Evaluate, correct to four significant figures, 573.06 × 184.25 573.06 \times 184.25 573.06 × 184.25
A 105600.00 B 105622.00 C 105500.00 D 105532.00
Worked solution (try it first) Multiply:
573.06 × 184.25 = 105 586.305 573.06 \times 184.25 = 105\,586.305 573.06 × 184.25 = 105 586.305 .
Four significant figures are 1, 0, 5 and 5.
The next figure is 8, so round up: 1056.
Fill the dropped places with zeros:
105 600 105\,600 105 600 , which is 105600.00, option A.
Watch out
Round, don't chop: the 8 after 1055 rounds it up to 1056. Cutting it off gives 105 500 (option C). Report a problem with this question
Change 432 five 432_{\text{five}} 43 2 five
A 10100 three 10100_{\text{three}} 1010 0 three B 11100 three 11100_{\text{three}} 1110 0 three C 11101 three 11101_{\text{three}} 1110 1 three D 10110 three 10110_{\text{three}} 1011 0 three
Worked solution (try it first) Change to base ten:
432 five = 4 × 25 + 3 × 5 + 2 432_{\text{five}} = 4 \times 25 + 3 \times 5 + 2 43 2 five = 4 × 25 + 3 × 5 + 2 Divide by 3 repeatedly:
117 = 3 × 39 + 0 117 = 3 \times 39 + 0 117 = 3 × 39 + 0 ,
39 = 3 × 13 + 0 39 = 3 \times 13 + 0 39 = 3 × 13 + 0 ,
13 = 3 × 4 + 1 13 = 3 \times 4 + 1 13 = 3 × 4 + 1 ,
4 = 3 × 1 + 1 4 = 3 \times 1 + 1 4 = 3 × 1 + 1 , and
1 = 3 × 0 + 1 1 = 3 \times 0 + 1 1 = 3 × 0 + 1 .
Read the remainders from the bottom up:
11100 three 11100_{\text{three}} 1110 0 three , option B.
Watch out
Zero remainders are digits too; keep them. Check by converting back: 11100 three = 81 + 27 + 9 = 117 11100_{\text{three}} = 81 + 27 + 9 = 117 1110 0 three = 81 + 27 + 9 = 117 11101 three 11101_{\text{three}} 1110 1 three Report a problem with this question
Given that A A A B B B n ( A ) = 8 n(A) = 8 n ( A ) = 8 n ( B ) = 12 n(B) = 12 n ( B ) = 12 n ( A ∩ B ) = 3 n(A \cap B) = 3 n ( A ∩ B ) = 3 n ( A ∪ B ) n(A \cup B) n ( A ∪ B )
Worked solution (try it first) Use
n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) n(A \cup B) = n(A) + n(B) - n(A \cap B) n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) .
So
n ( A ∪ B ) = 8 + 12 − 3 = 17 n(A \cup B) = 8 + 12 - 3 = 17 n ( A ∪ B ) = 8 + 12 − 3 = 17 , option B.
Watch out
Subtract the overlap: 8 + 12 = 20 8 + 12 = 20 8 + 12 = 20 Report a problem with this question
If 24 + 96 − 600 = y 6 \sqrt{24} + \sqrt{96} - \sqrt{600} = y\sqrt6 24 + 96 − 600 = y 6 y y y
Worked solution (try it first) Take out square factors:
24 = 2 6 \sqrt{24} = 2\sqrt6 24 = 2 6 ,
96 = 4 6 \sqrt{96} = 4\sqrt6 96 = 4 6 and
600 = 10 6 \sqrt{600} = 10\sqrt6 600 = 10 6 .
Combine:
2 6 + 4 6 − 10 6 = − 4 6 2\sqrt6 + 4\sqrt6 - 10\sqrt6 = -4\sqrt6 2 6 + 4 6 − 10 6 = − 4 6 .
So
y = − 4 y = -4 y = − 4 , option D.
Watch out
2 + 4 − 10 = − 4 2 + 4 - 10 = -4 2 + 4 − 10 = − 4 600 \sqrt{600} 600 y y y Report a problem with this question
Evaluate 23 × 54 ( m o d 7 ) 23 \times 54 \pmod 7 23 × 54 ( mod 7 )
Worked solution (try it first) Reduce each number:
23 = 3 × 7 + 2 23 = 3 \times 7 + 2 23 = 3 × 7 + 2 , so
23 ≡ 2 23 \equiv 2 23 ≡ 2 .
54 = 7 × 7 + 5 54 = 7 \times 7 + 5 54 = 7 × 7 + 5 , so
54 ≡ 5 ( m o d 7 ) 54 \equiv 5 \pmod 7 54 ≡ 5 ( mod 7 ) .
Multiply the remainders:
2 × 5 = 10 2 \times 5 = 10 2 × 5 = 10 .
Reduce again:
10 = 7 + 3 10 = 7 + 3 10 = 7 + 3 , so
23 × 54 ≡ 3 ( m o d 7 ) 23 \times 54 \equiv 3 \pmod 7 23 × 54 ≡ 3 ( mod 7 ) , option B.
Watch out
Multiply, don't add. Adding the remainders gives 2 + 5 = 7 ≡ 0 2 + 5 = 7 \equiv 0 2 + 5 = 7 ≡ 0 77 ≡ 0 77 \equiv 0 77 ≡ 0 Report a problem with this question
If 4 3 x = 16 x + 1 4^{3x} = 16^{x + 1} 4 3 x = 1 6 x + 1 x x x
Worked solution (try it first) Write 16 as
4 2 4^2 4 2 :
16 x + 1 = 4 2 ( x + 1 ) = 4 2 x + 2 16^{x + 1} = 4^{2(x + 1)} = 4^{2x + 2} 1 6 x + 1 = 4 2 ( x + 1 ) = 4 2 x + 2 .
The bases match, so
3 x = 2 x + 2 3x = 2x + 2 3 x = 2 x + 2 .
Subtract
2 x 2x 2 x :
x = 2 x = 2 x = 2 , option B.
Watch out
Multiply both terms of the index by 2: 2 ( x + 1 ) = 2 x + 2 2(x + 1) = 2x + 2 2 ( x + 1 ) = 2 x + 2 2 x + 1 2x + 1 2 x + 1 3 x = 2 x + 1 3x = 2x + 1 3 x = 2 x + 1 x = 1 x = 1 x = 1 Report a problem with this question
A weaver bought a bundle of grass for $50.00 from which he made 8 mats. If each mat was sold for $15.00, find the percentage profit.
A 240 % 240\% 240% B 140 % 140\% 140% C 120 % 120\% 120% D 40 % 40\% 40%
Worked solution (try it first) The 8 mats sell for
8 × 15 = 120 8 \times 15 = 120 8 × 15 = 120 dollars, $120.
The profit is
120 − 50 = 70 120 - 50 = 70 120 − 50 = 70 dollars, $70.
As a percentage of the cost:
70 50 × 100 % = 140 % \dfrac{70}{50} \times 100\% = 140\% 50 70 × 100% = 140% , option B.
