Evaluate and correct to two decimal places: 75.0785 − 34.624 + 9.83 75.0785 - 34.624 + 9.83 75.0785 − 34.624 + 9.83
Worked solution (try it first) Work from left to right:
75.0785 − 34.624 = 40.4545 75.0785 - 34.624 = 40.4545 75.0785 − 34.624 = 40.4545 .
Add:
40.4545 + 9.83 = 50.2845 40.4545 + 9.83 = 50.2845 40.4545 + 9.83 = 50.2845 .
To 2 decimal places, the third decimal is 4, so round down: 50.28, option C.
Watch out
Round once, from the full answer. Rounding in two stages (50.2845 to 50.285, then to 50.29) gives 50.29 (option B); only the third decimal, 4, matters. Report a problem with this question
If X = { x : x < 7 } X = \{x : x < 7\} X = { x : x < 7 } Y = { y : y is a factor of 24 } Y = \{y : y \text{ is a factor of } 24\} Y = { y : y is a factor of 24 } μ = { 1 , 2 , 3 , … , 10 } \mu = \{1, 2, 3, \ldots, 10\} μ = { 1 , 2 , 3 , … , 10 } X ∩ Y X \cap Y X ∩ Y
A { 2 , 3 , 4 , 6 } \{2, 3, 4, 6\} { 2 , 3 , 4 , 6 } B { 1 , 2 , 3 , 4 , 6 } \{1, 2, 3, 4, 6\} { 1 , 2 , 3 , 4 , 6 } C { 2 , 3 , 4 , 6 , 8 } \{2, 3, 4, 6, 8\} { 2 , 3 , 4 , 6 , 8 } D { 1 , 2 , 3 , 4 , 6 , 8 } \{1, 2, 3, 4, 6, 8\} { 1 , 2 , 3 , 4 , 6 , 8 }
Worked solution (try it first) X X X is the members of
μ \mu μ less than 7:
{ 1 , 2 , 3 , 4 , 5 , 6 } \{1, 2, 3, 4, 5, 6\} { 1 , 2 , 3 , 4 , 5 , 6 } .
Y Y Y is the factors of 24 in
μ \mu μ :
{ 1 , 2 , 3 , 4 , 6 , 8 } \{1, 2, 3, 4, 6, 8\} { 1 , 2 , 3 , 4 , 6 , 8 } .
The elements in both are
X ∩ Y = { 1 , 2 , 3 , 4 , 6 } X \cap Y = \{1, 2, 3, 4, 6\} X ∩ Y = { 1 , 2 , 3 , 4 , 6 } , option B.
Watch out
1 is a factor of every number, including 24. Leaving it out gives option A. Report a problem with this question
Simplify [ ( 16 9 ) − 3 2 × 16 − 3 4 ] 1 3 \left[\left(\frac{16}{9}\right)^{-\frac32} \times 16^{-\frac34}\right]^{\frac13} [ ( 9 16 ) − 2 3 × 1 6 − 4 3 ] 3 1
A 3 4 \frac34 4 3 B 9 16 \frac{9}{16} 16 9 C 3 8 \frac38 8 3 D 1 4 \frac14 4 1
Worked solution (try it first) The negative index flips the fraction:
( 16 9 ) − 3 2 = ( 9 16 ) 3 2 \left(\frac{16}{9}\right)^{-\frac32} = \left(\frac{9}{16}\right)^{\frac32} ( 9 16 ) − 2 3 = ( 16 9 ) 2 3 .
The square root is
3 4 \frac34 4 3 , and cubing gives
27 64 \frac{27}{64} 64 27 .
16 − 3 4 = 1 16 3 4 16^{-\frac34} = \frac{1}{16^{\frac34}} 1 6 − 4 3 = 1 6 4 3 1 .
The fourth root of 16 is 2, and
2 3 = 8 2^3 = 8 2 3 = 8 , so this is
1 8 \frac18 8 1 .
Multiply:
27 64 × 1 8 = 27 512 \frac{27}{64} \times \frac18 = \frac{27}{512} 64 27 × 8 1 = 512 27 .
Cube root:
27 3 = 3 \sqrt[3]{27} = 3 3 27 = 3 and
512 3 = 8 \sqrt[3]{512} = 8 3 512 = 8 , so the value is
3 8 \frac38 8 3 , option C.
Watch out
Apply the outside power 1 3 \frac13 3 1 27 512 \frac{27}{512} 512 27 Report a problem with this question
Find the least value of x x x 4 x ≡ 7 ( m o d 9 ) 4x \equiv 7 \pmod 9 4 x ≡ 7 ( mod 9 )
Worked solution (try it first) 4 x 4x 4 x must leave remainder 7 when divided by 9, so list the numbers
7 , 16 , 25 , 34 , … 7, 16, 25, 34, \ldots 7 , 16 , 25 , 34 , … (add 9 each time).
The first that is a multiple of 4 is 16, so
4 x = 16 4x = 16 4 x = 16 and
x = 4 x = 4 x = 4 .
Check:
4 × 4 = 16 = 9 + 7 4 \times 4 = 16 = 9 + 7 4 × 4 = 16 = 9 + 7 .
So
x = 4 x = 4 x = 4 , option D.
Watch out
Test each option by multiplying by 4 and reducing. Picking the 7 from the question (option A) gives 4 × 7 = 28 ≡ 1 ( m o d 9 ) 4 \times 7 = 28 \equiv 1 \pmod 9 4 × 7 = 28 ≡ 1 ( mod 9 ) Report a problem with this question
Express 1 + 2 log 10 3 1 + 2\log_{10} 3 1 + 2 log 10 3 log 10 q \log_{10} q log 10 q
A log 10 90 \log_{10} 90 log 10 90 B log 10 19 \log_{10} 19 log 10 19 C log 10 9 \log_{10} 9 log 10 9 D log 10 6 \log_{10} 6 log 10 6
Worked solution (try it first) Write 1 as a log:
1 = log 10 10 1 = \log_{10} 10 1 = log 10 10 .
Move the 2 up as a power:
2 log 10 3 = log 10 9 2\log_{10} 3 = \log_{10} 9 2 log 10 3 = log 10 9 .
Adding logs multiplies:
log 10 10 + log 10 9 = log 10 90 \log_{10} 10 + \log_{10} 9 = \log_{10} 90 log 10 10 + log 10 9 = log 10 90 , option A.
Watch out
Adding logs multiplies the numbers: 10 × 9 = 90 10 \times 9 = 90 10 × 9 = 90 log 10 19 \log_{10} 19 log 10 19 Report a problem with this question
If 101 two + 12 y = 23 five 101_{\text{two}} + 12_y = 23_{\text{five}} 10 1 two + 1 2 y = 2 3 five y y y
Worked solution (try it first) Change to base ten:
101 two = 4 + 1 = 5 101_{\text{two}} = 4 + 1 = 5 10 1 two = 4 + 1 = 5 and
23 five = 10 + 3 = 13 23_{\text{five}} = 10 + 3 = 13 2 3 five = 10 + 3 = 13 .
In base
y y y ,
12 y = y + 2 12_y = y + 2 1 2 y = y + 2 , so
5 + y + 2 = 13 5 + y + 2 = 13 5 + y + 2 = 13 .
So
y + 7 = 13 y + 7 = 13 y + 7 = 13 and
y = 6 y = 6 y = 6 , option C.
