Paper JAMB 2018 General Maths Objective
Objective paper · 24 questions · partial
JAMB 2018 · UTME Topics include Number bases, Surds, Expressions, formulae & change of subject, Polynomials & algebraic division, Coordinate geometry, Commercial arithmetic.
Our copy of this paper is missing questions 2, 6, 7, 8, 12, 14, 15, 18, 20, 24, 26, 31, 32, 33, 37, 40.
Sit this paper Answer every question in order, timed if you like (suggested 15 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 3 4 5 9 10 11 13 16 17 19 21 22 23 25 27 28 29 30 34 35 36 38 39 Find x x x y y y
4 2 4 3 − 1 3 x 4 y 3 4 4 \begin{array}{rcccc} & 4 & 2 & 4 & 3 \\ - & 1 & 3 & x & 4 \\ \hline & y & 3 & 4 & 4 \end{array} − 4 1 y 2 3 3 4 x 4 3 4 4
Worked solution (try it first) The digits are all below 5, so work in base 5.
Units:
3 < 4 3 < 4 3 < 4 , so borrow 5:
8 − 4 = 4 8 - 4 = 4 8 − 4 = 4 .
Fives:
4 − 1 = 3 4 - 1 = 3 4 − 1 = 3 is less than
x x x , so borrow again:
8 − x = 4 8 - x = 4 8 − x = 4 , giving
x = 4 x = 4 x = 4 .
25s:
2 − 1 = 1 < 3 2 - 1 = 1 < 3 2 − 1 = 1 < 3 , so borrow:
6 − 3 = 3 6 - 3 = 3 6 − 3 = 3 . 125s:
4 − 1 − 1 = 2 4 - 1 - 1 = 2 4 − 1 − 1 = 2 , so
y = 2 y = 2 y = 2 .
So
x = 4 x = 4 x = 4 and
y = 2 y = 2 y = 2 , option C.
Watch out
A borrow brings 5, not 10. Borrowing 10 in the units gives 13 − 4 = 9 13 - 4 = 9 13 − 4 = 9 Also set as JAMB 2004 · UME · Q1
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Simplify 2 2 − 3 2 + 3 \dfrac{2\sqrt2 - \sqrt3}{\sqrt2 + \sqrt3} 2 + 3 2 2 − 3
A 5 6 + 1 5\sqrt6 + 1 5 6 + 1 B 3 6 − 7 3\sqrt6 - 7 3 6 − 7 C 3 6 + 7 3\sqrt6 + 7 3 6 + 7 D 3 6 − 1 3\sqrt6 - 1 3 6 − 1
Worked solution (try it first) Multiply the top and bottom by
3 − 2 \sqrt3 - \sqrt2 3 − 2 , so the bottom becomes
( 3 + 2 ) ( 3 − 2 ) = 3 − 2 (\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2) = 3 - 2 ( 3 + 2 ) ( 3 − 2 ) = 3 − 2 , which is 1.
Top:
( 2 2 − 3 ) ( 3 − 2 ) = 2 6 − 4 − 3 + 6 (2\sqrt2 - \sqrt3)(\sqrt3 - \sqrt2) = 2\sqrt6 - 4 - 3 + \sqrt6 ( 2 2 − 3 ) ( 3 − 2 ) = 2 6 − 4 − 3 + 6 .
Collect like terms:
3 6 − 7 3\sqrt6 - 7 3 6 − 7 , option B.
Watch out
Both whole-number products are negative: 2 2 × ( − 2 ) = − 4 2\sqrt2 \times (-\sqrt2) = -4 2 2 × ( − 2 ) = − 4 − 3 × 3 = − 3 -\sqrt3 \times \sqrt3 = -3 − 3 × 3 = − 3 − 7 -7 − 7 3 6 − 1 3\sqrt6 - 1 3 6 − 1 Also set as JAMB 2014 · UTME · Q8
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If g t 2 − u − w = 0 gt^2 - u - w = 0 g t 2 − u − w = 0 g g g
A u − w t \dfrac{u - w}{t} t u − w B u + w t 2 \dfrac{u + w}{t^2} t 2 u + w C u − w t 2 \dfrac{u - w}{t^2} t 2 u − w D u + w t \dfrac{u + w}{t} t u + w
Worked solution (try it first) Add
u u u and
w w w to both sides:
g t 2 = u + w gt^2 = u + w g t 2 = u + w .