Watch out
Profit is takings minus cost. 120 50 × 100 % = 240 % \frac{120}{50} \times 100\% = 240\% 50 120 × 100% = 240% Report a problem with this question
Find the 17th term of the arithmetic progression whose first three terms are − 6 , − 1 , 4 -6, -1, 4 − 6 , − 1 , 4
A − 91 -91 − 91 B − 86 -86 − 86 C 74 D 79
Worked solution (try it first) This is an A.P. with
a = − 6 a = -6 a = − 6 and
d = − 1 − ( − 6 ) = 5 d = -1 - (-6) = 5 d = − 1 − ( − 6 ) = 5 .
The 17th term is
a + 16 d = − 6 + 16 × 5 a + 16d = -6 + 16 \times 5 a + 16 d = − 6 + 16 × 5 .
So
T 17 = − 6 + 80 = 74 T_{17} = -6 + 80 = 74 T 17 = − 6 + 80 = 74 , option C.
Watch out
The 17th term has 16 d 16d 16 d 17 d 17d 17 d 17 d 17d 17 d − 6 + 85 = 79 -6 + 85 = 79 − 6 + 85 = 79 Report a problem with this question
M M M n n n p p p M = 3 M = 3 M = 3 n = 2 n = 2 n = 2 p = 1 p = 1 p = 1 M M M n n n p p p
A 3 n 2 p 2 \dfrac{3n}{2p^2} 2 p 2 3 n B 2 n 3 p 2 \dfrac{2n}{3p^2} 3 p 2 2 n C 2 n 3 p \dfrac{2n}{3p} 3 p 2 n D 3 n 2 2 p 2 \dfrac{3n^2}{2p^2} 2 p 2 3 n 2
Worked solution (try it first) Directly as
n n n , inversely as
p 2 p^2 p 2 :
M = k n p 2 M = \dfrac{kn}{p^2} M = p 2 k n .
Put in
M = 3 M = 3 M = 3 ,
n = 2 n = 2 n = 2 ,
p = 1 p = 1 p = 1 :
3 = 2 k 3 = 2k 3 = 2 k , so
k = 3 2 k = \frac32 k = 2 3 .
So
M = 3 n 2 p 2 M = \dfrac{3n}{2p^2} M = 2 p 2 3 n , option A.
Watch out
From 3 = 2 k 3 = 2k 3 = 2 k k = 3 2 k = \frac32 k = 2 3 2 3 \frac23 3 2 Report a problem with this question
If a = 3 a = 3 a = 3 b = − 7 b = -7 b = − 7 5 b + ( a + b ) 2 ( a − b ) 2 \dfrac{5b + (a + b)^2}{(a - b)^2} ( a − b ) 2 5 b + ( a + b ) 2
A 0.51 B 0.19 C − 0.19 -0.19 − 0.19 D − 0.51 -0.51 − 0.51
Worked solution (try it first) Put in
a = 3 a = 3 a = 3 and
b = − 7 b = -7 b = − 7 .
The top is
5 ( − 7 ) + ( 3 − 7 ) 2 = − 35 + 16 5(-7) + (3 - 7)^2 = -35 + 16 5 ( − 7 ) + ( 3 − 7 ) 2 = − 35 + 16 , which is
− 19 -19 − 19 .
The bottom is
( 3 − ( − 7 ) ) 2 = 10 2 = 100 (3 - (-7))^2 = 10^2 = 100 ( 3 − ( − 7 ) ) 2 = 1 0 2 = 100 .
So the value is
− 19 100 = − 0.19 \dfrac{-19}{100} = -0.19 100 − 19 = − 0.19 , option C.
Watch out
A square is never negative: ( a + b ) 2 = ( − 4 ) 2 = + 16 (a + b)^2 = (-4)^2 = +16 ( a + b ) 2 = ( − 4 ) 2 = + 16 − 16 -16 − 16 − 51 -51 − 51 − 0.51 -0.51 − 0.51 Report a problem with this question
Three boys shared D10,500.00 in the ratio 6 : 7 : 8 6 : 7 : 8 6 : 7 : 8
A D4,000.00 B D5,000.00 C D4,500.00 D D3,500.00
Worked solution (try it first) Add the ratio:
6 + 7 + 8 = 21 6 + 7 + 8 = 21 6 + 7 + 8 = 21 parts.
One part is
10 500 ÷ 21 = 500 10\,500 \div 21 = 500 10 500 ÷ 21 = 500 .
The largest share is 8 parts:
8 × 500 = 4000 8 \times 500 = 4000 8 × 500 = 4000 , so D4,000.00, option A.
Watch out
The largest share is the 8 parts. Using 7 parts gives 3500 (option D). Report a problem with this question
The length of a piece of stick is 1.75 m 1.75\text{ m} 1.75 m 1.80 m 1.80\text{ m} 1.80 m
A 4 4 7 % 4\frac47\% 4 7 4 % B 2 6 7 % 2\frac67\% 2 7 6 % C 2 7 9 % 2\frac79\% 2 9 7 % D 4 7 9 % 4\frac79\% 4 9 7 %
Worked solution (try it first) The error is
1.80 − 1.75 = 0.05 1.80 - 1.75 = 0.05 1.80 − 1.75 = 0.05 m.
Divide by the true length and multiply by 100:
0.05 1.75 × 100 % = 5 1.75 % \frac{0.05}{1.75} \times 100\% = \frac{5}{1.75}\% 1.75 0.05 × 100% = 1.75 5 % .
5 1.75 = 20 7 = 2 6 7 \frac{5}{1.75} = \frac{20}{7} = 2\frac67 1.75 5 = 7 20 = 2 7 6 , so the percentage error is
2 6 7 % 2\frac67\% 2 7 6 % , option B.
Watch out
Divide by the true length, 1.75 m, not the measured 1.80 m. Using 1.80 gives 5 1.8 = 2 7 9 % \frac{5}{1.8} = 2\frac79\% 1.8 5 = 2 9 7 % Report a problem with this question
If 5 x + 3 y = 4 5x + 3y = 4 5 x + 3 y = 4 5 x − 3 y = 2 5x - 3y = 2 5 x − 3 y = 2 ( 25 x 2 − 9 y 2 ) (25x^2 - 9y^2) ( 25 x 2 − 9 y 2 )
Worked solution (try it first) 25 x 2 − 9 y 2 25x^2 - 9y^2 25 x 2 − 9 y 2 is a difference of two squares:
( 5 x ) 2 − ( 3 y ) 2 = ( 5 x + 3 y ) ( 5 x − 3 y ) (5x)^2 - (3y)^2 = (5x + 3y)(5x - 3y) ( 5 x ) 2 − ( 3 y ) 2 = ( 5 x + 3 y ) ( 5 x − 3 y ) .
Put in the two given values:
4 × 2 = 8 4 \times 2 = 8 4 × 2 = 8 , option D.
Watch out
Multiply the two brackets, don't square and add. 4 2 + 2 2 = 20 4^2 + 2^2 = 20 4 2 + 2 2 = 20 25 x 2 − 9 y 2 25x^2 - 9y^2 25 x 2 − 9 y 2 x x x y y y Report a problem with this question
Mary has $3.00 more than Ben but $5.00 less than Jane. If Mary has $x x x
A $ ( 2 x − 8 ) \text{\textdollar}(2x - 8) $ ( 2 x − 8 ) B $ ( 2 x + 8 ) \text{\textdollar}(2x + 8) $ ( 2 x + 8 ) C $ ( 2 x − 2 ) \text{\textdollar}(2x - 2) $ ( 2 x − 2 ) D $ ( 2 x + 2 ) \text{\textdollar}(2x + 2) $ ( 2 x + 2 )
Worked solution (try it first) Mary has $3.00 more than Ben, so Ben has $
( x − 3 ) (x - 3) ( x − 3 ) .
