 k = 1 k = 1 k = 1 E l = 6 l = 6 l = 6 k = 0 k = 0 k = 0
Worked solution (try it first) By the factor theorem,
x − 1 x - 1 x − 1 is a factor, so the expression is 0 at
x = 1 x = 1 x = 1 :
l + 2 k + 24 = 0 l + 2k + 24 = 0 l + 2 k + 24 = 0 .
x + 2 x + 2 x + 2 is a factor, so it is 0 at
x = − 2 x = -2 x = − 2 :
− 8 l + 8 k + 24 = 0 -8l + 8k + 24 = 0 − 8 l + 8 k + 24 = 0 .
Divide by 8:
k = l − 3 k = l - 3 k = l − 3 .
Put
k = l − 3 k = l - 3 k = l − 3 into the first equation:
l + 2 l − 6 + 24 = 0 l + 2l - 6 + 24 = 0 l + 2 l − 6 + 24 = 0 , so
3 l = − 18 3l = -18 3 l = − 18 and
l = − 6 l = -6 l = − 6 .
Then
k = − 6 − 3 = − 9 k = -6 - 3 = -9 k = − 6 − 3 = − 9 , option A.
Watch out
The factor x + 2 x + 2 x + 2 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2 ( − 2 ) 3 = − 8 (-2)^3 = -8 ( − 2 ) 3 = − 8 x = 2 x = 2 x = 2 l + k = − 3 l + k = -3 l + k = − 3 Report a problem with this question
In a class of 60 pupils, the numbers offering Biology, History, French, Geography and Additional Mathematics are shown in the pie chart. How many pupils offer Additional Mathematics?
Worked solution (try it first) The angles at the centre of a pie chart add up to
360 ∘ 360^\circ 36 0 ∘ :
( 2 x − 24 ) + ( 3 x − 18 ) + ( 2 x + 12 ) + ( x + 12 ) + x = 360 (2x - 24) + (3x - 18) + (2x + 12) + (x + 12) + x = 360 ( 2 x − 24 ) + ( 3 x − 18 ) + ( 2 x + 12 ) + ( x + 12 ) + x = 360 .
Collect terms:
9 x − 18 = 360 9x - 18 = 360 9 x − 18 = 360 , so
9 x = 378 9x = 378 9 x = 378 and
x = 42 x = 42 x = 42 .
Additional Mathematics has angle
2 ( 42 ) − 24 = 60 ∘ 2(42) - 24 = 60^\circ 2 ( 42 ) − 24 = 6 0 ∘ .
Its share of the 60 pupils is
60 360 × 60 = 10 \frac{60}{360} \times 60 = 10 360 60 × 60 = 10 , option B.
Watch out
Check which sector you are working out. Biology is 3 ( 42 ) − 18 = 108 ∘ 3(42) - 18 = 108^\circ 3 ( 42 ) − 18 = 10 8 ∘ ( 2 x − 24 ) ∘ (2x - 24)^\circ ( 2 x − 24 ) ∘ Report a problem with this question
y y y x x x x x x y y y y = 2 y = 2 y = 2 x = 1 x = 1 x = 1 y = 6 y = 6 y = 6 x = 4 x = 4 x = 4
A y = 10 x 2 31 + 52 31 x y = \dfrac{10x^2}{31} + \dfrac{52}{31\sqrt x} y = 31 10 x 2 + 31 x 52 B y = x 2 + 1 x y = x^2 + \dfrac{1}{\sqrt x} y = x 2 + x 1 C y = x 2 + 1 x y = x^2 + \dfrac1x y = x 2 + x 1 D y = x 2 31 + 1 31 x y = \dfrac{x^2}{31} + \dfrac{1}{31\sqrt x} y = 31 x 2 + 31 x 1 E y = 10 31 ( x 2 + 1 x ) y = \dfrac{10}{31}\left(x^2 + \dfrac{1}{\sqrt x}\right) y = 31 10 ( x 2 + x 1 )
Worked solution (try it first) "Partly … partly" means a sum of two parts with two constants:
y = a x 2 + b x y = ax^2 + \dfrac{b}{\sqrt x} y = a x 2 + x b .
x = 1 x = 1 x = 1 ,
y = 2 y = 2 y = 2 gives
a + b = 2 a + b = 2 a + b = 2 .
x = 4 x = 4 x = 4 ,
y = 6 y = 6 y = 6 gives
16 a + b 2 = 6 16a + \frac{b}{2} = 6 16 a + 2 b = 6 .
Put
a = 2 − b a = 2 - b a = 2 − b into the second equation:
32 − 16 b + b 2 = 6 32 - 16b + \frac b2 = 6 32 − 16 b + 2 b = 6 , so
31 b 2 = 26 \frac{31b}{2} = 26 2 31 b = 26 and
b = 52 31 b = \frac{52}{31} b = 31 52 .
Then
a = 2 − 52 31 = 10 31 a = 2 - \frac{52}{31} = \frac{10}{31} a = 2 − 31 52 = 31 10 .
So
y = 10 x 2 31 + 52 31 x y = \dfrac{10x^2}{31} + \dfrac{52}{31\sqrt x} y = 31 10 x 2 + 31 x 52 , option A.
Watch out
Check an answer against both pairs of values. Option B fits x = 1 x = 1 x = 1 1 + 1 = 2 1 + 1 = 2 1 + 1 = 2 x = 4 x = 4 x = 4 16 + 1 2 = 16 1 2 16 + \frac12 = 16\frac12 16 + 2 1 = 16 2 1 Report a problem with this question
Simplify x − 7 x 2 − 9 × x 2 − 3 x x 2 − 49 \dfrac{x - 7}{x^2 - 9} \times \dfrac{x^2 - 3x}{x^2 - 49} x 2 − 9 x − 7 × x 2 − 49 x 2 − 3 x
A x ( x − 3 ) ( x + 7 ) \dfrac{x}{(x - 3)(x + 7)} ( x − 3 ) ( x + 7 ) x B ( x + 3 ) ( x + 7 ) x \dfrac{(x + 3)(x + 7)}{x} x ( x + 3 ) ( x + 7 ) C x ( x − 3 ) ( x − 7 ) \dfrac{x}{(x - 3)(x - 7)} ( x − 3 ) ( x − 7 ) x D x ( x + 3 ) ( x + 7 ) \dfrac{x}{(x + 3)(x + 7)} ( x + 3 ) ( x + 7 ) x E x ( x + 4 ) ( x + 7 ) \dfrac{x}{(x + 4)(x + 7)} ( x + 4 ) ( x + 7 ) x
Worked solution (try it first) Factorise each part:
x 2 − 9 = ( x − 3 ) ( x + 3 ) x^2 - 9 = (x - 3)(x + 3) x 2 − 9 = ( x − 3 ) ( x + 3 ) ,
x 2 − 3 x = x ( x − 3 ) x^2 - 3x = x(x - 3) x 2 − 3 x = x ( x − 3 ) and
x 2 − 49 = ( x − 7 ) ( x + 7 ) x^2 - 49 = (x - 7)(x + 7) x 2 − 49 = ( x − 7 ) ( x + 7 ) .
