Express 0.0030752 to three significant figures.
A 0.003 B 0.003075 C 0.00308 D 0.0031 E 0.00375
Worked solution (try it first) The first significant figure is the first non-zero digit, 3.
The first three significant figures are 3, 0 and 7.
The next digit is 5, so round the 7 up to 8: 0.00308, option C.
Watch out
Zeros before the first non-zero digit are not significant, but the zero between 3 and 7 is. 0.0031 (option D) has only two significant figures. Report a problem with this question
Simplify 1 1 9 ÷ 1 1 3 × 3 3 5 1\frac{1}{9} \div 1\frac{1}{3} \times 3\frac{3}{5} 1 9 1 ÷ 1 3 1 × 3 5 3
A 1 3 \frac{1}{3} 3 1 B 1 2 \frac{1}{2} 2 1 C 1 D 2 E 3
Worked solution (try it first) Write the mixed numbers as improper fractions:
10 9 ÷ 4 3 × 18 5 \frac{10}{9} \div \frac{4}{3} \times \frac{18}{5} 9 10 ÷ 3 4 × 5 18 .
Divide first (work from left to right):
10 9 × 3 4 = 5 6 \frac{10}{9} \times \frac{3}{4} = \frac{5}{6} 9 10 × 4 3 = 6 5 .
Then multiply:
5 6 × 18 5 = 3 \frac{5}{6} \times \frac{18}{5} = 3 6 5 × 5 18 = 3 , option E.
Watch out
Division and multiplication go from left to right. Working out 1 1 3 × 3 3 5 1\frac{1}{3} \times 3\frac{3}{5} 1 3 1 × 3 5 3 25 108 \frac{25}{108} 108 25 Report a problem with this question
Find the 15th term of a sequence whose n n n T n = 5 n − 5 5 T_n = \dfrac{5n - 5}{5} T n = 5 5 n − 5
Worked solution (try it first) Put
n = 15 n = 15 n = 15 : the top is
5 ( 15 ) − 5 = 70 5(15) - 5 = 70 5 ( 15 ) − 5 = 70 .
Divide by 5:
T 15 = 14 T_{15} = 14 T 15 = 14 , option C.
Watch out
Subtract the 5 before dividing: 75 − 5 5 = 14 \frac{75 - 5}{5} = 14 5 75 − 5 = 14 − 5 -5 − 5 Report a problem with this question
If 3 ( 6 − 9 x ) = 27 ( 1 − 2 x ) 3^{(6 - 9x)} = 27^{(1 - 2x)} 3 ( 6 − 9 x ) = 2 7 ( 1 − 2 x ) x x x
A 1 3 \frac{1}{3} 3 1 B 1 2 \frac{1}{2} 2 1 C 1 D 3 E 5
Worked solution (try it first) Write 27 as
3 3 3^3 3 3 , so the right side is
3 3 ( 1 − 2 x ) = 3 3 − 6 x 3^{3(1 - 2x)} = 3^{3 - 6x} 3 3 ( 1 − 2 x ) = 3 3 − 6 x .
The bases are equal, so the powers are equal:
6 − 9 x = 3 − 6 x 6 - 9x = 3 - 6x 6 − 9 x = 3 − 6 x .
Collect terms:
3 = 3 x 3 = 3x 3 = 3 x , so
x = 1 x = 1 x = 1 , option C.
Watch out
Multiply the whole power by 3: 27 1 − 2 x = 3 3 − 6 x 27^{1 - 2x} = 3^{3 - 6x} 2 7 1 − 2 x = 3 3 − 6 x 3 3 − 2 x 3^{3 - 2x} 3 3 − 2 x x = 3 7 x = \frac{3}{7} x = 7 3 Report a problem with this question
Simplify 2 18 − 4 32 + 3 50 2\sqrt{18} - 4\sqrt{32} + 3\sqrt{50} 2 18 − 4 32 + 3 50
A 3 2 3\sqrt{2} 3 2 B 4 2 4\sqrt{2} 4 2 C 5 2 5\sqrt{2} 5 2 D 6 2 6\sqrt{2} 6 2 E 9 2 9\sqrt{2} 9 2
Worked solution (try it first) Take out the largest square factor of each:
18 = 3 2 \sqrt{18} = 3\sqrt{2} 18 = 3 2 ,
32 = 4 2 \sqrt{32} = 4\sqrt{2} 32 = 4 2 and
50 = 5 2 \sqrt{50} = 5\sqrt{2} 50 = 5 2 .
Multiply by the numbers in front:
6 2 − 16 2 + 15 2 6\sqrt{2} - 16\sqrt{2} + 15\sqrt{2} 6 2 − 16 2 + 15 2 .
Collect the like surds:
6 − 16 + 15 = 5 6 - 16 + 15 = 5 6 − 16 + 15 = 5 , so the answer is
5 2 5\sqrt{2} 5 2 , option C.
Watch out
Keep the minus sign with the middle term: − 4 32 = − 16 2 -4\sqrt{32} = -16\sqrt{2} − 4 32 = − 16 2 37 2 37\sqrt{2} 37 2 21 2 21\sqrt{2} 21 2 Report a problem with this question
Approximate 8.5326 to the nearest hundredth.
A 8.50 B 8.53 C 9.00 D 9.53 E 9.55
Worked solution (try it first) The hundredths digit is the second decimal place, 3.
The next digit is 2, which is less than 5, so round down: 8.53, option B.
Watch out
Hundredths means two decimal places. 8.50 (option A) is 8.5326 to the nearest tenth, written with a trailing zero. Report a problem with this question
Given that log 2 = 0.30103 \log 2 = 0.30103 log 2 = 0.30103 log 3 = 0.47712 \log 3 = 0.47712 log 3 = 0.47712 log 6 − log 0.3 \log 6 - \log 0.3 log 6 − log 0.3
A 0.25527 B 0.52288 C 0.55280 D 0.77815 E 1.30103
Worked solution (try it first) Subtracting logs divides the numbers:
log 6 − log 0.3 = log 6 0.3 \log 6 - \log 0.3 = \log \dfrac{6}{0.3} log 6 − log 0.3 = log 0.3 6 .
6 0.3 = 20 \dfrac{6}{0.3} = 20 0.3 6 = 20 , and
log 20 = log 2 + log 10 \log 20 = \log 2 + \log 10 log 20 = log 2 + log 10 .
log 10 = 1 \log 10 = 1 log 10 = 1 , so the value is
0.30103 + 1 = 1.30103 0.30103 + 1 = 1.30103 0.30103 + 1 = 1.30103 , option E.
Watch out
log 0.3 \log 0.3 log 0.3 log 3 − 1 = − 0.52288 \log 3 - 1 = -0.52288 log 3 − 1 = − 0.52288 log 6 = 0.77815 \log 6 = 0.77815 log 6 = 0.77815 Report a problem with this question
A toy train starts at the station and moves around a circular track of 220 cm. How many rotations will it move along the track when it has covered 440 cm?
Worked solution (try it first) One rotation is one lap of the track, 220 cm.
Divide the distance by one lap:
440 ÷ 220 = 2 440 \div 220 = 2 440 ÷ 220 = 2 rotations, option A.
Watch out
220 cm is already the distance round the track (the circumference). Don't treat it as a radius or diameter and multiply by π \pi π Report a problem with this question
A square of side 10 cm is measured by a student as 9.9 cm. If she uses her measurement to calculate the area of the square, calculate the percentage error.
A 0.15 B 0.99 C 1.15 D 1.19 E 1.99
Worked solution (try it first) The true area is
10 2 = 100 cm 2 10^2 = 100\text{ cm}^2 1 0 2 = 100 cm 2 and the calculated area is
9.9 2 = 98.01 cm 2 9.9^2 = 98.01\text{ cm}^2 9. 9 2 = 98.01 cm 2 .
The error in the area is
100 − 98.01 = 1.99 cm 2 100 - 98.01 = 1.99\text{ cm}^2 100 − 98.01 = 1.99 cm 2 .
Percentage error
= 1.99 100 × 100 % = 1.99 % = \dfrac{1.99}{100} \times 100\% = 1.99\% = 100 1.99 × 100% = 1.99% , option E.