Watch out
23 five 23_{\text{five}} 2 3 five y + 7 = 23 y + 7 = 23 y + 7 = 23 y = 16 y = 16 y = 16 Report a problem with this question
An amount of ₦550,000.00 was realised when a principal x x x 2 % 2\% 2% x x x
A ₦470,000.00 B ₦480,000.00 C ₦490,000.00 D ₦500,000.00
Worked solution (try it first) Simple interest for 5 years at
2 % 2\% 2% is
5 × 2 % = 10 % 5 \times 2\% = 10\% 5 × 2% = 10% of the principal
x x x .
The amount is principal plus interest, so
1.1 x = 550 000 1.1x = 550\,000 1.1 x = 550 000 .
Divide both sides by 1.1:
x = 500 000 x = 500\,000 x = 500 000 .
The principal is ₦500,000.00, option D.
Watch out
Interest is 10 % 10\% 10% 10 % 10\% 10% Report a problem with this question
Given that 3 + 5 5 = x + y 15 \dfrac{\sqrt3 + \sqrt5}{\sqrt5} = x + y\sqrt{15} 5 3 + 5 = x + y 15 ( x + y ) (x + y) ( x + y )
A 1 3 5 1\frac35 1 5 3 B 1 2 5 1\frac25 1 5 2 C 1 1 5 1\frac15 1 5 1 D 1 5 \frac15 5 1
Worked solution (try it first) Split the fraction:
3 + 5 5 = 3 5 + 1 \dfrac{\sqrt3 + \sqrt5}{\sqrt5} = \dfrac{\sqrt3}{\sqrt5} + 1 5 3 + 5 = 5 3 + 1 .
Rationalise:
3 5 = 3 × 5 5 \dfrac{\sqrt3}{\sqrt5} = \dfrac{\sqrt3 \times \sqrt5}{5} 5 3 = 5 3 × 5 , which is
1 5 15 \frac15\sqrt{15} 5 1 15 .
So
x = 1 x = 1 x = 1 and
y = 1 5 y = \frac15 y = 5 1 , and
x + y = 1 1 5 x + y = 1\frac15 x + y = 1 5 1 , option C.
Watch out
5 5 = 1 \frac{\sqrt5}{\sqrt5} = 1 5 5 = 1 x = 1 x = 1 x = 1 y y y 1 5 \frac15 5 1 Report a problem with this question
If x = 3 x = 3 x = 3 y = − 1 y = -1 y = − 1 2 ( x 2 − y 3 ) 2(x^2 - y^3) 2 ( x 2 − y 3 )
Worked solution (try it first) x 2 = 3 2 = 9 x^2 = 3^2 = 9 x 2 = 3 2 = 9 and
y 3 = ( − 1 ) 3 = − 1 y^3 = (-1)^3 = -1 y 3 = ( − 1 ) 3 = − 1 .
Inside the bracket:
9 − ( − 1 ) = 10 9 - (-1) = 10 9 − ( − 1 ) = 10 .
So
2 ( x 2 − y 3 ) = 2 × 10 = 20 2(x^2 - y^3) = 2 \times 10 = 20 2 ( x 2 − y 3 ) = 2 × 10 = 20 , option C.
Watch out
An odd power of − 1 -1 − 1 − 1 -1 − 1 y 3 y^3 y 3 + 1 +1 + 1 2 ( 9 − 1 ) = 16 2(9 - 1) = 16 2 ( 9 − 1 ) = 16 Report a problem with this question
Solve 3 x − 2 y = 10 3x - 2y = 10 3 x − 2 y = 10 x + 3 y = 7 x + 3y = 7 x + 3 y = 7
A x = − 4 x = -4 x = − 4 y = 1 y = 1 y = 1 B x = − 1 x = -1 x = − 1 y = − 4 y = -4 y = − 4 C x = 1 x = 1 x = 1 y = 4 y = 4 y = 4 D x = 4 x = 4 x = 4 y = 1 y = 1 y = 1
Worked solution (try it first) Make
x x x the subject of the second equation:
x = 7 − 3 y x = 7 - 3y x = 7 − 3 y .
Substitute into the first:
3 ( 7 − 3 y ) − 2 y = 10 3(7 - 3y) - 2y = 10 3 ( 7 − 3 y ) − 2 y = 10 , so
21 − 11 y = 10 21 - 11y = 10 21 − 11 y = 10 .
Take 21 from both sides:
− 11 y = − 11 -11y = -11 − 11 y = − 11 , so
y = 1 y = 1 y = 1 .
Then
x = 7 − 3 = 4 x = 7 - 3 = 4 x = 7 − 3 = 4 .
So
x = 4 x = 4 x = 4 and
y = 1 y = 1 y = 1 , option D.
Watch out
Option C swaps the values. Check it: x = 1 x = 1 x = 1 y = 4 y = 4 y = 4 3 x − 2 y = − 5 3x - 2y = -5 3 x − 2 y = − 5 Report a problem with this question
The implication x ⇒ y x \Rightarrow y x ⇒ y
A ∼ y ⇒ ∼ x \sim y \Rightarrow \sim x ∼ y ⇒∼ x B y ⇒ ∼ x y \Rightarrow \sim x y ⇒∼ x C ∼ x ⇒ ∼ y \sim x \Rightarrow \sim y ∼ x ⇒∼ y D y ⇒ x y \Rightarrow x y ⇒ x
Worked solution (try it first) ⇒ y \Rightarrow y ⇒ y is equivalent to its contrapositive: swap the two parts and negate both.
⇒ y \Rightarrow y ⇒ y is equivalent to
∼ y \sim y ∼ y ⇒ ∼ x \Rightarrow \sim x ⇒∼ x , option A.
Watch out
Negating both parts without swapping them gives the inverse, ∼ x ⇒ ∼ y \sim x \Rightarrow \sim y ∼ x ⇒∼ y x ⇒ y x \Rightarrow y x ⇒ y Report a problem with this question
The first term of a geometric progression is 3 and the 5th term is 48. Find the common ratio.
Worked solution (try it first) The 5th term of a G.P. is
a r 4 ar^4 a r 4 , so
3 r 4 = 48 3r^4 = 48 3 r 4 = 48 .
Divide by 3:
r 4 = 16 r^4 = 16 r 4 = 16 .
Take the fourth root:
2 4 = 16 2^4 = 16 2 4 = 16 , so
r = 2 r = 2 r = 2 , option A.
Watch out
From r 4 = 16 r^4 = 16 r 4 = 16 r 2 r^2 r 2 Report a problem with this question
Solve 1 3 ( 5 − 3 x ) < 2 5 ( 3 − 7 x ) \frac13(5 - 3x) < \frac25(3 - 7x) 3 1 ( 5 − 3 x ) < 5 2 ( 3 − 7 x )
A x > 7 22 x > \frac{7}{22} x > 22 7 B x < 7 22 x < \frac{7}{22} x < 22 7 C x > − 7 27 x > -\frac{7}{27} x > − 27 7 D x < − 7 27 x < -\frac{7}{27} x < − 27 7
Worked solution (try it first) Multiply both sides by 15, the LCM of 3 and 5:
5 ( 5 − 3 x ) < 6 ( 3 − 7 x ) 5(5 - 3x) < 6(3 - 7x) 5 ( 5 − 3 x ) < 6 ( 3 − 7 x ) .
Expand:
25 − 15 x < 18 − 42 x 25 - 15x < 18 - 42x 25 − 15 x < 18 − 42 x .