So the product is
x − 7 ( x − 3 ) ( x + 3 ) × x ( x − 3 ) ( x − 7 ) ( x + 7 ) \dfrac{x - 7}{(x - 3)(x + 3)} \times \dfrac{x(x - 3)}{(x - 7)(x + 7)} ( x − 3 ) ( x + 3 ) x − 7 × ( x − 7 ) ( x + 7 ) x ( x − 3 ) .
Cancel the common factors
x − 7 x - 7 x − 7 and
x − 3 x - 3 x − 3 , top and bottom.
This leaves
x ( x + 3 ) ( x + 7 ) \dfrac{x}{(x + 3)(x + 7)} ( x + 3 ) ( x + 7 ) x , option D.
Watch out
The factor that cancels with x ( x − 3 ) x(x - 3) x ( x − 3 ) x − 3 x - 3 x − 3 x 2 − 9 x^2 - 9 x 2 − 9 x + 3 x + 3 x + 3 x − 3 x - 3 x − 3 Also set as JAMB 2018 · UTME · Q13
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The lengths of the sides of a right-angled triangle are ( 3 x + 1 ) (3x + 1) ( 3 x + 1 ) ( 3 x − 1 ) (3x - 1) ( 3 x − 1 ) x x x x x x
Worked solution (try it first) The longest side,
3 x + 1 3x + 1 3 x + 1 , is the hypotenuse.
By Pythagoras,
( 3 x + 1 ) 2 = ( 3 x − 1 ) 2 + x 2 (3x + 1)^2 = (3x - 1)^2 + x^2 ( 3 x + 1 ) 2 = ( 3 x − 1 ) 2 + x 2 .
Expand both brackets:
9 x 2 + 6 x + 1 = 9 x 2 − 6 x + 1 + x 2 9x^2 + 6x + 1 = 9x^2 - 6x + 1 + x^2 9 x 2 + 6 x + 1 = 9 x 2 − 6 x + 1 + x 2 .
Cancel
9 x 2 9x^2 9 x 2 and 1 from both sides and collect the
x x x terms:
12 x = x 2 12x = x^2 12 x = x 2 , so
x ( x − 12 ) = 0 x(x - 12) = 0 x ( x − 12 ) = 0 .
A side can't be 0 cm, so
x = 12 x = 12 x = 12 (sides 12, 35 and 37), option D.
Watch out
Don't keep the root x = 0 x = 0 x = 0 x 2 = 12 x x^2 = 12x x 2 = 12 x Report a problem with this question
The scores of a set of final-year students in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.
A 47 B 48 1 2 48\frac12 48 2 1 C 50 D 48 E 49
Worked solution (try it first) Put the 14 scores in order: 21, 23, 29, 40, 41, 43, 47, 50, 50, 55, 56, 70, 70, 73.
With an even number of scores, the median is halfway between the 7th and 8th: 47 and 50.
So the median is
47 + 50 2 = 48 1 2 \frac{47 + 50}{2} = 48\frac12 2 47 + 50 = 48 2 1 , option B.
Watch out
With 14 scores there is no single middle score: average the 7th and 8th. Taking just the 7th gives 47 (option A), and 50 (option C) is the mode. Report a problem with this question
Which of the following equations represents the graph shown?
A y = 1 + 2 x + 3 x 2 y = 1 + 2x + 3x^2 y = 1 + 2 x + 3 x 2 B y = 1 − 2 x + 3 x 2 y = 1 - 2x + 3x^2 y = 1 − 2 x + 3 x 2 C y = 1 + 2 x − 3 x 2 y = 1 + 2x - 3x^2 y = 1 + 2 x − 3 x 2 D y = 1 − 2 x − 3 x 2 y = 1 - 2x - 3x^2 y = 1 − 2 x − 3 x 2 E y = 3 x 2 + 2 x − 1 y = 3x^2 + 2x - 1 y = 3 x 2 + 2 x − 1
Worked solution (try it first) The curve opens downward, so the
x 2 x^2 x 2 term is negative.
That leaves options C and D, which both cut the
y y y -axis at 1.
The curve crosses the
x x x -axis at
− 1 -1 − 1 and
1 3 \frac13 3 1 , so the factors are
( 1 + x ) (1 + x) ( 1 + x ) and
( 1 − 3 x ) (1 - 3x) ( 1 − 3 x ) .
Expand:
( 1 + x ) ( 1 − 3 x ) = 1 − 3 x + x − 3 x 2 (1 + x)(1 - 3x) = 1 - 3x + x - 3x^2 ( 1 + x ) ( 1 − 3 x ) = 1 − 3 x + x − 3 x 2 , which is
1 − 2 x − 3 x 2 1 - 2x - 3x^2 1 − 2 x − 3 x 2 .
So the graph is
y = 1 − 2 x − 3 x 2 y = 1 - 2x - 3x^2 y = 1 − 2 x − 3 x 2 , option D.