Watch out
Compare areas, not sides: the side is out by only 1%, but the area is out by 1.99%. Divide by the true area 100, not the measured 98.01. Report a problem with this question
The n n n T n = ( 2 n + 1 2 ) T_n = \left(2n + \frac{1}{2}\right) T n = ( 2 n + 2 1 )
A 11.5 B 23.0 C 57.0 D 115.0 E 230.0
Worked solution (try it first) The first term is
T 1 = 2 + 1 2 = 2.5 T_1 = 2 + \frac{1}{2} = 2.5 T 1 = 2 + 2 1 = 2.5 and the tenth is
T 10 = 20 + 1 2 = 20.5 T_{10} = 20 + \frac{1}{2} = 20.5 T 10 = 20 + 2 1 = 20.5 .
Use
S n = n 2 ( a + l ) S_n = \frac{n}{2}(a + l) S n = 2 n ( a + l ) :
S 10 = 5 × ( 2.5 + 20.5 ) S_{10} = 5 \times (2.5 + 20.5) S 10 = 5 × ( 2.5 + 20.5 ) .
So
S 10 = 5 × 23 = 115 S_{10} = 5 \times 23 = 115 S 10 = 5 × 23 = 115 , option D.
Watch out
The formula halves n n n 10 2 = 5 \frac{10}{2} = 5 2 10 = 5 ( a + l ) = 23 (a + l) = 23 ( a + l ) = 23 n n n Report a problem with this question
The 5th and 11th terms of an Arithmetic Progression (A.P.) are 13 and 31 respectively. Find the sum of its first term and common difference.
Worked solution (try it first) Write the terms:
a + 4 d = 13 a + 4d = 13 a + 4 d = 13 and
a + 10 d = 31 a + 10d = 31 a + 10 d = 31 .
Subtract:
6 d = 18 6d = 18 6 d = 18 , so
d = 3 d = 3 d = 3 .
Then
a = 13 − 12 = 1 a = 13 - 12 = 1 a = 13 − 12 = 1 , and
a + d = 1 + 3 = 4 a + d = 1 + 3 = 4 a + d = 1 + 3 = 4 , option A.
Watch out
The 5th term is a + 4 d a + 4d a + 4 d a + 5 d a + 5d a + 5 d a + 5 d = 13 a + 5d = 13 a + 5 d = 13 a = − 2 a = -2 a = − 2 Report a problem with this question
The 6th term of a Geometric Progression (G.P.) is 486. Find the common ratio if its first term is 2.
Worked solution (try it first) The 6th term is
a r 5 ar^5 a r 5 , so
2 r 5 = 486 2r^5 = 486 2 r 5 = 486 .
Divide by 2:
r 5 = 243 r^5 = 243 r 5 = 243 .
243 = 3 5 243 = 3^5 243 = 3 5 , so
r = 3 r = 3 r = 3 , option D.
Watch out
( − 3 ) 5 = − 243 (-3)^5 = -243 ( − 3 ) 5 = − 243 r = − 3 r = -3 r = − 3 − 486 -486 − 486 Report a problem with this question
Evaluate ( 256 81 ) − 3 4 \left(\dfrac{256}{81}\right)^{-\frac{3}{4}} ( 81 256 ) − 4 3
A 3 4 \frac{3}{4} 4 3 B 17 32 \frac{17}{32} 32 17 C 17 64 \frac{17}{64} 64 17 D 27 64 \frac{27}{64} 64 27 E 37 64 \frac{37}{64} 64 37
Worked solution (try it first) A negative power turns the fraction upside down:
( 81 256 ) 3 4 \left(\dfrac{81}{256}\right)^{\frac{3}{4}} ( 256 81 ) 4 3 .
Take the fourth root:
81 4 = 3 \sqrt[4]{81} = 3 4 81 = 3 and
256 4 = 4 \sqrt[4]{256} = 4 4 256 = 4 , giving
3 4 \frac{3}{4} 4 3 .
Cube it:
( 3 4 ) 3 = 27 64 \left(\frac{3}{4}\right)^3 = \frac{27}{64} ( 4 3 ) 3 = 64 27 , option D.
Watch out
The power 3 4 \frac{3}{4} 4 3 3 4 \frac{3}{4} 4 3 Report a problem with this question
Simplify 110011 two × 101 two 110011_{\text{two}} \times 101_{\text{two}} 11001 1 two × 10 1 two
A 11111111 two 11111111_{\text{two}} 1111111 1 two B 11100001 two 11100001_{\text{two}} 1110000 1 two C 10011111 two 10011111_{\text{two}} 1001111 1 two D 10011001 two 10011001_{\text{two}} 1001100 1 two E 10000111 two 10000111_{\text{two}} 1000011 1 two
Worked solution (try it first) Convert to base ten:
110011 two = 32 + 16 + 2 + 1 110011_{\text{two}} = 32 + 16 + 2 + 1 11001 1 two = 32 + 16 + 2 + 1 = 51 = 51 = 51 and
101 two = 5 101_{\text{two}} = 5 10 1 two = 5 .
Multiply:
51 × 5 = 255 51 \times 5 = 255 51 × 5 = 255 .
255 = 256 − 1 = 2 8 − 1 255 = 256 - 1 = 2^8 - 1 255 = 256 − 1 = 2 8 − 1 , which is eight 1s in base two:
11111111 two 11111111_{\text{two}} 1111111 1 two , option A.
Watch out
If you multiply in base two, add the shifted rows 110011 110011 110011 11001100 11001100 11001100 1 + 1 = 10 two 1 + 1 = 10_{\text{two}} 1 + 1 = 1 0 two Report a problem with this question
Simplify 3 3 + 6 \dfrac{3}{\sqrt{3} + 6} 3 + 6 3
A 1 33 ( − 15 3 ) \frac{1}{33}(-15\sqrt{3}) 33 1 ( − 15 3 ) B 1 11 ( 6 − 3 ) \frac{1}{11}(6 - \sqrt{3}) 11 1 ( 6 − 3 ) C 1 11 ( 6 + 3 ) \frac{1}{11}(6 + \sqrt{3}) 11 1 ( 6 + 3 ) D 1 11 ( 3 − 6 ) \frac{1}{11}(\sqrt{3} - 6) 11 1 ( 3 − 6 ) E 1 33 ( 21 3 ) \frac{1}{33}(21\sqrt{3}) 33 1 ( 21 3 )
Worked solution (try it first) Multiply the top and bottom by the conjugate
6 − 3 6 - \sqrt{3} 6 − 3 .
The bottom becomes
6 2 − ( 3 ) 2 = 36 − 3 = 33 6^2 - (\sqrt{3})^2 = 36 - 3 = 33 6 2 − ( 3 ) 2 = 36 − 3 = 33 and the top becomes
3 ( 6 − 3 ) 3(6 - \sqrt{3}) 3 ( 6 − 3 ) .
Cancel 3:
6 − 3 11 \dfrac{6 - \sqrt{3}}{11} 11 6 − 3 , option B.
Watch out
The top becomes 3 ( 6 − 3 ) 3(6 - \sqrt{3}) 3 ( 6 − 3 ) 1 11 ( 6 + 3 ) \frac{1}{11}(6 + \sqrt{3}) 11 1 ( 6 + 3 ) Report a problem with this question
The value of a car depreciates by 20% at the beginning of each year. If the car costs ₦1 million, find the value of the car after 2 years.
A ₦512,000.00 B ₦600,000.00 C ₦640,000.00 D ₦800,000.00 E ₦830,000.00
Worked solution (try it first) Each year the car keeps 80% of its value, so multiply by 0.8 each year.
After 1 year:
0.8 × 1 000 000 = 800 000 0.8 \times 1\,000\,000 = 800\,000 0.8 × 1 000 000 = 800 000 .
After 2 years:
0.8 × 800 000 = 640 000 0.8 \times 800\,000 = 640\,000 0.8 × 800 000 = 640 000 .
The value is ₦640,000.00, option C.
Watch out
The second 20% is taken off the new value (₦800,000), not the original. Taking 40% of ₦1 million gives ₦600,000 (option B). Report a problem with this question
Given that A = ( 2 − 1 3 3 4 ) A = \begin{pmatrix} 2 & -\frac{1}{3} \\ 3 & 4 \end{pmatrix} A = ( 2 3 − 3 1 4 ) B = ( 1 2 3 6 ) B = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} B = ( 1 3 2 6 ) A B AB A B
A ( 1 2 15 30 ) \begin{pmatrix} 1 & 2 \\ 15 & 30 \end{pmatrix} ( 1 15 2 30 ) B ( − 1 − 2 9 18 ) \begin{pmatrix} -1 & -2 \\ 9 & 18 \end{pmatrix} ( − 1 9 − 2 18 ) C ( − 1 2 5 15 ) \begin{pmatrix} -1 & 2 \\ 5 & 15 \end{pmatrix} ( − 1 5 2 15 ) D ( 1 − 2 15 30 ) \begin{pmatrix} 1 & -2 \\ 15 & 30 \end{pmatrix} ( 1 15 − 2 30 ) E ( 1 0 0 1 ) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 1 0 0 1 )
Worked solution (try it first) Row 1 of
A A A times the columns of
B B B :
2 ( 1 ) + ( − 1 3 ) ( 3 ) = 1 2(1) + \left(-\frac{1}{3}\right)(3) = 1 2 ( 1 ) + ( − 3 1 ) ( 3 ) = 1 and
2 ( 2 ) + ( − 1 3 ) ( 6 ) = 2 2(2) + \left(-\frac{1}{3}\right)(6) = 2 2 ( 2 ) + ( − 3 1 ) ( 6 ) = 2 .