Divide both sides by
t 2 t^2 t 2 :
g = u + w t 2 g = \dfrac{u + w}{t^2} g = t 2 u + w , option B.
Watch out
g g g t 2 t^2 t 2 t 2 t^2 t 2 t t t t t t u + w t \frac{u + w}{t} t u + w Report a problem with this question
Find the value of u u u y − 1 y - 1 y − 1 y 3 + 4 y 2 + u y − 6 y^3 + 4y^2 + uy - 6 y 3 + 4 y 2 + u y − 6
Worked solution (try it first) By the factor theorem,
y − 1 y - 1 y − 1 is a factor, so the expression is 0 at
y = 1 y = 1 y = 1 .
1 + 4 + u − 6 = 0 1 + 4 + u - 6 = 0 1 + 4 + u − 6 = 0 , so
u − 1 = 0 u - 1 = 0 u − 1 = 0 and
u = 1 u = 1 u = 1 , option D.
Watch out
y − 1 y - 1 y − 1 y = + 1 y = +1 y = + 1 y = − 1 y = -1 y = − 1 − 1 + 4 − u − 6 = 0 -1 + 4 - u - 6 = 0 − 1 + 4 − u − 6 = 0 u = − 3 u = -3 u = − 3 Report a problem with this question
What is the equation of the line that cuts the y y y ( 0 , 5 ) (0, 5) ( 0 , 5 ) x x x ( 5 , 0 ) (5, 0) ( 5 , 0 )
A y = − x − 5 y = -x - 5 y = − x − 5 B − y = x + 5 -y = x + 5 − y = x + 5 C y = − x + 5 y = -x + 5 y = − x + 5 D y = x − 5 y = x - 5 y = x − 5
Worked solution (try it first) Gradient:
0 − 5 5 − 0 = − 1 \dfrac{0 - 5}{5 - 0} = -1 5 − 0 0 − 5 = − 1 .
The line cuts the
y y y -axis at 5, so
c = 5 c = 5 c = 5 .
So
y = − x + 5 y = -x + 5 y = − x + 5 , option C.
Watch out
Check both points in your answer. y = x − 5 y = x - 5 y = x − 5 ( 5 , 0 ) (5, 0) ( 5 , 0 ) y y y − 5 -5 − 5 Also set as JAMB 2014 · UTME · Q34
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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 B ₦25,000 C ₦50,000 D ₦45,000
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 D.
Watch out
The question asks for the total. X X X Y Y Y Also set as JAMB 1983 · UME · Q2
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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
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 1983 · UME · Q5
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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
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 1983 · UME · Q15
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Correct each of 59.81798 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures.
Worked solution (try it first) To 3 significant figures, 59.81798 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.
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 1983 · UME · Q30
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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 ∘
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 1983 · UME · Q35
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Convert 241 5 241_5 24 1 5
A 71 8 71_8 7 1 8 B 107 8 107_8 10 7 8 C 176 8 176_8 17 6 8 D 241 8 241_8 24 1 8
Worked solution (try it first) Change to base ten:
241 5 = 2 × 25 + 4 × 5 + 1 = 71 241_5 = 2 \times 25 + 4 \times 5 + 1 = 71 24 1 5 = 2 × 25 + 4 × 5 + 1 = 71 .
Divide by 8:
71 ÷ 8 = 8 71 \div 8 = 8 71 ÷ 8 = 8 remainder 7, then
8 ÷ 8 = 1 8 \div 8 = 1 8 ÷ 8 = 1 remainder 0, then
1 ÷ 8 = 0 1 \div 8 = 0 1 ÷ 8 = 0 remainder 1.
Read the remainders from the bottom up:
107 8 107_8 10 7 8 , option B.
Watch out
Changing base changes the digits. 241 8 241_8 24 1 8 128 + 32 + 1 = 161 128 + 32 + 1 = 161 128 + 32 + 1 = 161 Also set as JAMB 1987 · UME · Q1
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By selling 20 oranges for ₦1.35 a trader makes a profit of 8 % 8\% 8%
A 8 % 8\% 8% B 10 % 10\% 10% C 12 % 12\% 12% D 15 % 15\% 15%
Worked solution (try it first) An
8 % 8\% 8% profit means ₦1.35 is
108 % 108\% 108% of the cost price.
