Find ( 101 2 ) 2 (101_2)^2 ( 10 1 2 ) 2
A 11101 2 11101_2 1110 1 2 B 10101 2 10101_2 1010 1 2 C 10010 2 10010_2 1001 0 2 D 11001 2 11001_2 1100 1 2
Worked solution (try it first) Change to base ten:
101 2 = 4 + 1 = 5 101_2 = 4 + 1 = 5 10 1 2 = 4 + 1 = 5 .
Square it:
5 2 = 25 5^2 = 25 5 2 = 25 .
Change back:
25 = 16 + 8 + 1 25 = 16 + 8 + 1 25 = 16 + 8 + 1 , so the answer is
11001 2 11001_2 1100 1 2 , option D.
Watch out
Squaring means 5 × 5 5 \times 5 5 × 5 5 × 2 5 \times 2 5 × 2 10 = 1010 2 10 = 1010_2 10 = 101 0 2 16 + 8 + 1 = 25 16 + 8 + 1 = 25 16 + 8 + 1 = 25 Report a problem with this question
If three children share ₦10.50 among themselves in the ratio 6 : 7 : 8 6 : 7 : 8 6 : 7 : 8
Worked solution (try it first) The ratio
6 : 7 : 8 6 : 7 : 8 6 : 7 : 8 has
6 + 7 + 8 = 21 6 + 7 + 8 = 21 6 + 7 + 8 = 21 parts, so one part is
10.50 ÷ 21 = 10.50 \div 21 = 10.50 ÷ 21 = ₦0.50.
The largest share is 8 parts:
8 × 0.50 = 8 \times 0.50 = 8 × 0.50 = ₦4.00, option B.
Watch out
The largest share goes with the largest number, 8. Using 7 parts gives ₦3.50 (option A), the middle share. Report a problem with this question
Express 0.000834 0.000834 0.000834
A 8.34 × 10 − 4 8.34 \times 10^{-4} 8.34 × 1 0 − 4 B 8.34 × 10 − 3 8.34 \times 10^{-3} 8.34 × 1 0 − 3 C 8.34 × 10 4 8.34 \times 10^{4} 8.34 × 1 0 4 D 8.34 × 10 3 8.34 \times 10^{3} 8.34 × 1 0 3
Worked solution (try it first) Move the point to just after the first non-zero digit: 0.000834 becomes 8.34.
The point moved 4 places to the right, and the number is less than 1, so the power is
− 4 -4 − 4 .
So
0.000834 = 8.34 × 10 − 4 0.000834 = 8.34 \times 10^{-4} 0.000834 = 8.34 × 1 0 − 4 , option A.
Watch out
Count the places the point moves, not the zeros after it: there are three zeros, but the point moves four places. Counting three gives 8.34 × 10 − 3 8.34 \times 10^{-3} 8.34 × 1 0 − 3 Report a problem with this question
Simplify 0.027 − 1 3 0.027^{-\frac13} 0.02 7 − 3 1
A 3 10 \frac{3}{10} 10 3 B 3 1 3 3\frac13 3 3 1 C 1 3 \frac13 3 1 D 3
Worked solution (try it first) 0.027 = 27 1000 0.027 = \frac{27}{1000} 0.027 = 1000 27 , and its cube root is
3 10 = 0.3 \frac{3}{10} = 0.3 10 3 = 0.3 .
The negative index means the reciprocal:
1 0.3 = 10 3 \frac{1}{0.3} = \frac{10}{3} 0.3 1 = 3 10 .
As a mixed number,
10 3 = 3 1 3 \frac{10}{3} = 3\frac13 3 10 = 3 3 1 , option B.
Watch out
Don't forget the minus sign in the index: it flips 0.3 to 10 3 \frac{10}{3} 3 10 3 10 \frac{3}{10} 10 3 Report a problem with this question
Given that log 2 a = log 8 4 \log_2 a = \log_8 4 log 2 a = log 8 4 a a a
A 2 1 2 2^{\frac12} 2 2 1 B 2 2 3 2^{\frac23} 2 3 2 C 4 2 3 4^{\frac23} 4 3 2 D 4 1 2 4^{\frac12} 4 2 1
Worked solution (try it first) Find
log 8 4 \log_8 4 log 8 4 :
8 = 2 3 8 = 2^3 8 = 2 3 and
4 = 2 2 4 = 2^2 4 = 2 2 , so
8 2 3 = 2 2 = 4 8^{\frac23} = 2^2 = 4 8 3 2 = 2 2 = 4 and
log 8 4 = 2 3 \log_8 4 = \frac23 log 8 4 = 3 2 .
So
log 2 a = 2 3 \log_2 a = \frac23 log 2 a = 3 2 .
Change to index form:
a = 2 2 3 a = 2^{\frac23} a = 2 3 2 , option B.
Watch out
8 1 2 8^{\frac12} 8 2 1 log 8 4 \log_8 4 log 8 4 2 3 \frac23 3 2 1 2 \frac12 2 1 1 2 \frac12 2 1 2 1 2 2^{\frac12} 2 2 1 Report a problem with this question
By selling some crates of soft drinks for ₦600.00, a dealer makes a profit of 50 % 50\% 50%
A ₦400.00 B ₦1200.00 C ₦900.00 D ₦450.00
Worked solution (try it first) A
50 % 50\% 50% profit means ₦600 is
150 % 150\% 150% of the cost:
1.5 × cost = 600 1.5 \times \text{cost} = 600 1.5 × cost = 600 .
Divide both sides by 1.5: cost
= 400 = 400 = 400 .
So the dealer paid ₦400.00, option A.
Watch out
₦600 already includes the profit, so divide by 1.5. Multiplying by 1.5 gives ₦900.00 (option C), which would be a selling price. Report a problem with this question
Find the n n n U n U_n U n 11 , 4 , − 3 , … 11, 4, -3, \dots 11 , 4 , − 3 , …
A U n = 19 − 7 n U_n = 19 - 7n U n = 19 − 7 n B U n = 19 + 7 n U_n = 19 + 7n U n = 19 + 7 n C U n = 18 − 7 n U_n = 18 - 7n U n = 18 − 7 n D U n = 18 + 7 n U_n = 18 + 7n U n = 18 + 7 n
Worked solution (try it first) This is an A.P. with
a = 11 a = 11 a = 11 and
d = 4 − 11 = − 7 d = 4 - 11 = -7 d = 4 − 11 = − 7 .
So
U n = a + ( n − 1 ) d = 11 − 7 ( n − 1 ) U_n = a + (n - 1)d = 11 - 7(n - 1) U n = a + ( n − 1 ) d = 11 − 7 ( n − 1 ) .
Expand:
11 − 7 n + 7 = 18 − 7 n 11 - 7n + 7 = 18 - 7n 11 − 7 n + 7 = 18 − 7 n , option C.