Watch out
Option C, y = 1 + 2 x − 3 x 2 = ( 1 + 3 x ) ( 1 − x ) y = 1 + 2x - 3x^2 = (1 + 3x)(1 - x) y = 1 + 2 x − 3 x 2 = ( 1 + 3 x ) ( 1 − x ) x = 1 x = 1 x = 1 x = − 1 3 x = -\frac13 x = − 3 1 x = − 1 x = -1 x = − 1 1 + 2 − 3 = 0 1 + 2 - 3 = 0 1 + 2 − 3 = 0 Report a problem with this question
In the figure, F G H K FGHK F G H K ∠ G K F = 30 ∘ \angle GKF = 30^\circ ∠ G K F = 3 0 ∘ x x x H H H
A 90 ∘ 90^\circ 9 0 ∘ B 30 ∘ 30^\circ 3 0 ∘ C 150 ∘ 150^\circ 15 0 ∘ D 120 ∘ 120^\circ 12 0 ∘ E 60 ∘ 60^\circ 6 0 ∘
Worked solution (try it first) A diagonal of a rhombus bisects the angles it passes through.
K G KG K G bisects
∠ H K F \angle HKF ∠ H K F , so
∠ H K F = 2 × 30 ∘ = 60 ∘ \angle HKF = 2 \times 30^\circ = 60^\circ ∠ H K F = 2 × 3 0 ∘ = 6 0 ∘ .
K F ∥ H G KF \parallel HG K F ∥ H G , so the angles at
K K K and
H H H are co-interior and add up to
180 ∘ 180^\circ 18 0 ∘ .
So
x = 180 ∘ − 60 ∘ = 120 ∘ x = 180^\circ - 60^\circ = 120^\circ x = 18 0 ∘ − 6 0 ∘ = 12 0 ∘ , option D.
Watch out
Double the 30 ∘ 30^\circ 3 0 ∘ K K K 180 ∘ − 30 ∘ 180^\circ - 30^\circ 18 0 ∘ − 3 0 ∘ 150 ∘ 150^\circ 15 0 ∘ Report a problem with this question
P Q R S PQRS P QR S 2 m × 0.8 m 2\text{ m} \times 0.8\text{ m} 2 m × 0.8 m 30 ∘ 30^\circ 3 0 ∘ P Q = 2 m PQ = 2\text{ m} P Q = 2 m 0.8 m 0.8\text{ m} 0.8 m 30 ∘ 30^\circ 3 0 ∘ P R PR P R
A 23 ∘ 35 ′ 23^\circ35' 2 3 ∘ 3 5 ′ B 30 ∘ 30^\circ 3 0 ∘ C 15 ∘ 36 ′ 15^\circ36' 1 5 ∘ 3 6 ′ D 10 ∘ 10^\circ 1 0 ∘ E 10 ∘ 42 ′ 10^\circ42' 1 0 ∘ 4 2 ′
Worked solution (try it first) The sloping edge
Q R = 0.8 QR = 0.8 QR = 0.8 m rises at
30 ∘ 30^\circ 3 0 ∘ , so
R R R is
0.8 sin 30 ∘ = 0.4 0.8\sin30^\circ = 0.4 0.8 sin 3 0 ∘ = 0.4 m above the horizontal edge
P Q PQ P Q .
The diagonal of the desk top, by Pythagoras:
P R = 2 2 + 0.8 2 PR = \sqrt{2^2 + 0.8^2} P R = 2 2 + 0. 8 2 ≈ 2.154 \approx 2.154 ≈ 2.154 m.
The angle
θ \theta θ of
P R PR P R to the horizontal has
sin θ = height P R \sin\theta = \frac{\text{height}}{PR} sin θ = P R height = 0.4 2.154 = \frac{0.4}{2.154} = 2.154 0.4 ≈ 0.1857 \approx 0.1857 ≈ 0.1857 .
So
θ ≈ 10.70 ∘ = 10 ∘ 42 ′ \theta \approx 10.70^\circ = 10^\circ42' θ ≈ 10.7 0 ∘ = 1 0 ∘ 4 2 ′ , option E.
Watch out
The diagonal rises only 0.4 m over its full length of 2.154 m, so it slopes much less than the 30 ∘ 30^\circ 3 0 ∘ 30 ∘ 30^\circ 3 0 ∘ P R PR P R Report a problem with this question
If w w w V V V u u u w 3 w^3 w 3 u u u V V V u = 1 u = 1 u = 1 V = 2 V = 2 V = 2
A u = 8 V 3 u = 8V^3 u = 8 V 3 B u = 2 V u = 2\sqrt V u = 2 V C V = 8 u 2 V = \dfrac{8}{u^2} V = u 2 8 D V = 8 u 2 V = 8u^2 V = 8 u 2 E u = 8 V 3 u = \dfrac{8}{V^3} u = V 3 8
Worked solution (try it first) w w w varies inversely as
V V V , so
w = c V w = \dfrac{c}{V} w = V c for some constant
c c c .
u u u varies as
w 3 w^3 w 3 , and
w 3 = c 3 V 3 w^3 = \dfrac{c^3}{V^3} w 3 = V 3 c 3 , so
u u u varies inversely as
V 3 V^3 V 3 :
u = k V 3 u = \dfrac{k}{V^3} u = V 3 k .
Put in
u = 1 u = 1 u = 1 ,
V = 2 V = 2 V = 2 :
1 = k 8 1 = \dfrac{k}{8} 1 = 8 k , so
k = 8 k = 8 k = 8 and
u = 8 V 3 u = \dfrac{8}{V^3} u = V 3 8 , option E.
Watch out
Cubing c V \frac{c}{V} V c V V V u u u V 3 V^3 V 3 u = 8 V 3 u = 8V^3 u = 8 V 3 V 3 V^3 V 3 u = 64 u = 64 u = 64 V = 2 V = 2 V = 2 Report a problem with this question
Solve the simultaneous equations x 2 + y − 8 = 0 x^2 + y - 8 = 0 x 2 + y − 8 = 0 y + 5 x − 2 = 0 y + 5x - 2 = 0 y + 5 x − 2 = 0 x x x
A − 28 , 7 -28, 7 − 28 , 7 B 6 , − 28 6, -28 6 , − 28 C 6 , − 1 6, -1 6 , − 1 D − 1 , 7 -1, 7 − 1 , 7 E 3 , 2 3, 2 3 , 2
Worked solution (try it first) Make
y y y the subject of the second equation:
y = 2 − 5 x y = 2 - 5x y = 2 − 5 x .
Substitute into the first:
x 2 + ( 2 − 5 x ) − 8 = 0 x^2 + (2 - 5x) - 8 = 0 x 2 + ( 2 − 5 x ) − 8 = 0 , so
x 2 − 5 x − 6 = 0 x^2 - 5x - 6 = 0 x 2 − 5 x − 6 = 0 .