Add
42 x 42x 42 x and subtract 25 from both sides:
27 x < − 7 27x < -7 27 x < − 7 .
Divide by 27, a positive number, so the sign stays:
x < − 7 27 x < -\frac{7}{27} x < − 27 7 , option D.
Watch out
You divide by + 27 +27 + 27 x > − 7 27 x > -\frac{7}{27} x > − 27 7 Report a problem with this question
Make m m m k = m − y m + 1 k = \sqrt{\dfrac{m - y}{m + 1}} k = m + 1 m − y
A m = y + k 2 k 2 + 1 m = \dfrac{y + k^2}{k^2 + 1} m = k 2 + 1 y + k 2 B m = y + k 2 1 − k 2 m = \dfrac{y + k^2}{1 - k^2} m = 1 − k 2 y + k 2 C m = y − k 2 k 2 + 1 m = \dfrac{y - k^2}{k^2 + 1} m = k 2 + 1 y − k 2 D m = y − k 2 1 − k 2 m = \dfrac{y - k^2}{1 - k^2} m = 1 − k 2 y − k 2
Worked solution (try it first) Square both sides and multiply by
m + 1 m + 1 m + 1 :
k 2 m + k 2 = m − y k^2m + k^2 = m - y k 2 m + k 2 = m − y .
Collect the
m m m terms on the right:
k 2 + y = m − k 2 m k^2 + y = m - k^2m k 2 + y = m − k 2 m , so
k 2 + y = m ( 1 − k 2 ) k^2 + y = m(1 - k^2) k 2 + y = m ( 1 − k 2 ) .
Divide by
1 − k 2 1 - k^2 1 − k 2 :
m = y + k 2 1 − k 2 m = \dfrac{y + k^2}{1 - k^2} m = 1 − k 2 y + k 2 , option B.
Watch out
k 2 m k^2m k 2 m − k 2 m -k^2m − k 2 m 1 − k 2 1 - k^2 1 − k 2 k 2 + 1 k^2 + 1 k 2 + 1 Report a problem with this question
Find the quadratic equation whose roots are 1 2 \frac12 2 1 − 1 3 -\frac13 − 3 1
A 3 x 2 + x + 1 = 0 3x^2 + x + 1 = 0 3 x 2 + x + 1 = 0 B 6 x 2 + x − 1 = 0 6x^2 + x - 1 = 0 6 x 2 + x − 1 = 0 C 3 x 2 + x − 1 = 0 3x^2 + x - 1 = 0 3 x 2 + x − 1 = 0 D 6 x 2 − x − 1 = 0 6x^2 - x - 1 = 0 6 x 2 − x − 1 = 0
Worked solution (try it first) Roots
1 2 \frac12 2 1 and
− 1 3 -\frac13 − 3 1 give the factors
( 2 x − 1 ) (2x - 1) ( 2 x − 1 ) and
( 3 x + 1 ) (3x + 1) ( 3 x + 1 ) , cleared of fractions.
Expand:
( 2 x − 1 ) ( 3 x + 1 ) = 6 x 2 + 2 x − 3 x − 1 (2x - 1)(3x + 1) = 6x^2 + 2x - 3x - 1 ( 2 x − 1 ) ( 3 x + 1 ) = 6 x 2 + 2 x − 3 x − 1 = 6 x 2 − x − 1 = 6x^2 - x - 1 = 6 x 2 − x − 1 .
So the equation is
6 x 2 − x − 1 = 0 6x^2 - x - 1 = 0 6 x 2 − x − 1 = 0 , option D.
Watch out
A root of 1 2 \frac12 2 1 2 x − 1 = 0 2x - 1 = 0 2 x − 1 = 0 2 x + 1 2x + 1 2 x + 1 ( 2 x + 1 ) ( 3 x − 1 ) = 6 x 2 + x − 1 (2x + 1)(3x - 1) = 6x^2 + x - 1 ( 2 x + 1 ) ( 3 x − 1 ) = 6 x 2 + x − 1 Report a problem with this question
Given that x x x y y y z z z x = 15 x = 15 x = 15 y = 10 y = 10 y = 10 z = 4 z = 4 z = 4 x x x y y y z z z
A x = 6 y z x = \dfrac{6y}{z} x = z 6 y B x = 12 y z x = \dfrac{12y}{z} x = z 12 y C x = 3 y z x = \dfrac{3y}{z} x = z 3 y D x = 3 y 2 z x = \dfrac{3y}{2z} x = 2 z 3 y
Worked solution (try it first) Directly as
y y y , inversely as
z z z :
x = k y z x = \dfrac{ky}{z} x = z k y .
Put in
x = 15 x = 15 x = 15 ,
y = 10 y = 10 y = 10 ,
z = 4 z = 4 z = 4 :
15 = 10 k 4 15 = \dfrac{10k}{4} 15 = 4 10 k , so
k = 60 10 = 6 k = \dfrac{60}{10} = 6 k = 10 60 = 6 .
So
x = 6 y z x = \dfrac{6y}{z} x = z 6 y , option A.
Watch out
Include z z z k k k 15 = 10 k 15 = 10k 15 = 10 k k = 3 2 k = \frac32 k = 2 3 Report a problem with this question
Two buses start from the same station at 9.00 am and travel in opposite directions along the same straight road. The first bus travels at 72 km/h 72\text{ km/h} 72 km/h 48 km/h 48\text{ km/h} 48 km/h 240 km 240\text{ km} 240 km
A 1:00 pm B 12:00 noon C 11:00 am D 10:00 am
Worked solution (try it first) The buses move in opposite directions, so the distance between them grows at
72 + 48 = 120 km/h 72 + 48 = 120\text{ km/h} 72 + 48 = 120 km/h .
Time to be 240 km apart:
240 ÷ 120 = 2 240 \div 120 = 2 240 ÷ 120 = 2 hours.
Two hours after 9.00 am is 11.00 am, option C.
Watch out
Add the speeds, don't average them. The average, 60 km/h 60\text{ km/h} 60 km/h Report a problem with this question
A solid cuboid has a length of 7 cm 7\text{ cm} 7 cm 5 cm 5\text{ cm} 5 cm 4 cm 4\text{ cm} 4 cm
A 280 cm 2 280\text{ cm}^2 280 cm 2 B 166 cm 2 166\text{ cm}^2 166 cm 2 C 140 cm 2 140\text{ cm}^2 140 cm 2 D 83 cm 2 83\text{ cm}^2 83 cm 2
Worked solution (try it first) A cuboid has three pairs of equal faces:
7 × 5 = 35 7 \times 5 = 35 7 × 5 = 35 ,
7 × 4 = 28 7 \times 4 = 28 7 × 4 = 28 and
5 × 4 = 20 5 \times 4 = 20 5 × 4 = 20 .
Total surface area:
2 ( 35 + 28 + 20 ) = 2 × 83 2(35 + 28 + 20) = 2 \times 83 2 ( 35 + 28 + 20 ) = 2 × 83 = 166 cm 2 = 166\text{ cm}^2 = 166 cm 2 , option B.
Watch out
Each face has a matching opposite face, so double the sum. 83 cm 2 83\text{ cm}^2 83 cm 2 Report a problem with this question
In the diagram, P Q ∥ S R PQ \parallel SR P Q ∥ S R x x x
Worked solution (try it first) Angles at a point: the angle between the two lines at the bend is
360 ∘ − 246 ∘ = 114 ∘ 360^\circ - 246^\circ = 114^\circ 36 0 ∘ − 24 6 ∘ = 11 4 ∘ .
