Evaluate, correct to three decimal places, 4.314 × 0.000056 0.0067 \dfrac{4.314 \times 0.000056}{0.0067} 0.0067 4.314 × 0.000056
Worked solution (try it first) Top:
4.314 × 0.000056 = 0.000241584 4.314 \times 0.000056 = 0.000241584 4.314 × 0.000056 = 0.000241584 .
Divide:
0.000241584 ÷ 0.0067 = 0.036057 … 0.000241584 \div 0.0067 = 0.036057\ldots 0.000241584 ÷ 0.0067 = 0.036057 … To 3 decimal places, the fourth decimal is 0, so round down: 0.036, option B.
Watch out
Keep track of the decimal point. A slip of one place gives 0.361 (option A), ten times too big. Report a problem with this question
There are 30 students in a class. 15 study woodwork and 13 study metalwork. 6 study neither of the two subjects. How many students study woodwork but not metalwork?
Worked solution (try it first) 30 − 6 = 24 30 - 6 = 24 30 − 6 = 24 students study at least one of the two subjects.
Add the two subject totals and take away 24 to find how many study both:
15 + 13 − 24 = 4 15 + 13 - 24 = 4 15 + 13 − 24 = 4 .
Woodwork but not metalwork:
15 − 4 = 11 15 - 4 = 11 15 − 4 = 11 , option C.
Watch out
Read which group is asked for. Woodwork but not metalwork is 15 − 4 = 11 15 - 4 = 11 15 − 4 = 11 13 − 4 = 9 13 - 4 = 9 13 − 4 = 9 Report a problem with this question
Solve 2 5 x ÷ 2 x = 2 10 5 2^{5x} \div 2^x = \sqrt[5]{2^{10}} 2 5 x ÷ 2 x = 5 2 10
A 1 3 \frac13 3 1 B 1 5 \frac15 5 1 C 1 2 \frac12 2 1 D 1 4 \frac14 4 1
Worked solution (try it first) Dividing powers of 2, subtract the indices:
2 5 x ÷ 2 x = 2 4 x 2^{5x} \div 2^x = 2^{4x} 2 5 x ÷ 2 x = 2 4 x .
A fifth root divides the index by 5:
2 10 5 = 2 2 \sqrt[5]{2^{10}} = 2^2 5 2 10 = 2 2 .
So
4 x = 2 4x = 2 4 x = 2 and
x = 1 2 x = \frac12 x = 2 1 , option C.
Watch out
Subtract the indices when dividing. Adding them gives 2 6 x = 2 2 2^{6x} = 2^2 2 6 x = 2 2 x = 1 3 x = \frac13 x = 3 1 Report a problem with this question
Solve 1 + x − 3 3 = 4 1 + \sqrt[3]{x - 3} = 4 1 + 3 x − 3 = 4
Worked solution (try it first) Subtract 1 from both sides:
x − 3 3 = 3 \sqrt[3]{x - 3} = 3 3 x − 3 = 3 .
Cube both sides to undo the cube root:
x − 3 = 27 x - 3 = 27 x − 3 = 27 .
Add 3:
x = 30 x = 30 x = 30 , option C.
Watch out
Undo a cube root by cubing, not squaring. Squaring gives x − 3 = 9 x - 3 = 9 x − 3 = 9 x = 12 x = 12 x = 12 Report a problem with this question
Express 413 7 413_7 41 3 7
A 1131 5 1131_5 113 1 5 B 1311 5 1311_5 131 1 5 C 2311 5 2311_5 231 1 5 D 2132 5 2132_5 213 2 5
Worked solution (try it first) Change to base ten:
413 7 = 4 × 49 + 7 + 3 = 206 413_7 = 4 \times 49 + 7 + 3 = 206 41 3 7 = 4 × 49 + 7 + 3 = 206 .
Divide by 5 repeatedly:
206 = 5 × 41 + 1 206 = 5 \times 41 + 1 206 = 5 × 41 + 1 ,
41 = 5 × 8 + 1 41 = 5 \times 8 + 1 41 = 5 × 8 + 1 ,
8 = 5 × 1 + 3 8 = 5 \times 1 + 3 8 = 5 × 1 + 3 , and
1 = 5 × 0 + 1 1 = 5 \times 0 + 1 1 = 5 × 0 + 1 .
Read the remainders from the bottom up:
1311 5 1311_5 131 1 5 , option B.
Watch out
Read the remainders from the last division to the first. Reading them top-down gives 1131 5 1131_5 113 1 5 Report a problem with this question
Solve log 3 x + log 3 ( x − 8 ) = 2 \log_3 x + \log_3(x - 8) = 2 log 3 x + log 3 ( x − 8 ) = 2
Worked solution (try it first) Adding logs multiplies:
log 3 [ x ( x − 8 ) ] = 2 \log_3 [x(x - 8)] = 2 log 3 [ x ( x − 8 )] = 2 .
Change to index form:
x ( x − 8 ) = 3 2 = 9 x(x - 8) = 3^2 = 9 x ( x − 8 ) = 3 2 = 9 , so
x 2 − 8 x − 9 = 0 x^2 - 8x - 9 = 0 x 2 − 8 x − 9 = 0 .
Factorise:
( x − 9 ) ( x + 1 ) = 0 (x - 9)(x + 1) = 0 ( x − 9 ) ( x + 1 ) = 0 , so
x = 9 x = 9 x = 9 or
x = − 1 x = -1 x = − 1 .
x = − 1 x = -1 x = − 1 would need the log of a negative number, so
x = 9 x = 9 x = 9 , option D.
Watch out
Check that each log is defined. x = 8 x = 8 x = 8 log 3 ( x − 8 ) = log 3 0 \log_3(x - 8) = \log_3 0 log 3 ( x − 8 ) = log 3 0 x = − 1 x = -1 x = − 1 Report a problem with this question
Mr Manu is 4 times as old as his son, Adu; 7 years ago the sum of their ages was 76. How old is Adu?
A 12 years B 15 years C 18 years D 22 years
Worked solution (try it first) Let Adu be
a a a years old now, so Mr Manu is
4 a 4a 4 a .
Seven years ago each was 7 years younger:
( 4 a − 7 ) + ( a − 7 ) = 76 (4a - 7) + (a - 7) = 76 ( 4 a − 7 ) + ( a − 7 ) = 76 , so
5 a − 14 = 76 5a - 14 = 76 5 a − 14 = 76 .
Add 14 to both sides:
5 a = 90 5a = 90 5 a = 90 .
Divide by 5:
a = 18 a = 18 a = 18 .
Adu is 18 years old, option C.
Watch out
Take 7 years from each person, 14 in all. Taking 7 only once gives 5 a = 83 5a = 83 5 a = 83 5 a = 76 5a = 76 5 a = 76 Report a problem with this question
Factorize completely: x 2 − ( y + z ) 2 x^2 - (y + z)^2 x 2 − ( y + z ) 2
A ( x − y + z ) ( x − y − z ) (x - y + z)(x - y - z) ( x − y + z ) ( x − y − z ) B ( x + y + z ) ( x − y − z ) (x + y + z)(x - y - z) ( x + y + z ) ( x − y − z ) C ( x + y − z ) ( x + y + z ) (x + y - z)(x + y + z) ( x + y − z ) ( x + y + z ) D ( x − y − z ) ( x − y − z ) (x - y - z)(x - y - z) ( x − y − z ) ( x − y − z )
Worked solution (try it first) Use the difference of two squares with
A = x A = x A = x and
B = y + z B = y + z B = y + z :
x 2 − ( y + z ) 2 = [ x + ( y + z ) ] [ x − ( y + z ) ] x^2 - (y + z)^2 = [x + (y + z)][x - (y + z)] x 2 − ( y + z ) 2 = [ x + ( y + z )] [ x − ( y + z )] .
