Decrease 120 by 25 % 25\% 25%
Worked solution (try it first) Decreasing by
25 % 25\% 25% leaves
100 % − 25 % = 75 % 100\% - 25\% = 75\% 100% − 25% = 75% of the number.
75 % 75\% 75% of 120 is
0.75 × 120 = 90 0.75 \times 120 = 90 0.75 × 120 = 90 , option C.
Watch out
25 % 25\% 25% 120 − 30 = 90 120 - 30 = 90 120 − 30 = 90 Report a problem with this question
Find the product of 10110 two 10110_{\text{two}} 1011 0 two 11 two 11_{\text{two}} 1 1 two
A 1000010 two 1000010_{\text{two}} 100001 0 two B 1000101 two 1000101_{\text{two}} 100010 1 two C 1000110 two 1000110_{\text{two}} 100011 0 two D 1001000 two 1001000_{\text{two}} 100100 0 two E 1001010 two 1001010_{\text{two}} 100101 0 two
Worked solution (try it first) Change to base ten:
10110 two = 16 + 4 + 2 = 22 10110_{\text{two}} = 16 + 4 + 2 = 22 1011 0 two = 16 + 4 + 2 = 22 and
11 two = 3 11_{\text{two}} = 3 1 1 two = 3 .
Multiply:
22 × 3 = 66 22 \times 3 = 66 22 × 3 = 66 .
Change back:
66 = 64 + 2 66 = 64 + 2 66 = 64 + 2 , so the product is
1000010 two 1000010_{\text{two}} 100001 0 two , option A.
Watch out
If you multiply in columns, 1 + 1 = 10 two 1 + 1 = 10_{\text{two}} 1 + 1 = 1 0 two Report a problem with this question
Express 5 + 2 100 + 3 1000 + 4 100000 5 + \dfrac{2}{100} + \dfrac{3}{1000} + \dfrac{4}{100000} 5 + 100 2 + 1000 3 + 100000 4
A 5.20304 B 5.02034 C 5.02304 D 5.00234 E 5.20034
Worked solution (try it first) Write each fraction as a decimal: the number of zeros tells you the place.
2 100 = 0.02 \frac{2}{100} = 0.02 100 2 = 0.02 ,
3 1000 = 0.003 \frac{3}{1000} = 0.003 1000 3 = 0.003 and
4 100 000 = 0.00004 \frac{4}{100\,000} = 0.00004 100 000 4 = 0.00004 .
Add them to 5, lining up the places:
5 + 0.02 + 0.003 + 0.00004 = 5.02304 5 + 0.02 + 0.003 + 0.00004 = 5.02304 5 + 0.02 + 0.003 + 0.00004 = 5.02304 , option C.
Watch out
2 100 \frac{2}{100} 100 2 Report a problem with this question
Simplify 2 5 10 \dfrac{2\sqrt5}{\sqrt{10}} 10 2 5
A 5 B 2 C 2 \sqrt2 2 D 5 \sqrt5 5 E 10 \sqrt{10} 10
Worked solution (try it first) Split the bottom:
10 = 5 × 2 \sqrt{10} = \sqrt5 \times \sqrt2 10 = 5 × 2 .
Cancel
5 \sqrt5 5 :
2 5 5 2 = 2 2 \dfrac{2\sqrt5}{\sqrt5\sqrt2} = \dfrac{2}{\sqrt2} 5 2 2 5 = 2 2 .
Rationalise:
2 2 = 2 2 2 \dfrac{2}{\sqrt2} = \dfrac{2\sqrt2}{2} 2 2 = 2 2 2 , which is
2 \sqrt2 2 .
So option C.
Watch out
After cancelling 5 \sqrt5 5 2 \sqrt2 2 2 2 = 2 \frac{2}{\sqrt2} = \sqrt2 2 2 = 2 Report a problem with this question
A boy walks 88 paces in a minute. If his average pace length is 0.55 m 0.55\text{ m} 0.55 m 1936 m 1936\text{ m} 1936 m
A 1 4 \frac14 4 1 B 1 3 \frac13 3 1 C 1 2 \frac12 2 1 D 2 3 \frac23 3 2 E 3 4 \frac34 4 3
Worked solution (try it first) Each minute he walks 88 paces of 0.55 m:
88 × 0.55 = 48.4 88 \times 0.55 = 48.4 88 × 0.55 = 48.4 m a minute.
Time is distance over speed:
1936 ÷ 48.4 = 40 1936 \div 48.4 = 40 1936 ÷ 48.4 = 40 minutes.
As a fraction of an hour:
40 60 = 2 3 \frac{40}{60} = \frac23 60 40 = 3 2 , option D.
Watch out
An hour has 60 minutes, not 100: 40 minutes is 40 60 = 2 3 \frac{40}{60} = \frac23 60 40 = 3 2 Report a problem with this question
Find x x x 3 × 8 ≡ x ( m o d 9 ) 3 \times 8 \equiv x \pmod 9 3 × 8 ≡ x ( mod 9 )
Worked solution (try it first) Multiply first:
3 × 8 = 24 3 \times 8 = 24 3 × 8 = 24 .
Divide by 9:
24 = 2 × 9 + 6 24 = 2 \times 9 + 6 24 = 2 × 9 + 6 , so the remainder is 6.
So
x = 6 x = 6 x = 6 , option C.
Watch out
The answer in modulo 9 is the remainder, not the quotient. 24 ÷ 9 24 \div 9 24 ÷ 9 Report a problem with this question
If log 10 3 = 0.4771 \log_{10}3 = 0.4771 log 10 3 = 0.4771 log 10 8.1 \log_{10}8.1 log 10 8.1
A 0.0916 B 0.4771 C 0.5229 D 0.9084 E 1.9084
Worked solution (try it first) Write 8.1 using 3 and 10:
8.1 = 81 10 = 3 4 10 8.1 = \frac{81}{10} = \frac{3^4}{10} 8.1 = 10 81 = 10 3 4 .
So
log 8.1 = 4 log 3 − log 10 \log 8.1 = 4\log 3 - \log 10 log 8.1 = 4 log 3 − log 10 = 4 log 3 − 1 = 4\log 3 - 1 = 4 log 3 − 1 .
That is
4 ( 0.4771 ) − 1 = 1.9084 − 1 = 0.9084 4(0.4771) - 1 = 1.9084 - 1 = 0.9084 4 ( 0.4771 ) − 1 = 1.9084 − 1 = 0.9084 , option D.
Watch out
Remember the division by 10, which subtracts 1. Stopping at 4 log 3 = 1.9084 4\log 3 = 1.9084 4 log 3 = 1.9084 log 81 \log 81 log 81 Report a problem with this question
Given that 2 log y = 8 log p + 4 log q 2\log y = 8\log p + 4\log q 2 log y = 8 log p + 4 log q y y y p p p q q q
A y = p 4 + q 2 y = p^4 + q^2 y = p 4 + q 2 B y = p 8 + q 4 y = p^8 + q^4 y = p 8 + q 4 C y = p 8 q 4 y = p^8q^4 y = p 8 q 4 D y = p 8 q 4 y = \dfrac{p^8}{q^4} y = q 4 p 8 E y = p 4 q 2 y = p^4q^2 y = p 4 q 2
Worked solution (try it first) Divide both sides by 2:
log y = 4 log p + 2 log q \log y = 4\log p + 2\log q log y = 4 log p + 2 log q .
Move the numbers up as powers and combine:
log y = log p 4 + log q 2 \log y = \log p^4 + \log q^2 log y = log p 4 + log q 2 = log ( p 4 q 2 ) = \log(p^4q^2) = log ( p 4 q 2 ) .
So
y = p 4 q 2 y = p^4q^2 y = p 4 q 2 , option E.
Watch out
The left side is 2 log y 2\log y 2 log y y = p 8 q 4 y = p^8q^4 y = p 8 q 4 y 2 y^2 y 2 Report a problem with this question
Calculate the compound interest on ₦1,200.00 for 4 years at 8 % 8\% 8%
A ₦120.90 B ₦384.00 C ₦432.59 D ₦1,511.65 E ₦1,632.59
Worked solution (try it first) At
8 % 8\% 8% compound interest the money is multiplied by
1.08 1.08 1.08 each year, so after 4 years it is
1200 × 1.08 4 1200 \times 1.08^4 1200 × 1.0 8 4 .
1.08 4 = 1.36049 1.08^4 = 1.36049 1.0 8 4 = 1.36049 , so the amount is
1200 × 1.36049 = 1200 \times 1.36049 = 1200 × 1.36049 = ₦1,632.59.
The interest is the amount minus the principal:
1632.59 − 1200 = 1632.59 - 1200 = 1632.59 − 1200 = ₦432.59, option C.