Mary has $5.00 less than Jane, so Jane has $
( x + 5 ) (x + 5) ( x + 5 ) .
Together:
( x + 5 ) + ( x − 3 ) = 2 x + 2 (x + 5) + (x - 3) = 2x + 2 ( x + 5 ) + ( x − 3 ) = 2 x + 2 , so $
( 2 x + 2 ) (2x + 2) ( 2 x + 2 ) , option D.
Watch out
Mary has more than Ben, so Ben has less than Mary: x − 3 x - 3 x − 3 = x + 3 = x + 3 = x + 3 = x − 5 = x - 5 = x − 5 ( 2 x − 2 ) (2x - 2) ( 2 x − 2 ) Report a problem with this question
Consider the statements p p p q q q p ⇒ q p \Rightarrow q p ⇒ q
A If Stephen is good at Mathematics, then he is intelligent B If Stephen is not good at Mathematics, then he is not intelligent C If Stephen is not intelligent, then he is not good at Mathematics D If Stephen is not good at Mathematics, then he is intelligent
Worked solution (try it first) ⇒ q \Rightarrow q ⇒ q is equivalent to its contrapositive
∼ q \sim q ∼ q ⇒ ∼ p \Rightarrow \sim p ⇒∼ p : swap the parts and negate both.
∼ q \sim q ∼ q is "Stephen is not good at Mathematics" and
∼ p \sim p ∼ p is "he is not intelligent".
So the valid conclusion is "If Stephen is not good at Mathematics, then he is not intelligent", option B.
Watch out
Negating both parts without swapping them gives the inverse, ∼ p ⇒ ∼ q \sim p \Rightarrow \sim q ∼ p ⇒∼ q p ⇒ q p \Rightarrow q p ⇒ q Report a problem with this question
What value of p p p ( x 2 − 4 x + p ) (x^2 - 4x + p) ( x 2 − 4 x + p )
Worked solution (try it first) To complete the square, add the square of half the coefficient of
x x x .
Half of
− 4 -4 − 4 is
− 2 -2 − 2 , and
( − 2 ) 2 = 4 (-2)^2 = 4 ( − 2 ) 2 = 4 .
So
x 2 − 4 x + 4 = ( x − 2 ) 2 x^2 - 4x + 4 = (x - 2)^2 x 2 − 4 x + 4 = ( x − 2 ) 2 and
p = 4 p = 4 p = 4 , option C.
Watch out
Halve the coefficient before squaring: ( − 2 ) 2 = 4 (-2)^2 = 4 ( − 2 ) 2 = 4 − 4 -4 − 4 Report a problem with this question
Find the value of x x x 1 x + 4 3 x − 5 6 x + 1 = 0 \frac1x + \frac{4}{3x} - \frac{5}{6x} + 1 = 0 x 1 + 3 x 4 − 6 x 5 + 1 = 0
A 1 6 \frac16 6 1 B 1 4 \frac14 4 1 C − 3 2 -\frac32 − 2 3 D − 7 6 -\frac76 − 6 7
Worked solution (try it first) The LCM of the denominators
x x x ,
3 x 3x 3 x and
6 x 6x 6 x is
6 x 6x 6 x .
Multiply every term by
6 x 6x 6 x , including the 1:
6 + 8 − 5 + 6 x = 0 6 + 8 - 5 + 6x = 0 6 + 8 − 5 + 6 x = 0 .
Collect the numbers:
9 + 6 x = 0 9 + 6x = 0 9 + 6 x = 0 , so
6 x = − 9 6x = -9 6 x = − 9 .
Divide by 6:
x = − 9 6 = − 3 2 x = -\frac96 = -\frac32 x = − 6 9 = − 2 3 , option C.
Watch out
The + 1 +1 + 1 6 x 6x 6 x 6 x 6x 6 x x x x 10 = 0 10 = 0 10 = 0 Report a problem with this question
Make t t t k = m t − p r k = m\sqrt{\dfrac{t - p}{r}} k = m r t − p
A k 2 r + p m 2 \dfrac{k^2r + p}{m^2} m 2 k 2 r + p B k 2 r + p m 2 m 2 \dfrac{k^2r + pm^2}{m^2} m 2 k 2 r + p m 2 C k 2 r − p m 2 \dfrac{k^2r - p}{m^2} m 2 k 2 r − p D k 2 r + p 2 m 2 \dfrac{k^2r + p^2}{m^2} m 2 k 2 r + p 2
Worked solution (try it first) Divide by
m m m and square both sides:
k 2 m 2 = t − p r \frac{k^2}{m^2} = \frac{t - p}{r} m 2 k 2 = r t − p .
Multiply by
r r r :
t − p = k 2 r m 2 t - p = \frac{k^2r}{m^2} t − p = m 2 k 2 r .
Add
p p p , writing it over
m 2 m^2 m 2 :
t = k 2 r m 2 + p m 2 m 2 t = \frac{k^2r}{m^2} + \frac{pm^2}{m^2} t = m 2 k 2 r + m 2 p m 2 = k 2 r + p m 2 m 2 = \dfrac{k^2r + pm^2}{m^2} = m 2 k 2 r + p m 2 , option B.
Watch out
To add p p p m 2 m^2 m 2 p m 2 m 2 \frac{pm^2}{m^2} m 2 p m 2 p p p k 2 r + p m 2 \frac{k^2r + p}{m^2} m 2 k 2 r + p Report a problem with this question
In the diagram, ∣ X Y ∣ = ∣ Y Z ∣ |XY| = |YZ| ∣ X Y ∣ = ∣ Y Z ∣ ∠ X Y Z = 130 ∘ \angle XYZ = 130^\circ ∠ X Y Z = 13 0 ∘ y y y
A 50 ∘ 50^\circ 5 0 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 25 ∘ 25^\circ 2 5 ∘ D 155 ∘ 155^\circ 15 5 ∘
Worked solution (try it first) ∣ X Y ∣ = ∣ Y Z ∣ |XY| = |YZ| ∣ X Y ∣ = ∣ Y Z ∣ , so triangle
X Y Z XYZ X Y Z is isosceles and its base angles at
X X X and
Z Z Z are equal.
The angles of the triangle add up to
180 ∘ 180^\circ 18 0 ∘ : each base angle is
180 ∘ − 130 ∘ 2 = 25 ∘ \frac{180^\circ - 130^\circ}{2} = 25^\circ 2 18 0 ∘ − 13 0 ∘ = 2 5 ∘ , so
∠ Y Z X = 25 ∘ \angle YZX = 25^\circ ∠ Y Z X = 2 5 ∘ .
W Z X WZX W Z X is a straight line, so
y = 180 ∘ − 25 ∘ = 155 ∘ y = 180^\circ - 25^\circ = 155^\circ y = 18 0 ∘ − 2 5 ∘ = 15 5 ∘ , option D.
