Paper JAMB 1994 General Maths Objective
Objective paper · 45 questions · partial
JAMB 1994 · UME Topics include Number foundations & fractions, Approximation & error, Indices & standard form, Logarithms, Surds, Sets & Venn diagrams.
Our copy of this paper is missing questions 10, 18, 26, 44, 48.
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.
1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 19 20 21 22 23 24 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 46 47 49 50 Evaluate 1 3 ÷ [ 5 7 ( 9 10 − 1 + 3 4 ) ] \frac13 \div \left[\frac57\left(\frac{9}{10} - 1 + \frac34\right)\right] 3 1 ÷ [ 7 5 ( 10 9 − 1 + 4 3 ) ]
A 28 39 \frac{28}{39} 39 28 B 13 84 \frac{13}{84} 84 13 C 39 28 \frac{39}{28} 28 39 D 84 13 \frac{84}{13} 13 84
Worked solution (try it first) Inside the bracket, use the LCD 20:
9 10 − 1 + 3 4 = 18 − 20 + 15 20 \frac{9}{10} - 1 + \frac34 = \frac{18 - 20 + 15}{20} 10 9 − 1 + 4 3 = 20 18 − 20 + 15 = 13 20 = \frac{13}{20} = 20 13 .
Multiply by
5 7 \frac57 7 5 :
5 7 × 13 20 = 13 28 \frac57 \times \frac{13}{20} = \frac{13}{28} 7 5 × 20 13 = 28 13 .
To divide, multiply by the reciprocal:
1 3 × 28 13 = 28 39 \frac13 \times \frac{28}{13} = \frac{28}{39} 3 1 × 13 28 = 39 28 , option A.
Watch out
Turn the second fraction upside down, not the first. Working out 13 28 ÷ 1 3 \frac{13}{28} \div \frac13 28 13 ÷ 3 1 39 28 \frac{39}{28} 28 39 Report a problem with this question
Evaluate ( 0.36 × 5.4 × 0.63 ) ÷ ( 4.2 × 9.0 × 2.4 ) (0.36 \times 5.4 \times 0.63) \div (4.2 \times 9.0 \times 2.4) ( 0.36 × 5.4 × 0.63 ) ÷ ( 4.2 × 9.0 × 2.4 )
Worked solution (try it first) Top:
0.36 × 5.4 = 1.944 0.36 \times 5.4 = 1.944 0.36 × 5.4 = 1.944 , and
1.944 × 0.63 = 1.22472 1.944 \times 0.63 = 1.22472 1.944 × 0.63 = 1.22472 .
Bottom:
4.2 × 9.0 = 37.8 4.2 \times 9.0 = 37.8 4.2 × 9.0 = 37.8 , and
37.8 × 2.4 = 90.72 37.8 \times 2.4 = 90.72 37.8 × 2.4 = 90.72 .
Divide:
1.22472 ÷ 90.72 = 0.0135 1.22472 \div 90.72 = 0.0135 1.22472 ÷ 90.72 = 0.0135 .
The third significant figure is 5, so round up: 0.014, option B.
Watch out
0.0135 to 2 significant figures rounds up to 0.014, not down to 0.013 (option A). Also count significant figures from the 1, not from the point, or you get 0.14 (option D). Report a problem with this question
Evaluate log 5 0.04 log 3 18 − log 3 2 \dfrac{\log_5 0.04}{\log_3 18 - \log_3 2} log 3 18 − log 3 2 log 5 0.04
A 1 B − 1 -1 − 1 C 2 3 \frac23 3 2 D − 2 3 -\frac23 − 3 2
Worked solution (try it first) 0.04 = 1 25 = 5 − 2 0.04 = \frac{1}{25} = 5^{-2} 0.04 = 25 1 = 5 − 2 , so
log 5 0.04 = − 2 \log_5 0.04 = -2 log 5 0.04 = − 2 .
Subtracting logs divides:
log 3 18 − log 3 2 = log 3 9 = 2 \log_3 18 - \log_3 2 = \log_3 9 = 2 log 3 18 − log 3 2 = log 3 9 = 2 .
So the value is
− 2 2 = − 1 \frac{-2}{2} = -1 2 − 2 = − 1 , option B.
Watch out
0.04 is less than 1, so its log is negative: log 5 0.04 = − 2 \log_5 0.04 = -2 log 5 0.04 = − 2 + 2 +2 + 2 Report a problem with this question
Without using tables, solve the equation 8 x − 2 = 2 25 8x^{-2} = \frac{2}{25} 8 x − 2 = 25 2
Worked solution (try it first) The index
− 2 -2 − 2 applies only to
x x x :
8 x − 2 = 8 x 2 8x^{-2} = \frac{8}{x^2} 8 x − 2 = x 2 8 , so
8 x 2 = 2 25 \frac{8}{x^2} = \frac{2}{25} x 2 8 = 25 2 .
Cross-multiply:
2 x 2 = 8 × 25 = 200 2x^2 = 8 \times 25 = 200 2 x 2 = 8 × 25 = 200 , so
x 2 = 100 x^2 = 100 x 2 = 100 .
Take the square root:
x = 10 x = 10 x = 10 , option D.
Watch out
8 x − 2 8x^{-2} 8 x − 2 8 x 2 \frac{8}{x^2} x 2 8 1 8 x 2 \frac{1}{8x^2} 8 x 2 1 x 2 = 25 16 x^2 = \frac{25}{16} x 2 = 16 25 Report a problem with this question
Simplify 48 − 9 3 + 75 \sqrt{48} - \dfrac{9}{\sqrt3} + \sqrt{75} 48 − 3 9 + 75
A 5 3 5\sqrt3 5 3 B 6 3 6\sqrt3 6 3 C 8 3 8\sqrt3 8 3 D 18 3 18\sqrt3 18 3
Worked solution (try it first) Take out square factors:
48 = 4 3 \sqrt{48} = 4\sqrt3 48 = 4 3 and
75 = 5 3 \sqrt{75} = 5\sqrt3 75 = 5 3 .
Rationalise:
9 3 = 9 3 3 \dfrac{9}{\sqrt3} = \dfrac{9\sqrt3}{3} 3 9 = 3 9 3 , which is
3 3 3\sqrt3 3 3 .
Combine:
4 3 − 3 3 + 5 3 = 6 3 4\sqrt3 - 3\sqrt3 + 5\sqrt3 = 6\sqrt3 4 3 − 3 3 + 5 3 = 6 3 , option B.
Watch out
9 3 = 3 3 \frac{9}{\sqrt3} = 3\sqrt3 3 9 = 3 3 3 \sqrt3 3 3 \sqrt3 3 4 − 1 + 5 = 8 4 - 1 + 5 = 8 4 − 1 + 5 = 8 8 3 8\sqrt3 8 3 Report a problem with this question
Given that 2 = 1.414 \sqrt2 = 1.414 2 = 1.414 1 2 \frac{1}{\sqrt2} 2 1
Worked solution (try it first) Rationalise:
1 2 = 2 2 \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2} 2 1 = 2 2 .