Watch out
Test your formula with n = 1 n = 1 n = 1 18 − 7 = 11 18 - 7 = 11 18 − 7 = 11 19 − 7 = 12 19 - 7 = 12 19 − 7 = 12 Report a problem with this question
If R = { 2 , 4 , 6 , 7 } R = \{2, 4, 6, 7\} R = { 2 , 4 , 6 , 7 } S = { 1 , 2 , 4 , 8 } S = \{1, 2, 4, 8\} S = { 1 , 2 , 4 , 8 } R ∪ S R \cup S R ∪ S
A { 1 , 2 , 4 , 6 , 7 , 8 } \{1, 2, 4, 6, 7, 8\} { 1 , 2 , 4 , 6 , 7 , 8 } B { 1 , 2 , 4 , 7 , 8 } \{1, 2, 4, 7, 8\} { 1 , 2 , 4 , 7 , 8 } C { 1 , 4 , 7 , 8 } \{1, 4, 7, 8\} { 1 , 4 , 7 , 8 } D { 2 , 6 , 7 } \{2, 6, 7\} { 2 , 6 , 7 }
Worked solution (try it first) The union holds every element that is in
R R R or
S S S or both, each listed once.
So
R ∪ S = { 1 , 2 , 4 , 6 , 7 , 8 } R \cup S = \{1, 2, 4, 6, 7, 8\} R ∪ S = { 1 , 2 , 4 , 6 , 7 , 8 } , option A.
Watch out
Go through both sets element by element so nothing is dropped. Option B leaves out 6, which is in R R R Report a problem with this question
If 16 9 , x , 1 , y \frac{16}{9}, x, 1, y 9 16 , x , 1 , y x y xy x y
A 16 9 \frac{16}{9} 9 16 B 3 4 \frac34 4 3 C 1 D 4 3 \frac43 3 4
Worked solution (try it first) For three consecutive terms of a G.P., the middle term squared equals the product of the outer two.
Take the three terms
x , 1 , y x, 1, y x , 1 , y :
1 2 = x y 1^2 = xy 1 2 = x y , so
x y = 1 xy = 1 x y = 1 , option C.
Check:
x 2 = 16 9 × 1 x^2 = \frac{16}{9} \times 1 x 2 = 9 16 × 1 , so
x = 4 3 x = \frac43 x = 3 4 and
r = 3 4 r = \frac34 r = 4 3 , which gives
y = 3 4 y = \frac34 y = 4 3 and
x y = 1 xy = 1 x y = 1 .
Watch out
4 3 \frac43 3 4 x x x 3 4 \frac34 4 3 y y y Report a problem with this question
In the Venn diagram, the shaded portion is represented by which set expression?
A ( P ∪ Q ) ′ ∩ R (P \cup Q)' \cap R ( P ∪ Q ) ′ ∩ R B P ′ ∩ Q ∩ R P' \cap Q \cap R P ′ ∩ Q ∩ R C P ′ ∩ R P' \cap R P ′ ∩ R D Q ′ ∩ R Q' \cap R Q ′ ∩ R
Worked solution (try it first) The shading covers the whole of circle
R R R except the part that overlaps
P P P .
That includes the piece of
R R R inside
Q Q Q , so
Q Q Q doesn't restrict it.
So the shaded region is
P ′ ∩ R P' \cap R P ′ ∩ R , option C.
Watch out
P ′ ∩ Q ∩ R P' \cap Q \cap R P ′ ∩ Q ∩ R R R R Q Q Q R R R P P P Q Q Q Report a problem with this question
Find the values of x x x 6 x − 1 x 2 + 4 x − 5 \dfrac{6x - 1}{x^2 + 4x - 5} x 2 + 4 x − 5 6 x − 1
A + 5 +5 + 5 − 1 -1 − 1 B − 5 -5 − 5 + 1 +1 + 1 C − 5 -5 − 5 − 1 -1 − 1 D + 4 +4 + 4 + 1 +1 + 1
Worked solution (try it first) A fraction is undefined when its bottom is zero:
x 2 + 4 x − 5 = 0 x^2 + 4x - 5 = 0 x 2 + 4 x − 5 = 0 .
Factorise:
( x + 5 ) ( x − 1 ) = 0 (x + 5)(x - 1) = 0 ( x + 5 ) ( x − 1 ) = 0 .
So
x = − 5 x = -5 x = − 5 or
x = 1 x = 1 x = 1 , option B.
Watch out
( x + 5 ) ( x − 1 ) = 0 (x + 5)(x - 1) = 0 ( x + 5 ) ( x − 1 ) = 0 x = − 5 x = -5 x = − 5 x = 1 x = 1 x = 1 Report a problem with this question
Solve the inequality 1 3 ( 2 x − 1 ) < 5 \frac13(2x - 1) < 5 3 1 ( 2 x − 1 ) < 5
A x < − 5 x < -5 x < − 5 B x < 8 x < 8 x < 8 C x < − 6 x < -6 x < − 6 D x < 7 x < 7 x < 7
Worked solution (try it first) Multiply both sides by 3:
2 x − 1 < 15 2x - 1 < 15 2 x − 1 < 15 .
Add 1 to both sides:
2 x < 16 2x < 16 2 x < 16 .
Divide by 2:
x < 8 x < 8 x < 8 , option B.
Watch out
To remove the − 1 -1 − 1 2 x < 16 2x < 16 2 x < 16 2 x < 14 2x < 14 2 x < 14 x < 7 x < 7 x < 7 Report a problem with this question
Find the smaller value of x x x x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0
Worked solution (try it first) Factorise:
( x − 1 ) ( x − 2 ) = 0 (x - 1)(x - 2) = 0 ( x − 1 ) ( x − 2 ) = 0 .
So
x = 1 x = 1 x = 1 or
x = 2 x = 2 x = 2 .
The smaller value is 1, option D.
Watch out
2 (option B) is the larger root. The question asks for the smaller one. Report a problem with this question
Factorize x ( a − c ) + y ( c − a ) x(a - c) + y(c - a) x ( a − c ) + y ( c − a )
A ( a − c ) ( y − x ) (a - c)(y - x) ( a − c ) ( y − x ) B ( a + c ) ( x + y ) (a + c)(x + y) ( a + c ) ( x + y ) C ( a + c ) ( x − y ) (a + c)(x - y) ( a + c ) ( x − y ) D ( a − c ) ( x − y ) (a - c)(x - y) ( a − c ) ( x − y )
Worked solution (try it first) Reverse the second bracket:
c − a = − ( a − c ) c - a = -(a - c) c − a = − ( a − c ) , so
y ( c − a ) = − y ( a − c ) y(c - a) = -y(a - c) y ( c − a ) = − y ( a − c ) .
The expression is
x ( a − c ) − y ( a − c ) x(a - c) - y(a - c) x ( a − c ) − y ( a − c ) .