Watch out
25 ∘ 25^\circ 2 5 ∘ ∠ Y Z X \angle YZX ∠ Y Z X y y y Z Y ZY Z Y 25 ∘ 25^\circ 2 5 ∘ 180 ∘ 180^\circ 18 0 ∘ Report a problem with this question
An exterior angle of a regular polygon is 22.5 ∘ 22.5^\circ 22. 5 ∘
Worked solution (try it first) The exterior angles of a regular polygon are equal and add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 22.5 360 \div 22.5 360 ÷ 22.5 .
Double both numbers to clear the decimal:
720 ÷ 45 = 16 720 \div 45 = 16 720 ÷ 45 = 16 , option D.
Watch out
Divide by the exact angle. Rounding 22.5 ∘ 22.5^\circ 22. 5 ∘ 24 ∘ 24^\circ 2 4 ∘ Report a problem with this question
In the diagram, ∠ P O Q = 150 ∘ \angle POQ = 150^\circ ∠ P O Q = 15 0 ∘ P S Q R PSQR P S QR 4.2 cm 4.2\text{ cm} 4.2 cm P R Q PRQ P R Q [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 11.00 cm 11.00\text{ cm} 11.00 cm B 15.40 cm 15.40\text{ cm} 15.40 cm C 17.64 cm 17.64\text{ cm} 17.64 cm D 23.10 cm 23.10\text{ cm} 23.10 cm
Worked solution (try it first) Arc length
= θ 360 × 2 π r = \frac{\theta}{360} \times 2\pi r = 360 θ × 2 π r .
The circumference is
2 × 22 7 × 4.2 = 26.4 2 \times \frac{22}{7} \times 4.2 = 26.4 2 × 7 22 × 4.2 = 26.4 cm.
The minor arc is
150 360 \frac{150}{360} 360 150 of it:
5 12 × 26.4 = 11 \frac{5}{12} \times 26.4 = 11 12 5 × 26.4 = 11 cm, option A.
Watch out
An arc is a length, so use 2 π r 2\pi r 2 π r π r 2 \pi r^2 π r 2 23.10 23.10 23.10 Report a problem with this question
In the diagram, ∠ P O Q = 150 ∘ \angle POQ = 150^\circ ∠ P O Q = 15 0 ∘ P S Q R PSQR P S QR 4.2 cm 4.2\text{ cm} 4.2 cm O P S Q OPSQ O P S Q [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 15.40 cm 2 15.40\text{ cm}^2 15.40 cm 2 B 17.64 cm 2 17.64\text{ cm}^2 17.64 cm 2 C 23.10 cm 2 23.10\text{ cm}^2 23.10 cm 2 D 32.34 cm 2 32.34\text{ cm}^2 32.34 cm 2
Worked solution (try it first) Sector
O P S Q OPSQ O P S Q goes round through
S S S , away from the
150 ∘ 150^\circ 15 0 ∘ angle, so it is the major sector with angle
360 ∘ − 150 ∘ = 210 ∘ 360^\circ - 150^\circ = 210^\circ 36 0 ∘ − 15 0 ∘ = 21 0 ∘ .
Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 .
Here
π r 2 = 22 7 × 4.2 2 \pi r^2 = \frac{22}{7} \times 4.2^2 π r 2 = 7 22 × 4. 2 2 = 55.44 cm 2 = 55.44\text{ cm}^2 = 55.44 cm 2 .
So the area is
210 360 × 55.44 = 32.34 cm 2 \frac{210}{360} \times 55.44 = 32.34\text{ cm}^2 360 210 × 55.44 = 32.34 cm 2 , option D.
Watch out
Check which way round the sector goes. Using the 150 ∘ 150^\circ 15 0 ∘ 23.10 cm 2 23.10\text{ cm}^2 23.10 cm 2 S S S Report a problem with this question
A ladder 6 m 6\text{ m} 6 m 53 ∘ 53^\circ 5 3 ∘
A 3.611 m 3.611\text{ m} 3.611 m B 4.521 m 4.521\text{ m} 4.521 m C 4.792 m 4.792\text{ m} 4.792 m D 3.962 m 3.962\text{ m} 3.962 m
Worked solution (try it first) The ladder (6 m) is the hypotenuse, and the height up the wall is opposite the
53 ∘ 53^\circ 5 3 ∘ angle at the ground.
So the height is
6 sin 53 ∘ = 6 × 0.7986 6\sin53^\circ = 6 \times 0.7986 6 sin 5 3 ∘ = 6 × 0.7986 ≈ 4.792 \approx 4.792 ≈ 4.792 m, option C.
Watch out
The height is opposite the angle, so use sine. Cosine gives the distance of the foot from the wall, 6 cos 53 ∘ ≈ 3.611 6\cos53^\circ \approx 3.611 6 cos 5 3 ∘ ≈ 3.611 Report a problem with this question
A cylinder, open at one end, has a radius of 3.5 cm 3.5\text{ cm} 3.5 cm 8 cm 8\text{ cm} 8 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 126.5 cm 2 126.5\text{ cm}^2 126.5 cm 2 B 165.0 cm 2 165.0\text{ cm}^2 165.0 cm 2 C 212.0 cm 2 212.0\text{ cm}^2 212.0 cm 2 D 214.5 cm 2 214.5\text{ cm}^2 214.5 cm 2
Worked solution (try it first) Open at one end: the curved surface plus one circle.
Curved surface:
2 π r h = 2 × 22 7 × 3.5 × 8 2\pi rh = 2 \times \frac{22}{7} \times 3.5 \times 8 2 π r h = 2 × 7 22 × 3.5 × 8 = 176 cm 2 = 176\text{ cm}^2 = 176 cm 2 .
One end:
22 7 × 3.5 2 = 38.5 cm 2 \frac{22}{7} \times 3.5^2 = 38.5\text{ cm}^2 7 22 × 3. 5 2 = 38.5 cm 2 .
Total:
176 + 38.5 = 214.5 cm 2 176 + 38.5 = 214.5\text{ cm}^2 176 + 38.5 = 214.5 cm 2 , option D.
Watch out
The curved surface is 2 π r h 2\pi rh 2 π r h π r h = 88 \pi rh = 88 π r h = 88 88 + 38.5 = 126.5 cm 2 88 + 38.5 = 126.5\text{ cm}^2 88 + 38.5 = 126.5 cm 2 Report a problem with this question
In the diagram, ∠ W Z Y \angle WZY ∠ W Z Y ∠ W Y X \angle WYX ∠ W Y X W X Y Z WXYZ W X Y Z
A 30 cm 30\text{ cm} 30 cm B 32 cm 32\text{ cm} 32 cm C 35 cm 35\text{ cm} 35 cm D 37 cm 37\text{ cm} 37 cm
Worked solution (try it first) Triangle
W Z Y WZY W Z Y is right-angled at
Z Z Z , so
W Y = 4 2 + 3 2 = 5 WY = \sqrt{4^2 + 3^2} = 5 W Y = 4 2 + 3 2 = 5 cm.