So the cost price is
1.35 1.08 = 1.25 \frac{1.35}{1.08} = 1.25 1.08 1.35 = 1.25 , that is ₦1.25.
Selling for ₦1.10 is less than ₦1.25, a loss of ₦0.15.
Percentage loss on the cost price:
0.15 1.25 × 100 = 12 % \frac{0.15}{1.25} \times 100 = 12\% 1.25 0.15 × 100 = 12% , option C.
Watch out
Divide by 1.08 to find the cost price; don't take 8 % 8\% 8% 0.15 1.10 \frac{0.15}{1.10} 1.10 0.15 13.6 % 13.6\% 13.6% Also set as JAMB 1987 · UME · Q10
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Simplify without using tables 2 14 × 3 21 7 24 × 2 98 \dfrac{2\sqrt{14} \times 3\sqrt{21}}{7\sqrt{24} \times 2\sqrt{98}} 7 24 × 2 98 2 14 × 3 21
A 3 14 4 \dfrac{3\sqrt{14}}{4} 4 3 14 B 3 2 4 \dfrac{3\sqrt2}{4} 4 3 2 C 3 14 28 \dfrac{3\sqrt{14}}{28} 28 3 14 D 3 2 28 \dfrac{3\sqrt2}{28} 28 3 2
Worked solution (try it first) Multiply out the top:
2 14 × 3 21 = 6 294 2\sqrt{14} \times 3\sqrt{21} = 6\sqrt{294} 2 14 × 3 21 = 6 294 .
Since
294 = 49 × 6 294 = 49 \times 6 294 = 49 × 6 , this is
42 6 42\sqrt6 42 6 .
Simplify the bottom surds:
7 24 = 14 6 7\sqrt{24} = 14\sqrt6 7 24 = 14 6 and
2 98 = 14 2 2\sqrt{98} = 14\sqrt2 2 98 = 14 2 .
Their product is
196 12 196\sqrt{12} 196 12 , which is
392 3 392\sqrt3 392 3 .
Divide:
42 6 392 3 = 42 392 2 \dfrac{42\sqrt6}{392\sqrt3} = \dfrac{42}{392}\sqrt2 392 3 42 6 = 392 42 2 , because
6 3 = 2 \frac{\sqrt6}{\sqrt3} = \sqrt2 3 6 = 2 .
Cancel by 14:
42 392 = 3 28 \frac{42}{392} = \frac{3}{28} 392 42 = 28 3 .
So the value is
3 2 28 \dfrac{3\sqrt2}{28} 28 3 2 , option D.
Watch out
6 3 = 6 ÷ 3 = 2 \frac{\sqrt6}{\sqrt3} = \sqrt{6 \div 3} = \sqrt2 3 6 = 6 ÷ 3 = 2 14 \sqrt{14} 14 14 \sqrt{14} 14 3 14 28 \frac{3\sqrt{14}}{28} 28 3 14 Also set as JAMB 1987 · UME · Q15
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Make y y y Z = x 2 + 1 y 3 Z = x^2 + \dfrac{1}{y^3} Z = x 2 + y 3 1
A y = 1 ( Z − x 2 ) 3 y = \dfrac{1}{(Z - x^2)^3} y = ( Z − x 2 ) 3 1 B y = 1 ( Z + x 2 ) 3 y = \dfrac{1}{(Z + x^2)^3} y = ( Z + x 2 ) 3 1 C y = 1 ( Z − x 2 ) 1 3 y = \dfrac{1}{(Z - x^2)^{\frac13}} y = ( Z − x 2 ) 3 1 1 D y = 1 Z 3 − x 2 3 y = \dfrac{1}{\sqrt[3]{Z} - \sqrt[3]{x^2}} y = 3 Z − 3 x 2 1
Worked solution (try it first) Subtract
x 2 x^2 x 2 :
1 y 3 = Z − x 2 \frac{1}{y^3} = Z - x^2 y 3 1 = Z − x 2 .