Now you only need to halve:
1.414 2 = 0.707 \dfrac{1.414}{2} = 0.707 2 1.414 = 0.707 , option D.
Watch out
Rationalise first so that you divide by 2, not by 1.414. Just moving the decimal point (0.141, option A) is not the same as taking the reciprocal. Report a problem with this question
In a science class of 42 students, each offers at least one of Mathematics and Physics. If 22 students offer Physics and 28 offer Mathematics, how many students offer Physics only?
Worked solution (try it first) Every student takes at least one subject, so
n ( M ∪ P ) = 42 n(M \cup P) = 42 n ( M ∪ P ) = 42 .
Both subjects: add the two subjects and take away the total,
28 + 22 − 42 = 8 28 + 22 - 42 = 8 28 + 22 − 42 = 8 .
Physics only:
22 − 8 = 14 22 - 8 = 14 22 − 8 = 14 , option D.
Watch out
8 (option B) is the number taking both subjects. Take it away from the 22 Physics students to get Physics only. Report a problem with this question
Given that for sets A A A B B B E E E A ⊆ B A \subseteq B A ⊆ B A ∩ ( A ∩ B ) ′ A \cap (A \cap B)' A ∩ ( A ∩ B ) ′
A A A A B ∅ \varnothing ∅ C B B B D E E E
Worked solution (try it first) A ⊆ B A \subseteq B A ⊆ B means every element of
A A A is in
B B B , so
A ∩ B = A A \cap B = A A ∩ B = A .
So the expression is
A ∩ A ′ A \cap A' A ∩ A ′ .
Nothing is in a set and in its complement at once, so
A ∩ A ′ = ∅ A \cap A' = \varnothing A ∩ A ′ = ∅ , option B.
Watch out
Notice the dash on the bracket: ( A ∩ B ) ′ (A \cap B)' ( A ∩ B ) ′ A ′ A' A ′ A A A A ∩ A = A A \cap A = A A ∩ A = A Report a problem with this question
Solve for x x x 25 x + 3 ( 5 x ) = 4 25^x + 3(5^x) = 4 2 5 x + 3 ( 5 x ) = 4
A 1 or − 4 -4 − 4 B 0 C 1 D − 4 -4 − 4
Worked solution (try it first) Then
25 x = ( 5 2 ) x = u 2 25^x = (5^2)^x = u^2 2 5 x = ( 5 2 ) x = u 2 , and the equation is
u 2 + 3 u − 4 = 0 u^2 + 3u - 4 = 0 u 2 + 3 u − 4 = 0 .
Factorise:
( u + 4 ) ( u − 1 ) = 0 (u + 4)(u - 1) = 0 ( u + 4 ) ( u − 1 ) = 0 , so
u = 1 u = 1 u = 1 or
u = − 4 u = -4 u = − 4 .
5 x 5^x 5 x is never negative, so reject
− 4 -4 − 4 .
From
5 x = 1 = 5 0 5^x = 1 = 5^0 5 x = 1 = 5 0 ,
x = 0 x = 0 x = 0 , option B.
Watch out
1 and − 4 -4 − 4 5 x 5^x 5 x x x x 5 x = 1 5^x = 1 5 x = 1 x = 0 x = 0 x = 0 5 x = − 4 5^x = -4 5 x = − 4 Report a problem with this question
Factorize a 2 x − b 2 y − b 2 x + a 2 y a^2x - b^2y - b^2x + a^2y a 2 x − b 2 y − b 2 x + a 2 y
A ( a − b ) ( x + y ) (a - b)(x + y) ( a − b ) ( x + y ) B ( y − x ) ( a − b ) ( a + b ) (y - x)(a - b)(a + b) ( y − x ) ( a − b ) ( a + b ) C ( x − y ) ( a − b ) ( a + b ) (x - y)(a - b)(a + b) ( x − y ) ( a − b ) ( a + b ) D ( x + y ) ( a − b ) ( a + b ) (x + y)(a - b)(a + b) ( x + y ) ( a − b ) ( a + b )
Worked solution (try it first) Group the terms with
a 2 a^2 a 2 and those with
b 2 b^2 b 2 :
( a 2 x + a 2 y ) − ( b 2 x + b 2 y ) (a^2x + a^2y) - (b^2x + b^2y) ( a 2 x + a 2 y ) − ( b 2 x + b 2 y ) .
Take out the common factors:
a 2 ( x + y ) − b 2 ( x + y ) = ( x + y ) ( a 2 − b 2 ) a^2(x + y) - b^2(x + y) = (x + y)(a^2 - b^2) a 2 ( x + y ) − b 2 ( x + y ) = ( x + y ) ( a 2 − b 2 ) .
Factorise the difference of two squares:
( x + y ) ( a − b ) ( a + b ) (x + y)(a - b)(a + b) ( x + y ) ( a − b ) ( a + b ) , option D.
Watch out
Taking out − b 2 -b^2 − b 2 − b 2 y − b 2 x -b^2y - b^2x − b 2 y − b 2 x − b 2 ( x + y ) -b^2(x + y) − b 2 ( x + y ) x − y x - y x − y Report a problem with this question
Find the values of p p p q q q ( x − 1 ) (x - 1) ( x − 1 ) ( x − 3 ) (x - 3) ( x − 3 ) p x 3 + q x 2 + 11 x − 6 px^3 + qx^2 + 11x - 6 p x 3 + q x 2 + 11 x − 6
A − 1 , − 6 -1, -6 − 1 , − 6 B 1 , − 6 1, -6 1 , − 6 C 1 , 6 1, 6 1 , 6 D 6 , − 1 6, -1 6 , − 1
Worked solution (try it first) By the factor theorem, the expression is 0 at
x = 1 x = 1 x = 1 :
p + q + 11 − 6 = 0 p + q + 11 - 6 = 0 p + q + 11 − 6 = 0 , so
p + q = − 5 p + q = -5 p + q = − 5 .
It is 0 at
x = 3 x = 3 x = 3 :
27 p + 9 q + 33 − 6 = 0 27p + 9q + 33 - 6 = 0 27 p + 9 q + 33 − 6 = 0 .
Divide by 9:
3 p + q = − 3 3p + q = -3 3 p + q = − 3 .
Subtract the first equation:
2 p = 2 2p = 2 2 p = 2 , so
p = 1 p = 1 p = 1 and
q = − 6 q = -6 q = − 6 , option B.
Watch out
Include the 11 x 11x 11 x x = 3 x = 3 x = 3 33 − 6 = 27 33 - 6 = 27 33 − 6 = 27 1 − 6 + 11 − 6 = 0 1 - 6 + 11 - 6 = 0 1 − 6 + 11 − 6 = 0 Report a problem with this question
The equation of the graph shown is
A y = ( x − 3 ) 3 y = (x - 3)^3 y = ( x − 3 ) 3 B y = ( x + 3 ) 3 y = (x + 3)^3 y = ( x + 3 ) 3 C y = x 3 − 27 y = x^3 - 27 y = x 3 − 27 D y = − x 3 + 27 y = -x^3 + 27 y = − x 3 + 27
Worked solution (try it first) The curve touches the
x x x -axis at
( 3 , 0 ) (3, 0) ( 3 , 0 ) and flattens out there, so
( x − 3 ) (x - 3) ( x − 3 ) is a repeated factor, three times over for a cubic.