Triangle
W Y X WYX W Y X is right-angled at
Y Y Y , so
W X = 5 2 + 12 2 = 13 WX = \sqrt{5^2 + 12^2} = 13 W X = 5 2 + 1 2 2 = 13 cm.
The perimeter goes round the outside only:
W X + X Y + Y Z + Z W = 13 + 12 + 3 + 4 = 32 WX + XY + YZ + ZW = 13 + 12 + 3 + 4 = 32 W X + X Y + Y Z + Z W = 13 + 12 + 3 + 4 = 32 cm, option B.
Watch out
W Y WY W Y 32 + 5 = 37 32 + 5 = 37 32 + 5 = 37 Report a problem with this question
The length of a rectangle is 10 cm 10\text{ cm} 10 cm 28 cm 28\text{ cm} 28 cm
A 30 cm 2 30\text{ cm}^2 30 cm 2 B 40 cm 2 40\text{ cm}^2 40 cm 2 C 60 cm 2 60\text{ cm}^2 60 cm 2 D 80 cm 2 80\text{ cm}^2 80 cm 2
Worked solution (try it first) Half the perimeter is length + width:
28 ÷ 2 = 14 28 \div 2 = 14 28 ÷ 2 = 14 cm.
So the width is
14 − 10 = 4 14 - 10 = 4 14 − 10 = 4 cm.
Area
= 10 × 4 = 40 cm 2 = 10 \times 4 = 40\text{ cm}^2 = 10 × 4 = 40 cm 2 , option B.
Watch out
28 − 2 × 10 = 8 28 - 2 \times 10 = 8 28 − 2 × 10 = 8 80 cm 2 80\text{ cm}^2 80 cm 2 Report a problem with this question
In the diagram, M R W MRW M R W M N S T MNST M N S T ∣ M N ∣ = ∣ N R ∣ |MN| = |NR| ∣ M N ∣ = ∣ N R ∣ ∠ M N R = 110 ∘ \angle MNR = 110^\circ ∠ M N R = 11 0 ∘ ∠ W R S = 86 ∘ \angle WRS = 86^\circ ∠ W R S = 8 6 ∘ x x x
A 86 ∘ 86^\circ 8 6 ∘ B 70 ∘ 70^\circ 7 0 ∘ C 51 ∘ 51^\circ 5 1 ∘ D 42 ∘ 42^\circ 4 2 ∘
Worked solution (try it first) ∣ M N ∣ = ∣ N R ∣ |MN| = |NR| ∣ M N ∣ = ∣ N R ∣ , so triangle
M N R MNR M N R is isosceles and
∠ N M R = ∠ N R M \angle NMR = \angle NRM ∠ N M R = ∠ N R M .
The angles of triangle
M N R MNR M N R add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ N M R = 180 ∘ − 110 ∘ 2 \angle NMR = \frac{180^\circ - 110^\circ}{2} ∠ N M R = 2 18 0 ∘ − 11 0 ∘ M R W MRW M R W is a straight line, so
∠ W R S \angle WRS ∠ W R S is an exterior angle of triangle
M R S MRS M R S .
It equals the two interior opposite angles added:
86 ∘ = 35 ∘ + x 86^\circ = 35^\circ + x 8 6 ∘ = 3 5 ∘ + x .
Subtract
35 ∘ 35^\circ 3 5 ∘ :
x = 51 ∘ x = 51^\circ x = 5 1 ∘ , option C.
Watch out
∠ W R S \angle WRS ∠ W R S M R S MRS M R S 35 ∘ + x 35^\circ + x 3 5 ∘ + x 86 ∘ 86^\circ 8 6 ∘ x = 180 ∘ − 86 ∘ − 35 ∘ = 59 ∘ x = 180^\circ - 86^\circ - 35^\circ = 59^\circ x = 18 0 ∘ − 8 6 ∘ − 3 5 ∘ = 5 9 ∘ Report a problem with this question
A boy 1.4 m 1.4\text{ m} 1.4 m 10 m 10\text{ m} 10 m 12 m 12\text{ m} 12 m
A 71 ∘ 71^\circ 7 1 ∘ B 47 ∘ 47^\circ 4 7 ∘ C 19 ∘ 19^\circ 1 9 ∘ D 8 ∘ 8^\circ 8 ∘
Worked solution (try it first) Measure from the boy's eyes: the top of the tree is
12 − 1.4 = 10.6 12 - 1.4 = 10.6 12 − 1.4 = 10.6 m higher, and 10 m away.
So
tan θ = 10.6 10 = 1.06 \tan\theta = \frac{10.6}{10} = 1.06 tan θ = 10 10.6 = 1.06 .
So
θ = tan − 1 1.06 ≈ 46.7 ∘ \theta = \tan^{-1}1.06 \approx 46.7^\circ θ = tan − 1 1.06 ≈ 46. 7 ∘ , which is
47 ∘ 47^\circ 4 7 ∘ to the nearest degree, option B.
Watch out
Take off the boy's height first: 12 − 1.4 = 10.6 12 - 1.4 = 10.6 12 − 1.4 = 10.6 50 ∘ 50^\circ 5 0 ∘ Report a problem with this question
Given that sin ( 5 x − 28 ) ∘ = cos ( 3 x − 50 ) ∘ \sin(5x - 28)^\circ = \cos(3x - 50)^\circ sin ( 5 x − 28 ) ∘ = cos ( 3 x − 50 ) ∘ 0 ∘ ≤ x ≤ 90 ∘ 0^\circ \le x \le 90^\circ 0 ∘ ≤ x ≤ 9 0 ∘ x x x
Worked solution (try it first) sin A = cos B \sin A = \cos B sin A = cos B when
A A A and
B B B are complementary,
A + B = 90 ∘ A + B = 90^\circ A + B = 9 0 ∘ .
So
( 5 x − 28 ) + ( 3 x − 50 ) = 90 (5x - 28) + (3x - 50) = 90 ( 5 x − 28 ) + ( 3 x − 50 ) = 90 .
Collect terms:
8 x − 78 = 90 8x - 78 = 90 8 x − 78 = 90 , so
8 x = 168 8x = 168 8 x = 168 .
Divide by 8:
x = 21 x = 21 x = 21 , option C.
(Check: the angles are
77 ∘ 77^\circ 7 7 ∘ and
13 ∘ 13^\circ 1 3 ∘ , which add to
90 ∘ 90^\circ 9 0 ∘ .)
Watch out
The two angles add up to 90 ∘ 90^\circ 9 0 ∘ 5 x − 28 = 3 x − 50 5x - 28 = 3x - 50 5 x − 28 = 3 x − 50 x = − 11 x = -11 x = − 11 Report a problem with this question
In the diagram, M N R MNR M N R N N N ∠ N O S = 108 ∘ \angle NOS = 108^\circ ∠ N O S = 10 8 ∘ ∠ O S N \angle OSN ∠ O S N
A 72 ∘ 72^\circ 7 2 ∘ B 42 ∘ 42^\circ 4 2 ∘ C 36 ∘ 36^\circ 3 6 ∘ D 18 ∘ 18^\circ 1 8 ∘
Worked solution (try it first) O N = O S ON = OS O N = O S (radii), so triangle
N O S NOS N O S is isosceles and its base angles at
N N N and
S S S are equal.
