Express, correct to three significant figures, 0.003597.
A 0.359 B 0.004 C 0.00360 D 0.00359
Worked solution (try it first) The zeros after the point are not significant, so the first three significant figures are 3, 5 and 9.
The next figure is 7, so round the 9 up.
That makes 359 into 360, so the number is 0.00360.
Keep the final zero to show three significant figures: 0.00360, option C.
Watch out
Round, don't chop: the 7 after the 9 rounds it up. Cutting it off gives 0.00359 (option D). Report a problem with this question
Evaluate ( 0.064 ) − 1 3 (0.064)^{-\frac13} ( 0.064 ) − 3 1
A 5 2 \frac52 2 5 B 2 5 \frac25 5 2 C − 2 5 -\frac25 − 5 2 D − 5 2 -\frac52 − 2 5
Worked solution (try it first) 0.064 = 64 1000 0.064 = \frac{64}{1000} 0.064 = 1000 64 , and its cube root is
4 10 = 0.4 \frac{4}{10} = 0.4 10 4 = 0.4 .
The negative index means the reciprocal:
( 0.064 ) − 1 3 = 1 0.4 (0.064)^{-\frac13} = \frac{1}{0.4} ( 0.064 ) − 3 1 = 0.4 1 So the value is
5 2 \frac52 2 5 , option A.
Watch out
A negative index gives the reciprocal, not a negative answer. Forgetting to flip gives 2 5 \frac25 5 2 Report a problem with this question
Solve y + 1 2 − 2 y − 1 3 = 4 \dfrac{y + 1}{2} - \dfrac{2y - 1}{3} = 4 2 y + 1 − 3 2 y − 1 = 4
A y = 19 y = 19 y = 19 B y = − 19 y = -19 y = − 19 C y = − 29 y = -29 y = − 29 D y = 29 y = 29 y = 29
Worked solution (try it first) Multiply every term by 6, the LCM of 2 and 3:
3 ( y + 1 ) − 2 ( 2 y − 1 ) = 24 3(y + 1) - 2(2y - 1) = 24 3 ( y + 1 ) − 2 ( 2 y − 1 ) = 24 .
Expand, taking care with the minus:
3 y + 3 − 4 y + 2 = 24 3y + 3 - 4y + 2 = 24 3 y + 3 − 4 y + 2 = 24 , so
− y + 5 = 24 -y + 5 = 24 − y + 5 = 24 .
Take 5 from both sides:
− y = 19 -y = 19 − y = 19 .
Multiply by
− 1 -1 − 1 :
y = − 19 y = -19 y = − 19 , option B.
Watch out
Don't stop at − y = 19 -y = 19 − y = 19 − 1 -1 − 1 y = − 19 y = -19 y = − 19 y = 19 y = 19 y = 19 Report a problem with this question
Simplify, correct to three significant figures, ( 27.63 ) 2 − ( 12.37 ) 2 (27.63)^2 - (12.37)^2 ( 27.63 ) 2 − ( 12.37 ) 2
Worked solution (try it first) Use the difference of two squares:
a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a - b)(a + b) a 2 − b 2 = ( a − b ) ( a + b ) .
Here
a − b = 27.63 − 12.37 = 15.26 a - b = 27.63 - 12.37 = 15.26 a − b = 27.63 − 12.37 = 15.26 and
a + b = 40 a + b = 40 a + b = 40 , so the value is
15.26 × 40 = 610.4 15.26 \times 40 = 610.4 15.26 × 40 = 610.4 .
To 3 significant figures the 0 is the third figure and the next figure is 4, so round down: 610, option D.
Watch out
The zero in 610 is the third significant figure, and the next figure, 4, rounds it down. Squaring each number and rounding along the way can leave you with 611 or 612 (options B and C). Report a problem with this question
If 7 + y ≡ 4 ( m o d 8 ) 7 + y \equiv 4 \pmod 8 7 + y ≡ 4 ( mod 8 ) y y y 10 ≤ y ≤ 30 10 \le y \le 30 10 ≤ y ≤ 30
Worked solution (try it first) Take 7 from both sides:
y ≡ 4 − 7 = − 3 ( m o d 8 ) y \equiv 4 - 7 = -3 \pmod 8 y ≡ 4 − 7 = − 3 ( mod 8 ) .
Add 8 to make it positive:
y ≡ 5 ( m o d 8 ) y \equiv 5 \pmod 8 y ≡ 5 ( mod 8 ) , so
y y y is one of 5, 13, 21, 29, ….
The least of these with
10 ≤ y ≤ 30 10 \le y \le 30 10 ≤ y ≤ 30 is 13, option B.
Watch out
Subtract the 7, don't add it. 4 + 7 = 11 4 + 7 = 11 4 + 7 = 11 7 + 11 = 18 ≡ 2 ( m o d 8 ) 7 + 11 = 18 \equiv 2 \pmod 8 7 + 11 = 18 ≡ 2 ( mod 8 ) Report a problem with this question
If T = { prime numbers } T = \{\text{prime numbers}\} T = { prime numbers } M = { odd numbers } M = \{\text{odd numbers}\} M = { odd numbers } μ = { x : 0 < x ≤ 10 } \mu = \{x : 0 < x \le 10\} μ = { x : 0 < x ≤ 10 } x x x T ′ ∩ M ′ T' \cap M' T ′ ∩ M ′
A { 4 , 6 , 8 , 10 } \{4, 6, 8, 10\} { 4 , 6 , 8 , 10 } B { 1 , 4 , 6 , 8 , 10 } \{1, 4, 6, 8, 10\} { 1 , 4 , 6 , 8 , 10 } C { 1 , 2 , 4 , 6 , 8 , 10 } \{1, 2, 4, 6, 8, 10\} { 1 , 2 , 4 , 6 , 8 , 10 } D { 1 , 2 , 3 , 5 , 7 , 8 , 9 } \{1, 2, 3, 5, 7, 8, 9\} { 1 , 2 , 3 , 5 , 7 , 8 , 9 }
Worked solution (try it first) μ = { 1 , 2 , … , 10 } \mu = \{1, 2, \dots, 10\} μ = { 1 , 2 , … , 10 } ,
T = { 2 , 3 , 5 , 7 } T = \{2, 3, 5, 7\} T = { 2 , 3 , 5 , 7 } and
M = { 1 , 3 , 5 , 7 , 9 } M = \{1, 3, 5, 7, 9\} M = { 1 , 3 , 5 , 7 , 9 } .
T ′ = { 1 , 4 , 6 , 8 , 9 , 10 } T' = \{1, 4, 6, 8, 9, 10\} T ′ = { 1 , 4 , 6 , 8 , 9 , 10 } and
M ′ = { 2 , 4 , 6 , 8 , 10 } M' = \{2, 4, 6, 8, 10\} M ′ = { 2 , 4 , 6 , 8 , 10 } .