Watch out
₦1,632.59 (option E) is the amount. The compound interest is what is added to the ₦1,200, so take the principal off. Report a problem with this question
Given sets A = { a , 1 , c , 4 , d } A = \{a, 1, c, 4, d\} A = { a , 1 , c , 4 , d } B = { b , 4 , 0 , 9 , 7 , 6 } B = \{b, 4, 0, 9, 7, 6\} B = { b , 4 , 0 , 9 , 7 , 6 } C = { a , 4 , 8 , 9 , d , 2 , 5 } C = \{a, 4, 8, 9, d, 2, 5\} C = { a , 4 , 8 , 9 , d , 2 , 5 } ( A ∪ B ) ∩ ( A ∪ C ) (A \cup B) \cap (A \cup C) ( A ∪ B ) ∩ ( A ∪ C )
A { a , 1 , 4 , 8 , 9 } \{a, 1, 4, 8, 9\} { a , 1 , 4 , 8 , 9 } B { 4 , 8 , 9 , 2 , 5 } \{4, 8, 9, 2, 5\} { 4 , 8 , 9 , 2 , 5 } C { b , 4 , 2 , 5 , 8 } \{b, 4, 2, 5, 8\} { b , 4 , 2 , 5 , 8 } D { a , b , c , d , 2 } \{a, b, c, d, 2\} { a , b , c , d , 2 } E { a , c , d , 1 , 4 , 9 } \{a, c, d, 1, 4, 9\} { a , c , d , 1 , 4 , 9 }
Worked solution (try it first) A ∪ B = { a , 1 , c , 4 , d , b , 0 , 9 , 7 , 6 } A \cup B = \{a, 1, c, 4, d, b, 0, 9, 7, 6\} A ∪ B = { a , 1 , c , 4 , d , b , 0 , 9 , 7 , 6 } .
A ∪ C = { a , 1 , c , 4 , d , 8 , 9 , 2 , 5 } A \cup C = \{a, 1, c, 4, d, 8, 9, 2, 5\} A ∪ C = { a , 1 , c , 4 , d , 8 , 9 , 2 , 5 } .
The elements in both are all of
A A A together with 9:
{ a , c , d , 1 , 4 , 9 } \{a, c, d, 1, 4, 9\} { a , c , d , 1 , 4 , 9 } , option E.
Watch out
Every element of A A A A A A c c c d d d Report a problem with this question
In a Chemistry class, a student recorded 21.23 cm 3 21.23\text{ cm}^3 21.23 cm 3 21.32 cm 3 21.32\text{ cm}^3 21.32 cm 3
A 0.04 B 0.40 C 0.80 D 1.40 E 1.80
Worked solution (try it first) The error is
21.32 − 21.23 = 0.09 cm 3 21.32 - 21.23 = 0.09\text{ cm}^3 21.32 − 21.23 = 0.09 cm 3 .
Divide by the true titre and multiply by 100:
0.09 21.32 × 100 % = 0.422 … % \frac{0.09}{21.32} \times 100\% = 0.422\ldots\% 21.32 0.09 × 100% = 0.422 … % .
To one decimal place this is 0.4%, which is 0.40, option B.
Watch out
Multiply by 100 to get a percentage: 0.09 21.32 = 0.0042 \frac{0.09}{21.32} = 0.0042 21.32 0.09 = 0.0042 Report a problem with this question
In an arithmetic progression, the first term is 3 and the sum of the 3rd and 12th terms is 38 1 2 38\frac12 38 2 1
Worked solution (try it first) With
a = 3 a = 3 a = 3 , the 3rd term is
3 + 2 d 3 + 2d 3 + 2 d and the 12th term is
3 + 11 d 3 + 11d 3 + 11 d .
Their sum gives
6 + 13 d = 38 1 2 6 + 13d = 38\frac12 6 + 13 d = 38 2 1 , so
13 d = 32 1 2 13d = 32\frac12 13 d = 32 2 1 and
d = 2 1 2 d = 2\frac12 d = 2 2 1 .
The 17th term is
a + 16 d a + 16d a + 16 d , which is
3 + 16 × 2 1 2 = 43 3 + 16 \times 2\frac12 = 43 3 + 16 × 2 2 1 = 43 , option B.
Watch out
Both terms contain a a a 3 + 13 d = 38 1 2 3 + 13d = 38\frac12 3 + 13 d = 38 2 1 d ≈ 2.73 d \approx 2.73 d ≈ 2.73 Report a problem with this question
The Venn diagram shows the number of students who wrote Biology, Physics and Mathematics in a school. Find the number of students who wrote at least two subjects and the total number of students in the school, respectively.
A 13, 32 B 17, 32 C 13, 36 D 17, 36 E 15, 36
Worked solution (try it first) "At least two subjects" means the three two-subject regions and the centre:
5 + 7 + 1 + 4 = 17 5 + 7 + 1 + 4 = 17 5 + 7 + 1 + 4 = 17 .
The total is every number in the diagram, including the 4 outside the circles:
6 + 5 + 3 + 7 + 4 + 1 + 6 + 4 = 36 6 + 5 + 3 + 7 + 4 + 1 + 6 + 4 = 36 6 + 5 + 3 + 7 + 4 + 1 + 6 + 4 = 36 .
So the answer is 17, 36, option D.
Watch out
Include the 4 students outside the circles in the total. Leaving them out gives 32 (option B). Report a problem with this question
The 3rd term of a geometric progression is 18 and the 6th term is 486. Find the first term.
Worked solution (try it first) The 3rd term is
a r 2 = 18 ar^2 = 18 a r 2 = 18 and the 6th is
a r 5 = 486 ar^5 = 486 a r 5 = 486 .
Divide:
r 3 = 27 r^3 = 27 r 3 = 27 , so
r = 3 r = 3 r = 3 .
Then
a × 3 2 = 18 a \times 3^2 = 18 a × 3 2 = 18 , so
9 a = 18 9a = 18 9 a = 18 .
So
a = 2 a = 2 a = 2 , option A.
Watch out
The 3rd term is a r 2 ar^2 a r 2 r 2 = 9 r^2 = 9 r 2 = 9 r = 3 r = 3 r = 3 Report a problem with this question
The area of a rectangular piece of cardboard is 104 cm 2 104\text{ cm}^2 104 cm 2 8 cm 8\text{ cm} 8 cm
A 52 cm 52\text{ cm} 52 cm B 42 cm 42\text{ cm} 42 cm C 32 cm 32\text{ cm} 32 cm D 26 cm 26\text{ cm} 26 cm E 21 cm 21\text{ cm} 21 cm
Worked solution (try it first) Length
= = = area
÷ \div ÷ width
= 104 ÷ 8 = 13 = 104 \div 8 = 13 = 104 ÷ 8 = 13 cm.
Perimeter
= 2 ( 13 + 8 ) = 42 = 2(13 + 8) = 42 = 2 ( 13 + 8 ) = 42 cm, option B.
Watch out
A rectangle has two lengths and two widths. 13 + 8 = 21 13 + 8 = 21 13 + 8 = 21 Report a problem with this question
Find the determinant of the matrix ( 2 3 1 1 0 2 0 2 3 ) \begin{pmatrix} 2 & 3 & 1 \\ 1 & 0 & 2 \\ 0 & 2 & 3 \end{pmatrix} 2 1 0 3 0 2 1 2 3
A − 15 -15 − 15 B − 8 -8 − 8 C − 1 -1 − 1 D 7 E 8
Worked solution (try it first) Expand along the first row, with signs
+ − + + \; - \; + + − + .
The first term is
2 × ( 0 × 3 − 2 × 2 ) = − 8 2 \times (0 \times 3 - 2 \times 2) = -8 2 × ( 0 × 3 − 2 × 2 ) = − 8 .
The second term is
− 3 × ( 1 × 3 − 2 × 0 ) = − 9 -3 \times (1 \times 3 - 2 \times 0) = -9 − 3 × ( 1 × 3 − 2 × 0 ) = − 9 .
The third term is
1 × ( 1 × 2 − 0 × 0 ) = 2 1 \times (1 \times 2 - 0 \times 0) = 2 1 × ( 1 × 2 − 0 × 0 ) = 2 .
Add them:
− 8 − 9 + 2 = − 15 -8 - 9 + 2 = -15 − 8 − 9 + 2 = − 15 , option A.
Watch out
− 8 -8 − 8 Report a problem with this question
A helicopter takes 3 hours from Kano to Lagos at a constant speed. How long does the same journey take another helicopter at a quarter of the speed of the first?
A 3 hrs B 6 hrs C 9 hrs D 12 hrs E 15 hrs
Worked solution (try it first) For a fixed distance, time varies inversely as speed.
A quarter of the speed means 4 times the time:
3 × 4 = 12 3 \times 4 = 12 3 × 4 = 12 hours, option D.
Watch out
The slower helicopter takes longer, so multiply by 4. Dividing, 3 ÷ 4 3 \div 4 3 ÷ 4 Report a problem with this question
If 1 is added to the denominator of a fraction, the fraction becomes 1 2 \frac12 2 1 3 4 \frac34 4 3
A 2 5 \frac25 5 2 B 1 2 \frac12 2 1 C 3 5 \frac35 5 3 D 3 4 \frac34 4 3 E 4 5 \frac45 5 4
Worked solution (try it first) Let the fraction be
x y \frac xy y x .