So
∠ O S N = 180 ∘ − 108 ∘ 2 \angle OSN = \dfrac{180^\circ - 108^\circ}{2} ∠ O S N = 2 18 0 ∘ − 10 8 ∘ = 36 ∘ = 36^\circ = 3 6 ∘ , option C.
The tangent is not needed.
Watch out
180 ∘ − 108 ∘ = 72 ∘ 180^\circ - 108^\circ = 72^\circ 18 0 ∘ − 10 8 ∘ = 7 2 ∘ Report a problem with this question
In the diagram, M N R MNR M N R O O O N N N ∠ N O S = 108 ∘ \angle NOS = 108^\circ ∠ N O S = 10 8 ∘ ∠ S N R \angle SNR ∠ S N R
A 36 ∘ 36^\circ 3 6 ∘ B 42 ∘ 42^\circ 4 2 ∘ C 54 ∘ 54^\circ 5 4 ∘ D 72 ∘ 72^\circ 7 2 ∘
Worked solution (try it first) O N = O S ON = OS O N = O S (radii), so triangle
N O S NOS N O S is isosceles:
∠ O N S = 180 ∘ − 108 ∘ 2 \angle ONS = \frac{180^\circ - 108^\circ}{2} ∠ O N S = 2 18 0 ∘ − 10 8 ∘ A tangent is perpendicular to the radius at the point of contact, so
∠ O N R = 90 ∘ \angle ONR = 90^\circ ∠ O N R = 9 0 ∘ .
So
∠ S N R = 90 ∘ − 36 ∘ \angle SNR = 90^\circ - 36^\circ ∠ S N R = 9 0 ∘ − 3 6 ∘ = 54 ∘ = 54^\circ = 5 4 ∘ , option C.
(Check: the angle between a tangent and a chord is half the angle at the centre,
1 2 × 108 ∘ = 54 ∘ \frac12 \times 108^\circ = 54^\circ 2 1 × 10 8 ∘ = 5 4 ∘ .)
Watch out
36 ∘ 36^\circ 3 6 ∘ ∠ O N S \angle ONS ∠ O N S ∠ S N R \angle SNR ∠ S N R O N ON O N 90 ∘ − 36 ∘ = 54 ∘ 90^\circ - 36^\circ = 54^\circ 9 0 ∘ − 3 6 ∘ = 5 4 ∘ Report a problem with this question
Mrs Gabriel is pregnant. The probability that she will give birth to a girl is 1 2 \frac12 2 1 1 4 \frac14 4 1
A 1 B 3 4 \frac34 4 3 C 1 8 \frac18 8 1 D 1 4 \frac14 4 1
Worked solution (try it first) The baby's sex and eye colour are independent.
For "a girl and blue eyes", multiply:
1 2 × 1 4 = 1 8 \frac12 \times \frac14 = \frac18 2 1 × 4 1 = 8 1 , option C.
Watch out
"And" means multiply. Adding gives 1 2 + 1 4 = 3 4 \frac12 + \frac14 = \frac34 2 1 + 4 1 = 4 3 Report a problem with this question
The mean of a set of 10 numbers is 56. If the mean of the first nine numbers is 55, find the 10th number.
Worked solution (try it first) Total of all 10 numbers:
10 × 56 = 560 10 \times 56 = 560 10 × 56 = 560 .
Total of the first nine:
9 × 55 = 495 9 \times 55 = 495 9 × 55 = 495 .
The 10th number is the difference:
560 − 495 = 65 560 - 495 = 65 560 − 495 = 65 , option B.
Watch out
Work with totals. The 10th number isn't just the higher mean: it must lift each of the first nine by 1 as well, so it is 56 + 9 = 65 56 + 9 = 65 56 + 9 = 65 Report a problem with this question
Simplify 2 − 18 m 2 1 + 3 m \dfrac{2 - 18m^2}{1 + 3m} 1 + 3 m 2 − 18 m 2
A 2 ( 1 + 3 m ) 2(1 + 3m) 2 ( 1 + 3 m ) B 2 ( 1 + 3 m 2 ) 2(1 + 3m^2) 2 ( 1 + 3 m 2 ) C 2 ( 1 − 3 m ) 2(1 - 3m) 2 ( 1 − 3 m ) D 2 ( 1 − 3 m 2 ) 2(1 - 3m^2) 2 ( 1 − 3 m 2 )
Worked solution (try it first) Take out 2 on top:
2 − 18 m 2 = 2 ( 1 − 9 m 2 ) 2 - 18m^2 = 2(1 - 9m^2) 2 − 18 m 2 = 2 ( 1 − 9 m 2 ) .
Use the difference of two squares:
1 − 9 m 2 = ( 1 − 3 m ) ( 1 + 3 m ) 1 - 9m^2 = (1 - 3m)(1 + 3m) 1 − 9 m 2 = ( 1 − 3 m ) ( 1 + 3 m ) .
Cancel
1 + 3 m 1 + 3m 1 + 3 m :
2 ( 1 − 3 m ) 2(1 - 3m) 2 ( 1 − 3 m ) , option C.
Watch out
The bracket that cancels is 1 + 3 m 1 + 3m 1 + 3 m 1 − 3 m 1 - 3m 1 − 3 m 1 + 3 m 1 + 3m 1 + 3 m Report a problem with this question
The diagram shows triangle P Q R PQR P QR P S PS P S P P P ∠ Q P R = 58 ∘ \angle QPR = 58^\circ ∠ QP R = 5 8 ∘ ∠ R P S = 73 ∘ \angle RPS = 73^\circ ∠ R P S = 7 3 ∘ ∠ P R Q \angle PRQ ∠ P R Q
A 49 ∘ 49^\circ 4 9 ∘ B 58 ∘ 58^\circ 5 8 ∘ C 73 ∘ 73^\circ 7 3 ∘ D 131 ∘ 131^\circ 13 1 ∘
Worked solution (try it first) The angle between a tangent and a chord equals the angle in the alternate segment.
So
∠ P Q R = ∠ R P S = 73 ∘ \angle PQR = \angle RPS = 73^\circ ∠ P QR = ∠ R P S = 7 3 ∘ .
The angles of triangle
P Q R PQR P QR add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P R Q = 180 ∘ − 58 ∘ − 73 ∘ \angle PRQ = 180^\circ - 58^\circ - 73^\circ ∠ P R Q = 18 0 ∘ − 5 8 ∘ − 7 3 ∘ .
So
∠ P R Q = 49 ∘ \angle PRQ = 49^\circ ∠ P R Q = 4 9 ∘ , option A.
Watch out
The 73 ∘ 73^\circ 7 3 ∘ P R PR P R Q Q Q P R PR P R R R R ∠ P R Q = 73 ∘ \angle PRQ = 73^\circ ∠ P R Q = 7 3 ∘ Report a problem with this question
In the diagram, triangle M N R MNR M N R M N R MNR M N R P Q PQ P Q ∠ M R N = 41 ∘ \angle MRN = 41^\circ ∠ M R N = 4 1 ∘ ∠ P M R = 141 ∘ \angle PMR = 141^\circ ∠ P M R = 14 1 ∘ ∠ Q N R \angle QNR ∠ QN R
A 39 ∘ 39^\circ 3 9 ∘ B 80 ∘ 80^\circ 8 0 ∘ C 110 ∘ 110^\circ 11 0 ∘ D 141 ∘ 141^\circ 14 1 ∘
Worked solution (try it first) P M N Q PMNQ P M N Q is a straight line:
∠ R M N = 180 ∘ − 141 ∘ \angle RMN = 180^\circ - 141^\circ ∠ R M N = 18 0 ∘ − 14 1 ∘ ∠ Q N R \angle QNR ∠ QN R is an exterior angle of triangle
M N R MNR M N R , so it equals the sum of the two interior opposite angles.