The elements in both are
T ′ ∩ M ′ = { 4 , 6 , 8 , 10 } T' \cap M' = \{4, 6, 8, 10\} T ′ ∩ M ′ = { 4 , 6 , 8 , 10 } , option A.
Watch out
1 is not prime, but it is odd, so it is not in M ′ M' M ′ Report a problem with this question
Evaluate log 3 9 − log 2 8 log 3 9 \dfrac{\log_3 9 - \log_2 8}{\log_3 9} log 3 9 log 3 9 − log 2 8
A − 1 3 -\frac13 − 3 1 B 1 2 \frac12 2 1 C 1 3 \frac13 3 1 D − 1 2 -\frac12 − 2 1
Worked solution (try it first) 3 2 = 9 3^2 = 9 3 2 = 9 , so
log 3 9 = 2 \log_3 9 = 2 log 3 9 = 2 .
And
2 3 = 8 2^3 = 8 2 3 = 8 , so
log 2 8 = 3 \log_2 8 = 3 log 2 8 = 3 .
The top is
2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 and the bottom is 2.
So the value is
− 1 2 -\frac12 − 2 1 , option D.
Watch out
Keep the order of the subtraction: 2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 3 − 2 3 - 2 3 − 2 1 2 \frac12 2 1 Report a problem with this question
If 23 y = 1111 two 23_y = 1111_{\text{two}} 2 3 y = 111 1 two y y y
Worked solution (try it first) Change the right side to base ten:
1111 two = 8 + 4 + 2 + 1 = 15 1111_{\text{two}} = 8 + 4 + 2 + 1 = 15 111 1 two = 8 + 4 + 2 + 1 = 15 .
In base
y y y ,
23 y = 2 y + 3 23_y = 2y + 3 2 3 y = 2 y + 3 , so
2 y + 3 = 15 2y + 3 = 15 2 y + 3 = 15 .
Subtract 3 and divide by 2:
y = 6 y = 6 y = 6 , option C.
Watch out
1111 two 1111_{\text{two}} 111 1 two Report a problem with this question
If 6, P P P P P P
Worked solution (try it first) In an A.P. the middle term is the average of its neighbours.
So
P = 6 + 14 2 = 10 P = \frac{6 + 14}{2} = 10 P = 2 6 + 14 = 10 , option B.
Watch out
8 (option D) is the gap 14 − 6 14 - 6 14 − 6 P = 6 + 4 = 10 P = 6 + 4 = 10 P = 6 + 4 = 10 Report a problem with this question
Evaluate 2 28 − 3 50 + 72 2\sqrt{28} - 3\sqrt{50} + \sqrt{72} 2 28 − 3 50 + 72
A 4 7 − 21 2 4\sqrt7 - 21\sqrt2 4 7 − 21 2 B 4 7 − 11 2 4\sqrt7 - 11\sqrt2 4 7 − 11 2 C 4 7 − 9 2 4\sqrt7 - 9\sqrt2 4 7 − 9 2 D 4 7 + 2 4\sqrt7 + \sqrt2 4 7 + 2
Worked solution (try it first) Take out square factors:
2 28 = 2 × 2 7 2\sqrt{28} = 2 \times 2\sqrt7 2 28 = 2 × 2 7 , which is
4 7 4\sqrt7 4 7 .
Likewise
3 50 = 3 × 5 2 = 15 2 3\sqrt{50} = 3 \times 5\sqrt2 = 15\sqrt2 3 50 = 3 × 5 2 = 15 2 and
72 = 6 2 \sqrt{72} = 6\sqrt2 72 = 6 2 .
Combine the
2 \sqrt2 2 terms:
− 15 2 + 6 2 = − 9 2 -15\sqrt2 + 6\sqrt2 = -9\sqrt2 − 15 2 + 6 2 = − 9 2 .
So the value is
4 7 − 9 2 4\sqrt7 - 9\sqrt2 4 7 − 9 2 , option C.
Watch out
The 72 \sqrt{72} 72 − 15 2 + 6 2 = − 9 2 -15\sqrt2 + 6\sqrt2 = -9\sqrt2 − 15 2 + 6 2 = − 9 2 − 21 2 -21\sqrt2 − 21 2 Report a problem with this question
If m : n = 2 : 1 m : n = 2 : 1 m : n = 2 : 1 3 m 2 − 2 n 2 m 2 + m n \dfrac{3m^2 - 2n^2}{m^2 + mn} m 2 + mn 3 m 2 − 2 n 2
A 4 3 \frac43 3 4 B 5 3 \frac53 3 5 C 3 4 \frac34 4 3 D 3 5 \frac35 5 3
Worked solution (try it first) m : n = 2 : 1 m : n = 2 : 1 m : n = 2 : 1 , so take
m = 2 m = 2 m = 2 and
n = 1 n = 1 n = 1 (any multiple gives the same value).
Top:
3 m 2 − 2 n 2 = 3 × 4 − 2 × 1 3m^2 - 2n^2 = 3 \times 4 - 2 \times 1 3 m 2 − 2 n 2 = 3 × 4 − 2 × 1 Bottom:
m 2 + m n = 4 + 2 = 6 m^2 + mn = 4 + 2 = 6 m 2 + mn = 4 + 2 = 6 .
So the value is
10 6 = 5 3 \frac{10}{6} = \frac53 6 10 = 3 5 , option B.
Watch out
Square before you multiply: 2 n 2 = 2 × 1 2 = 2 2n^2 = 2 \times 1^2 = 2 2 n 2 = 2 × 1 2 = 2 ( 2 × 1 ) 2 = 4 (2 \times 1)^2 = 4 ( 2 × 1 ) 2 = 4 8 6 = 4 3 \frac{8}{6} = \frac43 6 8 = 3 4 Report a problem with this question
H H H p p p y y y H = 1 H = 1 H = 1 p = 8 p = 8 p = 8 y = 2 y = 2 y = 2 H H H p p p y y y
A H = p 4 y 2 H = \dfrac{p}{4y^2} H = 4 y 2 p B H = 2 p y 2 H = \dfrac{2p}{y^2} H = y 2 2 p C H = p 2 y 2 H = \dfrac{p}{2y^2} H = 2 y 2 p D H = p 2 y 2 H = \dfrac{p^2}{y^2} H = y 2 p 2
Worked solution (try it first) Directly as
p p p , inversely as
y 2 y^2 y 2 :
H = k p y 2 H = \dfrac{kp}{y^2} H = y 2 k p .
Put in
H = 1 H = 1 H = 1 ,
p = 8 p = 8 p = 8 ,
y = 2 y = 2 y = 2 :
1 = 8 k 4 1 = \dfrac{8k}{4} 1 = 4 8 k , so
1 = 2 k 1 = 2k 1 = 2 k and
k = 1 2 k = \frac12 k = 2 1 .
So
H = p 2 y 2 H = \dfrac{p}{2y^2} H = 2 y 2 p , option C.