Take out the common bracket:
( a − c ) ( x − y ) (a - c)(x - y) ( a − c ) ( x − y ) , option D.
Watch out
Reversing c − a c - a c − a y y y x x x x x x ( a − c ) ( y − x ) (a - c)(y - x) ( a − c ) ( y − x ) Report a problem with this question
If y ∝ 1 x 2 y \propto \dfrac{1}{x^2} y ∝ x 2 1 x = 3 x = 3 x = 3 y = 4 y = 4 y = 4 y y y x = 2 x = 2 x = 2
Worked solution (try it first) y = k x 2 y = \dfrac{k}{x^2} y = x 2 k , so
k = x 2 y k = x^2y k = x 2 y , which is
9 × 4 = 36 9 \times 4 = 36 9 × 4 = 36 .
When
x = 2 x = 2 x = 2 :
y = 36 4 = 9 y = \dfrac{36}{4} = 9 y = 4 36 = 9 , option B.
Watch out
Square the new x x x 36 2 = 18 \frac{36}{2} = 18 2 36 = 18 x x x Report a problem with this question
Solve 3 x 2 + 25 x − 18 = 0 3x^2 + 25x - 18 = 0 3 x 2 + 25 x − 18 = 0
A − 3 , 2 -3, 2 − 3 , 2 B − 9 , 2 3 -9, \frac23 − 9 , 3 2 C − 2 , 9 -2, 9 − 2 , 9 D − 2 , 3 -2, 3 − 2 , 3
Worked solution (try it first) Find two numbers with product
3 × ( − 18 ) = − 54 3 \times (-18) = -54 3 × ( − 18 ) = − 54 and sum 25: they are 27 and
− 2 -2 − 2 .
Split and group:
3 x 2 + 27 x − 2 x − 18 = 3 x ( x + 9 ) − 2 ( x + 9 ) 3x^2 + 27x - 2x - 18 = 3x(x + 9) - 2(x + 9) 3 x 2 + 27 x − 2 x − 18 = 3 x ( x + 9 ) − 2 ( x + 9 ) = ( 3 x − 2 ) ( x + 9 ) = (3x - 2)(x + 9) = ( 3 x − 2 ) ( x + 9 ) .
So
x = 2 3 x = \frac23 x = 3 2 or
x = − 9 x = -9 x = − 9 , option B.
Watch out
Check a root in the equation. x = − 9 x = -9 x = − 9 243 − 225 − 18 = 0 243 - 225 - 18 = 0 243 − 225 − 18 = 0 x = 2 x = 2 x = 2 12 + 50 − 18 = 44 12 + 50 - 18 = 44 12 + 50 − 18 = 44 Report a problem with this question
Solve ( x + 2 ) ( x − 7 ) = 0 (x + 2)(x - 7) = 0 ( x + 2 ) ( x − 7 ) = 0
A − 1 -1 − 1 B − 4 -4 − 4 C − 3 -3 − 3 D − 2 -2 − 2
Worked solution (try it first) A product is zero only when one of its factors is zero.
x + 2 = 0 x + 2 = 0 x + 2 = 0 gives
x = − 2 x = -2 x = − 2 , and
x − 7 = 0 x - 7 = 0 x − 7 = 0 gives
x = 7 x = 7 x = 7 .
So
x = − 2 x = -2 x = − 2 or
x = 7 x = 7 x = 7 , option D.
Watch out
Each root has the opposite sign to the number in its bracket: x + 2 = 0 x + 2 = 0 x + 2 = 0 x = − 2 x = -2 x = − 2 Report a problem with this question
Solve x + y = 3 2 x + y = \frac32 x + y = 2 3 x − y = 5 2 x - y = \frac52 x − y = 2 5 2 y + x 2y + x 2 y + x
A − 2 -2 − 2 B 1 2 \frac12 2 1 C 1 D − 1 -1 − 1
Worked solution (try it first) Add the two equations so
y y y cancels:
2 x = 3 2 + 5 2 = 4 2x = \frac32 + \frac52 = 4 2 x = 2 3 + 2 5 = 4 , so
x = 2 x = 2 x = 2 .
Put
x = 2 x = 2 x = 2 into
x + y = 3 2 x + y = \frac32 x + y = 2 3 :
y = 3 2 − 2 = − 1 2 y = \frac32 - 2 = -\frac12 y = 2 3 − 2 = − 2 1 .
So
2 y + x = − 1 + 2 = 1 2y + x = -1 + 2 = 1 2 y + x = − 1 + 2 = 1 , option C.
Watch out
Finish the sum: 2 y = − 1 2y = -1 2 y = − 1 x = 2 x = 2 x = 2 2 y 2y 2 y − 1 -1 − 1 Report a problem with this question
Which inequality is illustrated by the sketch graph? The boundary line passes through ( 0 , 3 ) (0, 3) ( 0 , 3 ) ( 1 , 0 ) (1, 0) ( 1 , 0 )
A y ≤ − 3 x + 3 y \le -3x + 3 y ≤ − 3 x + 3 B y ≤ 3 x + 2 y \le 3x + 2 y ≤ 3 x + 2 C y ≥ − 2 x + 3 y \ge -2x + 3 y ≥ − 2 x + 3 D y ≤ x + 3 y \le x + 3 y ≤ x + 3
Worked solution (try it first) The gradient of the line through
( 0 , 3 ) (0, 3) ( 0 , 3 ) and
( 1 , 0 ) (1, 0) ( 1 , 0 ) is
0 − 3 1 − 0 = − 3 \frac{0 - 3}{1 - 0} = -3 1 − 0 0 − 3 = − 3 , and the
y y y -intercept is 3.
So the line is
y = − 3 x + 3 y = -3x + 3 y = − 3 x + 3 .
The region below a line is
y ≤ y \le y ≤ the line's expression.
So it is
y ≤ − 3 x + 3 y \le -3x + 3 y ≤ − 3 x + 3 , option A.
Watch out
Check the line with both points. Option D's y = x + 3 y = x + 3 y = x + 3 ( 0 , 3 ) (0, 3) ( 0 , 3 ) y = 4 y = 4 y = 4 x = 1 x = 1 x = 1 Report a problem with this question
The graph of y = 3 x 2 + x − 7 y = 3x^2 + x - 7 y = 3 x 2 + x − 7 y y y
A − 7 -7 − 7 B − 10 -10 − 10 C − 4 -4 − 4 D − 1 -1 − 1
Worked solution (try it first) The minimum value of
y y y is the
y y y -value of the lowest point of the curve.
On the graph the lowest point is just below
y = − 7 y = -7 y = − 7 , slightly left of the
y y y -axis.
Check: the turning point is at
x = − 1 6 x = -\frac16 x = − 6 1 , where
y = − 7 1 12 y = -7\frac{1}{12} y = − 7 12 1 .