Row 2 of
A A A times the columns of
B B B :
3 ( 1 ) + 4 ( 3 ) = 15 3(1) + 4(3) = 15 3 ( 1 ) + 4 ( 3 ) = 15 and
3 ( 2 ) + 4 ( 6 ) = 30 3(2) + 4(6) = 30 3 ( 2 ) + 4 ( 6 ) = 30 .
So
A B = ( 1 2 15 30 ) AB = \begin{pmatrix} 1 & 2 \\ 15 & 30 \end{pmatrix} A B = ( 1 15 2 30 ) , option A.
Watch out
Multiply rows by columns, not matching entries. Multiplying entry by entry gives ( 2 − 2 3 9 24 ) \begin{pmatrix} 2 & -\frac{2}{3} \\ 9 & 24 \end{pmatrix} ( 2 9 − 3 2 24 ) Report a problem with this question
Find the values of x x x 3 x 2 + 6 x − 2 ( x + 2 ) ( x − 4 ) \dfrac{3x^2 + 6x - 2}{(x + 2)(x - 4)} ( x + 2 ) ( x − 4 ) 3 x 2 + 6 x − 2
A x = 2 x = 2 x = 2 4 4 4 B x = 2 x = 2 x = 2 − 4 -4 − 4 C x = − 2 x = -2 x = − 2 4 4 4 D x = 0 x = 0 x = 0 4 4 4 E x = 0 x = 0 x = 0 2 2 2
Worked solution (try it first) A fraction is undefined when its denominator is zero:
( x + 2 ) ( x − 4 ) = 0 (x + 2)(x - 4) = 0 ( x + 2 ) ( x − 4 ) = 0 .
So
x + 2 = 0 x + 2 = 0 x + 2 = 0 or
x − 4 = 0 x - 4 = 0 x − 4 = 0 , giving
x = − 2 x = -2 x = − 2 or
x = 4 x = 4 x = 4 , option C.
Watch out
Each factor gives the value that makes it zero, so x + 2 x + 2 x + 2 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2 − 4 -4 − 4 Report a problem with this question
Find the quadratic equation whose roots are 2 and − 3 1 2 -3\frac{1}{2} − 3 2 1
A 2 x 2 + 3 x + 14 = 0 2x^2 + 3x + 14 = 0 2 x 2 + 3 x + 14 = 0 B 2 x 2 + 5 x + 7 = 0 2x^2 + 5x + 7 = 0 2 x 2 + 5 x + 7 = 0 C 2 x 2 + 5 x − 7 = 0 2x^2 + 5x - 7 = 0 2 x 2 + 5 x − 7 = 0 D 2 x 2 + 3 x − 14 = 0 2x^2 + 3x - 14 = 0 2 x 2 + 3 x − 14 = 0 E 2 x 2 − 3 x − 14 = 0 2x^2 - 3x - 14 = 0 2 x 2 − 3 x − 14 = 0
Worked solution (try it first) The roots give the factors
( x − 2 ) (x - 2) ( x − 2 ) and
( x + 7 2 ) \left(x + \frac{7}{2}\right) ( x + 2 7 ) .
Doubling the second gives
( 2 x + 7 ) (2x + 7) ( 2 x + 7 ) .
Multiply out:
( x − 2 ) ( 2 x + 7 ) = 2 x 2 + 7 x − 4 x − 14 (x - 2)(2x + 7) = 2x^2 + 7x - 4x - 14 ( x − 2 ) ( 2 x + 7 ) = 2 x 2 + 7 x − 4 x − 14 .
So the equation is
2 x 2 + 3 x − 14 = 0 2x^2 + 3x - 14 = 0 2 x 2 + 3 x − 14 = 0 , option D.
Watch out
A root of − 3 1 2 -3\frac{1}{2} − 3 2 1 ( 2 x + 7 ) (2x + 7) ( 2 x + 7 ) ( 2 x − 7 ) (2x - 7) ( 2 x − 7 ) ( x + 2 ) (x + 2) ( x + 2 ) 2 x 2 − 3 x − 14 = 0 2x^2 - 3x - 14 = 0 2 x 2 − 3 x − 14 = 0 Report a problem with this question
Simplify a 2 − 4 a 2 − a − 6 \dfrac{a^2 - 4}{a^2 - a - 6} a 2 − a − 6 a 2 − 4
A 2 a + 2 a + 3 \frac{2a + 2}{a + 3} a + 3 2 a + 2 B a + 2 a + 3 \frac{a + 2}{a + 3} a + 3 a + 2 C 2 a − 2 a + 3 \frac{2a - 2}{a + 3} a + 3 2 a − 2 D 2 a + 2 a − 3 \frac{2a + 2}{a - 3} a − 3 2 a + 2 E a − 2 a − 3 \frac{a - 2}{a - 3} a − 3 a − 2
Worked solution (try it first) Factorise the top as a difference of two squares:
a 2 − 4 = ( a − 2 ) ( a + 2 ) a^2 - 4 = (a - 2)(a + 2) a 2 − 4 = ( a − 2 ) ( a + 2 ) .
Factorise the bottom:
a 2 − a − 6 = ( a − 3 ) ( a + 2 ) a^2 - a - 6 = (a - 3)(a + 2) a 2 − a − 6 = ( a − 3 ) ( a + 2 ) .
Cancel the common factor
( a + 2 ) (a + 2) ( a + 2 ) :
a − 2 a − 3 \dfrac{a - 2}{a - 3} a − 3 a − 2 , option E.
Watch out
Cancel factors, not terms. The a 2 a^2 a 2 ( a + 2 ) (a + 2) ( a + 2 ) Report a problem with this question
Given the quadratic equation 3 x 2 − 5 x = − 2 3x^2 - 5x = -2 3 x 2 − 5 x = − 2
A 2 3 \frac{2}{3} 3 2 B 1 1 3 1\frac{1}{3} 1 3 1 C 1 4 9 1\frac{4}{9} 1 9 4 D 1 2 3 1\frac{2}{3} 1 3 2 E 2 7 9 2\frac{7}{9} 2 9 7
Worked solution (try it first) Rearrange to
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 :
3 x 2 − 5 x + 2 = 0 3x^2 - 5x + 2 = 0 3 x 2 − 5 x + 2 = 0 .
The product of the roots is
c a = 2 3 \dfrac{c}{a} = \dfrac{2}{3} a c = 3 2 , option A.
Watch out
The product is c a \frac{c}{a} a c − b a -\frac{b}{a} − a b 5 3 = 1 2 3 \frac{5}{3} = 1\frac{2}{3} 3 5 = 1 3 2 Report a problem with this question
Simplify m 2 + m m ( m + 1 ) ( m + 2 ) \dfrac{m^2 + m}{m(m + 1)(m + 2)} m ( m + 1 ) ( m + 2 ) m 2 + m
A 1 m ( m + 2 ) \frac{1}{m(m + 2)} m ( m + 2 ) 1 B 1 m ( m + 1 ) \frac{1}{m(m + 1)} m ( m + 1 ) 1 C 1 m + 2 \frac{1}{m + 2} m + 2 1 D 1 m + 1 \frac{1}{m + 1} m + 1 1 E 1 m \frac{1}{m} m 1
Worked solution (try it first) Factorise the top:
m 2 + m = m ( m + 1 ) m^2 + m = m(m + 1) m 2 + m = m ( m + 1 ) .
Cancel
m m m and
( m + 1 ) (m + 1) ( m + 1 ) from top and bottom, leaving
1 m + 2 \dfrac{1}{m + 2} m + 2 1 , option C.