Adding 1 to the denominator gives
x y + 1 = 1 2 \frac{x}{y + 1} = \frac12 y + 1 x = 2 1 , so
y + 1 = 2 x y + 1 = 2x y + 1 = 2 x and
y = 2 x − 1 y = 2x - 1 y = 2 x − 1 .
Adding 3 to both gives
x + 3 y + 3 = 3 4 \frac{x + 3}{y + 3} = \frac34 y + 3 x + 3 = 4 3 .
Cross-multiply:
4 x + 12 = 3 y + 9 4x + 12 = 3y + 9 4 x + 12 = 3 y + 9 .
Substitute
y = 2 x − 1 y = 2x - 1 y = 2 x − 1 :
4 x + 12 = 6 x + 6 4x + 12 = 6x + 6 4 x + 12 = 6 x + 6 , so
2 x = 6 2x = 6 2 x = 6 and
x = 3 x = 3 x = 3 .
Then
y = 5 y = 5 y = 5 , so the fraction is
3 5 \frac35 5 3 , option C.
Watch out
1 2 \frac12 2 1 3 4 \frac34 4 3 3 5 \frac35 5 3 3 6 = 1 2 \frac{3}{6} = \frac12 6 3 = 2 1 6 8 = 3 4 \frac{6}{8} = \frac34 8 6 = 4 3 Report a problem with this question
y y y x x x y = 2 y = 2 y = 2 x = 3 x = 3 x = 3 y = 5 y = 5 y = 5 x = 6 x = 6 x = 6 x x x y y y
A y = x + 1 y = x + 1 y = x + 1 B y = x − 1 y = x - 1 y = x − 1 C y = 1 − x y = 1 - x y = 1 − x D y = 2 x − 1 y = 2x - 1 y = 2 x − 1 E y = 2 x + 1 y = 2x + 1 y = 2 x + 1
Worked solution (try it first) Partly constant and partly varies as
x x x :
y = a + b x y = a + bx y = a + b x .
a + 3 b = 2 a + 3b = 2 a + 3 b = 2 and
a + 6 b = 5 a + 6b = 5 a + 6 b = 5 .
Subtract:
3 b = 3 3b = 3 3 b = 3 , so
b = 1 b = 1 b = 1 .
Then
a = 2 − 3 = − 1 a = 2 - 3 = -1 a = 2 − 3 = − 1 , so
y = x − 1 y = x - 1 y = x − 1 , option B.
Watch out
Watch the sign of the constant: a = 2 − 3 = − 1 a = 2 - 3 = -1 a = 2 − 3 = − 1 y = x + 1 y = x + 1 y = x + 1 y = 4 y = 4 y = 4 x = 3 x = 3 x = 3 Report a problem with this question
Find a quadratic equation whose roots are 2 and − 1 3 -\frac13 − 3 1
A 3 x 2 + 6 x − 1 = 0 3x^2 + 6x - 1 = 0 3 x 2 + 6 x − 1 = 0 B 3 x 2 − 5 x − 2 = 0 3x^2 - 5x - 2 = 0 3 x 2 − 5 x − 2 = 0 C 3 x 2 − 5 x + 2 = 0 3x^2 - 5x + 2 = 0 3 x 2 − 5 x + 2 = 0 D 6 x 2 − x + 15 = 0 6x^2 - x + 15 = 0 6 x 2 − x + 15 = 0 E 3 x 2 − 2 x − 5 = 0 3x^2 - 2x - 5 = 0 3 x 2 − 2 x − 5 = 0
Worked solution (try it first) Roots 2 and
− 1 3 -\frac13 − 3 1 give the factors
( x − 2 ) (x - 2) ( x − 2 ) and
( x + 1 3 ) \left(x + \frac13\right) ( x + 3 1 ) .
Multiply the second factor by 3 to clear the fraction:
( x − 2 ) ( 3 x + 1 ) = 0 (x - 2)(3x + 1) = 0 ( x − 2 ) ( 3 x + 1 ) = 0 .
Expand:
3 x 2 + x − 6 x − 2 = 3 x 2 − 5 x − 2 3x^2 + x - 6x - 2 = 3x^2 - 5x - 2 3 x 2 + x − 6 x − 2 = 3 x 2 − 5 x − 2 .
So the equation is
3 x 2 − 5 x − 2 = 0 3x^2 - 5x - 2 = 0 3 x 2 − 5 x − 2 = 0 , option B.
Watch out
The product of the roots, 2 × ( − 1 3 ) 2 \times \left(-\frac13\right) 2 × ( − 3 1 ) + 2 +2 + 2 2 3 \frac23 3 2 Report a problem with this question
Use the graph (a quadratic graph and a straight line) to answer: which of the following gives the points of intersection of the linear graph and the quadratic graph?
A ( 0 , 2 ) , ( − 1 , 0 ) (0, 2), (-1, 0) ( 0 , 2 ) , ( − 1 , 0 ) B ( − 1 , 0 ) , ( 7 , 3 ) (-1, 0), (7, 3) ( − 1 , 0 ) , ( 7 , 3 ) C ( − 1 , 0 ) , ( 0 , 2 ) (-1, 0), (0, 2) ( − 1 , 0 ) , ( 0 , 2 ) D ( 0 , − 1 ) , ( 3 , 7 ) (0, -1), (3, 7) ( 0 , − 1 ) , ( 3 , 7 ) E ( − 1 , 0 ) , ( 3 , 7 ) (-1, 0), (3, 7) ( − 1 , 0 ) , ( 3 , 7 )
Worked solution (try it first) The points of intersection are where the line meets the curve.
The graph shows two.
The first is on the
x x x -axis at
x = − 1 x = -1 x = − 1 : the point
( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) .
The dashed lines mark the second:
x = 3 x = 3 x = 3 and
y = 7 y = 7 y = 7 , the point
( 3 , 7 ) (3, 7) ( 3 , 7 ) .
So the answer is option E.
Watch out
Write the x x x ( 3 , 7 ) (3, 7) ( 3 , 7 ) ( 7 , 3 ) (7, 3) ( 7 , 3 ) Report a problem with this question
Use the graph (a quadratic graph and a straight line) to answer: the equation of the line of symmetry of the quadratic graph is
A x = − 1 x = -1 x = − 1 B x = 0 x = 0 x = 0 C x = 0.5 x = 0.5 x = 0.5 D x = 1 x = 1 x = 1 E x = 1.5 x = 1.5 x = 1.5
Worked solution (try it first) A parabola is symmetrical about the vertical line through its lowest point, halfway between its roots.
The curve cuts the
x x x -axis at
− 1 -1 − 1 and 2, so the line is at
x = − 1 + 2 2 = 0.5 x = \frac{-1 + 2}{2} = 0.5 x = 2 − 1 + 2 = 0.5 .
So the line of symmetry is
x = 0.5 x = 0.5 x = 0.5 , option C.
Watch out
Keep the minus sign on − 1 -1 − 1 − 1 + 2 2 = 0.5 \frac{-1 + 2}{2} = 0.5 2 − 1 + 2 = 0.5 1 + 2 2 \frac{1 + 2}{2} 2 1 + 2 x = 1.5 x = 1.5 x = 1.5 Report a problem with this question
Use the graph (a quadratic graph and a straight line) to answer: find the equation of the quadratic graph (the equation whose roots it shows).
A x 2 − x − 2 = 0 x^2 - x - 2 = 0 x 2 − x − 2 = 0 B x 2 − 3 x − 2 = 0 x^2 - 3x - 2 = 0 x 2 − 3 x − 2 = 0 C x 2 − 2 x − 3 = 0 x^2 - 2x - 3 = 0 x 2 − 2 x − 3 = 0 D x 2 − x + 2 = 0 x^2 - x + 2 = 0 x 2 − x + 2 = 0 E x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0
Worked solution (try it first) The curve cuts the
x x x -axis at
x = − 1 x = -1 x = − 1 and
x = 2 x = 2 x = 2 , so the factors are
( x + 1 ) (x + 1) ( x + 1 ) and
( x − 2 ) (x - 2) ( x − 2 ) .
Expand:
( x + 1 ) ( x − 2 ) = x 2 − 2 x + x − 2 (x + 1)(x - 2) = x^2 - 2x + x - 2 ( x + 1 ) ( x − 2 ) = x 2 − 2 x + x − 2 = x 2 − x − 2 = x^2 - x - 2 = x 2 − x − 2 .
So the equation is
x 2 − x − 2 = 0 x^2 - x - 2 = 0 x 2 − x − 2 = 0 , option A.
Watch out
A root of − 1 -1 − 1 x + 1 x + 1 x + 1 x − 1 x - 1 x − 1 ( x − 1 ) ( x − 2 ) (x - 1)(x - 2) ( x − 1 ) ( x − 2 ) x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0 Report a problem with this question
Which of the regions U, V, X, Y, Z shown satisfies the inequalities 0 < y < 2 0 < y < 2 0 < y < 2 y < 3 + x y < 3 + x y < 3 + x x < 0 x < 0 x < 0
Worked solution (try it first) 0 < y < 2 0 < y < 2 0 < y < 2 puts the region between the
x x x -axis and the line
y = 2 y = 2 y = 2 .