That suggests
y = ( x − 3 ) 3 y = (x - 3)^3 y = ( x − 3 ) 3 .
Check the
y y y -intercept: at
x = 0 x = 0 x = 0 ,
y = ( − 3 ) 3 = − 27 y = (-3)^3 = -27 y = ( − 3 ) 3 = − 27 , which matches the graph.
So the equation is
y = ( x − 3 ) 3 y = (x - 3)^3 y = ( x − 3 ) 3 , option A.
Watch out
y = x 3 − 27 y = x^3 - 27 y = x 3 − 27 ( 0 , − 27 ) (0, -27) ( 0 , − 27 ) ( 3 , 0 ) (3, 0) ( 3 , 0 ) ( 0 , − 27 ) (0, -27) ( 0 , − 27 ) ( 3 , 0 ) (3, 0) ( 3 , 0 ) ( x − 3 ) 3 (x - 3)^3 ( x − 3 ) 3 Report a problem with this question
If a = 1 a = 1 a = 1 b = 3 b = 3 b = 3 x x x a a − x = b x − b \dfrac{a}{a - x} = \dfrac{b}{x - b} a − x a = x − b b
A 4 3 \frac43 3 4 B 2 3 \frac23 3 2 C 3 2 \frac32 2 3 D 3 4 \frac34 4 3
Worked solution (try it first) Put in
a = 1 a = 1 a = 1 and
b = 3 b = 3 b = 3 :
1 1 − x = 3 x − 3 \dfrac{1}{1 - x} = \dfrac{3}{x - 3} 1 − x 1 = x − 3 3 .
Cross-multiply:
x − 3 = 3 ( 1 − x ) x - 3 = 3(1 - x) x − 3 = 3 ( 1 − x ) , so
x − 3 = 3 − 3 x x - 3 = 3 - 3x x − 3 = 3 − 3 x .
Add
3 x 3x 3 x and 3 to both sides:
4 x = 6 4x = 6 4 x = 6 .
Divide by 4:
x = 3 2 x = \frac32 x = 2 3 , option C.
Watch out
From 4 x = 6 4x = 6 4 x = 6 x = 6 4 = 3 2 x = \frac64 = \frac32 x = 4 6 = 2 3 2 3 \frac23 3 2 Report a problem with this question
Solve for r r r 1 r − 1 + 2 r + 1 = 3 r \dfrac{1}{r - 1} + \dfrac{2}{r + 1} = \dfrac3r r − 1 1 + r + 1 2 = r 3
Worked solution (try it first) Add the left side over the common denominator
( r − 1 ) ( r + 1 ) = r 2 − 1 (r - 1)(r + 1) = r^2 - 1 ( r − 1 ) ( r + 1 ) = r 2 − 1 : the top is
( r + 1 ) + 2 ( r − 1 ) = 3 r − 1 (r + 1) + 2(r - 1) = 3r - 1 ( r + 1 ) + 2 ( r − 1 ) = 3 r − 1 .
So
3 r − 1 r 2 − 1 = 3 r \dfrac{3r - 1}{r^2 - 1} = \dfrac3r r 2 − 1 3 r − 1 = r 3 .
Cross-multiply:
r ( 3 r − 1 ) = 3 ( r 2 − 1 ) r(3r - 1) = 3(r^2 - 1) r ( 3 r − 1 ) = 3 ( r 2 − 1 ) .
Expand:
3 r 2 − r = 3 r 2 − 3 3r^2 - r = 3r^2 - 3 3 r 2 − r = 3 r 2 − 3 .
The
3 r 2 3r^2 3 r 2 terms cancel, leaving
− r = − 3 -r = -3 − r = − 3 .
So
r = 3 r = 3 r = 3 , option A.
Watch out
Multiply the whole bracket: 2 ( r − 1 ) = 2 r − 2 2(r - 1) = 2r - 2 2 ( r − 1 ) = 2 r − 2 2 r − 1 2r - 1 2 r − 1 3 r 3r 3 r Report a problem with this question
Find P P P x − 3 ( 1 − x ) ( x + 2 ) = P 1 − x + Q x + 2 \dfrac{x - 3}{(1 - x)(x + 2)} = \dfrac{P}{1 - x} + \dfrac{Q}{x + 2} ( 1 − x ) ( x + 2 ) x − 3 = 1 − x P + x + 2 Q
A − 2 3 -\frac23 − 3 2 B − 5 3 -\frac53 − 3 5 C 5 3 \frac53 3 5 D 2 3 \frac23 3 2
Worked solution (try it first) Multiply through by
( 1 − x ) ( x + 2 ) (1 - x)(x + 2) ( 1 − x ) ( x + 2 ) :
x − 3 = P ( x + 2 ) + Q ( 1 − x ) x - 3 = P(x + 2) + Q(1 - x) x − 3 = P ( x + 2 ) + Q ( 1 − x ) .
Put
x = 1 x = 1 x = 1 so the
Q Q Q term vanishes:
1 − 3 = 3 P 1 - 3 = 3P 1 − 3 = 3 P , so
− 2 = 3 P -2 = 3P − 2 = 3 P .
So
P = − 2 3 P = -\frac23 P = − 3 2 , option A.
Watch out
To find P P P 1 − x = 0 1 - x = 0 1 − x = 0 x = 1 x = 1 x = 1 x = − 2 x = -2 x = − 2 Q Q Q − 5 = 3 Q -5 = 3Q − 5 = 3 Q − 5 3 -\frac53 − 3 5 Report a problem with this question
Find the range of values of x x x 1 x > 2 \frac1x > 2 x 1 > 2
A x < 1 2 x < \frac12 x < 2 1 B x < 0 x < 0 x < 0 x > 1 2 x > \frac12 x > 2 1 C 0 < x < 1 2 0 < x < \frac12 0 < x < 2 1 D 1 < x < 2 1 < x < 2 1 < x < 2
Worked solution (try it first) If
x x x were negative,
1 x \frac1x x 1 would be negative and could not be greater than 2.
With
x x x positive, multiply both sides by
x x x without reversing the sign:
1 > 2 x 1 > 2x 1 > 2 x , so
x < 1 2 x < \frac12 x < 2 1 .
Put the two together:
0 < x < 1 2 0 < x < \frac12 0 < x < 2 1 , option C.
Watch out
Option A, x < 1 2 x < \frac12 x < 2 1 x = − 1 x = -1 x = − 1 1 x = − 1 \frac1x = -1 x 1 = − 1 x x x Report a problem with this question
If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference.
A 12 5 \frac{12}{5} 5 12 B 5 3 \frac53 3 5 C − 2 -2 − 2 D 2
Worked solution (try it first) The 6th term of an A.P. is
a + 5 d a + 5d a + 5 d , so
1 + 5 d = 11 1 + 5d = 11 1 + 5 d = 11 .