Turn both sides upside down:
y 3 = 1 Z − x 2 y^3 = \frac{1}{Z - x^2} y 3 = Z − x 2 1 .
Undo the cube with a cube root, which is the power
1 3 \frac13 3 1 :
y = 1 ( Z − x 2 ) 1 3 y = \dfrac{1}{(Z - x^2)^{\frac13}} y = ( Z − x 2 ) 3 1 1 , option C.
Watch out
To undo a cube, take the cube root (power 1 3 \frac13 3 1 1 ( Z − x 2 ) 3 \frac{1}{(Z - x^2)^3} ( Z − x 2 ) 3 1 Also set as JAMB 1987 · UME · Q20
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Find the eleventh term of the progression 4 , 8 , 16 , … 4, 8, 16, \dots 4 , 8 , 16 , …
A 2 13 2^{13} 2 13 B 2 12 2^{12} 2 12 C 2 11 2^{11} 2 11 D 2 16 2^{16} 2 16
Worked solution (try it first) This is a G.P. with
a = 4 a = 4 a = 4 and
r = 8 ÷ 4 = 2 r = 8 \div 4 = 2 r = 8 ÷ 4 = 2 .
The
n n n th term is
a r n − 1 ar^{n - 1} a r n − 1 , so the eleventh term is
4 × 2 10 4 \times 2^{10} 4 × 2 10 .
Write 4 as
2 2 2^2 2 2 and add the powers:
2 2 × 2 10 = 2 12 2^2 \times 2^{10} = 2^{12} 2 2 × 2 10 = 2 12 , option B.
Watch out
The power is n − 1 = 10 n - 1 = 10 n − 1 = 10 r r r 2 11 2^{11} 2 11 4 × 2 11 = 2 13 4 \times 2^{11} = 2^{13} 4 × 2 11 = 2 13 Also set as JAMB 1987 · UME · Q30
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In how many ways can 6 subjects be selected from 10 subjects for an examination?
Worked solution (try it first) The order of the subjects doesn't matter, so this is
10 C 6 ^{10}C_6 10 C 6 , which is the same as
10 C 4 ^{10}C_4 10 C 4 .
10 C 4 = 10 × 9 × 8 × 7 4 × 3 × 2 × 1 ^{10}C_4 = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} 10 C 4 = 4 × 3 × 2 × 1 10 × 9 × 8 × 7 = 5040 24 = \dfrac{5040}{24} = 24 5040 .
That is 210, option D.
Watch out
The options are close together, so work carefully. Choosing 6 to take is the same as choosing 4 to leave out, and 10 C 4 ^{10}C_4 10 C 4 10 ! 6 ! 4 ! \frac{10!}{6!\,4!} 6 ! 4 ! 10 ! Report a problem with this question
Find the value of x x x f ( x ) = 2 x 3 − x 2 − 4 x + 4 f(x) = 2x^3 - x^2 - 4x + 4 f ( x ) = 2 x 3 − x 2 − 4 x + 4
A 2 3 \frac23 3 2 B 1 C − 2 3 -\frac23 − 3 2 D − 1 -1 − 1
Worked solution (try it first) At a turning point
f ′ ( x ) = 6 x 2 − 2 x − 4 = 0 f'(x) = 6x^2 - 2x - 4 = 0 f ′ ( x ) = 6 x 2 − 2 x − 4 = 0 .
Factorise:
2 ( 3 x + 2 ) ( x − 1 ) = 0 2(3x + 2)(x - 1) = 0 2 ( 3 x + 2 ) ( x − 1 ) = 0 , so
x = − 2 3 x = -\frac23 x = − 3 2 or
x = 1 x = 1 x = 1 .
f ′ ′ ( x ) = 12 x − 2 f''(x) = 12x - 2 f ′′ ( x ) = 12 x − 2 .
At
x = − 2 3 x = -\frac23 x = − 3 2 it is
− 10 < 0 -10 < 0 − 10 < 0 (maximum).
At
x = 1 x = 1 x = 1 it is
10 > 0 10 > 0 10 > 0 (minimum).
So the maximum is at
x = − 2 3 x = -\frac23 x = − 3 2 , option C.