Watch out
From 1 = 2 k 1 = 2k 1 = 2 k k = 1 2 k = \frac12 k = 2 1 k = 2 k = 2 k = 2 Report a problem with this question
Solve 4 x 2 − 16 x + 15 = 0 4x^2 - 16x + 15 = 0 4 x 2 − 16 x + 15 = 0
A x = 1 1 2 x = 1\frac12 x = 1 2 1 x = − 2 1 2 x = -2\frac12 x = − 2 2 1 B x = 1 1 2 x = 1\frac12 x = 1 2 1 x = 2 1 2 x = 2\frac12 x = 2 2 1 C x = 1 1 2 x = 1\frac12 x = 1 2 1 x = − 1 1 2 x = -1\frac12 x = − 1 2 1 D x = − 1 1 2 x = -1\frac12 x = − 1 2 1 x = − 2 1 2 x = -2\frac12 x = − 2 2 1
Worked solution (try it first) Find two numbers with product
4 × 15 = 60 4 \times 15 = 60 4 × 15 = 60 and sum
− 16 -16 − 16 : they are
− 6 -6 − 6 and
− 10 -10 − 10 .
Split and group:
4 x 2 − 6 x − 10 x + 15 = 2 x ( 2 x − 3 ) − 5 ( 2 x − 3 ) 4x^2 - 6x - 10x + 15 = 2x(2x - 3) - 5(2x - 3) 4 x 2 − 6 x − 10 x + 15 = 2 x ( 2 x − 3 ) − 5 ( 2 x − 3 ) = ( 2 x − 3 ) ( 2 x − 5 ) = (2x - 3)(2x - 5) = ( 2 x − 3 ) ( 2 x − 5 ) .
So
x = 3 2 = 1 1 2 x = \frac32 = 1\frac12 x = 2 3 = 1 2 1 or
x = 5 2 = 2 1 2 x = \frac52 = 2\frac12 x = 2 5 = 2 2 1 , option B.
Watch out
The constant + 15 +15 + 15 − 2 1 2 -2\frac12 − 2 2 1 2 x + 5 2x + 5 2 x + 5 Report a problem with this question
Simplify log 10 6 − 3 log 10 3 + 2 3 log 10 27 \log_{10} 6 - 3\log_{10} 3 + \frac23\log_{10} 27 log 10 6 − 3 log 10 3 + 3 2 log 10 27
A 3 log 10 2 3\log_{10} 2 3 log 10 2 B log 10 2 \log_{10} 2 log 10 2 C log 10 3 \log_{10} 3 log 10 3 D 2 log 10 3 2\log_{10} 3 2 log 10 3
Worked solution (try it first) Move the numbers up as powers:
3 log 3 = log 27 3\log 3 = \log 27 3 log 3 = log 27 , and
2 3 log 27 = log 27 2 3 \frac23\log 27 = \log 27^{\frac23} 3 2 log 27 = log 2 7 3 2 Combine:
log 6 − log 27 + log 9 = log 6 × 9 27 \log 6 - \log 27 + \log 9 = \log \frac{6 \times 9}{27} log 6 − log 27 + log 9 = log 27 6 × 9 .
54 27 = 2 \frac{54}{27} = 2 27 54 = 2 , so the answer is
log 10 2 \log_{10} 2 log 10 2 , option B.
Watch out
27 2 3 27^{\frac23} 2 7 3 2 2 3 × 27 = 18 \frac23 \times 27 = 18 3 2 × 27 = 18 log 4 \log 4 log 4 Report a problem with this question
Bala sold an article for ₦6,900.00 and made a profit of 15 % 15\% 15%
A 5 % 5\% 5% B 10 % 10\% 10% C 12 % 12\% 12% D 13 % 13\% 13%
Worked solution (try it first) ₦6,900 is
115 % 115\% 115% of the cost, so the cost is
6900 ÷ 1.15 = 6900 \div 1.15 = 6900 ÷ 1.15 = ₦6,000.
Selling for ₦6,600 gives a profit of
6600 − 6000 = 6600 - 6000 = 6600 − 6000 = ₦600.
As a percentage of the cost:
600 6000 × 100 % = 10 % \dfrac{600}{6000} \times 100\% = 10\% 6000 600 × 100% = 10% , option B.
Watch out
Find the cost price first. The ₦300 drop in price is 5 % 5\% 5% 15 % − 5 % = 10 % 15\% - 5\% = 10\% 15% − 5% = 10% Report a problem with this question
If 3 p = 4 q 3p = 4q 3 p = 4 q 9 p = 8 q − 12 9p = 8q - 12 9 p = 8 q − 12 p q pq pq
Worked solution (try it first) From
3 p = 4 q 3p = 4q 3 p = 4 q , divide by 4:
q = 3 p 4 q = \frac{3p}{4} q = 4 3 p , so
8 q = 6 p 8q = 6p 8 q = 6 p .
Substitute into the second equation:
9 p = 6 p − 12 9p = 6p - 12 9 p = 6 p − 12 , so
3 p = − 12 3p = -12 3 p = − 12 and
p = − 4 p = -4 p = − 4 .
Then
q = 3 × ( − 4 ) 4 = − 3 q = \frac{3 \times (-4)}{4} = -3 q = 4 3 × ( − 4 ) = − 3 .
So
p q = ( − 4 ) ( − 3 ) = 12 pq = (-4)(-3) = 12 pq = ( − 4 ) ( − 3 ) = 12 , option A.
Watch out
Both p p p q q q p q = 12 pq = 12 pq = 12 − 12 -12 − 12 Report a problem with this question
If ( 0.25 ) y = 32 (0.25)^y = 32 ( 0.25 ) y = 32 y y y
A y = − 5 2 y = -\frac52 y = − 2 5 B y = − 3 2 y = -\frac32 y = − 2 3 C y = 3 2 y = \frac32 y = 2 3 D y = 5 2 y = \frac52 y = 2 5
Worked solution (try it first) 0.25 = 1 4 = 2 − 2 0.25 = \frac14 = 2^{-2} 0.25 = 4 1 = 2 − 2 , so
( 0.25 ) y = 2 − 2 y (0.25)^y = 2^{-2y} ( 0.25 ) y = 2 − 2 y .
32 = 2 5 32 = 2^5 32 = 2 5 , so
2 − 2 y = 2 5 2^{-2y} = 2^5 2 − 2 y = 2 5 and
− 2 y = 5 -2y = 5 − 2 y = 5 .
Divide by
− 2 -2 − 2 :
y = − 5 2 y = -\frac52 y = − 2 5 , option A.
Watch out
0.25 is a fraction, so as a power of 2 its index is negative: 2 − 2 2^{-2} 2 − 2 4 = 2 2 4 = 2^2 4 = 2 2 y = 5 2 y = \frac52 y = 2 5 Report a problem with this question
There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?
