Multiply 3.4 × 10 − 5 3.4 \times 10^{-5} 3.4 × 1 0 − 5 7.1 × 10 8 7.1 \times 10^{8} 7.1 × 1 0 8
A 2.414 × 10 2 2.414 \times 10^{2} 2.414 × 1 0 2 B 2.414 × 10 3 2.414 \times 10^{3} 2.414 × 1 0 3 C 2.414 × 10 4 2.414 \times 10^{4} 2.414 × 1 0 4 D 2.414 × 10 5 2.414 \times 10^{5} 2.414 × 1 0 5
Worked solution (try it first) Multiply the numbers:
3.4 × 7.1 = 24.14 3.4 \times 7.1 = 24.14 3.4 × 7.1 = 24.14 .
Add the powers:
10 − 5 × 10 8 = 10 3 10^{-5} \times 10^8 = 10^3 1 0 − 5 × 1 0 8 = 1 0 3 .
So the product is
24.14 × 10 3 24.14 \times 10^3 24.14 × 1 0 3 .
24.14 is more than 10, so write it as
2.414 × 10 2.414 \times 10 2.414 × 10 .
The answer is
2.414 × 10 4 2.414 \times 10^4 2.414 × 1 0 4 , option C.
Watch out
24.14 is not in standard form: turning it into 2.414 adds one to the power. Stopping at 10 3 10^3 1 0 3 Report a problem with this question
Given that P = { p : 1 < p < 20 } P = \{p : 1 < p < 20\} P = { p : 1 < p < 20 } p p p R = { r : 0 ≤ r ≤ 25 R = \{r : 0 \le r \le 25 R = { r : 0 ≤ r ≤ 25 r r r 4 } 4\} 4 } P ∩ R P \cap R P ∩ R
A { 4 , 8 , 10 , 16 } \{4, 8, 10, 16\} { 4 , 8 , 10 , 16 } B { 4 , 8 , 12 , 16 } \{4, 8, 12, 16\} { 4 , 8 , 12 , 16 } C { 4 , 8 , 12 , 16 , 20 } \{4, 8, 12, 16, 20\} { 4 , 8 , 12 , 16 , 20 } D { 4 , 8 , 12 , 16 , 20 , 24 } \{4, 8, 12, 16, 20, 24\} { 4 , 8 , 12 , 16 , 20 , 24 }
Worked solution (try it first) P P P is the whole numbers strictly between 1 and 20:
{ 2 , 3 , … , 19 } \{2, 3, \dots, 19\} { 2 , 3 , … , 19 } .
R R R is the multiples of 4 from 0 to 25:
{ 0 , 4 , 8 , 12 , 16 , 20 , 24 } \{0, 4, 8, 12, 16, 20, 24\} { 0 , 4 , 8 , 12 , 16 , 20 , 24 } .
The common elements are
{ 4 , 8 , 12 , 16 } \{4, 8, 12, 16\} { 4 , 8 , 12 , 16 } , option B.
(0 and 20 are not in
P P P .)
Watch out
p < 20 p < 20 p < 20 P P P Report a problem with this question
The first term of an Arithmetic Progression (A.P.) is 2 and the last term is 29. If the common difference is 3, how many terms are in the A.P.?
Worked solution (try it first) The last term is the
n n n th term,
a + ( n − 1 ) d a + (n - 1)d a + ( n − 1 ) d , so
2 + 3 ( n − 1 ) = 29 2 + 3(n - 1) = 29 2 + 3 ( n − 1 ) = 29 .
Subtract 2:
3 ( n − 1 ) = 27 3(n - 1) = 27 3 ( n − 1 ) = 27 , so
n − 1 = 9 n - 1 = 9 n − 1 = 9 .
So
n = 10 n = 10 n = 10 , option C.
Watch out
n − 1 = 9 n - 1 = 9 n − 1 = 9 Report a problem with this question
Express in index form: log a x + log a y = 3 \log_a x + \log_a y = 3 log a x + log a y = 3
A x + y = 3 x + y = 3 x + y = 3 B x y = 3 xy = 3 x y = 3 C x + y = a 3 x + y = a^3 x + y = a 3 D x y = a 3 xy = a^3 x y = a 3
Worked solution (try it first) Adding logs multiplies, so the equation becomes
log a ( x y ) = 3 \log_a(xy) = 3 log a ( x y ) = 3 .
Change to index form:
log a N = 3 \log_a N = 3 log a N = 3 means
N = a 3 N = a^3 N = a 3 .
So
x y = a 3 xy = a^3 x y = a 3 , option D.
Watch out
When you remove the log, the base becomes the base of a power: x y = a 3 xy = a^3 x y = a 3 x y = 3 xy = 3 x y = 3 Report a problem with this question
Simplify: ( 2 p − q ) 2 − ( p + q ) 2 (2p - q)^2 - (p + q)^2 ( 2 p − q ) 2 − ( p + q ) 2
A 3 p ( p − 2 q ) 3p(p - 2q) 3 p ( p − 2 q ) B 2 p ( p − 3 q ) 2p(p - 3q) 2 p ( p − 3 q ) C 3 p ( 2 p − q ) 3p(2p - q) 3 p ( 2 p − q ) D 2 p ( 3 p − q ) 2p(3p - q) 2 p ( 3 p − q )
Worked solution (try it first) Use the difference of two squares,
A 2 − B 2 = ( A − B ) ( A + B ) A^2 - B^2 = (A - B)(A + B) A 2 − B 2 = ( A − B ) ( A + B ) , with
A = 2 p − q A = 2p - q A = 2 p − q and
B = p + q B = p + q B = p + q .
A − B = 2 p − q − p − q = p − 2 q A - B = 2p - q - p - q = p - 2q A − B = 2 p − q − p − q = p − 2 q , and
A + B = 3 p A + B = 3p A + B = 3 p .
So the expression is
3 p ( p − 2 q ) 3p(p - 2q) 3 p ( p − 2 q ) , option A.
Watch out
Subtract the whole of B B B ( 2 p − q ) − ( p + q ) = p − 2 q (2p - q) - (p + q) = p - 2q ( 2 p − q ) − ( p + q ) = p − 2 q q q q p p p 3 p 2 3p^2 3 p 2 Report a problem with this question
If ( 3 − 4 2 ) ( 1 + 3 2 ) = a + b 2 (3 - 4\sqrt2)(1 + 3\sqrt2) = a + b\sqrt2 ( 3 − 4 2 ) ( 1 + 3 2 ) = a + b 2 b b b
A 5 5 5 B − 5 -5 − 5 C − 21 -21 − 21 D 21 21 21
Worked solution (try it first) Multiply out each pair:
3 × 1 = 3 3 \times 1 = 3 3 × 1 = 3 ,
3 × 3 2 = 9 2 3 \times 3\sqrt2 = 9\sqrt2 3 × 3 2 = 9 2 ,
− 4 2 × 1 = − 4 2 -4\sqrt2 \times 1 = -4\sqrt2 − 4 2 × 1 = − 4 2 and
− 4 2 × 3 2 = − 24 -4\sqrt2 \times 3\sqrt2 = -24 − 4 2 × 3 2 = − 24 .
Collect the whole numbers:
3 − 24 = − 21 3 - 24 = -21 3 − 24 = − 21 .
Collect the surds:
9 2 − 4 2 = 5 2 9\sqrt2 - 4\sqrt2 = 5\sqrt2 9 2 − 4 2 = 5 2 .