That rules out U, V and X.
x < 0 x < 0 x < 0 puts it to the left of the
y y y -axis, and both Y and Z are there.
y < 3 + x y < 3 + x y < 3 + x puts it below, that is to the right of, the slanted line.
Only Z is on that side, so option E.
Watch out
Test a point in Y, such as ( − 3 , 1 ) (-3, 1) ( − 3 , 1 ) 3 + x = 0 3 + x = 0 3 + x = 0 1 < 0 1 < 0 1 < 0 y = 3 + x y = 3 + x y = 3 + x Report a problem with this question
Which of the following inequalities is represented by the number line shown?
A x < − 1 x < -1 x < − 1 B x ≤ 3.5 x \le 3.5 x ≤ 3.5 C x ≤ − 1 x \le -1 x ≤ − 1 D x ≥ 3.5 x \ge 3.5 x ≥ 3.5 E x ≥ − 1 x \ge -1 x ≥ − 1
Worked solution (try it first) The arrow starts at
− 1 -1 − 1 and points right, so
x x x takes
− 1 -1 − 1 and every value above it.
The dot at
− 1 -1 − 1 is filled in, so
− 1 -1 − 1 itself is included.
That gives
x ≥ − 1 x \ge -1 x ≥ − 1 , option E.
Watch out
The arrow's tip near 3.5 is not an end point; the arrowhead means the values carry on for ever. The only boundary is the dot at − 1 -1 − 1 Report a problem with this question
Solve the equation 2 x + 8 = 21 x 2 2x + 8 = 21x^2 2 x + 8 = 21 x 2
A x = 2 3 x = \frac23 x = 3 2 x = 4 7 x = \frac47 x = 7 4 B x = 2 3 x = \frac23 x = 3 2 x = − 4 7 x = -\frac47 x = − 7 4 C x = − 2 3 x = -\frac23 x = − 3 2 x = 4 7 x = \frac47 x = 7 4 D x = − 2 3 x = -\frac23 x = − 3 2 x = − 4 7 x = -\frac47 x = − 7 4 E x = 2 3 x = \frac23 x = 3 2 x = − 6 7 x = -\frac67 x = − 7 6
Worked solution (try it first) Bring everything to one side:
21 x 2 − 2 x − 8 = 0 21x^2 - 2x - 8 = 0 21 x 2 − 2 x − 8 = 0 .
Find two numbers with product
21 × ( − 8 ) = − 168 21 \times (-8) = -168 21 × ( − 8 ) = − 168 and sum
− 2 -2 − 2 : they are
− 14 -14 − 14 and 12.
Split and group:
21 x 2 − 14 x + 12 x − 8 = 7 x ( 3 x − 2 ) + 4 ( 3 x − 2 ) 21x^2 - 14x + 12x - 8 = 7x(3x - 2) + 4(3x - 2) 21 x 2 − 14 x + 12 x − 8 = 7 x ( 3 x − 2 ) + 4 ( 3 x − 2 ) = ( 3 x − 2 ) ( 7 x + 4 ) = (3x - 2)(7x + 4) = ( 3 x − 2 ) ( 7 x + 4 ) .
So
x = 2 3 x = \frac23 x = 3 2 or
x = − 4 7 x = -\frac47 x = − 7 4 , option B.
Watch out
3 x − 2 = 0 3x - 2 = 0 3 x − 2 = 0 x = + 2 3 x = +\frac23 x = + 3 2 7 x + 4 = 0 7x + 4 = 0 7 x + 4 = 0 x = − 4 7 x = -\frac47 x = − 7 4 Report a problem with this question
A candidate is asked to draw the graph of y = x 2 + 6 x − 27 y = x^2 + 6x - 27 y = x 2 + 6 x − 27 x 2 + 5 x − 29 = 0 x^2 + 5x - 29 = 0 x 2 + 5 x − 29 = 0
A y = 2 x − 1 y = 2x - 1 y = 2 x − 1 B y = x + 1 y = x + 1 y = x + 1 C y = x − 1 y = x - 1 y = x − 1 D y = x − 2 y = x - 2 y = x − 2 E y = x + 2 y = x + 2 y = x + 2
Worked solution (try it first) Setting the curve equal to the line must give the same equation as
x 2 + 5 x − 29 = 0 x^2 + 5x - 29 = 0 x 2 + 5 x − 29 = 0 .
Subtract the equation from the curve:
( x 2 + 6 x − 27 ) − ( x 2 + 5 x − 29 ) = x + 2 (x^2 + 6x - 27) - (x^2 + 5x - 29) = x + 2 ( x 2 + 6 x − 27 ) − ( x 2 + 5 x − 29 ) = x + 2 .
So the line is
y = x + 2 y = x + 2 y = x + 2 , option E.
Check:
x 2 + 6 x − 27 = x + 2 x^2 + 6x - 27 = x + 2 x 2 + 6 x − 27 = x + 2 rearranges to
x 2 + 5 x − 29 = 0 x^2 + 5x - 29 = 0 x 2 + 5 x − 29 = 0 .
Watch out
Subtracting a negative adds: − 27 − ( − 29 ) = + 2 -27 - (-29) = +2 − 27 − ( − 29 ) = + 2 − 2 -2 − 2 y = x − 2 y = x - 2 y = x − 2 Report a problem with this question
What must be added to 2 y 2 + 7 y 2y^2 + 7y 2 y 2 + 7 y
A 49 B 14 C 49 4 \frac{49}{4} 4 49 D 49 8 \frac{49}{8} 8 49 E 4
Worked solution (try it first) Take out the 2 so the
y 2 y^2 y 2 term has coefficient 1:
2 y 2 + 7 y = 2 ( y 2 + 7 2 y ) 2y^2 + 7y = 2\left(y^2 + \frac72y\right) 2 y 2 + 7 y = 2 ( y 2 + 2 7 y ) .
Inside the bracket, add the square of half of
7 2 \frac72 2 7 :
( 7 4 ) 2 = 49 16 \left(\frac74\right)^2 = \frac{49}{16} ( 4 7 ) 2 = 16 49 .
Outside the bracket that is
2 × 49 16 = 49 8 2 \times \frac{49}{16} = \frac{49}{8} 2 × 16 49 = 8 49 .
So add
49 8 \frac{49}{8} 8 49 , option D.
Watch out
"Add the square of half the coefficient" only works when the y 2 y^2 y 2 1 y 2 1y^2 1 y 2 49 4 \frac{49}{4} 4 49 Report a problem with this question
Solve the simultaneous equations x + 2 y = − 4 x + 2y = -4 x + 2 y = − 4 2 x + 3 y = − 5 2x + 3y = -5 2 x + 3 y = − 5
A x = − 2 , y = − 1 x = -2, y = -1 x = − 2 , y = − 1 B x = − 2 , y = 3 x = -2, y = 3 x = − 2 , y = 3 C x = 2 , y = − 3 x = 2, y = -3 x = 2 , y = − 3 D x = − 2 , y = − 3 x = -2, y = -3 x = − 2 , y = − 3 E x = 2 , y = 3 x = 2, y = 3 x = 2 , y = 3
Worked solution (try it first) Make
x x x the subject of the first equation:
x = − 4 − 2 y x = -4 - 2y x = − 4 − 2 y .
Substitute into the second:
2 ( − 4 − 2 y ) + 3 y = − 5 2(-4 - 2y) + 3y = -5 2 ( − 4 − 2 y ) + 3 y = − 5 , so
− 8 − y = − 5 -8 - y = -5 − 8 − y = − 5 .
Add 8 to both sides:
− y = 3 -y = 3 − y = 3 , so
y = − 3 y = -3 y = − 3 .
Then
x = − 4 + 6 = 2 x = -4 + 6 = 2 x = − 4 + 6 = 2 .
So
x = 2 x = 2 x = 2 ,
y = − 3 y = -3 y = − 3 , option C.
Watch out
Check both equations. Option A, x = − 2 x = -2 x = − 2 y = − 1 y = -1 y = − 1 x + 2 y = − 4 x + 2y = -4 x + 2 y = − 4 2 x + 3 y = − 7 2x + 3y = -7 2 x + 3 y = − 7 − 5 -5 − 5 Report a problem with this question
Factorize 12 a 2 − 3 ( a − 3 b ) 2 12a^2 - 3(a - 3b)^2 12 a 2 − 3 ( a − 3 b ) 2
A 9 ( a + 3 b ) ( a − b ) 9(a + 3b)(a - b) 9 ( a + 3 b ) ( a − b ) B 9 a ( a + 2 b ) − 27 b 2 9a(a + 2b) - 27b^2 9 a ( a + 2 b ) − 27 b 2 C 9 [ a ( a + 2 b ) − 3 b 2 ] 9[a(a + 2b) - 3b^2] 9 [ a ( a + 2 b ) − 3 b 2 ] D 9 a 2 + 9 b ( 2 a − 3 b ) 9a^2 + 9b(2a - 3b) 9 a 2 + 9 b ( 2 a − 3 b ) E 9 [ a 2 + b ( 2 a − 3 b ) ] 9[a^2 + b(2a - 3b)] 9 [ a 2 + b ( 2 a − 3 b )]
Worked solution (try it first) Take out the common factor 3:
3 [ 4 a 2 − ( a − 3 b ) 2 ] = 3 [ ( 2 a ) 2 − ( a − 3 b ) 2 ] 3[4a^2 - (a - 3b)^2] = 3[(2a)^2 - (a - 3b)^2] 3 [ 4 a 2 − ( a − 3 b ) 2 ] = 3 [( 2 a ) 2 − ( a − 3 b ) 2 ] .