Subtract 1 from both sides:
5 d = 10 5d = 10 5 d = 10 .
So
d = 2 d = 2 d = 2 , option D.
Watch out
The 6th term has 5 d 5d 5 d 6 d 6d 6 d 1 + 6 d = 11 1 + 6d = 11 1 + 6 d = 11 d = 10 6 = 5 3 d = \frac{10}{6} = \frac53 d = 6 10 = 3 5 Report a problem with this question
Find the value of r r r log 10 r + log 10 r 2 + log 10 r 4 + log 10 r 8 + log 10 r 16 + log 10 r 32 = 63 \log_{10} r + \log_{10} r^2 + \log_{10} r^4 + \log_{10} r^8 + \log_{10} r^{16} + \log_{10} r^{32} = 63 log 10 r + log 10 r 2 + log 10 r 4 + log 10 r 8 + log 10 r 16 + log 10 r 32 = 63
A 10 − 8 10^{-8} 1 0 − 8 B 10 0 10^0 1 0 0 C 10 D 10 2 10^2 1 0 2
Worked solution (try it first) Bring each power down: the left side is
( 1 + 2 + 4 + 8 + 16 + 32 ) log 10 r = 63 log 10 r (1 + 2 + 4 + 8 + 16 + 32)\log_{10} r = 63\log_{10} r ( 1 + 2 + 4 + 8 + 16 + 32 ) log 10 r = 63 log 10 r .
So
63 log 10 r = 63 63\log_{10} r = 63 63 log 10 r = 63 , and
log 10 r = 1 \log_{10} r = 1 log 10 r = 1 .
Change to index form:
r = 10 1 = 10 r = 10^1 = 10 r = 1 0 1 = 10 , option C.
Watch out
log 10 r = 1 \log_{10} r = 1 log 10 r = 1 r = 10 1 = 10 r = 10^1 = 10 r = 1 0 1 = 10 log 10 r = 0 \log_{10} r = 0 log 10 r = 0 r = 10 0 = 1 r = 10^0 = 1 r = 1 0 0 = 1 Report a problem with this question
Find the n n n 3 , 6 , 10 , 15 , 21 , … 3, 6, 10, 15, 21, \dots 3 , 6 , 10 , 15 , 21 , …
A n ( n − 1 2 ) n\left(n - \frac12\right) n ( n − 2 1 ) B n ( n + 1 2 ) n\left(n + \frac12\right) n ( n + 2 1 ) C ( n + 1 ) ( n + 2 ) 2 \frac{(n + 1)(n + 2)}{2} 2 ( n + 1 ) ( n + 2 ) D n ( 2 n + 1 ) n(2n + 1) n ( 2 n + 1 )
Worked solution (try it first) Each term is half the product of two consecutive numbers:
3 = 2 × 3 2 3 = \frac{2 \times 3}{2} 3 = 2 2 × 3 ,
6 = 3 × 4 2 6 = \frac{3 \times 4}{2} 6 = 2 3 × 4 ,
10 = 4 × 5 2 10 = \frac{4 \times 5}{2} 10 = 2 4 × 5 .
The first factor is always one more than the term number, so the
n n n th term is
( n + 1 ) ( n + 2 ) 2 \frac{(n + 1)(n + 2)}{2} 2 ( n + 1 ) ( n + 2 ) .
Check
n = 5 n = 5 n = 5 :
6 × 7 2 = 21 \frac{6 \times 7}{2} = 21 2 6 × 7 = 21 .
So the answer is option C.
Watch out
Check more than the first term. Option D, n ( 2 n + 1 ) n(2n + 1) n ( 2 n + 1 ) n = 1 n = 1 n = 1 n = 2 n = 2 n = 2 Report a problem with this question
A binary operation ∗ * ∗ a ∗ b = a b a * b = ab a ∗ b = ab
A Closure B Associativity C Identity D Inverse
Worked solution (try it first) Closure and associativity hold: the product of two positive integers is a positive integer, and
( a b ) c = a ( b c ) (ab)c = a(bc) ( ab ) c = a ( b c ) .
The identity is 1, because
a × 1 = a a \times 1 = a a × 1 = a , and 1 is a positive integer.
An inverse of 2 would need
2 b = 1 2b = 1 2 b = 1 , so
b = 1 2 b = \frac12 b = 2 1 , which is not a positive integer.
So inverse fails, option D.
Watch out
Check the identity is in the set before ruling it out: 1 is a positive integer, so the identity exists (option C holds). It is the inverses, such as 1 2 \frac12 2 1 Report a problem with this question
The table shows multiplication modulo 10 on the set S = { 2 , 4 , 6 , 8 } S = \{2, 4, 6, 8\} S = { 2 , 4 , 6 , 8 }
⊗ \otimes ⊗ 2
4
6
8
2
4
8
2
6
4
8
6
4
2
6
2
4
6
8
8
6
2
8
4
Worked solution (try it first) Find the identity first: the row for 6 reads 2, 4, 6, 8, the same as the heading, so 6 is the identity.
The inverse of 2 is the element that combines with 2 to give the identity 6.
In the row for 2, the 6 is under the column 8:
2 ⊗ 8 = 16 2 \otimes 8 = 16 2 ⊗ 8 = 16 , which is 6 modulo 10.
So the inverse of 2 is 8, option D.
Watch out
6 is the identity, not the inverse: 2 ⊗ 6 = 2 2 \otimes 6 = 2 2 ⊗ 6 = 2 Report a problem with this question
Solve for x x x y y y ( 1 1 3 y ) ( x 1 ) = ( 4 1 ) \begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\begin{pmatrix} x \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} ( 1 3 1 y ) ( x 1 ) = ( 4 1 )
A x = − 3 , y = 3 x = -3, y = 3 x = − 3 , y = 3 B x = 8 , y = 3 x = 8, y = 3 x = 8 , y = 3 C x = 3 , y = − 8 x = 3, y = -8 x = 3 , y = − 8 D x = 8 , y = − 3 x = 8, y = -3 x = 8 , y = − 3
Worked solution (try it first) Multiply each row of the square matrix by the column.
The first row gives
1 × x + 1 × 1 = x + 1 1 \times x + 1 \times 1 = x + 1 1 × x + 1 × 1 = x + 1 .
The second row gives
3 × x + y × 1 = 3 x + y 3 \times x + y \times 1 = 3x + y 3 × x + y × 1 = 3 x + y .
Match with
( 4 1 ) \begin{pmatrix} 4 \\ 1 \end{pmatrix} ( 4 1 ) :
x + 1 = 4 x + 1 = 4 x + 1 = 4 , so
x = 3 x = 3 x = 3 .
Then
3 ( 3 ) + y = 1 3(3) + y = 1 3 ( 3 ) + y = 1 , so
y = 1 − 9 = − 8 y = 1 - 9 = -8 y = 1 − 9 = − 8 .