Remove the inner brackets:
x − ( y + z ) = x − y − z x - (y + z) = x - y - z x − ( y + z ) = x − y − z .
So the factors are
( x + y + z ) ( x − y − z ) (x + y + z)(x - y - z) ( x + y + z ) ( x − y − z ) , option B.
Watch out
Keep y + z y + z y + z x − ( y + z ) = x − y − z x - (y + z) = x - y - z x − ( y + z ) = x − y − z ( x + y − z ) ( x + y + z ) (x + y - z)(x + y + z) ( x + y − z ) ( x + y + z ) ( x + y ) 2 − z 2 (x + y)^2 - z^2 ( x + y ) 2 − z 2 x 2 − ( y + z ) 2 x^2 - (y + z)^2 x 2 − ( y + z ) 2 Report a problem with this question
Find the roots of the equation 3 m 2 − 2 m − 65 = 0 3m^2 - 2m - 65 = 0 3 m 2 − 2 m − 65 = 0
A ( 13 3 , 5 ) \left(\frac{13}{3}, 5\right) ( 3 13 , 5 ) B ( − 13 3 , − 5 ) \left(-\frac{13}{3}, -5\right) ( − 3 13 , − 5 ) C ( − 13 3 , 5 ) \left(-\frac{13}{3}, 5\right) ( − 3 13 , 5 ) D ( 13 3 , − 5 ) \left(\frac{13}{3}, -5\right) ( 3 13 , − 5 )
Worked solution (try it first) Find two numbers with product
3 × ( − 65 ) = − 195 3 \times (-65) = -195 3 × ( − 65 ) = − 195 and sum
− 2 -2 − 2 : they are
− 15 -15 − 15 and 13.
Split and group:
3 m 2 − 15 m + 13 m − 65 = 3 m ( m − 5 ) + 13 ( m − 5 ) 3m^2 - 15m + 13m - 65 = 3m(m - 5) + 13(m - 5) 3 m 2 − 15 m + 13 m − 65 = 3 m ( m − 5 ) + 13 ( m − 5 ) = ( 3 m + 13 ) ( m − 5 ) = (3m + 13)(m - 5) = ( 3 m + 13 ) ( m − 5 ) .
So
m = 5 m = 5 m = 5 or
m = − 13 3 m = -\frac{13}{3} m = − 3 13 , option C.
Watch out
m − 5 = 0 m - 5 = 0 m − 5 = 0 + 5 +5 + 5 3 m + 13 = 0 3m + 13 = 0 3 m + 13 = 0 − 13 3 -\frac{13}{3} − 3 13 Report a problem with this question
M M M n n n q q q M = 24 M = 24 M = 24 n = 2 n = 2 n = 2 q = 4 q = 4 q = 4 M M M n = 5 n = 5 n = 5 q = 9 q = 9 q = 9
Worked solution (try it first) Jointly as
n 2 n^2 n 2 and
q \sqrt q q :
M = k n 2 q M = kn^2\sqrt q M = k n 2 q .
Put in
M = 24 M = 24 M = 24 ,
n = 2 n = 2 n = 2 ,
q = 4 q = 4 q = 4 :
24 = k × 4 × 2 = 8 k 24 = k \times 4 \times 2 = 8k 24 = k × 4 × 2 = 8 k , so
k = 3 k = 3 k = 3 .
When
n = 5 n = 5 n = 5 ,
q = 9 q = 9 q = 9 :
M = 3 × 25 × 3 = 225 M = 3 \times 25 \times 3 = 225 M = 3 × 25 × 3 = 225 , option D.
Watch out
Take the square root of q q q 4 = 2 \sqrt4 = 2 4 = 2 9 = 3 \sqrt9 = 3 9 = 3 q q q k = 1.5 k = 1.5 k = 1.5 M = 337.5 M = 337.5 M = 337.5 Report a problem with this question
If m : n = 2 1 3 : 1 1 5 m : n = 2\frac13 : 1\frac15 m : n = 2 3 1 : 1 5 1 n : q = 1 1 2 : 1 1 3 n : q = 1\frac12 : 1\frac13 n : q = 1 2 1 : 1 3 1 q : m q : m q : m
A 16 : 35 16 : 35 16 : 35 B 35 : 16 35 : 16 35 : 16 C 18 : 35 18 : 35 18 : 35 D 35 : 18 35 : 18 35 : 18
Worked solution (try it first) Clear the fractions.
m : n = 7 3 : 6 5 m : n = \frac73 : \frac65 m : n = 3 7 : 5 6 .
Multiply by 15 to get
35 : 18 35 : 18 35 : 18 .
n : q = 3 2 : 4 3 n : q = \frac32 : \frac43 n : q = 2 3 : 3 4 .
Multiply by 6 to get
9 : 8 9 : 8 9 : 8 .
Make the
n n n values the same:
9 : 8 = 18 : 16 9 : 8 = 18 : 16 9 : 8 = 18 : 16 .
So
m : n : q = 35 : 18 : 16 m : n : q = 35 : 18 : 16 m : n : q = 35 : 18 : 16 .
So
q : m = 16 : 35 q : m = 16 : 35 q : m = 16 : 35 , option A.
Watch out
Answer in the order asked, q : m q : m q : m 35 : 16 35 : 16 35 : 16 m : q m : q m : q Report a problem with this question
One-third of the sum of two numbers is 12, and twice their difference is 12. Find the numbers.
A 21 and 15 B 20 and 16 C 22 and 14 D 23 and 13
Worked solution (try it first) One-third of the sum is 12, so multiply by 3:
x + y = 36 x + y = 36 x + y = 36 .
Twice the difference is 12, so divide by 2:
x − y = 6 x - y = 6 x − y = 6 .
Add the equations:
2 x = 42 2x = 42 2 x = 42 , so
x = 21 x = 21 x = 21 .
Then
y = 36 − 21 = 15 y = 36 - 21 = 15 y = 36 − 21 = 15 .
So the numbers are 21 and 15, option A.
Watch out
Every option adds up to 36, so the difference decides. Twice the difference is 12, so the difference is 6; 22 and 14 (option C) differ by 8. Report a problem with this question
Find the quadratic equation whose roots are 2 3 \frac23 3 2 − 3 4 -\frac34 − 4 3
A y 2 + y − 6 = 0 y^2 + y - 6 = 0 y 2 + y − 6 = 0 B 12 y 2 − y − 6 = 0 12y^2 - y - 6 = 0 12 y 2 − y − 6 = 0 C 12 y 2 + y − 6 = 0 12y^2 + y - 6 = 0 12 y 2 + y − 6 = 0 D 12 y 2 − y + 6 = 0 12y^2 - y + 6 = 0 12 y 2 − y + 6 = 0
Worked solution (try it first) Roots
2 3 \frac23 3 2 and
− 3 4 -\frac34 − 4 3 give the factors
( 3 y − 2 ) (3y - 2) ( 3 y − 2 ) and
( 4 y + 3 ) (4y + 3) ( 4 y + 3 ) , cleared of fractions.