So the product is
− 21 + 5 2 -21 + 5\sqrt2 − 21 + 5 2 , and
b = 5 b = 5 b = 5 , option A.
Watch out
b b b 2 \sqrt2 2 − 21 -21 − 21 a a a Report a problem with this question
Find the time for which $1,250.00 will amount to $2,031.25 at 12.5 % 12.5\% 12.5%
A 2 years B 3 years C 4 years D 5 years
Worked solution (try it first) The interest is the amount minus the principal:
2031.25 − 1250 = 781.25 2031.25 - 1250 = 781.25 2031.25 − 1250 = 781.25 .
One year's interest at
12.5 % 12.5\% 12.5% is
0.125 × 1250 = 156.25 0.125 \times 1250 = 156.25 0.125 × 1250 = 156.25 .
So the time is
781.25 ÷ 156.25 = 5 781.25 \div 156.25 = 5 781.25 ÷ 156.25 = 5 years, option D.
Watch out
Use the interest, $781.25, not the amount. Dividing $2,031.25 by $156.25 gives 13 years, which is not an option. Report a problem with this question
If log 3 ( 2 x − 1 ) = 5 \log_3(2x - 1) = 5 log 3 ( 2 x − 1 ) = 5 x x x
Worked solution (try it first) Change to index form:
2 x − 1 = 3 5 = 243 2x - 1 = 3^5 = 243 2 x − 1 = 3 5 = 243 .
Add 1:
2 x = 244 2x = 244 2 x = 244 .
Divide by 2:
x = 122 x = 122 x = 122 , option D.
Watch out
3 5 3^5 3 5 3 × 3 × 3 × 3 × 3 = 243 3 \times 3 \times 3 \times 3 \times 3 = 243 3 × 3 × 3 × 3 × 3 = 243 3 × 5 = 15 3 \times 5 = 15 3 × 5 = 15 x = 8 x = 8 x = 8 Report a problem with this question
The population of a town increases by 3 % 3\% 3%
Worked solution (try it first) Each year the population is multiplied by
1.03 1.03 1.03 , and 2000 to 2003 is 3 years.
So the population is
3000 × 1.03 3 = 3000 × 1.092727 3000 \times 1.03^3 = 3000 \times 1.092727 3000 × 1.0 3 3 = 3000 × 1.092727 , which is 3278.18.
To the nearest person, 3,278, option B.
Watch out
Count the years: 2000 to 2003 is 3 growth steps, not 2. Using 1.03 2 1.03^2 1.0 3 2 Report a problem with this question
A trader gave a change of ₦540.00 instead of ₦570.00 to a customer. Calculate the percentage error.
A 5 5 19 % 5\frac{5}{19}\% 5 19 5 % B 5 5 9 % 5\frac{5}{9}\% 5 9 5 % C 5 7 19 % 5\frac{7}{19}\% 5 19 7 % D 5 7 9 % 5\frac{7}{9}\% 5 9 7 %
Worked solution (try it first) The error is ₦570.00 − ₦540.00 = ₦30.00.
Divide by the correct change and multiply by 100:
30 570 × 100 % = 100 19 % \frac{30}{570} \times 100\% = \frac{100}{19}\% 570 30 × 100% = 19 100 % .
100 ÷ 19 = 5 100 \div 19 = 5 100 ÷ 19 = 5 remainder 5, so the percentage error is
5 5 19 % 5\frac{5}{19}\% 5 19 5 % , option A.
Watch out
Divide by the correct amount, ₦570, not the amount given. Using 540 gives 30 540 × 100 % = 5 5 9 % \frac{30}{540} \times 100\% = 5\frac59\% 540 30 × 100% = 5 9 5 % Report a problem with this question
An 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
The exterior angles add up to 360 ∘ 360^\circ 36 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 180 180 180 Also set as WAEC 2020 · Paper 1 · Q29
Report a problem with this question
If 3 x − 2 y = − 5 3x - 2y = -5 3 x − 2 y = − 5 x + 2 y = 9 x + 2y = 9 x + 2 y = 9 x − y x + y \dfrac{x - y}{x + y} x + y x − y
A 5 3 \frac53 3 5 B 3 5 \frac35 5 3 C − 3 5 -\frac35 − 5 3 D − 5 3 -\frac53 − 3 5
Worked solution (try it first) Add the two equations so the
y y y terms cancel:
4 x = 4 4x = 4 4 x = 4 , so
x = 1 x = 1 x = 1 .
Put
x = 1 x = 1 x = 1 into
x + 2 y = 9 x + 2y = 9 x + 2 y = 9 :
2 y = 8 2y = 8 2 y = 8 , so
y = 4 y = 4 y = 4 .
Then
x − y = − 3 x - y = -3 x − y = − 3 and
x + y = 5 x + y = 5 x + y = 5 .
So
x − y x + y = − 3 5 \dfrac{x - y}{x + y} = -\frac35 x + y x − y = − 5 3 , option C.
Watch out
x − y = 1 − 4 x - y = 1 - 4 x − y = 1 − 4 − 3 5 -\frac35 − 5 3 3 5 \frac35 5 3 Report a problem with this question
A variable W W W M M M P P P correctly represents the relation with k 1 k_1 k 1 k 2 k_2 k 2
A W = k 1 M k 2 P W = \dfrac{k_1M}{k_2P} W = k 2 P k 1 M B W = ( k 1 + k 2 ) M P W = (k_1 + k_2)\dfrac{M}{P} W = ( k 1 + k 2 ) P M C W = k 1 M + k 2 P W = k_1M + \dfrac{k_2}{P} W = k 1 M + P k 2 D W = ( k 1 + k 2 ) M + P W = (k_1 + k_2)M + P W = ( k 1 + k 2 ) M + P
Worked solution (try it first) "Partly … partly" means the variable is a sum of two parts, each with its own constant.
The part that varies as
M M M is
k 1 M k_1M k 1 M , and the part that varies inversely as
P P P is
k 2 P \dfrac{k_2}{P} P k 2 .
So
W = k 1 M + k 2 P W = k_1M + \dfrac{k_2}{P} W = k 1 M + P k 2 , option C.
Watch out
Partial variation adds the parts. Options A and B combine M M M P P P Report a problem with this question
A cylindrical metallic barrel of height 2.5 m 2.5\text{ m} 2.5 m 0.245 m 0.245\text{ m} 0.245 m correct to one decimal place , the total surface area of the barrel. [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 2.1 m 2 2.1\text{ m}^2 2.1 m 2 B 3.5 m 2 3.5\text{ m}^2 3.5 m 2 C 4.0 m 2 4.0\text{ m}^2 4.0 m 2 D 9.4 m 2 9.4\text{ m}^2 9.4 m 2
Worked solution (try it first) Closed at one end: the curved surface plus one circle,
2 π r h + π r 2 2\pi rh + \pi r^2 2 π r h + π r 2 .
Curved surface:
2 × 22 7 × 0.245 × 2.5 = 3.85 m 2 2 \times \frac{22}{7} \times 0.245 \times 2.5 = 3.85\text{ m}^2 2 × 7 22 × 0.245 × 2.5 = 3.85 m 2 .
One end:
22 7 × 0.245 2 ≈ 0.19 m 2 \frac{22}{7} \times 0.245^2 \approx 0.19\text{ m}^2 7 22 × 0.24 5 2 ≈ 0.19 m 2 .