Draw a line through the bend parallel to
P Q PQ P Q .
By alternate angles, the upper line makes
x x x with it and the lower line makes
68 ∘ 68^\circ 6 8 ∘ with it.
So
x + 68 ∘ = 114 ∘ x + 68^\circ = 114^\circ x + 6 8 ∘ = 11 4 ∘ , which gives
x = 46 x = 46 x = 46 , option B.
Watch out
The 114 ∘ 114^\circ 11 4 ∘ x x x 68 ∘ 68^\circ 6 8 ∘ 57 ∘ 57^\circ 5 7 ∘ Report a problem with this question
Find the equation of the line parallel to 2 y = 3 ( x − 2 ) 2y = 3(x - 2) 2 y = 3 ( x − 2 ) ( 2 , 3 ) (2, 3) ( 2 , 3 )
A y = 3 2 x − 3 y = \frac32x - 3 y = 2 3 x − 3 B y = 2 3 x − 2 y = \frac23x - 2 y = 3 2 x − 2 C y = 3 2 x y = \frac32x y = 2 3 x D y = − 2 3 x y = -\frac23x y = − 3 2 x
Worked solution (try it first) 2 y = 3 ( x − 2 ) 2y = 3(x - 2) 2 y = 3 ( x − 2 ) gives
y = 3 2 x − 3 y = \frac32x - 3 y = 2 3 x − 3 , so its gradient is
3 2 \frac32 2 3 .
Parallel lines have the same gradient.
Through
( 2 , 3 ) (2, 3) ( 2 , 3 ) :
y − 3 = 3 2 ( x − 2 ) y - 3 = \frac32(x - 2) y − 3 = 2 3 ( x − 2 ) , so
y − 3 = 3 2 x − 3 y - 3 = \frac32x - 3 y − 3 = 2 3 x − 3 .
Add 3 to both sides:
y = 3 2 x y = \frac32x y = 2 3 x , option C.
Watch out
Option A has the right gradient but is the given line itself: at x = 2 x = 2 x = 2 y = 0 y = 0 y = 0 ( 2 , 3 ) (2, 3) ( 2 , 3 ) Report a problem with this question
The expression 5 x + 3 6 x ( x + 1 ) \dfrac{5x + 3}{6x(x + 1)} 6 x ( x + 1 ) 5 x + 3 x x x
A { 0 , 1 } \{0, 1\} { 0 , 1 } B { 0 , − 1 } \{0, -1\} { 0 , − 1 } C { − 3 , − 1 } \{-3, -1\} { − 3 , − 1 } D { − 3 , 0 } \{-3, 0\} { − 3 , 0 }
Worked solution (try it first) The expression is undefined when the bottom
6 x ( x + 1 ) 6x(x + 1) 6 x ( x + 1 ) is zero.
6 x = 0 6x = 0 6 x = 0 gives
x = 0 x = 0 x = 0 , and
x + 1 = 0 x + 1 = 0 x + 1 = 0 gives
x = − 1 x = -1 x = − 1 .
So
x ∈ { 0 , − 1 } x \in \{0, -1\} x ∈ { 0 , − 1 } , option B.
Watch out
Solve x + 1 = 0 x + 1 = 0 x + 1 = 0 x = − 1 x = -1 x = − 1 + 1 +1 + 1 { 0 , 1 } \{0, 1\} { 0 , 1 } Report a problem with this question
A man is five times as old as his son. In four years' time, the product of their ages would be 340. If the son's age is y y y y y y
A 5 y 2 − 16 y − 380 = 0 5y^2 - 16y - 380 = 0 5 y 2 − 16 y − 380 = 0 B 5 y 2 + 24 y − 308 = 0 5y^2 + 24y - 308 = 0 5 y 2 + 24 y − 308 = 0 C 5 y 2 − 16 y − 330 = 0 5y^2 - 16y - 330 = 0 5 y 2 − 16 y − 330 = 0 D 5 y 2 + 24 y − 324 = 0 5y^2 + 24y - 324 = 0 5 y 2 + 24 y − 324 = 0
Worked solution (try it first) In four years the son is
y + 4 y + 4 y + 4 and the man is
5 y + 4 5y + 4 5 y + 4 .
Their product then is 340:
( 5 y + 4 ) ( y + 4 ) = 340 (5y + 4)(y + 4) = 340 ( 5 y + 4 ) ( y + 4 ) = 340 .
Expand:
5 y 2 + 20 y + 4 y + 16 = 340 5y^2 + 20y + 4y + 16 = 340 5 y 2 + 20 y + 4 y + 16 = 340 , so
5 y 2 + 24 y + 16 = 340 5y^2 + 24y + 16 = 340 5 y 2 + 24 y + 16 = 340 .
Subtract 340:
5 y 2 + 24 y − 324 = 0 5y^2 + 24y - 324 = 0 5 y 2 + 24 y − 324 = 0 , option D.
Watch out
Add 4 years to each age separately: the man will be 5 y + 4 5y + 4 5 y + 4 5 ( y + 4 ) 5(y + 4) 5 ( y + 4 ) − 324 -324 − 324 Report a problem with this question
Simplify a b − b a − c b \dfrac ab - \dfrac ba - \dfrac cb b a − a b − b c
A a − b + c a b \dfrac{a - b + c}{ab} ab a − b + c B a b − b c − a c a b \dfrac{ab - bc - ac}{ab} ab ab − b c − a c C a 2 − b 2 + a c a b \dfrac{a^2 - b^2 + ac}{ab} ab a 2 − b 2 + a c D a 2 − b 2 − a c a b \dfrac{a^2 - b^2 - ac}{ab} ab a 2 − b 2 − a c
Worked solution (try it first) The LCD of
b b b ,
a a a and
b b b is
a b ab ab .
Rewrite each fraction over
a b ab ab :
a 2 a b − b 2 a b − a c a b \frac{a^2}{ab} - \frac{b^2}{ab} - \frac{ac}{ab} ab a 2 − ab b 2 − ab a c .
So the expression is
a 2 − b 2 − a c a b \dfrac{a^2 - b^2 - ac}{ab} ab a 2 − b 2 − a c , option D.
Watch out
The last fraction is subtracted, so a c ac a c + a c +ac + a c Report a problem with this question
In the diagram, X Y Z XYZ X Y Z 6 cm 6\text{ cm} 6 cm T T T X Y XY X Y tan ( ∠ X Z T ) \tan(\angle XZT) tan ( ∠ X Z T )
A 1 3 \frac{1}{\sqrt3} 3 1 B 3 2 \frac{\sqrt3}{2} 2 3 C 3 \sqrt3 3 D 1 2 \frac12 2 1
Worked solution (try it first) Every angle of an equilateral triangle is
60 ∘ 60^\circ 6 0 ∘ , so
∠ X Z Y = 60 ∘ \angle XZY = 60^\circ ∠ X Z Y = 6 0 ∘ .
Z T ZT Z T goes from
Z Z Z to the midpoint of
X Y XY X Y , so it is a line of symmetry and cuts the angle at
Z Z Z in half:
∠ X Z T = 30 ∘ \angle XZT = 30^\circ ∠ X Z T = 3 0 ∘ .