Watch out
Factorise the top fully. Taking out only m m m m + 1 ( m + 1 ) ( m + 2 ) \frac{m + 1}{(m + 1)(m + 2)} ( m + 1 ) ( m + 2 ) m + 1 ( m + 1 ) (m + 1) ( m + 1 ) 1 m ( m + 2 ) \frac{1}{m(m + 2)} m ( m + 2 ) 1 Report a problem with this question
Use the truth table below.
p p p q q q x x x y y y
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
F
Which of the following symbolic statements illustrate x x x
A p ∧ q p \wedge q p ∧ q B p ∨ q p \vee q p ∨ q C p ⇒ q p \Rightarrow q p ⇒ q D ∼ p ∧ ∼ q \sim p \wedge \sim q ∼ p ∧ ∼ q E ∼ p ∨ ∼ q \sim p \vee \sim q ∼ p ∨ ∼ q
Worked solution (try it first) Column
x x x is true only in the first row, where
p p p and
q q q are both true.
That is the conjunction "
p p p and
q q q ",
p ∧ q p \wedge q p ∧ q , option A.
Watch out
p ⇒ q p \Rightarrow q p ⇒ q p p p q q q x x x Report a problem with this question
Use the truth table below.
p p p q q q x x x y y y
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
F
Which of the following symbolic statements illustrate y y y
A ∼ p ∧ ∼ q \sim p \wedge \sim q ∼ p ∧ ∼ q B ∼ p ∨ ∼ q \sim p \vee \sim q ∼ p ∨ ∼ q C p ∨ q p \vee q p ∨ q D p ∧ q p \wedge q p ∧ q E p ⇒ q p \Rightarrow q p ⇒ q
Worked solution (try it first) Column
y y y is false only in the last row, where
p p p and
q q q are both false.
That is the disjunction "
p p p or
q q q ",
p ∨ q p \vee q p ∨ q , option C.
Watch out
p ⇒ q p \Rightarrow q p ⇒ q p p p q q q y y y Report a problem with this question
The perimeter of a rectangle is 44 m and its area is 120 m 2 120\text{ m}^2 120 m 2
A 11 m, 11 m B 12 m, 10 m C 15 m, 8 m D 16 m, 6 m E 18 m, 4 m
Worked solution (try it first) Half the perimeter is the length plus the width:
l + w = 22 l + w = 22 l + w = 22 .
The area gives
l w = 120 lw = 120 l w = 120 .
Two numbers that add to 22 and multiply to 120 are 12 and 10, so the rectangle is 12 m by 10 m, option B.
Watch out
Check both facts. 15 m by 8 m (option C) has area 120 but perimeter 46 m, and 11 m by 11 m (option A) has perimeter 44 but area 121. Report a problem with this question
Find the values of x x x 3 x 2 − 8 x − 3 = 0 3x^2 - 8x - 3 = 0 3 x 2 − 8 x − 3 = 0
A 1 3 , − 3 \frac{1}{3}, -3 3 1 , − 3 B − 1 3 , − 3 -\frac{1}{3}, -3 − 3 1 , − 3 C 1 3 , 3 \frac{1}{3}, 3 3 1 , 3 D − 1 3 , 3 -\frac{1}{3}, 3 − 3 1 , 3 E − 3 , 0 -3, 0 − 3 , 0
Worked solution (try it first) Factorise:
3 x 2 − 8 x − 3 = ( 3 x + 1 ) ( x − 3 ) 3x^2 - 8x - 3 = (3x + 1)(x - 3) 3 x 2 − 8 x − 3 = ( 3 x + 1 ) ( x − 3 ) .
Set each factor to zero:
3 x + 1 = 0 3x + 1 = 0 3 x + 1 = 0 gives
x = − 1 3 x = -\frac{1}{3} x = − 3 1 , and
x − 3 = 0 x - 3 = 0 x − 3 = 0 gives
x = 3 x = 3 x = 3 , option D.
Watch out
Check the middle term: ( 3 x + 1 ) ( x − 3 ) (3x + 1)(x - 3) ( 3 x + 1 ) ( x − 3 ) − 9 x + x = − 8 x -9x + x = -8x − 9 x + x = − 8 x ( 3 x − 1 ) ( x + 3 ) (3x - 1)(x + 3) ( 3 x − 1 ) ( x + 3 ) + 8 x +8x + 8 x 1 3 , − 3 \frac{1}{3}, -3 3 1 , − 3 Report a problem with this question
Factorize 5 y 2 + 2 a y − 3 a 2 5y^2 + 2ay - 3a^2 5 y 2 + 2 a y − 3 a 2
A ( 2 a − y ) ( 5 y − 3 a ) (2a - y)(5y - 3a) ( 2 a − y ) ( 5 y − 3 a ) B ( y − a ) ( 5 y − 3 a ) (y - a)(5y - 3a) ( y − a ) ( 5 y − 3 a ) C ( y − a ) ( 5 y + 3 a ) (y - a)(5y + 3a) ( y − a ) ( 5 y + 3 a ) D ( y + a ) ( 5 y − 3 a ) (y + a)(5y - 3a) ( y + a ) ( 5 y − 3 a ) E ( 2 y − a ) ( 5 y − 3 a ) (2y - a)(5y - 3a) ( 2 y − a ) ( 5 y − 3 a )
Worked solution (try it first) Look for two numbers with product
5 × ( − 3 ) = − 15 5 \times (-3) = -15 5 × ( − 3 ) = − 15 and sum 2: they are 5 and
− 3 -3 − 3 .
Split the middle term:
5 y 2 + 5 a y − 3 a y − 3 a 2 = 5 y ( y + a ) − 3 a ( y + a ) 5y^2 + 5ay - 3ay - 3a^2 = 5y(y + a) - 3a(y + a) 5 y 2 + 5 a y − 3 a y − 3 a 2 = 5 y ( y + a ) − 3 a ( y + a ) .
So
5 y 2 + 2 a y − 3 a 2 = ( y + a ) ( 5 y − 3 a ) 5y^2 + 2ay - 3a^2 = (y + a)(5y - 3a) 5 y 2 + 2 a y − 3 a 2 = ( y + a ) ( 5 y − 3 a ) , option D.
Watch out
Expand to check the middle term. ( y − a ) ( 5 y + 3 a ) (y - a)(5y + 3a) ( y − a ) ( 5 y + 3 a ) − 2 a y -2ay − 2 a y Report a problem with this question
Evaluate x y 2 − z 2 2 y z + x 2 2 y + z \dfrac{xy^2 - z^2}{2yz} + \dfrac{x^2}{2y + z} 2 y z x y 2 − z 2 + 2 y + z x 2 x = 2 x = 2 x = 2 y = − 3 y = -3 y = − 3 z = − 2 z = -2 z = − 2
A 7 6 \frac{7}{6} 6 7 B 2 3 \frac{2}{3} 3 2 C − 1 6 -\frac{1}{6} − 6 1 D − 1 2 -\frac{1}{2} − 2 1 E − 2 3 -\frac{2}{3} − 3 2
Worked solution (try it first) First fraction: the top is
2 ( 9 ) − 4 = 14 2(9) - 4 = 14 2 ( 9 ) − 4 = 14 and the bottom is
2 ( − 3 ) ( − 2 ) = 12 2(-3)(-2) = 12 2 ( − 3 ) ( − 2 ) = 12 , so it is
7 6 \frac{7}{6} 6 7 .
Second fraction: the top is
4 4 4 and the bottom is
− 6 − 2 = − 8 -6 - 2 = -8 − 6 − 2 = − 8 , so it is
− 1 2 -\frac{1}{2} − 2 1 .
Add:
7 6 − 3 6 = 4 6 \frac{7}{6} - \frac{3}{6} = \frac{4}{6} 6 7 − 6 3 = 6 4 = 2 3 = \frac{2}{3} = 3 2 , option B.
Watch out
2 y + z = − 8 2y + z = -8 2 y + z = − 8 7 6 \frac{7}{6} 6 7 Report a problem with this question
Let p p p q q q
Conclusion: I pass mathematics if and only if I pass physics. Translate the conclusion into symbol.
A p ⇒ q p \Rightarrow q p ⇒ q B p ⇐ q p \Leftarrow q p ⇐ q C p ⇔ q p \Leftrightarrow q p ⇔ q D p ⇔ ∼ q p \Leftrightarrow \sim q p ⇔∼ q E ∼ p ⇔ ∼ q \sim p \Leftrightarrow \sim q ∼ p ⇔∼ q
Worked solution (try it first) "If and only if" is the biconditional, written
⇔ \Leftrightarrow ⇔ .