Total:
3.85 + 0.19 = 4.04 3.85 + 0.19 = 4.04 3.85 + 0.19 = 4.04 , which is
4.0 m 2 4.0\text{ m}^2 4.0 m 2 to 1 decimal place, option C.
Watch out
The curved surface is 2 π r h 2\pi rh 2 π r h π r h \pi rh π r h 1.925 + 0.19 ≈ 2.1 m 2 1.925 + 0.19 \approx 2.1\text{ m}^2 1.925 + 0.19 ≈ 2.1 m 2 Report a problem with this question
Make R R R V = π l ( R 2 − r 2 ) V = \pi l(R^2 - r^2) V = π l ( R 2 − r 2 )
A R = V π l + r 2 R = \sqrt{\dfrac{V}{\pi l} + r^2} R = π l V + r 2 B R = V π l − r 2 R = \sqrt{\dfrac{V}{\pi l} - r^2} R = π l V − r 2 C R = V − π l r 2 R = \sqrt{V - \pi lr^2} R = V − π l r 2 D R = V + π l r 2 R = \sqrt{V + \pi lr^2} R = V + π l r 2
Worked solution (try it first) Divide both sides by
π l \pi l π l :
V π l = R 2 − r 2 \frac{V}{\pi l} = R^2 - r^2 π l V = R 2 − r 2 .
Add
r 2 r^2 r 2 :
R 2 = V π l + r 2 R^2 = \frac{V}{\pi l} + r^2 R 2 = π l V + r 2 .
Take the square root:
R = V π l + r 2 R = \sqrt{\dfrac{V}{\pi l} + r^2} R = π l V + r 2 , option A.
Watch out
r 2 r^2 r 2 + r 2 +r^2 + r 2 Report a problem with this question
Consider the following statements:m m m n n n m ⇒ n m \Rightarrow n m ⇒ n
A ∼ m ⇒ ∼ n \sim m \Rightarrow \sim n ∼ m ⇒∼ n B n ⇒ ∼ m n \Rightarrow \sim m n ⇒∼ m C ∼ n ⇒ ∼ m \sim n \Rightarrow \sim m ∼ n ⇒∼ m D n ⇒ m n \Rightarrow m n ⇒ m
Worked solution (try it first) ⇒ n \Rightarrow n ⇒ n is equivalent to its contrapositive: swap the parts and negate both.
⇒ ∼ m \Rightarrow \sim m ⇒∼ m is valid, option C.
Watch out
Negating both parts without swapping them gives the inverse, ∼ m ⇒ ∼ n \sim m \Rightarrow \sim n ∼ m ⇒∼ n m ⇒ n m \Rightarrow n m ⇒ n Report a problem with this question
A number is added to both the numerator and the denominator of the fraction 1 8 \frac18 8 1 1 2 \frac12 2 1
Worked solution (try it first) Call the number
x x x :
1 + x 8 + x = 1 2 \dfrac{1 + x}{8 + x} = \dfrac12 8 + x 1 + x = 2 1 .
Cross-multiply:
2 ( 1 + x ) = 8 + x 2(1 + x) = 8 + x 2 ( 1 + x ) = 8 + x , so
2 + 2 x = 8 + x 2 + 2x = 8 + x 2 + 2 x = 8 + x .
Take
x x x and 2 from both sides:
x = 6 x = 6 x = 6 , option D.
Check:
7 14 = 1 2 \frac{7}{14} = \frac12 14 7 = 2 1 .
Watch out
The number is added, not multiplied. Multiplying the top by 4 turns 1 8 \frac18 8 1 4 8 = 1 2 \frac48 = \frac12 8 4 = 2 1 5 12 \frac{5}{12} 12 5 Report a problem with this question
Gifty, Justina and Frank shared 60 oranges in the ratio 5 : 3 : 7 5 : 3 : 7 5 : 3 : 7
Worked solution (try it first) Add the ratio:
5 + 3 + 7 = 15 5 + 3 + 7 = 15 5 + 3 + 7 = 15 parts.
One part is
60 ÷ 15 = 4 60 \div 15 = 4 60 ÷ 15 = 4 oranges.
Justina is second, so she has 3 parts:
3 × 4 = 12 3 \times 4 = 12 3 × 4 = 12 oranges, option A.
Watch out
Match the names to the ratio in order: Gifty 5, Justina 3, Frank 7. Using 5 parts gives 20 (option C), which is Gifty's share. Report a problem with this question
Find the quadratic equation whose roots are 2 3 \frac23 3 2 − 1 -1 − 1
A 3 x 2 + x − 2 = 0 3x^2 + x - 2 = 0 3 x 2 + x − 2 = 0 B 3 x 2 − x − 2 = 0 3x^2 - x - 2 = 0 3 x 2 − x − 2 = 0 C 3 x 2 + x + 2 = 0 3x^2 + x + 2 = 0 3 x 2 + x + 2 = 0 D 3 x 2 + x − 1 = 0 3x^2 + x - 1 = 0 3 x 2 + x − 1 = 0
Worked solution (try it first) Roots
2 3 \frac23 3 2 and
− 1 -1 − 1 give the factors
( 3 x − 2 ) (3x - 2) ( 3 x − 2 ) and
( x + 1 ) (x + 1) ( x + 1 ) .
Expand:
( 3 x − 2 ) ( x + 1 ) = 3 x 2 + 3 x − 2 x − 2 (3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 ( 3 x − 2 ) ( x + 1 ) = 3 x 2 + 3 x − 2 x − 2 = 3 x 2 + x − 2 = 3x^2 + x - 2 = 3 x 2 + x − 2 .
So the equation is
3 x 2 + x − 2 = 0 3x^2 + x - 2 = 0 3 x 2 + x − 2 = 0 , option A.
Watch out
A root of − 1 -1 − 1 x + 1 x + 1 x + 1 ( 3 x + 2 ) ( x − 1 ) = 3 x 2 − x − 2 (3x + 2)(x - 1) = 3x^2 - x - 2 ( 3 x + 2 ) ( x − 1 ) = 3 x 2 − x − 2 Report a problem with this question
A piece of rod of length 44 m 44\text{ m} 44 m 7 : 4 7 : 4 7 : 4
A 8 m 8\text{ m} 8 m B 14 m 14\text{ m} 14 m C 16 m 16\text{ m} 16 m D 24 m 24\text{ m} 24 m
Worked solution (try it first) The rod makes the whole perimeter, so length + breadth
= 44 ÷ 2 = 22 = 44 \div 2 = 22 = 44 ÷ 2 = 22 m.
Share 22 in the ratio
7 : 4 7 : 4 7 : 4 , which has
7 + 4 = 11 7 + 4 = 11 7 + 4 = 11 parts.
The breadth is 4 parts:
4 11 × 22 = 8 \frac{4}{11} \times 22 = 8 11 4 × 22 = 8 m, option A.