Difference of two squares:
( 2 a − a + 3 b ) ( 2 a + a − 3 b ) = ( a + 3 b ) ( 3 a − 3 b ) (2a - a + 3b)(2a + a - 3b) = (a + 3b)(3a - 3b) ( 2 a − a + 3 b ) ( 2 a + a − 3 b ) = ( a + 3 b ) ( 3 a − 3 b ) .
Take out 3 from
3 a − 3 b 3a - 3b 3 a − 3 b :
3 × 3 ( a + 3 b ) ( a − b ) = 9 ( a + 3 b ) ( a − b ) 3 \times 3(a + 3b)(a - b) = 9(a + 3b)(a - b) 3 × 3 ( a + 3 b ) ( a − b ) = 9 ( a + 3 b ) ( a − b ) , option A.
Watch out
Options B to E are all equal to the expression, but none is a product of simple factors. "Completely" means keep going until you reach 9 ( a + 3 b ) ( a − b ) 9(a + 3b)(a - b) 9 ( a + 3 b ) ( a − b ) Report a problem with this question
Given that T = 2 π l g T = 2\pi\sqrt{\dfrac{l}{g}} T = 2 π g l T T T π = 22 7 \pi = \frac{22}{7} π = 7 22 l = 16 l = 16 l = 16 g = 10 g = 10 g = 10
A 1.26 B 1.60 C 3.14 D 7.95 E 10.00
Worked solution (try it first) Inside the root:
l g = 16 10 = 1.6 \frac lg = \frac{16}{10} = 1.6 g l = 10 16 = 1.6 , and
1.6 = 1.2649 \sqrt{1.6} = 1.2649 1.6 = 1.2649 .
Multiply by
2 π 2\pi 2 π :
2 × 22 7 = 44 7 2 \times \frac{22}{7} = \frac{44}{7} 2 × 7 22 = 7 44 So
T = 6.2857 × 1.2649 = 7.95 T = 6.2857 \times 1.2649 = 7.95 T = 6.2857 × 1.2649 = 7.95 , option D.
Watch out
1.6 ≈ 1.26 \sqrt{1.6} \approx 1.26 1.6 ≈ 1.26 2 π 2\pi 2 π Report a problem with this question
Expand ( x − 2 ) ( x + 6 ) (x - 2)(x + 6) ( x − 2 ) ( x + 6 )
A x 2 + 4 x − 12 x^2 + 4x - 12 x 2 + 4 x − 12 B x 2 + 4 x + 12 x^2 + 4x + 12 x 2 + 4 x + 12 C x 2 − 4 x + 12 x^2 - 4x + 12 x 2 − 4 x + 12 D x 2 − 8 x − 12 x^2 - 8x - 12 x 2 − 8 x − 12 E x 2 − 4 x − 12 x^2 - 4x - 12 x 2 − 4 x − 12
Worked solution (try it first) Multiply each term in the first bracket by each in the second:
x 2 + 6 x − 2 x − 12 x^2 + 6x - 2x - 12 x 2 + 6 x − 2 x − 12 .
Collect the
x x x terms:
6 x − 2 x = 4 x 6x - 2x = 4x 6 x − 2 x = 4 x .
So
( x − 2 ) ( x + 6 ) = x 2 + 4 x − 12 (x - 2)(x + 6) = x^2 + 4x - 12 ( x − 2 ) ( x + 6 ) = x 2 + 4 x − 12 , option A.
Watch out
The x x x + 6 x +6x + 6 x − 2 x -2x − 2 x + 4 x +4x + 4 x x 2 − 4 x − 12 x^2 - 4x - 12 x 2 − 4 x − 12 Report a problem with this question
Simplify x 2 − 8 x + 12 3 ( x 2 + x − 6 ) × 9 ( x + 3 ) \dfrac{x^2 - 8x + 12}{3(x^2 + x - 6)} \times 9(x + 3) 3 ( x 2 + x − 6 ) x 2 − 8 x + 12 × 9 ( x + 3 )
A x + 6 x + 6 x + 6 B 3 ( x − 2 ) 3(x - 2) 3 ( x − 2 ) C 3 ( x − 3 ) 3(x - 3) 3 ( x − 3 ) D 3 ( x − 6 ) 3(x - 6) 3 ( x − 6 ) E 3 ( x + 6 ) 3(x + 6) 3 ( x + 6 )
Worked solution (try it first) Factorise:
x 2 − 8 x + 12 = ( x − 2 ) ( x − 6 ) x^2 - 8x + 12 = (x - 2)(x - 6) x 2 − 8 x + 12 = ( x − 2 ) ( x − 6 ) and
x 2 + x − 6 = ( x + 3 ) ( x − 2 ) x^2 + x - 6 = (x + 3)(x - 2) x 2 + x − 6 = ( x + 3 ) ( x − 2 ) .
So the expression is
( x − 2 ) ( x − 6 ) 3 ( x + 3 ) ( x − 2 ) × 9 ( x + 3 ) \dfrac{(x - 2)(x - 6)}{3(x + 3)(x - 2)} \times 9(x + 3) 3 ( x + 3 ) ( x − 2 ) ( x − 2 ) ( x − 6 ) × 9 ( x + 3 ) .
Cancel
x − 2 x - 2 x − 2 and
x + 3 x + 3 x + 3 , and divide 9 by 3.
This leaves
3 ( x − 6 ) 3(x - 6) 3 ( x − 6 ) , option D.
Watch out
For x 2 − 8 x + 12 x^2 - 8x + 12 x 2 − 8 x + 12 − 2 -2 − 2 − 6 -6 − 6 − 8 -8 − 8 x − 6 x - 6 x − 6 + 6 +6 + 6 3 ( x + 6 ) 3(x + 6) 3 ( x + 6 ) Report a problem with this question
What is the angular difference in longitude between town A (lat. 47 ∘ 47^\circ 4 7 ∘ 54 ∘ 54^\circ 5 4 ∘ 47 ∘ 47^\circ 4 7 ∘ 147 ∘ 147^\circ 14 7 ∘
A 93 ∘ 93^\circ 9 3 ∘ B 94 ∘ 94^\circ 9 4 ∘ C 100 ∘ 100^\circ 10 0 ∘ D 101 ∘ 101^\circ 10 1 ∘ E 201 ∘ 201^\circ 20 1 ∘
Worked solution (try it first) Both longitudes are east of Greenwich, so subtract them.
The difference is
147 ∘ − 54 ∘ = 93 ∘ 147^\circ - 54^\circ = 93^\circ 14 7 ∘ − 5 4 ∘ = 9 3 ∘ , option A.
Watch out
Add only when one longitude is east and the other west. Adding here gives 201 ∘ 201^\circ 20 1 ∘ Report a problem with this question
In the diagram, O O O P Q R PQR P QR ∠ Q P R = 64 ∘ \angle QPR = 64^\circ ∠ QP R = 6 4 ∘ ∠ Q R P = 46 ∘ \angle QRP = 46^\circ ∠ QR P = 4 6 ∘ ∠ P O Q \angle POQ ∠ P O Q
A 44 ∘ 44^\circ 4 4 ∘ B 70 ∘ 70^\circ 7 0 ∘ C 92 ∘ 92^\circ 9 2 ∘ D 110 ∘ 110^\circ 11 0 ∘ E 140 ∘ 140^\circ 14 0 ∘
Worked solution (try it first) ∠ P O Q \angle POQ ∠ P O Q is at the centre on arc
P Q PQ P Q .
The angle at the circumference on the same arc is at
R R R :
∠ P R Q = 46 ∘ \angle PRQ = 46^\circ ∠ P R Q = 4 6 ∘ .
The angle at the centre is twice the angle at the circumference:
∠ P O Q = 2 × 46 ∘ = 92 ∘ \angle POQ = 2 \times 46^\circ = 92^\circ ∠ P O Q = 2 × 4 6 ∘ = 9 2 ∘ , option C.
Watch out
Double the angle that faces P Q PQ P Q R R R 64 ∘ 64^\circ 6 4 ∘ P P P 128 ∘ 128^\circ 12 8 ∘ ∠ Q O R \angle QOR ∠ QO R Report a problem with this question
In the figures, the two right-angled triangles have equal marked angles. Find the value of L L L
Worked solution (try it first) The triangles have equal angles, so they are similar and their matching sides are in the same ratio.
L L L and 18 cm are the vertical sides.
420 m and 24 cm are the bases.
So
L 420 = 18 24 \dfrac{L}{420} = \dfrac{18}{24} 420 L = 24 18 .