So
tan ( ∠ X Z T ) = tan 30 ∘ \tan(\angle XZT) = \tan30^\circ tan ( ∠ X Z T ) = tan 3 0 ∘ = 1 3 = \frac{1}{\sqrt3} = 3 1 , option A.
Watch out
∠ X Z T \angle XZT ∠ X Z T 60 ∘ 60^\circ 6 0 ∘ Z Z Z 60 ∘ 60^\circ 6 0 ∘ tan 60 ∘ = 3 \tan60^\circ = \sqrt3 tan 6 0 ∘ = 3 Report a problem with this question
A fence 2.4 m 2.4\text{ m} 2.4 m 10 m 10\text{ m} 10 m 16 m 16\text{ m} 16 m
A 76.11 ∘ 76.11^\circ 76.1 1 ∘ B 53.67 ∘ 53.67^\circ 53.6 7 ∘ C 52.40 ∘ 52.40^\circ 52.4 0 ∘ D 51.32 ∘ 51.32^\circ 51.3 2 ∘
Worked solution (try it first) Look from the top of the fence: the top of the tree is
16 − 2.4 = 13.6 16 - 2.4 = 13.6 16 − 2.4 = 13.6 m higher, and 10 m away.
So
tan θ = 13.6 10 = 1.36 \tan\theta = \frac{13.6}{10} = 1.36 tan θ = 10 13.6 = 1.36 .
So
θ = tan − 1 1.36 \theta = \tan^{-1}1.36 θ = tan − 1 1.36 ≈ 53.67 ∘ \approx 53.67^\circ ≈ 53.6 7 ∘ , option B.
Watch out
Measure the rise from the top of the fence: 16 − 2.4 = 13.6 16 - 2.4 = 13.6 16 − 2.4 = 13.6 58.0 ∘ 58.0^\circ 58. 0 ∘ Report a problem with this question
Fati buys milk at ₦x x x y y y
A ₦( x y + 10 ) (xy + 10) ( x y + 10 ) B ₦( x + 10 y ) (x + 10y) ( x + 10 y ) C ₦( 10 x + y ) (10x + y) ( 10 x + y ) D ₦10 ( x + y ) 10(x + y) 10 ( x + y )
Worked solution (try it first) She sells each tin for its cost plus the profit: ₦
( x + y ) (x + y) ( x + y ) a tin.
10 tins bring in 10 times that: ₦
10 ( x + y ) 10(x + y) 10 ( x + y ) , option D.
Watch out
The profit y y y x x x ( 10 x + y ) (10x + y) ( 10 x + y ) Report a problem with this question
If tan y \tan y tan y sin y \sin y sin y y y y
A First and third only B First and second only C Third only D Second only
Worked solution (try it first) Sine is negative in the third and fourth quadrants.
Tangent is positive in the first and third quadrants.
Only the third quadrant is on both lists, so the answer is the third only, option C.
Watch out
Use both conditions. "First and third" (option A) is where tangent is positive, but sine is positive in the first quadrant. Report a problem with this question
The dimensions of a rectangular base of a right pyramid are 9 cm 9\text{ cm} 9 cm 5 cm 5\text{ cm} 5 cm 105 cm 3 105\text{ cm}^3 105 cm 3
A 10 cm 10\text{ cm} 10 cm B 6 cm 6\text{ cm} 6 cm C 8 cm 8\text{ cm} 8 cm D 7 cm 7\text{ cm} 7 cm
Worked solution (try it first) Volume of a pyramid:
1 3 × base area × h \frac13 \times \text{base area} \times h 3 1 × base area × h .
The base area is
9 × 5 = 45 cm 2 9 \times 5 = 45\text{ cm}^2 9 × 5 = 45 cm 2 .
1 3 × 45 × h = 105 \frac13 \times 45 \times h = 105 3 1 × 45 × h = 105 , so
15 h = 105 15h = 105 15 h = 105 .
So
h = 7 h = 7 h = 7 cm, option D.
Watch out
Keep the 1 3 \frac13 3 1 45 h = 105 45h = 105 45 h = 105 h = 2 1 3 h = 2\frac13 h = 2 3 1 Report a problem with this question
Each interior angle of a regular polygon is 168 ∘ 168^\circ 16 8 ∘
Worked solution (try it first) An interior angle and its exterior angle add up to
180 ∘ 180^\circ 18 0 ∘ , so each exterior angle is
180 ∘ − 168 ∘ = 12 ∘ 180^\circ - 168^\circ = 12^\circ 18 0 ∘ − 16 8 ∘ = 1 2 ∘ .
The exterior angles add up to
360 ∘ 360^\circ 36 0 ∘ , so the number of sides is
360 ÷ 12 = 30 360 \div 12 = 30 360 ÷ 12 = 30 , option A.
Watch out
Divide 360 ∘ 360^\circ 36 0 ∘ 12 ∘ 12^\circ 1 2 ∘ 10 ∘ 10^\circ 1 0 ∘ Also set as WAEC 2024 · Paper 1 · Q11
Report a problem with this question
In the diagram, M N ∥ P Q MN \parallel PQ M N ∥ P Q ∠ M N P = 2 x \angle MNP = 2x ∠ M N P = 2 x ∠ N P Q = ( 3 x − 50 ) ∘ \angle NPQ = (3x - 50)^\circ ∠ N P Q = ( 3 x − 50 ) ∘ ∠ N P Q \angle NPQ ∠ N P Q
A 200 ∘ 200^\circ 20 0 ∘ B 150 ∘ 150^\circ 15 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) M N ∥ P Q MN \parallel PQ M N ∥ P Q and
∠ M N P \angle MNP ∠ M N P ,
∠ N P Q \angle NPQ ∠ N P Q are alternate angles, so they are equal:
2 x = 3 x − 50 2x = 3x - 50 2 x = 3 x − 50 .
Subtract
2 x 2x 2 x and add 50:
x = 50 x = 50 x = 50 .
So
∠ N P Q = 3 ( 50 ) − 50 = 100 ∘ \angle NPQ = 3(50) - 50 = 100^\circ ∠ N P Q = 3 ( 50 ) − 50 = 10 0 ∘ , option D.
Watch out
These are alternate (Z) angles, so they are equal. Treating them as co-interior (adding to 180 ∘ 180^\circ 18 0 ∘ x = 46 x = 46 x = 46 88 ∘ 88^\circ 8 8 ∘ Report a problem with this question
The length of an arc of a circle of radius 3.5 cm 3.5\text{ cm} 3.5 cm 1 19 36 cm 1\frac{19}{36}\text{ cm} 1 36 19 cm [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 55 ∘ 55^\circ 5 5 ∘ B 36 ∘ 36^\circ 3 6 ∘ C 25 ∘ 25^\circ 2 5 ∘ D 22 ∘ 22^\circ 2 2 ∘
Worked solution (try it first) Write the arc as an improper fraction:
1 19 36 = 55 36 1\frac{19}{36} = \frac{55}{36} 1 36 19 = 36 55 cm.
The circumference is
2 × 22 7 × 3.5 = 22 2 \times \frac{22}{7} \times 3.5 = 22 2 × 7 22 × 3.5 = 22 cm.
The angle is the arc's share of
360 ∘ 360^\circ 36 0 ∘ :
θ = 55 36 ÷ 22 × 360 \theta = \frac{55}{36} \div 22 \times 360 θ = 36 55 ÷ 22 × 360 .