Watch out
Share half the perimeter, not all of it. 4 11 × 44 = 16 \frac{4}{11} \times 44 = 16 11 4 × 44 = 16 Report a problem with this question
In the diagram, M N ‾ ∥ K L ‾ \overline{MN} \parallel \overline{KL} M N ∥ K L M L ‾ \overline{ML} M L K N ‾ \overline{KN} K N X X X ∣ M N ∣ = 12 cm |MN| = 12\text{ cm} ∣ M N ∣ = 12 cm ∣ M X ∣ = 10 cm |MX| = 10\text{ cm} ∣ M X ∣ = 10 cm ∣ K L ∣ = 9 cm |KL| = 9\text{ cm} ∣ K L ∣ = 9 cm △ M X N \triangle MXN △ M X N 16 cm 2 16\text{ cm}^2 16 cm 2 △ L X K \triangle LXK △ L X K
A 8 cm 2 8\text{ cm}^2 8 cm 2 B 9 cm 2 9\text{ cm}^2 9 cm 2 C 10 cm 2 10\text{ cm}^2 10 cm 2 D 12 cm 2 12\text{ cm}^2 12 cm 2
Worked solution (try it first) M N ∥ K L MN \parallel KL M N ∥ K L , so the alternate angles at
K K K and
N N N are equal, and so are those at
L L L and
M M M .
Triangles
L X K LXK L X K and
M X N MXN M X N are similar.
The matching sides
K L KL K L and
M N MN M N give the length scale factor
9 12 = 3 4 \frac{9}{12} = \frac34 12 9 = 4 3 .
Areas scale by the square of that:
( 3 4 ) 2 = 9 16 \left(\frac34\right)^2 = \frac{9}{16} ( 4 3 ) 2 = 16 9 .
So the area of
△ L X K \triangle LXK △ L X K is
16 × 9 16 = 9 cm 2 16 \times \frac{9}{16} = 9\text{ cm}^2 16 × 16 9 = 9 cm 2 , option B.
Watch out
Square the scale factor for areas. Using 3 4 \frac34 4 3 16 × 3 4 = 12 cm 2 16 \times \frac34 = 12\text{ cm}^2 16 × 4 3 = 12 cm 2 M X = 10 MX = 10 M X = 10 Report a problem with this question
A ladder 15 m 15\text{ m} 15 m 72 ∘ 72^\circ 7 2 ∘ correct to one decimal place , the distance between the foot of the ladder and the pole.
A 15.8 m 15.8\text{ m} 15.8 m B 14.3 m 14.3\text{ m} 14.3 m C 4.9 m 4.9\text{ m} 4.9 m D 4.6 m 4.6\text{ m} 4.6 m
Worked solution (try it first) The ladder (15 m) is the hypotenuse, and the distance from the foot to the pole is next to the
72 ∘ 72^\circ 7 2 ∘ angle at the ground.
So the distance is
15 cos 72 ∘ = 15 × 0.3090 15\cos72^\circ = 15 \times 0.3090 15 cos 7 2 ∘ = 15 × 0.3090 ≈ 4.635 \approx 4.635 ≈ 4.635 m.
To one decimal place, that is 4.6 m, option D.
Watch out
The distance along the ground is adjacent to the angle, so use cosine. Sine gives the height up the pole, 15 sin 72 ∘ ≈ 14.3 15\sin72^\circ \approx 14.3 15 sin 7 2 ∘ ≈ 14.3 Report a problem with this question
In the diagram, O O O ∣ O A ‾ ∣ = 25 cm |\overline{OA}| = 25\text{ cm} ∣ O A ∣ = 25 cm ∣ A B ‾ ∣ = 40 cm |\overline{AB}| = 40\text{ cm} ∣ A B ∣ = 40 cm ∣ O H ‾ ∣ |\overline{OH}| ∣ O H ∣
A 15 cm 15\text{ cm} 15 cm B 20 cm 20\text{ cm} 20 cm C 25 cm 25\text{ cm} 25 cm D 30 cm 30\text{ cm} 30 cm
Worked solution (try it first) The perpendicular from the centre to a chord bisects it, so
∣ A H ∣ = 40 ÷ 2 = 20 |AH| = 40 \div 2 = 20 ∣ A H ∣ = 40 ÷ 2 = 20 cm.
Triangle
O H A OHA O H A has a right angle at
H H H and hypotenuse
O A = 25 OA = 25 O A = 25 .
Pythagoras:
∣ O H ∣ 2 = 25 2 − 20 2 = 225 |OH|^2 = 25^2 - 20^2 = 225 ∣ O H ∣ 2 = 2 5 2 − 2 0 2 = 225 .
So
∣ O H ∣ = 15 |OH| = 15 ∣ O H ∣ = 15 cm, option A.
Watch out
20 20 20 ∣ A H ∣ |AH| ∣ A H ∣ ∣ O H ∣ |OH| ∣ O H ∣ Report a problem with this question
Given that P P P 25 m 25\text{ m} 25 m 330 ∘ 330^\circ 33 0 ∘ Q Q Q P P P Q Q Q
A 25.2 m 25.2\text{ m} 25.2 m B 21.7 m 21.7\text{ m} 21.7 m C 19.8 m 19.8\text{ m} 19.8 m D 18.5 m 18.5\text{ m} 18.5 m
Worked solution (try it first) A bearing of
330 ∘ 330^\circ 33 0 ∘ is
30 ∘ 30^\circ 3 0 ∘ west of north, so
Q P QP QP makes
30 ∘ 30^\circ 3 0 ∘ with the north line through
Q Q Q .
The distance north is adjacent to that
30 ∘ 30^\circ 3 0 ∘ angle:
25 cos 30 ∘ ≈ 21.7 25\cos30^\circ \approx 21.7 25 cos 3 0 ∘ ≈ 21.7 m.
So
P P P is 21.7 m north of
Q Q Q , and
Q Q Q is 21.7 m south of
P P P , option B.
Watch out
The north–south distance is next to the 30 ∘ 30^\circ 3 0 ∘ cos \cos cos 25 sin 30 ∘ = 12.5 25\sin30^\circ = 12.5 25 sin 3 0 ∘ = 12.5 Report a problem with this question
A car valued at $600,000.00 depreciates by 10 % 10\% 10% each year. What will be the value of the car at the end of two years?
A $120,000.00 B $480,000.00 C $486,000.00 D $540,000.00
Worked solution (try it first) Each year the car keeps
90 % 90\% 90% of its value, so multiply by 0.9 once a year.
After one year:
0.9 × 600 000 = 540 000 0.9 \times 600\,000 = 540\,000 0.9 × 600 000 = 540 000 .
After two years:
0.9 × 540 000 = 486 000 0.9 \times 540\,000 = 486\,000 0.9 × 540 000 = 486 000 .
So the car is worth $486,000.00, option C.
Watch out
The second 10 % 10\% 10% 10 % 10\% 10% Report a problem with this question
The length and breadth of a cuboid are 15 cm 15\text{ cm} 15 cm 8 cm 8\text{ cm} 8 cm 1,560 cm 3 1{,}560\text{ cm}^3 1 , 560 cm 3
A 976 cm 2 976\text{ cm}^2 976 cm 2 B 838 cm 2 838\text{ cm}^2 838 cm 2 C 792 cm 2 792\text{ cm}^2 792 cm 2 D 746 cm 2 746\text{ cm}^2 746 cm 2
Worked solution (try it first) Find the height first: the base is
15 × 8 = 120 cm 2 15 \times 8 = 120\text{ cm}^2 15 × 8 = 120 cm 2 , so
h = 1560 ÷ 120 = 13 h = 1560 \div 120 = 13 h = 1560 ÷ 120 = 13 cm.