Expand:
( 3 y − 2 ) ( 4 y + 3 ) = 12 y 2 + 9 y − 8 y − 6 (3y - 2)(4y + 3) = 12y^2 + 9y - 8y - 6 ( 3 y − 2 ) ( 4 y + 3 ) = 12 y 2 + 9 y − 8 y − 6 = 12 y 2 + y − 6 = 12y^2 + y - 6 = 12 y 2 + y − 6 .
So the equation is
12 y 2 + y − 6 = 0 12y^2 + y - 6 = 0 12 y 2 + y − 6 = 0 , option C.
Watch out
A root of 2 3 \frac23 3 2 3 y − 2 = 0 3y - 2 = 0 3 y − 2 = 0 ( 3 y + 2 ) ( 4 y − 3 ) = 12 y 2 − y − 6 (3y + 2)(4y - 3) = 12y^2 - y - 6 ( 3 y + 2 ) ( 4 y − 3 ) = 12 y 2 − y − 6 Report a problem with this question
Make x x x y = a x 3 − b 3 z y = \dfrac{ax^3 - b}{3z} y = 3 z a x 3 − b
A x = 3 y z b a 3 x = \sqrt[3]{\dfrac{3yzb}{a}} x = 3 a 3 y z b B x = 3 y z + b a 3 x = \sqrt[3]{\dfrac{3yz + b}{a}} x = 3 a 3 y z + b C x = 3 y z − b a 3 x = \sqrt[3]{\dfrac{3yz - b}{a}} x = 3 a 3 y z − b D x = 3 y 3 + b a z 3 x = \sqrt[3]{\dfrac{3y^3 + b}{az}} x = 3 a z 3 y 3 + b
Worked solution (try it first) Multiply both sides by
3 z 3z 3 z :
3 y z = a x 3 − b 3yz = ax^3 - b 3 y z = a x 3 − b .
Add
b b b and divide by
a a a :
x 3 = 3 y z + b a x^3 = \frac{3yz + b}{a} x 3 = a 3 y z + b .
Take the cube root:
x = 3 y z + b a 3 x = \sqrt[3]{\dfrac{3yz + b}{a}} x = 3 a 3 y z + b , option B.
Watch out
b b b + b +b + b − b -b − b Report a problem with this question
The price of a shoe was decreased by 22 % 22\% 22%
A $62.30 B $42.30 C $72.00 D $35.00
Worked solution (try it first) A
22 % 22\% 22% decrease leaves
78 % 78\% 78% of the original price
P P P :
0.78 P = 27.30 0.78P = 27.30 0.78 P = 27.30 .
Divide both sides by 0.78:
P = 35 P = 35 P = 35 .
So the original price was $35.00, option D.
Watch out
The 22 % 22\% 22% 22 % 22\% 22% Report a problem with this question
The radius and height of a solid cylinder are 8 cm 8\text{ cm} 8 cm 14 cm 14\text{ cm} 14 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 1,106.28 cm 2 1{,}106.28\text{ cm}^2 1 , 106.28 cm 2 B 1,016.29 cm 2 1{,}016.29\text{ cm}^2 1 , 016.29 cm 2 C 1,106.29 cm 2 1{,}106.29\text{ cm}^2 1 , 106.29 cm 2 D 1,206.27 cm 2 1{,}206.27\text{ cm}^2 1 , 206.27 cm 2
Worked solution (try it first) A solid cylinder has a curved surface and two ends:
2 π r ( r + h ) 2\pi r(r + h) 2 π r ( r + h ) .
2 × 22 7 × 8 × ( 8 + 14 ) = 7744 7 2 \times \frac{22}{7} \times 8 \times (8 + 14) = \frac{7744}{7} 2 × 7 22 × 8 × ( 8 + 14 ) = 7 7744 ≈ 1106.29 cm 2 \approx 1106.29\text{ cm}^2 ≈ 1106.29 cm 2 , option C.
Watch out
Include both ends. The curved surface alone is 2 × 22 7 × 8 × 14 = 704 cm 2 2 \times \frac{22}{7} \times 8 \times 14 = 704\text{ cm}^2 2 × 7 22 × 8 × 14 = 704 cm 2 Report a problem with this question
In the diagram, O O O ∣ N T ∣ = ∣ S T ∣ |NT| = |ST| ∣ N T ∣ = ∣ S T ∣ ∠ N T S = 36 ∘ \angle NTS = 36^\circ ∠ N T S = 3 6 ∘ R N RN R N R S RS R S N N N S S S t t t
A 36 ∘ 36^\circ 3 6 ∘ B 72 ∘ 72^\circ 7 2 ∘ C 108 ∘ 108^\circ 10 8 ∘ D 54 ∘ 54^\circ 5 4 ∘
Worked solution (try it first) t t t is between the tangent
S R SR S R and the chord
S N SN S N .
The angle between a tangent and a chord equals the angle in the alternate segment, which is the angle at
T T T opposite
S N SN S N .
So
t = ∠ N T S = 36 ∘ t = \angle NTS = 36^\circ t = ∠ N T S = 3 6 ∘ , option A.
Watch out
The base angles of isosceles triangle N T S NTS N T S 72 ∘ 72^\circ 7 2 ∘ t t t S N SN S N 36 ∘ 36^\circ 3 6 ∘ T T T Report a problem with this question
The radius of a sphere is 3 cm 3\text{ cm} 3 cm π \pi π
A 27 π cm 3 27\pi\text{ cm}^3 27 π cm 3 B 108 π cm 3 108\pi\text{ cm}^3 108 π cm 3 C 36 π cm 3 36\pi\text{ cm}^3 36 π cm 3 D 30 π cm 3 30\pi\text{ cm}^3 30 π 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 × 27 = 36 \frac43 \times 27 = 36 3 4 × 27 = 36 , so the volume is
36 π cm 3 36\pi\text{ cm}^3 36 π cm 3 , option C.
Watch out
Divide by 3 as well as multiplying by 4. 4 × 27 = 108 4 \times 27 = 108 4 × 27 = 108 108 π cm 3 108\pi\text{ cm}^3 108 π cm 3 Report a problem with this question
Arrange the following in ascending order of magnitude: 110 two 110_{\text{two}} 11 0 two 31 eight 31_{\text{eight}} 3 1 eight 42 five 42_{\text{five}} 4 2 five
A 110 two , 31 eight , 42 five 110_{\text{two}}, 31_{\text{eight}}, 42_{\text{five}} 11 0 two , 3 1 eight , 4 2 five B 110 two , 42 five , 31 eight 110_{\text{two}}, 42_{\text{five}}, 31_{\text{eight}} 11 0 two , 4 2 five , 3 1 eight C 31 eight , 42 five , 110 two 31_{\text{eight}}, 42_{\text{five}}, 110_{\text{two}} 3 1 eight , 4 2 five , 11 0 two D 42 five , 31 eight , 110 two 42_{\text{five}}, 31_{\text{eight}}, 110_{\text{two}} 4 2 five , 3 1 eight , 11 0 two
Worked solution (try it first) Change each to base ten:
110 two = 4 + 2 = 6 110_{\text{two}} = 4 + 2 = 6 11 0 two = 4 + 2 = 6 ,
31 eight = 24 + 1 = 25 31_{\text{eight}} = 24 + 1 = 25 3 1 eight = 24 + 1 = 25 and
42 five = 20 + 2 = 22 42_{\text{five}} = 20 + 2 = 22 4 2 five = 20 + 2 = 22 .