Both parts are stated positively, so the conclusion is
p ⇔ q p \Leftrightarrow q p ⇔ q , option C.
Watch out
"If" alone would be one arrow; "if and only if" means the arrow goes both ways. p ⇒ q p \Rightarrow q p ⇒ q Report a problem with this question
Given that V = 1 3 π r 2 ( 2 r + h ) V = \frac{1}{3}\pi r^2 (2r + h) V = 3 1 π r 2 ( 2 r + h ) h h h
A 3 V π r 2 + 2 r \frac{3V}{\pi r^2} + 2r π r 2 3 V + 2 r B 3 V π r 2 − 2 r \frac{3V}{\pi r^2} - 2r π r 2 3 V − 2 r C 3 V − 2 π 2 π r 2 \frac{3V - 2\pi^2}{\pi r^2} π r 2 3 V − 2 π 2 D 3 V + 2 π 2 π r 2 \frac{3V + 2\pi^2}{\pi r^2} π r 2 3 V + 2 π 2 E 3 V − 2 π r π r 2 \frac{3V - 2\pi r}{\pi r^2} π r 2 3 V − 2 π r
Worked solution (try it first) Multiply both sides by 3:
3 V = π r 2 ( 2 r + h ) 3V = \pi r^2 (2r + h) 3 V = π r 2 ( 2 r + h ) .
Divide both sides by
π r 2 \pi r^2 π r 2 :
3 V π r 2 = 2 r + h \dfrac{3V}{\pi r^2} = 2r + h π r 2 3 V = 2 r + h .
Subtract
2 r 2r 2 r :
h = 3 V π r 2 − 2 r h = \dfrac{3V}{\pi r^2} - 2r h = π r 2 3 V − 2 r , option B.
Watch out
Moving 2 r 2r 2 r + 2 r +2r + 2 r Report a problem with this question
A A A B B B B B B C C C A A A C C C K K K
A A = K C 2 A = KC^2 A = K C 2 B A = K C 2 A = \frac{K}{C^2} A = C 2 K C A = K C A = KC A = K C D A = K C 4 A = \frac{K}{C^4} A = C 4 K E A = K C 4 A = KC^4 A = K C 4
Worked solution (try it first) Write the two variations:
A = k B 2 A = \dfrac{k}{B^2} A = B 2 k and
B = m C 2 B = mC^2 B = m C 2 .
Square the second:
B 2 = m 2 C 4 B^2 = m^2 C^4 B 2 = m 2 C 4 .
Substitute:
A = k m 2 C 4 A = \dfrac{k}{m^2 C^4} A = m 2 C 4 k .
Call
k m 2 = K \frac{k}{m^2} = K m 2 k = K , so
A = K C 4 A = \dfrac{K}{C^4} A = C 4 K , option D.
Watch out
A A A B 2 B^2 B 2 B B B B = m C 2 B = mC^2 B = m C 2 A = k B A = \frac{k}{B} A = B k A = K C 2 A = \frac{K}{C^2} A = C 2 K Report a problem with this question
If M M M N N N M = 15 2 M = \frac{15}{2} M = 2 15 N = 4 N = 4 N = 4 N N N M = ( 1 20 ) − 1 M = \left(\frac{1}{20}\right)^{-1} M = ( 20 1 ) − 1
A 1 3 \frac{1}{3} 3 1 B 2 3 \frac{2}{3} 3 2 C 1 1 2 1\frac{1}{2} 1 2 1 D 2 2 3 2\frac{2}{3} 2 3 2 E 3 1 4 3\frac{1}{4} 3 4 1
Worked solution (try it first) Write
M = k N M = \dfrac{k}{N} M = N k , so
k = M N = 15 2 × 4 = 30 k = MN = \frac{15}{2} \times 4 = 30 k = M N = 2 15 × 4 = 30 .
A power of
− 1 -1 − 1 turns the fraction upside down:
M = ( 1 20 ) − 1 = 20 M = \left(\frac{1}{20}\right)^{-1} = 20 M = ( 20 1 ) − 1 = 20 .
So
N = k M N = \dfrac{k}{M} N = M k = 30 20 = \dfrac{30}{20} = 20 30 = 1 1 2 = 1\frac{1}{2} = 1 2 1 , option C.
Watch out
( 1 20 ) − 1 \left(\frac{1}{20}\right)^{-1} ( 20 1 ) − 1 1 20 \frac{1}{20} 20 1 − 1 20 -\frac{1}{20} − 20 1 M = 1 20 M = \frac{1}{20} M = 20 1 N = 600 N = 600 N = 600 Report a problem with this question
Find the value of k k k O O O
A 133 ∘ 133^\circ 13 3 ∘ B 67 ∘ 67^\circ 6 7 ∘ C 45 ∘ 45^\circ 4 5 ∘ D 30 ∘ 30^\circ 3 0 ∘ E 22 ∘ 22^\circ 2 2 ∘
Worked solution (try it first) The quadrilateral has all four vertices on the circle, so it is cyclic.
Opposite angles of a cyclic quadrilateral add up to
180 ∘ 180^\circ 18 0 ∘ :
k + 3 k = 180 ∘ k + 3k = 180^\circ k + 3 k = 18 0 ∘ .
So
4 k = 180 ∘ 4k = 180^\circ 4 k = 18 0 ∘ and
k = 45 ∘ k = 45^\circ k = 4 5 ∘ , option C.
Watch out
Opposite angles of a cyclic quadrilateral are supplementary (sum 180 ∘ 180^\circ 18 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
Given the line 2 x − 3 y = 9 2x - 3y = 9 2 x − 3 y = 9
A − 3 -3 − 3 B 2 3 \frac{2}{3} 3 2 C 3 2 \frac{3}{2} 2 3 D 2 E 3
Worked solution (try it first) Make
y y y the subject:
3 y = 2 x − 9 3y = 2x - 9 3 y = 2 x − 9 , so
y = 2 3 x − 3 y = \frac{2}{3}x - 3 y = 3 2 x − 3 .
The gradient is the coefficient of
x x x :
2 3 \frac{2}{3} 3 2 , option B.
Watch out
In y = m x + c y = mx + c y = m x + c m m m c c c − 3 -3 − 3 y y y 3 2 \frac{3}{2} 2 3 Report a problem with this question
In an isosceles triangle M N O MNO M N O M N ‾ = M O ‾ = 6 \overline{MN} = \overline{MO} = 6 M N = M O = 6 ∠ N M O = 112 ∘ \angle NMO = 112^\circ ∠ N M O = 11 2 ∘ N O ‾ \overline{NO} N O
A 10.0 cm B 9.9 cm C 7.9 cm D 5.0 cm E 4.9 cm
Worked solution (try it first) Use the cosine rule:
N O 2 = 6 2 + 6 2 − 2 ( 6 ) ( 6 ) cos 112 ∘ NO^2 = 6^2 + 6^2 - 2(6)(6)\cos 112^\circ N O 2 = 6 2 + 6 2 − 2 ( 6 ) ( 6 ) cos 11 2 ∘ .
cos 112 ∘ ≈ − 0.3746 \cos 112^\circ \approx -0.3746 cos 11 2 ∘ ≈ − 0.3746 , so
N O 2 ≈ 72 + 26.97 = 98.97 NO^2 \approx 72 + 26.97 = 98.97 N O 2 ≈ 72 + 26.97 = 98.97 .
N O ≈ 9.948 NO \approx 9.948 N O ≈ 9.948 cm, which is 9.9 cm to 2 significant figures, option B.
Watch out
Round once, from the full value: 9.948 is 9.9 to 2 significant figures. Rounding to 9.95 first and then again gives 10.0 (option A), which is also written to 3 significant figures. Report a problem with this question
Find the total surface area of the cuboid in the figure.
A 270 cm 2 270\text{ cm}^2 270 cm 2 B 298 cm 2 298\text{ cm}^2 298 cm 2 C 308 cm 2 308\text{ cm}^2 308 cm 2 D 318 cm 2 318\text{ cm}^2 318 cm 2 E 516 cm 2 516\text{ cm}^2 516 cm 2
Worked solution (try it first) A cuboid has three pairs of equal faces:
18 × 5 = 90 18 \times 5 = 90 18 × 5 = 90 ,
18 × 3 = 54 18 \times 3 = 54 18 × 3 = 54 and
5 × 3 = 15 cm 2 5 \times 3 = 15\text{ cm}^2 5 × 3 = 15 cm 2 .
One of each adds to
90 + 54 + 15 = 159 cm 2 90 + 54 + 15 = 159\text{ cm}^2 90 + 54 + 15 = 159 cm 2 .