The answer is
x = 3 x = 3 x = 3 ,
y = − 8 y = -8 y = − 8 , option C.
Watch out
Keep track of which value is which: x x x x = 8 x = 8 x = 8 y = − 3 y = -3 y = − 3 x + 1 = 4 x + 1 = 4 x + 1 = 4 Report a problem with this question
The determinant of the matrix ( 1 2 3 4 5 6 2 0 − 1 ) \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 0 & -1 \end{pmatrix} 1 4 2 2 5 0 3 6 − 1
A − 67 -67 − 67 B − 57 -57 − 57 C − 3 -3 − 3 D 3
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
1 × ( 5 × ( − 1 ) − 6 × 0 ) = − 5 1 \times (5 \times (-1) - 6 \times 0) = -5 1 × ( 5 × ( − 1 ) − 6 × 0 ) = − 5 .
The second term is
− 2 × ( 4 × ( − 1 ) − 6 × 2 ) = − 2 × ( − 16 ) -2 \times (4 \times (-1) - 6 \times 2) = -2 \times (-16) − 2 × ( 4 × ( − 1 ) − 6 × 2 ) = − 2 × ( − 16 ) The third term is
3 × ( 4 × 0 − 5 × 2 ) = 3 × ( − 10 ) 3 \times (4 \times 0 - 5 \times 2) = 3 \times (-10) 3 × ( 4 × 0 − 5 × 2 ) = 3 × ( − 10 ) Add them:
− 5 + 32 − 30 = − 3 -5 + 32 - 30 = -3 − 5 + 32 − 30 = − 3 , option C.
Watch out
The middle term of the expansion takes a minus sign. Adding it instead gives − 5 − 32 − 30 = − 67 -5 - 32 - 30 = -67 − 5 − 32 − 30 = − 67 Report a problem with this question
In the diagram, O O O S O Q SOQ S O Q ∠ P R S = 38 ∘ \angle PRS = 38^\circ ∠ P R S = 3 8 ∘ ∠ P S Q \angle PSQ ∠ P S Q
A 148 ∘ 148^\circ 14 8 ∘ B 104 ∘ 104^\circ 10 4 ∘ C 80 ∘ 80^\circ 8 0 ∘ D 52 ∘ 52^\circ 5 2 ∘
Worked solution (try it first) Angles in the same segment are equal.
∠ P Q S \angle PQS ∠ P QS and
∠ P R S \angle PRS ∠ P R S both stand on arc
P S PS P S , so
∠ P Q S = 38 ∘ \angle PQS = 38^\circ ∠ P QS = 3 8 ∘ .
S Q SQ S Q is a diameter, so
∠ S P Q = 90 ∘ \angle SPQ = 90^\circ ∠ S P Q = 9 0 ∘ (angle in a semicircle).
The angles of triangle
P S Q PSQ P S Q add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P S Q = 180 ∘ − 90 ∘ − 38 ∘ \angle PSQ = 180^\circ - 90^\circ - 38^\circ ∠ P S Q = 18 0 ∘ − 9 0 ∘ − 3 8 ∘ = 52 ∘ = 52^\circ = 5 2 ∘ , option D.
Watch out
Use the right angle at P P P P S Q PSQ P S Q 90 ∘ 90^\circ 9 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 180 ∘ − 38 ∘ = 142 ∘ 180^\circ - 38^\circ = 142^\circ 18 0 ∘ − 3 8 ∘ = 14 2 ∘ Report a problem with this question
If three angles of a quadrilateral are ( 3 y − x − z ) ∘ (3y - x - z)^\circ ( 3 y − x − z ) ∘ 3 x ∘ 3x^\circ 3 x ∘ ( 2 z − 2 y − x ) ∘ (2z - 2y - x)^\circ ( 2 z − 2 y − x ) ∘ x x x y y y z z z
A ( 360 − x − y − z ) ∘ (360 - x - y - z)^\circ ( 360 − x − y − z ) ∘ B ( 360 + x + y − z ) ∘ (360 + x + y - z)^\circ ( 360 + x + y − z ) ∘ C ( 180 − x + y + z ) ∘ (180 - x + y + z)^\circ ( 180 − x + y + z ) ∘ D ( 180 + x + y + z ) ∘ (180 + x + y + z)^\circ ( 180 + x + y + z ) ∘
Worked solution (try it first) The angles of a quadrilateral add up to
360 ∘ 360^\circ 36 0 ∘ .
Add the three given angles:
( 3 y − x − z ) + 3 x + ( 2 z − 2 y − x ) = x + y + z (3y - x - z) + 3x + (2z - 2y - x) = x + y + z ( 3 y − x − z ) + 3 x + ( 2 z − 2 y − x ) = x + y + z .
So the fourth angle is
360 − ( x + y + z ) = ( 360 − x − y − z ) ∘ 360 - (x + y + z) = (360 - x - y - z)^\circ 360 − ( x + y + z ) = ( 360 − x − y − z ) ∘ , option A.
Watch out
A quadrilateral's angles add up to 360 ∘ 360^\circ 36 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
An open rectangular box is made of wood 2 cm thick. If the internal dimensions of the box are 50 cm long, 36 cm wide and 20 cm deep, what is the volume of wood in the box?
A 11520 cm 3 11520\text{ cm}^3 11520 cm 3 B 36000 cm 3 36000\text{ cm}^3 36000 cm 3 C 38200 cm 3 38200\text{ cm}^3 38200 cm 3 D 47520 cm 3 47520\text{ cm}^3 47520 cm 3
Worked solution (try it first) External size: add 2 cm on each side of the length and width, but only the base on the height (the box is open):
54 × 40 × 22 = 47 520 cm 3 54 \times 40 \times 22 = 47\,520\text{ cm}^3 54 × 40 × 22 = 47 520 cm 3 .
Internal volume:
50 × 36 × 20 = 36 000 cm 3 50 \times 36 \times 20 = 36\,000\text{ cm}^3 50 × 36 × 20 = 36 000 cm 3 .
Wood:
47 520 − 36 000 = 11 520 cm 3 47\,520 - 36\,000 = 11\,520\text{ cm}^3 47 520 − 36 000 = 11 520 cm 3 , option A.
Watch out
Subtract the inside. 47 520 cm 3 47\,520\text{ cm}^3 47 520 cm 3 36 000 cm 3 36\,000\text{ cm}^3 36 000 cm 3 Report a problem with this question
Calculate the perimeter, in cm, of a sector of a circle of radius 8 cm and angle 45 ∘ 45^\circ 4 5 ∘
A 2 π 2\pi 2 π B 8 + 2 π 8 + 2\pi 8 + 2 π C 16 + 2 π 16 + 2\pi 16 + 2 π D 16 + 16 π 16 + 16\pi 16 + 16 π
Worked solution (try it first) The circumference is
2 π × 8 = 16 π 2\pi \times 8 = 16\pi 2 π × 8 = 16 π cm.
The arc is
45 360 = 1 8 \frac{45}{360} = \frac18 360 45 = 8 1 of it:
1 8 × 16 π = 2 π \frac18 \times 16\pi = 2\pi 8 1 × 16 π = 2 π cm.