In ascending order:
6 < 22 < 25 6 < 22 < 25 6 < 22 < 25 .
So the order is
110 two , 42 five , 31 eight 110_{\text{two}}, 42_{\text{five}}, 31_{\text{eight}} 11 0 two , 4 2 five , 3 1 eight , option B.
Watch out
Don't compare the digits as they stand: 42 looks bigger than 31, but 42 five = 22 42_{\text{five}} = 22 4 2 five = 22 31 eight = 25 31_{\text{eight}} = 25 3 1 eight = 25 Report a problem with this question
A notebook of length 15 cm 15\text{ cm} 15 cm 16.8 cm 16.8\text{ cm} 16.8 cm
A 12.00 % 12.00\% 12.00% B 11.71 % 11.71\% 11.71% C 11.00 % 11.00\% 11.00% D 10.71 % 10.71\% 10.71%
Worked solution (try it first) The error is
16.8 − 15 = 1.8 16.8 - 15 = 1.8 16.8 − 15 = 1.8 cm.
Divide by the true length and multiply by 100:
1.8 15 × 100 % = 0.12 × 100 % \frac{1.8}{15} \times 100\% = 0.12 \times 100\% 15 1.8 × 100% = 0.12 × 100% .
So the percentage error is 12.00%, option A.
Watch out
Divide by the true length, 15 cm, not the measured one. Using 16.8 gives 1.8 16.8 × 100 % = 10.71 % \frac{1.8}{16.8} \times 100\% = 10.71\% 16.8 1.8 × 100% = 10.71% Report a problem with this question
Find the value of m m m
A 140 ∘ 140^\circ 14 0 ∘ B 130 ∘ 130^\circ 13 0 ∘ C 50 ∘ 50^\circ 5 0 ∘ D 40 ∘ 40^\circ 4 0 ∘
Worked solution (try it first) Draw a line through the right-angled corner parallel to the other two.
By alternate angles, the lower arm makes
40 ∘ 40^\circ 4 0 ∘ with it.
The corner is
90 ∘ 90^\circ 9 0 ∘ , so the upper arm makes
90 ∘ − 40 ∘ = 50 ∘ 90^\circ - 40^\circ = 50^\circ 9 0 ∘ − 4 0 ∘ = 5 0 ∘ with the new line.
m m m and that
50 ∘ 50^\circ 5 0 ∘ are co-interior between the new line and the upper parallel:
m = 180 ∘ − 50 ∘ = 130 ∘ m = 180^\circ - 50^\circ = 130^\circ m = 18 0 ∘ − 5 0 ∘ = 13 0 ∘ , option B.
Watch out
50 ∘ 50^\circ 5 0 ∘ m m m 180 ∘ − 50 ∘ 180^\circ - 50^\circ 18 0 ∘ − 5 0 ∘ Report a problem with this question
A line L L L ( 6 , − 13 ) (6, -13) ( 6 , − 13 ) ( 7 , 4 ) (7, 4) ( 7 , 4 ) ( − 3 , 9 ) (-3, 9) ( − 3 , 9 ) L L L
A y = 1 2 x − 10 y = \frac12x - 10 y = 2 1 x − 10 B y = − 1 2 x + 10 y = -\frac12x + 10 y = − 2 1 x + 10 C y = − 1 2 x − 10 y = -\frac12x - 10 y = − 2 1 x − 10 D y = 1 2 x + 10 y = \frac12x + 10 y = 2 1 x + 10
Worked solution (try it first) Gradient of the given line:
9 − 4 − 3 − 7 = 5 − 10 \dfrac{9 - 4}{-3 - 7} = \frac{5}{-10} − 3 − 7 9 − 4 = − 10 5 Parallel lines share it.
Through
( 6 , − 13 ) (6, -13) ( 6 , − 13 ) :
y + 13 = − 1 2 ( x − 6 ) y + 13 = -\frac12(x - 6) y + 13 = − 2 1 ( x − 6 ) , so
y + 13 = − 1 2 x + 3 y + 13 = -\frac12x + 3 y + 13 = − 2 1 x + 3 .
Subtract 13 from both sides:
y = − 1 2 x − 10 y = -\frac12x - 10 y = − 2 1 x − 10 , option C.
Watch out
Check the constant with the point: − 1 2 ( 6 ) − 10 = − 13 -\frac12(6) - 10 = -13 − 2 1 ( 6 ) − 10 = − 13 − 3 + 10 = 7 -3 + 10 = 7 − 3 + 10 = 7 Report a problem with this question
An empty cylindrical tank is 140 cm 140\text{ cm} 140 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 91 cm 91\text{ cm} 91 cm B 57 cm 57\text{ cm} 57 cm C 13 cm 13\text{ cm} 13 cm D 7 cm 7\text{ cm} 7 cm
Worked solution (try it first) Change litres to cm³: 200 litres is
200 000 cm 3 200\,000\text{ cm}^3 200 000 cm 3 .
The radius is
140 ÷ 2 = 70 140 \div 2 = 70 140 ÷ 2 = 70 cm.
Base area:
22 7 × 70 2 = 15 400 cm 2 \frac{22}{7} \times 70^2 = 15\,400\text{ cm}^2 7 22 × 7 0 2 = 15 400 cm 2 .
Height
= 200 000 ÷ 15 400 ≈ 12.99 = 200\,000 \div 15\,400 \approx 12.99 = 200 000 ÷ 15 400 ≈ 12.99 , which is 13 cm to the nearest centimetre, option C.
Watch out
Halve the diameter. Using 140 cm as the radius makes the base four times too big and gives about 3 cm, which is not an option. Report a problem with this question
Mrs Kebeh stands at a distance of 110 m 110\text{ m} 110 m 58 m 58\text{ m} 58 m 2 m 2\text{ m} 2 m
A 28 ∘ 28^\circ 2 8 ∘ B 27 ∘ 27^\circ 2 7 ∘ C 26 ∘ 26^\circ 2 6 ∘ D 20 ∘ 20^\circ 2 0 ∘
Worked solution (try it first) Measure from her eye: the top of the building is
58 − 2 = 56 58 - 2 = 56 58 − 2 = 56 m higher, and 110 m away.
So
tan θ = 56 110 ≈ 0.509 \tan\theta = \frac{56}{110} \approx 0.509 tan θ = 110 56 ≈ 0.509 .
So
θ ≈ 26.98 ∘ \theta \approx 26.98^\circ θ ≈ 26.9 8 ∘ , which is
27 ∘ 27^\circ 2 7 ∘ to the nearest degree, option B.
Watch out
Take off her height first: 58 − 2 = 56 58 - 2 = 56 58 − 2 = 56 27.8 ∘ 27.8^\circ 27. 8 ∘ 28 ∘ 28^\circ 2 8 ∘ Report a problem with this question
Find the mean deviation of the numbers 14, 15, 16, 17, 18, 19.