A 1 6 \frac16 6 1 B 1 4 \frac14 4 1 C 2 3 \frac23 3 2 D 1 2 \frac12 2 1
Worked solution (try it first) There are
8 + 4 = 12 8 + 4 = 12 8 + 4 = 12 people, each equally likely to step out first.
8 of them are boys, so the probability is
8 12 = 2 3 \frac{8}{12} = \frac23 12 8 = 3 2 , option C.
Watch out
Boy and girl are not equally likely here: there are twice as many boys. Taking it as 1 2 \frac12 2 1 Report a problem with this question
Simplify x 2 − 5 x − 14 x 2 − 9 x + 14 \dfrac{x^2 - 5x - 14}{x^2 - 9x + 14} x 2 − 9 x + 14 x 2 − 5 x − 14
A x − 7 x + 7 \dfrac{x - 7}{x + 7} x + 7 x − 7 B x + 7 x − 7 \dfrac{x + 7}{x - 7} x − 7 x + 7 C x − 2 x + 4 \dfrac{x - 2}{x + 4} x + 4 x − 2 D x + 2 x − 2 \dfrac{x + 2}{x - 2} x − 2 x + 2
Worked solution (try it first) Top: two numbers multiplying to
− 14 -14 − 14 and adding to
− 5 -5 − 5 are
− 7 -7 − 7 and
2 2 2 , so
x 2 − 5 x − 14 = ( x − 7 ) ( x + 2 ) x^2 - 5x - 14 = (x - 7)(x + 2) x 2 − 5 x − 14 = ( x − 7 ) ( x + 2 ) .
Bottom:
− 7 -7 − 7 and
− 2 -2 − 2 multiply to 14 and add to
− 9 -9 − 9 , so
x 2 − 9 x + 14 = ( x − 7 ) ( x − 2 ) x^2 - 9x + 14 = (x - 7)(x - 2) x 2 − 9 x + 14 = ( x − 7 ) ( x − 2 ) .
Cancel
x − 7 x - 7 x − 7 :
x + 2 x − 2 \dfrac{x + 2}{x - 2} x − 2 x + 2 , option D.
Watch out
Check each factor pair adds to the middle number. For the top, + 7 +7 + 7 − 2 -2 − 2 + 5 +5 + 5 − 5 -5 − 5 x + 7 x + 7 x + 7 Report a problem with this question
Which of these values would make 3 p − 1 p 2 − p \dfrac{3p - 1}{p^2 - p} p 2 − p 3 p − 1
A 1 B 1 3 \frac13 3 1 C − 1 3 -\frac13 − 3 1 D − 1 -1 − 1
Worked solution (try it first) A fraction is undefined when its bottom is zero.
Factorise the bottom:
p 2 − p = p ( p − 1 ) p^2 - p = p(p - 1) p 2 − p = p ( p − 1 ) , which is zero when
p = 0 p = 0 p = 0 or
p = 1 p = 1 p = 1 .
Only 1 is an option, so the answer is option A.
Watch out
Setting the top to zero gives p = 1 3 p = \frac13 p = 3 1 Report a problem with this question
The total surface area of a solid cylinder is 165 cm 2 165\text{ cm}^2 165 cm 2 7 cm 7\text{ cm} 7 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 7.5 cm 7.5\text{ cm} 7.5 cm B 4.5 cm 4.5\text{ cm} 4.5 cm C 4.0 cm 4.0\text{ cm} 4.0 cm D 2.0 cm 2.0\text{ cm} 2.0 cm
Worked solution (try it first) Radius
= 7 ÷ 2 = 3.5 = 7 \div 2 = 3.5 = 7 ÷ 2 = 3.5 cm.
Total surface area:
2 π r ( r + h ) 2\pi r(r + h) 2 π r ( r + h ) , and
2 π r = 2 × 22 7 × 3.5 2\pi r = 2 \times \frac{22}{7} \times 3.5 2 π r = 2 × 7 22 × 3.5 So
22 ( 3.5 + h ) = 165 22(3.5 + h) = 165 22 ( 3.5 + h ) = 165 .
Divide by 22:
3.5 + h = 7.5 3.5 + h = 7.5 3.5 + h = 7.5 .
So
h = 4.0 h = 4.0 h = 4.0 cm, option C.
Watch out
7.5 7.5 7.5 r + h r + h r + h Report a problem with this question
If 2 a = 64 2^a = \sqrt{64} 2 a = 64 b a = 3 \frac ba = 3 a b = 3 a 2 + b 2 a^2 + b^2 a 2 + b 2
Worked solution (try it first) 64 = 8 = 2 3 \sqrt{64} = 8 = 2^3 64 = 8 = 2 3 , so
2 a = 2 3 2^a = 2^3 2 a = 2 3 and
a = 3 a = 3 a = 3 .
b a = 3 \frac ba = 3 a b = 3 means
b = 3 a = 9 b = 3a = 9 b = 3 a = 9 .
So
a 2 + b 2 = 9 + 81 = 90 a^2 + b^2 = 9 + 81 = 90 a 2 + b 2 = 9 + 81 = 90 , option C.
Watch out
Take the square root first: 2 a = 8 2^a = 8 2 a = 8 a = 6 a = 6 a = 6 b = 18 b = 18 b = 18 a 2 + b 2 = 360 a^2 + b^2 = 360 a 2 + b 2 = 360 Report a problem with this question
In △ X Y Z \triangle XYZ △ X Y Z ∣ Y Z ∣ = 32 cm |YZ| = 32\text{ cm} ∣ Y Z ∣ = 32 cm ∠ Y X Z = 52 ∘ \angle YXZ = 52^\circ ∠ Y X Z = 5 2 ∘ ∠ X Z Y = 90 ∘ \angle XZY = 90^\circ ∠ X Z Y = 9 0 ∘ ∣ X Z ∣ |XZ| ∣ X Z ∣
A 31 cm 31\text{ cm} 31 cm B 25 cm 25\text{ cm} 25 cm C 20 cm 20\text{ cm} 20 cm D 13 cm 13\text{ cm} 13 cm
Worked solution (try it first) Y Z = 32 YZ = 32 Y Z = 32 cm is opposite the
52 ∘ 52^\circ 5 2 ∘ angle at
X X X , and
X Z XZ X Z is adjacent to it, so use tangent:
tan 52 ∘ = 32 X Z \tan52^\circ = \dfrac{32}{XZ} tan 5 2 ∘ = X Z 32 .
Rearrange:
X Z = 32 tan 52 ∘ XZ = \dfrac{32}{\tan52^\circ} X Z = tan 5 2 ∘ 32 = 32 1.280 = \dfrac{32}{1.280} = 1.280 32 To the nearest centimetre,
X Z = 25 XZ = 25 X Z = 25 cm, option B.