The three pairs of faces are
15 × 8 = 120 15 \times 8 = 120 15 × 8 = 120 ,
15 × 13 = 195 15 \times 13 = 195 15 × 13 = 195 and
8 × 13 = 104 8 \times 13 = 104 8 × 13 = 104 .
Total surface area:
2 ( 120 + 195 + 104 ) = 2 × 419 2(120 + 195 + 104) = 2 \times 419 2 ( 120 + 195 + 104 ) = 2 × 419 = 838 cm 2 = 838\text{ cm}^2 = 838 cm 2 , option B.
Watch out
Double the sum: each face has an equal opposite face. Adding the three faces once gives 419, which is not an option. Report a problem with this question
The number 1621 1621 1621 6244 6244 6244 x x x 4323 4323 4323 x x x
Worked solution (try it first) Check by adding:
4323 + 1621 4323 + 1621 4323 + 1621 must give
6244 6244 6244 in base
x x x .
Units:
3 + 1 = 4 3 + 1 = 4 3 + 1 = 4 , and second column:
2 + 2 = 4 2 + 2 = 4 2 + 2 = 4 , with no carries.
Third column:
3 + 6 = 9 3 + 6 = 9 3 + 6 = 9 is written as 2 carry 1, so
9 = x + 2 9 = x + 2 9 = x + 2 .
So
x = 7 x = 7 x = 7 , option A.
The last column agrees:
4 + 1 + 1 = 6 4 + 1 + 1 = 6 4 + 1 + 1 = 6 .
Watch out
In base ten, 6244 − 1621 = 4623 6244 - 1621 = 4623 6244 − 1621 = 4623 Report a problem with this question
Factorize completely: 27 x 2 − 48 y 2 27x^2 - 48y^2 27 x 2 − 48 y 2
A 3 ( 3 x + 4 y ) ( 3 x + 4 y ) 3(3x + 4y)(3x + 4y) 3 ( 3 x + 4 y ) ( 3 x + 4 y ) B 3 ( 3 x + 4 y ) ( 3 x − 4 y ) 3(3x + 4y)(3x - 4y) 3 ( 3 x + 4 y ) ( 3 x − 4 y ) C 3 ( 9 x − 16 y ) ( 9 x + 16 y ) 3(9x - 16y)(9x + 16y) 3 ( 9 x − 16 y ) ( 9 x + 16 y ) D 3 ( 9 x − 16 y ) ( 9 x − 16 y ) 3(9x - 16y)(9x - 16y) 3 ( 9 x − 16 y ) ( 9 x − 16 y )
Worked solution (try it first) Take out the common factor 3:
27 x 2 − 48 y 2 = 3 ( 9 x 2 − 16 y 2 ) 27x^2 - 48y^2 = 3(9x^2 - 16y^2) 27 x 2 − 48 y 2 = 3 ( 9 x 2 − 16 y 2 ) .
The bracket is a difference of two squares:
9 x 2 = ( 3 x ) 2 9x^2 = (3x)^2 9 x 2 = ( 3 x ) 2 and
16 y 2 = ( 4 y ) 2 16y^2 = (4y)^2 16 y 2 = ( 4 y ) 2 .
So
27 x 2 − 48 y 2 = 3 ( 3 x + 4 y ) ( 3 x − 4 y ) 27x^2 - 48y^2 = 3(3x + 4y)(3x - 4y) 27 x 2 − 48 y 2 = 3 ( 3 x + 4 y ) ( 3 x − 4 y ) , option B.
Watch out
Take square roots: 9 x 2 = 3 x \sqrt{9x^2} = 3x 9 x 2 = 3 x 16 y 2 = 4 y \sqrt{16y^2} = 4y 16 y 2 = 4 y 9 x 9x 9 x 16 y 16y 16 y 81 x 2 81x^2 81 x 2 256 y 2 256y^2 256 y 2 Report a problem with this question
For what values of x x x x − 3 4 + x + 1 8 ≥ 2 \dfrac{x - 3}{4} + \dfrac{x + 1}{8} \ge 2 4 x − 3 + 8 x + 1 ≥ 2
A x ≥ 5 x \ge 5 x ≥ 5 B x ≥ 6 x \ge 6 x ≥ 6 C x ≥ 7 x \ge 7 x ≥ 7 D x ≥ 8 x \ge 8 x ≥ 8
Worked solution (try it first) Multiply every term by 8, the LCM of 4 and 8:
2 ( x − 3 ) + ( x + 1 ) ≥ 16 2(x - 3) + (x + 1) \ge 16 2 ( x − 3 ) + ( x + 1 ) ≥ 16 .
Expand and collect:
2 x − 6 + x + 1 ≥ 16 2x - 6 + x + 1 \ge 16 2 x − 6 + x + 1 ≥ 16 , so
3 x − 5 ≥ 16 3x - 5 \ge 16 3 x − 5 ≥ 16 .
Add 5 to both sides:
3 x ≥ 21 3x \ge 21 3 x ≥ 21 .
Divide by 3:
x ≥ 7 x \ge 7 x ≥ 7 , option C.
Watch out
Multiply the whole of x − 3 x - 3 x − 3 2 x − 6 2x - 6 2 x − 6 2 x − 3 2x - 3 2 x − 3 3 x − 2 ≥ 16 3x - 2 \ge 16 3 x − 2 ≥ 16 x ≥ 6 x \ge 6 x ≥ 6 Report a problem with this question
In the diagram, ∠ S Q R = 52 ∘ \angle SQR = 52^\circ ∠ S QR = 5 2 ∘ ∠ P R T = 16 ∘ \angle PRT = 16^\circ ∠ P R T = 1 6 ∘ y y y
A 64 ∘ 64^\circ 6 4 ∘ B 68 ∘ 68^\circ 6 8 ∘ C 112 ∘ 112^\circ 11 2 ∘ D 128 ∘ 128^\circ 12 8 ∘
Worked solution (try it first) P Q S T PQST P QS T is a cyclic quadrilateral.
Its exterior angle at
Q Q Q ,
∠ S Q R \angle SQR ∠ S QR , equals the interior opposite angle at
T T T :
∠ P T S = 52 ∘ \angle PTS = 52^\circ ∠ P T S = 5 2 ∘ .
The angles of triangle
P R T PRT P R T add up to
180 ∘ 180^\circ 18 0 ∘ :
y = 180 ∘ − 52 ∘ − 16 ∘ y = 180^\circ - 52^\circ - 16^\circ y = 18 0 ∘ − 5 2 ∘ − 1 6 ∘ = 112 ∘ = 112^\circ = 11 2 ∘ , option C.