Worked solution (try it first) The numbers add up to 99, so the mean is
99 6 = 16.5 \frac{99}{6} = 16.5 6 99 = 16.5 .
The distances from 16.5 are 2.5, 1.5, 0.5, 0.5, 1.5 and 2.5, which add up to 9.
The mean deviation is
9 6 = 1.5 \frac96 = 1.5 6 9 = 1.5 , option B.
Watch out
Mean deviation uses the distances, not their squares. Squaring them gives the standard deviation, about 1.7 (option A). Report a problem with this question
The interior angle of a regular polygon is 6 times its exterior angle. Find the number of sides of the polygon.
Worked solution (try it first) Let the exterior angle be
e e e .
The interior angle is
6 e 6e 6 e , and the two add up to
180 ∘ 180^\circ 18 0 ∘ :
7 e = 180 ∘ 7e = 180^\circ 7 e = 18 0 ∘ .
So
e = 180 ∘ 7 e = \frac{180^\circ}{7} e = 7 18 0 ∘ .
The number of sides is
360 ÷ 180 7 = 360 × 7 180 360 \div \frac{180}{7} = 360 \times \frac{7}{180} 360 ÷ 7 180 = 360 × 180 7 Watch out
The interior and exterior angles together make 180 ∘ 180^\circ 18 0 ∘ 7 e = 180 ∘ 7e = 180^\circ 7 e = 18 0 ∘ 6 e = 180 ∘ 6e = 180^\circ 6 e = 18 0 ∘ e = 30 ∘ e = 30^\circ e = 3 0 ∘ Report a problem with this question
The length of the diagonal of a square is 12 cm 12\text{ cm} 12 cm
A 36 cm 2 36\text{ cm}^2 36 cm 2 B 48 cm 2 48\text{ cm}^2 48 cm 2 C 72 cm 2 72\text{ cm}^2 72 cm 2 D 18 cm 2 18\text{ cm}^2 18 cm 2
Worked solution (try it first) The diagonal is the hypotenuse of a right-angled triangle whose legs are two sides
s s s :
s 2 + s 2 = 12 2 s^2 + s^2 = 12^2 s 2 + s 2 = 1 2 2 .
So
2 s 2 = 144 2s^2 = 144 2 s 2 = 144 , and the area is
s 2 = 72 cm 2 s^2 = 72\text{ cm}^2 s 2 = 72 cm 2 , option C.
Watch out
The side is not half the diagonal. Taking s = 6 s = 6 s = 6 36 cm 2 36\text{ cm}^2 36 cm 2 d 2 2 = 72 \frac{d^2}{2} = 72 2 d 2 = 72 Report a problem with this question
Consider the statements p p p q q q
A q ⇒ p q \Rightarrow p q ⇒ p B p ⇔ q p \Leftrightarrow q p ⇔ q C p ⇒ q p \Rightarrow q p ⇒ q D ∼ q ⇔ p \sim q \Leftrightarrow p ∼ q ⇔ p
Worked solution (try it first) "Siah is from Foya" is
p p p and "Foya is in Lofa" is
q q q .
"If
p p p , then
q q q " is written
p p p ⇒ q \Rightarrow q ⇒ q , option C.
Watch out
The "if" part goes first, before the arrow. Writing it the other way round gives the converse, q ⇒ p q \Rightarrow p q ⇒ p Report a problem with this question
Name the triangle with the vertices ( 1 , − 3 ) (1, -3) ( 1 , − 3 ) ( 6 , 2 ) (6, 2) ( 6 , 2 ) ( 0 , 4 ) (0, 4) ( 0 , 4 )
A Isosceles triangle B Equilateral triangle C Right triangle D Scalene triangle
Worked solution (try it first) Call the points
A ( 1 , − 3 ) A(1, -3) A ( 1 , − 3 ) ,
B ( 6 , 2 ) B(6, 2) B ( 6 , 2 ) and
C ( 0 , 4 ) C(0, 4) C ( 0 , 4 ) , and find the squared side lengths.
A B 2 = 5 2 + 5 2 = 50 AB^2 = 5^2 + 5^2 = 50 A B 2 = 5 2 + 5 2 = 50 ,
B C 2 = 6 2 + 2 2 = 40 BC^2 = 6^2 + 2^2 = 40 B C 2 = 6 2 + 2 2 = 40 and
A C 2 = 1 2 + 7 2 = 50 AC^2 = 1^2 + 7^2 = 50 A C 2 = 1 2 + 7 2 = 50 .
A B = A C AB = AC A B = A C , so two sides are equal.
It is not right-angled, since
40 + 50 ≠ 50 40 + 50 \ne 50 40 + 50 = 50 and
50 + 50 ≠ 40 50 + 50 \ne 40 50 + 50 = 40 .
So it is an isosceles triangle, option A.
Watch out
To test for a right angle, the two smaller squares must add up to the largest: here 40 + 50 = 90 40 + 50 = 90 40 + 50 = 90 Report a problem with this question
In the diagram, N R NR N R S R M SRM S R M ∠ M N R = x ∘ \angle MNR = x^\circ ∠ M N R = x ∘ ∠ S R N = ( 5 x + 20 ) ∘ \angle SRN = (5x + 20)^\circ ∠ S R N = ( 5 x + 20 ) ∘ 2 x ∘ 2x^\circ 2 x ∘
A 20 ∘ 20^\circ 2 0 ∘ B 35 ∘ 35^\circ 3 5 ∘ C 42 ∘ 42^\circ 4 2 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) N R NR N R is a diameter, so
∠ N M R = 90 ∘ \angle NMR = 90^\circ ∠ N M R = 9 0 ∘ .
In triangle
N M R NMR N M R ,
∠ M R N = 180 ∘ − 90 ∘ − x ∘ \angle MRN = 180^\circ - 90^\circ - x^\circ ∠ M R N = 18 0 ∘ − 9 0 ∘ − x ∘ = ( 90 − x ) ∘ = (90 - x)^\circ = ( 90 − x ) ∘ .
S R M SRM S R M is a straight line, so
∠ S R N + ∠ M R N = 180 ∘ \angle SRN + \angle MRN = 180^\circ ∠ S R N + ∠ M R N = 18 0 ∘ :
( 5 x + 20 ) + ( 90 − x ) = 180 (5x + 20) + (90 - x) = 180 ( 5 x + 20 ) + ( 90 − x ) = 180 .
Simplify:
4 x + 110 = 180 4x + 110 = 180 4 x + 110 = 180 , so
4 x = 70 4x = 70 4 x = 70 and
x = 17.5 x = 17.5 x = 17.5 .
So
2 x = 35 ∘ 2x = 35^\circ 2 x = 3 5 ∘ , option B.
Watch out
The question asks for 2 x 2x 2 x x x x x = 17.5 ∘ x = 17.5^\circ x = 17. 5 ∘ 35 ∘ 35^\circ 3 5 ∘ Report a problem with this question
Find the value of α \alpha α cos ( α + 14 ) ∘ = sin ( 4 α + 6 ) ∘ \cos(\alpha + 14)^\circ = \sin(4\alpha + 6)^\circ cos ( α + 14 ) ∘ = sin ( 4 α + 6 ) ∘
Worked solution (try it first) The cosine of an angle equals the sine of its complement, so the two angles add up to
90 ∘ 90^\circ 9 0 ∘ :
( α + 14 ) + ( 4 α + 6 ) = 90 (\alpha + 14) + (4\alpha + 6) = 90 ( α + 14 ) + ( 4 α + 6 ) = 90 .