Watch out
32 cm is a shorter side, not the hypotenuse, so use tangent. Treating it as the hypotenuse gives 32 cos 52 ∘ ≈ 20 32\cos52^\circ \approx 20 32 cos 5 2 ∘ ≈ 20 Report a problem with this question
If log x 2 = 0.3 \log_x 2 = 0.3 log x 2 = 0.3 log x 8 \log_x 8 log x 8
Worked solution (try it first) 8 = 2 3 8 = 2^3 8 = 2 3 , so
log x 8 = 3 log x 2 \log_x 8 = 3\log_x 2 log x 8 = 3 log x 2 .
3 × 0.3 = 0.9 3 \times 0.3 = 0.9 3 × 0.3 = 0.9 , option C.
Watch out
Multiply by the power 3, not by 8 itself. 8 × 0.3 = 2.4 8 \times 0.3 = 2.4 8 × 0.3 = 2.4 Report a problem with this question
An arc subtends an angle of 72 ∘ 72^\circ 7 2 ∘ 3.5 cm 3.5\text{ cm} 3.5 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 6.6 cm 6.6\text{ cm} 6.6 cm B 8.8 cm 8.8\text{ cm} 8.8 cm C 4.4 cm 4.4\text{ cm} 4.4 cm D 2.2 cm 2.2\text{ cm} 2.2 cm
Worked solution (try it first) The circumference is
2 × 22 7 × 3.5 = 22 2 \times \frac{22}{7} \times 3.5 = 22 2 × 7 22 × 3.5 = 22 cm.
The arc is
72 360 = 1 5 \frac{72}{360} = \frac15 360 72 = 5 1 of it:
22 ÷ 5 = 4.4 22 \div 5 = 4.4 22 ÷ 5 = 4.4 cm, option C.
Watch out
The circumference is 2 π r 2\pi r 2 π r Report a problem with this question
Make b b b l b = 1 2 ( a + b ) h lb = \frac12(a + b)h l b = 2 1 ( a + b ) h
A a h 2 l − h \dfrac{ah}{2l - h} 2 l − h ah B 2 l − h a l \dfrac{2l - h}{al} a l 2 l − h C a l 2 l − h \dfrac{al}{2l - h} 2 l − h a l D a l 2 − h \dfrac{al}{2 - h} 2 − h a l
Worked solution (try it first) Multiply by 2 and expand:
2 l b = a h + b h 2lb = ah + bh 2 l b = ah + bh .
Collect the
b b b terms:
2 l b − b h = a h 2lb - bh = ah 2 l b − bh = ah , so
b ( 2 l − h ) = a h b(2l - h) = ah b ( 2 l − h ) = ah .
Divide by
2 l − h 2l - h 2 l − h :
b = a h 2 l − h b = \dfrac{ah}{2l - h} b = 2 l − h ah , option A.
Watch out
After collecting the b b b a h ah ah 1 2 × 2 × a h \frac12 \times 2 \times ah 2 1 × 2 × ah a l al a l Report a problem with this question
Eric sold his house through an agent who charged 8 % 8\% 8%
A $130,000.00 B $128,000.00 C $125,000.00 D $120,000.00
Worked solution (try it first) The commission is
8 % 8\% 8% of the selling price
S S S , so Eric keeps
92 % 92\% 92% of it:
0.92 S = 117 760 0.92S = 117\,760 0.92 S = 117 760 .
Divide both sides by 0.92:
S = 128 000 S = 128\,000 S = 128 000 .
So the selling price was $128,000.00, option B.
Watch out
The 8 % 8\% 8% 8 % 8\% 8% Report a problem with this question
Find the angle which an arc of length 22 cm 22\text{ cm} 22 cm 15 cm 15\text{ cm} 15 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 70 ∘ 70^\circ 7 0 ∘ B 84 ∘ 84^\circ 8 4 ∘ C 96 ∘ 96^\circ 9 6 ∘ D 156 ∘ 156^\circ 15 6 ∘
Worked solution (try it first) The circumference is
2 × 22 7 × 15 = 660 7 2 \times \frac{22}{7} \times 15 = \frac{660}{7} 2 × 7 22 × 15 = 7 660 cm.
The angle is the arc's share of
360 ∘ 360^\circ 36 0 ∘ :
θ = 22 660 / 7 × 360 \theta = \frac{22}{660/7} \times 360 θ = 660/7 22 × 360 = 154 660 × 360 = \frac{154}{660} \times 360 = 660 154 × 360 .
So
θ = 84 ∘ \theta = 84^\circ θ = 8 4 ∘ , option B.
Watch out
Use 2 π r 2\pi r 2 π r π r \pi r π r 168 ∘ 168^\circ 16 8 ∘ Report a problem with this question
A rectangular board has a length of 15 cm 15\text{ cm} 15 cm x cm x\text{ cm} x cm
A 60 x cm 2 60x\text{ cm}^2 60 x cm 2 B 45 x cm 2 45x\text{ cm}^2 45 x cm 2 C 30 x cm 2 30x\text{ cm}^2 30 x cm 2 D 15 x cm 2 15x\text{ cm}^2 15 x cm 2
Worked solution (try it first) Doubling the sides makes the board
30 30 30 cm by
2 x 2x 2 x cm.
The new area is
30 × 2 x = 60 x cm 2 30 \times 2x = 60x\text{ cm}^2 30 × 2 x = 60 x cm 2 , option A.
Watch out
Double both sides, not just the length. Doubling only the 15 gives 30 x cm 2 30x\text{ cm}^2 30 x cm 2 Report a problem with this question
In the diagram, P O S POS P O S R O T ROT R O T O P Q R OPQR O P QR ∣ O S ∣ = ∣ O T ∣ |OS| = |OT| ∣ O S ∣ = ∣ O T ∣ ∠ O S T = 50 ∘ \angle OST = 50^\circ ∠ O S T = 5 0 ∘ ∠ O P Q \angle OPQ ∠ O P Q
A 100 ∘ 100^\circ 10 0 ∘ B 120 ∘ 120^\circ 12 0 ∘ C 140 ∘ 140^\circ 14 0 ∘ D 160 ∘ 160^\circ 16 0 ∘
Worked solution (try it first) ∣ O S ∣ = ∣ O T ∣ |OS| = |OT| ∣ O S ∣ = ∣ O T ∣ , so both base angles of triangle
O S T OST O S T are
50 ∘ 50^\circ 5 0 ∘ and
∠ S O T = 180 ∘ − 100 ∘ \angle SOT = 180^\circ - 100^\circ ∠ S O T = 18 0 ∘ − 10 0 ∘ P O S POS P O S and
R O T ROT R O T are straight lines, so
∠ P O R = ∠ S O T = 80 ∘ \angle POR = \angle SOT = 80^\circ ∠ P O R = ∠ S O T = 8 0 ∘ (vertically opposite).
In the parallelogram
O R ∥ P Q OR \parallel PQ O R ∥ P Q , so
∠ O P Q \angle OPQ ∠ O P Q and
∠ P O R \angle POR ∠ P O R are co-interior:
∠ O P Q = 180 ∘ − 80 ∘ \angle OPQ = 180^\circ - 80^\circ ∠ O P Q = 18 0 ∘ − 8 0 ∘ = 100 ∘ = 100^\circ = 10 0 ∘ , option A.