Double it for the pairs:
2 × 159 = 318 cm 2 2 \times 159 = 318\text{ cm}^2 2 × 159 = 318 cm 2 , option D.
Watch out
Surface area adds the areas of the six faces; multiplying the three edges, 18 × 5 × 3 = 270 18 \times 5 \times 3 = 270 18 × 5 × 3 = 270 cm 3 \text{cm}^3 cm 3 Report a problem with this question
Find the acute angle between the lines x + 4 y = 12 x + 4y = 12 x + 4 y = 12 2 y − x = − 6 2y - x = -6 2 y − x = − 6
A 39.4 ∘ 39.4^\circ 39. 4 ∘ B 40.6 ∘ 40.6^\circ 40. 6 ∘ C 42.4 ∘ 42.4^\circ 42. 4 ∘ D 49.4 ∘ 49.4^\circ 49. 4 ∘ E 51.6 ∘ 51.6^\circ 51. 6 ∘
Worked solution (try it first) Make
y y y the subject of each:
y = − 1 4 x + 3 y = -\frac{1}{4}x + 3 y = − 4 1 x + 3 and
y = 1 2 x − 3 y = \frac{1}{2}x - 3 y = 2 1 x − 3 , so the gradients are
m 1 = − 1 4 m_1 = -\frac{1}{4} m 1 = − 4 1 and
m 2 = 1 2 m_2 = \frac{1}{2} m 2 = 2 1 .
Use
tan θ = ∣ m 1 − m 2 1 + m 1 m 2 ∣ \tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right| tan θ = 1 + m 1 m 2 m 1 − m 2 : the top is
− 3 4 -\frac{3}{4} − 4 3 and the bottom is
1 − 1 8 = 7 8 1 - \frac{1}{8} = \frac{7}{8} 1 − 8 1 = 8 7 .
So
tan θ = 3 4 ÷ 7 8 \tan\theta = \frac{3}{4} \div \frac{7}{8} tan θ = 4 3 ÷ 8 7 = 6 7 = \frac{6}{7} = 7 6 , and
θ ≈ 40.6 ∘ \theta \approx 40.6^\circ θ ≈ 40. 6 ∘ , option B.
Watch out
Keep 1 + m 1 m 2 1 + m_1 m_2 1 + m 1 m 2 tan θ = 7 6 \tan\theta = \frac{7}{6} tan θ = 6 7 49.4 ∘ 49.4^\circ 49. 4 ∘ Report a problem with this question
Find the equation of a straight line passing through the point ( − 1 , 4 ) (-1, 4) ( − 1 , 4 ) y = 3 x + 2 y = 3x + 2 y = 3 x + 2
A y − 3 x = − 7 y - 3x = -7 y − 3 x = − 7 B y − 3 x = 7 y - 3x = 7 y − 3 x = 7 C y + 3 x = 7 y + 3x = 7 y + 3 x = 7 D y − 3 x = − 1 y - 3x = -1 y − 3 x = − 1 E y − 3 x = 1 y - 3x = 1 y − 3 x = 1
Worked solution (try it first) Parallel lines have the same gradient, so
m = 3 m = 3 m = 3 .
Use
y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) :
y − 4 = 3 ( x + 1 ) y - 4 = 3(x + 1) y − 4 = 3 ( x + 1 ) .
Expand:
y = 3 x + 7 y = 3x + 7 y = 3 x + 7 , so
y − 3 x = 7 y - 3x = 7 y − 3 x = 7 , option B.
Watch out
x 1 = − 1 x_1 = -1 x 1 = − 1 x − x 1 = x + 1 x - x_1 = x + 1 x − x 1 = x + 1 x − 1 x - 1 x − 1 y = 3 x + 1 y = 3x + 1 y = 3 x + 1 y − 3 x = 1 y - 3x = 1 y − 3 x = 1 Report a problem with this question
A sector of a circle subtends an angle of 90 ∘ 90^\circ 9 0 ∘ 11 cm 2 11\text{ cm}^2 11 cm 2
A 2.65 cm B 3.74 cm C 7.00 cm D 7.65 cm E 13.75 cm
Worked solution (try it first) A
90 ∘ 90^\circ 9 0 ∘ sector is a quarter of the circle:
90 360 × 22 7 r 2 = 11 \frac{90}{360} \times \frac{22}{7} r^2 = 11 360 90 × 7 22 r 2 = 11 .
So
22 28 r 2 = 11 \frac{22}{28} r^2 = 11 28 22 r 2 = 11 , which gives
r 2 = 14 r^2 = 14 r 2 = 14 .
r = 14 ≈ 3.74 r = \sqrt{14} \approx 3.74 r = 14 ≈ 3.74 cm, option B.
Watch out
Take the square root at the end: r 2 = 14 r^2 = 14 r 2 = 14 r ≈ 3.74 r \approx 3.74 r ≈ 3.74 Report a problem with this question
Find the curved surface area of a cone of height 4 cm and base radius 3 cm, correct to 2 decimal places.
A 5.00 cm 2 5.00\text{ cm}^2 5.00 cm 2 B 13.63 cm 2 13.63\text{ cm}^2 13.63 cm 2 C 47.14 cm 2 47.14\text{ cm}^2 47.14 cm 2 D 75.43 cm 2 75.43\text{ cm}^2 75.43 cm 2 E 108.51 cm 2 108.51\text{ cm}^2 108.51 cm 2
Worked solution (try it first) Find the slant height by Pythagoras:
l = 3 2 + 4 2 = 5 l = \sqrt{3^2 + 4^2} = 5 l = 3 2 + 4 2 = 5 cm.
The curved surface area is
π r l = 22 7 × 3 × 5 \pi r l = \frac{22}{7} \times 3 \times 5 π r l = 7 22 × 3 × 5 .
That is
330 7 ≈ 47.14 cm 2 \frac{330}{7} \approx 47.14\text{ cm}^2 7 330 ≈ 47.14 cm 2 , option C.
Watch out
The curved surface is π r l \pi r l π r l π r 2 \pi r^2 π r 2 75.43 cm 2 75.43\text{ cm}^2 75.43 cm 2 Report a problem with this question
The angle of elevation of the top B B B A A A 30 ∘ 30^\circ 3 0 ∘ A B = 50 AB = 50 A B = 50 A A A
A 25 m B 42 m C 43 m D 44 m E 45 m
Worked solution (try it first) A B AB A B is the hypotenuse and the distance to the foot of the wall is the side next to the
30 ∘ 30^\circ 3 0 ∘ angle, so use cosine.
Distance
= 50 cos 30 ∘ = 50\cos 30^\circ = 50 cos 3 0 ∘ ≈ 50 × 0.8660 \approx 50 \times 0.8660 ≈ 50 × 0.8660 To 2 significant figures that is 43 m, option C.
Watch out
The distance along the ground is adjacent to the angle, so use cosine. Sine gives 50 sin 30 ∘ = 25 50\sin 30^\circ = 25 50 sin 3 0 ∘ = 25 Report a problem with this question
Given that sin θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3 θ \theta θ sin θ + cos θ 1 − tan θ \dfrac{\sin\theta + \cos\theta}{1 - \tan\theta} 1 − tan θ sin θ + cos θ
A 1 2 5 1\frac{2}{5} 1 5 2 B 3 3 5 3\frac{3}{5} 3 5 3 C 4 1 2 4\frac{1}{2} 4 2 1 D 5 3 5 5\frac{3}{5} 5 5 3 E 8 3 4 8\frac{3}{4} 8 4 3
Worked solution (try it first) Use a 3–4–5 right triangle:
cos θ = 4 5 \cos\theta = \frac{4}{5} cos θ = 5 4 and
tan θ = 3 4 \tan\theta = \frac{3}{4} tan θ = 4 3 .
The top is
3 5 + 4 5 = 7 5 \frac{3}{5} + \frac{4}{5} = \frac{7}{5} 5 3 + 5 4 = 5 7 and the bottom is
1 − 3 4 = 1 4 1 - \frac{3}{4} = \frac{1}{4} 1 − 4 3 = 4 1 .
Dividing by
1 4 \frac{1}{4} 4 1 multiplies by 4:
28 5 = 5 3 5 \frac{28}{5} = 5\frac{3}{5} 5 28 = 5 5 3 , option D.