Collect terms:
5 α + 20 = 90 5\alpha + 20 = 90 5 α + 20 = 90 , so
5 α = 70 5\alpha = 70 5 α = 70 .
Divide by 5:
α = 14 \alpha = 14 α = 14 , option A.
(Check: the angles are
28 ∘ 28^\circ 2 8 ∘ and
62 ∘ 62^\circ 6 2 ∘ .)
Watch out
The two angles add up to 90 ∘ 90^\circ 9 0 ∘ α + 14 = 4 α + 6 \alpha + 14 = 4\alpha + 6 α + 14 = 4 α + 6 α = 8 3 \alpha = \frac83 α = 3 8 Report a problem with this question
A bag contains 4 white marbles and 3 blue marbles. Another bag contains 5 red marbles and 6 blue marbles. If a marble is picked at random from each bag, find the probability that they are of the same colour.
A 1 2 \frac12 2 1 B 18 77 \frac{18}{77} 77 18 C 11 12 \frac{11}{12} 12 11 D 9 11 \frac{9}{11} 11 9
Worked solution (try it first) The first bag has white and blue.
The second has red and blue.
So the only colour they share is blue.
Both blue:
3 7 × 6 11 \frac37 \times \frac{6}{11} 7 3 × 11 6 , since the picks are independent.
So the probability is
18 77 \frac{18}{77} 77 18 , option B.
Watch out
Both marbles must be blue, so multiply the two chances. Adding 3 7 + 6 11 \frac37 + \frac{6}{11} 7 3 + 11 6 9 11 \frac{9}{11} 11 9 Report a problem with this question
The angle of a sector of a circle of radius 3.4 cm 3.4\text{ cm} 3.4 cm 115 ∘ 115^\circ 11 5 ∘ [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 10.2 cm 2 10.2\text{ cm}^2 10.2 cm 2 B 11.6 cm 2 11.6\text{ cm}^2 11.6 cm 2 C 12.7 cm 2 12.7\text{ cm}^2 12.7 cm 2 D 9.4 cm 2 9.4\text{ cm}^2 9.4 cm 2
Worked solution (try it first) Area of a sector
= θ 360 × π r 2 = \frac{\theta}{360} \times \pi r^2 = 360 θ × π r 2 , with
r 2 = 3.4 2 = 11.56 r^2 = 3.4^2 = 11.56 r 2 = 3. 4 2 = 11.56 .
22 7 × 11.56 ≈ 36.33 \frac{22}{7} \times 11.56 \approx 36.33 7 22 × 11.56 ≈ 36.33 , so the area is
115 360 × 36.33 ≈ 11.6 cm 2 \frac{115}{360} \times 36.33 \approx 11.6\text{ cm}^2 360 115 × 36.33 ≈ 11.6 cm 2 , option B.
Watch out
Use π r 2 \pi r^2 π r 2 2 π r 2\pi r 2 π r Report a problem with this question
The diagonals of a rhombus are 16 cm 16\text{ cm} 16 cm 12 cm 12\text{ cm} 12 cm
A 8 cm 8\text{ cm} 8 cm B 10 cm 10\text{ cm} 10 cm C 14 cm 14\text{ cm} 14 cm D 20 cm 20\text{ cm} 20 cm
Worked solution (try it first) The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with legs 8 cm and 6 cm.
By Pythagoras, the side is
8 2 + 6 2 = 100 = 10 \sqrt{8^2 + 6^2} = \sqrt{100} = 10 8 2 + 6 2 = 100 = 10 cm, option B.
Watch out
Use half of each diagonal. With the full 16 and 12 you get 20 cm (option D), twice the side. Report a problem with this question
The angle of elevation of the top of a building from a point Z Z Z 50 ∘ 50^\circ 5 0 ∘ 124 m 124\text{ m} 124 m Z Z Z
A 104.05 m 104.05\text{ m} 104.05 m B 147.78 m 147.78\text{ m} 147.78 m C 161.87 m 161.87\text{ m} 161.87 m D 192.91 m 192.91\text{ m} 192.91 m
Worked solution (try it first) The height 124 m is opposite the
50 ∘ 50^\circ 5 0 ∘ angle and the distance
d d d is adjacent, so
tan 50 ∘ = 124 d \tan50^\circ = \frac{124}{d} tan 5 0 ∘ = d 124 .
Rearrange:
d = 124 tan 50 ∘ d = \frac{124}{\tan50^\circ} d = t a n 5 0 ∘ 124 = 124 1.1918 = \frac{124}{1.1918} = 1.1918 124 .
So
d ≈ 104.05 d \approx 104.05 d ≈ 104.05 m, option A.
Watch out
Divide the height by tan 50 ∘ \tan50^\circ tan 5 0 ∘ 124 tan 50 ∘ ≈ 147.78 124\tan50^\circ \approx 147.78 124 tan 5 0 ∘ ≈ 147.78 Report a problem with this question
A student measured the height of a pole as 5.98 m 5.98\text{ m} 5.98 m 5 % 5\% 5%
A 6.65 m 6.65\text{ m} 6.65 m B 7.67 m 7.67\text{ m} 7.67 m C 6.29 m 6.29\text{ m} 6.29 m D 7.18 m 7.18\text{ m} 7.18 m
Worked solution (try it first) Let the actual height be
h h h .
The percentage error is taken of the actual height, so the error is
0.05 h 0.05h 0.05 h .
The measurement is too small, so
h − 0.05 h = 5.98 h - 0.05h = 5.98 h − 0.05 h = 5.98 , that is
0.95 h = 5.98 0.95h = 5.98 0.95 h = 5.98 .
Divide by 0.95:
h = 6.2947 … h = 6.2947\ldots h = 6.2947 … , which is 6.29 m to two decimal places, option C.
Watch out
The 5% is of the actual height, not of the measurement. Adding 5% of 5.98 gives 5.98 × 1.05 = 6.28 5.98 \times 1.05 = 6.28 5.98 × 1.05 = 6.28 Report a problem with this question
In the diagram, O O O ∠ S Q R = 28 ∘ \angle SQR = 28^\circ ∠ S QR = 2 8 ∘ ∠ O R S \angle ORS ∠ O R S
A 76 ∘ 76^\circ 7 6 ∘ B 62 ∘ 62^\circ 6 2 ∘ C 56 ∘ 56^\circ 5 6 ∘ D 28 ∘ 28^\circ 2 8 ∘
Worked solution (try it first) ∠ S O R \angle SOR ∠ S O R is at the centre on the same arc
S R SR S R as
∠ S Q R \angle SQR ∠ S QR .
So
∠ S O R = 2 × 28 ∘ = 56 ∘ \angle SOR = 2 \times 28^\circ = 56^\circ ∠ S O R = 2 × 2 8 ∘ = 5 6 ∘ .