The perimeter adds both radii:
8 + 8 + 2 π = 16 + 2 π 8 + 8 + 2\pi = 16 + 2\pi 8 + 8 + 2 π = 16 + 2 π cm, option C.
Watch out
A sector has two straight edges. Adding only one radius gives 8 + 2 π 8 + 2\pi 8 + 2 π Report a problem with this question
In the diagram, P T S PTS P T S T Q R TQR T QR T T T ∠ Q R T = 60 ∘ \angle QRT = 60^\circ ∠ QR T = 6 0 ∘ ∠ Q T R = 50 ∘ \angle QTR = 50^\circ ∠ QT R = 5 0 ∘ ∠ R T S \angle RTS ∠ R T S
A 120 ∘ 120^\circ 12 0 ∘ B 70 ∘ 70^\circ 7 0 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 40 ∘ 40^\circ 4 0 ∘
Worked solution (try it first) The angles of triangle
Q R T QRT QR T add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ R Q T = 180 ∘ − 60 ∘ − 50 ∘ \angle RQT = 180^\circ - 60^\circ - 50^\circ ∠ R QT = 18 0 ∘ − 6 0 ∘ − 5 0 ∘ The angle between tangent
T S TS T S and chord
T R TR T R equals the angle in the alternate segment, which is the angle at
Q Q Q .
So
∠ R T S = ∠ R Q T = 70 ∘ \angle RTS = \angle RQT = 70^\circ ∠ R T S = ∠ R QT = 7 0 ∘ , option B.
Watch out
The tangent–chord angle at T T T T R TR T R T R TR T R Q Q Q ∠ Q R T = 60 ∘ \angle QRT = 60^\circ ∠ QR T = 6 0 ∘ Report a problem with this question
A triangle has base 7 cm and other sides 6 cm and 5 cm. Find its height h h h
A 12 7 \frac{12}{7} 7 12 B 12 7 6 \frac{12}{7}\sqrt6 7 12 6 C 7 12 \frac{7}{12} 12 7 D 1 2 51 \frac12\sqrt{51} 2 1 51
Worked solution (try it first) Find the area with Heron's formula.
The semi-perimeter is
s = 7 + 6 + 5 2 = 9 s = \frac{7 + 6 + 5}{2} = 9 s = 2 7 + 6 + 5 = 9 .
Area
= s ( s − a ) ( s − b ) ( s − c ) = \sqrt{s(s - a)(s - b)(s - c)} = s ( s − a ) ( s − b ) ( s − c ) , which is
9 × 2 × 3 × 4 = 216 \sqrt{9 \times 2 \times 3 \times 4} = \sqrt{216} 9 × 2 × 3 × 4 = 216 .
As
216 = 36 × 6 216 = 36 \times 6 216 = 36 × 6 , the area is
6 6 cm 2 6\sqrt6\text{ cm}^2 6 6 cm 2 .
Area
= 1 2 × = \frac12 \times = 2 1 × base
× \times × height, so
1 2 × 7 × h = 6 6 \frac12 \times 7 \times h = 6\sqrt6 2 1 × 7 × h = 6 6 and
h = 12 6 7 h = \frac{12\sqrt6}{7} h = 7 12 6 = 12 7 6 = \frac{12}{7}\sqrt6 = 7 12 6 cm, option B.
Watch out
Simplify 216 \sqrt{216} 216 36 × 6 = 6 6 \sqrt{36 \times 6} = 6\sqrt6 36 × 6 = 6 6 6 \sqrt6 6 12 7 \frac{12}{7} 7 12 Report a problem with this question
In a frustum of a cone (an upturned cup shape), the top diameter is twice the bottom diameter. If the height of the frustum is h h h
A 2 h 2h 2 h B 2 π h 2\pi h 2 π h C π h \pi h π h D π h 2 \frac{\pi h}{2} 2 π h
Worked solution (try it first) Let the full cone have height
H H H .
The small cone cut off has height
H − h H - h H − h .
Similar triangles: radius is proportional to distance from the vertex, so
2 r H = r H − h \dfrac{2r}{H} = \dfrac{r}{H - h} H 2 r = H − h r .
Cross-multiply:
2 ( H − h ) = H 2(H - h) = H 2 ( H − h ) = H , so
H = 2 h H = 2h H = 2 h , option A.
Watch out
A height comes from similar triangles, which are ratios of lengths, so no π \pi π Report a problem with this question
What is the locus of a point P P P X Y XY X Y X P Y XPY X P Y 90 ∘ 90^\circ 9 0 ∘
A The perpendicular bisector of X Y XY X Y B A right-angled triangle C A circle D A semicircle
Worked solution (try it first) The angle in a semicircle is a right angle, so if
∠ X P Y = 90 ∘ \angle XPY = 90^\circ ∠ X P Y = 9 0 ∘ then
P P P lies on the circle with
X Y XY X Y as diameter.
P P P stays on one side of
X Y XY X Y , so it traces only half of that circle.
The locus is a semicircle, option D.
Watch out
Read "on one side of X Y XY X Y Report a problem with this question
If M ( 4 , q ) M(4, q) M ( 4 , q ) L ( p , − 2 ) L(p, -2) L ( p , − 2 ) N ( q , p ) N(q, p) N ( q , p ) p p p q q q
A p = 2 , q = 4 p = 2, q = 4 p = 2 , q = 4 B p = 3 , q = 1 p = 3, q = 1 p = 3 , q = 1 C p = 5 , q = 3 p = 5, q = 3 p = 5 , q = 3 D p = 6 , q = 2 p = 6, q = 2 p = 6 , q = 2
Worked solution (try it first) The midpoint is the average of the ends.
For
x x x :
p + q 2 = 4 \frac{p + q}{2} = 4 2 p + q = 4 , so
p + q = 8 p + q = 8 p + q = 8 .
For
y y y :
− 2 + p 2 = q \frac{-2 + p}{2} = q 2 − 2 + p = q , so
p − 2 = 2 q p - 2 = 2q p − 2 = 2 q and
p = 2 q + 2 p = 2q + 2 p = 2 q + 2 .
Put this into
p + q = 8 p + q = 8 p + q = 8 :
3 q + 2 = 8 3q + 2 = 8 3 q + 2 = 8 , so
q = 2 q = 2 q = 2 and
p = 6 p = 6 p = 6 , option D.
Watch out
Divide the sums by 2. Forgetting to halve gives p + q = 4 p + q = 4 p + q = 4 p − 2 = q p - 2 = q p − 2 = q p = 3 p = 3 p = 3 q = 1 q = 1 q = 1 Report a problem with this question
The equation of the line in the graph shown is
A 3 y = 4 x + 12 3y = 4x + 12 3 y = 4 x + 12 B 3 y = 3 x + 12 3y = 3x + 12 3 y = 3 x + 12 C 3 y = − 4 x + 12 3y = -4x + 12 3 y = − 4 x + 12 D 3 y = − 4 x + 9 3y = -4x + 9 3 y = − 4 x + 9
Worked solution (try it first) The line cuts the
y y y -axis at
( 0 , 4 ) (0, 4) ( 0 , 4 ) and the
x x x -axis at
( 3 , 0 ) (3, 0) ( 3 , 0 ) .