Multiply by 420:
L = 420 × 3 4 = 315 L = 420 \times \frac34 = 315 L = 420 × 4 3 = 315 m, option E.
Watch out
Match the sides by position: L L L L = 420 × 24 18 = 560 L = 420 \times \frac{24}{18} = 560 L = 420 × 18 24 = 560 Report a problem with this question
Find the gradient of the curve y = 2 x 2 + 5 x − 1 y = 2x^2 + 5x - 1 y = 2 x 2 + 5 x − 1 x = 4 x = 4 x = 4
Worked solution (try it first) The gradient is
d y d x = 4 x + 5 \frac{dy}{dx} = 4x + 5 d x d y = 4 x + 5 .
At
x = 4 x = 4 x = 4 :
16 + 5 = 21 16 + 5 = 21 16 + 5 = 21 , option E.
Watch out
5 x 5x 5 x − 1 -1 − 1 Report a problem with this question
An interior angle of a regular polygon is 108 ∘ 108^\circ 10 8 ∘
Worked solution (try it first) An interior angle and its exterior angle add up to
180 ∘ 180^\circ 18 0 ∘ , so each exterior angle is
180 ∘ − 108 ∘ = 72 ∘ 180^\circ - 108^\circ = 72^\circ 18 0 ∘ − 10 8 ∘ = 7 2 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 72 = 5 360 \div 72 = 5 360 ÷ 72 = 5 , option B.
Watch out
Divide 360 ∘ 360^\circ 36 0 ∘ 72 ∘ 72^\circ 7 2 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question
A B C ABC A B C E E E D D D A C AC A C B C BC B C B E ⊥ A C BE \perp AC B E ⊥ A C E D ⊥ B C ED \perp BC E D ⊥ B C ∠ A B E = 68 ∘ \angle ABE = 68^\circ ∠ A B E = 6 8 ∘ ∠ A = ∠ C \angle A = \angle C ∠ A = ∠ C ∠ C E D \angle CED ∠ C E D
A 22 ∘ 22^\circ 2 2 ∘ B 34 ∘ 34^\circ 3 4 ∘ C 44 ∘ 44^\circ 4 4 ∘ D 52 ∘ 52^\circ 5 2 ∘ E 68 ∘ 68^\circ 6 8 ∘
Worked solution (try it first) B E ⊥ A C BE \perp AC B E ⊥ A C , so triangle
A B E ABE A B E has a right angle at
E E E :
∠ A = 90 ∘ − 68 ∘ = 22 ∘ \angle A = 90^\circ - 68^\circ = 22^\circ ∠ A = 9 0 ∘ − 6 8 ∘ = 2 2 ∘ .
∠ C = ∠ A = 22 ∘ \angle C = \angle A = 22^\circ ∠ C = ∠ A = 2 2 ∘ .
E D ⊥ B C ED \perp BC E D ⊥ B C , so triangle
C D E CDE C D E has a right angle at
D D D :
∠ C E D = 90 ∘ − 22 ∘ \angle CED = 90^\circ - 22^\circ ∠ C E D = 9 0 ∘ − 2 2 ∘ = 68 ∘ = 68^\circ = 6 8 ∘ , option E.
Watch out
22 ∘ 22^\circ 2 2 ∘ ∠ C \angle C ∠ C C D E CDE C D E ∠ C E D \angle CED ∠ C E D Report a problem with this question
In the figure, O O O D B DB D B A E D AED A E D A B C ABC A B C ∠ E A B = 34 ∘ \angle EAB = 34^\circ ∠ E A B = 3 4 ∘ ∠ E D B = 40 ∘ \angle EDB = 40^\circ ∠ E D B = 4 0 ∘ x x x
A 74 ∘ 74^\circ 7 4 ∘ B 66 ∘ 66^\circ 6 6 ∘ C 56 ∘ 56^\circ 5 6 ∘ D 50 ∘ 50^\circ 5 0 ∘ E 40 ∘ 40^\circ 4 0 ∘
Worked solution (try it first) D B DB D B is a diameter, so
∠ D E B = 90 ∘ \angle DEB = 90^\circ ∠ D E B = 9 0 ∘ and, on the straight line
A E D AED A E D ,
∠ A E B = 90 ∘ \angle AEB = 90^\circ ∠ A E B = 9 0 ∘ .
Triangle
A E B AEB A E B gives
∠ A B E = 180 ∘ − 90 ∘ − 34 ∘ \angle ABE = 180^\circ - 90^\circ - 34^\circ ∠ A B E = 18 0 ∘ − 9 0 ∘ − 3 4 ∘ Triangle
D E B DEB D E B gives
∠ E B D = 90 ∘ − 40 ∘ \angle EBD = 90^\circ - 40^\circ ∠ E B D = 9 0 ∘ − 4 0 ∘ On the straight line
A B C ABC A B C ,
∠ D B C = 180 ∘ − 56 ∘ − 50 ∘ \angle DBC = 180^\circ - 56^\circ - 50^\circ ∠ D B C = 18 0 ∘ − 5 6 ∘ − 5 0 ∘ Angles in the same segment:
∠ E C B \angle ECB ∠ E C B and
∠ E D B \angle EDB ∠ E D B both stand on arc
E B EB E B , so
∠ E C B = 40 ∘ \angle ECB = 40^\circ ∠ E C B = 4 0 ∘ .
In the triangle made by
B B B ,
C C C and the crossing of
E C EC E C with
D B DB D B :
x = 180 ∘ − 74 ∘ − 40 ∘ x = 180^\circ - 74^\circ - 40^\circ x = 18 0 ∘ − 7 4 ∘ − 4 0 ∘ = 66 ∘ = 66^\circ = 6 6 ∘ , option B.
Watch out
74 ∘ 74^\circ 7 4 ∘ ∠ D B C \angle DBC ∠ D B C B B B x x x E C EC E C D B DB D B Report a problem with this question
In the diagram, O O O A B AB A B ∣ D B ∣ = ∣ B C ∣ |DB| = |BC| ∣ D B ∣ = ∣ B C ∣ ∠ A B C = 54 ∘ \angle ABC = 54^\circ ∠ A B C = 5 4 ∘ ∠ A C D \angle ACD ∠ A C D
A 153 ∘ 153^\circ 15 3 ∘ B 117 ∘ 117^\circ 11 7 ∘ C 63 ∘ 63^\circ 6 3 ∘ D 36 ∘ 36^\circ 3 6 ∘ E 27 ∘ 27^\circ 2 7 ∘
Worked solution (try it first) D B = B C DB = BC D B = B C , so triangle
D B C DBC D B C is isosceles with apex
54 ∘ 54^\circ 5 4 ∘ at
B B B :
∠ B C D = 180 ∘ − 54 ∘ 2 \angle BCD = \dfrac{180^\circ - 54^\circ}{2} ∠ B C D = 2 18 0 ∘ − 5 4 ∘ A B AB A B is a diameter, so
∠ A C B = 90 ∘ \angle ACB = 90^\circ ∠ A C B = 9 0 ∘ (angle in a semicircle).
∠ A C D \angle ACD ∠ A C D is the rest of that right angle:
∠ A C D = 90 ∘ − 63 ∘ \angle ACD = 90^\circ - 63^\circ ∠ A C D = 9 0 ∘ − 6 3 ∘ = 27 ∘ = 27^\circ = 2 7 ∘ , option E.
Watch out
63 ∘ 63^\circ 6 3 ∘ ∠ B C D \angle BCD ∠ B C D ∠ A C D \angle ACD ∠ A C D 90 ∘ 90^\circ 9 0 ∘ C C C Report a problem with this question
Determine the value of 7 x 7x 7 x 5 x 5x 5 x 2 x 2x 2 x
A 20 ∘ 20^\circ 2 0 ∘ B 40 ∘ 40^\circ 4 0 ∘ C 80 ∘ 80^\circ 8 0 ∘ D 100 ∘ 100^\circ 10 0 ∘ E 140 ∘ 140^\circ 14 0 ∘
Worked solution (try it first) The angles of a triangle add up to
180 ∘ 180^\circ 18 0 ∘ :
5 x + 2 x + 2 x = 180 ∘ 5x + 2x + 2x = 180^\circ 5 x + 2 x + 2 x = 18 0 ∘ .
So
9 x = 180 ∘ 9x = 180^\circ 9 x = 18 0 ∘ and
x = 20 ∘ x = 20^\circ x = 2 0 ∘ .
Then
7 x = 7 × 20 ∘ = 140 ∘ 7x = 7 \times 20^\circ = 140^\circ 7 x = 7 × 2 0 ∘ = 14 0 ∘ , option E.
Watch out
20 ∘ 20^\circ 2 0 ∘ x x x 7 x 7x 7 x Report a problem with this question
In the diagram, K T N KTN K T N T T T ∠ A B T = 65 ∘ \angle ABT = 65^\circ ∠ A B T = 6 5 ∘ ∠ N T A \angle NTA ∠ N T A
A 25 ∘ 25^\circ 2 5 ∘ B 32 ∘ 32^\circ 3 2 ∘ C 65 ∘ 65^\circ 6 5 ∘ D 90 ∘ 90^\circ 9 0 ∘ E 115 ∘ 115^\circ 11 5 ∘
Worked solution (try it first) ∠ N T A \angle NTA ∠ N T A is between the tangent
T N TN T N and the chord
T A TA T A .