The best estimate is
− 7 -7 − 7 , option A.
Watch out
Read the y y y − 1.7 -1.7 − 1.7 y = 0 y = 0 y = 0 Report a problem with this question
Using the graph of y = 3 x 2 + x − 7 y = 3x^2 + x - 7 y = 3 x 2 + x − 7 3 x 2 + x − 7 = 0 3x^2 + x - 7 = 0 3 x 2 + x − 7 = 0
A 1.4 and − 1.7 -1.7 − 1.7 B 2.0 and − 1.9 -1.9 − 1.9 C 1.0 and − 1.2 -1.2 − 1.2 D 1.1 and − 1.3 -1.3 − 1.3
Worked solution (try it first) The roots of
3 x 2 + x − 7 = 0 3x^2 + x - 7 = 0 3 x 2 + x − 7 = 0 are where
y = 0 y = 0 y = 0 , so read where the curve crosses the
x x x -axis.
It crosses at about
x = 1.4 x = 1.4 x = 1.4 and
x = − 1.7 x = -1.7 x = − 1.7 .
So the roots are 1.4 and
− 1.7 -1.7 − 1.7 , option A.
Watch out
Read the x x x x x x y y y − 7 -7 − 7 3 ( 1.4 ) 2 + 1.4 − 7 = 0.28 3(1.4)^2 + 1.4 - 7 = 0.28 3 ( 1.4 ) 2 + 1.4 − 7 = 0.28 Report a problem with this question
The diagonals A C AC A C B D BD B D A B C D ABCD A B C D 16 cm 16\text{ cm} 16 cm 12 cm 12\text{ cm} 12 cm
A 96 cm 2 96\text{ cm}^2 96 cm 2 B 36 cm 2 36\text{ cm}^2 36 cm 2 C 48 cm 2 48\text{ cm}^2 48 cm 2 D 24 cm 2 24\text{ cm}^2 24 cm 2
Worked solution (try it first) The diagonals cut the rhombus into four right-angled triangles, each with legs 8 cm and 6 cm, so each has area
1 2 × 8 × 6 = 24 cm 2 \frac12 \times 8 \times 6 = 24\text{ cm}^2 2 1 × 8 × 6 = 24 cm 2 .
The rhombus is four of them:
4 × 24 = 96 cm 2 4 \times 24 = 96\text{ cm}^2 4 × 24 = 96 cm 2 .
This is the same as
1 2 × 16 × 12 \frac12 \times 16 \times 12 2 1 × 16 × 12 , option A.
Watch out
24 cm 2 24\text{ cm}^2 24 cm 2 Report a problem with this question
A cylindrical container, closed at both ends, has radius 7 cm 7\text{ cm} 7 cm 5 cm 5\text{ cm} 5 cm [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 35 cm 2 35\text{ cm}^2 35 cm 2 B 528 cm 2 528\text{ cm}^2 528 cm 2 C 154 cm 2 154\text{ cm}^2 154 cm 2 D 220 cm 2 220\text{ cm}^2 220 cm 2
Worked solution (try it first) Closed at both ends: the curved surface plus two circles,
2 π r ( r + h ) 2\pi r(r + h) 2 π r ( r + h ) .
2 × 22 7 × 7 × ( 7 + 5 ) = 44 × 12 2 \times \frac{22}{7} \times 7 \times (7 + 5) = 44 \times 12 2 × 7 22 × 7 × ( 7 + 5 ) = 44 × 12 = 528 cm 2 = 528\text{ cm}^2 = 528 cm 2 , option B.
Watch out
Add the two ends. The curved surface alone is 2 × 22 7 × 7 × 5 = 220 cm 2 2 \times \frac{22}{7} \times 7 \times 5 = 220\text{ cm}^2 2 × 7 22 × 7 × 5 = 220 cm 2 Report a problem with this question
Find the total surface area of a solid circular cone with base radius 3 cm 3\text{ cm} 3 cm 4 cm 4\text{ cm} 4 cm [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 66 cm 2 66\text{ cm}^2 66 cm 2 B 37 5 7 cm 2 37\frac57\text{ cm}^2 37 7 5 cm 2 C 75 3 7 cm 2 75\frac37\text{ cm}^2 75 7 3 cm 2 D 78 6 7 cm 2 78\frac67\text{ cm}^2 78 7 6 cm 2
Worked solution (try it first) A solid cone has a curved surface
π r l \pi rl π r l and a base
π r 2 \pi r^2 π r 2 : a total of
π r ( r + l ) \pi r(r + l) π r ( r + l ) .
22 7 × 3 × ( 3 + 4 ) = 22 × 3 \frac{22}{7} \times 3 \times (3 + 4) = 22 \times 3 7 22 × 3 × ( 3 + 4 ) = 22 × 3 = 66 cm 2 = 66\text{ cm}^2 = 66 cm 2 , option A.
Watch out
A solid cone includes its base. The curved surface alone is 22 7 × 3 × 4 = 37 5 7 cm 2 \frac{22}{7} \times 3 \times 4 = 37\frac57\text{ cm}^2 7 22 × 3 × 4 = 37 7 5 cm 2 Report a problem with this question
A water tank of height 1 2 m \frac12\text{ m} 2 1 m 1 1 2 m 1\frac12\text{ m} 1 2 1 m
A 375 litres B 3750 litres C 37.5 litres D 37500 litres
Worked solution (try it first) Volume of the tank:
1.5 × 1.5 × 0.5 = 1.125 m 3 1.5 \times 1.5 \times 0.5 = 1.125\text{ m}^3 1.5 × 1.5 × 0.5 = 1.125 m 3 .
1 m 3 1\text{ m}^3 1 m 3 is 1000 litres, so the tank takes 1125 litres.
Left in the tanker:
1500 − 1125 = 375 1500 - 1125 = 375 1500 − 1125 = 375 litres, option A.
Watch out
1 m 3 = 1000 1\text{ m}^3 = 1000 1 m 3 = 1000 Report a problem with this question
A sphere has volume k cm 3 k\text{ cm}^3 k cm 3 k cm 2 k\text{ cm}^2 k cm 2
A 3 cm 3\text{ cm} 3 cm B 9 cm 9\text{ cm} 9 cm C More information needed D 6 cm 6\text{ cm} 6 cm
Worked solution (try it first) The volume and the surface area have the same number:
4 3 π r 3 = 4 π r 2 \frac43\pi r^3 = 4\pi r^2 3 4 π r 3 = 4 π r 2 .
Divide both sides by
4 π r 2 4\pi r^2 4 π r 2 :
r 3 = 1 \frac r3 = 1 3 r = 1 , so
r = 3 r = 3 r = 3 cm.
The diameter is
2 × 3 = 6 2 \times 3 = 6 2 × 3 = 6 cm, option D.