Watch out
Divide by the whole bottom, 1 4 \frac{1}{4} 4 1 7 5 = 1 2 5 \frac{7}{5} = 1\frac{2}{5} 5 7 = 1 5 2 Report a problem with this question
In the figure, O O O ∠ C A F \angle CAF ∠ C A F
A 30 ∘ 30^\circ 3 0 ∘ B 40 ∘ 40^\circ 4 0 ∘ C 50 ∘ 50^\circ 5 0 ∘ D 60 ∘ 60^\circ 6 0 ∘ E 80 ∘ 80^\circ 8 0 ∘
Worked solution (try it first) A C B E ACBE A C B E is a cyclic quadrilateral, so opposite angles add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ A C B = 180 ∘ − 120 ∘ \angle ACB = 180^\circ - 120^\circ ∠ A C B = 18 0 ∘ − 12 0 ∘ Angles on a straight line add up to
180 ∘ 180^\circ 18 0 ∘ , so
∠ A F C = 180 ∘ − 100 ∘ \angle AFC = 180^\circ - 100^\circ ∠ A F C = 18 0 ∘ − 10 0 ∘ The angles of triangle
A C F ACF A C F add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ C A F = 180 ∘ − 60 ∘ − 80 ∘ \angle CAF = 180^\circ - 60^\circ - 80^\circ ∠ C A F = 18 0 ∘ − 6 0 ∘ − 8 0 ∘ = 40 ∘ = 40^\circ = 4 0 ∘ , option B.
Watch out
60 ∘ 60^\circ 6 0 ∘ ∠ A C B \angle ACB ∠ A C B ∠ C A F \angle CAF ∠ C A F A C F ACF A C F Report a problem with this question
X X X Y Y Y 165 ∘ 165^\circ 16 5 ∘ 67 ∘ 67^\circ 6 7 ∘ R = 6400 R = 6400 R = 6400
A 10951.1 km B 14303.5 km C 25925.1 km D 30925.1 km E 44303.5 km
Worked solution (try it first) One place is east and one west, so add:
165 ∘ + 67 ∘ = 232 ∘ 165^\circ + 67^\circ = 232^\circ 16 5 ∘ + 6 7 ∘ = 23 2 ∘ .
That is more than
180 ∘ 180^\circ 18 0 ∘ , so the shorter way round is
360 ∘ − 232 ∘ = 128 ∘ 360^\circ - 232^\circ = 128^\circ 36 0 ∘ − 23 2 ∘ = 12 8 ∘ .
The equator is a great circle of radius 6400 km, so the arc is
128 360 × 2 × 22 7 × 6400 \frac{128}{360} \times 2 \times \frac{22}{7} \times 6400 360 128 × 2 × 7 22 × 6400 .
That is about 14303.5 km, option B.
Watch out
232 ∘ 232^\circ 23 2 ∘ 360 ∘ − 232 ∘ = 128 ∘ 360^\circ - 232^\circ = 128^\circ 36 0 ∘ − 23 2 ∘ = 12 8 ∘ Report a problem with this question
In the figure, P Q PQ P Q P R PR P R P P P O O O ∠ Q P R = 34 ∘ \angle QPR = 34^\circ ∠ QP R = 3 4 ∘ x x x
A 136 ∘ 136^\circ 13 6 ∘ B 140 ∘ 140^\circ 14 0 ∘ C 146 ∘ 146^\circ 14 6 ∘ D 148 ∘ 148^\circ 14 8 ∘ E 150 ∘ 150^\circ 15 0 ∘
Worked solution (try it first) A tangent meets the radius at
90 ∘ 90^\circ 9 0 ∘ , so
∠ O Q P = ∠ O R P = 90 ∘ \angle OQP = \angle ORP = 90^\circ ∠ O QP = ∠ O R P = 9 0 ∘ .
The angles of quadrilateral
O Q P R OQPR O QP R add up to
360 ∘ 360^\circ 36 0 ∘ :
x + 90 ∘ + 90 ∘ + 34 ∘ = 360 ∘ x + 90^\circ + 90^\circ + 34^\circ = 360^\circ x + 9 0 ∘ + 9 0 ∘ + 3 4 ∘ = 36 0 ∘ .
So
x = 360 ∘ − 214 ∘ = 146 ∘ x = 360^\circ - 214^\circ = 146^\circ x = 36 0 ∘ − 21 4 ∘ = 14 6 ∘ , option C.
Watch out
The two right angles take 180 ∘ 180^\circ 18 0 ∘ 360 ∘ 360^\circ 36 0 ∘ x x x 34 ∘ 34^\circ 3 4 ∘ 180 ∘ 180^\circ 18 0 ∘ P P P 17 ∘ 17^\circ 1 7 ∘ 163 ∘ 163^\circ 16 3 ∘ Report a problem with this question
Find the gradient of a straight line joining the points A ( 3 , − 5 ) A(3, -5) A ( 3 , − 5 ) B ( 1 , 3 ) B(1, 3) B ( 1 , 3 )
A − 4 -4 − 4 B − 1 4 -\frac{1}{4} − 4 1 C 0 D 1 4 \frac{1}{4} 4 1 E 4
Worked solution (try it first) Gradient
= y 2 − y 1 x 2 − x 1 = \dfrac{y_2 - y_1}{x_2 - x_1} = x 2 − x 1 y 2 − y 1 .
The rise is
3 − ( − 5 ) = 8 3 - (-5) = 8 3 − ( − 5 ) = 8 and the run is
1 − 3 = − 2 1 - 3 = -2 1 − 3 = − 2 .
So the gradient is
8 − 2 = − 4 \frac{8}{-2} = -4 − 2 8 = − 4 , option A.
Watch out
Subtract in the same order top and bottom. Using 3 − ( − 5 ) 3 - (-5) 3 − ( − 5 ) 3 − 1 3 - 1 3 − 1 − 1 4 -\frac{1}{4} − 4 1 Report a problem with this question
A plastic spherical ball of radius 6 cm was melted and reshaped into a cube. Find the length of the cube, correct to the nearest centimetre.
Worked solution (try it first) The volume of the ball is
4 3 × 22 7 × 6 3 ≈ 905.1 cm 3 \frac{4}{3} \times \frac{22}{7} \times 6^3 \approx 905.1\text{ cm}^3 3 4 × 7 22 × 6 3 ≈ 905.1 cm 3 .
The cube has the same volume, so its edge is
905.1 3 ≈ 9.67 \sqrt[3]{905.1} \approx 9.67 3 905.1 ≈ 9.67 cm.
To the nearest centimetre that is 10 cm, option D.
Watch out
Round 9.67 up, not down: cutting off the decimals gives 9 (option C). Report a problem with this question
The mean age of seven children is 10 years. The ages (in years) of six of the children are 3, 6, 8, 14, 15 and 16. Calculate the age of the seventh child.
A 6 years B 7 years C 8 years D 10 years E 14 years
Worked solution (try it first) The total of the seven ages is
7 × 10 = 70 7 \times 10 = 70 7 × 10 = 70 years.
The six known ages add up to
3 + 6 + 8 + 14 + 15 + 16 = 62 3 + 6 + 8 + 14 + 15 + 16 = 62 3 + 6 + 8 + 14 + 15 + 16 = 62 .
So the seventh child is
70 − 62 = 8 70 - 62 = 8 70 − 62 = 8 years old, option C.
Watch out
Multiply the mean by 7, the number of children, to get the total 70. 10 years (option D) is the mean, not the missing age. Report a problem with this question
The mean age of seven children is 10 years. The ages (in years) of six of the children are 3, 6, 8, 14, 15 and 16. Find the modal age.
A 6 years B 8 years C 14 years D 15 years E 16 years
Worked solution (try it first) The seventh child is
7 × 10 − 62 = 8 7 \times 10 - 62 = 8 7 × 10 − 62 = 8 years old, so the ages are 3, 6, 8, 8, 14, 15, 16.
The mode is the most common age: 8 appears twice, so the modal age is 8 years, option B.
Watch out
Include the seventh child's age. Without it every age appears once and there is no mode. Report a problem with this question
The mean age of seven children is 10 years. The ages (in years) of six of the children are 3, 6, 8, 14, 15 and 16. What is the median age?
A 6 years B 7 years C 8 years D 10 years E 14 years
Worked solution (try it first) The seventh child is 8 years old.
In order, the ages are 3, 6, 8, 8, 14, 15, 16.
With 7 values the median is the 4th: 8 years, option C.
Watch out
Put the ages in order first. Adding 8 to the end of the list as given and taking the middle value gives 14 (option E). Report a problem with this question
The mean of the following set of data: 5, 7, 9, x x x x x x
Worked solution (try it first) There are 5 numbers with mean 6, so they add up to
5 × 6 = 30 5 \times 6 = 30 5 × 6 = 30 .