Factorise:
( x − 6 ) ( x + 1 ) = 0 (x - 6)(x + 1) = 0 ( x − 6 ) ( x + 1 ) = 0 .
So
x = 6 x = 6 x = 6 or
x = − 1 x = -1 x = − 1 , option C.
Watch out
Collect the numbers carefully: 2 − 8 = − 6 2 - 8 = -6 2 − 8 = − 6 + 6 +6 + 6 x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) x^2 - 5x + 6 = (x - 2)(x - 3) x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) Report a problem with this question
Find the missing value for x = − 2 x = -2 x = − 2 y = x 3 − x + 3 y = x^3 - x + 3 y = x 3 − x + 3
Worked solution (try it first) Put
x = − 2 x = -2 x = − 2 into
y = x 3 − x + 3 y = x^3 - x + 3 y = x 3 − x + 3 .
The cube of a negative is negative:
( − 2 ) 3 = − 8 (-2)^3 = -8 ( − 2 ) 3 = − 8 .
Subtracting
− 2 -2 − 2 adds 2.
So
y = − 8 + 2 + 3 = − 3 y = -8 + 2 + 3 = -3 y = − 8 + 2 + 3 = − 3 , option A.
Watch out
( − 2 ) 3 = − 8 (-2)^3 = -8 ( − 2 ) 3 = − 8 + 8 +8 + 8 + 8 +8 + 8 8 + 2 + 3 = 13 8 + 2 + 3 = 13 8 + 2 + 3 = 13 Report a problem with this question
If O O O x x x
Worked solution (try it first) C C C is on the minor arc, so
∠ A C B = 130 ∘ \angle ACB = 130^\circ ∠ A C B = 13 0 ∘ stands on the major arc
A B AB A B .
The angle at the centre is twice the angle at the circumference, so the reflex angle
A O B AOB A O B is
2 × 130 ∘ = 260 ∘ 2 \times 130^\circ = 260^\circ 2 × 13 0 ∘ = 26 0 ∘ .
Angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ , so
x = 360 ∘ − 260 ∘ = 100 ∘ x = 360^\circ - 260^\circ = 100^\circ x = 36 0 ∘ − 26 0 ∘ = 10 0 ∘ , option C.
Watch out
2 × 130 ∘ = 260 ∘ 2 \times 130^\circ = 260^\circ 2 × 13 0 ∘ = 26 0 ∘ x x x O O O 360 ∘ 360^\circ 36 0 ∘ Report a problem with this question
Find the angles of the sectors representing each item in a pie chart of the data 6, 10, 14, 16, 26.
A 15 ∘ , 25 ∘ , 35 ∘ , 40 ∘ , 65 ∘ 15^\circ, 25^\circ, 35^\circ, 40^\circ, 65^\circ 1 5 ∘ , 2 5 ∘ , 3 5 ∘ , 4 0 ∘ , 6 5 ∘ B 60 ∘ , 100 ∘ , 140 ∘ , 160 ∘ , 260 ∘ 60^\circ, 100^\circ, 140^\circ, 160^\circ, 260^\circ 6 0 ∘ , 10 0 ∘ , 14 0 ∘ , 16 0 ∘ , 26 0 ∘ C 6 ∘ , 10 ∘ , 14 ∘ , 16 ∘ , 26 ∘ 6^\circ, 10^\circ, 14^\circ, 16^\circ, 26^\circ 6 ∘ , 1 0 ∘ , 1 4 ∘ , 1 6 ∘ , 2 6 ∘ D 30 ∘ , 50 ∘ , 70 ∘ , 80 ∘ , 130 ∘ 30^\circ, 50^\circ, 70^\circ, 80^\circ, 130^\circ 3 0 ∘ , 5 0 ∘ , 7 0 ∘ , 8 0 ∘ , 13 0 ∘ E None of the above
Worked solution (try it first) Add the data:
6 + 10 + 14 + 16 + 26 = 72 6 + 10 + 14 + 16 + 26 = 72 6 + 10 + 14 + 16 + 26 = 72 .
The whole circle is
360 ∘ 360^\circ 36 0 ∘ , so each unit gets
360 ∘ 72 = 5 ∘ \frac{360^\circ}{72} = 5^\circ 72 36 0 ∘ = 5 ∘ .
Multiply each item by 5:
30 ∘ , 50 ∘ , 70 ∘ , 80 ∘ , 130 ∘ 30^\circ, 50^\circ, 70^\circ, 80^\circ, 130^\circ 3 0 ∘ , 5 0 ∘ , 7 0 ∘ , 8 0 ∘ , 13 0 ∘ , option D.
Watch out
The sectors must fill 360 ∘ 360^\circ 36 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 720 ∘ 720^\circ 72 0 ∘ 360 ∘ 360^\circ 36 0 ∘ Report a problem with this question
The scores of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65, 70. What is the sum of the median and modal scores?
Worked solution (try it first) Count each score: 55 once, 60 four times, 65 six times, 70 four times and 75 once (16 in all).
With 16 scores the median is halfway between the 8th and 9th.
The first 5 are 55 and 60, so the 6th to 11th are all 65: the median is 65.
65 occurs most often, so the mode is also 65.
So the sum is
65 + 65 = 130 65 + 65 = 130 65 + 65 = 130 , option B.
Watch out
Put the scores in order before finding the median. The 8th and 9th scores in the list as written are 70 and 75, which gives a median of 72.5 and a sum of 137.5 (option E). Report a problem with this question
The letters of the word MATRICULATION are cut and put into a box. One letter is drawn at random. Find the probability of drawing a vowel.
A 2 13 \frac2{13} 13 2 B 5 13 \frac5{13} 13 5 C 6 13 \frac6{13} 13 6 D 8 13 \frac8{13} 13 8 E 4 13 \frac4{13} 13 4
Worked solution (try it first) MATRICULATION has 13 letters, so there are 13 equally likely outcomes.
Count every vowel, repeats included: A, I, U, A, I, O.
That is 6 vowels.
So the probability is
6 13 \frac{6}{13} 13 6 , option C.
Watch out
Each card is a separate letter, so the two As and two Is count twice. Counting only the different vowels A, I, O, U gives 4 13 \frac{4}{13} 13 4 Report a problem with this question
Correct each of the numbers 59.81789 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures.