O S = O R OS = OR O S = O R (radii), so triangle
S O R SOR S O R is isosceles:
∠ O R S = 180 ∘ − 56 ∘ 2 \angle ORS = \dfrac{180^\circ - 56^\circ}{2} ∠ O R S = 2 18 0 ∘ − 5 6 ∘ = 62 ∘ = 62^\circ = 6 2 ∘ , option B.
Watch out
56 ∘ 56^\circ 5 6 ∘ ∠ S O R \angle SOR ∠ S O R ∠ O R S \angle ORS ∠ O R S S O R SOR S O R Report a problem with this question
John was facing S35°E. If he turned 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) Change S
35 ∘ 35^\circ 3 5 ∘ E to a three-figure bearing:
180 ∘ − 35 ∘ = 145 ∘ 180^\circ - 35^\circ = 145^\circ 18 0 ∘ − 3 5 ∘ = 14 5 ∘ .
Anticlockwise turns reduce the bearing:
145 ∘ − 90 ∘ = 55 ∘ 145^\circ - 90^\circ = 55^\circ 14 5 ∘ − 9 0 ∘ = 5 5 ∘ .
A bearing of
055 ∘ 055^\circ 05 5 ∘ is
55 ∘ 55^\circ 5 5 ∘ east of north, N
55 ∘ 55^\circ 5 5 ∘ E, option A.
Watch out
Anticlockwise means subtract from the bearing. Adding 90 ∘ 90^\circ 9 0 ∘ 235 ∘ 235^\circ 23 5 ∘ 55 ∘ 55^\circ 5 5 ∘ Report a problem with this question
If 2 x − 3 y = − 11 2x - 3y = -11 2 x − 3 y = − 11 3 x + 2 y = 3 3x + 2y = 3 3 x + 2 y = 3 ( y − x ) 2 (y - x)^2 ( y − x ) 2
Worked solution (try it first) Make the
y y y terms opposite: multiply the first by 2 and the second by 3, giving
4 x − 6 y = − 22 4x - 6y = -22 4 x − 6 y = − 22 and
9 x + 6 y = 9 9x + 6y = 9 9 x + 6 y = 9 .
Add them:
13 x = − 13 13x = -13 13 x = − 13 , so
x = − 1 x = -1 x = − 1 .
Put
x = − 1 x = -1 x = − 1 into
3 x + 2 y = 3 3x + 2y = 3 3 x + 2 y = 3 :
− 3 + 2 y = 3 -3 + 2y = 3 − 3 + 2 y = 3 , so
y = 3 y = 3 y = 3 .
So
y − x = 3 − ( − 1 ) = 4 y - x = 3 - (-1) = 4 y − x = 3 − ( − 1 ) = 4 , and
( y − x ) 2 = 16 (y - x)^2 = 16 ( y − x ) 2 = 16 , option C.
Watch out
Subtracting a negative adds: 3 − ( − 1 ) = 4 3 - (-1) = 4 3 − ( − 1 ) = 4 3 − 1 = 2 3 - 1 = 2 3 − 1 = 2 ( y − x ) 2 = 4 (y - x)^2 = 4 ( y − x ) 2 = 4 Report a problem with this question
An equilateral triangle has a side of 2 cm 2\text{ cm} 2 cm
Worked solution (try it first) The height from the top meets the base at its midpoint, making a right-angled triangle with hypotenuse 2 cm and base 1 cm.
Pythagoras:
h 2 = 2 2 − 1 2 = 3 h^2 = 2^2 - 1^2 = 3 h 2 = 2 2 − 1 2 = 3 .
So
h = 3 h = \sqrt3 h = 3 cm, option A.
Watch out
The side of 2 cm is the hypotenuse, so subtract: h 2 = 4 − 1 h^2 = 4 - 1 h 2 = 4 − 1 5 \sqrt5 5 Report a problem with this question
A number is chosen at random from 40 to 50 inclusive. Find the probability that the number is prime.
A 3 11 \frac{3}{11} 11 3 B 4 11 \frac{4}{11} 11 4 C 5 11 \frac{5}{11} 11 5 D 8 11 \frac{8}{11} 11 8
Worked solution (try it first) From 40 to 50 inclusive there are 11 numbers.
The primes among them are 41, 43 and 47.
So the probability is
3 11 \frac{3}{11} 11 3 , option A.
Watch out
49 looks prime but is 7 × 7 7 \times 7 7 × 7 4 11 \frac{4}{11} 11 4 Report a problem with this question
The bar chart represents the distribution of marks scored by students in an economics examination. If the fail mark was 4, what is the probability that a student selected at random passed?
Worked solution (try it first) Read the bars and add:
3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 students.
A fail mark of 4 means a pass needs more than 4 marks: marks 5 to 8, which is
12 + 7 + 5 + 10 = 34 12 + 7 + 5 + 10 = 34 12 + 7 + 5 + 10 = 34 students.
So the probability of a pass is
34 53 ≈ 0.64 \frac{34}{53} \approx 0.64 53 34 ≈ 0.64 , option A.
Watch out
Check which group the question asks for: 19 53 ≈ 0.36 \frac{19}{53} \approx 0.36 53 19 ≈ 0.36 Report a problem with this question
Using the bar chart in the previous question (marks 1–8 with frequencies 3, 6, 9, 1, 12, 7, 5, 10), what percentage of the students scored at most 5 marks?
A 63.2 % 63.2\% 63.2% B 58.5 % 58.5\% 58.5% C 41.5 % 41.5\% 41.5% D 38.3 % 38.3\% 38.3%
Worked solution (try it first) "At most 5" means 5 or fewer: marks 1 to 5, which is
3 + 6 + 9 + 1 + 12 = 31 3 + 6 + 9 + 1 + 12 = 31 3 + 6 + 9 + 1 + 12 = 31 students.
Out of 53 students:
31 53 × 100 ≈ 58.5 % \frac{31}{53} \times 100 \approx 58.5\% 53 31 × 100 ≈ 58.5% , option B.
Watch out
"At most 5" includes 5 and everything below. 22 53 ≈ 41.5 % \frac{22}{53} \approx 41.5\% 53 22 ≈ 41.5% Report a problem with this question
Using the same bar chart, how many students scored at least 3 marks?
Worked solution (try it first) "At least 3" means 3 or more: marks 3 to 8.
Add those bars:
9 + 1 + 12 + 7 + 5 + 10 = 44 9 + 1 + 12 + 7 + 5 + 10 = 44 9 + 1 + 12 + 7 + 5 + 10 = 44 students, option C.
Watch out
"At least 3" includes the 9 students who scored exactly 3. Leaving them out gives 35, which is not an option. Report a problem with this question
If log a 3 = m \log_a 3 = m log a 3 = m log a 5 = p \log_a 5 = p log a 5 = p log a 75 \log_a 75 log a 75
A m + p 2 m + p^2 m + p 2 B 2 m + p 2m + p 2 m + p C m + 2 p m + 2p m + 2 p D m 2 + p m^2 + p m 2 + p
Worked solution (try it first) Write 75 using 3 and 5:
75 = 3 × 25 = 3 × 5 2 75 = 3 \times 25 = 3 \times 5^2 75 = 3 × 25 = 3 × 5 2 .
So
log a 75 = log a 3 + 2 log a 5 \log_a 75 = \log_a 3 + 2\log_a 5 log a 75 = log a 3 + 2 log a 5 .