The known numbers add up to
5 + 7 + 9 + 6 = 27 5 + 7 + 9 + 6 = 27 5 + 7 + 9 + 6 = 27 , so
x = 30 − 27 = 3 x = 30 - 27 = 3 x = 30 − 27 = 3 , option B.
Watch out
Count x x x 5 × 6 5 \times 6 5 × 6 4 × 6 = 24 4 \times 6 = 24 4 × 6 = 24 x x x Report a problem with this question
The mean of the following set of data: 5, 7, 9, x x x
Worked solution (try it first) From the mean,
x = 3 x = 3 x = 3 , so the data are 5, 7, 9, 3, 6 with mean 6.
The deviations from 6 are
− 1 , 1 , 3 , − 3 , 0 -1, 1, 3, -3, 0 − 1 , 1 , 3 , − 3 , 0 .
Their squares add up to
1 + 1 + 9 + 9 + 0 = 20 1 + 1 + 9 + 9 + 0 = 20 1 + 1 + 9 + 9 + 0 = 20 .
Variance
= 20 5 = 4.0 = \dfrac{20}{5} = 4.0 = 5 20 = 4.0 , option B.
Watch out
Variance is the mean of the squared deviations, before any square root. Taking 4 = 2 \sqrt{4} = 2 4 = 2 Report a problem with this question
Two fair dice are thrown together. Calculate the probability that the sum obtained is less than 8.
A 5 18 \frac{5}{18} 18 5 B 5 12 \frac{5}{12} 12 5 C 4 9 \frac{4}{9} 9 4 D 5 9 \frac{5}{9} 9 5 E 7 12 \frac{7}{12} 12 7
Worked solution (try it first) There are
6 × 6 = 36 6 \times 6 = 36 6 × 6 = 36 equally likely outcomes.
Count the sums from 2 to 7:
1 + 2 + 3 + 4 + 5 + 6 = 21 1 + 2 + 3 + 4 + 5 + 6 = 21 1 + 2 + 3 + 4 + 5 + 6 = 21 outcomes.
So the probability is
21 36 = 7 12 \frac{21}{36} = \frac{7}{12} 36 21 = 12 7 , option E.
Watch out
"Less than 8" stops at 7. The other 15 outcomes (sums 8 to 12) give 15 36 = 5 12 \frac{15}{36} = \frac{5}{12} 36 15 = 12 5 Report a problem with this question
Two fair dice are thrown together. Find the probability that the sum obtained is a multiple of 6.
A 1 6 \frac{1}{6} 6 1 B 1 3 \frac{1}{3} 3 1 C 2 3 \frac{2}{3} 3 2 D 3 4 \frac{3}{4} 4 3 E 5 6 \frac{5}{6} 6 5
Worked solution (try it first) The sums that are multiples of 6 are 6 and 12.
A sum of 6 comes 5 ways (1+5, 2+4, 3+3, 4+2, 5+1) and a sum of 12 comes 1 way (6+6): 6 outcomes out of 36.
So the probability is
6 36 = 1 6 \frac{6}{36} = \frac{1}{6} 36 6 = 6 1 , option A.
Watch out
12 is a multiple of 6 as well. Counting only the sum of 6 gives 5 36 \frac{5}{36} 36 5 Report a problem with this question
Find the mode of the distribution below, correct to one decimal place.
Class
20 – 24
25 – 29
30 – 34
Frequency
3
7
2
A 24.2 B 24.5 C 25.7 D 26.2 E 26.7
Worked solution (try it first) The modal class is 25 – 29 (frequency 7).
Its lower class boundary is
L = 24.5 L = 24.5 L = 24.5 and the class width is
c = 5 c = 5 c = 5 .
The differences are
Δ 1 = 7 − 3 = 4 \Delta_1 = 7 - 3 = 4 Δ 1 = 7 − 3 = 4 and
Δ 2 = 7 − 2 = 5 \Delta_2 = 7 - 2 = 5 Δ 2 = 7 − 2 = 5 .
Mode
= L + Δ 1 Δ 1 + Δ 2 × c = L + \dfrac{\Delta_1}{\Delta_1 + \Delta_2} \times c = L + Δ 1 + Δ 2 Δ 1 × c = 24.5 + 4 9 × 5 = 24.5 + \frac{4}{9} \times 5 = 24.5 + 9 4 × 5 ≈ 26.7 \approx 26.7 ≈ 26.7 , option E.
Watch out
Use the class boundary 24.5, not the class limit 25, and width 5, not 4. Starting from 25 gives 27.2, which is not an option. Report a problem with this question
Find the range of the following set of numbers: 8, 10, 2, 6, 16, 24, 12, 20.
Worked solution (try it first) The largest number is 24 and the smallest is 2.
Range
= 24 − 2 = 22 = 24 - 2 = 22 = 24 − 2 = 22 , option E.
Watch out
The range uses the largest and smallest values, not the last and first in the list: 20 − 8 = 12 20 - 8 = 12 20 − 8 = 12 Report a problem with this question
What is the derivative of f ( x ) = C f(x) = C f ( x ) = C C C C
Worked solution (try it first) A constant does not change as
x x x changes, so its rate of change is zero.
So
f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 , option B.
Watch out
The derivative of C x Cx C x C C C C C C C C C Report a problem with this question
When a stone is dropped into water, the radius of a circular wave is increasing at the rate of 0.2 m s − 1 0.2\text{ m s}^{-1} 0.2 m s − 1
A 44.000 m 2 s − 1 44.000\text{ m}^2\text{s}^{-1} 44.000 m 2 s − 1 B 8.800 m 2 s − 1 8.800\text{ m}^2\text{s}^{-1} 8.800 m 2 s − 1 C 0.440 m 2 s − 1 0.440\text{ m}^2\text{s}^{-1} 0.440 m 2 s − 1 D 0.088 m 2 s − 1 0.088\text{ m}^2\text{s}^{-1} 0.088 m 2 s − 1 E 0.014 m 2 s − 1 0.014\text{ m}^2\text{s}^{-1} 0.014 m 2 s − 1
Worked solution (try it first) The area is
A = π r 2 A = \pi r^2 A = π r 2 , so
d A d r = 2 π r \dfrac{dA}{dr} = 2\pi r d r d A = 2 π r .
Chain rule:
d A d t = d A d r × d r d t \dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt} d t d A = d r d A × d t d r = 2 π r × 0.2 = 2\pi r \times 0.2 = 2 π r × 0.2 .
Put in
r = 0.07 r = 0.07 r = 0.07 :
2 × 22 7 × 0.07 × 0.2 = 0.088 m 2 s − 1 2 \times \frac{22}{7} \times 0.07 \times 0.2 = 0.088\text{ m}^2\text{s}^{-1} 2 × 7 22 × 0.07 × 0.2 = 0.088 m 2 s − 1 , option D.
Watch out
Multiply by the rate d r d t = 0.2 \frac{dr}{dt} = 0.2 d t d r = 0.2 2 π r = 0.44 2\pi r = 0.44 2 π r = 0.44 Report a problem with this question
Evaluate ∫ 3 x d x \int 3\sqrt{x}\,dx ∫ 3 x d x
A 4 x 3 + c 4\sqrt{x^3} + c 4 x 3 + c B 3 x 3 + c 3\sqrt{x^3} + c 3 x 3 + c C 2 x 3 + c 2\sqrt{x^3} + c 2 x 3 + c D 2 x 2 + c 2\sqrt{x^2} + c 2 x 2 + c E x 3 + c \sqrt{x^3} + c x 3 + c
Worked solution (try it first) Write
3 x 3\sqrt{x} 3 x as
3 x 1 2 3x^{\frac{1}{2}} 3 x 2 1 .
Add 1 to the power and divide by the new power:
3 × x 3 2 3 2 = 2 x 3 2 3 \times \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} = 2x^{\frac{3}{2}} 3 × 2 3 x 2 3 = 2 x 2 3 .
x 3 2 = x 3 x^{\frac{3}{2}} = \sqrt{x^3} x 2 3 = x 3 , so the answer is
2 x 3 + c 2\sqrt{x^3} + c 2 x 3 + c , option C.
Watch out
Divide by the new power 3 2 \frac{3}{2} 2 3 2 3 \frac{2}{3} 3 2 3 x 3 + c 3\sqrt{x^3} + c 3 x 3 + c Report a problem with this question