A 4.46 B 4.48 C 4.47 D 4.49 E 4.50
Worked solution (try it first) To 3 significant figures, 59.81789 is 59.8 (the fourth figure is 1, so round down).
0.0746829 is 0.0747: the leading zeros don't count, and the fourth figure is 8, so the 6 rounds up to 7.
Multiply:
59.8 × 0.0747 = 4.46706 59.8 \times 0.0747 = 4.46706 59.8 × 0.0747 = 4.46706 .
The fourth figure is 7, so round up: 4.47, option C.
Watch out
Round, don't chop: 4.46706 has 7 after the 6, so it becomes 4.47. Cutting it off gives 4.46 (option A). Also set as JAMB 2018 · UTME · Q16
Report a problem with this question
A rod of length 250 cm is measured as 255 cm. What is the percentage error in the measurement?
Worked solution (try it first) The error is the difference between the measured and true lengths:
255 − 250 = 5 255 - 250 = 5 255 − 250 = 5 cm.
Percentage error is the error as a percentage of the true value:
5 250 × 100 % \frac{5}{250} \times 100\% 250 5 × 100% .
5 250 = 0.02 \frac{5}{250} = 0.02 250 5 = 0.02 , so the percentage error is 2%, option E.
Watch out
The error of 5 cm is not the percentage error. Picking 5 (option C) skips dividing by the true length, 250 cm. Report a problem with this question
If ( 2 3 ) m ( 3 4 ) n = 256 729 \left(\frac23\right)^m\left(\frac34\right)^n = \frac{256}{729} ( 3 2 ) m ( 4 3 ) n = 729 256 m m m n n n
A m = 4 m = 4 m = 4 n = 2 n = 2 n = 2 B m = − 4 m = -4 m = − 4 n = − 2 n = -2 n = − 2 C m = − 4 m = -4 m = − 4 n = 2 n = 2 n = 2 D m = 4 m = 4 m = 4 n = − 2 n = -2 n = − 2 E m = − 2 m = -2 m = − 2 n = 4 n = 4 n = 4
Worked solution (try it first) Split each fraction into primes:
( 2 3 ) m = 2 m × 3 − m \left(\frac23\right)^m = 2^m \times 3^{-m} ( 3 2 ) m = 2 m × 3 − m and
( 3 4 ) n = 3 n × 2 − 2 n \left(\frac34\right)^n = 3^n \times 2^{-2n} ( 4 3 ) n = 3 n × 2 − 2 n , because
4 = 2 2 4 = 2^2 4 = 2 2 .
Collect the powers: the left side is
2 m − 2 n × 3 n − m 2^{m - 2n} \times 3^{n - m} 2 m − 2 n × 3 n − m .
The right side is
256 729 = 2 8 × 3 − 6 \frac{256}{729} = 2^8 \times 3^{-6} 729 256 = 2 8 × 3 − 6 .
Match the powers:
m − 2 n = 8 m - 2n = 8 m − 2 n = 8 and
n − m = − 6 n - m = -6 n − m = − 6 .
Add the two equations:
− n = 2 -n = 2 − n = 2 , so
n = − 2 n = -2 n = − 2 .
Then
m = 8 + 2 n = 8 − 4 = 4 m = 8 + 2n = 8 - 4 = 4 m = 8 + 2 n = 8 − 4 = 4 .
So
m = 4 m = 4 m = 4 ,
n = − 2 n = -2 n = − 2 , option D.
Watch out
In ( 3 4 ) n \left(\frac34\right)^n ( 4 3 ) n n − m n - m n − m m − n m - n m − n n = 2 n = 2 n = 2 ( 2 3 ) 4 ( 3 4 ) 2 = 1 9 \left(\frac23\right)^4\left(\frac34\right)^2 = \frac19 ( 3 2 ) 4 ( 4 3 ) 2 = 9 1 256 729 \frac{256}{729} 729 256 Report a problem with this question
Without using tables, find the numerical value of log 7 49 + log 7 1 7 \log_7 49 + \log_7\frac17 log 7 49 + log 7 7 1
Worked solution (try it first) 49 = 7 2 49 = 7^2 49 = 7 2 , so
log 7 49 = 2 \log_7 49 = 2 log 7 49 = 2 .
1 7 = 7 − 1 \frac17 = 7^{-1} 7 1 = 7 − 1 , so
log 7 1 7 = − 1 \log_7 \frac17 = -1 log 7 7 1 = − 1 .
So the value is
2 + ( − 1 ) = 1 2 + (-1) = 1 2 + ( − 1 ) = 1 , option A.
Watch out
The log of a fraction less than 1 is negative: log 7 1 7 = − 1 \log_7 \frac17 = -1 log 7 7 1 = − 1 + 1 +1 + 1 Report a problem with this question
One interior angle of a convex hexagon is 170 ∘ 170^\circ 17 0 ∘ x ∘ x^\circ x ∘ x x x
A 120 ∘ 120^\circ 12 0 ∘ B 110 ∘ 110^\circ 11 0 ∘ C 105 ∘ 105^\circ 10 5 ∘ D 102 ∘ 102^\circ 10 2 ∘ E 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) The interior angles of a hexagon add up to
( 6 − 2 ) × 180 ∘ = 720 ∘ (6 - 2) \times 180^\circ = 720^\circ ( 6 − 2 ) × 18 0 ∘ = 72 0 ∘ .
One angle is
170 ∘ 170^\circ 17 0 ∘ and the other five are
x x x each, so
170 + 5 x = 720 170 + 5x = 720 170 + 5 x = 720 .
Subtract 170:
5 x = 550 5x = 550 5 x = 550 .
Divide by 5:
x = 110 x = 110 x = 110 , option B.
Watch out
The hexagon isn't regular, so don't use 120 ∘ 120^\circ 12 0 ∘ 720 ∘ 720^\circ 72 0 ∘ Also set as JAMB 2018 · UTME · Q17
Report a problem with this question
P Q R S PQRS P QR S P Q = P S PQ = PS P Q = P S P T PT P T P Q PQ P Q 50 ∘ 50^\circ 5 0 ∘ ∠ Q R S \angle QRS ∠ QR S
A 50 ∘ 50^\circ 5 0 ∘ B 40 ∘ 40^\circ 4 0 ∘ C 110 ∘ 110^\circ 11 0 ∘ D 80 ∘ 80^\circ 8 0 ∘ E 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) The angle between a tangent and a chord equals the angle in the alternate segment.