55 36 × 360 = 550 \frac{55}{36} \times 360 = 550 36 55 × 360 = 550 , and
550 ÷ 22 = 25 550 \div 22 = 25 550 ÷ 22 = 25 .
So
θ = 25 ∘ \theta = 25^\circ θ = 2 5 ∘ , option C.
Watch out
1 19 36 1\frac{19}{36} 1 36 19 55 36 ≈ 1.53 \frac{55}{36} \approx 1.53 36 55 ≈ 1.53 19 ∘ 19^\circ 1 9 ∘ Report a problem with this question
In the diagram, P U ∥ S R PU \parallel SR P U ∥ S R P S ∥ T R PS \parallel TR P S ∥ T R Q S ∥ U R QS \parallel UR QS ∥ U R ∣ U R ∣ = 15 cm |UR| = 15\text{ cm} ∣ U R ∣ = 15 cm ∣ S R ∣ = 8 cm |SR| = 8\text{ cm} ∣ S R ∣ = 8 cm ∣ P S ∣ = 10 cm |PS| = 10\text{ cm} ∣ P S ∣ = 10 cm S U R SUR S U R 24 cm 2 24\text{ cm}^2 24 cm 2 P T R S PTRS P T R S
A 40 cm 2 40\text{ cm}^2 40 cm 2 B 48 cm 2 48\text{ cm}^2 48 cm 2 C 80 cm 2 80\text{ cm}^2 80 cm 2 D 120 cm 2 120\text{ cm}^2 120 cm 2
Worked solution (try it first) △ S U R \triangle SUR △ S U R has base
S R = 8 SR = 8 S R = 8 cm and its height is the distance between the parallel lines:
1 2 × 8 × h = 24 \frac12 \times 8 \times h = 24 2 1 × 8 × h = 24 , so
h = 6 h = 6 h = 6 cm.
P T ∥ S R PT \parallel SR P T ∥ S R and
P S ∥ T R PS \parallel TR P S ∥ T R , so
P T R S PTRS P T R S is a parallelogram on base
S R SR S R with the same height.
Area
= 8 × 6 = 48 cm 2 = 8 \times 6 = 48\text{ cm}^2 = 8 × 6 = 48 cm 2 , option B.
Watch out
P S = 10 PS = 10 P S = 10 8 × 10 = 80 cm 2 8 \times 10 = 80\text{ cm}^2 8 × 10 = 80 cm 2 Report a problem with this question
In the diagram, O O O R P RP R P ∠ O P Q = 48 ∘ \angle OPQ = 48^\circ ∠ O P Q = 4 8 ∘ m m m
A 96 ∘ 96^\circ 9 6 ∘ B 90 ∘ 90^\circ 9 0 ∘ C 68 ∘ 68^\circ 6 8 ∘ D 42 ∘ 42^\circ 4 2 ∘
Worked solution (try it first) O P = O Q OP = OQ O P = O Q (radii), so triangle
O P Q OPQ O P Q is isosceles and
∠ O Q P = ∠ O P Q = 48 ∘ \angle OQP = \angle OPQ = 48^\circ ∠ O QP = ∠ O P Q = 4 8 ∘ .
R O P ROP R O P is a straight line, so
m = ∠ R O Q m = \angle ROQ m = ∠ R O Q is an exterior angle of triangle
O P Q OPQ O P Q .
An exterior angle equals the sum of the two interior opposite angles:
m = 48 ∘ + 48 ∘ = 96 ∘ m = 48^\circ + 48^\circ = 96^\circ m = 4 8 ∘ + 4 8 ∘ = 9 6 ∘ , option A.
Watch out
Triangle O P Q OPQ O P Q Q Q Q 90 ∘ 90^\circ 9 0 ∘ 42 ∘ 42^\circ 4 2 ∘ Report a problem with this question
The pie chart shows the population of men, women and children in a city. If the population of the city is 1,800,000, how many men are in the city?
A 845,000 B 600,000 C 355,000 D 250,000
Worked solution (try it first) The angles add up to
360 ∘ 360^\circ 36 0 ∘ , so Men is
360 ∘ − ( 120 ∘ + 169 ∘ ) = 71 ∘ 360^\circ - (120^\circ + 169^\circ) = 71^\circ 36 0 ∘ − ( 12 0 ∘ + 16 9 ∘ ) = 7 1 ∘ .
Men's share of the city:
71 360 × 1 800 000 \frac{71}{360} \times 1\,800\,000 360 71 × 1 800 000 .
Each degree is
1 800 000 ÷ 360 = 5000 1\,800\,000 \div 360 = 5000 1 800 000 ÷ 360 = 5000 people, so there are
71 × 5000 = 355 000 71 \times 5000 = 355\,000 71 × 5000 = 355 000 men, option C.
Watch out
Find the unlabelled Men sector first. 845,000 (option A) is the Children sector, 169 × 5000 169 \times 5000 169 × 5000 Report a problem with this question
The mean of the numbers 15, 21, 17, 26, 18 and 29 is 21. Calculate the standard deviation.
Worked solution (try it first) The deviations from the mean 21 are
− 6 -6 − 6 , 0,
− 4 -4 − 4 , 5,
− 3 -3 − 3 , 8.
Their squares are 36, 0, 16, 25, 9, 64, which add up to 150.
The variance is
150 6 = 25 \frac{150}{6} = 25 6 150 = 25 , so the standard deviation is
25 = 5 \sqrt{25} = 5 25 = 5 , option C.
Watch out
Square the deviations before adding. The signed deviations add up to 0 (option D) for any data. Report a problem with this question
In the diagram, O O O S O Q SOQ S O Q ∠ S R P = 37 ∘ \angle SRP = 37^\circ ∠ S R P = 3 7 ∘ ∠ P S Q \angle PSQ ∠ P S Q
A 127 ∘ 127^\circ 12 7 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 53 ∘ 53^\circ 5 3 ∘ D 37 ∘ 37^\circ 3 7 ∘
Worked solution (try it first) ∠ S Q P \angle SQP ∠ S QP and
∠ S R P \angle SRP ∠ S R P stand on the same arc
S P SP S P , so they are equal:
∠ S Q P = 37 ∘ \angle SQP = 37^\circ ∠ S QP = 3 7 ∘ .
S Q SQ S Q is a diameter, so the angle in the semicircle is a right angle:
∠ S P Q = 90 ∘ \angle SPQ = 90^\circ ∠ S P Q = 9 0 ∘ .
The angles of triangle
S P Q SPQ S P Q add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P S Q = 180 ∘ − 90 ∘ − 37 ∘ \angle PSQ = 180^\circ - 90^\circ - 37^\circ ∠ P S Q = 18 0 ∘ − 9 0 ∘ − 3 7 ∘ = 53 ∘ = 53^\circ = 5 3 ∘ , option C.
Watch out
37 ∘ 37^\circ 3 7 ∘ ∠ S Q P \angle SQP ∠ S QP ∠ S R P \angle SRP ∠ S R P ∠ P S Q \angle PSQ ∠ P S Q S P Q SPQ S P Q 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
Find the sum of the interior angles of a pentagon.
A 340 ∘ 340^\circ 34 0 ∘ B 350 ∘ 350^\circ 35 0 ∘ C 540 ∘ 540^\circ 54 0 ∘ D 550 ∘ 550^\circ 55 0 ∘
Worked solution (try it first) The interior angles of an
n n n -sided polygon add up to
( n − 2 ) × 180 ∘ (n - 2) \times 180^\circ ( n − 2 ) × 18 0 ∘ .