So
∠ Q N R = 39 ∘ + 41 ∘ \angle QNR = 39^\circ + 41^\circ ∠ QN R = 3 9 ∘ + 4 1 ∘ = 80 ∘ = 80^\circ = 8 0 ∘ , option B.
Watch out
39 ∘ 39^\circ 3 9 ∘ ∠ R M N \angle RMN ∠ R M N ∠ Q N R \angle QNR ∠ QN R N N N 39 ∘ 39^\circ 3 9 ∘ 41 ∘ 41^\circ 4 1 ∘ Report a problem with this question
Solve y + 2 4 − y − 1 3 > 1 \dfrac{y + 2}{4} - \dfrac{y - 1}{3} > 1 4 y + 2 − 3 y − 1 > 1
A y < − 10 y < -10 y < − 10 B y < − 2 y < -2 y < − 2 C y < 2 y < 2 y < 2 D y < 10 y < 10 y < 10
Worked solution (try it first) Multiply every term by 12, the LCM of 4 and 3:
3 ( y + 2 ) − 4 ( y − 1 ) > 12 3(y + 2) - 4(y - 1) > 12 3 ( y + 2 ) − 4 ( y − 1 ) > 12 .
Expand, taking care with the minus:
3 y + 6 − 4 y + 4 > 12 3y + 6 - 4y + 4 > 12 3 y + 6 − 4 y + 4 > 12 , so
− y + 10 > 12 -y + 10 > 12 − y + 10 > 12 .
Subtract 10 from both sides:
− y > 2 -y > 2 − y > 2 .
Multiply by
− 1 -1 − 1 and reverse the sign:
y < − 2 y < -2 y < − 2 , option B.
Watch out
The minus multiplies the whole bracket: − 4 ( y − 1 ) = − 4 y + 4 -4(y - 1) = -4y + 4 − 4 ( y − 1 ) = − 4 y + 4 − 4 y − 4 -4y - 4 − 4 y − 4 − y > 10 -y > 10 − y > 10 y < − 10 y < -10 y < − 10 Report a problem with this question
The ages in years of some members of a singing group are 12, 47, 49, 15, 43, 41, 13, 39, 43, 41 and 36. Find the lower quartile.
Worked solution (try it first) Put the 11 ages in order: 12, 13, 15, 36, 39, 41, 41, 43, 43, 47, 49.
Q 1 Q_1 Q 1 is at position
11 4 = 2.75 \frac{11}{4} = 2.75 4 11 = 2.75 .
At .75, round up to the 3rd value.
So
Q 1 = 15 Q_1 = 15 Q 1 = 15 , option C.
Watch out
Round the position 2.75 up, not down. Rounding down takes the 2nd value, 13 (option B). Report a problem with this question
The ages in years of some members of a singing group are 12, 47, 49, 15, 43, 41, 13, 39, 43, 41 and 36. Find the mean.
Worked solution (try it first) Add the 11 ages: the total is 379.
Divide by the 11 members:
379 11 = 34.4545 … \frac{379}{11} = 34.4545\ldots 11 379 = 34.4545 … .
To two decimal places the mean is 34.45, option C.
Watch out
Count the ages: there are 11, not 10. Dividing 379 by 10 gives 37.9, which is not an option. Report a problem with this question
Find, correct to two decimal places, the volume of a sphere whose radius is 3 cm 3\text{ cm} 3 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 72.57 cm 3 72.57\text{ cm}^3 72.57 cm 3 B 88.12 cm 3 88.12\text{ cm}^3 88.12 cm 3 C 105.29 cm 3 105.29\text{ cm}^3 105.29 cm 3 D 113.14 cm 3 113.14\text{ cm}^3 113.14 cm 3
Worked solution (try it first) Volume of a sphere:
4 3 π r 3 \frac43\pi r^3 3 4 π r 3 , and
3 3 = 27 3^3 = 27 3 3 = 27 .
4 3 × 22 7 × 27 = 792 7 \frac43 \times \frac{22}{7} \times 27 = \frac{792}{7} 3 4 × 7 22 × 27 = 7 792 ≈ 113.14 cm 3 \approx 113.14\text{ cm}^3 ≈ 113.14 cm 3 , option D.
Watch out
Cube the radius. With 3 2 = 9 3^2 = 9 3 2 = 9 37.71 cm 3 37.71\text{ cm}^3 37.71 cm 3 Report a problem with this question
The lengths of the parallel sides of a trapezium are 9 cm 9\text{ cm} 9 cm 12 cm 12\text{ cm} 12 cm 105 cm 2 105\text{ cm}^2 105 cm 2
A 5 cm 5\text{ cm} 5 cm B 7 cm 7\text{ cm} 7 cm C 10 cm 10\text{ cm} 10 cm D 15 cm 15\text{ cm} 15 cm
Worked solution (try it first) Area of a trapezium
= 1 2 ( a + b ) h = \frac12(a + b)h = 2 1 ( a + b ) h :
1 2 ( 9 + 12 ) h = 105 \frac12(9 + 12)h = 105 2 1 ( 9 + 12 ) h = 105 .
So
10.5 h = 105 10.5h = 105 10.5 h = 105 and
h = 10 h = 10 h = 10 cm, option C.
Watch out
Keep the 1 2 \frac12 2 1 21 h = 105 21h = 105 21 h = 105 h = 5 h = 5 h = 5 Report a problem with this question
Find the volume of a cone of radius 3.5 cm 3.5\text{ cm} 3.5 cm 12 cm 12\text{ cm} 12 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 15.5 cm 3 15.5\text{ cm}^3 15.5 cm 3 B 21.0 cm 3 21.0\text{ cm}^3 21.0 cm 3 C 142.0 cm 3 142.0\text{ cm}^3 142.0 cm 3 D 154.0 cm 3 154.0\text{ cm}^3 154.0 cm 3
Worked solution (try it first) Volume of a cone:
1 3 π r 2 h \frac13\pi r^2h 3 1 π r 2 h , and
3.5 2 = 12.25 3.5^2 = 12.25 3. 5 2 = 12.25 .
1 3 × 22 7 × 12.25 × 12 = 22 7 × 49 \frac13 \times \frac{22}{7} \times 12.25 \times 12 = \frac{22}{7} \times 49 3 1 × 7 22 × 12.25 × 12 = 7 22 × 49 = 154 cm 3 = 154\text{ cm}^3 = 154 cm 3 , option D.