Watch out
52 ∘ + 16 ∘ = 68 ∘ 52^\circ + 16^\circ = 68^\circ 5 2 ∘ + 1 6 ∘ = 6 8 ∘ P R T PRT P R T y y y 180 ∘ − 68 ∘ 180^\circ - 68^\circ 18 0 ∘ − 6 8 ∘ Report a problem with this question
In the diagram, J K L JKL J K L G H I K GHIK G H I K K K K ∠ L K G = 38 ∘ \angle LKG = 38^\circ ∠ L K G = 3 8 ∘ ∠ H I K = 87 ∘ \angle HIK = 87^\circ ∠ H I K = 8 7 ∘ x x x
A 93 ∘ 93^\circ 9 3 ∘ B 55 ∘ 55^\circ 5 5 ∘ C 42 ∘ 42^\circ 4 2 ∘ D 23 ∘ 23^\circ 2 3 ∘
Worked solution (try it first) H I K G HIKG H I K G is a cyclic quadrilateral, so opposite angles add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ H G K = 180 ∘ − 87 ∘ \angle HGK = 180^\circ - 87^\circ ∠ H G K = 18 0 ∘ − 8 7 ∘ H H H ,
G G G and
L L L are on a straight line, so
∠ K G L = 180 ∘ − 93 ∘ \angle KGL = 180^\circ - 93^\circ ∠ K G L = 18 0 ∘ − 9 3 ∘ The angles of triangle
K G L KGL K G L add up to
180 ∘ 180^\circ 18 0 ∘ :
x = 180 ∘ − 38 ∘ − 87 ∘ x = 180^\circ - 38^\circ - 87^\circ x = 18 0 ∘ − 3 8 ∘ − 8 7 ∘ = 55 ∘ = 55^\circ = 5 5 ∘ , option B.
Watch out
Inside triangle K G L KGL K G L G G G 87 ∘ 87^\circ 8 7 ∘ ∠ H G K = 93 ∘ \angle HGK = 93^\circ ∠ H G K = 9 3 ∘ 93 ∘ 93^\circ 9 3 ∘ 49 ∘ 49^\circ 4 9 ∘ Report a problem with this question
A cone and a cylinder are of equal volume. The base radius of the cone is twice the radius of the cylinder. What is the ratio of the height of the cylinder to that of the cone?
A 5 : 4 5 : 4 5 : 4 B 4 : 3 4 : 3 4 : 3 C 3 : 2 3 : 2 3 : 2 D 3 : 4 3 : 4 3 : 4
Worked solution (try it first) Let the cylinder have radius
r r r and height
h 2 h_2 h 2 , and the cone radius
2 r 2r 2 r and height
h 1 h_1 h 1 .
Equal volumes:
1 3 π ( 2 r ) 2 h 1 = π r 2 h 2 \frac13\pi(2r)^2h_1 = \pi r^2h_2 3 1 π ( 2 r ) 2 h 1 = π r 2 h 2 .
Divide both sides by
π r 2 \pi r^2 π r 2 :
4 3 h 1 = h 2 \frac43h_1 = h_2 3 4 h 1 = h 2 .
So
h 2 : h 1 = 4 : 3 h_2 : h_1 = 4 : 3 h 2 : h 1 = 4 : 3 , option B.
Watch out
The question asks for cylinder to cone. 3 : 4 3 : 4 3 : 4 Report a problem with this question
Find, correct to the nearest whole number, the value of h h h
A 16 m 16\text{ m} 16 m B 22 m 22\text{ m} 22 m C 23 m 23\text{ m} 23 m D 18 m 18\text{ m} 18 m
Worked solution (try it first) Draw a horizontal line from the top of the 17 m side to the right side.
It is 19 m long and makes a right-angled triangle with the 20 m slant as hypotenuse.
By Pythagoras, the triangle's height is
20 2 − 19 2 = 39 \sqrt{20^2 - 19^2} = \sqrt{39} 2 0 2 − 1 9 2 = 39 ≈ 6.24 \approx 6.24 ≈ 6.24 m.
So
h ≈ 17 + 6.24 = 23.24 h \approx 17 + 6.24 = 23.24 h ≈ 17 + 6.24 = 23.24 m, which is 23 m to the nearest whole number, option C.
Watch out
The base of the small triangle is the 19 m width, not 17 m. Using 17 gives 20 2 − 17 2 ≈ 10.5 \sqrt{20^2 - 17^2} \approx 10.5 2 0 2 − 1 7 2 ≈ 10.5 h ≈ 27.5 h \approx 27.5 h ≈ 27.5 Report a problem with this question
The gradient of the line joining the points P ( 2 , − 8 ) P(2, -8) P ( 2 , − 8 ) Q ( 1 , y ) Q(1, y) Q ( 1 , y ) − 4 -4 − 4 y y y
Worked solution (try it first) Gradient from
P P P to
Q Q Q :
y − ( − 8 ) 1 − 2 = − 4 \dfrac{y - (-8)}{1 - 2} = -4 1 − 2 y − ( − 8 ) = − 4 , so
y + 8 − 1 = − 4 \dfrac{y + 8}{-1} = -4 − 1 y + 8 = − 4 .
Multiply both sides by
− 1 -1 − 1 :
y + 8 = 4 y + 8 = 4 y + 8 = 4 .
So
y = − 4 y = -4 y = − 4 , option C.
Watch out
The run is 1 − 2 = − 1 1 - 2 = -1 1 − 2 = − 1 + 1 +1 + 1 y + 8 = − 4 y + 8 = -4 y + 8 = − 4 y = − 12 y = -12 y = − 12 Report a problem with this question
In the diagram, P Q ‾ ∥ R S ‾ \overline{PQ} \parallel \overline{RS} P Q ∥ R S ∠ W Y Z = 44 ∘ \angle WYZ = 44^\circ ∠ W Y Z = 4 4 ∘ ∠ W X Z = 50 ∘ \angle WXZ = 50^\circ ∠ W X Z = 5 0 ∘ ∠ W T X \angle WTX ∠ W T X
A 65 ∘ 65^\circ 6 5 ∘ B 68 ∘ 68^\circ 6 8 ∘ C 86 ∘ 86^\circ 8 6 ∘ D 90 ∘ 90^\circ 9 0 ∘
Worked solution (try it first) P Q ∥ R S PQ \parallel RS P Q ∥ R S , so
∠ T W X = ∠ W Y Z = 44 ∘ \angle TWX = \angle WYZ = 44^\circ ∠ T W X = ∠ W Y Z = 4 4 ∘ (alternate angles on the transversal
W Y WY W Y ).
The angles of triangle
W T X WTX W T X add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ W T X = 180 ∘ − 44 ∘ − 50 ∘ \angle WTX = 180^\circ - 44^\circ - 50^\circ ∠ W T X = 18 0 ∘ − 4 4 ∘ − 5 0 ∘ = 86 ∘ = 86^\circ = 8 6 ∘ , option C.
Watch out
The diagram is not drawn to scale, so don't assume the lines cross at 90 ∘ 90^\circ 9 0 ∘ Report a problem with this question
The perimeter of a rectangular garden is 90 m 90\text{ m} 90 m 7 m 7\text{ m} 7 m
A 19 m 19\text{ m} 19 m B 23 m 23\text{ m} 23 m C 24 m 24\text{ m} 24 m D 26 m 26\text{ m} 26 m
Worked solution (try it first) The width is
L − 7 L - 7 L − 7 .
Half the perimeter is
L + ( L − 7 ) = 45 L + (L - 7) = 45 L + ( L − 7 ) = 45 .
So
2 L − 7 = 45 2L - 7 = 45 2 L − 7 = 45 ,
2 L = 52 2L = 52 2 L = 52 and
L = 26 L = 26 L = 26 m, option D.
Watch out
19 m (option A) is the width, 26 − 7 26 - 7 26 − 7 Report a problem with this question
Four of the angles of a hexagon sum up to 420 ∘ 420^\circ 42 0 ∘ each of the angles.