Watch out
∠ O P Q \angle OPQ ∠ O P Q ∠ P O R \angle POR ∠ P O R 180 ∘ − 80 ∘ 180^\circ - 80^\circ 18 0 ∘ − 8 0 ∘ 80 ∘ 80^\circ 8 0 ∘ Report a problem with this question
Factorize completely ( 2 x + 2 y ) ( x − y ) + ( 2 x − 2 y ) ( x + y ) (2x + 2y)(x - y) + (2x - 2y)(x + y) ( 2 x + 2 y ) ( x − y ) + ( 2 x − 2 y ) ( x + y )
A 4 ( x − y ) ( x + y ) 4(x - y)(x + y) 4 ( x − y ) ( x + y ) B 4 ( x − y ) 4(x - y) 4 ( x − y ) C 2 ( x − y ) ( x + y ) 2(x - y)(x + y) 2 ( x − y ) ( x + y ) D 2 ( x − y ) 2(x - y) 2 ( x − y )
Worked solution (try it first) Take out 2 from the first bracket of each term:
2 ( x + y ) ( x − y ) + 2 ( x − y ) ( x + y ) 2(x + y)(x - y) + 2(x - y)(x + y) 2 ( x + y ) ( x − y ) + 2 ( x − y ) ( x + y ) .
The two terms are equal, so together they make
4 ( x + y ) ( x − y ) 4(x + y)(x - y) 4 ( x + y ) ( x − y ) .
So the expression is
4 ( x − y ) ( x + y ) 4(x - y)(x + y) 4 ( x − y ) ( x + y ) , option A.
Watch out
There are two equal terms, each 2 ( x − y ) ( x + y ) 2(x - y)(x + y) 2 ( x − y ) ( x + y ) Report a problem with this question
The interior angles of a polygon are 3 x ∘ 3x^\circ 3 x ∘ 2 x ∘ 2x^\circ 2 x ∘ 4 x ∘ 4x^\circ 4 x ∘ 3 x ∘ 3x^\circ 3 x ∘ 6 x ∘ 6x^\circ 6 x ∘
A 80 ∘ 80^\circ 8 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 40 ∘ 40^\circ 4 0 ∘ D 30 ∘ 30^\circ 3 0 ∘
Worked solution (try it first) There are five angles, so the polygon is a pentagon.
Its angles add up to
( 5 − 2 ) × 180 ∘ = 540 ∘ (5 - 2) \times 180^\circ = 540^\circ ( 5 − 2 ) × 18 0 ∘ = 54 0 ∘ .
3 x + 2 x + 4 x + 3 x + 6 x = 540 3x + 2x + 4x + 3x + 6x = 540 3 x + 2 x + 4 x + 3 x + 6 x = 540 , so
18 x = 540 18x = 540 18 x = 540 and
x = 30 x = 30 x = 30 .
The smallest angle is
2 x = 60 ∘ 2x = 60^\circ 2 x = 6 0 ∘ , option B.
Watch out
30 ∘ 30^\circ 3 0 ∘ x x x 2 x 2x 2 x Report a problem with this question
A box contains 2 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours?
A 2 3 \frac23 3 2 B 3 5 \frac35 5 3 C 7 20 \frac{7}{20} 20 7 D 12 25 \frac{12}{25} 25 12
Worked solution (try it first) With replacement, each pick is white with probability
2 5 \frac25 5 2 and blue with probability
3 5 \frac35 5 3 .
White then blue:
2 5 × 3 5 = 6 25 \frac25 \times \frac35 = \frac{6}{25} 5 2 × 5 3 = 25 6 , and blue then white is the same.
Add the two orders:
12 25 \frac{12}{25} 25 12 , option D.
Watch out
The ball is put back, so the second pick is still out of 5. Using 4 for the second pick gives 2 × 2 5 × 3 4 = 3 5 2 \times \frac25 \times \frac34 = \frac35 2 × 5 2 × 4 3 = 5 3 Report a problem with this question
Find the equation of a straight line passing through the point ( 1 , − 5 ) (1, -5) ( 1 , − 5 ) 3 4 \frac34 4 3
A 3 x + 4 y − 23 = 0 3x + 4y - 23 = 0 3 x + 4 y − 23 = 0 B 3 x + 4 y + 23 = 0 3x + 4y + 23 = 0 3 x + 4 y + 23 = 0 C 3 x − 4 y + 23 = 0 3x - 4y + 23 = 0 3 x − 4 y + 23 = 0 D 3 x − 4 y − 23 = 0 3x - 4y - 23 = 0 3 x − 4 y − 23 = 0
Worked solution (try it first) Use
y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) with
( 1 , − 5 ) (1, -5) ( 1 , − 5 ) :
y + 5 = 3 4 ( x − 1 ) y + 5 = \frac34(x - 1) y + 5 = 4 3 ( x − 1 ) .
Multiply both sides by 4:
4 y + 20 = 3 x − 3 4y + 20 = 3x - 3 4 y + 20 = 3 x − 3 .
Collect terms on one side:
3 x − 4 y − 23 = 0 3x - 4y - 23 = 0 3 x − 4 y − 23 = 0 , option D.
Watch out
Watch the sign of the constant when you move terms. Check with the point: 3 ( 1 ) − 4 ( − 5 ) − 23 = 0 3(1) - 4(-5) - 23 = 0 3 ( 1 ) − 4 ( − 5 ) − 23 = 0 Report a problem with this question
The foot of a ladder is 6 m 6\text{ m} 6 m 8 m 8\text{ m} 8 m
A 14 m 14\text{ m} 14 m B 12 m 12\text{ m} 12 m C 10 m 10\text{ m} 10 m D 7 m 7\text{ m} 7 m
Worked solution (try it first) The ground, the pole and the ladder make a right-angled triangle with the ladder as the hypotenuse.
Pythagoras: the ladder is
6 2 + 8 2 = 100 = 10 \sqrt{6^2 + 8^2} = \sqrt{100} = 10 6 2 + 8 2 = 100 = 10 m, option C.
Watch out
Square, add, then take the square root. Adding the sides directly gives 6 + 8 = 14 6 + 8 = 14 6 + 8 = 14 Report a problem with this question
If tan x = 3 4 \tan x = \frac34 tan x = 4 3 0 ∘ < x < 90 ∘ 0^\circ < x < 90^\circ 0 ∘ < x < 9 0 ∘ cos x 2 sin x \dfrac{\cos x}{2\sin x} 2 sin x cos x
A 8 3 \frac83 3 8 B 3 2 \frac32 2 3 C 4 3 \frac43 3 4 D 2 3 \frac23 3 2
Worked solution (try it first) cos x sin x = 1 tan x \dfrac{\cos x}{\sin x} = \dfrac{1}{\tan x} sin x cos x = tan x 1 , so
cos x 2 sin x = 1 2 tan x \dfrac{\cos x}{2\sin x} = \dfrac{1}{2\tan x} 2 sin x cos x = 2 tan x 1 .