Watch out
The question asks for the diameter. 3 3 3 Report a problem with this question
A cone is cut to leave a frustum with radii 12 cm 12\text{ cm} 12 cm 6 cm 6\text{ cm} 6 cm 12 cm 12\text{ cm} 12 cm
A 24 cm 24\text{ cm} 24 cm B 15 cm 15\text{ cm} 15 cm C 8 cm 8\text{ cm} 8 cm D 12 cm 12\text{ cm} 12 cm
Worked solution (try it first) Let the small cone cut off the top have slant height
s s s .
The full cone's slant height is
s + 12 s + 12 s + 12 .
Similar cones:
s s + 12 = 6 12 \dfrac{s}{s + 12} = \dfrac{6}{12} s + 12 s = 12 6 , so
2 s = s + 12 2s = s + 12 2 s = s + 12 and
s = 12 s = 12 s = 12 .
The full cone's slant height is
12 + 12 = 24 12 + 12 = 24 12 + 12 = 24 cm, option A.
Watch out
s = 12 s = 12 s = 12 Report a problem with this question
The positions of two towns P P P Q Q Q ( 15 ∘ N , 12 ∘ E ) (15^\circ\text{N}, 12^\circ\text{E}) ( 1 5 ∘ N , 1 2 ∘ E ) ( 65 ∘ N , 12 ∘ E ) (65^\circ\text{N}, 12^\circ\text{E}) ( 6 5 ∘ N , 1 2 ∘ E )
A 50 ∘ 50^\circ 5 0 ∘ B 104 ∘ 104^\circ 10 4 ∘ C 80 ∘ 80^\circ 8 0 ∘ D 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) Both towns are north of the equator, so subtract the latitudes.
The difference is
65 ∘ − 15 ∘ = 50 ∘ 65^\circ - 15^\circ = 50^\circ 6 5 ∘ − 1 5 ∘ = 5 0 ∘ , option A.
Watch out
Add only when one town is north and the other south. Adding here gives 80 ∘ 80^\circ 8 0 ∘ Report a problem with this question
A 120 ∘ 120^\circ 12 0 ∘ 21 cm 21\text{ cm} 21 cm
A 3 1 2 cm 3\frac12\text{ cm} 3 2 1 cm B 10 1 2 cm 10\frac12\text{ cm} 10 2 1 cm C 14 cm 14\text{ cm} 14 cm D 7 cm 7\text{ cm} 7 cm
Worked solution (try it first) The arc of the sector becomes the circumference of the base:
120 360 × 2 π × 21 = 14 π \frac{120}{360} \times 2\pi \times 21 = 14\pi 360 120 × 2 π × 21 = 14 π .
So
2 π r = 14 π 2\pi r = 14\pi 2 π r = 14 π .
Divide both sides by
2 π 2\pi 2 π :
r = 7 r = 7 r = 7 cm, option D.
Watch out
14 π 14\pi 14 π 2 π r 2\pi r 2 π r Report a problem with this question
A cylindrical container, closed at both ends, has radius 7 cm 7\text{ cm} 7 cm 5 cm 5\text{ cm} 5 cm [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 770 cm 3 770\text{ cm}^3 770 cm 3 B 35 cm 3 35\text{ cm}^3 35 cm 3 C 220 cm 3 220\text{ cm}^3 220 cm 3 D 154 cm 3 154\text{ cm}^3 154 cm 3
Worked solution (try it first) Volume of a cylinder:
π r 2 h \pi r^2h π r 2 h .
22 7 × 7 2 × 5 = 154 × 5 \frac{22}{7} \times 7^2 \times 5 = 154 \times 5 7 22 × 7 2 × 5 = 154 × 5 = 770 cm 3 = 770\text{ cm}^3 = 770 cm 3 , option A.
Watch out
Multiply the base area by the height. 154 cm 2 154\text{ cm}^2 154 cm 2 Report a problem with this question
If log 10 x = 2 ˉ .3675 \log_{10} x = \bar{2}.3675 log 10 x = 2 ˉ .3675 log 10 y = 2 ˉ .9738 \log_{10} y = \bar{2}.9738 log 10 y = 2 ˉ .9738 x + y x + y x + y
Worked solution (try it first) 2 ˉ .3675 \bar{2}.3675 2 ˉ .3675 means
− 2 + 0.3675 -2 + 0.3675 − 2 + 0.3675 .
The antilog of 0.3675 is 2.331, so
x = 2.331 × 10 − 2 = 0.02331 x = 2.331 \times 10^{-2} = 0.02331 x = 2.331 × 1 0 − 2 = 0.02331 .
In the same way, the antilog of 0.9738 is 9.415, so
y = 9.415 × 10 − 2 = 0.09415 y = 9.415 \times 10^{-2} = 0.09415 y = 9.415 × 1 0 − 2 = 0.09415 .
Add:
x + y = 0.11746 x + y = 0.11746 x + y = 0.11746 , which is 0.117 to three significant figures, option C.
Watch out
Round only at the end. The fourth significant figure of 0.11746 is 4, so it rounds down to 0.117; rounding up gives 0.118 (option A). Report a problem with this question
A sector has angle 108 ∘ 108^\circ 10 8 ∘ 3 1 2 cm 3\frac12\text{ cm} 3 2 1 cm [ π = 22 7 ] \left[\pi = \frac{22}{7}\right] [ π = 7 22 ]
A 6 3 5 cm 6\frac35\text{ cm} 6 5 3 cm B 7 2 10 cm 7\frac{2}{10}\text{ cm} 7 10 2 cm C 6 4 5 cm 6\frac45\text{ cm} 6 5 4 cm D 13 3 5 cm 13\frac35\text{ cm} 13 5 3 cm
Worked solution (try it first) Arc length
= 108 360 × 2 × 22 7 × 3.5 = \frac{108}{360} \times 2 \times \frac{22}{7} \times 3.5 = 360 108 × 2 × 7 22 × 3.5 .
The circumference is 22 cm, and
108 360 = 3 10 \frac{108}{360} = \frac{3}{10} 360 108 = 10 3 , so the arc is
6.6 6.6 6.6 cm.
The perimeter adds the two radii:
6.6 + 2 × 3.5 = 13.6 = 13 3 5 6.6 + 2 \times 3.5 = 13.6 = 13\frac35 6.6 + 2 × 3.5 = 13.6 = 13 5 3 cm, option D.
Watch out
6 3 5 6\frac35 6 5 3 Report a problem with this question
In the diagram, A O ⊥ O B AO \perp OB A O ⊥ O B x x x
A 15 ∘ 15^\circ 1 5 ∘ B 22.5 ∘ 22.5^\circ 22. 5 ∘ C 7.5 ∘ 7.5^\circ 7. 5 ∘ D 30 ∘ 30^\circ 3 0 ∘
Worked solution (try it first) Angles at a point add up to
360 ∘ 360^\circ 36 0 ∘ , and the right angle takes
90 ∘ 90^\circ 9 0 ∘ .