The angle between a tangent and a chord equals the angle in the alternate segment, here
∠ A B T \angle ABT ∠ A B T .
So
∠ N T A = 65 ∘ \angle NTA = 65^\circ ∠ N T A = 6 5 ∘ , option C.
Watch out
115 ∘ 115^\circ 11 5 ∘ ∠ K T A \angle KTA ∠ K T A T A TA T A ∠ A B T \angle ABT ∠ A B T B B B ∠ N T A \angle NTA ∠ N T A Report a problem with this question
An aeroplane flies from a town P P P 045 ∘ 045^\circ 04 5 ∘ Q Q Q 200 km 200\text{ km} 200 km R R R 120 ∘ 120^\circ 12 0 ∘ R R R P P P ∣ P R ∣ |PR| ∣ P R ∣
Worked solution (try it first) On
045 ∘ 045^\circ 04 5 ∘ ,
Q Q Q is
200 sin 45 ∘ ≈ 141.4 200\sin45^\circ \approx 141.4 200 sin 4 5 ∘ ≈ 141.4 km east and
200 cos 45 ∘ ≈ 141.4 200\cos45^\circ \approx 141.4 200 cos 4 5 ∘ ≈ 141.4 km north of
P P P .
R R R is due east of
P P P , so the second leg must come back 141.4 km south.
The bearing
120 ∘ 120^\circ 12 0 ∘ is
60 ∘ 60^\circ 6 0 ∘ from due south, so
Q R cos 60 ∘ = 141.4 QR\cos60^\circ = 141.4 QR cos 6 0 ∘ = 141.4 , giving
Q R ≈ 282.8 QR \approx 282.8 QR ≈ 282.8 km.
That leg goes
282.8 sin 60 ∘ ≈ 244.9 282.8\sin60^\circ \approx 244.9 282.8 sin 6 0 ∘ ≈ 244.9 km east.
So
∣ P R ∣ ≈ 141.4 + 244.9 ≈ 386 |PR| \approx 141.4 + 244.9 \approx 386 ∣ P R ∣ ≈ 141.4 + 244.9 ≈ 386 km, option B.
Watch out
Bearing 120 ∘ 120^\circ 12 0 ∘ 60 ∘ 60^\circ 6 0 ∘ 30 ∘ 30^\circ 3 0 ∘ 30 ∘ 30^\circ 3 0 ∘ 141.4 tan 30 ∘ ≈ 81.6 141.4\tan30^\circ \approx 81.6 141.4 tan 3 0 ∘ ≈ 81.6 ∣ P R ∣ ≈ 223 |PR| \approx 223 ∣ P R ∣ ≈ 223 Report a problem with this question
If the angle of depression of a boy standing on the ground from the top of a house is 72 ∘ 72^\circ 7 2 ∘
A 18 ∘ 18^\circ 1 8 ∘ B 36 ∘ 36^\circ 3 6 ∘ C 72 ∘ 72^\circ 7 2 ∘ D 90 ∘ 90^\circ 9 0 ∘ E 108 ∘ 108^\circ 10 8 ∘
Worked solution (try it first) The horizontal line at the top of the house is parallel to the ground.
The line of sight cuts both, so the angle of depression and the angle of elevation are alternate angles and are equal.
So the angle of elevation is
72 ∘ 72^\circ 7 2 ∘ , option C.
Watch out
The two angles are equal, not complementary. 90 ∘ − 72 ∘ = 18 ∘ 90^\circ - 72^\circ = 18^\circ 9 0 ∘ − 7 2 ∘ = 1 8 ∘ Report a problem with this question
If cos θ = 0.8 \cos\theta = 0.8 cos θ = 0.8 0 ∘ < θ < 90 ∘ 0^\circ < \theta < 90^\circ 0 ∘ < θ < 9 0 ∘ tan θ \tan\theta tan θ
A 3 5 \frac35 5 3 B 3 4 \frac34 4 3 C 4 5 \frac45 5 4 D 4 3 \frac43 3 4 E 5 3 \frac53 3 5
Worked solution (try it first) cos θ = 0.8 = 4 5 \cos\theta = 0.8 = \frac45 cos θ = 0.8 = 5 4 , so draw a right-angled triangle with adjacent side 4 and hypotenuse 5.
Pythagoras gives the opposite side:
25 − 16 = 3 \sqrt{25 - 16} = 3 25 − 16 = 3 .
So
tan θ = opposite adjacent \tan\theta = \frac{\text{opposite}}{\text{adjacent}} tan θ = adjacent opposite = 3 4 = \frac34 = 4 3 , option B.
Watch out
Tangent is opposite over adjacent. Turning it upside down gives 4 3 \frac43 3 4 Report a problem with this question
A ladder x x x 12 m 12\text{ m} 12 m 54 ∘ 54^\circ 5 4 ∘ x x x
A 14.7 B 14.8 C 20.6 D 147.0 E 206.0
Worked solution (try it first) The ladder is the hypotenuse, and the 12 m pole is opposite the
54 ∘ 54^\circ 5 4 ∘ angle at the ground.
So
sin 54 ∘ = 12 x \sin54^\circ = \dfrac{12}{x} sin 5 4 ∘ = x 12 .
Rearrange:
x = 12 sin 54 ∘ x = \dfrac{12}{\sin54^\circ} x = sin 5 4 ∘ 12 = 12 0.8090 = \dfrac{12}{0.8090} = 0.8090 12 ≈ 14.83 \approx 14.83 ≈ 14.83 .
To three significant figures,
x = 14.8 x = 14.8 x = 14.8 m, option B.
Watch out
Round 14.83 14.83 14.83 sin 54 ∘ = 0.8090 \sin54^\circ = 0.8090 sin 5 4 ∘ = 0.8090 Report a problem with this question
If 1.109 litres of water is poured into a cylindrical container of base radius 4.2 cm 4.2\text{ cm} 4.2 cm
A 0.02 cm 0.02\text{ cm} 0.02 cm B 0.20 cm 0.20\text{ cm} 0.20 cm C 2.00 cm 2.00\text{ cm} 2.00 cm D 20.00 cm 20.00\text{ cm} 20.00 cm E 200.00 cm 200.00\text{ cm} 200.00 cm
Worked solution (try it first) Change litres to cm³: 1 litre is
1000 cm 3 1000\text{ cm}^3 1000 cm 3 , so the water is
1109 cm 3 1109\text{ cm}^3 1109 cm 3 .
Base area:
22 7 × 4.2 2 = 55.44 cm 2 \frac{22}{7} \times 4.2^2 = 55.44\text{ cm}^2 7 22 × 4. 2 2 = 55.44 cm 2 .
Level
= 1109 ÷ 55.44 ≈ 20.00 = 1109 \div 55.44 \approx 20.00 = 1109 ÷ 55.44 ≈ 20.00 cm, option D.
Watch out
Convert litres to cm³ first. Dividing 1.109 by 55.44 gives 0.02 cm (option A). Report a problem with this question
Calculate the mean deviation of the scores 4, 5, 3, 2, 1.
Worked solution (try it first) The scores add up to 15, so the mean is 3.
The distances from 3 are 1, 2, 0, 1 and 2, which add up to 6.
The mean deviation is
6 5 = 1.2 \frac65 = 1.2 5 6 = 1.2 , option A.
Watch out
Mean deviation uses the distances themselves, not their squares. Squaring gives the standard deviation 2 ≈ 1.4 \sqrt2 \approx 1.4 2 ≈ 1.4 Report a problem with this question
What is the probability that an integer selected from { 1 , 2 , 3 , … , 29 , 30 } \{1, 2, 3, \dots, 29, 30\} { 1 , 2 , 3 , … , 29 , 30 }
A 1 6 \frac16 6 1 B 1 5 \frac15 5 1 C 4 15 \frac4{15} 15 4 D 3 10 \frac3{10} 10 3 E 1 3 \frac13 3 1
Worked solution (try it first) There are 30 numbers to choose from.
The primes up to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, which is 10 numbers.
So the probability is
10 30 = 1 3 \frac{10}{30} = \frac13 30 10 = 3 1 , option E.
Watch out
2 is a prime (the only even one). Leaving it out gives 9 30 = 3 10 \frac{9}{30} = \frac{3}{10} 30 9 = 10 3 Report a problem with this question
Two balls are taken one after the other without replacement from a bag containing 4 red and 6 blue balls. What is the probability that they are of different colours?
A 3 5 \frac35 5 3 B 8 15 \frac8{15} 15 8 C 2 5 \frac25 5 2 D 4 15 \frac4{15} 15 4 E 8 225 \frac8{225} 225 8
Worked solution (try it first) Red then blue:
4 10 × 6 9 = 24 90 \frac{4}{10} \times \frac69 = \frac{24}{90} 10 4 × 9 6 = 90 24 , since 9 balls are left after the first.