Put in
tan x = 3 4 \tan x = \frac34 tan x = 4 3 :
1 2 × 3 4 = 1 3 / 2 \dfrac{1}{2 \times \frac34} = \dfrac{1}{3/2} 2 × 4 3 1 = 3/2 1 = 2 3 = \dfrac23 = 3 2 , option D.
Watch out
Keep the 2 in the bottom: cos x sin x \frac{\cos x}{\sin x} s i n x c o s x 4 3 \frac43 3 4 Report a problem with this question
From the top of a vertical cliff 20 m 20\text{ m} 20 m 75 m 75\text{ m} 75 m
A 56 ∘ 56^\circ 5 6 ∘ B 75 ∘ 75^\circ 7 5 ∘ C 16 ∘ 16^\circ 1 6 ∘ D 15 ∘ 15^\circ 1 5 ∘
Worked solution (try it first) The angle of depression equals the angle of elevation of the cliff top from the boat (alternate angles).
The height 20 m is opposite the angle and 75 m is adjacent:
tan θ = 20 75 \tan\theta = \frac{20}{75} tan θ = 75 20 .
So
θ = tan − 1 0.2667 \theta = \tan^{-1}0.2667 θ = tan − 1 0.2667 ≈ 14.9 ∘ \approx 14.9^\circ ≈ 14. 9 ∘ , which is
15 ∘ 15^\circ 1 5 ∘ to the nearest degree, option D.
Watch out
Put the height on top: 20 75 \frac{20}{75} 75 20 75 20 \frac{75}{20} 20 75 75 ∘ 75^\circ 7 5 ∘ Report a problem with this question
In the diagram, O O O 18 cm 18\text{ cm} 18 cm ∠ Z X Y = 70 ∘ \angle ZXY = 70^\circ ∠ Z X Y = 7 0 ∘ Z Y ZY Z Y [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 11 cm 11\text{ cm} 11 cm B 22 cm 22\text{ cm} 22 cm C 44 cm 44\text{ cm} 44 cm D 80 cm 80\text{ cm} 80 cm
Worked solution (try it first) ∠ Z X Y \angle ZXY ∠ Z X Y is at the circumference on arc
Z Y ZY Z Y .
The angle at the centre is twice it:
∠ Z O Y = 140 ∘ \angle ZOY = 140^\circ ∠ Z O Y = 14 0 ∘ .
Arc length is
θ 360 ∘ × 2 π r \frac{\theta}{360^\circ} \times 2\pi r 36 0 ∘ θ × 2 π r : arc
Z Y = 140 360 × 2 × 22 7 × 18 ZY = \frac{140}{360} \times 2 \times \frac{22}{7} \times 18 Z Y = 360 140 × 2 × 7 22 × 18 .
That is 44 cm, option C.
Watch out
The arc formula needs the angle at the centre. Using the 70 ∘ 70^\circ 7 0 ∘ X X X Report a problem with this question
In the diagram, R T RT R T R R R ∠ P Q R = 70 ∘ \angle PQR = 70^\circ ∠ P QR = 7 0 ∘ ∠ Q R T = 52 ∘ \angle QRT = 52^\circ ∠ QR T = 5 2 ∘ ∠ Q S R = y \angle QSR = y ∠ QS R = y ∠ P R Q = x \angle PRQ = x ∠ P R Q = x y y y
A 70 ∘ 70^\circ 7 0 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 52 ∘ 52^\circ 5 2 ∘ D 18 ∘ 18^\circ 1 8 ∘
Worked solution (try it first) ∠ Q R T \angle QRT ∠ QR T is between the tangent
R T RT R T and the chord
R Q RQ R Q .
The angle between a tangent and a chord equals the angle in the alternate segment, which is at
S S S .
So
y = ∠ Q S R = 52 ∘ y = \angle QSR = 52^\circ y = ∠ QS R = 5 2 ∘ , option C.
Watch out
y y y Q R QR QR Q R QR QR 52 ∘ 52^\circ 5 2 ∘ ∠ P Q R = 70 ∘ \angle PQR = 70^\circ ∠ P QR = 7 0 ∘ Report a problem with this question
In the same diagram (R T RT R T R R R ∠ P Q R = 70 ∘ \angle PQR = 70^\circ ∠ P QR = 7 0 ∘ ∠ Q R T = 52 ∘ \angle QRT = 52^\circ ∠ QR T = 5 2 ∘ x = ∠ P R Q x = \angle PRQ x = ∠ P R Q
A 70 ∘ 70^\circ 7 0 ∘ B 58 ∘ 58^\circ 5 8 ∘ C 52 ∘ 52^\circ 5 2 ∘ D 18 ∘ 18^\circ 1 8 ∘
Worked solution (try it first) Alternate segment:
∠ Q P R \angle QPR ∠ QP R also stands on chord
Q R QR QR , so
∠ Q P R = ∠ Q R T = 52 ∘ \angle QPR = \angle QRT = 52^\circ ∠ QP R = ∠ QR T = 5 2 ∘ .
The angles of triangle
P Q R PQR P QR add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 70 ∘ − 52 ∘ x = 180^\circ - 70^\circ - 52^\circ x = 18 0 ∘ − 7 0 ∘ − 5 2 ∘ = 58 ∘ = 58^\circ = 5 8 ∘ , option B.
Watch out
Find the third angle of triangle P Q R PQR P QR 70 ∘ − 52 ∘ 70^\circ - 52^\circ 7 0 ∘ − 5 2 ∘ 18 ∘ 18^\circ 1 8 ∘ Report a problem with this question
Calculate the variance of 2, 4, 7, 8 and 9.
Worked solution (try it first) The numbers add up to 30, so the mean is 6.
The squared deviations are 16, 4, 1, 4, 9, which add up to 34.
The variance is
34 5 = 6.8 \frac{34}{5} = 6.8 5 34 = 6.8 , option B.
Watch out
The variance is not square-rooted. 6.8 ≈ 2.6 \sqrt{6.8} \approx 2.6 6.8 ≈ 2.6 Report a problem with this question
The fourth term of an arithmetic progression is 37 and the first term is − 20 -20 − 20
Worked solution (try it first) The 4th term of an A.P. is
a + 3 d a + 3d a + 3 d , so
− 20 + 3 d = 37 -20 + 3d = 37 − 20 + 3 d = 37 .
Add 20 to both sides:
3 d = 57 3d = 57 3 d = 57 .
Divide by 3:
d = 19 d = 19 d = 19 , option C.