A 60 ∘ 60^\circ 6 0 ∘ B 100 ∘ 100^\circ 10 0 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 150 ∘ 150^\circ 15 0 ∘
Worked solution (try it first) The interior angles of a hexagon add up to
( 6 − 2 ) × 180 ∘ = 720 ∘ (6 - 2) \times 180^\circ = 720^\circ ( 6 − 2 ) × 18 0 ∘ = 72 0 ∘ .
The two remaining angles share
720 ∘ − 420 ∘ = 300 ∘ 720^\circ - 420^\circ = 300^\circ 72 0 ∘ − 42 0 ∘ = 30 0 ∘ .
So each is
300 ∘ ÷ 2 = 150 ∘ 300^\circ \div 2 = 150^\circ 30 0 ∘ ÷ 2 = 15 0 ∘ , option D.
Watch out
The hexagon isn't regular, so its angles aren't all 120 ∘ 120^\circ 12 0 ∘ 720 ∘ 720^\circ 72 0 ∘ Report a problem with this question
Find the value of x x x
A 60 ∘ 60^\circ 6 0 ∘ B 65 ∘ 65^\circ 6 5 ∘ C 120 ∘ 120^\circ 12 0 ∘ D 125 ∘ 125^\circ 12 5 ∘
Worked solution (try it first) Q R ∥ M P QR \parallel MP QR ∥ M P , so
∠ N R Q = ∠ N P M = 55 ∘ \angle NRQ = \angle NPM = 55^\circ ∠ N R Q = ∠ N P M = 5 5 ∘ (corresponding angles).
The angles of triangle
N Q R NQR N QR add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ N Q R = 180 ∘ − 65 ∘ − 55 ∘ \angle NQR = 180^\circ - 65^\circ - 55^\circ ∠ N QR = 18 0 ∘ − 6 5 ∘ − 5 5 ∘ M Q N MQN M QN is a straight line, so
x = ∠ M Q R x = \angle MQR x = ∠ M QR = 180 ∘ − 60 ∘ = 180^\circ - 60^\circ = 18 0 ∘ − 6 0 ∘ = 120 ∘ = 120^\circ = 12 0 ∘ , option C.
Watch out
60 ∘ 60^\circ 6 0 ∘ ∠ N Q R \angle NQR ∠ N QR Q R QR QR x x x Q R QR QR M N MN M N Report a problem with this question
The following are the masses (in kg) of members in a club: 59, 44, 53, 57, 49, 40, 48 and 50. Use the information to answer questions 39 and 40.
A 40 kg 40\text{ kg} 40 kg B 44 kg 44\text{ kg} 44 kg C 50 kg 50\text{ kg} 50 kg D 53 kg 53\text{ kg} 53 kg
Worked solution (try it first) Add the eight masses:
59 + 44 + 53 + 57 + 49 + 40 + 48 + 50 = 400 59 + 44 + 53 + 57 + 49 + 40 + 48 + 50 = 400 59 + 44 + 53 + 57 + 49 + 40 + 48 + 50 = 400 kg.
Divide by the 8 members:
400 8 = 50 \frac{400}{8} = 50 8 400 = 50 kg, option C.
Watch out
Divide by the number of members, 8. Dividing by 10 gives 40 kg (option A). Report a problem with this question
The masses (in kg) of members in a club are 59, 44, 53, 57, 49, 40, 48 and 50. Calculate the variance of the distribution.
Worked solution (try it first) The masses add up to 400 and there are 8 of them, so the mean is 50 kg.
The deviations are 9,
− 6 -6 − 6 , 3, 7,
− 1 -1 − 1 ,
− 10 -10 − 10 ,
− 2 -2 − 2 , 0.
Their squares add up to
81 + 36 + 9 + 49 + 1 + 100 + 4 + 0 = 280 81 + 36 + 9 + 49 + 1 + 100 + 4 + 0 = 280 81 + 36 + 9 + 49 + 1 + 100 + 4 + 0 = 280 .
The variance is
280 8 = 35 \frac{280}{8} = 35 8 280 = 35 , option A.
Watch out
Divide by the number of members, 8. Dividing by 7 gives 40 (option C). Report a problem with this question
Two opposite sides of a rectangle are ( 5 x + 3 ) m (5x + 3)\text{ m} ( 5 x + 3 ) m ( 2 x + 9 ) m (2x + 9)\text{ m} ( 2 x + 9 ) m ( 6 x − 7 ) m (6x - 7)\text{ m} ( 6 x − 7 ) m m 2 \text{m}^2 m 2
Worked solution (try it first) Opposite sides of a rectangle are equal:
5 x + 3 = 2 x + 9 5x + 3 = 2x + 9 5 x + 3 = 2 x + 9 , so
3 x = 6 3x = 6 3 x = 6 and
x = 2 x = 2 x = 2 .
The sides are
5 ( 2 ) + 3 = 13 5(2) + 3 = 13 5 ( 2 ) + 3 = 13 m and
6 ( 2 ) − 7 = 5 6(2) - 7 = 5 6 ( 2 ) − 7 = 5 m.
Area
= 13 × 5 = 65 m 2 = 13 \times 5 = 65\text{ m}^2 = 13 × 5 = 65 m 2 , option B.
Watch out
The two opposite sides are the same side, 13 m. Multiply by the adjacent side, 5 m; multiplying the two opposite sides gives 169 m 2 169\text{ m}^2 169 m 2 Report a problem with this question
A die is tossed once. Find the probability of getting a prime number.
A 1 2 \frac12 2 1 B 1 6 \frac16 6 1 C 1 3 \frac13 3 1 D 2 3 \frac23 3 2
Worked solution (try it first) A die has 6 equally likely faces.
The primes on it are 2, 3 and 5.
1 is not a prime.
So the probability is
3 6 = 1 2 \frac36 = \frac12 6 3 = 2 1 , option A.
Watch out
1 is not a prime, since it has only one factor. Counting it gives 4 6 = 2 3 \frac46 = \frac23 6 4 = 3 2 Report a problem with this question
The area of a sector of a circle with radius 7 cm 7\text{ cm} 7 cm 51.3 cm 2 51.3\text{ cm}^2 51.3 cm 2 nearest whole number, the angle of the sector. [ Take π = 22 7 ] \left[\text{Take }\pi = \frac{22}{7}\right] [ Take π = 7 22 ]
A 60 ∘ 60^\circ 6 0 ∘ B 120 ∘ 120^\circ 12 0 ∘ C 150 ∘ 150^\circ 15 0 ∘ D 180 ∘ 180^\circ 18 0 ∘
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 .
Here
π r 2 = 22 7 × 49 \pi r^2 = \frac{22}{7} \times 49 π r 2 = 7 22 × 49 = 154 cm 2 = 154\text{ cm}^2 = 154 cm 2 .
So
θ 360 × 154 = 51.3 \frac{\theta}{360} \times 154 = 51.3 360 θ × 154 = 51.3 , and
θ = 51.3 × 360 154 \theta = \frac{51.3 \times 360}{154} θ = 154 51.3 × 360 ≈ 119.9 ∘ \approx 119.9^\circ ≈ 119. 9 ∘ .
To the nearest whole number,
θ = 120 ∘ \theta = 120^\circ θ = 12 0 ∘ , option B.