So
∠ P S Q = ∠ Q P T = 50 ∘ \angle PSQ = \angle QPT = 50^\circ ∠ P S Q = ∠ QP T = 5 0 ∘ .
P Q = P S PQ = PS P Q = P S , so triangle
P Q S PQS P QS is isosceles and
∠ P Q S = ∠ P S Q = 50 ∘ \angle PQS = \angle PSQ = 50^\circ ∠ P QS = ∠ P S Q = 5 0 ∘ .
Then
∠ Q P S = 180 ∘ − 50 ∘ − 50 ∘ \angle QPS = 180^\circ - 50^\circ - 50^\circ ∠ QP S = 18 0 ∘ − 5 0 ∘ − 5 0 ∘ Opposite angles of a cyclic quadrilateral add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ Q R S = 180 ∘ − 80 ∘ \angle QRS = 180^\circ - 80^\circ ∠ QR S = 18 0 ∘ − 8 0 ∘ = 100 ∘ = 100^\circ = 10 0 ∘ , option E.
Watch out
80 ∘ 80^\circ 8 0 ∘ ∠ Q P S \angle QPS ∠ QP S ∠ Q R S \angle QRS ∠ QR S 180 ∘ − 80 ∘ 180^\circ - 80^\circ 18 0 ∘ − 8 0 ∘ Report a problem with this question
A ship H H H P P P H H H P P P
A 026 ∘ 34 ′ 026^\circ34' 02 6 ∘ 3 4 ′ B 243 ∘ 26 ′ 243^\circ26' 24 3 ∘ 2 6 ′ C 116 ∘ 34 ′ 116^\circ34' 11 6 ∘ 3 4 ′ D 063 ∘ 26 ′ 063^\circ26' 06 3 ∘ 2 6 ′ E 240 ∘ 240^\circ 24 0 ∘
Worked solution (try it first) Sketch it:
H H H ends up 30 km south and 60 km west of
P P P , making a right-angled triangle.
The angle west of due south at
P P P is
tan − 1 60 30 = tan − 1 2 \tan^{-1}\frac{60}{30} = \tan^{-1}2 tan − 1 30 60 = tan − 1 2 = 63 ∘ 26 ′ = 63^\circ26' = 6 3 ∘ 2 6 ′ .
Bearings are measured clockwise from north, and due south is
180 ∘ 180^\circ 18 0 ∘ .
So the bearing is
180 ∘ + 63 ∘ 26 ′ = 243 ∘ 26 ′ 180^\circ + 63^\circ26' = 243^\circ26' 18 0 ∘ + 6 3 ∘ 2 6 ′ = 24 3 ∘ 2 6 ′ , option B.
Watch out
63 ∘ 26 ′ 63^\circ26' 6 3 ∘ 2 6 ′ P P P 180 ∘ 180^\circ 18 0 ∘ H H H P P P Report a problem with this question
A square of side 0.524375 cm is drawn on a square paper of side 2.524375 cm. Find the area of the paper not covered by the drawing, correct to 3 significant figures.
A 6.00 cm 2 6.00\text{ cm}^2 6.00 cm 2 B 6.10 cm 2 6.10\text{ cm}^2 6.10 cm 2 C 6 cm 2 6\text{ cm}^2 6 cm 2 D 6.09 cm 2 6.09\text{ cm}^2 6.09 cm 2 E 4.00 cm 2 4.00\text{ cm}^2 4.00 cm 2
Worked solution (try it first) The uncovered area is the big square minus the small one:
2.524375 2 − 0.524375 2 2.524375^2 - 0.524375^2 2.52437 5 2 − 0.52437 5 2 .
Use the difference of two squares,
a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a - b)(a + b) a 2 − b 2 = ( a − b ) ( a + b ) : this is
2 × 3.04875 = 6.0975 cm 2 2 \times 3.04875 = 6.0975\text{ cm}^2 2 × 3.04875 = 6.0975 cm 2 .
To 3 significant figures, the fourth figure is 7, so round up:
6.10 cm 2 6.10\text{ cm}^2 6.10 cm 2 , option B.
Watch out
Round, don't chop: 6.0975 to 3 significant figures is 6.10, because the next figure (7) is 5 or more. Cutting it off gives 6.09 (option D). Report a problem with this question
If f ( x ) = 1 x − 1 + x − 1 x 2 − 1 f(x) = \dfrac{1}{x - 1} + \dfrac{x - 1}{x^2 - 1} f ( x ) = x − 1 1 + x 2 − 1 x − 1 f ( 1 − x ) f(1 - x) f ( 1 − x )
A 1 x + 1 x + 2 \dfrac1x + \dfrac{1}{x + 2} x 1 + x + 2 1 B x + 1 2 x − 1 x + \dfrac{1}{2x - 1} x + 2 x − 1 1 C − 1 x − 1 x − 2 -\dfrac1x - \dfrac{1}{x - 2} − x 1 − x − 2 1 D − 1 x + 1 x 2 − 1 -\dfrac1x + \dfrac{1}{x^2 - 1} − x 1 + x 2 − 1 1
Worked solution (try it first) Factorise
x 2 − 1 = ( x − 1 ) ( x + 1 ) x^2 - 1 = (x - 1)(x + 1) x 2 − 1 = ( x − 1 ) ( x + 1 ) and cancel
x − 1 x - 1 x − 1 : the second fraction is
1 x + 1 \frac{1}{x + 1} x + 1 1 , so
f ( x ) = 1 x − 1 + 1 x + 1 f(x) = \frac{1}{x - 1} + \frac{1}{x + 1} f ( x ) = x − 1 1 + x + 1 1 .
Replace every
x x x by
1 − x 1 - x 1 − x :
f ( 1 − x ) = 1 ( 1 − x ) − 1 + 1 ( 1 − x ) + 1 f(1 - x) = \frac{1}{(1 - x) - 1} + \frac{1}{(1 - x) + 1} f ( 1 − x ) = ( 1 − x ) − 1 1 + ( 1 − x ) + 1 1 , which is
1 − x + 1 2 − x \frac{1}{-x} + \frac{1}{2 - x} − x 1 + 2 − x 1 .