So
2 x + 3 x + 4 x = 360 ∘ − 90 ∘ 2x + 3x + 4x = 360^\circ - 90^\circ 2 x + 3 x + 4 x = 36 0 ∘ − 9 0 ∘ So
9 x = 270 ∘ 9x = 270^\circ 9 x = 27 0 ∘ and
x = 30 ∘ x = 30^\circ x = 3 0 ∘ , option D.
Watch out
Take the right angle out of 360 ∘ 360^\circ 36 0 ∘ 9 x = 360 ∘ 9x = 360^\circ 9 x = 36 0 ∘ 40 ∘ 40^\circ 4 0 ∘ Report a problem with this question
In the diagram, O O O A O D AOD A O D C C C ∣ B D ∣ = ∣ D C ∣ |BD| = |DC| ∣ B D ∣ = ∣ D C ∣ ∠ D C B = 35 ∘ \angle DCB = 35^\circ ∠ D C B = 3 5 ∘ ∠ B A O \angle BAO ∠ B A O
A 25 ∘ 25^\circ 2 5 ∘ B 30 ∘ 30^\circ 3 0 ∘ C 20 ∘ 20^\circ 2 0 ∘ D 35 ∘ 35^\circ 3 5 ∘
Worked solution (try it first) ∣ B D ∣ = ∣ D C ∣ |BD| = |DC| ∣ B D ∣ = ∣ D C ∣ , so triangle
B D C BDC B D C is isosceles and
∠ D B C = ∠ D C B = 35 ∘ \angle DBC = \angle DCB = 35^\circ ∠ D B C = ∠ D C B = 3 5 ∘ .
∠ B D A \angle BDA ∠ B D A is an exterior angle of triangle
B D C BDC B D C :
∠ B D A = 35 ∘ + 35 ∘ \angle BDA = 35^\circ + 35^\circ ∠ B D A = 3 5 ∘ + 3 5 ∘ A D AD A D is a diameter, so
∠ A B D = 90 ∘ \angle ABD = 90^\circ ∠ A B D = 9 0 ∘ .
In triangle
A B D ABD A B D ,
∠ B A O = 180 ∘ − 90 ∘ − 70 ∘ \angle BAO = 180^\circ - 90^\circ - 70^\circ ∠ B A O = 18 0 ∘ − 9 0 ∘ − 7 0 ∘ = 20 ∘ = 20^\circ = 2 0 ∘ , option C.
Watch out
The exterior angle at D D D 35 ∘ + 35 ∘ = 70 ∘ 35^\circ + 35^\circ = 70^\circ 3 5 ∘ + 3 5 ∘ = 7 0 ∘ 35 ∘ 35^\circ 3 5 ∘ ∠ B A O = 55 ∘ \angle BAO = 55^\circ ∠ B A O = 5 5 ∘ 35 ∘ 35^\circ 3 5 ∘ ∠ D C B \angle DCB ∠ D C B A A A Report a problem with this question
Which of the following angles is an exterior angle of a regular polygon?
A 95 ∘ 95^\circ 9 5 ∘ B 78 ∘ 78^\circ 7 8 ∘ C 72 ∘ 72^\circ 7 2 ∘ D 85 ∘ 85^\circ 8 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 an exterior angle must divide
360 ∘ 360^\circ 36 0 ∘ a whole number of times.
360 ÷ 95 360 \div 95 360 ÷ 95 ,
360 ÷ 78 360 \div 78 360 ÷ 78 and
360 ÷ 85 360 \div 85 360 ÷ 85 are not whole numbers.
360 ÷ 72 = 5 360 \div 72 = 5 360 ÷ 72 = 5 , a regular pentagon.
So the answer is
72 ∘ 72^\circ 7 2 ∘ , option C.
Watch out
Don't pick by feel: test every option by dividing it into 360 ∘ 360^\circ 36 0 ∘ 72 ∘ 72^\circ 7 2 ∘ Report a problem with this question
The locus of a point equidistant from two fixed points is the
A angle bisector of the straight lines joining them B parallel line to the straight line joining them C perpendicular to the straight line joining them D perpendicular bisector of the straight line joining them
Worked solution (try it first) A point the same distance from two fixed points lies on the line that cuts the segment joining them in half at right angles.
So the locus is the perpendicular bisector of the line joining them, option D.
Watch out
"Perpendicular to the line joining them" (option C) is not enough: the line must also pass through the mid-point, so it is the perpendicular bisector. Report a problem with this question
In the diagram, A C AC A C B D BD B D E E E A B ∥ D C AB \parallel DC A B ∥ D C ∣ A B ∣ = 12 cm |AB| = 12\text{ cm} ∣ A B ∣ = 12 cm ∣ A E ∣ = 8 cm |AE| = 8\text{ cm} ∣ A E ∣ = 8 cm ∣ D C ∣ = 9 cm |DC| = 9\text{ cm} ∣ D C ∣ = 9 cm ∣ E C ∣ |EC| ∣ E C ∣
A 6 cm 6\text{ cm} 6 cm B 5 cm 5\text{ cm} 5 cm C 8 cm 8\text{ cm} 8 cm D 9 cm 9\text{ cm} 9 cm
Worked solution (try it first) A B ∥ D C AB \parallel DC A B ∥ D C , so the alternate angles at
A A A and
C C C are equal, and so are those at
B B B and
D D D .
Triangles
A B E ABE A B E and
C D E CDE C D E are similar.
Matching sides are in the same ratio:
E C A E = D C A B \dfrac{EC}{AE} = \dfrac{DC}{AB} A E E C = A B D C , so
E C 8 = 9 12 \dfrac{EC}{8} = \dfrac{9}{12} 8 E C = 12 9 .
E C = 8 × 3 4 = 6 EC = 8 \times \frac34 = 6 E C = 8 × 4 3 = 6 cm, option A.
Watch out
The small triangle C D E CDE C D E E C EC E C 8 × 12 9 ≈ 10.7 8 \times \frac{12}{9} \approx 10.7 8 × 9 12 ≈ 10.7 Report a problem with this question
In the diagram, P Q ∥ T U PQ \parallel TU P Q ∥ T U ∠ P Q R = 50 ∘ \angle PQR = 50^\circ ∠ P QR = 5 0 ∘ ∠ Q R S = 86 ∘ \angle QRS = 86^\circ ∠ QR S = 8 6 ∘ ∠ S T U = 64 ∘ \angle STU = 64^\circ ∠ S T U = 6 4 ∘ x x x
A 100 ∘ 100^\circ 10 0 ∘ B 108 ∘ 108^\circ 10 8 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 136 ∘ 136^\circ 13 6 ∘
Worked solution (try it first) Draw a line through
R R R parallel to
P Q PQ P Q .