Blue then red:
6 10 × 4 9 = 24 90 \frac{6}{10} \times \frac49 = \frac{24}{90} 10 6 × 9 4 = 90 24 .
Add the two orders:
48 90 = 8 15 \frac{48}{90} = \frac{8}{15} 90 48 = 15 8 , option B.
Watch out
Different colours can happen in two orders. Using only red then blue gives 24 90 = 4 15 \frac{24}{90} = \frac{4}{15} 90 24 = 15 4 Report a problem with this question
The pie chart shows the distribution of candidates who sat for certain subjects in a school certificate examination. What angle represents the students who sat for Economics?
A 60 ∘ 60^\circ 6 0 ∘ B 72 ∘ 72^\circ 7 2 ∘ C 108 ∘ 108^\circ 10 8 ∘ D 110 ∘ 110^\circ 11 0 ∘ E 120 ∘ 120^\circ 12 0 ∘
Worked solution (try it first) The angles at the centre add up to
360 ∘ 360^\circ 36 0 ∘ .
Economics is what is left:
360 ∘ − ( 70 ∘ + 20 ∘ + 150 ∘ ) = 120 ∘ 360^\circ - (70^\circ + 20^\circ + 150^\circ) = 120^\circ 36 0 ∘ − ( 7 0 ∘ + 2 0 ∘ + 15 0 ∘ ) = 12 0 ∘ , option E.
Watch out
Take all three labelled angles from 360 ∘ 360^\circ 36 0 ∘ 20 ∘ 20^\circ 2 0 ∘ 140 ∘ 140^\circ 14 0 ∘ Report a problem with this question
Using the same pie chart, what percentage of the students sat for English Language, correct to the nearest whole number?
Worked solution (try it first) English has
150 ∘ 150^\circ 15 0 ∘ of the
360 ∘ 360^\circ 36 0 ∘ .
As a percentage:
150 360 × 100 = 41.7 \frac{150}{360} \times 100 = 41.7 360 150 × 100 = 41.7 , which is 42 to the nearest whole number, option E.
Watch out
Divide the angle by 360, not by 100 or by the sum of the other angles. 120 360 × 100 = 33 \frac{120}{360} \times 100 = 33 360 120 × 100 = 33 Report a problem with this question
Find the mean of the frequency distribution, correct to one decimal place.
Marks
2
5
7
8
9
10
Frequency
9
4
3
7
8
2
Worked solution (try it first) Multiply each mark by its frequency and add:
18 + 20 + 21 + 56 + 72 + 20 = 207 18 + 20 + 21 + 56 + 72 + 20 = 207 18 + 20 + 21 + 56 + 72 + 20 = 207 .
Total frequency:
9 + 4 + 3 + 7 + 8 + 2 = 33 9 + 4 + 3 + 7 + 8 + 2 = 33 9 + 4 + 3 + 7 + 8 + 2 = 33 .
Mean
= 207 33 = 6.27 = \frac{207}{33} = 6.27 = 33 207 = 6.27 , which is 6.3 to one decimal place, option C.
Watch out
Divide by the total frequency, 33, not by the 6 marks in the table. Averaging the marks row alone gives 6.8, which ignores how often each mark occurs. Report a problem with this question
The table shows the scores of applicants in an interview. If an applicant is chosen at random, what is the probability that the applicant scored at most 8 marks?
Scores
6
7
8
9
10
Frequency
2
4
2
5
3
A 1 8 \frac18 8 1 B 1 4 \frac14 4 1 C 3 8 \frac38 8 3 D 7 16 \frac7{16} 16 7 E 1 2 \frac12 2 1
Worked solution (try it first) There are
2 + 4 + 2 + 5 + 3 = 16 2 + 4 + 2 + 5 + 3 = 16 2 + 4 + 2 + 5 + 3 = 16 applicants.
At most 8 means 6, 7 or 8:
2 + 4 + 2 = 8 2 + 4 + 2 = 8 2 + 4 + 2 = 8 applicants.
So the probability is
8 16 = 1 2 \frac{8}{16} = \frac12 16 8 = 2 1 , option E.
Watch out
"At most 8" includes 8 itself. Leaving out the score of 8 gives 6 16 = 3 8 \frac{6}{16} = \frac38 16 6 = 8 3 Report a problem with this question
A box contains 20 oranges, 14 of them ripe and 6 unripe. If two oranges are taken one after the other with replacement, find the probability that one is ripe and the other unripe.
A 21 100 \frac{21}{100} 100 21 B 3 10 \frac3{10} 10 3 C 21 50 \frac{21}{50} 50 21 D 7 10 \frac7{10} 10 7 E 21 25 \frac{21}{25} 25 21
Worked solution (try it first) With replacement, each draw is ripe with probability
14 20 \frac{14}{20} 20 14 and unripe with probability
6 20 \frac{6}{20} 20 6 .
Ripe then unripe:
14 20 × 6 20 = 84 400 \frac{14}{20} \times \frac{6}{20} = \frac{84}{400} 20 14 × 20 6 = 400 84 , and unripe then ripe is the same.
Add the two orders:
168 400 = 21 50 \frac{168}{400} = \frac{21}{50} 400 168 = 50 21 , option C.
Watch out
"One ripe and the other unripe" can happen in two orders. Using only one order gives 84 400 = 21 100 \frac{84}{400} = \frac{21}{100} 400 84 = 100 21 Report a problem with this question
The mean of the numbers 2, 5, x x x x x x
Worked solution (try it first) Four numbers with mean 4 add up to
4 × 4 = 16 4 \times 4 = 16 4 × 4 = 16 .
So
2 + 5 + x + 6 = 16 2 + 5 + x + 6 = 16 2 + 5 + x + 6 = 16 , which gives
13 + x = 16 13 + x = 16 13 + x = 16 .
So
x = 3 x = 3 x = 3 , option E.
Watch out
Use the total: the mean is 4, but x x x Report a problem with this question
Evaluate ∫ 0 2 ( 2 x − x 2 ) d x \displaystyle\int_0^2 (2x - x^2)\,dx ∫ 0 2 ( 2 x − x 2 ) d x
A − 2 -2 − 2 B 3 4 \frac34 4 3 C 1 1 3 1\frac13 1 3 1 D 2 E 6 2 3 6\frac23 6 3 2
Worked solution (try it first) Integrate:
[ x 2 − x 3 3 ] 0 2 \left[x^2 - \frac{x^3}{3}\right]_0^2 [ x 2 − 3 x 3 ] 0 2 .
At
x = 2 x = 2 x = 2 :
4 − 8 3 = 4 3 4 - \frac83 = \frac43 4 − 3 8 = 3 4 .
So the integral is
4 3 = 1 1 3 \frac43 = 1\frac13 3 4 = 1 3 1 , option C.
Watch out
− x 2 -x^2 − x 2 − x 3 3 -\frac{x^3}{3} − 3 x 3 8 3 \frac83 3 8 6 2 3 6\frac23 6 3 2 Report a problem with this question
If y = 3 x 2 − 4 x − 12 y = 3x^2 - 4x - 12 y = 3 x 2 − 4 x − 12 x x x d y d x = 0 \dfrac{dy}{dx} = 0 d x d y = 0
A 1 3 \frac13 3 1 B 1 2 \frac12 2 1 C 2 3 \frac23 3 2 D 1 E 2
Worked solution (try it first) Differentiate:
d y d x = 6 x − 4 \frac{dy}{dx} = 6x - 4 d x d y = 6 x − 4 .
Set it to zero:
6 x = 4 6x = 4 6 x = 4 , so
x = 4 6 = 2 3 x = \frac46 = \frac23 x = 6 4 = 3 2 , option C.
Watch out
6 x = 4 6x = 4 6 x = 4 x = 4 6 x = \frac46 x = 6 4 6 4 \frac64 4 6 3 2 \frac32 2 3 Report a problem with this question
A particle moves in a straight line such that its velocity after t t t ( 3 t + 4 ) m/s (3t + 4)\text{ m/s} ( 3 t + 4 ) m/s
A 16 m 16\text{ m} 16 m B 24 m 24\text{ m} 24 m C 40 m 40\text{ m} 40 m D 48 m 48\text{ m} 48 m E 64 m 64\text{ m} 64 m
Worked solution (try it first) Distance is the integral of velocity:
∫ 0 4 ( 3 t + 4 ) d t = [ 3 2 t 2 + 4 t ] 0 4 \int_0^4 (3t + 4)\,dt = \left[\frac32t^2 + 4t\right]_0^4 ∫ 0 4 ( 3 t + 4 ) d t = [ 2 3 t 2 + 4 t ] 0 4 .
At
t = 4 t = 4 t = 4 :
3 2 × 16 + 16 = 24 + 16 = 40 \frac32 \times 16 + 16 = 24 + 16 = 40 2 3 × 16 + 16 = 24 + 16 = 40 .
So the distance is 40 m, option C.
Watch out
The velocity changes, so don't multiply the final velocity by the time: 16 × 4 = 64 16 \times 4 = 64 16 × 4 = 64 Report a problem with this question