Gradient:
0 − 4 3 − 0 = − 4 3 \dfrac{0 - 4}{3 - 0} = -\frac43 3 − 0 0 − 4 = − 3 4 , and the
y y y -intercept is 4, so
y = − 4 3 x + 4 y = -\frac43x + 4 y = − 3 4 x + 4 .
Multiply every term by 3:
3 y = − 4 x + 12 3y = -4x + 12 3 y = − 4 x + 12 , option C.
Watch out
The line falls from left to right, so its gradient is negative, which rules out option A. Remember to multiply the intercept by 3 as well: keeping it as 4 or using the x x x Report a problem with this question
The angle of depression of a boat from the top of a cliff 10 m high is 30 ∘ 30^\circ 3 0 ∘
A 5 3 3 \frac{5\sqrt3}{3} 3 5 3 B 5 3 5\sqrt3 5 3 C 10 3 10\sqrt3 10 3 D 10 3 3 \frac{10\sqrt3}{3} 3 10 3
Worked solution (try it first) The angle of elevation of the cliff top from the boat is also
30 ∘ 30^\circ 3 0 ∘ (alternate angles).
So
tan 30 ∘ = 10 d \tan30^\circ = \frac{10}{d} tan 3 0 ∘ = d 10 , giving
d = 10 tan 30 ∘ d = \frac{10}{\tan30^\circ} d = t a n 3 0 ∘ 10 .
Dividing by
1 3 \frac{1}{\sqrt3} 3 1 multiplies by
3 \sqrt3 3 :
d = 10 3 d = 10\sqrt3 d = 10 3 m, option C.
Watch out
Divide the height by tan 30 ∘ \tan30^\circ tan 3 0 ∘ 10 tan 30 ∘ = 10 3 3 10\tan30^\circ = \frac{10\sqrt3}{3} 10 tan 3 0 ∘ = 3 10 3 Report a problem with this question
What is the value of sin ( − 690 ∘ ) \sin(-690^\circ) sin ( − 69 0 ∘ )
A 3 2 \frac{\sqrt3}{2} 2 3 B − 3 2 -\frac{\sqrt3}{2} − 2 3 C − 1 2 -\frac12 − 2 1 D 1 2 \frac12 2 1
Worked solution (try it first) Adding whole turns of
360 ∘ 360^\circ 36 0 ∘ does not change the sine.
Add two turns:
− 690 ∘ + 720 ∘ = 30 ∘ -690^\circ + 720^\circ = 30^\circ − 69 0 ∘ + 72 0 ∘ = 3 0 ∘ .
So
sin ( − 690 ∘ ) = sin 30 ∘ = 1 2 \sin(-690^\circ) = \sin30^\circ = \frac12 sin ( − 69 0 ∘ ) = sin 3 0 ∘ = 2 1 , option D.
Watch out
Adding 720 ∘ 720^\circ 72 0 ∘ − 690 ∘ -690^\circ − 69 0 ∘ + 30 ∘ +30^\circ + 3 0 ∘ − 30 ∘ -30^\circ − 3 0 ∘ − 30 ∘ -30^\circ − 3 0 ∘ − 1 2 -\frac12 − 2 1 Report a problem with this question
If y = 3 t 3 + 2 t 2 − 7 t + 3 y = 3t^3 + 2t^2 - 7t + 3 y = 3 t 3 + 2 t 2 − 7 t + 3 d y d t \frac{dy}{dt} d t d y t = − 1 t = -1 t = − 1
Worked solution (try it first) Differentiate term by term:
d y d t = 9 t 2 + 4 t − 7 \frac{dy}{dt} = 9t^2 + 4t - 7 d t d y = 9 t 2 + 4 t − 7 .
Put in
t = − 1 t = -1 t = − 1 :
9 ( 1 ) + 4 ( − 1 ) − 7 = 9 − 4 − 7 = − 2 9(1) + 4(-1) - 7 = 9 - 4 - 7 = -2 9 ( 1 ) + 4 ( − 1 ) − 7 = 9 − 4 − 7 = − 2 , option C.
Watch out
( − 1 ) 2 = 1 (-1)^2 = 1 ( − 1 ) 2 = 1 9 t 2 = 9 9t^2 = 9 9 t 2 = 9 4 t = − 4 4t = -4 4 t = − 4 − 20 -20 − 20 Report a problem with this question
Find the point ( x , y ) (x, y) ( x , y ) y = 2 x 2 − 2 x + 3 y = 2x^2 - 2x + 3 y = 2 x 2 − 2 x + 3
A ( 1 , 3 ) (1, 3) ( 1 , 3 ) B ( 2 , 7 ) (2, 7) ( 2 , 7 ) C ( 0 , 3 ) (0, 3) ( 0 , 3 ) D ( 3 , 15 ) (3, 15) ( 3 , 15 )
Worked solution (try it first) The gradient is
d y d x = 4 x − 2 \frac{dy}{dx} = 4x - 2 d x d y = 4 x − 2 .
Set it equal to 2:
4 x − 2 = 2 4x - 2 = 2 4 x − 2 = 2 , so
4 x = 4 4x = 4 4 x = 4 and
x = 1 x = 1 x = 1 .
Then
y = 2 − 2 + 3 = 3 y = 2 - 2 + 3 = 3 y = 2 − 2 + 3 = 3 , so the point is
( 1 , 3 ) (1, 3) ( 1 , 3 ) , option A.
Watch out
The gradient is d y d x \frac{dy}{dx} d x d y y y y ( 0 , 3 ) (0, 3) ( 0 , 3 ) 4 ( 0 ) − 2 = − 2 4(0) - 2 = -2 4 ( 0 ) − 2 = − 2 Report a problem with this question
Integrate 1 − x x 3 \dfrac{1 - x}{x^3} x 3 1 − x x x x
A x − x 2 x 4 + k \dfrac{x - x^2}{x^4 + k} x 4 + k x − x 2 B 4 x 4 − 3 x 3 + k \frac{4}{x^4} - \frac{3}{x^3} + k x 4 4 − x 3 3 + k C 1 x − 1 2 x 2 + k \frac1x - \frac{1}{2x^2} + k x 1 − 2 x 2 1 + k D 1 3 x 3 − 1 2 x + k \frac{1}{3x^3} - \frac{1}{2x} + k 3 x 3 1 − 2 x 1 + k
Worked solution (try it first) Split the fraction into powers of
x x x :
1 − x x 3 = x − 3 − x − 2 \dfrac{1 - x}{x^3} = x^{-3} - x^{-2} x 3 1 − x = x − 3 − x − 2 .