Watch out
Use the second derivative to tell the turning points apart. x = 1 x = 1 x = 1 Report a problem with this question
Make L L L d = 42 w 5 L 2 d = \sqrt{\dfrac{42w}{5L^2}} d = 5 L 2 42 w
A 42 w 5 d \sqrt{\dfrac{42w}{5d}} 5 d 42 w B 42 w 5 d 2 \sqrt{\dfrac{42w}{5d^2}} 5 d 2 42 w C 42 5 d 2 \sqrt{\dfrac{42}{5d^2}} 5 d 2 42 D 1 2 42 w 5 \frac12\sqrt{\dfrac{42w}{5}} 2 1 5 42 w
Worked solution (try it first) Square both sides to remove the root:
d 2 = 42 w 5 L 2 d^2 = \dfrac{42w}{5L^2} d 2 = 5 L 2 42 w .
Multiply both sides by
5 L 2 5L^2 5 L 2 :
5 L 2 d 2 = 42 w 5L^2d^2 = 42w 5 L 2 d 2 = 42 w .
Divide by
5 d 2 5d^2 5 d 2 :
L 2 = 42 w 5 d 2 L^2 = \dfrac{42w}{5d^2} L 2 = 5 d 2 42 w .
Take the square root:
L = 42 w 5 d 2 L = \sqrt{\dfrac{42w}{5d^2}} L = 5 d 2 42 w , option B.
Watch out
Squaring the whole equation squares d d d d d d 42 w 5 d \sqrt{\frac{42w}{5d}} 5 d 42 w Report a problem with this question
Calculate the simple interest on ₦1,500 for 8 years at 5 % 5\% 5%
Worked solution (try it first) Use
I = P R T 100 I = \dfrac{PRT}{100} I = 100 P R T with
P = 1500 P = 1500 P = 1500 ,
R = 5 R = 5 R = 5 and
T = 8 T = 8 T = 8 .
One year's interest is
5 100 × 1500 = \frac{5}{100} \times 1500 = 100 5 × 1500 = ₦75, so 8 years give
8 × 75 = 600 8 \times 75 = 600 8 × 75 = 600 .
The simple interest is ₦600, option B.
Watch out
Multiply by all 8 years. Two years' interest is ₦150 (option D); one year's is only ₦75. Report a problem with this question
Evaluate ∫ − π / 2 π / 2 cos x d x \displaystyle\int_{-\pi/2}^{\pi/2} \cos x\,dx ∫ − π /2 π /2 cos x d x
Worked solution (try it first) cos x \cos x cos x integrates to
sin x \sin x sin x .
[ sin x ] − π / 2 π / 2 = sin π 2 − sin ( − π 2 ) \left[\sin x\right]_{-\pi/2}^{\pi/2} = \sin\frac\pi2 - \sin\left(-\frac\pi2\right) [ sin x ] − π /2 π /2 = sin 2 π − sin ( − 2 π ) Watch out
sin ( − π 2 ) = − 1 \sin\left(-\frac\pi2\right) = -1 sin ( − 2 π ) = − 1 + 1 +1 + 1 cos x ≥ 0 \cos x \ge 0 cos x ≥ 0 Report a problem with this question
If the lines 3 y = 4 x − 1 3y = 4x - 1 3 y = 4 x − 1 q y = x + 3 qy = x + 3 q y = x + 3 q q q
A − 4 3 -\frac43 − 3 4 B − 5 4 -\frac54 − 4 5 C 4 5 \frac45 5 4 D 3 4 \frac34 4 3
Worked solution (try it first) 3 y = 4 x − 1 3y = 4x - 1 3 y = 4 x − 1 gives
y = 4 3 x − 1 3 y = \frac43x - \frac13 y = 3 4 x − 3 1 , so its gradient is
4 3 \frac43 3 4 .
q y = x + 3 qy = x + 3 q y = x + 3 gives
y = 1 q x + 3 q y = \frac1qx + \frac3q y = q 1 x + q 3 , so its gradient is
1 q \frac1q q 1 .
Parallel lines have equal gradients:
1 q = 4 3 \frac1q = \frac43 q 1 = 3 4 , so
q = 3 4 q = \frac34 q = 4 3 , option D.