By alternate angles,
R Q RQ R Q makes
50 ∘ 50^\circ 5 0 ∘ with it, so
R S RS R S makes
86 ∘ − 50 ∘ = 36 ∘ 86^\circ - 50^\circ = 36^\circ 8 6 ∘ − 5 0 ∘ = 3 6 ∘ with it.
Draw a line through
S S S parallel to
T U TU T U .
By alternate angles,
S R SR S R makes
36 ∘ 36^\circ 3 6 ∘ with it and
S T ST S T makes
64 ∘ 64^\circ 6 4 ∘ with it.
So
x = 36 ∘ + 64 ∘ = 100 ∘ x = 36^\circ + 64^\circ = 100^\circ x = 3 6 ∘ + 6 4 ∘ = 10 0 ∘ , option A.
Watch out
In a zigzag between parallels, the angles pointing one way add up to those pointing the other way: 50 + x = 86 + 64 50 + x = 86 + 64 50 + x = 86 + 64 50 + 86 50 + 86 50 + 86 136 ∘ 136^\circ 13 6 ∘ Report a problem with this question
In the diagram, A B ∥ C D AB \parallel CD A B ∥ C D ∠ B A C \angle BAC ∠ B A C ∠ A C D \angle ACD ∠ A C D E E E ∠ A E C \angle AEC ∠ A E C
A 45 ∘ 45^\circ 4 5 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 30 ∘ 30^\circ 3 0 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) A B ∥ C D AB \parallel CD A B ∥ C D , so
∠ B A C \angle BAC ∠ B A C and
∠ A C D \angle ACD ∠ A C D are co-interior and add up to
180 ∘ 180^\circ 18 0 ∘ .
A E AE A E and
C E CE C E bisect them, so
∠ E A C + ∠ E C A = 180 ∘ ÷ 2 \angle EAC + \angle ECA = 180^\circ \div 2 ∠ E A C + ∠ E C A = 18 0 ∘ ÷ 2 The angles of triangle
A E C AEC A E C add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ A E C = 180 ∘ − 90 ∘ \angle AEC = 180^\circ - 90^\circ ∠ A E C = 18 0 ∘ − 9 0 ∘ = 90 ∘ = 90^\circ = 9 0 ∘ , option D.
Watch out
Halve the 180 ∘ 180^\circ 18 0 ∘ ∠ A E C \angle AEC ∠ A E C 90 ∘ 90^\circ 9 0 ∘ 45 ∘ 45^\circ 4 5 ∘ Report a problem with this question
In the diagram, A B ∥ C D AB \parallel CD A B ∥ C D x x x
A 93 ∘ 93^\circ 9 3 ∘ B 77 ∘ 77^\circ 7 7 ∘ C 103 ∘ 103^\circ 10 3 ∘ D 62 ∘ 62^\circ 6 2 ∘
Worked solution (try it first) Vertically opposite angles:
A B AB A B also makes
52 ∘ 52^\circ 5 2 ∘ with the upper line on the other side of the crossing.
C D ∥ A B CD \parallel AB C D ∥ A B , so
C D CD C D makes the same
52 ∘ 52^\circ 5 2 ∘ with the upper line (corresponding angles).
This angle is inside the small triangle cut off by
C D CD C D at the
35 ∘ 35^\circ 3 5 ∘ vertex.
The angles of that triangle add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 52 ∘ − 35 ∘ x = 180^\circ - 52^\circ - 35^\circ x = 18 0 ∘ − 5 2 ∘ − 3 5 ∘ = 93 ∘ = 93^\circ = 9 3 ∘ , option A.
Watch out
Move the 52 ∘ 52^\circ 5 2 ∘ C D CD C D 52 ∘ 52^\circ 5 2 ∘ 35 ∘ 35^\circ 3 5 ∘ x x x Report a problem with this question
In the diagram, B B B A D AD A D C C C A E AE A E D C DC D C B E BE B E F F F a ∘ a^\circ a ∘ b + c b + c b + c
A a ∘ a^\circ a ∘ B 90 ∘ 90^\circ 9 0 ∘ C 180 ∘ 180^\circ 18 0 ∘ D d ∘ d^\circ d ∘
Worked solution (try it first) B F E BFE B F E is a straight line, so
∠ B F C = 180 ∘ − a \angle BFC = 180^\circ - a ∠ B F C = 18 0 ∘ − a (it sits next to the
a a a at
F F F ).
The angles of quadrilateral
A B F C ABFC A B F C add up to
360 ∘ 360^\circ 36 0 ∘ :
a + b + ( 180 − a ) + c = 360 a + b + (180 - a) + c = 360 a + b + ( 180 − a ) + c = 360 .
The
a a a s cancel:
b + c + 180 = 360 b + c + 180 = 360 b + c + 180 = 360 , so
b + c = 180 ∘ b + c = 180^\circ b + c = 18 0 ∘ , option C.
Watch out
∠ B F C \angle BFC ∠ B F C a a a B F E BFE B F E a a a 180 ∘ − a 180^\circ - a 18 0 ∘ − a a a a b + c = 360 − 2 a b + c = 360 - 2a b + c = 360 − 2 a Report a problem with this question
In △ A B C \triangle ABC △ A B C B C BC B C D D D ∣ A B ∣ = ∣ A C ∣ |AB| = |AC| ∣ A B ∣ = ∣ A C ∣ ∠ B A C = 50 ∘ \angle BAC = 50^\circ ∠ B A C = 5 0 ∘ ∠ A C D \angle ACD ∠ A C D
A 115 ∘ 115^\circ 11 5 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 60 ∘ 60^\circ 6 0 ∘ D 50 ∘ 50^\circ 5 0 ∘
Worked solution (try it first) ∣ A B ∣ = ∣ A C ∣ |AB| = |AC| ∣ A B ∣ = ∣ A C ∣ , so the base angles at
B B B and
C C C are equal: each is
180 ∘ − 50 ∘ 2 = 65 ∘ \frac{180^\circ - 50^\circ}{2} = 65^\circ 2 18 0 ∘ − 5 0 ∘ = 6 5 ∘ .
B C D BCD B C D is a straight line, so
∠ A C D = 180 ∘ − 65 ∘ \angle ACD = 180^\circ - 65^\circ ∠ A C D = 18 0 ∘ − 6 5 ∘ = 115 ∘ = 115^\circ = 11 5 ∘ , option A.