For a pentagon,
n = 5 n = 5 n = 5 :
( 5 − 2 ) × 180 ∘ = 540 ∘ (5 - 2) \times 180^\circ = 540^\circ ( 5 − 2 ) × 18 0 ∘ = 54 0 ∘ , option C.
Watch out
Take 2 from the number of sides before multiplying by 180 ∘ 180^\circ 18 0 ∘ 3 × 180 = 540 3 \times 180 = 540 3 × 180 = 540 360 ∘ 360^\circ 36 0 ∘ Report a problem with this question
The diameter of a sphere is 12 cm 12\text{ cm} 12 cm cm 3 \text{cm}^3 cm 3 [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 903 cm 3 903\text{ cm}^3 903 cm 3 B 904 cm 3 904\text{ cm}^3 904 cm 3 C 905 cm 3 905\text{ cm}^3 905 cm 3 D 906 cm 3 906\text{ cm}^3 906 cm 3
Worked solution (try it first) Radius
= 12 ÷ 2 = 6 = 12 \div 2 = 6 = 12 ÷ 2 = 6 cm, so
r 3 = 216 r^3 = 216 r 3 = 216 .
Volume:
4 3 × 22 7 × 216 ≈ 905.14 cm 3 \frac43 \times \frac{22}{7} \times 216 \approx 905.14\text{ cm}^3 3 4 × 7 22 × 216 ≈ 905.14 cm 3 .
To the nearest cm³,
905 cm 3 905\text{ cm}^3 905 cm 3 , option C.
Watch out
Use π = 22 7 \pi = \frac{22}{7} π = 7 22 Report a problem with this question
A box contains 12 identical balls of which 5 are red, 4 blue and the rest green. If a ball is selected at random, what is the probability that it is green?
A 3 4 \frac34 4 3 B 1 2 \frac12 2 1 C 1 3 \frac13 3 1 D 1 4 \frac14 4 1
Worked solution (try it first) The green balls are the rest:
12 − 5 − 4 = 3 12 - 5 - 4 = 3 12 − 5 − 4 = 3 .
So the probability of green is
3 12 = 1 4 \frac{3}{12} = \frac14 12 3 = 4 1 , option D.
Watch out
9 12 = 3 4 \frac{9}{12} = \frac34 12 9 = 4 3 Report a problem with this question
A box contains 12 identical balls of which 5 are red, 4 blue and the rest green. If two balls are selected at random one after the other with replacement, what is the probability that both are red?
A 25 144 \frac{25}{144} 144 25 B 5 33 \frac{5}{33} 33 5 C 5 6 \frac56 6 5 D 103 132 \frac{103}{132} 132 103
Worked solution (try it first) Each pick is red with probability
5 12 \frac{5}{12} 12 5 .
With replacement the picks are independent, so multiply:
5 12 × 5 12 = 25 144 \frac{5}{12} \times \frac{5}{12} = \frac{25}{144} 12 5 × 12 5 = 144 25 , option A.
Watch out
The ball is put back, so the second pick is also out of 12 with 5 red. Using 4 11 \frac{4}{11} 11 4 5 33 \frac{5}{33} 33 5 Report a problem with this question
In the diagram, P Q PQ P Q m = 1 2 ( x + y + z ) m = \frac12(x + y + z) m = 2 1 ( x + y + z ) m m m
A 45 ∘ 45^\circ 4 5 ∘ B 60 ∘ 60^\circ 6 0 ∘ C 90 ∘ 90^\circ 9 0 ∘ D 100 ∘ 100^\circ 10 0 ∘
Worked solution (try it first) Angles on a straight line add up to
180 ∘ 180^\circ 18 0 ∘ :
x + y + z + m = 180 ∘ x + y + z + m = 180^\circ x + y + z + m = 18 0 ∘ .
m = 1 2 ( x + y + z ) m = \frac12(x + y + z) m = 2 1 ( x + y + z ) means
x + y + z = 2 m x + y + z = 2m x + y + z = 2 m .
So
2 m + m = 180 ∘ 2m + m = 180^\circ 2 m + m = 18 0 ∘ .
So
3 m = 180 ∘ 3m = 180^\circ 3 m = 18 0 ∘ and
m = 60 ∘ m = 60^\circ m = 6 0 ∘ , option B.
Watch out
The angles lie on a straight line, so they add up to 180 ∘ 180^\circ 18 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 360 ∘ 360^\circ 36 0 ∘ m = 120 ∘ m = 120^\circ m = 12 0 ∘ Report a problem with this question
x x x 6.20
6.85
7.50
y y y 3.90
5.20
6.50
The points on a linear graph are as shown in the table. Find the gradient of the line.
A 2 1 2 2\frac12 2 2 1 B 2 C 1 D 1 2 \frac12 2 1
Worked solution (try it first) Use any two points, for example the first two: the change in
y y y is
5.20 − 3.90 = 1.30 5.20 - 3.90 = 1.30 5.20 − 3.90 = 1.30 .
The change in
x x x is
6.85 − 6.20 = 0.65 6.85 - 6.20 = 0.65 6.85 − 6.20 = 0.65 .
Gradient
= 1.30 0.65 = 2 = \frac{1.30}{0.65} = 2 = 0.65 1.30 = 2 , option B.
Watch out
Put the change in y y y x x x y y y 1 2 \frac12 2 1 Report a problem with this question
In the diagram, O O O P Q PQ P Q R S RS R S ( m + n ) ∘ (m + n)^\circ ( m + n ) ∘
A 120 ∘ 120^\circ 12 0 ∘ B 90 ∘ 90^\circ 9 0 ∘ C 75 ∘ 75^\circ 7 5 ∘ D 60 ∘ 60^\circ 6 0 ∘
Worked solution (try it first) Each of
m m m and
n n n is an angle between a tangent and a chord, so each equals the angle in the alternate segment: they are the two angles of the triangle at the ends of the diameter, taken crosswise.
The third angle of the triangle stands on the diameter, so it is
90 ∘ 90^\circ 9 0 ∘ (angle in a semicircle).
So the two angles at the ends of the diameter add up to
180 ∘ − 90 ∘ = 90 ∘ 180^\circ - 90^\circ = 90^\circ 18 0 ∘ − 9 0 ∘ = 9 0 ∘ , and
m + n = 90 ∘ m + n = 90^\circ m + n = 9 0 ∘ , option B.
Watch out
You don't need m m m n n n 60 ∘ 60^\circ 6 0 ∘ 120 ∘ 120^\circ 12 0 ∘ 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
In the diagram, O O O ∠ N L M = 74 ∘ \angle NLM = 74^\circ ∠ N L M = 7 4 ∘ ∠ L M N = 39 ∘ \angle LMN = 39^\circ ∠ L M N = 3 9 ∘ ∠ L O M = x \angle LOM = x ∠ L O M = x x x x
A 134 ∘ 134^\circ 13 4 ∘ B 126 ∘ 126^\circ 12 6 ∘ C 113 ∘ 113^\circ 11 3 ∘ D 106 ∘ 106^\circ 10 6 ∘
Worked solution (try it first) The angles of triangle
L M N LMN L M N add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ L N M = 180 ∘ − 74 ∘ − 39 ∘ \angle LNM = 180^\circ - 74^\circ - 39^\circ ∠ L N M = 18 0 ∘ − 7 4 ∘ − 3 9 ∘ ∠ L N M \angle LNM ∠ L N M is at the circumference on arc
L M LM L M , the same arc as
x x x at the centre.