That is
m + 2 p m + 2p m + 2 p , option C.
Watch out
A power comes down in front as a multiplier: log a 5 2 = 2 p \log_a 5^2 = 2p log a 5 2 = 2 p p 2 p^2 p 2 Report a problem with this question
In the diagram, M M M N N N R R R O O O ∠ O R N = 48 ∘ \angle ORN = 48^\circ ∠ O R N = 4 8 ∘ ∠ R N M = 124 ∘ \angle RNM = 124^\circ ∠ R N M = 12 4 ∘ ∠ O M N \angle OMN ∠ O M N
A 76 ∘ 76^\circ 7 6 ∘ B 64 ∘ 64^\circ 6 4 ∘ C 58 ∘ 58^\circ 5 8 ∘ D 48 ∘ 48^\circ 4 8 ∘
Worked solution (try it first) ∠ M N R = 124 ∘ \angle MNR = 124^\circ ∠ M N R = 12 4 ∘ stands on the major arc
M R MR M R , so the reflex angle
M O R MOR M O R is
2 × 124 ∘ = 248 ∘ 2 \times 124^\circ = 248^\circ 2 × 12 4 ∘ = 24 8 ∘ .
So the angle
M O R MOR M O R inside quadrilateral
O M N R OMNR O M N R is
360 ∘ − 248 ∘ = 112 ∘ 360^\circ - 248^\circ = 112^\circ 36 0 ∘ − 24 8 ∘ = 11 2 ∘ .
The angles of a quadrilateral add up to
360 ∘ 360^\circ 36 0 ∘ :
∠ O M N = 360 ∘ − 124 ∘ − 48 ∘ − 112 ∘ \angle OMN = 360^\circ - 124^\circ - 48^\circ - 112^\circ ∠ O M N = 36 0 ∘ − 12 4 ∘ − 4 8 ∘ − 11 2 ∘ = 76 ∘ = 76^\circ = 7 6 ∘ , option A.
Watch out
O M = O R OM = OR O M = O R M N MN M N N R NR N R ∠ O M N \angle OMN ∠ O M N ∠ O R N = 48 ∘ \angle ORN = 48^\circ ∠ O R N = 4 8 ∘ Report a problem with this question
Simplify 3 12 + 10 3 − 6 3 3\sqrt{12} + 10\sqrt3 - \dfrac{6}{\sqrt3} 3 12 + 10 3 − 3 6
A 14 3 14\sqrt3 14 3 B 18 3 18\sqrt3 18 3 C 10 3 10\sqrt3 10 3 D 7 3 7\sqrt3 7 3
Worked solution (try it first) Take out the square factor:
3 12 = 3 × 2 3 3\sqrt{12} = 3 \times 2\sqrt3 3 12 = 3 × 2 3 , which is
6 3 6\sqrt3 6 3 .
Rationalise:
6 3 = 6 3 3 \dfrac{6}{\sqrt3} = \dfrac{6\sqrt3}{3} 3 6 = 3 6 3 , which is
2 3 2\sqrt3 2 3 .
Combine:
6 3 + 10 3 − 2 3 = 14 3 6\sqrt3 + 10\sqrt3 - 2\sqrt3 = 14\sqrt3 6 3 + 10 3 − 2 3 = 14 3 , option A.
Watch out
The last term is subtracted. Adding 2 3 2\sqrt3 2 3 18 3 18\sqrt3 18 3 Report a problem with this question
The truth set of 8 + 2 x − x 2 = 0 8 + 2x - x^2 = 0 8 + 2 x − x 2 = 0 { p , q } \{p, q\} { p , q } p + q p + q p + q
Worked solution (try it first) Multiply by
− 1 -1 − 1 so the
x 2 x^2 x 2 term is positive:
x 2 − 2 x − 8 = 0 x^2 - 2x - 8 = 0 x 2 − 2 x − 8 = 0 .
Factorise:
( x − 4 ) ( x + 2 ) = 0 (x - 4)(x + 2) = 0 ( x − 4 ) ( x + 2 ) = 0 , so the roots are 4 and
− 2 -2 − 2 .
So
p + q = 4 + ( − 2 ) = 2 p + q = 4 + (-2) = 2 p + q = 4 + ( − 2 ) = 2 , option C.
Watch out
In 8 + 2 x − x 2 8 + 2x - x^2 8 + 2 x − x 2 x 2 x^2 x 2 − 1 -1 − 1 − 2 − 1 = 2 -\frac{2}{-1} = 2 − − 1 2 = 2 a = 1 a = 1 a = 1 − 2 -2 − 2 Report a problem with this question
Find the gradient of the line passing through the points ( 1 2 , − 1 3 ) \left(\frac12, -\frac13\right) ( 2 1 , − 3 1 ) ( 3 , 2 3 ) \left(3, \frac23\right) ( 3 , 3 2 )
A 7 2 \frac72 2 7 B 5 2 \frac52 2 5 C 2 7 \frac27 7 2 D 2 5 \frac25 5 2
Worked solution (try it first) The change in
y y y is
2 3 − ( − 1 3 ) = 1 \frac23 - \left(-\frac13\right) = 1 3 2 − ( − 3 1 ) = 1 .
The change in
x x x is
3 − 1 2 = 5 2 3 - \frac12 = \frac52 3 − 2 1 = 2 5 .
Gradient
= 1 ÷ 5 2 = 2 5 = 1 \div \frac52 = \frac25 = 1 ÷ 2 5 = 5 2 , option D.
Watch out
Put the change in y y y x x x y y y 5 2 \frac52 2 5 Report a problem with this question
For what values of x x x x 2 + 2 10 x 2 − 13 x − 3 \dfrac{x^2 + 2}{10x^2 - 13x - 3} 10 x 2 − 13 x − 3 x 2 + 2
A 3 2 , 1 5 \frac32, \frac15 2 3 , 5 1 B − 3 2 , − 1 5 -\frac32, -\frac15 − 2 3 , − 5 1 C 3 2 , − 1 5 \frac32, -\frac15 2 3 , − 5 1 D − 3 2 , 1 5 -\frac32, \frac15 − 2 3 , 5 1
Worked solution (try it first) The fraction is undefined when the bottom is zero:
10 x 2 − 13 x − 3 = 0 10x^2 - 13x - 3 = 0 10 x 2 − 13 x − 3 = 0 .
Factorise (numbers
− 15 -15 − 15 and
2 2 2 multiply to
− 30 -30 − 30 and add to
− 13 -13 − 13 ):
( 5 x + 1 ) ( 2 x − 3 ) = 0 (5x + 1)(2x - 3) = 0 ( 5 x + 1 ) ( 2 x − 3 ) = 0 .
So
x = − 1 5 x = -\frac15 x = − 5 1 or
x = 3 2 x = \frac32 x = 2 3 , option C.
Watch out
5 x + 1 = 0 5x + 1 = 0 5 x + 1 = 0 x = − 1 5 x = -\frac15 x = − 5 1 2 x − 3 = 0 2x - 3 = 0 2 x − 3 = 0 x = 3 2 x = \frac32 x = 2 3 1 5 \frac15 5 1 − 3 2 -\frac32 − 2 3 Report a problem with this question