Watch out
Use the area of the circle, π r 2 = 154 \pi r^2 = 154 π r 2 = 154 2 π r = 44 2\pi r = 44 2 π r = 44 400 ∘ 400^\circ 40 0 ∘ Report a problem with this question
A cliff on the bank of a river is 87 m 87\text{ m} 87 m 22 m 22\text{ m} 22 m
A 76 ∘ 76^\circ 7 6 ∘ B 64 ∘ 64^\circ 6 4 ∘ C 36 ∘ 36^\circ 3 6 ∘ D 24 ∘ 24^\circ 2 4 ∘
Worked solution (try it first) The angle of depression equals the angle of elevation of the cliff top from the boat (alternate angles).
The height 87 m is opposite the angle and 22 m is adjacent:
tan θ = 87 22 ≈ 3.955 \tan\theta = \frac{87}{22} \approx 3.955 tan θ = 22 87 ≈ 3.955 .
So
θ ≈ 75.8 ∘ \theta \approx 75.8^\circ θ ≈ 75. 8 ∘ , which is
76 ∘ 76^\circ 7 6 ∘ to the nearest degree, option A.
Watch out
Put the height on top: 87 22 \frac{87}{22} 22 87 22 87 \frac{22}{87} 87 22 14 ∘ 14^\circ 1 4 ∘ Report a problem with this question
In the diagram, T U TU T U P P P ∠ P T S = 44 ∘ \angle PTS = 44^\circ ∠ P T S = 4 4 ∘ ∠ S Q P = 35 ∘ \angle SQP = 35^\circ ∠ S QP = 3 5 ∘ ∠ P S T \angle PST ∠ P S T
A 101 ∘ 101^\circ 10 1 ∘ B 125 ∘ 125^\circ 12 5 ∘ C 130 ∘ 130^\circ 13 0 ∘ D 135 ∘ 135^\circ 13 5 ∘
Worked solution (try it first) ∠ S P T \angle SPT ∠ S P T is between the tangent
P T PT P T and the chord
P S PS P S , so it equals the angle in the alternate segment:
∠ S P T = ∠ S Q P = 35 ∘ \angle SPT = \angle SQP = 35^\circ ∠ S P T = ∠ S QP = 3 5 ∘ .
The angles of triangle
P S T PST P S T add up to
180 ∘ 180^\circ 18 0 ∘ :
∠ P S T = 180 ∘ − 44 ∘ − 35 ∘ \angle PST = 180^\circ - 44^\circ - 35^\circ ∠ P S T = 18 0 ∘ − 4 4 ∘ − 3 5 ∘ = 101 ∘ = 101^\circ = 10 1 ∘ , option A.
Watch out
The tangent–chord angle ∠ S P T \angle SPT ∠ S P T ∠ S Q P = 35 ∘ \angle SQP = 35^\circ ∠ S QP = 3 5 ∘ P P P 46 ∘ 46^\circ 4 6 ∘ ∠ P S T \angle PST ∠ P S T Report a problem with this question
The probability that Amaka will pass an examination is 3 7 \frac37 7 3 4 9 \frac49 9 4
A 2 21 \frac{2}{21} 21 2 B 4 21 \frac{4}{21} 21 4 C 5 21 \frac{5}{21} 21 5 D 9 21 \frac{9}{21} 21 9
Worked solution (try it first) Amaka's and Bala's results are independent.
For both to pass, multiply:
3 7 × 4 9 = 12 63 \frac37 \times \frac49 = \frac{12}{63} 7 3 × 9 4 = 63 12 .
So the probability is
4 21 \frac{4}{21} 21 4 , option B.
Watch out
"Both" means multiply. Adding the chances gives 55 63 \frac{55}{63} 63 55 Report a problem with this question
Which of the following points lies on the line 3 x − 8 y = 11 3x - 8y = 11 3 x − 8 y = 11
A ( 1 , 1 ) (1, 1) ( 1 , 1 ) B ( 1 , − 1 ) (1, -1) ( 1 , − 1 ) C ( − 1 , 1 ) (-1, 1) ( − 1 , 1 ) D ( − 1 , − 1 ) (-1, -1) ( − 1 , − 1 )
Worked solution (try it first) A point lies on the line if its coordinates make
3 x − 8 y 3x - 8y 3 x − 8 y equal to 11.
( 1 , − 1 ) (1, -1) ( 1 , − 1 ) :
3 ( 1 ) − 8 ( − 1 ) = 3 + 8 = 11 3(1) - 8(-1) = 3 + 8 = 11 3 ( 1 ) − 8 ( − 1 ) = 3 + 8 = 11 .
It works.
The others give
3 − 8 = − 5 3 - 8 = -5 3 − 8 = − 5 ,
− 3 − 8 = − 11 -3 - 8 = -11 − 3 − 8 = − 11 and
− 3 + 8 = 5 -3 + 8 = 5 − 3 + 8 = 5 .
So the answer is
( 1 , − 1 ) (1, -1) ( 1 , − 1 ) , option B.
Watch out
− 8 × ( − 1 ) = + 8 -8 \times (-1) = +8 − 8 × ( − 1 ) = + 8 − 8 -8 − 8 ( 1 , − 1 ) (1, -1) ( 1 , − 1 ) − 5 -5 − 5 Report a problem with this question
Find the range of the following set of numbers: 28, 29, 39, 38, 33, 37, 26, 20, 15 and 25.
Worked solution (try it first) The largest number is 39 and the smallest is 15.
The range is
39 − 15 = 24 39 - 15 = 24 39 − 15 = 24 , option B.
Watch out
Scan the whole list for the largest number: it is 39, not 37. Using 37 gives 37 − 15 = 22 37 - 15 = 22 37 − 15 = 22 Report a problem with this question
The fourth and eighth terms of an arithmetic progression are 16 and 40 respectively. Find the common difference.
Worked solution (try it first) From the 4th term to the 8th is 4 steps of
d d d .
So
4 d = 40 − 16 = 24 4d = 40 - 16 = 24 4 d = 40 − 16 = 24 .
Divide by 4:
d = 6 d = 6 d = 6 , option B.
Watch out
Subtract the earlier term from the later one: 40 − 16 40 - 16 40 − 16 − 24 -24 − 24 d = − 6 d = -6 d = − 6 d d d Report a problem with this question
For what values of y y y y + 2 8 y 2 − 10 y + 3 \dfrac{y + 2}{8y^2 - 10y + 3} 8 y 2 − 10 y + 3 y + 2
A − 3 4 , 1 2 -\frac34, \frac12 − 4 3 , 2 1 B − 3 4 , − 1 2 -\frac34, -\frac12 − 4 3 , − 2 1 C 3 4 , 1 2 \frac34, \frac12 4 3 , 2 1 D 3 4 , − 1 2 \frac34, -\frac12 4 3 , − 2 1
Worked solution (try it first) The fraction is not defined when the bottom is zero:
8 y 2 − 10 y + 3 = 0 8y^2 - 10y + 3 = 0 8 y 2 − 10 y + 3 = 0 .
Factorise (numbers
− 6 -6 − 6 and
− 4 -4 − 4 multiply to 24 and add to
− 10 -10 − 10 ):
( 4 y − 3 ) ( 2 y − 1 ) = 0 (4y - 3)(2y - 1) = 0 ( 4 y − 3 ) ( 2 y − 1 ) = 0 .
So
y = 3 4 y = \frac34 y = 4 3 or
y = 1 2 y = \frac12 y = 2 1 , option C.
Watch out
Both brackets have a minus, so both roots are positive. A sign slip in either bracket gives a negative root, as in options A, B and D. Report a problem with this question