The angle at the centre is twice the angle at the circumference:
x = 2 × 67 ∘ = 134 ∘ x = 2 \times 67^\circ = 134^\circ x = 2 × 6 7 ∘ = 13 4 ∘ , option A.
Watch out
Double the angle facing L M LM L M N N N 113 ∘ 113^\circ 11 3 ∘ 180 ∘ − 67 ∘ 180^\circ - 67^\circ 18 0 ∘ − 6 7 ∘ Report a problem with this question
Which of the following is not a sufficient condition for two triangles to be congruent?
Worked solution (try it first) SSS, SAS (the angle between the two sides) and AAS each fix a triangle completely, so they prove congruence.
In SSA the angle is not between the two sides.
The third side can swing to two positions, so two different triangles can fit the same data.
So SSA is not a sufficient condition, option D.
Watch out
The order of the letters matters: SAS has the angle between the two sides and works; SSA has it outside and doesn't. AAS works because the third angle follows. Report a problem with this question
A woman received a discount of 20 % 20\% 20%
A $675.25 B $660.25 C $656.25 D $616.25
Worked solution (try it first) A
20 % 20\% 20% discount means she paid
80 % 80\% 80% of the original price
P P P :
0.8 P = 525 0.8P = 525 0.8 P = 525 .
Divide both sides by 0.8:
P = 656.25 P = 656.25 P = 656.25 .
So the original price was $656.25, option C.
Watch out
The discount is 20 % 20\% 20% 20 % 20\% 20% Report a problem with this question
The interquartile range of a distribution is 7. If the 25th percentile is 16, find the upper quartile.
Worked solution (try it first) The 25th percentile is the lower quartile, so
Q 1 = 16 Q_1 = 16 Q 1 = 16 .
The interquartile range is
Q 3 − Q 1 = 7 Q_3 - Q_1 = 7 Q 3 − Q 1 = 7 , so
Q 3 = 16 + 7 = 23 Q_3 = 16 + 7 = 23 Q 3 = 16 + 7 = 23 , option C.
Watch out
The upper quartile is above the lower one, so add the 7. Subtracting gives 16 − 7 = 9 16 - 7 = 9 16 − 7 = 9 Report a problem with this question
The graphs of y = 2 x + 5 y = 2x + 5 y = 2 x + 5 y = 2 x 2 + x − 1 y = 2x^2 + x - 1 y = 2 x 2 + x − 1
A ( 2.0 , 9.0 ) (2.0, 9.0) ( 2.0 , 9.0 ) ( − 1.5 , 2.0 ) (-1.5, 2.0) ( − 1.5 , 2.0 ) B ( 2.0 , 8.5 ) (2.0, 8.5) ( 2.0 , 8.5 ) ( − 1.5 , 2.0 ) (-1.5, 2.0) ( − 1.5 , 2.0 ) C ( 2.0 , 8.0 ) (2.0, 8.0) ( 2.0 , 8.0 ) ( − 1.5 , 2.5 ) (-1.5, 2.5) ( − 1.5 , 2.5 ) D ( 2.0 , 7.5 ) (2.0, 7.5) ( 2.0 , 7.5 ) ( − 1.5 , 2.5 ) (-1.5, 2.5) ( − 1.5 , 2.5 )
Try it on a graph Where do the line and the curve meet?
Open the interactive graph Worked solution (try it first) Where the graphs meet,
2 x 2 + x − 1 = 2 x + 5 2x^2 + x - 1 = 2x + 5 2 x 2 + x − 1 = 2 x + 5 .
Bring everything to one side:
2 x 2 − x − 6 = 0 2x^2 - x - 6 = 0 2 x 2 − x − 6 = 0 .
Factorise:
( 2 x + 3 ) ( x − 2 ) = 0 (2x + 3)(x - 2) = 0 ( 2 x + 3 ) ( x − 2 ) = 0 , so
x = 2 x = 2 x = 2 or
x = − 1.5 x = -1.5 x = − 1.5 .
Find
y y y from the line
y = 2 x + 5 y = 2x + 5 y = 2 x + 5 :
x = 2 x = 2 x = 2 gives
y = 9 y = 9 y = 9 , and
x = − 1.5 x = -1.5 x = − 1.5 gives
y = 2 y = 2 y = 2 .
So the points are
( 2.0 , 9.0 ) (2.0, 9.0) ( 2.0 , 9.0 ) and
( − 1.5 , 2.0 ) (-1.5, 2.0) ( − 1.5 , 2.0 ) , option A.
Watch out
Check readings from the graph with the line's equation. At x = 2 x = 2 x = 2 y = 2 ( 2 ) + 5 = 9 y = 2(2) + 5 = 9 y = 2 ( 2 ) + 5 = 9 Report a problem with this question
Using the graph of y = 2 x 2 + x − 1 y = 2x^2 + x - 1 y = 2 x 2 + x − 1 x = − 2.5 x = -2.5 x = − 2.5 y y y
A y = 8.0 y = 8.0 y = 8.0 B y = 8.5 y = 8.5 y = 8.5 C y = 9.0 y = 9.0 y = 9.0 D y = 9.5 y = 9.5 y = 9.5
Try it on a graph Where do the line and the curve meet?
Open the interactive graph Worked solution (try it first) Put
x = − 2.5 x = -2.5 x = − 2.5 into
y = 2 x 2 + x − 1 y = 2x^2 + x - 1 y = 2 x 2 + x − 1 .
Square first:
( − 2.5 ) 2 = 6.25 (-2.5)^2 = 6.25 ( − 2.5 ) 2 = 6.25 , so
2 x 2 = 12.5 2x^2 = 12.5 2 x 2 = 12.5 .
So
y = 12.5 − 2.5 − 1 = 9.0 y = 12.5 - 2.5 - 1 = 9.0 y = 12.5 − 2.5 − 1 = 9.0 , option C.
Watch out
In 2 x 2 2x^2 2 x 2 x x x 2 × 6.25 = 12.5 2 \times 6.25 = 12.5 2 × 6.25 = 12.5 2 x 2x 2 x ( − 5 ) 2 = 25 (-5)^2 = 25 ( − 5 ) 2 = 25 Report a problem with this question
If ( x + 2 ) (x + 2) ( x + 2 ) x 2 + p x − 10 x^2 + px - 10 x 2 + p x − 10 p p p
Worked solution (try it first) By the factor theorem,
x + 2 x + 2 x + 2 is a factor, so the expression is 0 at
x = − 2 x = -2 x = − 2 .
( − 2 ) 2 + p ( − 2 ) − 10 = 0 (-2)^2 + p(-2) - 10 = 0 ( − 2 ) 2 + p ( − 2 ) − 10 = 0 , so
4 − 2 p − 10 = 0 4 - 2p - 10 = 0 4 − 2 p − 10 = 0 and
− 2 p = 6 -2p = 6 − 2 p = 6 .
So
p = − 3 p = -3 p = − 3 , option B.
Watch out
x + 2 x + 2 x + 2 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2 x = 2 x = 2 x = 2 4 + 2 p − 10 = 0 4 + 2p - 10 = 0 4 + 2 p − 10 = 0 p = 3 p = 3 p = 3 Report a problem with this question