Watch out
Parallel means equal gradients. Using the perpendicular rule instead, 1 q = − 3 4 \frac1q = -\frac34 q 1 = − 4 3 q = − 4 3 q = -\frac43 q = − 3 4 Report a problem with this question
The volume of a hemispherical bowl is 718 2 3 cm 3 718\frac23\text{ cm}^3 718 3 2 cm 3 [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 4.0 cm 4.0\text{ cm} 4.0 cm B 5.6 cm 5.6\text{ cm} 5.6 cm C 7.0 cm 7.0\text{ cm} 7.0 cm D 3.6 cm 3.6\text{ cm} 3.6 cm
Worked solution (try it first) A hemisphere is half a sphere, so its volume is
2 3 π r 3 \frac23\pi r^3 3 2 π r 3 .
Write
718 2 3 718\frac23 718 3 2 as
2156 3 \frac{2156}{3} 3 2156 .
2 3 × 22 7 × r 3 = 2156 3 \frac23 \times \frac{22}{7} \times r^3 = \frac{2156}{3} 3 2 × 7 22 × r 3 = 3 2156 , so
r 3 = 2156 × 7 44 = 343 r^3 = \dfrac{2156 \times 7}{44} = 343 r 3 = 44 2156 × 7 = 343 .
Take the cube root:
r = 7.0 r = 7.0 r = 7.0 cm, option C.
Watch out
Use half the sphere's volume. With 4 3 π r 3 \frac43\pi r^3 3 4 π r 3 r 3 = 171.5 r^3 = 171.5 r 3 = 171.5 r ≈ 5.6 r \approx 5.6 r ≈ 5.6 Report a problem with this question
A baking recipe calls for 2.5 kg 2.5\text{ kg} 2.5 kg 4.5 kg 4.5\text{ kg} 4.5 kg 24.5 kg 24.5\text{ kg} 24.5 kg
A 12.25 kg 12.25\text{ kg} 12.25 kg B 6.75 kg 6.75\text{ kg} 6.75 kg C 8.75 kg 8.75\text{ kg} 8.75 kg D 15.75 kg 15.75\text{ kg} 15.75 kg
Worked solution (try it first) The recipe uses sugar and flour in the ratio
2.5 : 4.5 2.5 : 4.5 2.5 : 4.5 , which is 7 kg of mixture in all.
So sugar is
2.5 7 \frac{2.5}{7} 7 2.5 of any mixture:
2.5 7 × 24.5 = 2.5 × 3.5 \frac{2.5}{7} \times 24.5 = 2.5 \times 3.5 7 2.5 × 24.5 = 2.5 × 3.5 = 8.75 = 8.75 = 8.75 kg, option C.
Watch out
The question asks for sugar. 4.5 7 × 24.5 = 15.75 \frac{4.5}{7} \times 24.5 = 15.75 7 4.5 × 24.5 = 15.75 Report a problem with this question
Two cars X X X Y Y Y P P P 150 km 150\text{ km} 150 km Y Y Y 60 km/h 60\text{ km/h} 60 km/h X X X P P P Y Y Y X X X
A 51 3 9 km/h 51\frac39\text{ km/h} 51 9 3 km/h B 72 km/h 72\text{ km/h} 72 km/h C 66 km/h 66\text{ km/h} 66 km/h D 37 1 2 km/h 37\frac12\text{ km/h} 37 2 1 km/h
Worked solution (try it first) Time is distance over speed:
Y Y Y takes
150 60 = 2 1 2 \dfrac{150}{60} = 2\frac12 60 150 = 2 2 1 hours.
X X X arrives 25 minutes earlier. 25 minutes is
25 60 = 5 12 \frac{25}{60} = \frac{5}{12} 60 25 = 12 5 hour, so
X X X takes
2 1 2 − 5 12 = 25 12 2\frac12 - \frac{5}{12} = \frac{25}{12} 2 2 1 − 12 5 = 12 25 hours.
Speed of
X X X = 150 ÷ 25 12 = 150 \div \dfrac{25}{12} = 150 ÷ 12 25 = 150 × 12 25 = 150 \times \dfrac{12}{25} = 150 × 25 12 = 72 = 72 = 72 km/h, option B.
Watch out
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