Take the minus sign out of each bottom:
1 2 − x = − 1 x − 2 \frac{1}{2 - x} = -\frac{1}{x - 2} 2 − x 1 = − x − 2 1 .
So
f ( 1 − x ) = − 1 x − 1 x − 2 f(1 - x) = -\dfrac1x - \dfrac{1}{x - 2} f ( 1 − x ) = − x 1 − x − 2 1 , option C.
Watch out
Reversing a bottom changes the sign: 1 2 − x = − 1 x − 2 \frac{1}{2 - x} = -\frac{1}{x - 2} 2 − x 1 = − x − 2 1 + 1 x − 2 +\frac{1}{x - 2} + x − 2 1 Report a problem with this question
In the figure, O O O P O Q POQ P O Q 235 ∘ 235^\circ 23 5 ∘ ∠ P R Q \angle PRQ ∠ P R Q
A 66 1 2 ∘ 66\frac12^\circ 66 2 1 ∘ B 62 1 2 ∘ 62\frac12^\circ 62 2 1 ∘ C 125 ∘ 125^\circ 12 5 ∘ D 105 ∘ 105^\circ 10 5 ∘ E 65 ∘ 65^\circ 6 5 ∘
Worked solution (try it first) Angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ , so the non-reflex angle
P O Q POQ P O Q is
360 ∘ − 235 ∘ = 125 ∘ 360^\circ - 235^\circ = 125^\circ 36 0 ∘ − 23 5 ∘ = 12 5 ∘ .
R R R is on the major arc, so
∠ P R Q \angle PRQ ∠ P R Q stands on the minor arc
P Q PQ P Q , the same arc as the
125 ∘ 125^\circ 12 5 ∘ at
O O O .
The angle at the circumference is half the angle at the centre:
∠ P R Q = 1 2 × 125 ∘ \angle PRQ = \frac12 \times 125^\circ ∠ P R Q = 2 1 × 12 5 ∘ = 62 1 2 ∘ = 62\frac12^\circ = 62 2 1 ∘ , option B.
Watch out
Halve the angle at the centre: 125 ∘ 125^\circ 12 5 ∘ ∠ P O Q \angle POQ ∠ P O Q R R R Report a problem with this question
Simplify 27 a 9 8 3 \sqrt[3]{\dfrac{27a^9}{8}} 3 8 27 a 9
A 9 a 2 2 \dfrac{9a^2}{2} 2 9 a 2 B 9 a 3 2 \dfrac{9a^3}{2} 2 9 a 3 C 2 3 a 2 \dfrac{2}{3a^2} 3 a 2 2 D 2 3 a 2 \dfrac{2}{3a^2} 3 a 2 2 E 3 a 3 2 \dfrac{3a^3}{2} 2 3 a 3
Worked solution (try it first) A cube root can be taken of each part separately:
27 3 = 3 \sqrt[3]{27} = 3 3 27 = 3 , because
3 3 = 27 3^3 = 27 3 3 = 27 , and
8 3 = 2 \sqrt[3]{8} = 2 3 8 = 2 .
For the power, divide the index by 3:
a 9 3 = a 9 ÷ 3 = a 3 \sqrt[3]{a^9} = a^{9 \div 3} = a^3 3 a 9 = a 9 ÷ 3 = a 3 .
So the result is
3 a 3 2 \dfrac{3a^3}{2} 2 3 a 3 , option E.
Watch out
The cube root of 27 is 3, not 27 ÷ 3 = 9 27 \div 3 = 9 27 ÷ 3 = 9 9 a 3 2 \frac{9a^3}{2} 2 9 a 3 Report a problem with this question
P Q R PQR P QR R S P RSP R S P Q Q Q ∠ Q P T = ∠ Q R T = 60 ∘ \angle QPT = \angle QRT = 60^\circ ∠ QP T = ∠ QR T = 6 0 ∘ [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 25 cm B 18 cm C 36 cm D 29 cm E 25.5 cm
Worked solution (try it first) In triangle
P T R PTR P T R the angles at
P P P and
R R R are both
60 ∘ 60^\circ 6 0 ∘ , so the third angle is
60 ∘ 60^\circ 6 0 ∘ too: the triangle is equilateral with side
P R = 2 × 3.5 = 7 PR = 2 \times 3.5 = 7 P R = 2 × 3.5 = 7 cm.
The semicircular arc is half the circumference:
1 2 × 2 × 22 7 × 3.5 = 11 \frac12 \times 2 \times \frac{22}{7} \times 3.5 = 11 2 1 × 2 × 7 22 × 3.5 = 11 cm.
The perimeter is the arc plus
P T PT P T and
T R TR T R :
11 + 7 + 7 = 25 11 + 7 + 7 = 25 11 + 7 + 7 = 25 cm, option A.
Watch out
The boundary has only half a circle, 11 cm. Using the full circumference, 22 cm, gives 22 + 14 = 36 22 + 14 = 36 22 + 14 = 36 P R PR P R Report a problem with this question
In a triangle P Q R PQR P QR Q R = 3 QR = 3 QR = 3 P R = 3 PR = 3 P R = 3 P Q = 3 3 PQ = 3\sqrt3 P Q = 3 3 ∠ P Q R = 30 ∘ \angle PQR = 30^\circ ∠ P QR = 3 0 ∘ P P P R R R
A P = 60 ∘ P = 60^\circ P = 6 0 ∘ R = 90 ∘ R = 90^\circ R = 9 0 ∘ B P = 30 ∘ P = 30^\circ P = 3 0 ∘ R = 120 ∘ R = 120^\circ R = 12 0 ∘ C P = 90 ∘ P = 90^\circ P = 9 0 ∘ R = 60 ∘ R = 60^\circ R = 6 0 ∘ D P = 60 ∘ P = 60^\circ P = 6 0 ∘ R = 60 ∘ R = 60^\circ R = 6 0 ∘ E P = 45 ∘ P = 45^\circ P = 4 5 ∘ R = 105 ∘ R = 105^\circ R = 10 5 ∘
Worked solution (try it first) P R = Q R = 3 PR = QR = 3 P R = QR = 3 cm, so the triangle is isosceles.