Watch out
Keep the 1 3 \frac13 3 1 462 cm 3 462\text{ cm}^3 462 cm 3 Report a problem with this question
A local community has two newspapers, the Morning Times and the Evening Dispatch. The Morning Times is read by 45 % 45\% 45% 60 % 60\% 60%
Worked solution (try it first) As probabilities:
P ( M ) = 0.45 P(M) = 0.45 P ( M ) = 0.45 ,
P ( E ) = 0.60 P(E) = 0.60 P ( E ) = 0.60 and
P ( M ∩ E ) = 0.20 P(M \cap E) = 0.20 P ( M ∩ E ) = 0.20 .
At least one paper is
P ( M ∪ E ) = P ( M ) + P ( E ) − P ( M ∩ E ) P(M \cup E) = P(M) + P(E) - P(M \cap E) P ( M ∪ E ) = P ( M ) + P ( E ) − P ( M ∩ E ) , which takes off the overlap counted twice.
So
0.45 + 0.60 − 0.20 = 0.85 0.45 + 0.60 - 0.20 = 0.85 0.45 + 0.60 − 0.20 = 0.85 , option C.
Watch out
"At least one" includes households that read both. Counting only one paper, 0.25 + 0.40 0.25 + 0.40 0.25 + 0.40 Report a problem with this question
A rectangle has width 3 4 cm \frac34\text{ cm} 4 3 cm 3 3 8 cm 2 3\frac38\text{ cm}^2 3 8 3 cm 2
A 6 cm 6\text{ cm} 6 cm B 4 1 2 cm 4\frac12\text{ cm} 4 2 1 cm C 2 5 8 cm 2\frac58\text{ cm} 2 8 5 cm D 12 cm 12\text{ cm} 12 cm
Worked solution (try it first) Length
= = = area
÷ \div ÷ width.
Write
3 3 8 3\frac38 3 8 3 as
27 8 \frac{27}{8} 8 27 .
Dividing by
3 4 \frac34 4 3 means multiplying by
4 3 \frac43 3 4 :
27 8 × 4 3 = 9 2 \frac{27}{8} \times \frac43 = \frac{9}{2} 8 27 × 3 4 = 2 9 .
So the length is
4 1 2 4\frac12 4 2 1 cm, option B.
Watch out
Divide the area by the width, don't multiply. 27 8 × 3 4 = 81 32 ≈ 2.53 \frac{27}{8} \times \frac34 = \frac{81}{32} \approx 2.53 8 27 × 4 3 = 32 81 ≈ 2.53 Report a problem with this question
The mean of two numbers x x x y y y x x x 2 x 2x 2 x y y y 2 y 2y 2 y
Worked solution (try it first) The mean of
x x x and
y y y is 4, so
x + y = 2 × 4 = 8 x + y = 2 \times 4 = 8 x + y = 2 × 4 = 8 .
The four numbers add up to
x + 2 x + y + 2 y = 3 ( x + y ) = 24 x + 2x + y + 2y = 3(x + y) = 24 x + 2 x + y + 2 y = 3 ( x + y ) = 24 .
Their mean is
24 4 = 6 \frac{24}{4} = 6 4 24 = 6 , option C.
Watch out
Divide by 4, the number of numbers. 8 (option D) is x + y x + y x + y Report a problem with this question
The straight line y = m x − 4 y = mx - 4 y = m x − 4 ( − 4 , 16 ) (-4, 16) ( − 4 , 16 )
Worked solution (try it first) The point is on the line, so put
x = − 4 x = -4 x = − 4 and
y = 16 y = 16 y = 16 into
y = m x − 4 y = mx - 4 y = m x − 4 :
16 = − 4 m − 4 16 = -4m - 4 16 = − 4 m − 4 .
Add 4 to both sides:
20 = − 4 m 20 = -4m 20 = − 4 m .
Divide by
− 4 -4 − 4 :
m = − 5 m = -5 m = − 5 , option A.
Watch out
Dividing 20 20 20 − 4 -4 − 4 Report a problem with this question
If the equations x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 x 2 + p x + 6 = 0 x^2 + px + 6 = 0 x 2 + p x + 6 = 0 p p p
Worked solution (try it first) Factorise the first equation:
( x − 2 ) ( x − 3 ) = 0 (x - 2)(x - 3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so the roots are 2 and 3.
For
x 2 + p x + 6 = 0 x^2 + px + 6 = 0 x 2 + p x + 6 = 0 , the sum of the roots is
− p -p − p .
So
− p = 2 + 3 = 5 -p = 2 + 3 = 5 − p = 2 + 3 = 5 .
So
p = − 5 p = -5 p = − 5 , option D.
Watch out
The sum of the roots is − p -p − p p p p p = 5 p = 5 p = 5 − 2 -2 − 2 − 3 -3 − 3 Report a problem with this question
A trader made a loss of 15 % 15\% 15%
A 3 : 20 3 : 20 3 : 20 B 3 : 17 3 : 17 3 : 17 C 17 : 20 17 : 20 17 : 20 D 20 : 23 20 : 23 20 : 23
Worked solution (try it first) A
15 % 15\% 15% loss means the selling price is
100 % − 15 % = 85 % 100\% - 15\% = 85\% 100% − 15% = 85% of the cost price.
So selling price : cost price
= 85 : 100 = 85 : 100 = 85 : 100 .
Divide both by 5:
17 : 20 17 : 20 17 : 20 , option C.
Watch out
15 : 100 = 3 : 20 15 : 100 = 3 : 20 15 : 100 = 3 : 20 85 % 85\% 85% Report a problem with this question
Given that log 3 27 = 2 x + 1 \log_3 27 = 2x + 1 log 3 27 = 2 x + 1 x x x
Worked solution (try it first) 27 = 3 3 27 = 3^3 27 = 3 3 , so
log 3 27 = 3 \log_3 27 = 3 log 3 27 = 3 .
So
2 x + 1 = 3 2x + 1 = 3 2 x + 1 = 3 .
Subtract 1 and divide by 2:
x = 1 x = 1 x = 1 , option B.
Watch out
3 is the value of log 3 27 \log_3 27 log 3 27 2 x + 1 2x + 1 2 x + 1 x = 3 x = 3 x = 3 2 x + 1 = 3 2x + 1 = 3 2 x + 1 = 3 Report a problem with this question
Solve 6 x 2 = 5 x − 1 6x^2 = 5x - 1 6 x 2 = 5 x − 1
A x = 2 , 3 x = 2, 3 x = 2 , 3 B x = 0 , 3 x = 0, 3 x = 0 , 3 C x = 1 2 , 1 3 x = \frac12, \frac13 x = 2 1 , 3 1 D x = 1 2 , − 1 3 x = \frac12, -\frac13 x = 2 1 , − 3 1
Worked solution (try it first) Bring everything to one side:
6 x 2 − 5 x + 1 = 0 6x^2 - 5x + 1 = 0 6 x 2 − 5 x + 1 = 0 .
Factorise:
( 2 x − 1 ) ( 3 x − 1 ) = 0 (2x - 1)(3x - 1) = 0 ( 2 x − 1 ) ( 3 x − 1 ) = 0 .
So
x = 1 2 x = \frac12 x = 2 1 or
x = 1 3 x = \frac13 x = 3 1 , option C.
Watch out
2 x − 1 = 0 2x - 1 = 0 2 x − 1 = 0 x = 1 2 x = \frac12 x = 2 1 Report a problem with this question