Watch out
65 ∘ 65^\circ 6 5 ∘ ∠ A C B \angle ACB ∠ A C B ∠ A C D \angle ACD ∠ A C D Report a problem with this question
In right-angled triangle P Q R PQR P QR ∠ P R Q = 90 ∘ \angle PRQ = 90^\circ ∠ P R Q = 9 0 ∘ ∠ Q P R = 30 ∘ \angle QPR = 30^\circ ∠ QP R = 3 0 ∘ ∣ P Q ∣ = 10 cm |PQ| = 10\text{ cm} ∣ P Q ∣ = 10 cm ∣ Q R ∣ = y |QR| = y ∣ QR ∣ = y
A 6 cm 6\text{ cm} 6 cm B 4 cm 4\text{ cm} 4 cm C 5 cm 5\text{ cm} 5 cm D 3 cm 3\text{ cm} 3 cm
Worked solution (try it first) The right angle is at
R R R , so
P Q = 10 PQ = 10 P Q = 10 cm is the hypotenuse, and
Q R = y QR = y QR = y is opposite the
30 ∘ 30^\circ 3 0 ∘ angle at
P P P .
So
y = 10 sin 30 ∘ y = 10\sin30^\circ y = 10 sin 3 0 ∘ = 10 × 1 2 = 10 \times \frac12 = 10 × 2 1 Watch out
y y y 30 ∘ 30^\circ 3 0 ∘ P R = 10 cos 30 ∘ = 5 3 ≈ 8.7 PR = 10\cos30^\circ = 5\sqrt3 \approx 8.7 P R = 10 cos 3 0 ∘ = 5 3 ≈ 8.7 y y y Report a problem with this question
Find the mode of 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8, 14.
Worked solution (try it first) Count each number: 8 appears four times, 9 and 10 three times each, and 11 and 14 once.
The mode is the number that occurs most often: 8, option A.
Watch out
Count carefully: 9 and 10 each appear three times, but 8 appears four times. Ticking off each value as you count avoids the slip. Report a problem with this question
The bearings of Q Q Q R R R P P P 030 ∘ 030^\circ 03 0 ∘ 120 ∘ 120^\circ 12 0 ∘ ∣ P Q ∣ = 12 m |PQ| = 12\text{ m} ∣ P Q ∣ = 12 m ∣ P R ∣ = 5 m |PR| = 5\text{ m} ∣ P R ∣ = 5 m ∣ Q R ∣ |QR| ∣ QR ∣
A 11 m 11\text{ m} 11 m B 13 m 13\text{ m} 13 m C 9 m 9\text{ m} 9 m D 7 m 7\text{ m} 7 m
Worked solution (try it first) The angle between the bearings at
P P P is
∠ Q P R = 120 ∘ − 30 ∘ \angle QPR = 120^\circ - 30^\circ ∠ QP R = 12 0 ∘ − 3 0 ∘ So
Q R QR QR is the hypotenuse of a right-angled triangle:
Q R 2 = 12 2 + 5 2 = 169 QR^2 = 12^2 + 5^2 = 169 Q R 2 = 1 2 2 + 5 2 = 169 .
So
∣ Q R ∣ = 13 |QR| = 13 ∣ QR ∣ = 13 m, option B.
Watch out
Use Pythagoras for the right angle at P P P 12 − 5 = 7 12 - 5 = 7 12 − 5 = 7 Q Q Q R R R Report a problem with this question
Without using tables, evaluate sin 20 ∘ cos 70 ∘ + cos 25 ∘ sin 65 ∘ \dfrac{\sin 20^\circ}{\cos 70^\circ} + \dfrac{\cos 25^\circ}{\sin 65^\circ} cos 7 0 ∘ sin 2 0 ∘ + sin 6 5 ∘ cos 2 5 ∘
Worked solution (try it first) The sine of an angle equals the cosine of its complement:
cos 70 ∘ = sin 20 ∘ \cos70^\circ = \sin20^\circ cos 7 0 ∘ = sin 2 0 ∘ and
sin 65 ∘ = cos 25 ∘ \sin65^\circ = \cos25^\circ sin 6 5 ∘ = cos 2 5 ∘ .
So
sin 20 ∘ cos 70 ∘ = 1 \dfrac{\sin20^\circ}{\cos70^\circ} = 1 cos 7 0 ∘ sin 2 0 ∘ = 1 and
cos 25 ∘ sin 65 ∘ = 1 \dfrac{\cos25^\circ}{\sin65^\circ} = 1 sin 6 5 ∘ cos 2 5 ∘ = 1 .
The total is
1 + 1 = 2 1 + 1 = 2 1 + 1 = 2 , option A.
Watch out
Each fraction is 1 because its top and bottom are equal; the fractions add, they don't cancel to 0 (option B). Report a problem with this question
The mean of 20 observations is 4. If the largest observation, 23, is removed, find the mean of the remaining observations.
Worked solution (try it first) Total of the 20 observations:
20 × 4 = 80 20 \times 4 = 80 20 × 4 = 80 .
Remove 23: the total is
80 − 23 = 57 80 - 23 = 57 80 − 23 = 57 for 19 observations.
The new mean is
57 19 = 3 \frac{57}{19} = 3 19 57 = 3 , option D.
Watch out
After removing one observation there are 19 left, not 20. Dividing 57 by 20 gives 2.85 (option C). Report a problem with this question
Two fair dice are tossed once. Find the probability that the sum is at least 10.
A 1 6 \frac16 6 1 B 1 12 \frac1{12} 12 1 C 1 4 \frac14 4 1 D 5 36 \frac5{36} 36 5
Worked solution (try it first) Two dice give 36 equally likely outcomes.
A sum of at least 10 comes from
( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) , ( 5 , 6 ) , ( 6 , 5 ) , ( 6 , 6 ) (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6) ( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) , ( 5 , 6 ) , ( 6 , 5 ) , ( 6 , 6 ) , which is 6 outcomes.
So the probability is
6 36 = 1 6 \frac{6}{36} = \frac16 36 6 = 6 1 , option A.
Watch out
"At least 10" includes 10 itself. Counting only 11 and 12 gives 3 36 = 1 12 \frac{3}{36} = \frac{1}{12} 36 3 = 12 1 Report a problem with this question
Find the median of 2.64, 2.50, 2.72, 2.91, 2.35.
Worked solution (try it first) Put the 5 numbers in order: 2.35, 2.50, 2.64, 2.72, 2.91.
With 5 numbers the median is the 3rd: 2.64, option B.
Watch out
Order the list first. The middle number as written is 2.72 (option D). Report a problem with this question
A box contains 2 red, 6 white and 5 black balls. A ball is selected at random. What is the probability that it is black?
A 5 13 \frac5{13} 13 5 B 5 11 \frac5{11} 11 5 C 2 13 \frac2{13} 13 2 D 5 6 \frac56 6 5
Worked solution (try it first) There are
2 + 6 + 5 = 13 2 + 6 + 5 = 13 2 + 6 + 5 = 13 balls.
5 of them are black, so the probability is
5 13 \frac{5}{13} 13 5 , option A.
Watch out
Divide by all 13 balls, including the 2 red. Leaving them out gives 5 11 \frac{5}{11} 11 5 Report a problem with this question