Equal sides face equal angles:
Q R QR QR faces
∠ P \angle P ∠ P and
P R PR P R faces
∠ Q \angle Q ∠ Q , so
∠ P = ∠ Q = 30 ∘ \angle P = \angle Q = 30^\circ ∠ P = ∠ Q = 3 0 ∘ .
The angles of a triangle add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ R = 180 ∘ − 30 ∘ − 30 ∘ \angle R = 180^\circ - 30^\circ - 30^\circ ∠ R = 18 0 ∘ − 3 0 ∘ − 3 0 ∘ Check with the cosine rule:
P Q 2 = 3 2 + 3 2 − 2 ( 3 ) ( 3 ) cos 120 ∘ PQ^2 = 3^2 + 3^2 - 2(3)(3)\cos120^\circ P Q 2 = 3 2 + 3 2 − 2 ( 3 ) ( 3 ) cos 12 0 ∘ , which is
18 + 9 = 27 18 + 9 = 27 18 + 9 = 27 , and
27 = 3 3 \sqrt{27} = 3\sqrt3 27 = 3 3 ✓.
So
P = 30 ∘ P = 30^\circ P = 3 0 ∘ and
R = 120 ∘ R = 120^\circ R = 12 0 ∘ , option B.
Watch out
The 3 \sqrt3 3 30 ∘ 30^\circ 3 0 ∘ 60 ∘ 60^\circ 6 0 ∘ 90 ∘ 90^\circ 9 0 ∘ P R = Q R PR = QR P R = QR ∠ P = ∠ Q \angle P = \angle Q ∠ P = ∠ Q 60 ∘ 60^\circ 6 0 ∘ 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
In the diagram, P S = S R PS = SR P S = S R P Q ∥ S R PQ \parallel SR P Q ∥ S R ∠ P S R = 130 ∘ \angle PSR = 130^\circ ∠ P S R = 13 0 ∘ ∠ P R Q = 100 ∘ \angle PRQ = 100^\circ ∠ P R Q = 10 0 ∘ ∠ P Q R \angle PQR ∠ P QR
A 25 ∘ 25^\circ 2 5 ∘ B 50 ∘ 50^\circ 5 0 ∘ C 55 ∘ 55^\circ 5 5 ∘ D 65 ∘ 65^\circ 6 5 ∘ E 75 ∘ 75^\circ 7 5 ∘
Worked solution (try it first) P S = S R PS = SR P S = S R , so triangle
P S R PSR P S R is isosceles and its base angles
∠ S P R \angle SPR ∠ S P R and
∠ S R P \angle SRP ∠ S R P are equal.
Each is
180 ∘ − 130 ∘ 2 = 25 ∘ \frac{180^\circ - 130^\circ}{2} = 25^\circ 2 18 0 ∘ − 13 0 ∘ = 2 5 ∘ .
P Q ∥ S R PQ \parallel SR P Q ∥ S R , so
∠ R P Q = ∠ S R P = 25 ∘ \angle RPQ = \angle SRP = 25^\circ ∠ R P Q = ∠ S R P = 2 5 ∘ (alternate angles).
The angles of triangle
P Q R PQR P QR add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P Q R = 180 ∘ − 25 ∘ − 100 ∘ \angle PQR = 180^\circ - 25^\circ - 100^\circ ∠ P QR = 18 0 ∘ − 2 5 ∘ − 10 0 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option C.
Watch out
Don't stop at 25 ∘ 25^\circ 2 5 ∘ ∠ R P Q \angle RPQ ∠ R P Q P Q R PQR P QR ∠ P Q R \angle PQR ∠ P QR Report a problem with this question
In the figure, P T PT P T O O O ∠ P Q T = 30 ∘ \angle PQT = 30^\circ ∠ P QT = 3 0 ∘ ∠ P T O \angle PTO ∠ P T O
A 30 ∘ 30^\circ 3 0 ∘ B 15 ∘ 15^\circ 1 5 ∘ C 24 ∘ 24^\circ 2 4 ∘ D 12 ∘ 12^\circ 1 2 ∘ E 60 ∘ 60^\circ 6 0 ∘
Worked solution (try it first) A tangent is perpendicular to the radius at the point of contact, so
∠ O P T = 90 ∘ \angle OPT = 90^\circ ∠ O P T = 9 0 ∘ and
∠ Q P T = x + 90 ∘ \angle QPT = x + 90^\circ ∠ QP T = x + 9 0 ∘ .
At
T T T the two marked angles make
∠ Q T P = x + 2 x = 3 x \angle QTP = x + 2x = 3x ∠ QT P = x + 2 x = 3 x .
Angles in triangle
P Q T PQT P QT add up to
180 ∘ 180^\circ 18 0 ∘ :
30 ∘ + ( x + 90 ∘ ) + 3 x = 180 ∘ 30^\circ + (x + 90^\circ) + 3x = 180^\circ 3 0 ∘ + ( x + 9 0 ∘ ) + 3 x = 18 0 ∘ , so
4 x = 60 ∘ 4x = 60^\circ 4 x = 6 0 ∘ and
x = 15 ∘ x = 15^\circ x = 1 5 ∘ .
So
∠ P T O = 2 x = 30 ∘ \angle PTO = 2x = 30^\circ ∠ P T O = 2 x = 3 0 ∘ , option A.
Watch out
The question asks for ∠ P T O \angle PTO ∠ P T O 2 x 2x 2 x x = 15 ∘ x = 15^\circ x = 1 5 ∘ Report a problem with this question
A man drove for 4 hours at a certain speed; he then doubled his speed and drove for another 3 hours. Altogether he covered 600 km. At what speed did he drive for the last 3 hours?
A 120 km/h B 60 km/h C 600 7 \frac{600}{7} 7 600 D 50 km/h E 100 km/h
Worked solution (try it first) Let the first speed be
v v v km/h.
Distance is speed times time, so the first part is
4 v 4v 4 v km and the second, at
2 v 2v 2 v for 3 hours, is
6 v 6v 6 v km.
Altogether
4 v + 6 v = 600 4v + 6v = 600 4 v + 6 v = 600 , so
10 v = 600 10v = 600 10 v = 600 and
v = 60 v = 60 v = 60 .
The last 3 hours were at double speed:
2 × 60 = 120 2 \times 60 = 120 2 × 60 = 120 km/h, option A.
Watch out
v = 60 v = 60 v = 60 Report a problem with this question