Add one to each power and divide by the new power:
x − 3 x^{-3} x − 3 gives
x − 2 − 2 = − 1 2 x 2 \frac{x^{-2}}{-2} = -\frac{1}{2x^2} − 2 x − 2 = − 2 x 2 1 , and
− x − 2 -x^{-2} − x − 2 gives
− x − 1 − 1 = 1 x -\frac{x^{-1}}{-1} = \frac1x − − 1 x − 1 = x 1 .
So the integral is
1 x − 1 2 x 2 + k \frac1x - \frac{1}{2x^2} + k x 1 − 2 x 2 1 + k , option C.
Watch out
Split the fraction first; you can't integrate the top and bottom separately. Then take care with negative powers: x − 3 x^{-3} x − 3 x − 2 x^{-2} x − 2 x − 4 x^{-4} x − 4 Report a problem with this question
Evaluate ∫ − 1 1 ( 2 x + 1 ) 2 d x \displaystyle\int_{-1}^{1} (2x + 1)^2\,dx ∫ − 1 1 ( 2 x + 1 ) 2 d x
A 3 2 3 3\frac23 3 3 2 B 4 C 4 1 3 4\frac13 4 3 1 D 4 2 3 4\frac23 4 3 2
Worked solution (try it first) Expand the square:
( 2 x + 1 ) 2 = 4 x 2 + 4 x + 1 (2x + 1)^2 = 4x^2 + 4x + 1 ( 2 x + 1 ) 2 = 4 x 2 + 4 x + 1 .
Integrate:
[ 4 3 x 3 + 2 x 2 + x ] − 1 1 \left[\frac43x^3 + 2x^2 + x\right]_{-1}^{1} [ 3 4 x 3 + 2 x 2 + x ] − 1 1 .
At
x = 1 x = 1 x = 1 this is
4 3 + 3 = 13 3 \frac43 + 3 = \frac{13}{3} 3 4 + 3 = 3 13 .
At
x = − 1 x = -1 x = − 1 it is
− 4 3 + 2 − 1 = − 1 3 -\frac43 + 2 - 1 = -\frac13 − 3 4 + 2 − 1 = − 3 1 .
Subtract:
13 3 − ( − 1 3 ) = 14 3 \frac{13}{3} - \left(-\frac13\right) = \frac{14}{3} 3 13 − ( − 3 1 ) = 3 14 = 4 2 3 = 4\frac23 = 4 3 2 , option D.
Watch out
Subtracting a negative lower value adds it: 13 3 + 1 3 \frac{13}{3} + \frac13 3 13 + 3 1 + 1 3 +\frac13 + 3 1 4 4 4 Report a problem with this question
The grades A1, A2, A3, C4 and F earned by students in a course are shown in the pie chart. What percentage of the students obtained a C4 grade?
Worked solution (try it first) The C4 sector has angle
43.2 ∘ 43.2^\circ 43. 2 ∘ out of
360 ∘ 360^\circ 36 0 ∘ .
As a percentage:
43.2 360 × 100 = 12.0 \frac{43.2}{360} \times 100 = 12.0 360 43.2 × 100 = 12.0 , option D.
Watch out
43.2 (option B) is the angle, not the percentage; divide it by 360 and multiply by 100. Report a problem with this question
The mean of twelve positive numbers is 3. When another number is added, the mean becomes 5. Find the thirteenth number.
Worked solution (try it first) Total of the twelve numbers:
12 × 3 = 36 12 \times 3 = 36 12 × 3 = 36 .
Total of all thirteen:
13 × 5 = 65 13 \times 5 = 65 13 × 5 = 65 .
The thirteenth number is the difference:
65 − 36 = 29 65 - 36 = 29 65 − 36 = 29 , option A.
Watch out
After adding a number there are 13 numbers. Using 12 × 5 = 60 12 \times 5 = 60 12 × 5 = 60 60 − 36 = 24 60 - 36 = 24 60 − 36 = 24 Report a problem with this question
Find the mean deviation of the numbers 4, 5, 9.
Worked solution (try it first) The mean is
4 + 5 + 9 3 = 6 \frac{4 + 5 + 9}{3} = 6 3 4 + 5 + 9 = 6 .
The distances from 6, ignoring signs, are 2, 1 and 3, which add up to 6.
The mean deviation is
6 3 = 2 \frac63 = 2 3 6 = 2 , option B.
Watch out
Use the absolute deviations. The signed deviations − 2 -2 − 2 − 1 -1 − 1 Report a problem with this question
Estimate the median of the frequency distribution below.
Class interval
1–5
6–10
11–15
16–20
21–25
Frequency
6
15
20
7
2
A 10 1 2 10\frac12 10 2 1 B 11 1 2 11\frac12 11 2 1 C 12 1 2 12\frac12 12 2 1 D 13
Worked solution (try it first) The cumulative frequencies are 6, 21, 41, 48, 50, so
N = 50 N = 50 N = 50 and the median is the 25th value.
The 25th value lies in the class 11–15 (running total 21 to 41).
Its lower boundary is 10.5, its frequency is 20 and its width is 5.
Median
= 10.5 + 25 − 21 20 × 5 = 10.5 + \frac{25 - 21}{20} \times 5 = 10.5 + 20 25 − 21 × 5 , which is
10.5 + 1 = 11 1 2 10.5 + 1 = 11\frac12 10.5 + 1 = 11 2 1 , option B.
Watch out
The median is not the mid-point of the median class (13, option D) or its lower boundary (option A). Add the fraction 25 − 21 20 \frac{25 - 21}{20} 20 25 − 21 Report a problem with this question
The table shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
Age in years
10
11
12
Number of pupils
6
27
7
A 27 40 \frac{27}{40} 40 27 B 17 20 \frac{17}{20} 20 17 C 33 40 \frac{33}{40} 40 33 D 3 20 \frac{3}{20} 20 3
Worked solution (try it first) There are
6 + 27 + 7 = 40 6 + 27 + 7 = 40 6 + 27 + 7 = 40 pupils.
At least 11 means 11 or 12:
27 + 7 = 34 27 + 7 = 34 27 + 7 = 34 pupils.
So the probability is
34 40 = 17 20 \frac{34}{40} = \frac{17}{20} 40 34 = 20 17 , option B.
Watch out
"At least 11" includes the 12-year-olds too. Using only the 27 eleven-year-olds gives 27 40 \frac{27}{40} 40 27 Report a problem with this question
In a survey, 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?
A 1 2 \frac12 2 1 B 2 3 \frac23 3 2 C 3 8 \frac38 8 3 D 3 11 \frac3{11} 11 3
Worked solution (try it first) Those counted in both groups are counted twice, so both
= 20 + 35 − 40 = 15 = 20 + 35 - 40 = 15 = 20 + 35 − 40 = 15 .
The total number of students is the 40 who read either.
So the probability is
15 40 = 3 8 \frac{15}{40} = \frac38 40 15 = 8 3 , option C.
Watch out
Divide by the 40 students, not by 20 + 35 = 55 20 + 35 = 55 20 + 35 = 55 15 55 = 3 11 \frac{15}{55} = \frac{3}{11} 55 15 = 11 3 Report a problem with this question