Watch out
57 (option B) is 3 d 3d 3 d d d d Report a problem with this question
In the diagram, P Q PQ P Q R S RS R S ∠ Q F G = 105 ∘ \angle QFG = 105^\circ ∠ QF G = 10 5 ∘ ∠ F E G = 50 ∘ \angle FEG = 50^\circ ∠ F E G = 5 0 ∘ m m m
A 130 ∘ 130^\circ 13 0 ∘ B 105 ∘ 105^\circ 10 5 ∘ C 75 ∘ 75^\circ 7 5 ∘ D 55 ∘ 55^\circ 5 5 ∘
Worked solution (try it first) Angles on the straight line
P Q PQ P Q :
∠ E F G = 180 ∘ − 105 ∘ \angle EFG = 180^\circ - 105^\circ ∠ E F G = 18 0 ∘ − 10 5 ∘ The angles of triangle
E F G EFG E F G add up to
180 ∘ 180^\circ 18 0 ∘ :
m = 180 ∘ − 50 ∘ − 75 ∘ m = 180^\circ - 50^\circ - 75^\circ m = 18 0 ∘ − 5 0 ∘ − 7 5 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option D.
Watch out
75 ∘ 75^\circ 7 5 ∘ ∠ E F G \angle EFG ∠ E F G m m m G G G Report a problem with this question
In the same diagram (P Q ∥ R S PQ \parallel RS P Q ∥ R S ∠ Q F G = 105 ∘ \angle QFG = 105^\circ ∠ QF G = 10 5 ∘ ∠ F E G = 50 ∘ \angle FEG = 50^\circ ∠ F E G = 5 0 ∘ n n n
A 40 ∘ 40^\circ 4 0 ∘ B 55 ∘ 55^\circ 5 5 ∘ C 75 ∘ 75^\circ 7 5 ∘ D 130 ∘ 130^\circ 13 0 ∘
Worked solution (try it first) P Q ∥ R S PQ \parallel RS P Q ∥ R S , so
∠ Q F G \angle QFG ∠ QF G and
∠ F G S \angle FGS ∠ F GS are co-interior:
∠ F G S = 180 ∘ − 105 ∘ \angle FGS = 180^\circ - 105^\circ ∠ F GS = 18 0 ∘ − 10 5 ∘ n n n is vertically opposite
∠ F G S \angle FGS ∠ F GS , so
n = 75 ∘ n = 75^\circ n = 7 5 ∘ , option C.
Watch out
55 ∘ 55^\circ 5 5 ∘ m m m E F G EFG E F G n n n R S RS R S ∠ F G S \angle FGS ∠ F GS Report a problem with this question
A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?
A 1 6 \frac16 6 1 B 1 4 \frac14 4 1 C 1 3 \frac13 3 1 D 1 2 \frac12 2 1
Worked solution (try it first) There are
5 + 6 + 7 = 18 5 + 6 + 7 = 18 5 + 6 + 7 = 18 pencils.
6 of them are green, so the probability is
6 18 = 1 3 \frac{6}{18} = \frac13 18 6 = 3 1 , option C.
Watch out
Divide by all the pencils, not by the others: 6 12 = 1 2 \frac{6}{12} = \frac12 12 6 = 2 1 Report a problem with this question
The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?
Worked solution (try it first) The angles add up to
360 ∘ 360^\circ 36 0 ∘ , so Orange is
360 ∘ − ( 60 ∘ + 100 ∘ + 120 ∘ ) = 80 ∘ 360^\circ - (60^\circ + 100^\circ + 120^\circ) = 80^\circ 36 0 ∘ − ( 6 0 ∘ + 10 0 ∘ + 12 0 ∘ ) = 8 0 ∘ .
80 ∘ 80^\circ 8 0 ∘ stands for 60 oranges, so each degree is
60 80 = 3 4 \frac{60}{80} = \frac34 80 60 = 4 3 of a fruit.
Apple is
120 ∘ 120^\circ 12 0 ∘ :
120 × 3 4 = 90 120 \times \frac34 = 90 120 × 4 3 = 90 apples, option A.
Watch out
80 ∘ 80^\circ 8 0 ∘ 80 ∘ 80^\circ 8 0 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16 and 13. Find the mode of the distribution.
Worked solution (try it first) Count each score: 13 appears three times.
8, 14 and 18 twice each.
The rest once.
The mode is the score that occurs most often: 13, option C.
Watch out
Count every score carefully: 8, 14 and 18 each appear twice, but 13 appears three times. 18 (option A) is the highest score, not the mode. Report a problem with this question
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16 and 13. Find the median score.
Worked solution (try it first) Put the 14 scores in order: 8, 8, 10, 11, 13, 13, 13, 14, 14, 15, 16, 17, 18, 18.
With an even count, the median is halfway between the 7th and 8th: 13 and 14.
So the median is
13 + 14 2 = 13.5 \frac{13 + 14}{2} = 13.5 2 13 + 14 = 13.5 , option C.
Watch out
With 14 scores, average the 7th and 8th. Taking only the 7th gives 13.0 (option D), and taking only the 8th gives 14.0 (option B). Report a problem with this question
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16 and 13. How many students scored above the mean score?
Worked solution (try it first) The 14 scores add up to 188, so the mean is
188 14 ≈ 13.43 \frac{188}{14} \approx 13.43 14 188 ≈ 13.43 .
Scores above 13.43 are 14, 14, 15, 16, 17, 18 and 18.
So 7 students scored above the mean, option D.
Watch out
The three scores of 13 are below the mean, 13.43, so leave them out. Counting them gives 10 (option A). Report a problem with this question
Evaluate 0.42 ÷ 2.5 0.5 × 2.05 \dfrac{0.42 \div 2.5}{0.5 \times 2.05} 0.5 × 2.05 0.42 ÷ 2.5
A 1.639 × 10 2 1.639 \times 10^2 1.639 × 1 0 2 B 1.639 × 10 1 1.639 \times 10^1 1.639 × 1 0 1 C 1.639 × 10 − 1 1.639 \times 10^{-1} 1.639 × 1 0 − 1 D 1.639 × 10 − 2 1.639 \times 10^{-2} 1.639 × 1 0 − 2
Worked solution (try it first) Top:
0.42 ÷ 2.5 = 0.168 0.42 \div 2.5 = 0.168 0.42 ÷ 2.5 = 0.168 .
Bottom:
0.5 × 2.05 = 1.025 0.5 \times 2.05 = 1.025 0.5 × 2.05 = 1.025 .
Divide:
0.168 ÷ 1.025 = 0.16390 … 0.168 \div 1.025 = 0.16390\ldots 0.168 ÷ 1.025 = 0.16390 … , which is
1.639 × 10 − 1 1.639 \times 10^{-1} 1.639 × 1 0 − 1 , option C.
Watch out
The answer is less than 1, so the power of 10 is negative: 0.1639 = 1.639 × 10 − 1 0.1639 = 1.639 \times 10^{-1} 0.1639 = 1.639 × 1 0 − 1 1.639 × 10 1 1.639 \times 10^1 1.639 × 1 0 1 Report a problem with this question